Principles of Time and SpaceNew Octonions and Relativity
Third Edition

Hiroshige Goto

KIYOTAKI
Fukuoka Japan

This is the third edition of the Japanese book (ISBN 978-4-9906664-0-8) publishedin September 2012. The text can be read at http://gotoclinic.org/space-time.pdf.
Download is free for personal use.
The author’s e-mail address is gotoclinic@able.ocn.ne.jp.
Copyright c 2014 by Hiroshige Goto
Published by KIYOTAKI
1-5-2-1301 Kiyotaki, Moji, Kitakyushu, Fukuoka 801-0833, Japane-mail: kiyotaki@gotoclinic.org
This book can be purchased only at Amazon.co.jp (www.amazon.co.jp).
ISBN 978-4-9906664-2-2
Printed in Japan on acid-free paper

This book is dedicated to my beloved wife Rieko

Preface to the Revised Edition

The quaternion is a complex number that extends into four dimensions. In thisbook, the quaternion is transformed into a number in the curved four-dimensional
space–time and the transformed quaternion is called the new octonion. Because
the new octonion consists of four real numbers and four imaginary numbers, fourkinds are thought to exist (similar to the real and imaginary numbers), each inour space–time. In addition, if the structure of the four-dimensional space–time isexamined using the new octonion, space is considered to have a double structure.
In other words, two four-dimensional space–time structures overlap. Furthermore,when using the new octonion, the conclusions of special relativity can be explained
without contradiction if the axiom that mass is the time component of the unit
world line is imposed. This leads to the conclusion that in our world, mass andenergy are expressed as imaginary numbers.
An algebraic theorem states that a numerical system with the algebraic operations of addition, subtraction, multiplication, and division is composed only of real
numbers, complex numbers, quaternions, and octonions. Therefore, it is natural
to assume that the physical laws of four-dimensional space–time must be describedwith quaternions. William Rowan Hamilton, who discovered the quaternion in 1843,
dedicated the latter half of his life to applying quaternions to physics but without
success. However, the concept of vectors came to light through his eﬀorts; theconcept of vector led to the development of the concept of tensors. Tensors are indispensable to mathematics in modern physics and engineering; in general relativity,
black holes were predicted and described using tensors.
Because the tensor is a mathematical tool by which curved four-dimensional
space–time is described, it is widely thought that using the quaternion to study
space–time would represent a backward motion—an unnecessary and fruitless endeavor. However, tensor has a complex matrix form. The purpose of natural scienceis to obtain simple laws from complex phenomena. I feel that the usage of a tensor
is significantly complex for describing physical laws. This study starts by proposingthat a simple algebra such as the quaternion may indeed be suitable to describe
the laws of space–time. Therefore, I have tried to create an algebra using only addition, subtraction, multiplication, and division to study curved four-dimensional
space–time.

vii

Hamilton did not succeed in applying quaternions to physics because the theoryof relativity did not exist at that time. Hamilton’s quaternion is the mathematics
of flat space–time; however, space–time curves in the theory of relativity. Physicallaws must be described with mathematics that matches the curved space–time. In
other words, if the quaternion is not altered to suit the demands of the curved
space–time, physical laws will not be described by quaternions.
First, the complex number is transformed to fit the mathematics of Lorentz transformations and is called the new complex number. Similar to the complex number, the quaternion is transformed into the curved space–time quaternion. Thisis dubbed an octonion because it includes eight numbers. The Graves’ octonion
consists of one real number and seven imaginary numbers, but the octonion of
the curved space–time consists of four real numbers and four imaginary numbers.
Therefore, it is termed a new octonion. If the space–time structure is examinedwith the new octonion and the mechanics of special relativity is recalculated, newinterpretations of the physical principles and double structure of space–time aresuggested.
However, the space–time diagram suggested by the new octonion may disagreewith actual space–time even if the new octonion is mathematically consistent. It
is well known that we cannot use Euclidean geometry, which is consistent in flatspace, to study curved space–time. The actual space–time may not have a double
structure even if it is suggested by the new octonion. Future problems include
whether the new octonion is mathematics without contradiction and whether theactual space-time is identical to that suggested by the new octonion.
The desire to learn the truth is a motivational force of science. Steven Paul Jobs,
one of the founders of Apple Computer, Inc., was quoted as saying “Stay hungry,
stay foolish,” which means that we should always seek better things and not be
tethered to preconceived notions. The physics described by tensors is accomplished.
However, it is a scientific action to seek a diﬀerent mathematics to describe curved
space–time.
Because the new octonion is simple mathematics, it can be understood even bya high school student. It is my wish that people who are interested in physics andmathematics and who want to learn the truth must acknowledge this book and use
the new octonion in their studies.
Hiroshige Goto
August 2012viii

Abstract

The contents of this book are summarized here. Because the technical terms are
explained in the text, they are used here without explanation.
Complex numbers have been expanded to Hamilton’s quaternion in four dimensions. This study aims to transform the quaternion into mathematics suitable fordescribing curved four-dimensional space–time. An algebraic theorem states thata numerical system with the algebraic operations of addition, subtraction, multiplication, and division has only real numbers, complex numbers, quaternions, andoctonions. Therefore, the physical principles of four-dimensional space–time should
be describable using quaternions.
First, it is found that the division
A/ |A|
of a complex number is the coordinate transformation if A is a complex number and
A is its complex conjugate. (Chap. 2)
Therefore, it is considered that the Lorentz transformations for special relativityt − (v/c2 )xt0 = √,
1 − v 2 /c2
x − vtx0 = √
1 − v 2 /c2
can be obtained if A/ |A| is applied in two-dimensional space–time. The result isthat the sign of the velocity of light squared of the Lorentz transformations changes
from minus to plus, presumably because complex numbers are the mathematics
of flat space–time. However, the space–time is curved. Therefore, new complex
numbers must be created to describe the curved space–time. It is assumed that a
fourth imaginary number h exists in addition to the three imaginary numbers i, j, k
of Hamilton’s quaternions. The algorithms are as follows:
i2 = j 2 = k 2 = h2 = −1,
ij = −ji = k, jk = −kj = i, ki = −ik = j,hi = ih, hj = jh, hk = kh.
ix

A new complex number ah + bi is created from the complex number a + bi andequations identical to the Lorentz transformations are obtained when the coordinatetransformation A/ |A| is done with this new complex number. This shows that the
new complex number gives mathematics of the curved space–time. (Chap. 3)
In addition, it is considered that time advances only in the positive direction
because a contradiction occurs if time t is assumed as negative when we use thecoordinate transformation A/ |A| to obtain the Lorentz transformations. (Chap.
3)
Herein, we examine the new complex number to determine whether it leads to accurate mathematics. Time dilation, proper time, length contraction, world distance,
and the twin paradox of special relativity in two-dimensional space–time are provenwith the new complex numbers and new complex planes. Without light emission
and examples of a clock and ruler, results identical to those of special relativity are
obtained using only the new complex number and the equation of the world line.
(Chap. 6, Chap. 7, Chap. 8, Chap. 9)
Next, by using the fourth imaginary number h, a new quaternion ah + bi + cj + dk
in the curved space–time is created from Hamilton’s quaternion a + bi + cj + dk.
Furthermore, the new quaternion is used in the coordinate transformation A/ |A|and the Lorentz transformations in the curved four-dimensional space–time are obtained. These equations are called the new Lorentz transformations. If observer B
moves with constant velocity v in the x-direction with respect to observer A at rest,
y 0 = y,z0 = zin special relativity. However, y 0 and z 0 are functions of the three variables y, z, andv, and the x motion changes the y and z distances if these are calculated using the
new quaternion mathematics without assuming the isotropy of space–time. Theseequations are as follows:y + (v/c)zh
y0 = √,
1 − v 2 /c2z − (v/c)yhz0 = √.
1 − v 2 /c2
Because of the homogeneity of space and the fact that y 0 = f (v)y and z 0 = g(v)zwere assumed in special relativity, these results were not obtained. Because a constant velocity of light is obtained by the velocity-transformation equations andbecause the world distance calculated by these transformations is invariant, theaccuracy of the new Lorentz transformations is proven. (Chap. 10)xIn addition, unlike the definition of world distance of special relativity, i.e.,s2 = (ct)2 − x2 − y 2 − z 2 ,the definition of world distance calculated by the new quaternion iss2 = −(ct)2 + x2 + y 2 + z 2 .

(Chap.10)

Here, we prove that the new quaternion has eight numbers that consist of areal number a, imaginary numbers h, i, j, k, and products hi, hj, hk of two
imaginary numbers. We also show that hi, hj, and hk are new real numbers that are
independent of other numbers. In other words, the new quaternion is the octonion
that consists of four real numbers and four imaginary numbers. This is called thenew octonion because it is the mathematics of curved space–time unlike Graves’octonion (Cayley number) in flat space, which is already known. Cayley numbers
consist of one real number and seven imaginary numbers. (Chap. 11)
The space–time structure is examined using the new octonion. This examinationsuggests that four-dimensional space–time has a double structure. We live in aworld where the world point is expressed as cth + xi + yj + zk and the number ofthe world point in an alternative world is ct + xhi + yhj + zhk. Strictly speaking,our world is a world of imaginary numbers, while the other world is that of real
numbers. However, our world is called the positive world and the other world iscalled the negative world until a time when this space–time theory becomes moreprevalent. The positive and negative worlds do not exist in parallel but rather
overlap. In other words, each coordinate axis of the four-dimensional space–timehas a coordinate that is part of the positive world as well as one that is part of
the negative world; the temporal-axis parts are ct0 h and ct1 , the x-axis parts are
x0 i and x1 hi, the y-axis parts are y0 j and y1 hj, and the z-axis parts are z0 k andz1 hk. If we assume only one four-dimensional space-time, each coordinate part of
the four-dimensional space–time is a complex number. The temporal axis part is
ct0 h + ct1 , the x-axis part is (x0 + x1 h)i, the y-axis part is (y0 + y1 h)j, and thez-axis part is (z0 + z1 h)k. (Chap. 11)
In addition, by using the number ct+xhi+yhj +zhk to express the world point in
the negative world, the Lorentz transformations in the negative world are obtained.
The results indicate that the physical laws in the negative world are the same as
those in the positive world. (Chap. 11)
Furthermore, it is proven that the coordinate transformation A/ |A| creates obliquecoordinate axes. (Chap. 12)
To examine whether the new octonion is correct, its axioms and theorems arechosen. It is suﬃcient to examine the erratum of axioms and theorems to seexiwhether the new octonion causes a contradiction. In addition, it is known that theassociative law(AB)C = A(BC)is not possible with Graves’ octonion, but it is proven to be possible with thenew octonion. Based on this, the new octonion is thought to be a more complete
number than Graves’ octonion, which has been used in a recent study of fundamental
particles. However, researchers may not obtain precise conclusions without the newoctonion. (Chap. 13)
In addition, because the new octonion space–time is bent but parallel lines do
not cross there, the new octonion geometry is a non-Euclidean geometry and nonRiemannian geometry, respectively. (Chap. 13)
We can rewrite a vector with the new octonion if we accept the double structureof four-dimensional space–time. If we assume that there are two vectors A and B
and new octonions A and B express each vector, the relationships between the newquaternion, scalar product, and vector product are as follows:
BA = A·B + A × B,
A·B = (BA + AB)/2,
A × B = (BA − AB)/2.
Furthermore, triple and quadruple vector products can be rewritten using the newoctonion. (Chap. 14)
In addition, the metric tensor and Kronecker δ are not required because these tensors can be calculated with the new octonion in straight coordinate systems. The
diﬀerences between a tensor and the new octonion are examined. Tensor mathematics calculates physical quantities in four-dimensional space–time by confining them
to a cross section of four-dimensional space–time. In contrast, the new octonion
mathematics calculates physical quantities throughout the entire four-dimensional
space–time. (Chap. 15)
Synchrotron radiation is also explained with the new octonion. According tocalculations based on Lorentz transformations, the light laterally released by anelectron moving with the velocity of light angles forward, but using the new Lorentz
transformations, it becomes bidirectional light, i.e., a combination of lights in thefront and lateral directions. At present, the interpretation of this result is unknown.
(Chap. 16)
In addition, momentum conservation and the change of mass due to velocity effects under the new Lorentz transformations are proven. That these results are

xii

identical to the conclusions of special relativity shows that the new Lorentz transformations found by the new octonion are correct. (Chap. 17)
Furthermore, it is shown that the results of special relativity can be explained
without contradiction if mass and energy are viewed as the time part of the unit
world line and momentum as the space parts of the unit world line. Unlike theresult of general relativity, mass does not bend space–time and the time part of
the unit world line of a particle itself is mass. In addition, mass and energy in thepositive world are described with imaginary numbers if we assume that mass is thetime part of the unit world line. At that time, mass and energy in the negative
world are described with real numbers and the sum of the squares of mass or energy
in the positive and negative worlds becomes zero. In general, it is thought that theaddition of mass or energy of matters and antimatters yields zero. However, this is
not correct. (Chap. 18)
Because the world distance of a point on x = ct is zero, light possesses no rest
mass. The world distance of the point on x = vt, on the other hand, is non-zero;hence, substance has mass. That is, mass is not given; however, it is finite or zerodepending on the locus in four-dimensional space–time. In this picture, no Higgs
boson is required. (Chap. 18)
It is proven that the energy conservation law and the momentum conservation
law are laws of reflection of world lines. This is proven under the assumption thatforce does not exist and the curve of the world line by collision and reflection of theworld lines is force. (Chap. 18)
In addition, Hiroyuki Kamada demonstrated that Dirac’s γ matrix and the newoctonion are mathematically equivalent (personal communication). (Chap. 18)
Although it is not the original purpose of this book, the new Lorentz transformations with superluminal velocity is found by coordinate transformations using thenew octonion. The result shows that a particle with superluminal velocity is in thenegative world and we cannot observe it from the positive world. In addition, timecannot be reversed from the equation of proper time. In other words, we cannot
return to the past even by moving at superluminal velocity. It is also shown thatspace–time is discontinuous. (Chap. 19)
Other problems related to space–time theories that can be solved by the newoctonion are also shown. Maxwell’s equations of electromagnetic waves satisfy the
Lorentz transformations. However, how do they change if the new Lorentz transformations are applied? Can we find negative-world parts of the electromagnetic
wave like the y 0 - and z 0 -axis parts of the new Lorentz transformations? In addition,
if it is assumed that diﬀerences between the world lines of light and matter arexiii

only diﬀerences between the places of passage in four-dimensional space–time, theworld line of a point mass has the property of a wave. In that case, what will theMaxwell’s equations of a matter wave be? Is there a possibility that the equations
lead to the unified field theory? (Chap. 20)
It is revealed that biquaternion, which was discovered by Hamilton in 1844, andthe new octonion are equivalent. However, their interpretation and application
methods markedly diﬀer. In the biquaternion, h is treated as an attached imaginary
number rather than a fourth imaginary number. On the other hand, in the newoctonion, h is the fourth imaginary number and is more important than i, j, andk. The important physical quantity of special relativity (proper time) is denoted by
the imaginary number h. Mass and energy are also related to h. (Chap. 20)
Although the velocity of light is not constant in an accelerated system in general relativity, can we obtain an identical conclusion using the new octonion? The
Lorentz transformations in an accelerated system do not exist, but equations equivalent to the new Lorentz transformations can be obtained by coordinate transformations using the new octonion. The results of such calculations are shown. (Chap.
20)
Finally, it is shown that the new octonion and the current string theory may haveidentical contents. According to the space–time theory using the new octonion,
all fundamental particles are constrained by five variables: (ct, x, y, z) express
the space–time position in four dimensions and the variable Ψ (psi) expresses theamplitude of the world line. In addition, the size of Ψ is smaller than that of afundamental particle because Ψ is the thickness of the world line. In this reckoning,
there is something akin to the five dimensions of the space–time proposed by Kaluza–
Klein theory, which precedes string theory. In that theory, the fifth dimension has
the shape of a ring that is too miniscule to be observed. This fifth dimension may
be identical to the amplitude Ψ of the world line. Because all particles have fivevariables (ct, x, y, z, Ψ), five dimensions of space–time follow easily. However, only
four-dimensional space–time exists because Ψ is an amplitude. In addition, because
the positive and negative worlds have an overlapping double structure, there areten variables between the two worlds—(ct, x, y, z, Ψ) for the positive world and(ct0 , x0 , y 0 , z 0 , Ψ0 ) for the negative world. Because the string theory developed from
Kaluza–Klein theory exists only in ten-dimensional space–time, proponents of thetheory may think that there also must be ten dimensions in the world. However, only
four-dimensional space–time with a double structure exists. Although all particles
are expressed with oscillating strings by string theory, the world line of a particle
vibrates in a way similar to light according to space–time theory using the newxiv

octonion. The smallest unit of the world line is smaller than a fundamental particle
from a space–time discontinuity. Because space–time theory using the new octonion
and string theory may have the same contents, the problem is whether we can rewrite
string theory with the new octonion. (Chap. 20)

xv

xvi

Acknowledgements

I would like to thank Mr. Osami Ikeda, who advised me to enter Tokyo University;
Dr. Kenzo Yoshida, who supported me during my school days at Tokyo University;and Dr. Ken Kuroki, who taught me the mental attitude of a doctor during myschool days at Kyushu University School of Medicine. In addition, I would like tothank Enago (www.enago.jp) and its editors for the English language review.

xvii

xviiiContents

Preface to the revised edition, vii
Abstract, ix
Acknowledgements, xvii
Contents, xix
1

Basic knowledge of complex numbers, 1
1.1 Definition and formulae of complex numbers, 1
1.2 Complex numbers and rotation, 4

2

Coordinate transformation by complex numbers, 7
2.1 About coordinate transformation, 7
2.2 General formulae of coordinate transformation by complex numbers, 8
2.3 Constancy of the physical quantity through coordinate transformation,
10
2.4 Inverse transformation of coordinate transformation, 113Complex numbers and Lorentz transformations, 13
3.1 Standard derivation methods of Lorentz transformations, 13
3.2 Postulates for coordinate transformation using complex numbers, 14
3.3 Derivation of Lorentz transformations using complex numbers, 16
3.4 The fourth imaginary number h, 18
3.5 Derivation of Lorentz transformations using new complex numbers, 19
3.6 The reason why time advances only in the positive direction, 22

4

New complex plane and oblique coordinate axes, 25
4.1 New complex number to express the rotation of coordinate axes, 25
4.2 Proof of the oblique coordinate axes, 26
4.3 A simple method to obtain oblique coordinate axes, 28

xix

5

Length in the new complex plane and Lorentz
transformations, 29
5.1 Method of obtaining Lorentz transformations from length in the newcomplex plane, 29
5.2 Equation of a straight line through new complex numbers, 37
5.3 Method to express coordinate parts through new complex numbers, 38
5.4 Correctness of the imaginary number h, 38

6Time dilation and length contraction, 41
6.1 Proper time and time dilation, 41
6.2 Special relativity and length contraction, 44
6.3 Coordinates of the node of the world lines and length contraction, 45

7The new quaternion and world distance, 49
7.1 Definition of world distance in special relativity, 49
7.2 Derivation of world distance by the new quaternion, 50
7.3 Interpretation of the negativity of the square of the world distance, 53
7.4 Least-squares theory, 54

8The new quaternion and the world distance of light, 55
8.1 Proof of the world distance of light by special relativity, 55
8.2 Proof of the world distance of light by the new quaternion, 56
8.3 Interpretation of the zero world distance of light and the constancy ofthe velocity of light, 57

9

The twin paradox, 59
9.1 Contents of the twin paradox, 59
9.2 Calculation in the frame of a stationary observer, 60
9.3 Calculation in the frame of a moving observer, 62
9.4 Curve of the world line and a force, 67

10 The new Lorentz transformations, 69
10.1 Derivation of the new Lorentz transformations by the new quaternion,69
10.2 Diﬀerences between the new and original Lorentz transformations, 72
10.3 The new Lorentz transformations and world distance, 73
10.4 Transformation of velocities under Lorentz transformations and theconstancy of the velocity of light, 75xx

10.5 Transformation of velocities under the new Lorentz transformations andthe constancy of the velocity of light, 78
10.6 Independency of the imaginary number h, 80
10.7 The new Lorentz transformations and inverse transformations, 82
10.8 General formulae of the new Lorentz transformations, 84

11 Double structure of four-dimensional space-time, 87
11.1 The algebraic theorem, 87
11.2 The octonion, 88
11.3 The new octonion, 89
11.4 Double structure of four-dimensional space–time, 90
11.5 New octonions and new Lorentz transformations in the negative
world, 92
11.6 World distance in the negative world, 97
11.7 World point in double-structured four-dimensional space–time, 98
11.8 Four real numbers and four imaginary numbers in four-dimensional
space–time, 100

12 Oblique coordinates of the four-dimensional space-time,103
12.1 Oblique coordinates of motion along x-axis, 103
12.2 Oblique coordinates of motion in any direction, 109
12.3 Interpretation of y 0 and z 0 using new Lorentz transformations, 111

13 Axioms and theorems of the new octonion, 117
13.1 Necessity of axioms and theorems, 117
13.2 Axioms of the new octonion, 118
13.3 Theorems of the new octonion, 121

14 New octonion and vectors, 155
14.1 Basic properties of vectors, 155
14.2 Vectors and coordinate transformations, 159
14.3 New octonions and direction of outer products of vectors, 163
14.4 Calculation of three-vector products by the new octonion, 164
14.5 Calculation of four-vector products by the new octonion, 167
14.6 New octonion and the four-dimensional vector, 168
14.7 New octonion and rotation vector, 172xxi

15 New octonions and tensors, 177
15.1 Reasons why tensors were made, 177
15.2 Diﬀerences between the tensor and the new octonion, 181
15.3 Kronecker δ, 192
15.4 Metric tensor, 194

16 Synchrotron radiation, 197
16.1 About synchrotron radiation, 197
16.2 Proof under Lorentz transformations, 197
16.3 Proof under the new Lorentz transformations, 199
16.4 Synchrotron radiation and the constancy of the velocity of light, 201
16.5 Slant of synchrotron radiation, 202
16.6 Observation of the perpendicular light of synchrotron radiation, 205
16.7 Light and inertial law, 207
16.8 Upthrow of a ball, 208

17 Dynamics by the new octonion, 213
17.1 Formulae of the physical quantity that is constant by coordinatetransformation, 213
17.2 Mass and coordinate transformations of velocity and momentum of thex-axial motion, 215
17.3 Mass and coordinate transformations of velocity and momentum ofarbitrary axial motion, 220
17.4 Four-velocity and four-momentum, 223
17.5 Defects of four-velocity, 227
17.6 Conservation of momentum, 231
17.7 Acceleration and four-acceleration, 239

18 New octonion and mass, 245
18.1 Mass is a time component of the unit world line, 245
18.2 Energy and momentum, 249
18.3 The sum of mass or energy in four-dimensional space–time, 252
18.4 Meaning of c in E = mc2 , 253
18.5 Mass of light and the Higgs boson, 255
18.6 Constancy of the coordinate transformation of energy conservation,258
18.7 Force and reflection of the world line, 264
18.8 Extinction of the world line, 267
xxii

18.9 Energy–momentum equation and Dirac’s γ matrix, 269
18.10 An easy method for obtaining E = mc2 , 272

19 Special relativity at the superluminal velocity, 275
19.1 New Lorentz transformations at the superluminal velocity, 275
19.2 Proper time at the superluminal velocity, 277
19.3 Zeno’s paradox and a discontinuity axiom, 279

20 Future problems, 281
20.1 Electromagnetism and biquaternion, 281
20.2 General relativity, 286
20.3 Five-dimensional space–time and string theory, 292

Bibliography, 295
Index, 297

xxiii

xxiv

1
Basic Knowledge ofComplex Numbers

1.1 Definition and formulae of complex numbers
In this book, the complex number and the quaternion are changed into mathematicalentities suitable for curved four-dimensional space–time: new complex numbers,
new quaternions, and new octonions. There may be readers who think they cannot
understand the contents of this book if they have not studied the quaternion and theoctonion, or even the complex number. However, only the basic knowledge of thecomplex number is enough to understand the contents of this book, and necessarycomponents are explained in this chapter.
A complex number is the combination of a real number and an imaginary number.
We define the imaginary number i as a number that, when squared, becomes −1, ori2 = −1. This is a departure from our common experience because when a numberthat is used every day is squared, it does not become negative, even if it itself isa negative number. This type of number is called a real number. Putting them
together yields a complex number. For example, if a and b are real numbers, acomplex number A can be represented as A = a + bi. We call a the real part and b
the imaginary part.
Typically, the imaginary unit is written at the beginning of a term as ib. However,in this book, the order is reversed so that a number is followed by the imaginary
unit, i.e., bi. This practice serves the customary formalism enhances comprehension.
Given two complex numbers A = a + bi and B = c + di, the formulae are
A + B = (a + bi) + (c + di)= (a + c) + (b + d)i,
A × B = (a + bi)(c + di)= ac + adi + bci + bdi2

1

= (ac − bd) + (ad + bc)i.
A ÷ B is explained later in this section. In addition, if A = B, we can writea = c, b = d, which we call as coeﬃcient comparison.
Concerning complex numbers A, B, and C, we have natural results:
A + B = B + A,
AB = BA,(A + B) + C = A + (B + C),(AB)C = A(BC),A(B + C) = AB + AC.
As a quick look ahead, note that in Hamilton’s quaternion, which will be explainedlater, the commutative property of multiplication (second equation from above)
does not hold. That is
AB 6= BA.
In other words, multiplication order matters in Hamilton’s quaternion.
If complex number is A = a + bi, then its complex conjugate is defined as A =a − bi. To make a complex conjugate out of a complex number, simply switch thesign on the imaginary part. We call the size of A as the magnitude of A and write
it as |A|. Furthermore, we find2

|A| = AA= (a + bi)(a − bi)= a2 − abi + bai − b2 i2= a2 + b2 .

2

(1.1)

To represent complex numbers on a two-dimensional coordinate grid, the horizontal axis is taken to be the real axis and the vertical axis is the imaginary axis,
as in Figure 1.1. Then, the real axial part of a complex number A = a + bi is a andthe imaginary axial part is b. A is expressed as the point (a, b). In addition, |A|
expresses the length from origin O to the point A. A formula2|A| = a2 + b2
is obvious by the Pythagorean theorem and we can clearly see that the definition2|A| = AA in (1.1) is correct.
Because division 1/A is calculated by multiplying the numerator and denominator
by A, we have
1
A
=
A
AA
A
=
2.
|A|

(1.2)

Thus, if B = c + di and B 6= 0, the equations are
A
B
AB
=
BB
AB
=
2
|B|
(a + bi)(c − di)=c2 + d2(ac + bd) + (bc − ad)i=.
c2 + d2

A÷B =In addition, we can also write
A + B = A + B,
AB = A B.
However, because the placement of A and B is reversed in the quaternion and theoctonion, the equation becomes
AB = B A.
With complex numbers,
A B = B A.
3

Thus, if we write
AB = B A,
even in the complex number as in the quaternion, the new quaternion, and theoctonion, there may be no confusion. We will explain this in more detail in Theorem
3 of Section 13.3.

1.2 Complex numbers and rotation
In this section, we explain the relation between the complex number and rotation.
This is rarely taught in high schools, but we can explore it here using simple examples.
√First, we consider multiplication. Figure 1.2 is drawn assuming that A = 3 +√i, B = 2 + 2 3i.

From (1.1), the magnitudes squared of A and B are√√
2|A| = ( 3 + i)( 3 − i)√
= ( 3)2 − i2
=3+1
= 4,

√
√
|B| = (2 + 2 3i)(2 − 2 3i)√= 4 − (2 3i)22

= 4 + 12= 16.

4

Thus, the magnitudes of A and B are|A| = 2, |B| = 4.
The angle that the real axis makes with the line segment OA or OB is called theargument. The arguments of A and B are 30◦ and 60◦ , respectively, with each value
arrived at using the ratios of triangular sides. If A × B is calculated, multiplying
gives
√
√
A × B = ( 3 + i)(2 + 2 3i)√√√= 2 3 + 2( 3)2 i + 2i + 2 3i2√√= 2 3 + 6i + 2i − 2 3= 8i.
Because A×B has only an imaginary part, it solely lies on the imaginary axis and its
argument is 90◦ . This argument of 90◦ means the addition of the 30◦ argument of A
and the 60◦ argument of B. In addition, magnitude 8 of A × B is the multiplicationof magnitude 2 of A and magnitude 4 of B.
As we infer from this example, we have in generalarg(A × B) = argA + argB,|A × B| = |A| × |B| .
In other words, we have A × B if we rotate A counterclockwise on origin O by theargument of B and multiply the magnitude of A by the magnitude of B.
√
Next, we consider division. Given A = 3 + i and C = 8i, from (1.2), C ÷ A is
C
A
CA
=
AA √
8i( 3 − i)√= √( 3 + i)( 3 − i)√
8 3i − 8i2=
3 − i2√8 + 8 3i=
4√
= 2 + 2 3i.

C ÷A=Thus, C ÷ A is identical to B. Because the magnitude squared of C ÷ A is√√2
|C ÷ A| = (2 + 2 3i)(2 − 2 3i)5√= 4 − 4( 3)2 i2= 4 + 12= 16,
we have|C ÷ A| = 4.
We can draw Figure 1.2.

By the ratio of the sides, the argument of C ÷ A is 60◦ . It is the number that
we have if an argument of 30◦ of A is subtractd from an argument of 90◦ of C.
Furthermore, the magnitude of C ÷ A is the number that we have if magnitude 8
of C is divided by magnitude 2 of A.
As we infer from this example, we have in generalarg(C ÷ A) = argC − argA,|C ÷ A| = |C| ÷ |A| .
In other words, we have C ÷ A if we rotate C clockwise on origin O by the argumentof A and divide the magnitude of C by the magnitude of A.
We now see that multiplication and division of complex numbers are rotations on
origin O. We add the arguments if we multiply and subtract them if we divide.
We can generally discuss the rotations using the trigonometric function sin θ(theta) and cos θ. However, it is unknown as to whether angle θ changes by rotation in curved space–time. Therefore, sin θ and cos θ are not used in this book.
This is explained in Theorem 10 of Section 13.3.

6

2
Coordinate Transformation byComplex Numbers

2.1 About coordinate transformation
It is necessary to make a general formula of coordinate transformation by a complex number to find the Lorentz transformations of special relativity using complex
numbers. First, we explain coordinate transformation. We consider a case in which
two observers P and Q observe a certain point R. We assume that P is at rest andQ moves. The time t and position x of R observed by P will be diﬀerent from thoseobserved by Q. Because coordinate means how to be seen, coordinates (t, x) of R
observed by P and coordinates (t0 , x0 ) of R observed by Q must be diﬀerent.

The horizontal axis is defined as the temporal axis and the vertical axis is definedas the length axis in a two-dimensional coordinate plane, as shown in Figure 2.1.
If the position of R is fixed, each coordinate part is the coordinate of the point
where the perpendicular lines drawn to each coordinate axis from R cross with thatcoordinate axis. Therefore, the t0 -axis and the x0 -axis of Q turn around origin O
for the t-axis and the x-axis of P .
7

Because the term is coordinate transformation, the coordinates (t, x) of R observed by P are transformed into the coordinates (t0 , x0 ) of R observed by Q.
Mathematically, equations to express relations between (t0 , x0 ) and (t, x) are theequations of coordinate transformation. Lorentz transformations of special relativity are the equations of coordinate transformation. Though Lorentz transformations
were obtained using the constancy of the velocity of light, we will obtain them bycoordinate transformations using complex numbers in this book.
A Minkowski coordinate diagram, or Minkowski space–time diagram, uses theseprinciples to describe the coordinate plane of special relativity, in which length x
lies on the horizontal axis and time t on the vertical axis. However, in this book, wedetermine that the horizontal axis will be the temporal axis and the vertical axis
will be the length axis. There are advantages of calculations and understandingsin this coordinate diagram because the slope of a straight line expresses velocity,
unlike that in the case of Minkowski space–time diagrams.

2.2 General formulae of coordinate transformation bycomplex numbers
We consider a two-dimensional complex plane, such as the one in Figure 2.2.

We assume that the magnitudes of A and B are |A| and |B|, and the angles thatcomplex numbers A and B make with the real axis are argA = θ1 and argB = θ2 .
As for complex number B/A, which is the same as B ÷ A, the magnitude andargument are as follows:
|B|
B
,
(2.1)
=
A
|A|
arg(B/A) = argB − argA = θ2 − θ1 ,
8

(2.2)

from (1.3) and (1.4), as explained in Section 1.2. Because ∠AOB = θ2 − θ1 in
Figure 2.2, B/A is piled on B, as shown in Figure 2.3, if the real axis is turned
counterclockwise around origin O and is put on A.

The magnitude of (B/A) |A|, which we find by multiplying the complex number
B/A by |A|, is
B
|B|
|A|
|A| =
A
|A|
= |B|
from (2.1). Therefore, (B/A) |A| expresses the real number part and the imaginary
number part of B if we assume that A is the new real axis. In other words, if we
assume that D is the intersection of the perpendicular line drawn to A from B, we
can write
B|A| = |OD| + |BD| i,(2.3)
A
where i is an imaginary number. |OD| and |BD| express lengths between two
points and do not express the magnitudes of complex numbers. From (2.3), we
can understand that (B/A)|A| expresses how B is seen by A. Therefore, it is thecoordinate transformation of B by A.
There may be readers who think that in Figure 2.3, the i of the imaginary axis
before the rotation and the i of the imaginary axis after the rotation are not identicalbecause they are in a diﬀerent position. However, each i is identical because strictly
speaking, the real axis does not rotate; rather, A and B rotate to the real axis.
We only rotate the coordinate axis so that it is easy to be understood. Figures aredescribed using this form from here on. In Theorem 9 of Section 13.3, we examine
the cases, in which we can perform coeﬃcient comparison and where we cannot.

9

If we multiply the numerator and the denominator of a fraction by complex conjugate A to change an equation of division (2.3) into an equation of the multiplication,
we haveB
B |A| A
|A| =A
AA
B |A| A
=
2
|A|
=

BA
.
|A|

In other words,
BA
|A|

(2.4)

is an equation to express the coordinate transformation of B by A.
If observers P, Q, and observed point R in the last section are applied, complex
number B expresses the coordinates (x, t) of R observed by P at rest. Complex
number A expresses the coordinates of moving observer Q and (2.4) expresses thecoordinates (x0 , t0 ) of R observed by Q. Relations of (x, t) and (x0 , t0 ) are calculatedusing (2.4) in Section 3.3.
From the above results, if we assume that A is a complex number expressing thecoordinates of Q, the equation of the coordinate transformation by moving observer
Q is
A
.
(2.5)
|A|

2.3 Constancy of the physical quantity through coordinatetransformation
We explain an important property of (2.5) here. A quantity that does not change
in magnitude under Lorentz transformation is called an invariant quantity undercoordinate transformation in special relativity. Researchers have defined world distance, four-velocity, and four-momentum as the physical quantities that do notchange under coordinate transformation and consider the motion of point mass infour-dimensional space–time. In other words, coordinate transformation that does
not change magnitudes is important in physics. From the calculations mentioned
above and in Figure 2.3, the coordinate transformation (2.4) by a complex number
only changes the coordinate parts of B, i.e., how B is seen, and it does not change
the magnitude of B. We will prove this in the following text.
10If F is the complex number result of B following coordinate transformation by
A, we can write
BA
= F.
|A|
Thus, |F |2 is given as follows:2|F | = F F
BA AB|A| |A|=2

2|A| |B|

=

2

|A|
2

= |B| .
In other words, because the magnitude of BA/|A| is identical to that of B, coordinate transformation A/|A| does not change the magnitude of B in the complex
plane, which is the length of B from origin O.

2.4 Inverse transformation of coordinate transformation
We find the inverse transformation of A/ |A|, which will enable us to change BA/ |A|back into B. If we multiply both sides of BA/ |A| = F by A/ |A|, the equations
becomeBA A
A=F,|A| |A|
|A|
B |A|

2

A
,
|A|
|A|
A
B=F
.
|A|
2

=F

Thus, the inverse transformation of A/ |A| isA.
|A|
We can also obtain this by
2

A A
|A|
=
2
|A| |A|
|A|
= 1.

11

12

3
Complex Numbers andLorentz Transformations

3.1 Standard derivation methods of Lorentz transformations
We explain the method to find Lorentz transformations in special relativity before
obtaining Lorentz transformations using the coordinate transformation A/ |A| bycomplex numbers. This is a standard method that can be found in any book about
relativity theory. In the following explanation, the symbol A represents observer A,
not the complex number A of A/ |A| from coordinate transformations.
We assume that the coordinates in four-dimensional space–time of the tip of lightreleased at time t = 0 from origin O are (ct, x, y, z) as seen from observer A and(ct0 , x0 , y 0 , z 0 ) as seen from observer B. The variable c represents the velocity oflight, t expresses time, and x, y, and z express lengths. Because the tip of the lightmakes a spherical surface of radius ct in the three-dimensional space at time t, theequation is given as follows:
x2 + y 2 + z 2 = (ct)2 .

(3.1)

A coincides with B at origin O at time t = 0. The constancy of the velocity oflight means that the velocity of light is c regardless of whether the light is observedby an inertial observer or by an observer at rest. The equation of the sphericalsurface seen by B is as follows:
x02 + y 02 + z 02 = (ct0 )2 ,(3.2)

wherever B is. We have y = y 0 = 0 and z = z 0 = 0 if B moves in the x-direction
with respect to observer A at rest with a constant velocity v. Under conditions
x = 0, t = 0, (3.1), and (3.2), we havex = ct,13

x0 = ct0 .
We assume that functions satisfying these two equations using fixed number a are
x0 = a(x − vt),x = a(x0 + vt0 ).
From the four equations mentioned above, we can find the Lorentz transformations:t − (v/c2 )xt0 = √,
1 − v 2 /c2
x − vtx0 = √.
1 − v 2 /c2

(3.3)
(3.4)

These calculations are omitted for brevity.
In special relativity, it is assumed thaty 0 = y,(3.5)

z 0 = z.

(3.6)

We discuss the problems with such assumptions in Section 10.1.

3.2 Postulates for coordinate transformation using complex
numbers
The following three items are premised to find Lorentz transformations of special
relativity using the coordinate transformation A/ |A| with complex numbers.
(1) To add or subtract time t and length x, each unit has to be made identical. For
this purpose, time is calculated in ct, which is obtained by multiplying time t by
the velocity of light c. Because the unit of velocity is obtained by dividing lengthby time, the unit of the velocity of light becomes length, if velocity is multipliedby time.
(2) Unlike the Minkowski space–time diagram of special relativity, in which it isassumed that x is the horizontal axis and ct is the vertical axis, we assume acomplex plane in which ct is the horizontal axis and x is the vertical axis. Therefore, coordinates are expressed as (ct, x). Because the slope of a straight line
expresses velocity by this method, there are advantages to drawing figures thisway and calculations are easy.
(3) A complex number expressed in coordinates following the form (ct, x) is ct+xi. Typically, the imaginary unit is written at the beginning of a term as ib.
14However, in this book, the order is reversed so that a number is followed by
the imaginary unit, i.e., bi. This practice serves the customary formalism enhances comprehension.

On the basis of these three preconditions, let us consider motion in the x-axis. As
shown in Figure 3.1, observer A is at rest at origin O and observer B moves with auniform velocity v in the positive x-direction. In addition, observed point D is at
rest at a position of distance x and B coincides with A at origin O at time t = 0.
After t seconds, A is at time t at distance 0; B is at time t at distance vt; and D is
at time t at distance x. The coordinates are A(ct, 0), B(ct, vt), and D(ct, x) andthe complex numbers are A = ct, B = ct + vti, and D = ct + xi. Figure 3.2 depicts
this representation if it is drawn in the complex plane. From now on, we call theimaginary axis the ct-axis and the real axis the xi-axis.

Observer A moves in the ct-axis because time t passes but distance x remains 0.
The equation of the straight line is x = 0. Because the slope of the straight line ofobserver B is vt/(ct) = v/c, the equation of the straight line isx=v(ct) = vt.
c
15

And B moves in the straight line x = vt. Note that though the slope of the straightline of B is v/c, the equation is not
x=

v
t.
c

Because the unit of the ct-axis is ct, the correct equation is as follows:x=v(ct) = vt.
c

Stationary point D moves in a parallel straight line on the ct-axis because x does notchange; only ct changes. Distance x of D is temporarily transformed to x0 because
it is a fixed number. And the equation of the straight line in which D moves isx = x0 . In addition, as seen from A at rest, the times of A, B, and D are t.
Thus, A, B, and D form one straight line, which is parallel to the xi-axis. A point
moving in four-dimensional space–time is called a world point and the line which thepoint traces out is called a world line or four-dimensional line in relativity theory.
Thus, x = 0, x = vt, and x = x0 are the equations of world lines of A, B, andD, respectively. In addition, the complex plane is equivalent to four-dimensional
space–time, in which y = z = 0.

3.3 Derivation of Lorentz transformations using complex
numbers
Using A/ |A| for the coordinate transformation by complex numbers, DB/ |B| means
the coordinate transformation of D by B. In other words, DB/ |B| represents thecoordinates of D described by B in uniform motion.

As shown in Figure 3.2, if we assume that the straight line x = vt of B is a newreal axis, i.e., ct0 -axis, and that the straight line that meets x = vt at a right angle
16through origin O is the new imaginary axis, i.e., the x0 i-axis, the coordinates of D
as seen from B are (ct0 , x0 ). However, we have ct0 = x0 = 0 if ct = x = 0 because
B coincides with A at origin O at time t = 0. If we assume that E is the point ofintersection of the ct0 -axis and the perpendicular line that is drawn to x = vt fromD, we can write
DB= |OE| + |DE| i|B|= ct0 + x0 i.

(3.7)

However, |OE| and |DE| express distances between two points and they are notmagnitudes of complex numbers, as explained before.
If DB/ |B| is calculated using the complex numbers B = ct + vti and D = ct + xi,
the equations are
DB
DB
=√
|B|
2
|B|
(ct + xi)(ct − vti)=√
(ct + vti)(ct − vti)
c2 t2 − cvt2 i + xcti − xvti2√
c2 t2 − v 2 t2 i2
c2 t2 + xvt − cvt2 i + xcti√.
=
c2 t2 + v 2 t2=

Under the condition that ct > 0, we haveDB
c2 t(t + vx/c2 ) + ct(x − vt)i√=
|B|
ct 1 + v 2 /c2=c(t + vx/c2 ) + (x − vt)i√.
1 + v 2 /c2(3.8)

Because (3.7) and (3.8) are both equal representations of DB/ |B|, we can write
ct0 + x0 i =c(t + vx/c2 ) + (x − vt)i√.
1 + v 2 /c2

If we compare the coeﬃcients of the imaginary parts and the real parts, we havet + (v/c2 )x,t0 = √
1 + v 2 /c2
x − vtx0 = √.
1 + v 2 /c2

17

(3.9)
(3.10)

Equations (3.9) and (3.10) express the relations of coordinates (ct, x) of D observed
from A at rest and coordinates (ct0 , x0 ) of D observed from B with uniform velocity
v. In other words, they are the equations of coordinate transformations.
Next, we compare (3.9) and (3.10) with Lorentz transformations of special relativity. As explained in Section 3.1, Lorentz transformations aret − (v/c2 )xt0 = √,
1 − v 2 /c2
x − vtx0 = √.
1 − v 2 /c2

(3.3)
(3.4)

By comparing (3.3) and (3.4) with (3.9) and (3.10), we can understand that (3.9)
and (3.10) will become identical with (3.3) and (3.4) if the sign of c2 is changed into(−) from (+) in (3.9) and (3.10). The complex plane used to derive (3.9) and (3.10)
is a flat plane. However, according to general relativity, space–time curves. In otherwords, (3.9) and (3.10) are Lorentz transformations in flat space–time and (3.3) and(3.4) are those in curved space–time. Diﬀerences occur depending on whether thesign of c2 is (+) or (−). In addition, though mass bends space in general relativity,
we can tell by our previous results that space–time is already curved in special
relativity, in which mass does not yet exist.

3.4 The fourth imaginary number hHamilton expanded a complex number a + bi into a quaternion a + bi + cj + dk andexpressed the position of a point in four-dimensional space–time. Variables a, b, c,and d are real numbers and i, j, and k are imaginary numbers. The algorithms are
as follows:
i2 = j 2 = k 2 = −1,
ij = −ji = k, jk = −kj = i, ki = −ik = j.
Hamilton and his successor tried to describe physical laws using the quaternion in
the latter half of the 19th century, but they did not succeed. The reason for theirfailure is thought to be that the four-dimensional space–time, where Hamilton’squaternion can be used, is flat space–time. If we intend to use the quaternion for
coordinate transformations in curved space–time, an introduction of a new imaginary number is necessary. Therefore, we insert a fourth imaginary number h. The
algorithms are as follows:
h2 = −1,18hi = ih, hj = jh, hk = kh.
The algorithms of i, j, and k are identical to Hamilton’s algorithms.
If we assume hi = −ih, calculations become complicated and we cannot findLorentz transformations later. One can see this oneself. In addition, though i, j,
and k have mutual relations as ij = k, h has a single relation with each imaginary
number as hi = ih. The reason is that though we can move from space to space,
we cannot move from space to time. Because h expresses time and i, j, k express
space, h is unrelated to i, j, and k. In other words, we cannot mathematically movefrom space to time because hi = ih, hj = jh, and hk = kh. Time travel is notpossible.
From now on, the position of a point in four-dimensional space–time is describedby a new quaternion ah + bi + cj + dk. However, c does not represent the velocity oflight any more. In addition, a is the time part and b, c, and d are the space parts.
The number that is made by assuming c = d = 0 in the new quaternion is the newcomplex number ah + bi. ct is multiplied by imaginary number h and is calculated
in cth if the new complex number is used. However, c in this case is the velocity oflight. We expect that the signs of c2 of (3.9) and (3.10) become (−) from (+) if we
perform coordinate transformation DB/ |B| using this new time unit cth because(cth)2 = −c2 t2 .

3.5 Derivation of Lorentz transformations using new complex
numbers
The horizontal axis becomes the cth-axis and the vertical axis becomes the xi-axis
in the new complex plane if the new complex number ah+bi is used. As for observerA, observer B, and stationary point D in the new complex plane, the coordinatesA(ct, 0), B(ct, vt), and D(ct, x) do not change; however, the new complex numbers
to express their positions become A = cth, B = cth + vti, and D = cth + xi. If wecalculate DB/ |B| again, we have
DB
DB
=√
|B|
2
|B|
(cth + xi)(cth − vti)=√
(cth + vti)(cth − vti)
c2 t2 h2 − cvt2 hi + xcthi − xvti2√
c2 t2 h2 − v 2 t2 i2
2 2−c t + xvt − cvt2 hi + xcthi√=.
c2 t2 h2 − v 2 t2 i2=19(3.11)

The denominator of (3.11) is√√
c2 t2 h2 − v 2 t2 i2 = −c2 t2 + v 2 t2√
= −c2 t2 (1 − v 2 /c2 ),(3.12)and the sign inside the square root symbol becomes negative in the condition wherev < c, which means that the velocity v of observer B is under the velocity of light c.
Therefore, if we form an axiom: place the negative sign outside of the square rootwhen the quantity inside the square root is negative, we find√√c2 t2 h2 − v 2 t2 i2 = cth 1 − (v 2 t2 i2 )/(c2 t2 h2 )√= cth 1 − v 2 /c2 ,(3.13)√
because c2 t2 h2 = −c2 t2 < 0 and ct > 0. In addition, h2 = h is an axiom. An
axiom is a law that is thought to be right, but cannot be proved. The two previous
axioms are termed Axiom 9 and Axiom 10 in Section 13.2.
In addition, if the velocity v is beyond the velocity of light c, i.e., v > c, (3.12)
becomes√√c2 t2 h2 − v 2 t2 i2 = −c2 t2 + v 2 t2√
= c2 t2 (v 2 /c2 − 1)√
= ct v 2 /c2 − 1.
Therefore, the equations of Lorentz transformations with the superluminal velocity
can be obtained if coordinate transformations are performed using this denominator.
We will calculate this in Chapter 19.
Then, if we apply (3.13) to (3.11) and calculate again, the equations become
DB
DB
=√
|B|
2
|B|
−c2 t2 + xvt − cvt2 hi + xcthi√
c2 t2 h2 − v 2 t2 i22−c t(t − vx/c2 ) + ct(x − vt)hi√=
cth 1 − v 2 /c2==−(c/h)(t − vx/c2 ) + (x − vt)i√
1 − v 2 /c2=−(ch/h2 )(t − vx/c2 ) + (x − vt)i√
1 − v 2 /c2=c(t − vx/c2 )h + (x − vt)i√.
1 − v 2 /c2
20(3.14)

Because the new complex number used to express the position of D as seen from B
is ct0 h + x0 i, if we use the new complex number, we can write it using (3.14) as
ct0 h + x0 i =c(t − vx/c2 )h + (x − vt)i√.
1 − v 2 /c2

If we compare the coeﬃcients of the imaginary parts and the real parts, we havet − (v/c2 )xt0 = √,
1 − v 2 /c2
x − vt.
x0 = √
1 − v 2 /c2(3.15)(3.16)

Equations (3.15) and (3.16) are identical to Lorentz transformations, i.e.,t − (v/c2 )xt0 = √,
1 − v 2 /c2
x − vtx0 = √,
1 − v 2 /c2

(3.3)
(3.4)

which was explained in Section 3.1.
We can obtain Lorentz transformations after performing coordinate transformation DB/ |B| using the new complex number cth + xi, as shown above. In otherwords, the fourth imaginary number h is a number necessary to describe the curved
four-dimensional space–time and is not a number of suppositions. From now, we
express the position of a point in four-dimensional space–time with the new quaternion cth + xi + yj + zk and the position in the new complex plane by the conditiony = z = 0.
In special relativity, there is a case in which time ct is multiplied by imaginary
number i and calculated as cti. However, in that case, lengths x, y, and z are
calculated as real numbers. Therefore, cti is a number unlike cth of the new quaternion. Because cti in relativity theory is used for calculation convenience, cti does
not express the actual condition of space–time. On the contrary, cth of the newquaternion cth + xi + yj + zk expresses the property of curved space–time and cth
is the number that is necessary for investigations of space–time.
In addition, though we use the condition ct > 0 to obtain (3.14), we do not makeit necessary that t > 0. We find ct > 0 if c < 0 and t < 0. Therefore, we canobtain Lorentz transformations under the condition of t < 0. However, because
the universality of Lorentz transformations is lost if the sign of the velocity of lightc changes in the case of positive or negative time t, we will continue under thecondition that c is always positive.
21

3.6 The reason why time advances only in the positive
direction
In the last section, we assumed that the velocity of light c is always positive. If
that assumption is used, we can explain the reason why time advances only in thepositive direction in our world.
If we had calculated:(cth + xi)(cth − vti)
DB
=√
|B|
(cth + vti)(cth − vti)=

−c2 t2 + xvt − cvt2 hi + xcthi√
c2 t2 h2 − v 2 t2 i2(3.11)

in the last section under the conditions of c > 0, t < 0, i.e., ct < 0, we would have−c2 t2 + xvt − cvt2 hi + xcthi
DB
√
=
|B|
c2 t2 h2 − v 2 t2 i22−c t(t − vx/c2 ) + ct(x − vt)hi√=−cth 1 − v 2 /c2=(c/h)(t − vx/c2 ) − (x − vt)i√
1 − v 2 /c2=c(vx/c2 − t)h + (vt − x)i√.
1 − v 2 /c2(3.17)

Because the new complex number expressing the position of D as seen from B is
ct0 h + x0 i, we can write it using equation (3.17) as
ct0 h + x0 i =c(vx/c2 − t)h + (vt − x)i√.
1 − v 2 /c2

If we compare coeﬃcients of the imaginary parts and real parts, we have(v/c2 )x − tt0 = √,
1 − v 2 /c2
vt − xx0 = √.
1 − v 2 /c2(3.18)√
The axiom h2 = h is used here.
In the case of x = 0 in (3.18), we have−tt0 = √.
1 − v 2 /c2
From this equation, the sign of time t of D as seen from A is reversed compared tothe sign of time t0 of D as seen from B. Because B coincides with A if the velocity
22

of B is 0, it is unnatural that the sign of t should be reversed with respect to t0 .
Thus, the first suppositions of c > 0, t < 0 are thought to be incorrect. In otherwords, according to the calculations, time t can move only in the positive directionin our world. This is the reason why we cannot return to the past.
In addition, we will discuss whether we can go against time in the superluminal
velocity region in Section 19.2.

23

24

4
New Complex Plane andOblique Coordinate Axes

4.1 New complex number to express the rotation of
coordinate axes
Because the complex plane is a plane without a curve, an argument does not change
by rotation. Moreove, if we assume that a straight line x = vt is the new real axis,
i.e., the ct0 -axis, as shown in Figure 3.2, the new imaginary axis, i.e., the x0 i-axis,
is a straight line that is perpendicular to the straight line x = vt through origin O.

However, because in reality the new complex plane is a plane that curves, it isunknown where the new imaginary axis, i.e., the x0 i-axis, moves to. It is diﬃcult
to use a figure if we do not know the placement of the x0 i-axis. In this section,
we calculate the place where the x0 i-axis moves to. This calculation is necessary to
examine whether the new imaginary number h is a fictitious number or an actual
number.
If observer B moves in the x-direction with respect to observer A with uniform
velocity v, the new complex numbers A and B are A = cth and B = cth + vti,
25

respectively. We considere a case, in which the time of A is t0 . Because the time t0of A and B are the same, we have Figure 4.1.

At this time, by coordinate transformations, point A(ct0 , 0) in the cth-axis moves
to point B(ct0 , vt0 ) in the ct0 h-axis and point D(0, x1 ) in the xi-axis moves to pointE(ct2 , x2 ) in the x0 i-axis. However, all the coordinates of A, B, D, and E are thecoordinates seen from A. If the coordinates of B are the coordinates seen fromB, imaginary number h of new complex number A = ct0 h and h of new complex
number B = ct0 h + vt0 i are not in the same axis. In that case, the following
discussions are not realized.
Because the movement from A to B is a rotation around origin O, it can be
expressed by new complex number H. In addition, if it is not a rotation aroundorigin O, the movement is not expressed by a new complex number.Because the
new complex numbers of A and B are A = ct0 h and B = ct0 h + vt0 i, respectively,
using AH = B, we have
ct0 hH = ct0 h + vt0 i,
ct0 h + vt0 i
ct0 h
vi
=1+
ch
vih
=1+ 2
ch
v
= 1 − hi.
c

H=(: t0 6= 0)(4.1)(4.1) changes the cth-axis into the ct0 h-axis of the new complex number system.

4.2 Proof of the oblique coordinate axes
New complex numbers expressing the positions of D and E in Figure 4.1 are D = x1 i
and E = ct2 h + x2 i, respectively. Because D is moved to E by transformation H,
26

which moves A to B, using DH = E we havex1 iH = ct2 h + x2 i.
If (4.1) is substituted for this equation, we findv
ct2 h + x2 i = x1 i(1 − hi)cvx1 2= x1 i −hicv= x1 i + x1 hcv
= x1 h + x1 i.
cAfter comparing the coeﬃcients, we havect2 =v
x1 , x2 = x1 .
c

And if x1 is eliminated from the two formulae, we can writex2 =c(ct2 ).
v(4.2)

Equation (4.2) is an equation of the x0 i-axis. Because the unit of the temporal axis
is ct, the slope of the x0 i-axis is not c2 /v but c/v. The equation of the ct0 h-axis, towhich the cth-axis moves, is as follows:
x = vt =

v
(ct).
c

Thus, the slope of the ct0 h-axis becomes v/c and as shown in Figure 4.2, the x0 iand the ct0 h-axes lean inward at the same arguments from the xi- and cth-axes. In
other words, the x0 i- and ct0 h-axes are oblique coordinate axes.

27

From the above results, if we move one axis of the coordinate axes that makea couple, a pair of new coordinate axes become oblique coordinate axes in the
new complex plane. Although coordinate axes become oblique coordinate axes in
special relativity as well, a time part is calculated by temporarily turning it into
imaginary number cti because the axes become rectangular coordinate axes throughcalculations using ct. However, as stated above, the special operation for making
oblique coordinate axes is not necessary with the new complex numbers.
Because the Minkowski space–time diagram is a flat surface, if the coordinate
axes rotate, they become rectangular coordinate axes. On the other hand, in the
new complex plane, the rotations of coordinate axes make oblique coordinate axes.
The new complex plane improves upon the Minkowski space–time diagram in thisregard as well.

4.3 A simple method to obtain oblique coordinate axes
We have come to understand that the coordinate axes in the new complex plane are
oblique coordinate axes. Now we consider the method by which we can obtain theequation of a coordinate system that makes a pair when the equation of anothercoordinate axis is given.
The equation of the world line of light is x = ct, which means that light isa straight line equidistant from both coordinate axes. Because oblique coordinate
axes inwardly declining from the orthogonal coordinate axes are linearly symmetric,
we have the same result even if we replace the cth- and xi-axes. Therefore, we canhave the equations of the coordinate axes that make a pair if we switch x and ct ina given equation. For example, if we switch ct and x in the equationx = vt =v(ct)
cof the ct0 h-axis in Figure 4.2, we have
ct =

v
x.
c

If the equation is transformed, it becomes
x=

c
(ct),
v

and it corresponds with (4.2).
If we use this method, we can easily have the equations of coordinate axes thatmake a pair if the coordinate axes are curved. Curved coordinate axes mean thatobserver B is accelerating. We argue this point in Section 20.2.
28

5Length in
the New Complex Plane andLorentz Transformations

5.1 Method of obtaining Lorentz transformations fromlength in the new complex plane
As proven in Section 3.5, the new complex numbers of observer A at rest, observer
B moving along a straight line with constant velocity v, and observed point D are
A = cth, B = cth+vti, and D = cth+xi, respectively. The Lorentz transformations
can be found by coordinate transformation DB/ |B|.
There exists another method to find the Lorentz transformations by the coordinates of the nodes of the world lines in the new complex plane. This is the second
method to find the Lorentz transformations using the new complex number. By
this method, neither emission of light, nor the coordinate transformation by the
new complex number, is needed.

29

It is assumed that observer A is at rest at the origin O, and observer B moves withconstant velocity v in the positive direction along the x-axis. The observed point D
is at rest at position x0 . In addition, it is assumed that A coincides B at O at time
t = 0. The coordinates of A, B, and D after t0 seconds are A(ct0 , 0), B(ct0 , vt0 ),and D(ct0 , x0 ), respectively. Figure 5.1 shows these coordinates drawn in a newcomplex plane.
A moves along the cth-axis. B moves along the straight line x = vt, i.e., thect0 h-axis. The node where the perpendicular line from D meets the xi-axis is G,
and D moves along the straight line GD. The node where the straight line x = vt
meets the straight line drawn from D and is parallel to the x0 i-axis, is E. The
node where the x0 i-axis meets the straight line drawn from D and is parallel to thect0 h-axis, is F . As explained in Section 4.3, if x and ct are switched in the equationx = vt =of the ct0 h-axis, we have

v
(ct)
c

ct =

v
x.
c

x=

c2
t
v

The transformed equation

corresponds to the x0 i-axis.
The coordinates (ct0 , x0 ) of D, seen from the ct0 h- and x0 i-axes, which are theoblique coordinate axes of observer B, are the lengths of nodes E and F from O.
Then, we can write
ct0 = |OE| = |DF | , x0 = |OF | = |DE| .

(5.1)

However, |OE| , |DF | , |OF |, and |DE| do not express the magnitudes of the complex numbers but the lengths between the two points. As is shown below, we findt0 and x0 from the coordinates of the nodes of two straight lines.

(1) t0 is obtained from the coordinates of point E
The slope of the x0 i-axis is c/v, as can be seen from the equationc2tv
c= (ct).
v

x=30

Because straight line DE is parallel to the x0 i-axis, its argument is c/v. Also,
because straight line DE passes through point D(ct0 , x0 ), its equation isc(ct − ct0 ),v
c2
x = (t − t0 ) + x0 .
v

x − x0 =(5.2)

In addition, the equation of straight line OB isx = vt.

(5.3)

Thus, if x is eliminated from (5.2) and (5.3), we havec2(t − t0 ) + x0 = vt,v
c2c2t − vt = t0 − x0 ,vv
(c2 − v 2 )c2
t = t0 − x 0 ,vv
22(c − v )t = c2 t0 − vx0 ,t=

c2 t0 − vx0. (: c 6= v)
c2 − v 2(5.4)

If t is eliminated from (5.3) and (5.4), we findx=v(c2 t0 − vx0 ).
c2 − v 2(5.5)

If the coordinates of E are (ct1 , x1 ), then from (5.4) and (5.5), we havect1 =v(c2 t0 − vx0 )c(c2 t0 − vx0 ), x1 =.
22c −v
c2 − v 2(5.6)

However, (ct1 , x1 ) are the coordinates as seen from the cth- and xi-axes.
From (5.6), as well as the definition of length by the new complex number, length|OE| is obtained by the definition of magnitude. We find2

|OE| = EE= (ct1 h + x1 i)(ct1 h − x1 i)= −c2 t21 + x21
c2 (c2 t0 − vx0 )2v 2 (c2 t0 − vx0 )2+(c2 − v 2 )2(c2 − v 2 )22
2
(c t0 − vx0 )=−c2 − v 22
c (t0 − vx0 /c2 )2=−.
1 − v 2 /c2

=−

31

(5.7)

Sincect0 = |OE| = |DF | , x0 = |OF | = |DE|

(5.1)

and since the new complex number of point E as seen from B after the coordinatetransformation is ct0 h, we have|OE|2 = (ct0 h)2= −c2 t02 .

(5.8)

Because (5.7) and (5.8) are the same, we can writec2 (t0 − vx0 /c2 )2,
1 − v 2 /c2(t0 − vx0 /c2 )2t02 =.
1 − v 2 /c2− c2 t02 = −(5.9)

To find t0 from (5.9), we consider the signs of the numerator and denominator of(5.9). If c > v, the sign of the denominator of (5.9) is
1 − v 2 /c2 > 0.

(5.10)

Calculations to determine the sign of the numerator are somewhat complex.

As shown in Figure 5.2, the coordinates of the node where straight line AD meetsthe x0 i-axis is H. The x coordinate is obtained by substituting t for t0 in theequation of the x0 -axis, i.e.,c2
x = t,v
and we findc2
x = t0 .
v
32Because point D(ct0 , x0 ) lies vertically below point H, we havec2
t0 = x0 .
v
Therefore, from the numerator of (5.9), we can writet0 − vx0 /c2 = t0 −c2
c2 vt0t0 (v/c2 ) = t0 − 2 = 0v
c v

or in short,t0 − vx0 /c2 = 0.

(5.11)

From (5.10) and (5.11), (5.9) becomest0 − (v/c2 )x0
t0 = √.
1 − v 2 /c2
If t0 and x0 are replaced with t and x in order to use generalized notation, thisbecomest − (v/c2 )xt0 = √,
1 − v 2 /c2

which is a standard equation of the Lorentz transformations, i.e.,t − (v/c2 )xt0 = √.
1 − v 2 /c2(3.3)

As shown above, the equation for t0 in the Lorentz transformations can be foundfrom the coordinates of the node of the world lines in the new complex plane.
(2) x0 is obtained from the coordinates of point F

33

In Figure 5.1, from the equationx = vtv= (ct)
cof the ct0 h-axis, the slope is v/c. Because straight line DF is parallel to the ct0 h-axis,
the slope is v/c. Thus, because straight line DF passes through point D(ct0 , x0 ),its equation becomesv(ct − ct0 ),c
x = v(t − t0 ) + x0 .

x − x0 =(5.12)

Because straight line OF lies along the x0 i-axis, its equation becomes
x=

c2
t.
v

(5.13)

If x is eliminated from (5.12) and (5.13), we havec2
t = v(t − t0 ) + x0 ,v
(c2 − v 2 )t = −vt0 + x0 ,v
v(x0 − vt0 )t=. (: c 6= v)
c2 − v 2(5.14)

If t is eliminated from (5.13) and (5.14), we havex=c2 (x0 − vt0 ).
c2 − v 2(5.15)

If the coordinates of point F are (ct2 , x2 ), then from (5.14) and (5.15), we findct2 =c2 (x0 − vt0 )cv(x0 − vt0 )
,
x
=
.
2
c2 − v 2c2 − v 2(5.16)

However, (ct2 , x2 ) are also the coordinates as seen from the cth- and xi-axes.
From the definition of the square magnitude by the new complex number andfrom (5.16), we can write
2|OF | = F F= (ct2 h + x2 i)(ct2 h − x2 i)= −c2 t22 + x22

34

c2 v 2 (x0 − vt0 )2
c4 (x0 − vt0 )2+(c2 − v 2 )2(c2 − v 2 )222
c (x0 − vt0 )=c2 − v 2(x0 − vt0 )2=.
1 − v 2 /c2=−(5.17)

Because the equations
ct0 = |OE| = |DF | , x0 = |OF | = |DE|

(5.1)

show that the new complex number of point F as seen from B after the coordinatetransformation is x0 i, we have
|OF | = (x0 i)(−x0 i)
2= x02 .

(5.18)

Because (5.17) and (5.18) are the same, we can writex02 =(x0 − vt0 )2.
1 − v 2 /c2(5.19)

If c > v, the sign of the denominator of (5.19) is(1 − v 2 /c2 ) > 0.

(5.20)

Because point D lies above point B in Figure 5.2, we have x0 = vt0 .

Therefore, the sign of the numerator of (5.19) becomes
x0 − vt0 = 0.
35

(5.21)

From (5.20) and (5.21), (5.19) becomes
x0 − vt0.
x0 = √
1 − v 2 /c2
If t0 and x0 are replaced with t and x in order to use generalized notation, thisbecomesx − vtx0 = √,
1 − v 2 /c2
which is the standard equation of the Lorentz transformations, i.e.,x − vt.
x0 = √
1 − v 2 /c2(3.4)

As shown above, the equation for x0 in the Lorentz transformations can be foundfrom the coordinates of the node of the world lines in the new complex plane.
In the above verifications, the Lorentz transformations were obtained by the following methods without emission of light and without using the coordinate transformation by the new complex number.
1) The coeﬃcient, which changes time into length, is c, and time is calculated byct.
2) The rest frame of observer A is made into a rectangular frame and the world
line of moving observer B is set as the temporal axis of the oblique frame. The
equation of the length axis of the oblique frame is obtained by switching x andct in the equation of the temporal axis in this frame.
3) The coordinates of nodes E or F of each oblique coordinate axis and the straight
line, which is drawn from observed point D and parallel to the oblique coordinate
axis, are calculated from the equations of the world lines.
4) If the coordinates of nodes E or F are (ct, x), the world distance is calculatedby√(cth + xi)(cth − xi),

because the new complex number of that point is cth + xi.
5) The length from O to point E or F on the oblique coordinate axis becomes thecoordinates (ct0 , x0 ) of the point at rest in the oblique frame.
This result shows that the Lorentz transformations can be obtained using only the
new complex number and the new complex plane. In contrast, in Minkowski spacetime diagrams, Lorentz transformations cannot be obtained from the coordinates of
the nodes of world lines.

36

5.2 Equation of a straight line through new complex numbers
Though the new complex plane is a plane consisting of two imaginary axes, why
can we obtain the nodes using only the equations of real numbers? To answer this,
we investigate whether the equation of the straight line in the new complex plane
may be written with real numbers.

As shown in Figure 5.3, there are two points A and B in the new complex plane
having coordinates (ct0 , x0 ) and (ct1 , x1 ), respectively. If the equation corresponding to the straight line that passes through them is calculated with real numbers,
the slope is
x1 − x0
ct1 − ct0
and it passes through point A(ct0 , x0 ); hence, the equation of the straight line isx − x0 =The equation becomesx=x1 − x0(ct − ct0 ).
ct1 − ct0x1 − x0(t − t0 ) + x0 .
t1 − t0(5.22)

Then, the equation corresponding to the straight line is calculated by the newcomplex number. The points A and B are displayed as A(ct0 h, x0 i) and B(ct1 h, x1 i),
respectively. Because the slope is
x1 i − x0 i
ct1 h − ct0 h
and it passes through point A(ct0 h, x0 i), the equation of the straight line isxi − x0 i =x1 i − x0 i
(cth − ct0 h).
ct1 h − ct0 h

These equations become(x − x0 )i =(x1 − x0 )i(t − t0 )ch,(t1 − t0 )ch37

x1 − x0(t − t0 ),t1 − t0x1 − x0x=(t − t0 ) + x0 .
t1 − t0x − x0 =(5.23)(5.23) is the same as (5.22). In other words, even if the equation of the straightline is calculated by new complex numbers, it will be the same equation as thatobtained by real numbers. Thus, the straight line in the new complex plane can becalculated by equations of real numbers.

5.3 Method to express coordinate parts through newcomplex numbers
Hamilton invented the method of denoting coordinate parts by the combination ofreal numbers (a, b) if the complex number a + bi is drawn in the complex plane. If
we see the coordinate axes in the complex plane, we can easily determine whether aand b express real or imaginary parts. However, Hamilton’s quaternion consists of
one real number and three imaginary numbers, and the new quaternion consists offour imaginary numbers. Thus, we must determine whether the four real numbers(ct, x, y, z) are real or imaginary by placing them on the coordinate axes. To solve
this inconvenience, real and imaginary numbers are written in ( , , , ) form fromnow on. We can see that no contradictions occur if we change into this notation,
because the coordinates of point A(ct0 , x0 ) can be calculated by writing them as
A(ct0 h, x0 i), as explained in the last section.
The reader who begins with Chapter 6 without first reading Chapters 1-5 may
think that the notation of writing imaginary numbers in ( , ) is wrong. However,from Chapter 6, imaginary numbers are written in this way because of the reason
explained above. Since the meaning can be easily understood if imaginary numbers
are written in ( , ), we could use the new notation from Chapter 1. However, because
the explanation of the new complex number was not completed, Hamilton’s notation
was used until Chapter 5.

5.4 Correctness of the imaginary number h
As explained in Section 3.4, in order for the equations of the coordinate transformations by the complex number and the equations of Lorentz transformations of special
relativity to coincide, a fourth imaginary number h was introduced, in addition to
Hamilton’s three imaginary numbers i, j, and k. Because calculations in curved
38

space–time can be performed without contradictions by h, it is appropriate to thinkof h as having the ability to express the property of this curved four-dimensional
space–time.
Typical mathematics and physics explain the phenomenon in flat space–time.
Though tensors are used to apply the mathematics and physics to curved spacetime, it is sometimes diﬃcult to master the tensor since the symbols are complex.
However, as shown in Section 3.5 and Section 5.1, if the time part is multiplied byh and length parts by i, j, and k, the mathematics and physics required for curved
space–time can be derived by ordinary methods. This is a result of the fact that the
new quaternion itself contains curved properties. Curved space–time becomes flat
when we apply the new quaternion. Thus, we could precisely explain the physicalphenomenon in four-dimensional space–time using the new quaternion.

39

40

6
Time Dilation andLength Contraction

6.1 Proper time and time dilation
Proper time τ (tau) is an important concept next to Lorentz transformations in
special relativity. Most of the conclusions of special relativity can be obtained using
Lorentz transformations and the proper time. Proper time is the time measured
by the inertial frame. The time of observer B and observed point D is proper
time if B comes next to and coincides with D. Because it is diﬃcult to express
the proper time in this explanation, the Minkowski space-time diagram is used to
explain it visually. However, because the slope of the world line is the inverse of thevelocity in the Minkowski space-time diagram, it is diﬃcult to understand proper
time. In addition, in the diagram, the length of the world line is incalculable fromthe coordinates of the nodes of world lines. However, because the slope of the world
line expresses the velocity, and the length can be calculated from the coordinatesof the node in the new complex plane, we explain the proper time using the newcomplex plane in this section.

41

Although observed point mass D was at rest prior to now, we currently assume
that D is in a state of uniform linear motion, similar to observer B. Also, as shownin Figure 6.1, it is assumed that observer B coincides with D. At this time, thevelocity of D becomes v, and the straight line x = vt becomes the temporal axis,
i.e., the ct0 h-axis, of D. In other words, the ct0 h-axis is the world line of B and
D. As seen from B and D, their length x0 is always 0, but time t0 passes. Because|OD| is the length from origin O to point D in the ct0 h axis in Figure 6.1, |OD| isthe proper time τ of D. In other words,|OD| = cτ h.
The reason why we cannot remove h, which is attached to |OD| in this equation, is
explained as follows. Using the new complex conjugate D = cτ h and the definitionof magnitude, i.e., |A| = AA, we have2

|OD| = DD= (cτ h)2
= −c2 τ 2 .

(6.1)

Because −c2 τ 2 < 0, using the axiom that place the negative sign outside of thesquare root when the quantity inside the square root is negative, we can write√|OD| = −c2 τ 2= cτ h.
As this result shows, in curved space–time, the magnitude of time is an imaginary
number, and thus, its square is negative. In flat space–time, the magnitude |A| is apositive real number, whereas |A| is a positive real number or a positive imaginary
number in curved space-time.
Then, we obtain the relations between the proper time τ of moving point mass
D and the time t of observer A at rest. Because the coordinates of D as seen from
A are (cth, vti) in Figure 6.1 and the new complex number of D is D = cth + vti,
we have2|OD| = DD= (cth + vti)(cth − vti)
= (cth)2 − (vti)2= −c2 t2 + v 2 t2
= −c2 t2 (1 − v 2 /c2 ).
42

(6.2)

From (6.1) and (6.2), we have−c2 τ 2 = −c2 t2 (1 − v 2 /c2 ),τ 2 = t2 (1 − v 2 /c2 ).
If c > v, t > 0, and τ > 0, we find√
τ = t 1 − v 2 /c2 .
In (6.3), we have
1>

√

(6.3)

1 − v 2 /c2 .

Thus, we find τ < t, and it has therefore been shown that the proper time τ ofmoving point mass D is less than the time t of observer A at rest. This is thefamouse time-dilation theory for a moving clock in special relativity. In addition,
because time is an imaginary quantity in our world as previously explained, it iscorrect to write (6.3) as√τ h = th 1 − v 2 /c2 .
However, it is usually written in the form of (6.3), because only a real part can beobserved.
In Figure 6.1, it appears that |OD| > |OA|. In other words, we can see that
τ > t. However, τ < t in (6.3); thus, there appears to be a contradiction. However,
because the new complex plane represents curved two-dimensional space–time, aline segment parallel to a coordinate axis is the longest, and it becomes shorter
when inclined. The straight line, which is maximally inclined most from both thecth- and xi-axes, is a light ray, and the length of the ray from the origin is 0. These
results can be confirmed by calculating the magnitudes |OP |, |OQ|, and |OR| after
arranging the points P (ch, 0), Q(ch, i), and R(ch, ci) in the new complex plane.
These calculations are not presented here.
Many researchers in the field of special relativity think that the proper time τof the point mass is invariant and they transform (6.3) and express the time t ofobserver A by
τ
t= √
.
(6.4)
1 − v 2 /c2
If v changes, t changes in (6.4). However, τ is invariant. Although moving point
mass D is only one example, many observers exist with its velocity. In other words,
because the invariant time is τ only, it is called the proper time of the point mass.
From now on, we consider physical phenomena with respect to the moving pointmass. As measured by an observer co-moving with the point mass, the length iscalled proper length, and the mass is called relativistic mass or rest mass.
43

The trace of a point mass moving in the four-dimensional space–time is called
the world line in relativity theory, and the length of the world line is called theworld distance or four-dimensional distance. Because |OD| is the world distance sof point mass D, we can write s = cτ h. In Section 7.1, it will be explained that theworld distance is invariant in Lorentz transformations in special relativity. Thus,
the proper time is also invariant.

6.2 Special relativity and length contraction
The length of a rod moving with velocity v shortens in special relativity. The proofin special relativity is explained here before being proven with the method using
the new complex number. This is the proof written in the book of relativity theory.
As shown in Figure 6.2, a rod D with length l moves with uniform, linear motion
along the x-axis. It is assumed that the two-dimensional space–time coordinates ofthe ends of the rod as seen from observer A at rest are (ct1 , x1 ) and (ct2 , x2 ), andthose from observer B moving with uniform velocity v are (ct01 , x01 ) and (ct02 , x02 ).

From the Lorentz transformationx − vtx0 = √,
1 − v 2 /c2
we can write

x1 − vt1
x01 = √,
1 − v 2 /c2
x2 − vt2.
x02 = √
1 − v 2 /c2

(3.4)

(6.5)
(6.6)

If (6.5) is subtracted from (6.6), we findx02 − x01 =(x2 − x1 ) − v(t2 − t1 )√.
1 − v 2 /c2(6.7)

Because the times of the both ends of the rod are the same as seen from observer
A, we can write t1 = t2 . Therefore, (6.7) becomesx2 − x1
x02 − x01 = √.
1 − v 2 /c2
44

(6.8)

If the velocity of B coincides with D, the length of the rod measured by B iscalled the proper length. Thus, x02 − x01 is the proper length l0 . In addition, thelength x2 − x1 of the rod measured by A is the length l of the rod that changes with
diﬀerent observers. Thus, from (6.8), we havel0 = √The equation becomesl = l0l
1 − v 2 /c2.

√
1 − v 2 /c2 .

(6.9)√
Since 1 − v 2 /c2 < 1, (6.9) shows that l < l0 . In other words, if a rod with length
l0 is at rest at A and has velocity v after leaving from A, the length l shortens if it
is measured by A. In addition, the proper length l0 measured by B, who is moving
with rod D, does not change.

6.3 Coordinates of the node of the world lines and length
contraction
In this section, (6.9) is found by using the length of the world line to show thatthe conclusion of special relativity is obtained from the coordinates of the nodeof the world lines. For that purpose, it is considered to be suﬃcient to apply thesame method used to calculate the proper time. However, the method to obtain theproper length is somewhat complex. Figure 6.3 shows the new complex plane.

We assume that the position of observer B coincides with the near edge of the rod,and the tip of the rod is point D. Because B coincides with the rod, the velocities

45

of B and the rod are both v. The straight line x = vt becomes the temporal axis,
i.e., the ct0 h-axis, of B and the rod. The straight line
x=

c2
t
v

that is the linear symmetry for a light ray of the straight line x = vt becomes thex0 i-axis. The equation of the straight line of linear symmetry for a light ray is
obtained by switching x and ct in the equation x = vt as explained in Section 4.3.
The slope of the rod is the problem in Figure 6.3. If the times of both ends ofthe rod measured by B are t0 , the rod is parallel to the x0 i axis in the new complex
plane. In other words, the tip D0 of the rod is inclined in the new complex plane.
In this case, because the times of each end of the rod measured by observer A are
diﬀerent, this observer cannot measure the length of the rod. Therefore, because
the times of each end of the rod must be the same time t as seen from A, tip Dof the rod and the near edge of the rod, i.e., B , lies on the same line of the sametime t in the new complex plane. In other words, the rod becomes perpendicular.
An important point is that the times of D and B, i.e., the ends of the rod, as seenfrom observer A at rest must be the same.

In Figure 6.4, E is the node where the ct0 h-axis meets the straight line, which isparallel to the x0 i-axis, and drawn from point D. In this case, |DE| is the proper
length l0 of the rod as seen from B. This is due to the fact that the times of D
and E are the same as seen from B. In addition, |BD| is the length of the rod l
as seen from A. Point B is on x = vt, and the coordinates of B are (ct1 h, vt1 i) if
the coordinates (ct1 h, 0) of point A are assumed. Furthermore, because point D
is at a distance l higher than point B in the xi-direction, the coordinates of D are(ct1 h, vt1 i + li). Since straight line DE is parallel to the x0 i-axis, i.e., straight line46

x = (c/v)ct, and since it passes through the point D(ct1 h, vt1 i + li), the equationof DE becomescx − (vt1 + l) = (ct − ct1 ).
(6.10)v
The real part (ct1 , vt1 + l) of point D(ct1 h, vt1 i + li) is used here, because even
if new complex numbers or real numbers are used, the equations are the same as
those presented in Section 5.2.
Because the coordinates of point E are the coordinates of the point of the node
where straight line DE meets the straight line x = vt, we can eliminate x from
(6.10) and x = vt, leading toc(ct − ct1 ),v(c2 /v − v)t = (c2 /v − v)t1 − l,vt − (vt1 + l) =t = t1 − l/(c2 /v − v)
= t1 − lv/(c2 − v 2 ).

(6.11)

By eliminating t from (6.11) and x = vt, we findx = vt1 − lv 2 /(c2 − v 2 ).

(6.12)

If we assume that (ct2 h, x2 i) are the coordinates of point E, from (6.11) and (6.12),
we can write
ct2 = ct1 − clv/(c2 − v 2 ),(6.13)

x2 = vt1 − lv 2 /(c2 − v 2 ).

(6.14)

If the new complex number expressing DE is written as (DE), from D(ct1 h, vt1 i +li), (6.13), and (6.14), the equations become(DE) = D − E= [ct1 h + (vt1 + l)i]
{[
]
[
] }
− ct1 − clv/(c2 − v 2 ) h + vt1 − lv 2 /(c2 − v 2 ) i[]= clvh/(c2 − v 2 ) + l + lv 2 /(c2 − v 2 ) i,
and
{
[
] }
2
|DE| = clvh/(c2 − v 2 ) + l + lv 2 /(c2 − v 2 ) i{[] }× clvh/(c2 − v 2 ) − l + lv 2 /(c2 − v 2 ) i[]2= c2 l2 v 2 h2 /(c2 − v 2 )2 − l + lv 2 /(c2 − v 2 ) i2
47

[
]2
= −c2 l2 v 2 /(c2 − v 2 )2 + l + lv 2 /(c2 − v 2 )−c2 l2 v 2 + l2 (c2 − v 2 )2 + 2l2 v 2 (c2 − v 2 ) + l2 v 4(c2 − v 2 )222 2
4l (−c v + c − 2c2 v 2 + v 4 + 2c2 v 2 − 2v 4 + v 4 )=(c2 − v 2 )222 2
4l (−c v + c )=(c2 − v 2 )22 2 2
l c (c − v 2 )=(c2 − v 2 )2
l2 c2
= 2
c − v2l2=.
1 − v 2 /c2=(6.15)

If we assume that the new complex number expressing DE as seen from B is (DE)B ,we have(DE)B = l0 i.

(6.16)

Thus, the equations become
2|DE| = (DE)B (DE)B= l0 i(−l0 i)= l02 .

(6.17)

From l0 > 0, l > 0, (6.15), and (6.17), we can writel2,
1 − v 2 /c2ll0 = √,
1 − v 2 /c2√
l = l0 1 − v 2 /c2 .

(6.18)√
1 − v 2 /c2 ,

(6.9)

l02 =

Because (6.18) is the same asl = l0which is obtained in special relativity, it is clear that the equation of length contraction in special relativity can be proven using the new complex plane.
In addition, l and l0 are real numbers in (6.18), but as shown in (6.16), we live
in a world where the distances are expressed with imaginary numbers. The correct
distances are li and l0 i, but l and l0 can be observed.
48

7
The New Quaternion andWorld Distance

7.1 Definition of world distance in special relativity
In relativity theory, the interval of a world line is called world distance. In thischapter, the world distance is found using the new complex number. However,before coming to that, we explain how to find the world distance, which is written
in standard books on relativity theory.
The light emitted at time t = 0 from origin O forms the spherical surface of radius
ct at time t. The equation of the sphere is
x2 + y 2 + z 2 = (ct)2 ,which can be rewritten as(ct)2 − x2 − y 2 − z 2 = 0.
In special relativity, the value ofs2 = (ct)2 − x2 − y 2 − z 2(7.1)

does not change, even if the Lorentz transformationst − (v/c2 )xt0 = √,
1 − v 2 /c2
x − vtx0 = √
1 − v 2 /c2

(3.3)
(3.4)

are substituted. Thus, (7.1) is defined as world distance.
Calculations can then be performed. The Lorentz transformations (3.3), (3.4)
andy 0 = y,49

(3.5)

z0 = z

(3.6)

are substituted for (ct0 )2 − x02 − y 02 − z 02 , and the equations become[(ct0 )2 − x02 − y 02 − z 02 = c2t − (v/c2 )x√
1 − v 2 /c2]2[

x − vt− √1 − v 2 /c2]2− y2 − z2

c2 t2 − 2c2 t(v/c2 )x + c2 v 2 x2 /c4 − x2 + 2vtx − v 2 t2
− y2 − z2
1 − v 2 /c2
c2 t2 − 2vtx + v 2 x2 /c2 − x2 + 2vtx − v 2 t2
− y2 − z2=
1 − v 2 /c2
c2 t2 + v 2 x2 /c2 − x2 − v 2 t2=
− y2 − z2
1 − v 2 /c2
c2 t2 (1 − v 2 /c2 ) − x2 (1 − v 2 /c2 )=− y2 − z2
1 − v 2 /c2== c2 t2 − x2 − y 2 − z 2 .
In other words, even if the Lorentz transformations are performed,
s2 = (ct)2 − x2 − y 2 − z 2
is invariant.
In addition,s2 = −(ct)2 + x2 + y 2 + z 2(7.2)

is constant in the Lorentz transformations, even if the signs of (ct)2 and x2 + y 2 + z 2
are reversed. However, because the velocity v(vx , vy , vz ) of a point mass does notexceed the velocity of light, s2 becomes negative in (7.2). The reason is thatx2 + y 2 + z 2 = vx2 t2 + vy2 t2 + vz2 t2 = (vt)2 < (ct)2 .
Thus, generally,s2 = (ct)2 − x2 − y 2 − z 2(7.1)

is defined as the world distance s.

7.2 Derivation of world distance by the new quaternion
The theory that defines world distance in special relativity has some defects. Because the world distance s is defined by searching an invariant quantity under
Lorentz transformations, it is unknown whether other definitions are possible. One
might be able to choose other definitions of the world distance that fulfill the Lorentz
50invariance. In addition, it has not been mathematically proven which of (7.1) and(7.2) is correct. When the world distance is calculated using only the new quaternion, a unique formula of world distance is found.
As explained in Section 1.1, the magnitude squared of a complex number A =a + bi on a flat surface is2|A| = AA= (a + bi)(a − bi)= a2 − abi + bai − b2 i2= a2 + b2 .

(1.1)

If the same calculation is performed on the new complex number A = ah + bi, we
find
2|A| = AA= (ah + bi)(ah − bi)= a2 h2 − b2 i2= −a2 + b2 .

(7.3)

The sign of a2 in (7.3) is opposite to that of (1.1). In other words, (7.3) arises
because space–time bends in reality, although the new complex plane written on
paper looks like a flat surface.
In Hamilton’s quaternion which extends the complex number in four dimensions,
we have
A = a + bi + cj + dk,
A = a − bi − cj − dk.
The algorithms are
i2 = j 2 = k 2 = −1,
ij = −ji = k, jk = −kj = i, ki = −ik = j.
Thus, the equations become|A|2 = AA= (a + bi + cj + dk)(a − bi − cj − dk)= a2 − abi − acj − adk + bai − b2 i2 − bcij − bdik
+ caj − cbji − c2 j 2 − cdjk + dak − dbki − dckj − d2 k 251

= a2 − abi − acj − adk + abi + b2 − bck + bdj+ acj + bck + c2 − cdi + adk − bdj + cdi + d2
= a2 + b2 + c2 + d2 .

(7.4)

In the new quaternion, which extends the new complex number in four dimensions,
we have
A = ah + bi + cj + dk,
A = ah − bi − cj − dk.
The algorithms
h2 = −1, hi = ih, hj = jh, hk = kh
are added to the algorithms of Hamilton’s quaternion, and the equations become2|A| = AA= (ah + bi + cj + dk)(ah − bi − cj − dk)= a2 h2 − abhi − achj − adhk + baih − b2 i2 − bcij − bdik
+ cajh − cbji − c2 j 2 − cdjk + dakh − dbki − dckj − d2 k 2= −a2 − abhi − achj − adhk + abhi + b2 − bck + bdj+ achj + bck + c2 − cdi + adhk − bdj + cdi + d2
= −a2 + b2 + c2 + d2 .
In other words,
2|A| = AA= (ah + bi + cj + dk)(ah − bi − cj − dk)= −a2 + b2 + c2 + d2 .

(7.5)

If (7.5) is compared with (7.4), we can see that the sign of a2 is opposite as comparedwith that in (1.1) and (7.3). This result shows that the four-dimensional space–timeexpressed by the new quaternion is curved.
In four-dimensional space–time, the new quaternion, which expresses the locationof the point mass at time t, which was at origin O at time t = 0, is A = cth + xi +yj + zk. Substituting this into (7.5), we have2

|A| = AA= (cth + xi + yj + zk)(cth − xi − yj − zk)= −(ct)2 + x2 + y 2 + z 2 .
52

Because |A| is the world distance s, i.e., the distance between origin O and point
A, the equation becomess2 = −(ct)2 + x2 + y 2 + z 2 .

(7.6)

Equation (7.6) is the mathematically calculated formula of the world distance. Because this distance in four-dimensional space–time was found mathematically, thisformula is unique. In the last section, it was shown thats2 = (ct)2 − x2 − y 2 − z 2(7.1)

is the general definition of world distance in special relativity. However, (7.6) is themathematically correct formula. Unlike (7.1), s2 becomes negative in (7.6). Thereis no clear basis for choosing (7.1) or (7.6) in relativity theory texts. However, by
the new quaternion, the mathematically correct definition of the world distance is
s2 = −(ct)2 + x2 + y 2 + z 2 .

(7.6)

7.3 Interpretation of the negativity of the square of theworld distance
Using the new quaternion, we can explain what s becomes if s2 is negative. In thecalculation, by which (7.6) was obtained, the equation containing h, i, j, and k is
s2 = (cth)2 − (xi)2 − (yj)2 − (zk)2 .

(7.7)

If s is calculated from (7.7), under the condition that (ct)2 > x2 + y 2 + z 2 , we find√(xi)2 + (yj)2 + (zk)2s = cth 1 −(cth)2√x2 + y 2 + z 2= cth 1 −
.
(7.8)
(ct)2
The imaginary number h appearing in (7.8) signifies that the world distance s is
the time part, and it is expressed by an imaginary number in four-dimensional
space–time, just as it is in two-dimensional space-time.
Next, using the proper time, we again confirm that s is the time part. The
coordinates (cth, xi, yj, zk) of (7.7) are coordinates of point mass D as seen fromobserver A at rest. Because the coordinates of D as seen from observer B in uniform
motion, are (ct0 h, x0 i, y 0 j, z 0 k). Because s2 is invariant in all frames, we can write
s2 = (ct0 h)2 − (x0 i)2 − (y 0 j)2 − (z 0 k)2 .
53

Therefore, we find√s = ct0 h

1−x02 + y 02 + z 02.
(ct0 )2(7.9)

If B coincides with D, time t0 is the proper time τ of D and x0 = y 0 = z 0 = 0 in
(7.9). Thus, we finds = cτ h.

(7.10)

Equation (7.10) shows that the world distance s is an imaginary quantity obtainedby multiplying proper time τ by c.

7.4 Least-squares theory
As explained in Section 7.1, the definitions2 = (ct)2 − x2 − y 2 − z 2of world distance in special relativity becomes positive. Because the world distances2 becomes maximum when x2 + y 2 + z 2 is a minimum, it is clear that s2 is a maximum when the world line connecting two points is a straight line in four-dimensional
space–time. Note that since the square of the distance between two points is a minimum when the world line is a straight line in two- and three-dimensional spaces,
it is a unique property of the world distance that s2 is a maximum when the world
line is a straight line in four-dimensional space–time.
However, the formula of the world distance found mathematically by the newquaternion iss2 = −(ct)2 + x2 + y 2 + z 2 .
In this formula, s2 is a minimum when x2 + y 2 + z 2 is a minimum. Thus, thesquare of the distance between two points is a minimum along a straight line infour-dimensional space–time, as it is in two- and three-dimensional spaces. Thisshows the accuracy of the formulas2 = −(ct)2 + x2 + y 2 + z 2of the world distance by the new quaternion.
In addition, because s2 is positive in special relativity, the world distance s isa real number. However, since s2 is negative in the space–time theory using the
new quaternion, the world distance s is an imaginary number. From this diﬀerence,
although it is currently assumed that our universe lies in the four real axes inrelativity theory, it is shown here using the new quaternion that our universe lies in
the four imaginary axes in new quaternion space–time theory.
54

8
The New Quaternion andthe World Distance of Light

8.1 Proof of the world distance of light by special relativity
According to special relativity, the world distance of light is 0. In other words, whenlight travels in four-dimensional space–time, the distance from the original point is
0. We regard this as very mysterious. However, if the new complex plane is used,
it can be understood visually.
Before verification using the new complex plane, it is shown from the definitionof the world distance in special relativity that the world distance of light is 0. Thisis the proof written in relativity theory texts.
The point mass, which leaves origin O at time t = 0 and moves in a straight linewith uniform velocity v(vx , vy , vz ), has a world distance s. Then, by the definitionof world distance in special relativity, i.e.,s2 = (ct)2 − x2 − y 2 − z 2 ,(7.1)

the equations become
s2 = (ct)2 − x2 − y 2 − z 2= (ct)2 − vx2 t2 − vy2 t2 − vz2 t2= (ct)2 − (vt)2 .

(8.1)

Because the velocity v of light is c, from (8.1), the world distance s of light becomes
as follows;
s2 = (ct)2 − (ct)2
= 0.

55

In other words, no matter how far the light travels in four-dimensional space–time,
the world distance is 0. Although it is a mysterious result, it can be easily understood using the new quaternion and the new complex plane in the next section.

8.2 Proof of the world distance of light by the new quaternion
Because the equation of the world line of light emitted in the x-direction fromorigin O at time t = 0 in four-dimensional space–time is x = ct, the coordinatesare (cth, cti, 0, 0) by substituting x = ct in (cth, xi, 0, 0). In addition, the newquaternion is A = cth + cti. Because y = z = 0, and thus considering only twodimensional space–time, the world line of light becomes a set of points equidistantfrom the cth- and xi-axes in the new complex plane, as shown in Figure 8.1. Weshould be cautious not to say that this world line makes a 45◦ angle on these axes,
because there is a possibility that the angle changes under rotation in the newcomplex plane, since this plane is curved, even though it appears to be flat. We will
discuss this in Chapter 13.

Because |A| is the world distance s of light, we haves2 = |A|2= AA= (cth + cti)(cth − cti)
= (cth)2 − (cti)2= −(ct)2 + (ct)2= 0.
In other words, in the new complex plane, the distance between O and a point thatis equidistant from the two coordinate axes is 0. This suggests that the new complex
plane is a curved surface. In addition, it suggests that the new complex number
56

and the new quaternion are mathematical quantities capable of describing curved
space–time.
It is generally considered that a curved surface cannot be drawn on a flat paper,and the length of the world line cannot be directly measured on the paper. However,it can be calculated on a flat plane using the new complex number. This wasproven in Section 5.1 and Section 6.3. The same method is also applied to the
new quaternion. A node and a distance can be found through calculations using theequations of the world lines in four-dimensional space–time and the new quaternion.
The full theory of relativity includes both special and general relativity. The casein which the observer is at rest or undergoing linear uniform motion falls under
special relativity, while the case in which the observer is accelerated falls undergeneral relativity. It is assumed that space–time is flat in special relativity and it iscurved in general relativity.
However, as shown here, space–time is curved even under uniform linear motion.
Space–time bends doubly if the curvature by acceleration is also considered. This isa key diﬀerence between Einstein’s relativity theory and the new quaternion space–time theory.

8.3 Interpretation of the zero world distance of light and theconstancy of the velocity of light
As shown in Figure 8.1, in two-dimensional space–time, it can be said that theworld line of light is equidistant from the two coordinate axes. However, a reverse
aspect is possible. That is, the world line equidistant from the two coordinate axes
is light. In the former aspect, light exists as an absolute object and its properties are
explained. In the latter aspect, in many world lines, a world line equidistant fromthe two coordinate axes is light. In this aspect, the world line of a particle, whichconstitutes a substance, and the world line of light, have the same properties. The
world line equidistant from the coordinate axes is light, and the world line, which
is not equidistant, is the substance. If this theory is extended to four-dimensional
space–time, light is not necessarily equidistant from each coordinate axis like it is in
two-dimensional space–time. It passes only through the place whose world distances2 = −(ct)2 + x2 + y 2 + z 2(7.6)

is zero. In other words, the diﬀerence between the substance and the light rayexists only at a particular place in four-dimensional space–time. The reason why

57

light can be emitted in every direction in three-dimensional space is that the point(ct, x, y, z), which satisfies s = 0 in (7.6), exists everywhere in the space.
In classical physics, light has properties of both waves and particles. On theother hand, in quantum mechanics, a substance exists as a wave and only becomesa particle via observation. Although it may seem odd that a substance can be awave, it becomes clear when noted that the diﬀerence between the substance anda light ray is only a diﬀerence at a particular place in four-dimensional space–time.
Because they are initially the same, it can be thought that light and substance haveproperties of both waves and particles. In this book, an axiom has been developed
which states that the diﬀerence between substance and light is only a diﬀerence ata particular place in four-dimensional space–time, and discussions are carrried out
below.
If this theory is continued further, we can consider the following. If the coeﬃcientthat changes time t into distance is c, then time is denoted as ct. Because theequation of the world line equidistant from the temporal and distance axes is x = ct
in this case, the velocity becomes c. This world line then corresponds to a lightray. In this theory, c is only a coeﬃcient that changes time to distance. Because
the path of the light ray lies on a set of points equidistant from the two coordinate
axes, light has a velocity c. Then, the principle of the constancy of the velocity oflight is no longer a principle or an axiom that cannot be proven. The principle isonly a theorem drawn from the new quaternion.
If we assume that only four coordinate axes and world lines exist in four-dimensional
space–time and that the world lines correspond to light rays or substances depending on the path, we can avoid the mistake that occurs when we assume that natural
phenomena are only those that can be observed. According to the assumption thata natural phenomenon corresponds to a substance in special relativity, researchers
have explained that mass becomes energy. However, they have not explained massitself. In addition, in general relativity, the theory that mass bends space and produces gravity is based on the premise that mass exists independent of space. In
other words, mass exists as an absolute and it is not assumed that it is a shadow ofsomething else. In Chapter 18, we will show that the results of special relativity canbe explained without contradiction if mass and energy are viewed as the time partof the unit world line and momentums as the space parts of the unit world line.

58

9
The Twin Paradox

9.1 Contents of the twin paradox
Because this book is not a commentary book of relativity theory but a book thatproves physical laws using the new complex number, new quaternion, and newoctonion, the paradoxes of special relativity are not a matter of concern in thisbook. However, the relations between the world line and force can be understood
if the famous twin paradox is calculated using the new complex number.
At first, we explain contents of the twin paradox. We consider twins A and B. It
is assumed that A is at rest and B departs using a rocket with uniform velocity v inthe positive direction along the x-axis as seen from A. B accelerates suddenly after
a sudden slowdown, turns around, and heads toward A with velocity −v. A and B
synchronize their clocks at t = 0 when they depart. The times of the clocks of A
and B are t and t0 , respectively, upon arrival. Because the formula of time dilationof the moving clock is√τ = t 1 − v 2 /c2 ,(6.3)
we have√t0 = t 1 − v 2 /c2 .

Since 1 − v 2 /c2 < 1, the time of B is slower than that of A.
Switching to the reference frame of B, it is assumed that B is at rest and A
leaves with velocity −v in the negative direction along the x-axis as seen from B.
It also seems that A accelerates suddenly after a sudden slowdown, turns around,and heads towards B with velocity v. If the times of the clocks of A and B are t
and t0 , respectively, as mentioned above, thent = t0√1 − v 2 /c2

because the clock of A moves. A ages slower than B.

59

This phenomenon is called the twin paradox. However, the premise that A and
B are in the same conditions is incorrect. Since A is at rest, no force acts on A.
However, a force acts on B if B accelerates suddenly after the sudden slowdown.
On the other hand, no force acts on A when A appears to suddenly accelerateaway from B after the sudden slowdown. B experiences a force during acceleration.
Even if A seems to have accelerated suddenly after the sudden slowdown, A does
not experinces a force. The forces acting on A and B appear to be indistinguishable
if we consider only the change in the distance between A and B; however, the force
acts only on B and not on A.
This diﬀerence is only explained conceptually in many books because the factor of
the force cannot be included when the Lorentz transformations are used. However,the factor of the force can be added if we apply the new complex plane. The
calculations are shown below.

9.2 Calculation in the frame of a stationary observerFirst, we assume that B moves in the x-direction along A with uniform velocity v
and then returns with velocity −v. The new complex plane emerges as shown in
Figure 9.1. Acceleration a is constant.

Although A is at rest, time passes. Thus, A moves in the cth-axis. In the reference
frame of A, B moves in the straight line x = vt, i.e., the ct0 h-axis. The world lineof B becomes the straight line with gradient −v at time t0 and B meets A on the
cth-axis at time 2t0 . This node is called F . If the location of A at time t0 is D, thenthe coordinates are D(ct0 h, 0). In addition, if the node, where the velocity of B
changes from v to −v, is E, then the coordinates are E(ct0 h, vt0 i) by substitutingt = t0 for x = vt.

60

We assume that the elapsed times, during which the velocity of B changes from0 to v and from v to −v, are relatively short so that these times can be neglected.
On the basis of the new complex number, point E is ct0 h + vt0 i. Using thedefinition of distances, the magnitude squared becomes as follows:
2|OE| = (ct0 h + vt0 i)(ct0 h − vt0 i)= c2 t20 h2 − v 2 t20 i2 .

(9.1)

In addition, if the new complex number showing the line-segment EF is written as(EF ), then(EF ) = F − E= 2ct0 h − (ct0 h + vt0 i)= ct0 h − vt0 i.
Thus, we find2

|EF | = (ct0 h − vt0 i)(ct0 h + vt0 i)= c2 t20 h2 − v 2 t20 i2 .

(9.2)

From (9.1) and (9.2), we can write|OE| + |EF | = 2√
c2 t20 h2 − v 2 t20 i2 .

When c > v, the term inside the square root is
c2 t20 h2 − v 2 t20 i2 = −c2 t20 + v 2 t20 < 0.
Thus, using the axiom that place the negative sign outside of the square root whenthe quantity inside the square root is negative, we can write√|OE| + |EF | = 2ct0 h 1 − v 2 /c2 .
(9.3)
In addition, if we use the new complex number F = 2ct0 h of F , then we find2

|OF | = (2ct0 h + 0i)(2ct0 h − 0i)= (2ct0 h)2 .
Thus, the equation becomes|OF | = 2ct0 h.
From (9.3) and (9.4), we find|OE| + |EF | < |OF | .
61

(9.4)

This inequality implies that the elapsed time of B is shorter than that of A, and B
is younger than A. The fact that 2ct0 h contains the imaginary number h in (9.4)
indicates that |OF | is the time part. Though |OE| + |EF | appears to be longer
than |OF | in Figure 9.1, the line segment parallel to the cth- or x-axis is the longest
and becomes shorter when it inclines from the coordinate axes, because the newcomplex plane curves in reality.

9.3 Calculation in the frame of a moving observer
Next, we consider the new complex plane, in which B is at rest and A moves. If weassume that a world line, which is the position of a point mass in four-dimensional
space–time, travels like light with velocity v, then the connection of the world line
between two points becomes a straight line. In other words, if no force acts, then
the world line becomes a straight line. If a force acts, then the world line curves. If
the force disappears, then the curve becomes a straight line. If the force continues
to act, then the path is curved. We can say that the world line is curved when theforce acts on the object. Also we can say that the curve of a world line is the force.
The law of inertia states that a point mass moves in linear uniform motion if noforce acts on it. However, if we consider the world line, then the law implies thatno force results in a straight world line.
One may consider that because the space–time curves, a world line cannot be astraight line. However, in the new complex plane, a world line becomes a straightline if the equation of the world line is a linear function. It is thought that even if
the world line in four-dimensional space–time is drawn using the new quaternion,
the result is the same because the new quaternion itself contains the elements ofcurves. If the new quaternion is applied to a curved space–time, curves disappear
and the world line becomes a straight line. This was proven in Section 5.1.

62

If the twin paradox is seen from the viewpoint of B, B is at rest and A separates
from B with velocity −v in the negative direction along the x-axis. Then, A accelerates suddenly after the sudden slowdown and returns to B with velocity v. We
tend to think that the world line of B becomes the cth-axis because B is at rest,
as shown in Figure 9.2. Also, we tend to think that the world line of A becomes astraight line x = −vt, changes to the straight line x = vt − a at time t0 , and crosses
the cth-axis at time 2t0 . However, from the relation of the world line and force
described previously, the force did not act in this figure, because B is on a straightline. The fact that the world line of A is curved means that force acted on A. This
result conflicts with the precondition that the force acted only on B.
A graphical interpretation is not suﬃcient and can be misleading. Thus, weinvestigate using coordinate transformations. Because the multiplication of the newcomplex number implies a rotation about the origin O, a migration from point E
to point D is denoted by a new complex number in Figure 9.1, which is drawn fromthe viewpoint of A.

We assume that the denoted new complex number is H. Because the new complex
numbers of D and E are
D = ct0 h, E = ct0 h + vt0 i
in Figure 9.1, using EH = D, we find(ct0 h + vt0 i)H = ct0 h,
ct0 h
ct0 h + vt0 i
ct0 h(ct0 h − vt0 i)=(ct0 h + vt0 i)(ct0 h − vt0 i)−c2 t20 − cvt20 hi=−c2 t20 + v 2 t20

H=63

c2 + cvhi
c2 − v 2
1 + vhi/c=.
1 − v 2 /c2
=

(9.5)

If the issue is considered using the coordinate transformation, then the fact that
A is seen from B means that E moves to the location of D, which is in the rest
frame expressed by the new complex number H, implying a rotation. The original
D moves to another location. If we assume that E and D migrated by H are E 0
and D0 , respectively, and F migrates to F 0 , then the new complex plane becomesas shown in Figure 9.3.

The coordinates of E 0 are the same as those of D in Figure 9.1: E 0 (ct0 h, 0).
Because the coordinates of F before migration are two times those of D, those of
F 0 after the coordinate transformation by H are two times those of D0 . Thus, if thecoordinates of D0 are D0 (ct1 h, x1 i), then those of F 0 are F 0 (2ct1 h, 2x1 i). This is
the mathematical proof that when no force acts on A, D0 and F 0 are on the sameworld line. The world line of B is the cth axis at first and becomes a straight linefrom E 0 (ct0 h, 0) at time t0 and crosses the straight line OA at point F 0 at time 2t0 .
Using the new complex number H, which implies a rotation, the transformation,
which moves D to D0 , isDH = D0 .
Thus, using (9.5), D(ct0 h, 0), and D0 (ct1 h, x1 i), we have
ct1 h + x1 i = ct0 hH
1 + vhi/c)
1 − v 2 /c2
ct0 h + vt0 h2 i=
1 − v 2 /c2= ct0 h(64=

ct0 h − vt0 i.
1 − v 2 /c2

By comparing coeﬃcients, we findt0,
1 − v 2 /c2
−vt0
x1 =
.
1 − v 2 /c2
t1 =

(9.6)
(9.7)

The slope of straight line OD0 calculated using (9.6) and (9.7) is
x1
−vt0
v
=
=− .
ct1
ct0
c
Thus, the equation of the straight line OD0 isv
x = − (ct) = −vt.
cThis result confirms that the initial relative velocity of A as seen from B is −v.
The length of the world line is calculated as follows. From (9.6) and (9.7), wehave|OF 0 | = (2ct1 h + 2x1 i)(2ct1 h − 2x1 i)
2= 4c2 t21 h2 − 4x21 i2
4c2 t20 h2
4v 2 t20 i2−222(1 − v /c )(1 − v 2 /c2 )2[]4c2 t20 h2 1 − v 2 i2 /(c2 h2 )=(1 − v 2 /c2 )24c2 t20 h2 (1 − v 2 /c2 )=(1 − v 2 /c2 )22 2 2
4c t0 h=.
1 − v 2 /c2=

Since c > v, we find2ct0 h|OF 0 | = √.
1 − v 2 /c2(9.8)

The imaginary number h shows that world distance |OF 0 | is the temporal part.
If the new complex number describing line-segment E 0 F 0 is written as (E 0 F 0 ),then from (9.6) and (9.7), we find
(E 0 F 0 ) = F 0 − E 0= 2ct1 h + 2x1 i − ct0 h=

2vt0 i2ct0 h−− ct0 h1 − v 2 /c2
1 − v 2 /c265

2ct0 h2vt0 ict0 h(1 − v 2 /c2 )−−1 − v 2 /c2
1 − v 2 /c2
1 − v 2 /c2
2 2
ct0 h(2 − 1 + v /c )2vt0 i=−22
1 − v /c
1 − v 2 /c2
2 2
ct0 h(1 + v /c )2vt0 i=−22
1 − v /c
1 − v 2 /c2222
ct0 h(c + v ) 2c vt0 i− 2.
=c2 − v 2c − v2=Thus, we have][]
ct0 h(c2 + v 2 ) 2c2 vt0 i ct0 h(c2 + v 2 ) 2c2 vt0 i−+
c2 − v 2c2 − v 2c2 − v 2c2 − v 2
c2 t20 h2 (c2 + v 2 )2 − 4c4 v 2 t20 i2=(c2 − v 2 )2]
c2 t2 h2 [= 2 0 2 2 (c2 + v 2 )2 − 4c2 v 2(c − v )c2 t20 h2 (c2 − v 2 )2=(c2 − v 2 )2|E 0 F 0 | =2[= c2 t20 h2 .
Finally, we find
|E 0 F 0 | = ct0 h.

(9.9)

In addition, because E 0 = ct0 h, we have
|OE 0 | = ct0 h.

(9.10)

From (9.9) and (9.10), we can write|OE 0 | + |E 0 F 0 | = 2ct0 h.
Since

2ct0 h|OF 0 | = √1 − v 2 /c2from (9.8), we have
|OE 0 | + |E 0 F 0 | < |OF 0 | .
In other words, the elapsed time of B is shorter than that of A, and B is younger
than A. This result is the same as that from the viewpoint of A. The twin paradox
has now been solved.

66Though |OE 0 | and |E 0 F 0 | do not appear to be of the same length in Figure 9.3,
they are the same when (9.9) and (9.10) are applied. This indicates that the newcomplex plane curves. In addition, because we performed the calculations using the
new complex plane, we were able to obtain this result. Such a calculation cannot
be performed in a regular Minkowski space-time diagram.

9.4 Curve of the world line and a force
In the discussion of the twin paradox as seen from B, some readers may consider
calculating the new slope of B after point E 0 , however, the world line of B is astraight line from the departure to arrival points, whereas the world line of A iscurved. This is contrary to the axiom that if no force acts on A, then the world
line is a straight line and if a force acts, then the world line curves. In other words,
the curved world line of an accelerated point mass cannot be changed to a straightline by the coordinate transformations. This can be mathematically proven fromthe fact that a quadratic equation cannot be changed to a linear equation.
Consider Einstein’s Gedankenexperiment. The people who are in an elevator, inwhich the supporting tie breaks, do not notice that they are falling, because they are
falling at the same velocity as that of the elevator. In other words, it is the same as
being at rest. However, because a free-falling object accelerates under the influenceof gravity, the equation of the world line expresses a curve. It is clear that a force actson the object and the objects are not at rest. The reference frames of the observer,
who moves in linear uniform motion, are called inertial frames and the world line isa straight line. Thus, a free-fall movement of the elevator is not an inertial system
if we consider using the world line. The fallacy of this Gedankenexperiment lies inconsidering the physical phenomenon using objects. We can avoid this error if the67

same phenomenon is considered using the world line. In this book, it is an axiom
that if no force acts, then the world line becomes a straight line, which connects
two points in four-dimensional space–time.
If we assume that observers A and B are on the same curved world line, then it isincorrect for them to think that they are at rest or move in linear uniform motion
because of the constant distance between them. If we only consider the distance
between two points, we tend to think that being at rest is relative, depending on
the observer. However, if we think that the movement in a straight line between
two points in four-dimensional space–time is that of an object at rest or moving inlinear uniform motion, acceleration movement and linear uniform motion are distinguishable regardless of the observer. The most suitable example for understanding
these concepts is the twin paradox.

68

10
The New Lorentz Transformations

10.1 Derivation of the new Lorentz transformations by the
new quaternion
The Lorentz transformations using the coordinate transformation A/ |A| by the newcomplex number are derived in the new complex plane in two-dimensional space–time. However, because we live in four-dimensional space–time, we assume that thecoordinate transformation A/ |A| by the new complex number is also realized by
the new quaternion. Thus, we find the Lorentz transformations in four-dimensional
space–time containing the y- and z-coordinates. Using the Lorentz transformationsof special relativity, the y- and z-coordinates do not change and the equations are
as follows:
y 0 = y,(3.5)
z 0 = z.

(3.6)

However, we cannot find the same results using the coordinate transformation A/ |A|by the new quaternion.

69

First, we considered the three-dimensional space, where time does not exist. As
shown in Figure 10.1, we assume that observer A is at rest at origin O and observer
B moves along a straight line with uniform velocity v in the positive direction along
the x-axis. In addition, point D is at rest at point (xi, yj, zk) in three-dimensional
space. At time t = 0, A and B coincide at O. After t seconds, A is at (0, 0, 0), B
is at (vti, 0, 0), and D is at (xi, yj, zk) at time t. In four-dimensional space-time,
the coordinates are A(cth, 0, 0, 0), B(cth, vti, 0, 0), and D(cth, xi, yj, zk), andtheir new quaternions are A = cth, B = cth + vti, and D = cth + xi + yj + zk.
If positions of A and B are drawn in a four-dimensional space–time diagram where
the cth-axis is the horizontal axis and the spatial coordinate axes, which intersect
perpendicularly with the cth-axis, are the xi-, yj-, and zk-axes, then the diagram
becomes Figure 10.2. However, because the zk-axis, which is the fourth coordinate
axis, cannot be drawn with a perspective on the flat surface, it is drawn with thedashed line. In addition, B is on a straight line x = vt and since A, B, and D sharethe same time t, they are on the same flat plane perpendicular to the cth-axis.

As the Lorentz transformations in the new complex plane are obtained, the equation of the coordinate transformation of D by B, i.e., how D is seen from B, is
DB
.
|B|

(10.1)The new quaternions of B and D are substituted in (10.1) and calculations are
performed. The algorithms of the new quaternion are as follows:
h2 = i2 = j 2 = k 2 = −1,
ij = −ji = k, jk = −kj = i, ki = −ik = j,hi = ih, hj = jh, hk = kh,
70

1/h = h/h2 = −h.
If c > v, we haveDB(cth + xi + yj + zk)(cth − vti)√=
|B|
(cth + vti)(cth − vti)
1
=√
2
2
2
c t h − v 2 t2 i2× (c2 t2 h2 − cvt2 hi + xcthi − xvti2 + ycthj − yvtji + zcthk − zvtki)1√=
cth 1 − v 2 /c2
× (−c2 t2 − cvt2 hi + xcthi + xvt + ycthj + yvtk + zcthk − zvtj)
1√=
cth 1 − v 2 /c2[]
× (−c2 t2 + xvt) + (−cvt2 + xct)hi + (ycth − zvt)j + (zcth + yvt)k1√=
cth 1 − v 2 /c2× [(−ct + xv/c)ct + (−vt + x)cthi + (yh − zv/c)ctj + (zh + yv/c)ctk]
1=√
1 − v 2 /c2× [(−ct + xv/c)/h + (−vt + x)i + (yh − zv/c)j/h + (zh + yv/c)k/h]
1=√
1 − v 2 /c2× [(ct − vx/c)h + (x − vt)i + (y + vzh/c)j + (z − vyh/c)k] .

(10.2)

If the coordinates of D after the coordinate transformation by B are (ct0 h, x0 i, y 0 j, z 0 k),
the new quaternion is ct0 h + x0 i + y 0 j + z 0 k. If the coeﬃcients are compared with
(10.2), then the h part isct − vx/c
ct0 = √.
1 − v 2 /c2
Dividing both sides by c, we havet − (v/c2 )xt0 = √.
1 − v 2 /c2(10.3)The i, j, and k parts, respectively, arex − vtx0 = √,
1 − v 2 /c2(10.4)y + (v/c)zh
y0 = √,
1 − v 2 /c2(10.5)

71

z − (v/c)yhz0 = √.
1 − v 2 /c2(10.6)

Equations (10.3) and (10.4), respectively, coincide witht − (v/c2 )xt0 = √,
1 − v 2 /c2
x − vtx0 = √
1 − v 2 /c2

(3.3)
(3.4)

of the Lorentz transformations. However, (10.5) and (10.6), respectively, do notcoincide with
y 0 = y,
0

z =z

(3.5)
(3.6)

of the Lorentz transformations. From now on, (10.3), (10.4), (10.5), and (10.6) arereferred to as new Lorentz transformations.

10.2 Diﬀerences between the new and original Lorentz
transformations
Because y 0 = y and z 0 = z in the Lorentz transformations of special relativity, the yand the z-coordinates do not change after the coordinate transformations because of
the assumption of isotropy that space and time are uniform. Because our right-side
space has the same property as the left-side space, even if we distinguish between
the positive and negative x-direction, we can find the same equationx − vtx0 = √.
1 − v 2 /c2(3.4)

This is called the isotropy of space. However, the isotropy of space does not meanthat a movement in the x-direction does not influence the y- and the z-coordinates.
A movement in the x-direction changes the y- and z-coordinates, respectively, as
can be understood byy + (v/c)zh
y0 = √,
1 − v 2 /c2z − (v/c)yhz0 = √
1 − v 2 /c2(10.5)(10.6)of the new Lorentz transformations. In other words, linear uniform motion in the xdirection bends the space in the y- and the z-directions. This is the new conclusion
obtained from the coordinate transformations by the new quaternion.
72

There exist some texts in which the equation y 0 = y of the Lorentz transformations
is obtained by assuming that y 0 is a function of velocity v and coordinate y, i.e.,y 0 = f (v)y. However, both y 0 and z 0 are functions of three variables, i.e., y, z,
and v, and the correct equations are y 0 = f (y, z, v) and z 0 = g(y, z, v) as
understood by (10.5) and (10.6), respectively, if calculations are performed using
the new quaternion. Because it is assumed that y 0 is a function of two variables, i.e.,y and v, in special relativity, these results cannot be obtained. If this calculationis performed without assuming the property of space, then y 0 is a function of threevariables, i.e., y, z, and v. The same result can be obtained for z 0 .
Newtonian mechanics assumes absolute time and absolute space. This meansthat time and distance are constant regardless of an observer’s velocity. Special
relativity emerged by challenging these assumptions. Time and distance changewith an observer’s velocity in that theory. However, Einstein, who denied Newton’s
absolute time and absolute space, assumed the isotropy of space and concluded thaty 0 = y,z 0 = z.
Because (10.3), (10.4), (10.5), and (10.6) are the formulae obtained without assuming the property of space, it is thought that they express the true property ofspace-time rather than the formulae of the Lorentz transformations, which assume
the isotropy of space. The correctness of the new Lorentz transformations, whichtake the place of the Lorentz transformations, will be proven by considering thatthe space-time interval is invariant when using these new transformations in Section
10.3 and the velocity of light is constant using the new velocity transformation in
Section 10.5. In addition, in Section 12.3, the implications of the imaginary number
h in (10.5) and (10.6) are considered.

10.3 The new Lorentz transformations and world distanceTo verify the correctness of the new Lorentz transformations, we examine whether
the world distance becomes invariant under the new Lorentz transformations. As
explained in Section 7.2, the equation of the world distance by the new quaternioniss2 = −(ct)2 + x2 + y 2 + z 2 .
(7.6)
If we assume that the coordinates before the coordinate transformation are (cth, xi,
yj, zk) and those after the coordinate transformation are (ct0 h, x0 i, y 0 j, z 0 k), then73

we can write−(ct)2 + x2 + y 2 + z 2 = −(ct0 )2 + x02 + y 02 + z 02 ,(10.7)

because the world distance is invariant using the coordinate transformation. We
investigate whether the left side of (10.7) can be found by substituting the equationsof the new Lorentz transformations for the right side of (10.7). The equations
become− (ct0 )2 + x02 + y 02 + z 02
[
]2 [
]2
2
x − vt2 t − (v/c )x= −c √+ √1 − v 2 /c2
1 − v 2 /c2
[
]2 [
]2
y + (v/c)zhz − (v/c)yh
+ √
+ √
.
1 − v 2 /c2
1 − v 2 /c2(10.8)

Because the calculations are complex, we divide the right side of (10.8) into half.
The first part is
[
]2 [
]2
2
x − vt2 t − (v/c )x−c √+ √1 − v 2 /c2
1 − v 2 /c2−c2
1(t2 − 2vtx/c2 + v 2 x2 /c4 ) +(x2 − 2vtx + v 2 t2 )2
2
1 − v /c
1 − v 2 /c2
1=(−c2 t2 + 2vtx − v 2 x2 /c2 + x2 − 2vtx + v 2 t2 )1 − v 2 /c2
1(−c2 t2 + v 2 t2 − v 2 x2 /c2 + x2 )=1 − v 2 /c2[]
1=−(1 − v 2 /c2 )c2 t2 + (1 − v 2 /c2 )x222
1 − v /c== −(ct)2 + x2 .

(10.9)

Next, the second part of (10.8) is
[
]2 [
]2
y + (v/c)zhz − (v/c)yh√+ √1 − v 2 /c2
1 − v 2 /c2
1=(y 2 + 2vyzh/c + v 2 z 2 h2 /c2 + z 2 − 2vyzh/c + v 2 y 2 h2 /c2 )1 − v 2 /c2
1(y 2 − v 2 y 2 /c2 + z 2 − v 2 z 2 /c2 )=1 − v 2 /c2[]
1=(1 − v 2 /c2 )y 2 + (1 − v 2 /c2 )z 2
1 − v 2 /c2= y2 + z2 .

(10.10)74

From (10.8), (10.9), and (10.10), we find−(ct0 )2 + x02 + y 02 + z 02 = −(ct)2 + x2 + y 2 + z 2 .
Thus, the coordinate transformations under the new Lorentz transformations makethe world distance invariant. In other words, the y 0 - and z 0 -coordinates of the new
Lorentz transformations, i.e.,y + (v/c)zh
y0 = √,
1 − v 2 /c2z − (v/c)yhz0 = √
1 − v 2 /c2(10.5)(10.6)

do not produce a mathematical contradiction.

10.4 Transformation of velocities under Lorentz
transformations and the constancy of the velocity
of light
Next, we investigate whether the constancy of the velocity of light is maintained under the new Lorentz transformations. However, before that, the well-known Lorentz
transformations for velocity are explained using special relativity.
First, we explain the diﬀerential form of velocity for those who have not learned
the calculus. If the velocity does not change, the relation between the distance x,time t, and velocity v is v = x/t. However, if the velocity changes, the generalformula for velocity becomesdx.
(10.11)dt
Here, dx and dt imply an infinitesimally small distance and small period of time,
respectively. It is thought that even if the velocity changes with time, it is constantv=

between each infinitesimally small period of time. By dividing dx by dt, (10.11) is
considered to be the general formula of the velocity. It is enough to think that dmeans infinitesimally small.
Because the equations of the Lorentz transformations are as follows:t − (v/c2 )xt0 = √,
1 − v 2 /c2
x − vt,x0 = √
1 − v 2 /c2
y 0 = y,

(3.3)
(3.4)
(3.5)

75

z 0 = z,(3.6)

the diﬀerentials are
dt − (v/c2 )dx
dt0 = √,
1 − v 2 /c2dx − vdt,
dx0 = √
1 − v 2 /c2
dy 0 = dy,(10.12)(10.13)(10.14)0dz = dz.

(10.15)

It is assumed that the velocity of point mass D is V as seen from observer A
at rest and is V 0 as seen from observer B who is moving with constant velocity
v. We assume that the coordinates belonging to V and V 0 are (Vx , Vy , Vz ) and(Vx0 , Vy0 , Vz0 ), respectively. Thus, from (10.12) and (10.13), we finddx0
dt0
dx − vdt=.
dt − (v/c2 )dx

Vx0 =By dividing the numerator and denominator on the right side by dt, we havedx/dt − v
1 − (v/c2 )dx/dt
Vx − v=.
1 − (v/c2 )Vx

Vx0 =(10.16)

Similarly, from (10.12), (10.14), and (10.15), we find
dy 0
dt0√
dy 1 − v 2 /c2=
dt − (v/c2 )dx√
dy/dt 1 − v 2 /c2=
1 − (v/c2 )dx/dt√
Vy 1 − v 2 /c2=,
1 − (v/c2 )Vx
dz 0
Vz0 = 0dt √dz 1 − v 2 /c2=
dt − (v/c2 )dx√
dz/dt 1 − v 2 /c2=
1 − (v/c2 )dx/dt
Vy0 =

76

(10.17)√Vz 1 − v 2 /c2=.
1 − (v/c2 )Vx(10.18)Equations (10.16), (10.17), and (10.18) are the transformation equations of thevelocities obtained from the Lorentz transformations.
In the case that observed subject D is light emitted in the x-direction from A atrest, we calculate how the velocity of the light is observed as seen from B, which ismoving in the x-direction with a constant velocity v. By substituting the velocity
components of the light, i.e., Vx = c, Vy = 0, and Vz = 0, in (10.16), (10.17), and(10.18), we havec−v
1 − (v/c2 )cc−v=
1 − v/cc(c − v)
=
c−v

Vx0 =

= c,

(10.19)0
Vy0 =1 − (v/c2 )c
= 0,
Vz0 =

(10.20)0
1 − (v/c2 )c= 0.

(10.21)Equation (10.19) shows that the velocity of light is always a constant value c, regardless of the velocity v of observer B. This is a confirmation of the constancy ofthe velocity of light under the Lorentz transformations.
It is not correct here to use the word proof instead of confirmation because it
is mathematically natural to obtain the constancy of the velocity of light from thetransformation equations of the velocity using the Lorentz transformations. Afterall, the Lorentz transformations are obtained on the basis of the assumption thatthe velocity of light is constant as explained in Section 3.1. Therefore, (10.19) is
the confirmation of the correctness of the transformation equations of the velocity
calculated under the Lorentz transformations and it is not the proof of the constancyof the velocity of light. Thus, it should not be thought that the constancy of thevelocity of light has been proven using this calculation. Because verification is
impossible but a contradiction does not occur, the constancy of the velocity of lightis considered to be correct. In mathematical terms, it is an axiom. However, in thisbook, it is established that the constancy of the velocity of light is not an axiom; it77

is rather a theorem obtained from other axioms in Section 8.3.

10.5 Transformations of velocities under the new Lorentz
transformations and the constancy of the velocity oflight
We will confirm that the new Lorentz transformations are the correct equationsof transformation, if we also obtain the constancy of the velocity of light fromthe transformation equations of the velocities calculated using the new Lorentz
transformations and the new quaternion.
As explained in Section 10.1, the equations of the new Lorentz transformations,which are obtained using the coordinate transformation A/ |A| by the new quaternion, aret − (v/c2 )xt0 = √,
1 − v 2 /c2
x − vtx0 = √,
1 − v 2 /c2
y + (v/c)zh,
y0 = √
1 − v 2 /c2z − (v/c)yhz0 = √.
1 − v 2 /c2(10.3)(10.4)(10.5)(10.6)The diﬀerentials are
dt − (v/c2 )dx
dt0 = √,
1 − v 2 /c2dx − vdt
dx0 = √,
1 − v 2 /c2
dy + (v/c)dzh
dy 0 = √,1 − v 2 /c2dz − (v/c)dyh
dz 0 = √.
1 − v 2 /c2(10.21)(10.22)(10.23)(10.24)

If we consider the assumptions similar to those in the Lorentz transformations, from
(10.21), (10.22), (10.23), and (10.24), we finddx0
dt0
dx − vdt=
dt − (v/c2 )dx

Vx0 =

78

dx/dt − v
1 − (v/c2 )dx/dt
Vx − v=,
1 − (v/c2 )Vx
dy 0
Vy0 = 0dt
dy + (v/c)dzh=
dt − (v/c2 )dx
dy/dt + (v/c) hdz/dt=
1 − (v/c2 )dx/dt
Vy + (v/c)Vz h,=
1 − (v/c2 )Vx
dz 0
Vz0 = 0dt
dz − (v/c)dyh=
dt − (v/c2 )dx
dz/dt − (v/c)hdy/dt=
1 − (v/c2 )dx/dt
Vz − (v/c)Vy h=.
1 − (v/c2 )Vx=(10.25)(10.26)(10.27)

These are the transformation equations for velocity using the new Lorentz transformations.
Next, we investigate whether the constancy of the velocity of light is obtainedusing (10.25), (10.26), and (10.27). If observed subject D is light emitted in the xdirection from A at rest, the velocities of the light measured by A are Vx = c, Vy = 0,
and Vz = 0. Thus, if Vx = c, Vy = 0, and Vz = 0 are substituted in (10.25), (10.26),
and (10.27), we havec−v
1 − (v/c2 )cc−v=
1 − v/cc(c − v)
=
= c,
c−v
0
Vy0 =
= 0,
1 − (v/c2 )c
0
= 0.
Vz0 =
1 − (v/c2 )cVx0 =(10.28)Equation (10.28) indicates that the velocity of light is always a constant value c,regardless of velocity v of observer B, which is confirmed by the above results using
the new Lorentz transformations.
7910.6 Independency of the imaginary number h
Because the velocity transformations using the new Lorentz transformations havebeen explained, we prove the independency of the imaginary number h in this section. In Section 3.4, to make the coordinate transformations by the complex number similar to the Lorentz transformations, calculations were performed using cth
by multiplying ct with the imaginary number h. By performing these calculations,the signs of c2 oft + (v/c2 )xt0 = √,
1 + v 2 /c2
x − vtx0 = √
1 + v 2 /c2(3.9)(3.10)

changed from (+) to (-) and the Lorentz transformations could be found.
Because the fourth imaginary number h was introduced to change velocity c oflight into ch as mentioned above, a question arises whether ch itself has any phisicalmeaning. In other words, can h be used independently? This independency of theimaginary number h can be proven using the velocity transformation equations(10.25), (10.26), and (10.27) using the new Lorentz transformations.
In addition, this was not proven in Section 3.4 as the velocity transformations
using the new Lorentz transformations are explained first in this chapter.
We calculate how the velocity of the light, which is emitted along the y-direction
from observer A at rest, is observed from observer B, who moves in the x-direction
with velocity v. If Vx = 0, Vy = c, and Vz = 0 are substituted in the velocity
transformation equations using the new Lorentz transformations, i.e.,
Vx − v,
1 − (v/c2 )Vx
Vy + (v/c)Vz h
Vy0 =,1 − (v/c2 )Vx
Vz − (v/c)Vy hVz0 =,1 − (v/c2 )Vx
Vx0 =(10.25)(10.26)(10.27)

we findVx0 =0−v
1 − (v/c2 )0

= −v,
Vy0 =

(10.29)c + (v/c)0h
1 − (v/c2 )0= c,(10.30)80

Vz0 =

0 − (v/c)ch
1 − (v/c2 )0= −vh.

(10.31)

In (10.31), h has been separated from c and combined with v. In other words, theimaginary number h is a number independent of c. Therefore, there is no problem
if we multiply or divide by h alone.
The correctness of (10.29), (10.30), and (10.31) is proven because they satisfy theprinciple of the velocity of light from√√
Vx02 + Vy02 + Vz02 = (−v)2 + c2 + (−vh)2√= v 2 + c2 − v 2= c.
In addition, (10.29), (10.30), and (10.31) diﬀer from the conclusions obtained
from the Lorentz transformations. If Vx = 0, Vy = c, and Vz = 0 are substituted
for the velocity transformations
Vx − v,
1 − (v/c2 )Vx√
Vy 1 − v 2 /c2
Vy0 =,
1 − (v/c2 )Vx√
Vz 1 − v 2 /c20Vz =1 − (v/c2 )Vx

Vx0 =(10.16)(10.17)(10.18)

obtained from the Lorentz transformations, then we findVx0 = −v,√
Vy0 = c 1 − v 2 /c2 ,Vz0 = 0.
These results diﬀer from (10.29), (10.30), and (10.31). However, also in this case,they agree with the principle of the velocity of light because√√
Vx02 + Vy02 + Vz02 = (−v)2 + c2 − v 2= c.
As proven in Section 10.5, the same conclusions can be obtained if light is emitted
in the direction of moving observer B from observer A at rest under the Lorentz
and new Lorentz transformations. If light is emitted at a right angle in the direction81along the movement of B, as seen from observer A at rest, the principle of the velocity of light is realized in the Lorentz and new Lorentz transformations as mentioned
above. However, although Vx0 is same, the values of Vy0 and Vz0 are diﬀerent. In theLorentz transformations, the vertical component of the velocity of light, i.e.,√
Vy0 = c 1 − v 2 /c2 ,decreases as v approaches c. However, Vy0 is the constant value c in the new Lorentz
transformations. This diﬀerence is investigated in Chapter 16.

10.7 The new Lorentz transformations and inverse
transformations
Using the third proof, we show that the new Lorentz transformations are mathematically consistent, which has been already proven by two methods, i.e., invarianceof the world distance using coordinate transformations and the constancy of thevelocity of light. The reader may feel that the third proof is unnecessary; however,
because the new quaternion is the new mathematics, which does not exist anywhere,
it is necessary to verify its correctness by various methods.
First, we consider the Lorentz transformations. The equations for t0 and x0 of theLorentz transformations are as follows:t − (v/c2 )xt0 = √,
1 − v 2 /c2
x − vtx0 = √.
1 − v 2 /c2

(3.3)
(3.4)

If observer B moves with velocity v as seen from observer A, then A moves withvelocity −v as seen from B. Thus, the equation can be rewritten by substituting v →−v, t0 → t, t → t0 , and x → x0 in (3.3). This becomes the Lorentz transformation
under the condition that B is at rest and A moves with velocity −v. This is called
inverse transformation. If we substitute v → −v, t0 → t, t → t0 , and x → x0 in(3.3), we findt0 + (v/c2 )x0
t= √.
(10.32)
1 − v 2 /c2
If the equation of this inverse transformation can be obtained by subtracting x
from (3.3) and (3.4), we can see that the correctness of the Lorentz transformation
is proven.
By multiplying both sides of (3.4) by (v/c2 ), we have82

(v/c2 )x0 =(v/c2 )x − (v 2 /c2 )t√.
1 − v 2 /c2(10.33)

By adding both sides of (3.3) and (10.33), we havet − (v 2 /c2 )t(v/c2 )x0 + t0 = √1 − v 2 /c2√= t 1 − v 2 /c2 .
Then we findt0 + (v/c2 )x0
t= √.
1 − v 2 /c2(10.34)Equation (10.34) is the same as (10.32). In other words, (10.34) is the inverse
transformation equation. The fact that this expression is obtained by subtracting
x from (3.3) and (3.4) shows that (3.3) and (3.4) are mathematically consistent.
Similarly, if we can show that the equation obtained by subtracting z from theequations for y 0 and z 0 of the new Lorentz transformations, i.e.,y + (v/c)zh
y0 = √,
1 − v 2 /c2(10.5)z − (v/c)yhz0 = √,
1 − v 2 /c2(10.6)

agrees with the equation obtained by substituting v → −v, y 0 → y, y → y 0 ,and z → z 0 in (10.5), then (10.5) and (10.6) are confirmed to be mathematicallyconsistent.
By multiplying both sides of (10.6) by −(v/c)h, we have−(v/c)z 0 h =−(v/c)zh − (v 2 /c2 )y√.
1 − v 2 /c2(10.35)

By adding both sides of (10.5) and (10.35) and subtracting (v/c)zh, we find thaty − (v 2 /c2 )yy 0 − (v/c)z 0 h = √1 − v 2 /c2√= y 1 − v 2 /c2 .
Thus, we havey 0 − (v/c)z 0 h
y= √.
1 − v 2 /c2(10.36)Equation (10.36) is the equation in which we substituted v → −v, y 0 → y, y → y 0 ,and z → z 0 in (10.5). In other words, it is the equation of the inverse transformation of (10.5). This result shows that (10.5) and (10.6) are not mathematicallycontradictory.
83

As mentioned above, it has been proven that no contradiction originates fromthe new Lorentz transformations using invariance of the coordinate transformationof the world distance, the constancy of the velocity of light, and the inverse transformation. The structure of space–time, which has not been known until now, issuggested by the fact that the transformation equations of y 0 and z 0 of the new
Lorentz transformations contain the imaginary number h unlike the Lorentz transformations. We will consider it in Chapter 11.

10.8 General formulae of the new Lorentz transformations
When we obtained the new Lorentz transformations in Section 10.1, the directionof velocity v of observer B was limited to the x-direction along observer A. Because
the coordinate transformation A/ |A| by the new quaternion is a mathematical calculation, we can obtain the general formulae of the new Lorentz transformations in
the case that the direction of velocity v of B is arbitrary. On the other hand, because
researchers consider light graphically in special relativity, they cannot imagine thegeneral formulae of the Lorentz transformations. In fact, there is no book of thetheory of relativity in which the general formulae are written.
To obtain the general formulae describing the new Lorentz transformations, it isenough for us to perform a coordinate transformation by changing the coordinates(cth, vti, 0, 0) of B to (cth, vx ti, vy tj, vz tk) in Section 10.1. First, we consider
a three-dimensional space. we assume that observer A is at rest at the origin O
and observer B moves in a straight line with constant velocity v. The componentsof velocity v are (vx i, vy j, vz k). In addition, the observed point D is at rest at(xi, yj, zk), and A and B are coincident at the origin O at time t = 0. After t
seconds, A is at (0, 0, 0) at time t, B is at (vx ti, vy tj, vz tk) at time t, and D
is at (xi, yj, zk) at time t so that in four-dimensional space-time, the coordinates
bescome A(cth, 0, 0, 0), B(cth, vx ti, vy tj, vz tk), and D(cth, xi, yj, zk). Their
new quaternions are
A = cth, B = cth + vx ti + vy tj + vz tk, D = cth + xi + yj + zk.
The equation of the coordinate transformation of D by B, i.e., the equation
showing how D is seen by B, is DB/ |B|. If c > v, we find thatDB(cth + xi + yj + zk)(cth − vx ti − vy tj − vz tk)=√|B|(cth + vx ti + vy tj + vz tk)(cth − vx ti − vy tj − vz tk)

84

=√

1
c2 t2 h2 − vx2 t2 i2 − vy2 t2 j 2 − vz2 t2 k 2

× [cth(cth − vx ti − vy tj − vz tk) + xi(cth − vx ti − vy tj − vz tk)=√+ yj(cth − vx ti − vy tj − vz tk) + zk(cth − vx ti − vy tj − vz tk)]1
c2 t2 h2 − vx2 t2 i2 − vy2 t2 j 2 − vz2 t2 k 2

× [−c2 t2 − cvx t2 hi − cvy t2 hj − cvz t2 hk + ctxhi + vx tx − vy txk + vz txj+ ctyhj + vx tyk + vy ty − vz tyi + ctzhk − vx tzj + vy tzi + vz tz]
1√=
cth 1 − v 2 /c2× [(−c2 t + vx x + vy y + vz z)t + (x − vx t)cthi + (y − vy t)cthj + (z − vz t)cthk=√+ (vy z − vz y)ti + (vz x − vx z)tj + (vx y − vy x)tk]
1
1 − v 2 /c2× [(ct − vx x/c − vy y/c − vz z/c)h + (x − vx t)i + (y − vy t)j + (z − vz t)k+ (vz y/c − vy z/c)hi + (vx z/c − vz x/c)hj + (vy x/c − vx y/c)hk].

(10.37)

If we assume that the coordinates of D after the coordinate transformation by B
are (ct0 h, x0 i, y 0 j, z 0 k), then the new quaternion becomes ct0 h + x0 i + y 0 j + z 0 k.
Thus, by comparing the coeﬃcients, the h part is1(ct − vx x/c − vy y/c − vz z/c).
ct0 = √
1 − v 2 /c2
By dividing both sides by c, we havet0 = √

1(t − vx x/c2 − vy y/c2 − vz z/c2 ).

(10.38)(x − vx t + vz yh/c − vy zh/c),(10.39)1y0 = √(y − vy t + vx zh/c − vz xh/c),
1 − v 2 /c2(10.40)1
z0 = √(z − vz t + vy xh/c − vx yh/c),
1 − v 2 /c2(10.41)1 − v 2 /c2

The i, j, and k parts are
x0 = √

1
1 − v 2 /c2

respectively. (10.38), (10.39), (10.40), and (10.41) are the general formulae of the
new Lorentz transformations.
85

The case that B moves in the x-direction corresponds the case of vy = vz = 0
in the above equations. Thus, if vy = vz = 0 are substituted in (10.38), (10.39),(10.40), and (10.41), we findt − vx x/c2
t0 = √,
1 − v 2 /c2
x − vx t
x0 = √,
1 − v 2 /c2
y + vx zh/cy0 = √,
1 − v 2 /c2z − vx yh/cz0 = √.
1 − v 2 /c2(10.42)(10.43)(10.44)(10.45)Equations (10.42), (10.43), (10.44), and (10.45) agree with the new Lorentz transformations, i.e., (10.3), (10.4), (10.5), and (10.6), respectively, which were obtainedin Section 10.1.
It may seem odd that the signs of the h terms are opposite in the equations fory 0 and z 0 of the new Lorentz transformations, i.e.,y + vx zh/cy0 = √,
1 − v 2 /c2(10.5)z − vx yh/cz0 = √,
1 − v 2 /c2(10.6)

which were obtained in Section 10.1. However, since there is no polarization in
the general formulae (10.40) and (10.41) of the new Lorentz transformations, there
is no mathematical contradiction. If the conditions vy = vz = 0 are imposed in(10.40) and (10.41), polarizations occur in y 0 and z 0 , respectively. We tend to
think that the selection of the y- and z-axes must be arbitrary and neither axis isdominant. Therefore, y 0 and z 0 must not be polarized. However, after observer A
determines the directions of the y- and z-axes, the y 0 - and z 0 -axes become uniqueand polarizations occur.

86

11
Double Structure of
Four-Dimensional Space-Time

11.1 The algebraic theorem
A wise reader has probably noticed that, because the new quaternion consists offive numbers, one real and four imaginary numbers designated h, i, j, and k, it
must be a five-element number. An algebraic theorem states that the algebraic
calculations of addition, subtraction, multiplication, and division can be operated
only on real numbers, complex numbers, quaternions, and octonions. In otherwords, five-element numbers are mathematically implausible.
We explain the above algebraic theorem with an example. If a complex number
comprises two real numbers a and b and an imaginary number i, no new number
forms can exist; for example, multiplying i by b yields new imaginary number bi.
Even if we multiply two or more imaginary numbers, the result is either a real
number or an imaginary number; for instance, bi × i = −b and bi × i × i = −bi. In
other words, regardless of the calculation, real and imaginary are the only possible
number forms, and complex numbers are regarded as two-element numbers. The
same conclusion is drawn from Hamilton’s quaternions. The quaternion comprises
one real number and three imaginary numbers, denoted i, j, and k, wherei2 = j 2 = k 2 = −1,
ij = k, jk = i, ki = j,ijk = k 2 = −1.
All calculations performed on quaternions yield one real number and three imaginary
numbers i, j, and k. Therefore, quaternions are four-element numbers.
The new quaternion contains three numbers hi, hj, and hk in addition to fivenumbers (one real and four imaginary, designated h, i, j, and k). If hi, hj, and87hk remain after a set of calculations, they cannot be reduced to easier forms. Thus,
they are independent elements in the new quaternion. In other words, because the
new quaternion consists of eight fundamental numbers, one real number and sevennumbers h, i, j, k, hi, hj, and hk, it is called an octonion. Furthermore, hi, hj,
and hk are obtained by multiplying two imaginary numbers whose square is, e.g.,(hi)(hi) = h2 i2 = 1. Thus, they are considered as new real numbers. In otherwords, the new quaternion is an octonion composed of four real and four imaginary
numbers.
The existing octonions, known as Cayley number, operate in flat space–time. In
contrast, the new quaternion operates in curved space–time. In addition, unlike the
new quaternion, Cayley numbers consist of one real number and seven imaginary
numbers. Considering this diﬀerence, we hereafter refer to the new numbers as
new octonions. Because the new octonion is the highest number class admitted
by the algebraic theorem, it may be able to calculate all phenomena in curved
four-dimensional space–time.

11.2 The octonion
Before describing the properties of the new octonion, we introduce Graves’ octonion
discovered by John T. Graves in 1844. For details of Graves’ octonion, the reader
is referred to John H. Conway and Derec A. Smith’s work On Quaternions andOctonions. Graves was a friend of Hamilton, who just a year earlier had discovered
the quaternion. Thus, the quaternion and octonion were discovered essentially at
the same time. Graves informed Hamilton of his discovery and ongoing investigationof the octonion by letter. However, Arthur Cayley formally published his study in
1845. Although Hamilton communicated to the academic journal that Graves wasits first discoverer, octonions had become widely known as Cayley numbers, a namethat has persisted today.
The octonion consists of one real number and seven imaginary numbers. Assumingthat a and bn are real numbers and in is an imaginary number, the octonion can be
written as
a + b1 i1 + b2 i2 + b3 i3 + b4 i4 + b5 i5 + b6 i6 + b7 i7 .
Its algorithms are as follows:
i2n = −1,
in in+1 = in+3 = −in+1 in ,in+1 in+3 = in = −in+3 in+1 ,88

in+3 in = in+1 = −in in+3 .
These algorithms are too complicated for a concrete interpretation.
Recently, Graves’ octonion has been applied to the study of fundamental particles. However, because the calculations are complicated and octonion mathematics
operates in flat space-time, Graves’ octonion cannot yield precise conclusions on
the fundamental nature of particles. On the contrary, because the new octonion
performs simple operations in curved space–time, it may become a powerful tool for
fundamental particle study.

11.3 The new octonion
Unlike Graves’ octonion, the algorithms of the new quaternion, i.e., the new octonion, are obtained by adding
h2 = −1, hi = ih, hj = jh, hk = kh
to the algorithms of Hamilton’s quaternion, i.e.,
i2 = j 2 = k 2 = −1, ij = −ji = k, jk = −kj = i, ki = −ik = j.
Thus, the new octonion is easily interpretable. In addition, the properties of curved
four-dimensional space–time, to date explained by special relativity, can be mathematically demonstrated by the new octonion. Furthermore, whereas the seven
imaginary numbers square to −1 in Graves’ octonion, in the new octonion, h, i, j,
and k square to −1 while the square of real number a and the multiples hi, hj,
and hk become +1. This diﬀerence originates from the fact that the new octonion
operates in curved space–time.
Graves’ octonion is diﬃcult to understand because it operates in flat space–time,
whereas real space–time is curved. When curved space–time phenomena are calculated by the standard octonion, the results of the calculation may become complicated. On the other hand, calculations are easily performed in curved space–timeusing the new octonion. That is, the new octonion underlies the mathematics ofreal-world phenomena.
Next, we consider magnitude |A| of a new octonion A, where A is described as
A = ah + bi + cj + dk + p + qhi + rhj + shk.
This number contains real components a, b, c, d, p, q, r, and s and imaginary
components h, i, j, and k. From the definition of the complex conjugate, new89

complex conjugate, quaternion conjugate, and new quaternion conjugate, we define
a new octonion conjugate as
A = ah − bi − cj − dk + p − qhi − rhj − shk,
where all numbers other than imaginary number ah and real number p are negative.
2

In addition, if the the squares of complex magnitudes |A| = AA are also realized
by the new octonion, we can write2|A| = AA= (ah + bi + cj + dk + p + qhi + rhj + shk)× (ah − bi − cj − dk + p − qhi − rhj − shk).

(11.1)

By analogy to the complex number, new complex number, quaternion, and newquaternion, the reader might consider that (11.1) equates to2|A| = (ah)2 − (bi)2 − (cj)2 − (dk)2 + p2 − (qhi)2 − (rhj)2 − (shk)2= −a2 + b2 + c2 + d2 + p2 − q 2 − r2 − s2 .

(11.2)

However, this is not the case.
Instead, we obtain
2|A| = AA= (ah + bi + cj + dk + p + qhi + rhj + shk)× (ah − bi − cj − dk + p − qhi − rhj − shk)
= [(ah + p) + (b + qh)i + (c + rh)j + (d + sh)k]× [(ah + p) − (b + qh)i − (c + rh)j − (d + sh)k]= (ah + p)2 − (b + qh)2 i2 − (c + rh)2 j 2 − (d + sh)2 k 2= (ah + p)2 + (b + qh)2 + (c + rh)2 + (d + sh)2 .

(11.3)The diﬀerence between (11.3) and (11.2) is
2aph + 2bqh + 2crh + 2dsh.
2Thus, |A| does not equate to (11.2).

11.4 Double structure of four-dimensional space-time
To interpret (11.3), we can state that the new octonion expresses a four-dimensional
space–time formed by four coordinate axes, namely, the (ah+p)-, (b+qh)i-, (c+rh)j, and (d + sh)k-axes. Alternatively, each axis of four-dimensional space-time is a90

complex number composed of a real and an imaginary number. To ensure that newquaternion calculations are consistent with special relativity, the (ah + p)-axis is
assigned as the temporal axis while the (b + qh)i-, (c + rh)j-, and (d + sh)k-axes
become space axes.
If p is multiplied by imaginary number h and the (ah + p)-axis is transformed intothe (a+ph)-axis, the equation is simplified apparently. However, this transformation
restores Hamilton quaternion a + bi + cj + dk, which no longer operates in curved
space-time. In this book, we have sequentially advanced from the new complex
number to new octonion
ah + bi + cj + dk + p + qhi + rhj + shk.
When developing the mathematics of four-dimensional space–time, the most esthetically pleasing octonion isa + bi + cj + dk + ph + qhi + rhj + shk,
which leads us astray from the new octonion.
The mathematical conclusion that each coordinate axis of the four-dimensional
space–time is a complex number may be diﬃcult to accept per se, but one interpretation is readily acceptable. Four-dimensional space–time can be regarded as two overlapping space–times. The coordinates of our own space–time are (ah, bi, cj, dk),
and those of alternative space–time are (p, qhi, rhj, shk), which we cannot observe. At the same time, the inhabitants of the alternative coordinate system(p, qhi, rhj, shk) cannot observe the coordinates (ah, bi, cj, dk) of our space–time.
We call this interpretation the double structure of four-dimensional space–time. In
the future, reseachers may accept that each coordinate axis of four-dimensional
space–time is a complex number. Currently, we assume more intuitive double structure of four-dimensional space-time. That is, we consider that each coordinate axisof two four-dimensional space–times overlaps and coexists. In simple terms, we
regard our world and the alternative world as the positive and negative world, respectively. Some researchers advocate the parallel world scenario in which many
universes exist side-by-side. Mathematically, we can demonstrate that two fourdimensional space–times overlap; equivalently, that four-dimensional space–timepossesses four complex axes.
At this point, readers may be wondering why the octonion, as an eight-element
number, does not represent an eight-dimensional space–time. However, as indicatedin (11.3), the new octonion expresses a four-dimensional space–time, whose coordinate axes are complex numbers. Thus, the new octonion expresses two overlapped91four-dimensional space–times. Because the new octonion is the highest number
class permitted by the algebraic theorem, a larger dimensional space–time morethan four-dimensions cannot exist. While higher dimensions are frequently adopted
in cosmology, such as sixteen-dimensional cosmology, these cosmologies are inconsistent with the algebraic theorem.
The world of imaginary numbers is often regarded as the imaginary world. However, the new octonion permits a world in which four coordinate axes are expressedby the imaginary numbers ah, bi, cj, and dk while the real numbers p, qhi, rhj,
and shk are assigned to the cooordinates of the negative world. In the previous discussion, we arbitrarily referred to our world as the positive world. Strictly speaking,our world and the alternative world are worlds of imaginary and real numbers, respectively.

11.5 New octonions and new Lorentz transformations in
the negative world
In this section, we investigate the structure of the negative world in terms of the
double structure theory of four-dimensional space–time. The coordinates of thepositive world, in which we live, are (cth, xi, yj, zk) and a world point is represented
by new octonion cth + xi + yj + zk. Therefore, in new octonion
A = ah + bi + cj + dk + p + qhi + rhj + shk,
the variables ah, bi, cj, and dk are equivalent to cth, xi, yj, and zk, respectively.
Then, what is the physical meaning of the remaining variables p, qhi, rhj, andshk?
Multiplying the new octonion cth + xi + yj + zk, specifying a world point in thepositive world, by h, we obtain −ct + xhi + yhj + zhk, which contains the sameterms as the remaining variables p, qhi, rhj, and shk of the new octonion. Thus,
it is mathematically natural to replace p, qhi, rhj, and shk by −ct, xhi, yhj, andzhk. The new octonion −ct + xhi + yhj + zhk may specify a world point in thenegative world. Replacing −ct of this new octonion with ct, a world point in thenegative world can also be designated as ct + xhi + yhj + zhk. The correct form
is established by investigating the Lorentz transformations in the negative world,
assuming that each new octonion is proper.

92

(1) If the new octonion in the negative worlds is −ct + xhi + yhj + zhk
We replace the new octonion cth + xi + yj + zk describing a world point in thepositive world with a new octonion −ct + xhi + yhj + zhk describing a world pointin the negative world, and apply the calculations in Section 10.1 to obtain theLorentz transformations. We assume that motion occurs along the x-axis; thatis, y = z = 0. In the negative world, we assume that observer A is at rest at
the origin O, while observer B moves alomg the x-axis at constant velocity v.
In addition, a fixed observed point D lies at distance x, and observers A and B
coincide at O at time t = 0. After t seconds, A remains at the origin, while B has
traveled distance vt. D remains fixed at distance x. The coordinates of the threeobjects are A(−ct, 0), B(−ct, vthi), and D(−ct, xhi), and their new octonions are
A = −ct, B = −ct + vthi, and D = −ct + xhi. Observer A notes that time t has
passed but his distance x remains 0. Thus, A moves along the −ct-axis. Observer
B moves along the straight line x = vt. A space–time diagram of this situation will
be illustrated later.
The transformation formula DB/ |B| expresses the coordinate transformation of
D observed by B. In other words, it expresses the coordinates of D from thereference frame of B, moving along a straight line with uniform velocity. In termsof the new octonions, DB/ |B| is calculated using the new octonions B = −ct + vthi
and D = −ct + xhi to obtainDB(−ct + xhi)(−ct − vthi)=√|B|(−ct + vthi)(−ct − vthi)=

c2 t2 + cvt2 hi − xcthi − xvth2 i2√(−ct)2 − (vthi)2=

c2 t2 − xvt + cvt2 hi − xcthi√.
(−ct)2 − (vthi)2(11.4)

Before removing ct from the square root in the denominator of (11.4), we investigate its sign. If ct = x = y = z = 1 in the new octonion cth + xi + yj + zk describing
a world point in the positive world, the new octonion becomes h+i+j +k, the basic
new octonion in the positive world. Multiplying the new octonion cth + xi + yj + zk
by h, we obtain the new octonion −ct + xhi + yhj + zhk, describing a world point in
the negative world. Thus, the basic new octonion −1 + hi + hj + hk of the negative
world can be obtained by multiplying h + i + j + k by h. From this result, the timecomponent of the negative world seems to be negative; that is, −ct < 0, or ct > 0.
Since(hi)2 = h2 i2 = (−1)(−1) = 1,93

removing ct from the square root of the denominator of (11.4) under the condition
ct > 0, we obtainDB
c2 t2 − xvt + cvt2 hi − xcthi√=|B|(−ct)2 − (vthi)2=

c2 t2 − xvt + cvt2 hi − xcthi√
ct 1 − (vthi)2 /(−ct)2=c2 t(t − vx/c2 ) − ct(x − vt)hi√
ct 1 − v 2 /c2=c(t − vx/c2 ) − (x − vt)hi√.
1 − v 2 /c2(11.5)

Since (11.5) describes the coordinates −ct0 + x0 hi of D as seen from B, we have−ct0 + x0 hi =c(t − vx/c2 ) − (x − vt)hi√.
1 − v 2 /c2Comparing the coeﬃcients, we can write−t + (v/c2 )xt0 = √,
1 − v 2 /c2−x + vtx0 = √.
1 − v 2 /c2(11.6)(11.7)The signs of (11.6) and (11.7) are contrary to the equations of the Lorentz transformations of special relativityt − (v/c2 )x,t0 = √
1 − v 2 /c2
x − vtx0 = √.
1 − v 2 /c2

(3.3)
(3.4)

If x = 0 is assumed in (11.6), we find thatt0 = −t /√1 − v 2 /c2

and t0 is of opposite sign to t. In other words, stationary observer A and moving
observer B observe D at times of oppsosite sign. Thus, if we assume that the
new octonion of a world point in the negative world is −ct + xhi + yhj + zhk, acontradiction occurs. Therefore, we exclude the possibility of −ct + xhi + yhj + zhk
and ct > 0 in the negative world.

94

(2) If the new octonion of the negative world is ct + xhi + yhj + zhk
In terms of the new octonions B = ct + vthi and D = ct + xhi, DB/ |B| becomesDB(ct + xhi)(ct − vthi)=√|B|(ct + vthi)(ct − vthi)=

c2 t2 − cvt2 hi + xcthi − xvth2 i2√(ct)2 − (vthi)2=

c2 t2 − xvt − cvt2 hi + xcthi√.
(ct)2 − (vt)2(11.8)

Before removing ct from the square root of the denominator of (11.8), we investigate its sign. In this case, −ct is replaced by ct in the new octonion −ct + xhi +yhj + zhk. Replacing −1 in the basic new octonion −1 + hi + hj + hk of case (1)with 1, the basic new octonion in the negative world becomes 1 + hi + hj + hk. In
this scenario, the time component ct of the new octonion ct + xhi + yhj + zhk is positive, i.e., ct > 0. Therefore, removing ct from the square root of the denominatorof (11.8), we obtainDB
c2 t2 − xvt − cvt2 hi + xcthi√=|B|(ct)2 − (vt)2=c2 t(t − vx/c2 ) + ct(x − vt)hi√
ct 1 − v 2 /c2=c(t − vx/c2 ) + (x − vt)hi√.
1 − v 2 /c2(11.9)The new octonion, expressing the coordinates of D as seen from B, is ct0 + x0 hi.
Thus, from (11.9), we have
ct0 + x0 hi =c(t − vx/c2 ) + (x − vt)hi√.
1 − v 2 /c2Comparing the coeﬃcients, we can writet − (v/c2 )xt0 = √,
1 − v 2 /c2
x − vtx0 = √.
1 − v 2 /c2(11.10)(11.11)Equations (11.10) and (11.11) are exactly the equations of the Lorentz transformations of special relativityt − (v/c2 )xt0 = √,
1 − v 2 /c2
x − vt.
x0 = √
1 − v 2 /c2
95

(3.3)
(3.4)

In addition, if x = 0, (11.10) reduces to√t0 = t/ 1 − v 2 /c2 .
Thus, unlike case (1), time does not change sign when observed by diﬀerent observers. The new octonion ct + xhi + yhj + zhk with ct > 0 appropriately represents
a world point in the negative world.
Before constructing the four-dimensional space–time diagram in case (2), we elucidate the meaning of ct > 0. Because y = z = 0, the diagram reduces to a newcomplex plane. The expression ct > 0 permits two cases: if c > 0, we have t > 0,
and if c < 0, we have t < 0. Because y 0 and z 0 in the new positive-world Lorentz
transformations involve h, then y 0 and z 0 reside in the negative world. If the velocity
of light c changes sign in the positive or negative world, the new Lorentz transformations will not be realized. Thus, we may appropriately consider c > 0 and t > 0
in both positive and negative worlds. When c > 0, we have v > 0. When v > 0 andt > 0, we have x > 0. Thus, the world line x = vt of observer B in the negative
world is a straight line progressing in the upper-right direction from the origin O
in the new complex plane. The world line of light similarly progresses as a straightline in the upper-right direction, but remains equidistant from the ct- and xhi-axes.
Both world lines are illustrated in Figure 11.1. If this diagram is compared to the
new complex plane in the positive world, the reader will appreciate that the positive
and negative worlds overlap.

From (11.10) and (11.11), we draw an important conclusion that the same physicallaw, described by the same formula, applies in both positive and negative worlds.
Consequently, it is thought that matter and gravity exist in both worlds. Thisfinding is important to fundamental physics.

9611.6 World distance in the negative world
We calculate the world distance s given that the new octonion of a world point A
in the negative world is A = ct + xhi + yhj + zhk. The algorithms are as follows:
h2 = i2 = j 2 = k 2 = −1,
ij = −ji = k, jk = −kj = i, ki = −ik = j,hi = ih, hj = jh, hk = kh.
Thus, we finds2 = AA= (ct + xhi + yhj + zhk)(ct − xhi − yhj − zhk)
= (ct)2 − ctxhi − ctyhj − ctzhk+ xcthi − (xhi)2 − xyh2 ij − xzh2 ik+ ycthj − yxh2 ji − (yhj)2 − yzh2 jk+ ctzhk − zxh2 ki − zyh2 kj − (zhk)2= c2 t2 − ctxhi − ctyhj − ctzhk + ctxhi − x2 + xyk − zxj+ ctyhj − xyk − y 2 + yzi + ctzhk + zxj − yzi − z 2= c2 t2 − x2 − y 2 − z 2 .

(11.12)

Reconsider that, in Section 7.2, we derived the world distance in the positive
world ass2 = −c2 t2 + x2 + y 2 + z 2 .
(7.6)
The sole diﬀerence between (11.12) and (7.6) is reversal of sign. Furthermore, if
c > v, the square of the world distance in the positive and negative worlds, isnegative and positive, respectively.
According to this result, our world is strictly the negative world while the so-called
negative world is the positive world. However, since this concept is unacceptable tomany individuals, we refer our world as the positive world until the new octonion
gains wider acceptance.
In addition, the world distance derived from special relativity (see Section 7.1)
s2 = (ct)2 − x2 − y 2 − z 2
expresses the world distance in the negative world.

97

(7.1)

11.7 World point in double-structured four-dimensional
space-time
In the following discussions, we assume that the new octonion representing the world
point in the negative world is ct + xhi + yhj + zhk. Thus, the new octonion of aworld point in double structured four-dimensional space–time is
A = ct0 h + x0 i + y0 j + z0 k + ct1 + x1 hi + y1 hj + z1 hk.

(11.13)Equation (11.13) can be written as
A = (ct0 h + ct1 ) + (x0 + x1 h)i + (y0 + y1 h)j + (z0 + z1 h)k.
Assumed conditions are c > 0, t0 > 0, t1 > 0 and the algorithms are as follows:
h2 = i2 = j 2 = k 2 = −1,
ij = −ji = k, jk = −kj = i, ki = −ik = j,hi = ih, hj = jh, hk = kh.

Figure 11.2 is a four-dimensional space–time diagram in which each coordinate
axis is specified by a complex number. The four-dimensional space–time diagramin Figure 11.3 assumes that four-dimensional space–time has a double structure.
Because the zk- and zhk-axes cannot be projected onto the diagram, they are indicated by dashed lines. In addition, when the positive and negative worlds aresynchronously illustrated, we denote the coordinate axes in the negative world by[ct] , [xhi] , [yhj], and [hkz]. Even if the coordinate elements include components ofboth worlds, confusion is prevented by enclosing imaginary numbers in parenthesis,
e.g., (cth, xi, yj, zhk) in Chapter 6. The coordinates in the positive and negative
worlds are diﬀerent ((ah, bi) and (a, bhi), respectively). In contrast, in Hamilton’snotation, real numbers are expressed in (a, b) format, allowing no distinction between positive and negative worlds. Thus, Hamilton’s notation is inconvenient forour purpose.
98Describing a world point in the whole four-dimensional space–time by (11.13),
can we determine whether t0 and t1 are identical? Similarly, are x0 and x1 , y0 and
y1 , and z0 and z1 identical? To answer this question, we rewrite (11.13) as
B = cth + xi + yj + zk + ct + xhi + yhj + zhk= (cth + ct) + (xi + xhi) + (yj + yhj) + (zk + zhk),(11.14)and investigate whether the coordinate transformation DB/ |B| in terms of (11.14)
yields the Lorentz transformations.
The new octonion of observer B moving with uniform velocity v in the positive
x-direction away from stationary observer A is
B = (cth + ct) + (vti + vthi)= ct(1 + h) + vti(1 + h)
= (1 + h)(ct + vti).
The new octonion conjugate is
B = (cth + ct) − (vti + vthi)= (1 + h)(ct − vti)
and the new octonion of observed point mass D is
D = (cth + ct) + (xi + xhi)
= ct(1 + h) + xi(1 + h)
= (1 + h)(ct + xi).
Thus, we obtainDB(1 + h)(ct + xi)(1 + h)(ct − vti)=√
|B|
(1 + h)(ct + vti)(1 + h)(ct − vti)=

(1 + h)(ct + xi)(ct − vti)√
(ct + vti)(ct − vti)

(1 + h)(c2 t2 − cvt2 i + xcti − xvti2 )√c2 t2 − v 2 t2 i2
2 2(1 + h)(c t + xvt − cvt2 i + xcti)√=
c2 t2 + v 2 t2
[ 2]
(1 + h) c t(t + vx/c2 ) + ct(x − vt)i√=
ct 1 + v 2 /c2
=

99

[
]
(1 + h) c(t + vx/c2 ) + (x − vt)i√=
1 + v 2 /c2c(t + vx/c2 )x − vtc(t + vx/c2 )x − vt
= √
h+ √
i+ √
+√
hi.
2
2
2
2
2
2
1 + v /c
1 + v /c
1 + v /c
1 + v 2 /c2This expression is the new octonion D0 = ct0 h + x0 i + ct0 + x0 hi of D observed fromthe reference frame of B. Clearly, this expression is inconsistent with the Lorentz
transformations. Therefore, the positive and negative worlds cannot share common√
t, x, y, and z, as assumed in (11.14). In addition, because c2 of 1 + v 2 /c2 is
non-negative, (11.14) is thought to revert the new octonion to the mathematics offlat space–time.

11.8 Four real numbers and four imaginary numbers infour-dimensional space-time
In this section, we expand the concepts introduced in Section 11.1. More specifically,
we investigate whether the hi, hj, and hk components of a new octonion ah + bi +cj + dk + p + qhi + rhj + shk are imaginary or real.
Real and imaginary numbers are defined by the sign of their squares; a squared
real number is positive, while a squared imaginary number is negative. The newoctonion algorithm gives
h2 = i2 = j 2 = k 2 = −1
implying that h, i, j, and k are imaginary. On the other hand, sincehi = ih, hj = jh, hk = kh,
we have(hi)2 = hihi
= h2 i2= (−1)(−1)= 1,2(hj) = 1,(hk)2 = 1.
Thus, hi, hj, and hk are considered to be real numbers. In contrast, squaring ij, jk,

100

and ki, we get(ij)(ij) = k 2 = −1,(jk)(jk) = i2 = −1,(ki)(ki) = j 2 = −1.
Therefore, ij, jk, and ki are imaginary numbers and are distinct from their realcounterparts hi, hj, and hk.
As proven in Section 11.5, the new octonion of a world point in the positive
world, in which we live, is cth + xi + yj + zk, while the negative world equivalentis ct + xhi + yhj + zhk. Thus, h is a time-imaginary number and i, j, k are spaceimaginary numbers. Similarly, the real numbers a used in our daily routines aretime-real numbers, and hi, hj, and hk are space-real numbers. In addition, allcoordinate axes in the positive and negative worlds are represented by imaginary
and real numbers, respectively. The real number a is also the time-real numberin the negative world. In addition, a can be called a scale-real number because it
scales as a × i = ai.
At present, while multiple imaginary numbers are recognized, namely, i, j, andk in Hamilton’s quaternion
a + bi + cj + dk
and i1 , i2 , i3 , i4 , i5 , i6 , and i7 in Graves’ octoniona + b1 i1 + b2 i2 + b3 i3 + b4 i4 + b5 i5 + b6 i6 + b7 i7 ,it appears that a single real number exists. However, as mentioned above, we candefine three space-real numbers hi, hj, and hk in addition to the scalable time-real
number. If this is true, only the scale real number and the imaginary number i havebeen adopted in mathematics and physics, although four real and four imaginary
numbers exist in four-dimensional space–time. Quantum mechanics has exploited
only the imaginary number i, while Hamilton’s quaternion has been largely unused.
Given the rich content of the new octonion, we may naturally consider that, unless all eight numbers are adopted, phenomena in four-dimensional space–time will
never be understood completely. The eight numbers contained in the new octonions
allow precise calculation of four-dimensional space–time phenomena. Supporting
this conclusion, coordinate transformation A/ |A|, which does not consider light,

101retrieves the equations of special relativityt − (v/c2 )xt0 = √,
1 − v 2 /c2
x − vtx0 = √,
1 − v 2 /c2

(3.3)
(3.4)

plus two new equationsy + (v/c)zh
y0 = √,
1 − v 2 /c2(10.5)z − (v/c)yhz0 = √,
1 − v 2 /c2(10.6)

which have been overlooked in special relativity. Equations (10.5) and (10.6) emerged
only after applying the new octonion.

102

12
Oblique Coordinates ofFour-Dimensional Space-Time

12.1 Oblique coordinates of motion along x-axis
To verify the correctness of a coordinate transformation, we must investigate how
coordinate axes change after a coordinate transformation. In Chapter 4, it wasproven that, after a coordinate transformation, coordinate axes become obliquecoordinate axes in two-dimensional space–time. In this chapter, we investigate therelationship between the cth-, xi-, yj-, and zk-axes of observer A and the ct0 h-, x0 i-,
y 0 j-, and z 0 k-axes of observer B in four-dimensional space–time.
Conclusions obtained in this chapter may be unnecessary for the reader whose
specialty is neither physics nor mathematics. However, the following conclusions areimportant: A coordinate transformation of the yj- and zk-axes transforms them into
the y 0 j-axis in the positive world and the z 0 hk-axis in the negative world. Therefore,
the yj-zk plane becomes the y 0 j-[z 0 hk] plane, which contains each coordinate axis in
the positive and negative worlds. In other words, the positive and negative worlds
do not exist independently but are mixed like a mosaic. These conclusions areproven below.
It is assumed that, for observer A, observer B moves in linear uniform motion
with velocity v in the x-direction. The four-dimensional space–time diagram forthis motion is shown in Figure 12.1. The thick solid lines are the cth-, xi-, andyj-axes, respectively, and the thick dashed line is the zk-axis for observer A. The
thin full lines are the ct0 h-, x0 i-, and y 0 j-axes, respectively, and the thin dashed lineis the z 0 k-axis for observer B. In addition, the ct0 h-axis is the world line x = vt ofobserver B, and both the y and z coordinates are zero for B.

103

Since the time ct0 for A and B observed by A is the same, the point A(ct0 h, 0, 0, 0)
on the cth-axis moves to the point B(ct0 h, vt0 i, 0, 0) on the ct0 h-axis by a coordinate transformation that changes the cth-axis into the ct0 h-axis. In addition,
it is assumed that the point D(0, x1 i, 0, 0) on the xi-axis moves to the pointE(ct2 h, x2 i, y2 j, z2 k) on the x0 i-axis. The point F (0, 0, y3 j, 0) on the yj-axis
moves to the point G(ct4 h, x4 i, y4 j, z4 k) on the y 0 j-axis. Similarly, the point
P (0, 0, 0, z5 k) on the zk-axis moves to the point Q(ct6 h, x6 i, y6 j, z6 k) on thez 0 k-axis.
Because the migration from A to B constitutes a rotation about the origin, it canbe denoted by a new octonion H. Since the new octonions of point A and point B
are A = ct0 h and B = ct0 h + vt0 i , respectively, from AH = B, we have
ct0 hH = ct0 h + vt0 i,
ct0 h + vt0 i
ct0 h
vi
=1+
ch
vih
=1+ 2
ch
v
= 1 − hi.
c

H=(: t0 6= 0)(12.1)

Equation (12.1) transforms the cth-axis into the ct0 h-axis.
The new octonions of D and E are D = x1 i and E = ct2 h + x2 i + y2 j + z2 k,
respectively. Since D is moved to E by the transformation H, from DH = E, we
can write
x1 iH = ct2 h + x2 i + y2 j + z2 k.

104

Substituting (12.1) into this equation givesv
ct2 h + x2 i + y2 j + z2 k = x1 i(1 − hi)cvx1 2= x1 i −hicv= x1 i + x1 hcv
= x1 h + x1 i.
c
By comparing the coeﬃcients, we havect2 =v
x1 , x2 = x1 , y2 = 0, z2 = 0.
c

Eliminating x1 from these equations, we findx2 =c(ct2 ), y2 = z2 = 0.
v

(12.2)(12.2) describes the x0 i-axis. Because the unit of the temporal axis is ct, the gradientof the x0 i-axis is not c2 /v but c/v. Because the gradient of the ct0 h-axis, which is
the result of the transformation of the cth-axis, is v/c from x = vt = (v/c)ct, thex0 i-axis and the ct0 h-axis lean inward at the same angle with respect to the xiand cth-axes, respectively; i.e., the x0 i- and ct0 h-axes are oblique coordinate axes.
However, since y2 = z2 = 0, oblique coordinate axes lie in the two-dimensional
cth-xi plane.
Next, the y 0 -axis is obtained from the y-axis. The new octonions of F and G are
F = y3 j and G = ct4 h + x4 i + y4 j + z4 k, respectively. From F H = G, we can write
y3 jH = ct4 h + x4 i + y4 j + z4 k.
Substituting (12.1) into this equation givesv
ct4 h + x4 i + y4 j + z4 k = y3 j(1 − hi)cv= y3 j − y3 hjicv= y3 j + y3 hk.
c
By comparing the coeﬃcients, we find
ct4 = 0, x4 = 0, y4 = y3 , z4 =

v
y3 h.
c

If y3 is eliminated from these equations, we are left withz4 =v
y4 h, ct4 = x4 = 0.
c
105(12.3)

(12.3) describes the y 0 j-axis. If this is drawn on the yj-[zhk] plane with the conditionct = x = 0, we obtain the result shown in Figure 12.2. Since z has h, it becomes
the [zhk]-axial ingredient in the negative world.

As explained in Section 11.7, the zk-axis is the coordinate axis in the positive
world and the [zhk]-axis is the coordinate axis in the negative world. Thus, as
shown in Figure 12.2, the horizontal axis is the yj-axis in the positive world, andthe longitudinal axis is the [zhk]-axis in the negative world. From this figure, we seethat the positive and negative worlds do not exist independently but are mixed likea mosaic. To better understand, note that one four-dimensional space–time, wherecoordinate components are complex numbers, has tentatively been divided into a
positive world and a negative world. Thus, it is not surprising that the coordinate
axes in the positive and negative worlds are intermingled. Strictly speaking, one
four-dimensional space–time with complex coordinates exists and the axes of thepositive and negative world are observed as dictated by conditions.
Here, we reconfirm that the positive world and the negative world are incorrectterms—rather, we should speak of a world of imaginary numbers and a world ofreal numbers. The terms positive world and negative world are used to make thingseasier to understand. However, these terms should not be taken to imply that all
numbers are positive in the positive world and negative in the negative world. These
terms are fallbacks because the more appropriate terminology, i.e., the world andantiworld, was already used in the physics vocabulary, i.e., matter and antimatter,and so could not be adopted. When the contents of this book become more widely
known, the terms positive world and negative world should be changed to the worldof imaginary numbers and the world of real numbers.
Because, from (12.3), the gradient of the y 0 j-axis with respect to the yj-axis isv/c, the slope of the y 0 j-axis is v/c with respect to the original coordinate axis as
well as the ct0 h- and x0 i-axes. Therefore, the y 0 j-axis becomes an oblique coordinateaxis.
We now perform the same calculations for the z 0 k-axis. The point P (0, 0, 0, z5 k)
106

on the zk-axis transforms to the point Q(ct6 h, x6 i, y6 j, z6 k) on the z 0 k-axis.
Because their new octonions are P = z5 k and Q = ct6 h+x6 i+y6 j+z6 k, respectively,
we find the following from P H = Q:
z5 kH = ct6 h + x6 i + y6 j + z6 k.
If (12.1) is substituted into this equation, we findv
ct6 h + x6 i + y6 j + z6 k = z5 k(1 − hi)cv= z5 k − z5 hkicv= − z5 hj + z5 k.
cComparing the coeﬃcients, we havev
t6 = 0, x6 = 0, y6 = − z5 h, z6 = z5 .
cIf z5 is eliminated from this formula, we findv
y6 = − z6 h, t6 = x6 = 0.
c(12.4)(12.4) describes the z 0 k-axis. As opposed to (12.3), (12.4) refers to the plane thatconsists of the [yhj]-axis in the negative world and the zk-axis in the positive world.
If it is illustrated, it becomes as shown in Figure 12.3.

Because Figures 12.2 and 12.3 have diﬀerent coordinate axes, they cannot be
stacked. Thus, if both sides of the left equation from (12.4), i.e.,v
y6 = − z6 h,care multiplied by ch/v, we havevccy6 h = − z6 h( h)vcv= z6 .
107

Thus, the equation of point Q on the z 0 k-axis becomesz6 =cy6 h, t6 = x6 = 0v

on the yj-[zhk] plane. From the equation of z6 , we find that the z 0 k-axis becomesa straight line whose slope with respect to the [zhk]-axis is v/c. Thus, the z 0 k-axis
becomes [z 0 hk]-axis because it has h, i.e., the [z 0 hk]-axis forms an oblique coordinate
axis with respect to the y 0 j-axis on the yj-[zhk] plane under the conditions ct = x = 0
as shown in Figure 12.4.

The above results prove that the coordinate transformation, that changes the cthaxis into the ct0 h-axis in four-dimensional space–time, makes four oblique coordinate
axes. In addition, when the velocity v of observer B is in the x-direction withrespect to observer A, events in four-dimensional space–time can be illustrated andcalculated only in two two-dimensional planes: the cth-xi and yj-[zhk] planes. In
other words, there is no need to consider a complicated four-dimensional space–timediagram.
The fact that the cth-xi and yj-[zhk] planes are independent can be proven bystarting with the equation−c2 t02 + x02 + y 02 + z 02 = −c2 t2 + x2 + y 2 + z 2 ,which shows the invariance of the world distance, which was explained in Section
10.3. This equation can be solved by dividing it into two parts: −c2 t02 + x02 =−c2 t2 + x2 and y 02 + z 02 = y 2 + z 2 . The independence of the cth-xi and yj-[zhk]planes follows from the fact that the new and original Lorentz transformations
containing ct0 and x0 do not contain y or z and the equations containing y 0 and z 0
do not contain ct or x.

108

12.2 Oblique coordinates of motion in any direction
When the velocity v of observer B is in the x-direction of observer A, only thecth-xi and the yj-[zhk] planes need be considered. However, when the velocity vof observer B is in an arbitrary direction, we must consider a complicated fourdimensional space–time diagram. The diagram can be understood if the generalformula of each coordinate axis after the coordinate transformation, which changes
the cth-axis into the ct0 h-axis, is obtained. In this section, we investigate motionof coordinate axes, in case the velocity v of observer B is in an arbitrary direction.
However, because the number of equations increase, only motion of the cth- andxi-axes is calculated.
Assume that the components of the velocity v of observer B as seen by observer
A in three-dimensional space are (vx i, vy j, vz k). Because v = vx i + vy j + vz k, wehavev 2 = vv
= (vx i + vy j + vz k)(−vx i − vy j − vz k)= vx2 + vy2 + vz2 .
The four-dimensional space–time diagram can be obtained by changing the coordinates of point B into (ct0 h, vx t0 i, vy t0 j, vz t0 k) in Figure 12.1. Since thetime ct0 of A and B is the same, the point A(ct0 h, 0, 0, 0) on the cth-axis
moves to the point B(ct0 h, vx t0 i, vy t0 j, vz t0 k) on the ct0 h-axis by the coordinate transformation H. The new octonions of the points A and B are A = ct0 h and
B = ct0 h + vx t0 i + vy t0 j + vz t0 k, respectively, then from AH = B, we find
ct0 hH = ct0 h + vx t0 i + vy t0 j + vz t0 k,
ct0 h + vx t0 i + vy t0 j + vz t0 k
ct0 h
1
= 1 + (vx i + vy j + vz k)chh= 1 + 2 (vx i + vy j + vz k)chh= 1 − (vx i + vy j + vz k).
cH=(: t0 6= 0)(12.5)

Equation (12.5) transforms the cth-axis into the ct0 h-axis.
We now calculate the gradient of the ct0 h-axis. The coordinate of the cth-directionof point B is ct0 h. The magnitude of the ingredient vx t0 i + vy t0 j + vz t0 k perpendicular to the cth-axis is109

√(vx t0 i + vy t0 j + vz t0 k)(−vx t0 i − vy t0 j − vz t0 k)√= t0 −vx2 i2 − vy2 j 2 − vz2 k 2 (: t0 > 0)√
= t0 vx2 + vy2 + vz2= vt0 .
From these calculations, the magnitude of the coordinate in the cth-direction ofpoint B is ct0 , and
vt0 /ct0 = v/c(12.6)

is the gradient of the ct0 h-axis with respect to the cth-axis because the magnitude of
the coordinate in the direction perpendicular to the cth-axis is vt0 ; i.e., even whenthe velocity v of B is in an arbitrary direction, the gradient of the ct0 h-axis after acoordinate transformation is v/c. This result is the same as when the velocity v of
B is in the x-direction of A.
For D, the new octonion is D = x1 i. Since D is moved to E by the transformationH, we have
E = x1 iH.
If (12.5) is substituted into this equation, we find[]h
E = x1 i 1 − (vx i + vy j + vz k)cx1 h= x1 i −(vx i2 + vy ij + vz ik)cx1 h(−vx + vy k − vz j)= x1 i −
c
vx x1
vz x1
vy x1
=
h + x1 i +
hj −
hk.
c
c
c
(12.7) is the new octonion of the point E on the x0 i-axis.

(12.7)

We now calculate the gradient of the x0 i-axis with respect to the xi-axis. The
coordinate of point E in the xi-direction is x1 i. With the help of the other coordinates from (12.7), the magnitude of a coordinate in the direction perpendicular toxi-axis is√(vx x1 h/c + vz x1 hj/c − vy x1 hk/c)(vx x1 h/c − vz x1 hj/c + vy x1 hk/c)√= (vx x1 h/c)2 − (vz x1 hj/c)2 − (vy x1 hk/c)2√= (h/c) vx2 x21 + vz2 x21 + vy2 x21= vx1 h/c. (: x1 > 0)110

From the above results, the magnitude of the coordinate of point E in the xidirection is x1 and the magnitude of the coordinate in the direction perpendicular
to the xi-axis is vx1 /c. Thus, the gradient of the x0 i-axis with respect to the xi-axis
is(vx1 /c)/x1 = v/c.

(12.8)

Since (12.8) is the same as (12.6), even when the velocity v of B is in an arbitrary
direction, the ct0 h-axis and the x0 i-axis form oblique coordinate axes.
From the same calculations, the new octonions of point G on the y 0 j-axis andpoint I on the z 0 k-axis are
vy y2
vz y 2vx y 2h−hi + y2 j +
hk,
c
c
c
vy z3
vx z3
vz z3
h+
hi −
hj + z3 k.
I=
c
c
c

G=

The gradients of the y 0 j- and z 0 k-axes can be obtained by using the same methods
that led to (12.8). However, this calculation is omitted here because it involves
more equations.

12.3 Interpretation of y 0 and z 0 using new Lorentz
transformations
As obtained in Section 10.1, the new Lorentz transformations in four-dimensional
space–time aret − (v/c2 )xt0 = √,
1 − v 2 /c2
x − vt,x0 = √
1 − v 2 /c2
y + (v/c)zh
y0 = √,
1 − v 2 /c2z − (v/c)yhz0 = √.
1 − v 2 /c2(10.3)(10.4)(10.5)(10.6)

Although (10.3) and (10.4) are the same as the Lorentz transformations of special
relativity, (10.5) and (10.6) diﬀer from the results
y 0 = y,
0

z =z

111

(3.5)
(3.6)

of the Lorentz transformations. The new transformations include the coordinates
hj and hk in the negative world, because (10.5) and (10.6) are
yj + (v/c)zhj
y0 j = √,1 − v 2 /c2
zk − (v/c)yhk
z0k = √,
1 − v 2 /c2
if written out exactly.
Because the yj-zk plane changes to the yj-[zhk] plane after a coordinate transformation, as proven in Section 12.1, (10.5) and (10.6) are considered to containhj and hk. Thus, we will prove that (10.5) and (10.6) can be obtained from thecoordinates of the nodes of straight lines in the yj-[zhk] plane.
We have proven in Section 12.1 that when the velocity v of observer B is in the xdirection for observer A, events in four-dimensional space–time can be calculated byconsidering two two-dimensional planes, i.e., the cth-xi and yj-[zhk] planes. Thus,
we now consider the point D(y0 j, z0 hk) on the yj-[zhk] plane, which is shown in
Figure 12.5.

The straight lines parallel to the y 0 j- and [z 0 hk]-axes, which are oblique coordinate
axes, are drawn from D. They intersect each coordinate axis at F (y2 j, z2 hk) andE(y1 j, z1 hk). However, if the yj- and zk-axes of stationary observer A are not first
replaced by the yj- and [zhk]-axes, the calculations do not work. The reason forthis is unknown.
(1) New Lorentz transformation formula for y 0
The gradient of the straight line DE is c/v, which is the same as the gradient of
the [z 0 hk]-axis, and it passes through the point D(y0 j, z0 hk). Thus, its equation isc(12.9)z − z0 = (y − y0 )h.
v
112

In addition, the equation of the y 0 j-axis is
z=

v
yh.
c

(12.10)Because (12.9) and (12.10) describe lines that pass through the point E(y1 j, z1 hk),
we havev
y1 h,cc
z1 − z0 = (y1 − y0 )h.
v
z1 =(12.11)(12.12)

If (12.12) is subtracted from (12.11), we findvcy1 h − (y1 − y0 )h
c
v
v
c
c
= ( − )y1 h + y0 h,
c
v
v
c
v
c
( − )y1 h = y0 h − z0 ,
v
c
v
c
c
2 2
(1 − v /c )y1 h = y0 h − z0 ,v
v
1v
y1 h =(y0 h − z0 ),22
1 − v /cc
1v
y1 =(y0 + z0 h).
1 − v 2 /c2cz0 =(12.13)

Substituting (12.13) into (12.11) gives
z1 =

vh/c
v
(y0 + z0 h).
1 − v 2 /c2c

Let |OE| be the magnitude of the line segment OE, then we have2

|OE| = (y1 j + z1 k)(−y1 j − z1 k)= −y12 j 2 − z12 k 2= y12 + z12 .
If (12.13) and (12.14) are substituted into this equation, the result is2v
v 2 h2 /c2
v
1
2
(y
+
z
h)
+
(y0 + z0 h)200(1 − v 2 /c2 )2c(1 − v 2 /c2 )2c2 2
1 − v /cv=
(y0 + z0 h)22
2
2(1 − v /c )c1v=(y0 + z0 h)2 .
1 − v 2 /c2
c|OE| =

113(12.14)

Since c > v, y0 > 0, and z0 > 0, we have
|OE| = √1
1 − v 2 /c2v(y0 + z0 h).
c

Since |OE| is the y 0 coordinate of B, we findy0 = √

1
1 − v 2 /c2v(y0 + z0 h).
c

To generalize, y0 is substituted for y, which givesy + (v/c)zh.
y0 = √
1 − v 2 /c2(12.15)(12.15) is the same as the new Lorentz transformationy + (v/c)zh
y0 = √.
1 − v 2 /c2(10.5)We have obtained it from the coordinate of the node of straight lines in the yj-[zhk]plane.

(2) New Lorentz transformation formula for z 0
The new Lorentz transformation formula for z 0 can be obtained from the coordinateof the node of the straight lines, like (1). However, we obtain the formula by anothermethod. As proven in Section 4.3, in the oblique coordinate system, the formula forthe reverse transformation can be obtained by switching y and y 0 with zh and z 0 h,
respectively, in the formula in the original coordinates. If this is done in (12.15), we
find

zh + (v/c)yz0h = √.
1 − v 2 /c2

If both sides of this equation are multiplied by −h, we have−hzh − h(v/c)y√,
1 − v 2 /c2z − (v/c)yhz0 = √.
1 − v 2 /c2−hz 0 h =(12.16)(12.16) is the same as the new Lorentz transformationz − (v/c)yhz0 = √.
1 − v 2 /c2
114(10.6)

Based on (12.15) and (12.16), the coordinate axes after the coordinate transformation in the yj-zk plane are the yj- and [zhk]-axes, and the positive world andnegative world intermingle. In addition, the same conclusion is obtained even if
the coordinate axes are the [yhj]-axis in the negative world and the zk-axis in thepositive world. However, if the yj- and zk-axes of stationary observer A are not
set as the yj- and [zhk]-axes at the outset, the calculations do not work for an
unknown reason. Instead of thinking that positive and negative four-dimensional
space–times overlap, it may be better to envision one four-dimensional space–time,
whose coordinate axes are complex numbers.

115

116

13
Axioms and Theorems of
the New Octonion

13.1 Necessity of axioms and theorems
The new complex number introduced in this book enables a complex representationof the Lorentz transformations. The new numbers have been extended to fourdimensional space–time and the new quaternion, i.e., the new octonion, has been
proposed for the first time. To properly advance our theory and ensure accurate
future calculations, we must establish axioms and theorems of the new mathematics;
in particular, of the new octonion. This is the aim of the current chapter.
First, we briefly discuss the semantic of an axiom and define a theorem. An
axiom is a rule that, while unable to be proven, is consistent with fundamental
mathematical laws. For instance, 1 + 1 = 2 is an axiom, since it cannot be provenbut is regarded as correct. A theorem is a fundamental law that can be provenusing axioms. For instance, 1 + 2 = 3 is a theorem because it is a composite of two
axioms; namely, 1 + 1 = 2 and that the same quantity can be added to both sidesof an equation. A theorem is suspected when its proof is contradictory. When thetheorem is incorrect, it is necessary to suspect a single axiom. In other words, all
the conclusions of a certain mathematics can ultimately be traced to an axiom. In
this sense, an axiom is the fundamental in mathematical unit.
The axiom as the starting point of mathematical discourse was pioneered by
the Greeks around 2300 years ago. Among the most famous Greek scholars was
Euclid, whose seminal work Elements introduces five axioms on geometry and fiveaxioms on the whole mathematics. Correctly speaking, Euclid selected rather thandetermined his axioms. That is, Euclid’s axioms were regarded as correct but could
not be proven.
Of Euclid’s five axioms of geometry, the most well-known is the fifth axiom,
i.e., the fifth postulate, which states that parallel lines do not intersect. Though117intuitively correct, endeavors to prove this axiom using the other four axioms led tothe birth of non-Euclidean geometry. Parallel lines that do not intersect in Eulcieanspace can be made to intersect in non-Euclidean, curved-space geometry. Such nonEuclidean geometry, known as Riemannian geometry, was adopted by Einstein in
his theory of general relativity. As exemplified by the parallel axiom, once an axiom
is determined, other researchers can verify whether the mathematics is correct andcan develop new mathematics.
Unlike mathematics, axioms are not defined in physics. In addition, the word
axiom does not appear in physics textbooks. However, one premise of Newtonianmechanics, the existence of absolute rest, is an axiom because it is considered correct
but cannot be proven. On the other hand, the non-existence of absolute rest is anaxiom of special relativity. By merely suspecting an axiom of Newtonian physics,
Einstein pioneered a new physics. Therefore, if a physical theory is described by a setof axioms at the outset, contradictory conclusion may be resolvable by suspecting
one of these axioms. Here, by establishing the axioms and theorems of the newoctonion, any erroneous conclusions identified in this book can be traced to theirsource. However, since the following axioms and theorems are defined only at thistime, they are likely to be rewritten by mathematicians once the new octonion isproven to be correct.
The inclusion of time in the following axioms and theorems may appear incongruous to many readers. However, these concepts present a natural description of
the new octonion as a mathematics of four-dimensional space–time. By contrast,
Euclid’s Elements treats three-dimensional space in the absence of any temporaleﬀects.

13.2 Axioms of the new octonion
Axiom 1
The new octonion A is denoted as
A = ah + bi + cj + dk + p + qhi + rhj + shk,
where a, b, c, d, p, q, r, and s are real, and h, i, j, and k are imaginary
numbers. The algorithms of A are
h2 = i2 = j 2 = k2 = −1,
ij = −ji = k, jk = −kj = i, ki = −ik = j,118hi = ih, hj = jh, hk = kh.
In the strictest terms, because k 2 = −1 can be obtained from i2 = j 2 = −1 andij = −ji = k, Axiom 1 includes a theorem. Performing the algebra, we find thatk 2 = ijij
= i(−ij)j= −i2 j 2= −(−1)(−1)= −1.
Therefore, k 2 = −1 must be deleted from Axiom 1. However, we retain the above
definition of Axiom 1 for convenience and ease of understanding.

Axiom 2The new octonion conjugate A of the new octonion A is
A = ah − bi − cj − dk + p − qhi − rhj − shk.

Axiom 3
If the magnitude of the new octonion A is |A|, we have2

|A| = AA.

Axiom 4The division A ÷ B of two new octonions A and B is
A÷B =

AB
|B|

2.

Axiom 5The new octonion BA/ |A| is the coordinate transformation of B by A in
four-dimensional space-time, where A = ah + bi + cj + dk or A = p + qhi +rhj + shk.

119

In Section 2.2, we proved that BA/ |A| becomes a coordinate transformation in twodimensional space–time. However, we have not proven whether this transformation
is extendible to four-dimensional space–time. We further demonstrated, in Section
11.7, that BA/ |A| does not become a coordinate transformation when A = ah +bi + cj + dk + p + qhi + rhj + shk.

Axiom 6In the form ct (the product of time t and velocity of light c), t can undergo
addition and subtraction operations with the spatial dimensions x, y, andz.
Note that since the velocity of light c has unit [distance]/[time], the unit of ct is[distance]
× [time] = [distance].
[time]

Axiom 7
The new octonion, describing the world point in four-dimensional spacetime, is
A = ct0 h + x0 i + y0 j + z0 k + ct1 + x1 hi + y1 hj + z1 hk.
However, as deduced in Section 11.5, ct0 and ct1 are both positive.
That ct > 0 is correct in the negative world is an axiom. Moreover, the condition
ct > 0 implies that c > 0, t > 0 or c < 0, t < 0. The discussions advancedin this book have assumed that c > 0 in both positive and negative worlds. A
consistent sign of c is required to satisfy the y 0 and z 0 formulae of the new Lorentz
transformations in the negative world.

Axiom 8The new octonion describing the world point in the positive world (inwhich we live) is
ct0 h + x0 i + y0 j + z0 k, (ct0 > 0),
while the world point in the synchronously existing negative world is
described by
ct1 + x1 hi + y1 hj + z1 hk. (ct1 > 0)120

We can regard four-dimensional space–time as a single entity with complex coordinate axes disregarding its double structure. In this construction, the new octonion
describing the world point is
A = (ct0 h + ct1 ) + (x0 + x1 h)i + (y0 + y1 h)j + (z0 + z1 h)k.

Axiom 9When the quantity inside the square root is negative, the negative sign
can be placed outside of the square root.
As explained in Section 3.5, if c > v, we havec2 t2 h2 − v 2 t2 i2 = −c2 t2 + v 2 t2 < 0.
Thus, if ct > 0, cth is taken outside of the square root as follows:√√
c2 t2 h2 − v 2 t2 i2 = cth 1 − (v 2 t2 i2 )/(c2 t2 h2 )√= cth 1 − v 2 /c2 .
By this manipulation, the quantity in the square root is rendered positive. Since its
correctness has yet to be proven, this method constitutes an axiom.

Axiom 10√h2 = h.

√As shown in Section 3.5, setting h2 = −h does not recover the Lorentz transfor√
mations. Thus, we assume that h2 = h is correct.

13.3 Theorems of the new octonion
We now prove the theorems of the new octonion. If the content of the theorem denies
a certain proposition, the theorem is verified by an example that is not applied tothe proposition. In addition, a semantic diﬀers between two-dimensional space–timeand two-dimensional space. The former comprises one temporal and one spatial
dimension, while the latter comprises two spatial dimensions. The theorems that
also hold for real numbers, i.e., (A+B)+C = A+(B+C) and A(B+C) = AB+AC,
are omitted. Theorems that diﬀer from those of real and complex numbers are121

included. Most of the illustrated theorems are realized in two dimensions; theorems
realized in four dimensions are not shown.

Theorem 1
Given two new octonions A and B, we have
AB 6= BA.
By contrast, two complex numbers A = a + bi and B = c + di are multiplied as
AB = (a + bi)(c + di)= ac + adi + bci + bdi2
= (ac − bd) + (ad + bc)i,
BA = (c + di)(a + bi)= ca + cbi + dai + dbi2
= (ac − bd) + (ad + bc)i.
Thus, for complex numbers, we have
AB = BA.
Next, we consider two new octonions A and B. Because the imaginary number k
can be obtained from the imaginary numbers i and j, as explained in Axiom 1, it isthought that, without loss of generality, we can omit k when proving the theoremsof the new octonions A and B. In addition, since this theorem is a denial, thecomponents of the negative world are also omitted. If two new octonions A and B
are written as A = ah + bi + cj and B = dh + ei + f j, respectively, we find
AB = (ah + bi + cj)(dh + ei + f j)
= adh2 + aehi + af hj + bdhi + bei2 + bf ij + cdhj + ceji + cf j 2= (−ad − be − cf ) + (ae + bd)hi + (af + cd)hj + (bf − ce)k,
BA = (dh + ei + f j)(ah + bi + cj)= dah2 + dbhi + dchj + eahi + ebi2 + ecij + f ahj + f bji + f cj 2
= (−ad − be − cf ) + (ae + bd)hi + (af + cd)hj − (bf − ce)k.
Comparing AB with BA, the sign of k is reversed. Thus, in the new octonion, we
have proven that
AB 6= BA.
122

Hamilton’s quaternion follows the same law.

Theorem 2
Like complex numbers, a new octonion A and its new octonion conjugate
A satisfy
AA = AA.
Note that although AB 6= BA for two general new octonions A and B ( Theorem1), AA = AA is realized when a new octonion A is multiplied by its new octonion
conjugate A (or vice versa). Similar to Theorem 1, we omit the k part of the newoctonion in the proof of Theorem 2. First, we consider a new octonion in the positive
world.
Given a new octonion A = ah + bi + cj and its conjugate A = ah − bi − cj, theproducts are
AA = (ah + bi + cj)(ah − bi − cj)= a2 h2 − abhi − achj + bahi − b2 i2 − bcij + cahj − cbji − c2 j 2= (−a2 + b2 + c2 ) − (ab − ba)hi − (ac − ca)hj − (bc − cb)k
= −a2 + b2 + c2 ,AA = (ah − bi − cj)(ah + bi + cj)= a2 h2 + abhi + achj − bahi − b2 i2 − bcij − cahj − cbji − c2 j 2= (−a2 + b2 + c2 ) + (ab − ba)hi + (ac − ca)hj − (bc − cb)k
= −a2 + b2 + c2 ,whereby we have
AA = AA.
The whole new octonion, containing the new octonion in the negative world, canbe similarly proved. However, since the calculations are lengthy and more tedious,
they are omitted here.
2
In Axiom 3, the magnitude |A| of the new octonion A was defined as |A| = AA.
2

Similarly, we could have stated |A| = AA because AA = AA.

Theorem 3
Unlike complex number, given two new octonions A and B, we have
AB 6= A B, AB = B A.
123

The complex conjugates of two complex numbers A = a + bi and B = c + di are
A = a − bi and B = c − di, respectively. Thus, we have
A B = (a − bi)(c − di)= ac − adi − bci + bdi2
= (ac − bd) − (ad + bc)i,
B A = (c − di)(a − bi)= (ac − bd) − (ad + bc)i.
Performing the calculations used in the proof of Theorem 1, we find thatAB = (ac − bd) + (ad + bc)i.
Thus, for complex numbers,
AB = A B = B A.
By contrast, the complex conjugates of two new octonions A = ah + bi + cj andB = dh + ei + f j are A = ah − bi − cj and B = dh − ei − f j, respectively. In thiscase, we have
A B = (ah − bi − cj)(dh − ei − f j)
= adh2 − aehi − af hj − bdhi + bei2 + bf ij − cdhj + ceji + cf j 2= (−ad − be − cf ) − (ae + bd)hi − (af + cd)hj + (bf − ce)k.
Performing the calculations used in the proof of Theorem 1, we obtain
AB = (−ad − be − cf ) + (ae + bd)hi + (af + cd)hj + (bf − ce)k.
Thus, we find thatAB = (−ad − be − cf ) − (ae + bd)hi − (af + cd)hj − (bf − ce)k.
Comparing AB with A B, the sign of k is reversed. Therefore, we have demonstrated
that
AB 6= A B.
Now, calculating B A, we obtain
B A = (dh − ei − f j)(ah − bi − cj)= dah2 − dbhi − dchj − eahi + ebi2 + ecij − f ahj + f bji + f cj 2
= (−ad − be − cf ) − (ae + bd)hi − (af + cd)hj − (bf − ce)k,124

which is identical to AB, the product of two new octonions in the positive world.
Thus, we have demonstrated thatAB = B A.
The whole new octonion, containing the new octonion in the negative world, canbe proved similarly, but the equations bocome more complex so the calculations areomitted.

Theorem 4Given three new octonions A, B, and C, the associative law (AB)C =A(BC) holds, as for complex numbers.
To prove this theorem, we calculate (AB)C and A(BC) assuming that A = ah +bi + cj, B = dh + ei + f j, and C = lh + mi + nj (the calculations are lengthy andare hence omitted). Similarly, the associative law (AB)C = A(BC) can be proved
for the whole new octonion
ah + bi + cj + dk + p + qhi + rhj + shk(containing the new octonion in the negative world), by rewriting it in the form
(ah + p) + (b + qh)i + (c + rh)j + (d + sh)k.

Theorem 5
Although the associative law (AB)C = A(BC) is not realized in theGraves’ octonion, it is realized in the new octonion.
This theorem paraphrases Theorem 4. Rewriting Theorem 4 in this way emphasizes
the diﬀerence between the new octonion and the Graves’ octonion. As is well-known,
the Graves’ octoniona + b1 i1 + b2 i2 + b3 i3 + b4 i4 + b5 i5 + b6 i6 + b7 i7does not satisfy the associative law (AB)C = A(BC). Since new octonions do
satisfy the associative law, they may be considered as natural numbers showing theproperties of four-dimensional space–time.

125

Theorem 6
Like complex numbers, two new octonions A and B satisfy
2

2

2|AB| = |A| |B| .
2

From |A| = AA in Axiom 3, AB = B A in Theorem 3, and (AB)C = A(BC) in
Theorem 4, we obtain2|AB| = ABAB
= (AB)(B A)= A(BB A)
2= A |B| A= AA |B|
2

2
2= |A| |B| .
2

2

2

We must not consider AB = A B from |AB| = |A| |B| .

Theorem 7
The magnitude |A| of the new octonion A is either a positive real number
or a positive imaginary number.
In calculations involving real and complex numbers, the magnitude is a positive real
number or zero. However, the magnitude of a new octonion can also be a positive2

imaginary number. In complex numbers, if A = i, since |A| = AA by Axiom 2, we
have√|A| = AA√= i(−i)√= 1= 1.
However, in the new octonion, if A = h, we have√
|A| = AA
√
= hh
= h.

126

That is, |A| is a positive imaginary number. We should discard the prejudice thatthe magnitude must be a positive real number or zero, and instead regard the2magnitude only as |A| = AA.

Theorem 82

Numeric solutions exist for the equation |A| = α.
√2This theorem implies that A = ± α does not follow from |A| = α. We consider22
the case of |A| = 4. When A is a real number a satisfying |A| = 4, we have2

|A| = AA
= aa
= a2
2

from Axiom 2. Given that |A| = 4, and a is a real number, we can write
a2 = 4,√a = ± 4,a = ±2.
Thus, A = ±2.
Now let A be an imaginary number a + bi, where a and b are real. From Axiom
2, we have2

|A| = AA= (a + bi)(a − bi)= a2 + b2 .
2

Since |A| = 4, we can write
a2 + b2 = 4.
Clearly, this formula will be satisfied by some combination of a and b. Therefore,2
when A is a complex number, there exists a numeric solution to |A| = α.
We next consider that A is a new octonion. For simplicity, we assume thatA = ah + bi. From Axiom 2, we have2

|A| = AA= (ah + bi)(ah − bi)= −a2 + b2 .
127

2Since |A| = 4, we can write−a2 + b2 = 4.
Again, this formula will be satisfied by some combination of the real numbers a and2
b. Thus, when A is a new octonion, there exists a numeric solution to |A| = α.

Theorem 9
In a given frame of reference, the relationships between two new octonions A and B can be investigated by comparing their coeﬃcients.
However, when viewed from diﬀerent reference frames, we must instead
compare the magnitudes of their coeﬃcients.
Assume that the coordinates of a point mass D seen by a stationary observer A areD(cth, xi). To observer B, moving in a straight line relative to the x-axial directionof A at uniform velocity v, the coordinates of D are D(ct0 h, x0 i). As explained in
Section 3.5, the new complex plane transforms as shown in Figure 13.1.

The equation x = vt of the world line of B defines the ct0 h-axis of B. The x0 iaxis of B is the straight line symmetric about the world line of light, x = ct. Theintercept of the ct0 h-axis and straight line extending parallel to the x0 i-axis from thepoint D, is designated point E. Similarly, the intercept of the x0 i-axis and straightline extending parallel to the ct0 h-axis from point D is designated point F .
The new octonion that describes D seen by observer A is DA = cth + xi whilethat describing D from observer B’s viewpoint is DB = ct0 h + x0 i. If DA = DB , wehave
cth + xi = ct0 h + x0 i.
Comparing the coeﬃcients, we obtain
ct = ct0 , x = x0 ,128which are contradictory. The reason for this anomaly can be understood from Figure13.1. From that figure, we have
cth = |OA| , x = |DA| ,ct0 h = |OE| , x0 = |OF | ,from which it is clear that, in general|OA| 6= |OE| , |DA| 6= |OF | .
Since the imaginary numbers h of DA = cth + xi and DB = ct0 h + x0 i lie on
separate coordinate axes, simply comparing their coeﬃcients is inappropriate. The
same conclusion can be drawn regarding the imaginary number i.
Although the coeﬃcinets of DA and DB are not directly comparable, the relationship|DA | = |DB |holds, as clearly seen in Figure 13.1. Thus, this formula is appropriate for investigating the relationship between the coeﬃcients of DA and DB .
When obtaining the new Lorentz transformations in Section 3.5, the coeﬃcientsof the imaginary numbers h and i were compared with no contradictions because
the numbers resided on the the same coordinate axes.

Theorem 10
In two-dimensional space-time, rotation does not preserve similarity andcongruence of figures
In Figure 13.2, the horizontal and vertical axes are designated the cth-axis andxi-axis, respectively. The respective coordinates of points A and B are (h, 0) and(h, i). The point B lies on the world line of light x = ct.

129

To realize Euclidean geometry in two-dimensional space–time, point A moves topoint C along the line x = ct when the triangle OAB is rotated 45◦ counterclockwise, while point B moves to D on the xi-axis. In Euclidean geometry, Pythagoras’√√
theorem gives |OB| = 2. Thus, the coordinates of point D are (0, 2i). In addi√√tion, since |OC| = 1, the coordinates of point C are ( 2h/2, 2i/2). Here, |OB|and |OC| express the distance between the origin and points B and C, respectively.
We next examine whether 4OAB and 4OCD are congruent, which seems to be
the case in Euclidean geometry. The new octonions of each point are
√
√
√
2
2
h+
i, D = 2i.
A = h, B = h + i, C =2
2
The squares of each world distance are
2|OA| = h2= −1,
2|OB| = (h + i)(h − i)
= h2 − i2= −1 + 1
= 0,
√
√
√
√
2
2
2
2
2
|OC| = (
h+
i)(
h−
i)
2
2
2
2
1
= (h + i)(h − i)21
= (h2 − i2 )21
= (−1 + 1)
2
= 0,
√
√
2
|OD| = 2i(− 2i)= −2i2= 2.
Thus, since
|OB|
|OA|

0

2 = −1 ,

|OD|
|OC|

2

2

2

2 = 0,

130we have

|OB|

2

2 6=

|OD|

2

2.

|OA|
|OC|
This result reveals that 4OAB and 4OCD are neither congruent nor similar. Thus,
if a figure is rotated in two-dimensional space–time, the length of a side changes.
In other words, the figures are not congruent despite their congruent appearance.
This peculiar phenomenon is attributable to the curvature of two-dimensional space–time. In Einstein’s general relativity, space is bent by mass. However, the above
analysis suggests that four-dimensional space–time is inherently curved, even in theabsence of mass. By the same method, the similarity of figures can be proven to belost under rotation.
The space–time curve manifests from the relationship between time and space.
We now demonstrate that congruence and similarity of figures are preserved after
rotation in two-dimensional space. In Figure 13.3, the horizontal and vertical axes
are designated the xi-axis and the yj-axis, respectively.

If the points A, B, C, and D are assigned as in Figure 13.2, their coordinates are
√
√
√
2
2
A(i, 0), B(i, j), C(i,j), D(0, 2j).
2
2
The new octonions are

√

√
√
2
2
A = i, B = i + j, C =i+j, D = 2j.
2
2

If the new octonions specifying sides AB and CD are written as (AB) and (CD),respectively, we have(AB) = B − A=i+j−i= j,131

(CD) = D − C

√
√
√
2
2
= 2j −
i−
j
2√
2
√
2
2
i+
j.
=−
2
2
Thus, because the squares of the world distances are
2|OA| = i(−i)= 1,2|OB| = (i + j)(−i − j)
= −i2 − ij − ji − j 2=1−k+k+1= 2,
2|AB| = j(−j)= −j 2
= 1,
√
√
√
√
2
2
2
2
2
i+
j)(−
i−
j)
|OC| = (22221= − (i + j)(i + j)2
1
= − (i2 + ij + ji + j 2 )21= − (i2 + k − k + j 2 )
2
= 1,
√
√
2
|OD| = 2j(− 2j)= −2j 2
= 2,

√

√
√
√
2
2
2
2
|CD| = (−
i+
j)(
i−
j)
2
2
2
2
1
= − (i − j)(i − j)21
= − (i2 − k + k + j 2 )
2
= 1,
2

we can write
2

2

2

2

2

2

|OA| = |OC| , |OB| = |OD| , |AB| = |CD| .
132

This result shows that 4OAB and 4OCD are congruent. In two-dimensional
space, figures remain congruent after rotation in the xi-yj, yj-zk, and zk-xi planes.
Similarity can be proven in the same manner. Congruency of figures is also realized
under two-dimensional rotation in the negative world, i.e., under rotation in the
xhi-yhj, yhj-zhk, and zhk-xhi planes.
The above results imply that the relationship between time and space is curved
from the outset, and that space itself is flat. However, as will subsequently be provenin Theorem 15, if the positive and negative worlds are simultaneously considered,
the space–space relationship becomes intrinsically curved.

Theorem 11
In two-dimensional space-time, congruency and similarity of figures are
realized under parallel translation.
This theorem supplements Theorem 10. Since Theorem 11 is not a denial of a
proposition, we prove it by a general approach.

In Figure 13.4, the horizontal and vertical axes represent the cth-axis and xi-axis,
respectively. The respective coordinates of points A and point B are (a1 h, a2 i) and(b1 h, b2 i). Now assume that 4OAB is moved through a distance sh in the cthdirection and distance ti in the xi-direction (this movement constitutes a paralleltranslation). The translated figure is designated 4CDE. The new octonions ofeach point are
A = a1 h + a2 i,
B = b1 h + b2 i,
C = sh + ti,
D = a1 h + a2 i + sh + ti,133

E = b1 h + b2 i + sh + ti.
In terms of the new octonions, sides OA, OB, CD, and CE are defined as (OA), (OB), (CD),and (CE), respectively. Explicitly, we have
(OA) = A = a1 h + a2 i,(OB) = B = b1 h + b2 i,(CD) = D − C= a1 h + a2 i + sh + ti − sh − ti
= a1 h + a2 i,(CE) = E − C= b1 h + b2 i + sh + ti − sh − ti
= b1 h + b2 i.
Thus, we can write(OA) = (CD), (OB) = (CE).
Clearly from these relationships, 4OAB and 4CDE are congruent. Similarity canbe proven by the same method.

Theorem 12
In two-dimensional space-time, if the gradients of the straight lines andcurves that constitute a figure are invariant under translation, the translated figure is congruent and similar to the original figure. If the gradients change, congruency and similarity are violated.
This theorem summarizes Theorems 10 and 11.

Theorem 13
In two-dimensional space-time, arguments are not preserved under rotation.
In verifying Theorem 10, we found that
|OB|
|OA|

2

2 6=

|OD|

2

2.

|OC|

Thus, we can state∠AOB 6= ∠COD
134

demonstrating that arguments are not preserved under rotation.
Therefore, we have avoided the word angle throughout this book (as mentioned
in Section 1.2). Conformal mapping is a frequently used term in complex function theory. However, the curved nature of four-dimensional space–time renders it
unsuitable for conformal mapping. Conformal mapping is entirely appropriate incomplex number theory because the mathematics is constructed in two-dimensional
non-curved spaces.

Theorem 14
Trigonometric function formulae are not realized in two-dimensional spacetime.

In Figure 13.2, used to prove Theorem 10, angles ∠AOB and ∠COD are denotedθ1 and θ2 , respectively. In Euclidean space, θ1 = θ2 . However, since
|OA|
h
= ,
|OB|
0
|OC|
0
cos θ2 =
=√ ,
|OD|
2
cos θ1 =the trigonometric functions are not realized in two-dimensional space–time. These
functions are realized only in the three-dimensional space of the positive world. The
adoption of triangular numbers in the mathematics and physics of four-dimensional
space–time may lead to imprecise results. Similarly, a key equation in complex
function theory, Euler’s famous formula
eix = cos x + i sin x(13.1)

comprises trigonometric functions, and is hence realized only in Euclidean flat space.

135

Replacing x with π in (13.1) yields one of the most ideal mathematical formulae
eiπ + 1 = 0.

(13.2)

Equation (13.2) is ideal because it links five fundamental mathematical constantse, i, π, 0, 1
through a simple relationship. Although (13.2) is derived from trigonometric functions, it does not contain these functions, and is thus realizable in curved space–time.
In Theorem 17, we shall prove that Pythagoras’ theorem, which is applicable to flatspace, is also realizable in curved two-dimensional space–time. It is expected that
simple and ideal formulae will allow a general representation.

Theorem 15
Combining the positive and negative worlds admits two-dimensional spacecurves.
In verifying Theorem 10, we mentioned that when only the positive world is considered, distance is invariant under translation in two-dimensional space. However, if
the positive and negative worlds are combined in two-dimesional space, this property
is violated, as proven here.

In Figure 13.5, we assume that the horizontal axis is the yj-axis in the positive
world, while the vertical axis is the [zhk]-axis in the negative world. This coordinate
plane was used in Section 12.3. If the coordinates of point A are (aj, bhk), A
becomes a point in the yj-[zhk] plane described by the new octonion A = aj + bhk.
The square of the length of A’s world line is2|A| = (aj + bhk)(−aj − bhk)
136

= −(aj)2 − ajbhk − bhkaj − (bhk)2= a2 − abhi + abhi − b2
= a2 − b2 .

(13.3)

Suppose that point A0 in the positive world exists in the same location as point
A with coordinates (aj, bk). In this world, A0 is described by the new octonion
A0 = aj + bk. The square of the world distance is|A0 | = (aj + bk)(−aj − bk)= −(aj)2 − abjk − bakj − (bk)2= a2 − abi + abi + b2
= a2 + b2 .

(13.4)In Equation (13.4), the negative world is disregarded. By contrast, (13.3) is thesquare of the world distance when the positive and negative worlds are considered
together, or when each coordinate axis in four-dimensional space–time is regarded as
a complex number. In other words, considering the whole four-dimensional space–time as a combination of positive and negative worlds, we achieve curvature in
two-dimensional space.
As proven in Section 12.1, the yj-zk plane can be transformed into the yj-[zhk]plane. Thus, while space does not curve in the stationary observer’s world, it curvesin the moving observer’s world.

Theorem 16
If the case of realizing Pythagoras’ theorem is defined as a rectangularcross, the rectangular condition is m1 m2 = 1, where m1 and m2 are
the gradients of two straight lines in two-dimensional space-time. By
contrast, in the Euclidean plane, the rectangular condition is m1 m2 = −1.
Pythagoras’ theorem states that the squares of the shorter lengths, a and b, of aright-angled triangle sum to the square of the longest length d. Mathematically,
Pythagoras’ theorem is stated as a2 + b2 = d2 . As proven in Theorem 13, an argument is not preserved under rotation in two-dimensional space–time. Therefore, thephrase right-angle does not have a semantic. Consequently, if the intersection nodeof two straight lines and a point on each straight line can be connected to satisfy
Pythagoras’ theorem, we say that the two straight lines perpendicularly intersect.
That is, rather than stating that a right-angled triangle satisfies Pythagoras’ theorem, we consider that since Pythagoras’ theorem is satisfied, two straight lines
137perpendicularly intersect. Because the argument lacks a semantic, we adopt thephrase rectangular cross.
First, we review the relationship between two straight lines perpendicularly intersecting on a flat two-dimensional surface. The straight lines perpendicularlyintersecting through the origin are described by the equation y = m1 x and y = m2 x
(see Figure 13.6).

Since ∠AOB is right-angled, if m1 = b/a, m2 = −a/b. Therefore, we obtainba× (− )ab= −1.

m1 m2 =

(13.5)(13.5) is the condition under which two straight lines perpendicularly intersect ona flat two-dimensional surface.

Next, we consider two-dimensional space–time. The situation is illustrated in
Figure 13.7. Even under parallel translation, an argument is invariant from Theorem
11. Thus, to simplify the calculations, we assume the intersection node of two

138

straight lines as the origin O. We denote the equation of a straight line by x = m1 ct
and equation of the perpendicularly intersecting line by x = m2 ct. It is important tonote that the equation of the straight line of gradient m is not x = mt but x = mctbecause the unit of the horizontal axis is ct, not t.
We assume that the coordinates of point A on x = m1 ct are (ct1 h, m1 ct1 i), while
those of point B on x = m2 ct are (ct2 h, m2 ct2 i). Points A and B are expressed by
the new ocotnions A = ct1 h + m1 ct1 i and B = ct2 h + m2 ct2 i. In addition, if the
new octonion describing the side AB is written as (AB), we have(AB) = B − A= ct2 h + m2 ct2 i − ct1 h − m1 ct1 i= c(t2 − t1 )h + c(m2 t2 − m1 t1 )i.
The squares of the world lengths are
2|OA| = (ct1 h + m1 ct1 i)(ct1 h − m1 ct1 i)= c2 t21 h2 − m21 c2 t21 i2= −c2 t21 + m21 c2 t21 ,(13.6)

2|OB| = (ct2 h + m2 ct2 i)(ct2 h − m2 ct2 i)= c2 t22 h2 − m22 c2 t22 i2= −c2 t22 + m22 c2 t22 ,(13.7)

2|AB| = [c(t2 − t1 )h + c(m2 t2 − m1 t1 )i]× [c(t2 − t1 )h − c(m2 t2 − m1 t1 )i]= c2 (t2 − t1 )2 h2 − c2 (m2 t2 − m1 t1 )2 i2= −c2 (t2 − t1 )2 + c2 (m2 t2 − m1 t1 )2 .
If Pythagoras’ theorem holds, we can write
2

2

2|OA| + |OB| = |AB| .
Substituting (13.6), (13.7), and (13.8) into this formula, we obtain−c2 t21 + m21 c2 t21 − c2 t22 + m22 c2 t22 = −c2 (t2 − t1 )2 + c2 (m2 t2 − m1 t1 )2 .
If this equation is solved further, we find−c2 t21 + m21 c2 t21 − c2 t22 + m22 c2 t22 = −c2 t22 + 2c2 t2 t1 − c2 t21+ c2 m22 t22 − 2c2 m2 m1 t2 t1 + c2 m21 t21 ,139(13.8)

which simplifies to
0 = 2c2 t2 t1 − 2c2 m2 m1 t2 t1 .
If t2 6= 0 and t1 6= 0, both sides can be divided by 2c2 t2 t1 to yield
m1 m2 = 1.
Therefore, when two straight lines of gradients m1 and m2 perpendicularly intersect
in two-dimensional space–time, they are related by
m1 m2 = 1.

(13.9)

This result reveals that Pythagoras’ theorem is also realized in curved two-dimensional
space–time. Equation (13.9) diﬀers from Equation (13.5) only by a sign reversal.

Theorem 17
In two-dimensional space-time, if the gradients of two straight lines areline symmetric to the world line x = ct or x = −ct, the intersection node
and a point on each straight line can be connected to satisfy Pythagoras’
theorem.
We now investigate the two straight lines that fulfill (13.9). As shown in Figure
13.8, if the coordinates of point A on the straight line x = m1 ct are A(ah, bi), wehavebm1 = .
a

Substituting this relation into the rectangular condition
m1 m2 = 1

140

(13.9)

in two-dimensional space–time, the equation becomesbm2 = 1,aam2 = .
b
This indicates that point B(bh, ai) lies on the straight line x = m2 ct. From Figure
13.8, the gradient of the straight line x = m1 ct with respect to the cth-axis is clearly
that of the straight line x = m2 ct with respect to the xi-axis. That is, the straightlines x = m1 ct and x = m2 ct are line-symmetric to the world line x = ct and form
oblique coordinate axes. If the two straight lines do not intersect at the origin O in
two-dimensional space–time but if their gradients are line-symmetric to the world
line x = ct, the intersection node and a point on each straight line can be connected
to realize Pythagoras’ theorem. This result shows the generality of Pythagoras’
theorem. In addition, by the same method, it can be proven that two straight linesthat are line-symmetric to the straight line x = −ct perpendicularly intersect.

Theorem 18
Consider a two-dimensional plane, i.e., the yj-[zhk] or [yhj]-zk plane, constructed from space-coordinate axes of both positive and negative worlds.
If the gradients of the two straight lines are m1 and m2 , the rectangularcondition is m1 m2 = 1.
Since this theorem can be proven by the same method as Theorem 16, its verificationis omitted. As shown in (13.5), m1 m2 = −1 in the space frame of the positive world.
However, in the yj-[zhk] or [yhj]-zk plane, in which space coordinates in the positive
and negative worlds are mixed like a mosaic, m1 m2 = 1.

Theorem 19
An absolute rectangular frame does not exist.
From Theorem 17, Pythagoras’ theorem is realized in both oblique and rectangularframes. Therefore, we cannot know whether our frame is right-angled or oblique.
That is, oblique and right angles are relative.
In addition, because rectangular coordinates are considered as a static system,
the fact that no absolute rectangular coordinates exist implies that no absolute rest
frame exists. If points A and B are linearly moving at uniform velocities, their
moving frames can be regarded as the rest frames.
141

Theorem 20
A straight line that is parallel to the straight line equidistant from bothcoordinate axes in the cth-xi plane in two-dimensional space-time has aworld distance of zero.

The equation of the world line of light, which is equidistant from the cth-axis andxi-axis (see Figure 13.9) is x = ct. The coordinates of point A on this line are(cah, cai), and the new octonion describing A is A = cah + cai. Thus, we have2

|A| = (cah + cai)(cah − cai)= (cah)2 − (cai)2= −c2 a2 + c2 a2= 0,
as explained in Section 8.2.
Next, the origin O and point A are moved to points D and E, respectively, which
lie on the straight line x = ct − b that is parallel to x = ct (i.e., has the same
gradient). If the coordinates of point D are (dh, ei), the coordinates of point E are(cah + dh, cai + ei). The new octonions of points D and E are
D = dh + ei, E = (ca + d)h + (ca + e)i.
If the new octonion describing the line segment DE is written (DE), we have(DE) = E − D= (ca + d)h + (ca + e)i − (dh + ei)= cah + cai.

142

Since the same octonion describes A, we can write2

2|DE| = |A|= 0.

Therefore, in the cth-xi plane in two-dimensional space–time, the world distance ofa straight line parellel to the straight line equidistant from both positive coordinate
axes is zero.
We now calculate the world distance of point B on the straight line x = −ct,
which is equidistant from both the −cth-axis and xi-axis in Figure 13.9. Since thenew octonion of B is B = −cah + cai, we obtain
2|B| = (−cah + cai)(−cah − cai)= (cah)2 − (cai)2= −c2 a2 + c2 a2= 0.
Therefore, Theorem 20 is also realized in the negative region of the coordinate axis.

Theorem 21
In the two-dimensional mosaic space, comprising space-coordinate axesin the positive and negative worlds, such as the yj-[zhk] and [yhj]-zk
planes, the straight line parallel to the line equidistant from both coordinate axes has a world distance of zero.

To prove this theorem, we consider point A on the straight line z = yh that is
equidistant from both coordinate axes in the yj-[zhk] plane. The situation is illustrated in Figure 13.10. Since the new octonion is A = aj + ahk, we obtain
2|A| = (aj + ahk)(−aj − ahk)
= −(aj + ahk)(aj + ahk)
143

= −a2 j 2 − 2a2 hjk − 2a2 hkj − a2 h2 k 2= a2 − 2a2 hi + 2a2 hi − a2
= 0.
Therefore, similarly to Theorem 20, it is proven that the world distance of thestraight line parallel to the line equidistant from both coordinate axes is zero.

Theorem 22
Multiplying by i is identical to a counterclockwise 90◦ rotation in twodimensional space-time.
As explained in Section 1.2, multiplying by i in the complex plane performs acounterclockwise 90◦ rotation. We now prove that this operation genertes the same
result in the new complex plane, the cth-xi plane. In this proof, we must assume
the double structure of four-dimensional space–time.
We consider the two-dimensional space–time, in which the positive and negative
worlds overlap, as shown in Figure 13.11. If the coordinates of point A in thepositive world are (ah, bi), the new octonion describing A is A = ah + bi.

We systematically multiply each term in the new octonion by i. The symbol [ ]denotes a point in the negative world. We find that
Ai = (ah + bi)i = ahi + bi2 = −b + ahi = [B],Ai2 = [B]i = (−b + ahi)i = −ah − bi = D = −A,
Ai3 = [B]i2 = Di = (−ah − bi)i = b − ahi = [E] = −[B],Ai4 = [B]i3 = Di2 = [E]i = (b − ahi)i = ah + bi = A.

144

According to this result, multiplication by i is equivalent to a 90◦ counterclockwise
rotation of a point. In addition, the point cyclically enters and exits the positive
and negative worlds. This theorem is realized only if the double structure of fourdimensional space–time is assumed.
Above, we proved the theorem in two-dimensional space–time. We now examine
the result of multiplying by i in four-dimensional space–time.

We consider point A = aj + bk on the yj-zk plane, as shown in Figure 13.12.
Multiplying A by i, we obtain
Ai = (aj + bk)i
= aji + bki= −ak + bj
= bj − ak.
Denote the resultant point by B. B is the result of rotating point A by 90◦ in
the reverse direction of a direction to that of a right-hand screw advancing in thepositive direction of the xi-axis. Although not illustrated, if A lies on the xi-yj
plane, i.e., A = ai + bj, multiplication by i gives
Ai = (ai + bj)i
= −a − bk.
The time component −a of this resultant lies in the negative world, while the z-axis
component −b lies in the positive world.

Theorem 23
Multiplying by h is equivalent to a line-symmetric translation along thecoordinate axes in two-dimensional space-time.
145

To prove this theorem, we require the two-dimensional space–time in which thepositive and negative worlds overlap. This situation is illustrated in Figure 13.13.
If the coordinates of point A in the positive world are given by (ah, bi), the newoctonion is A = ah + bi. We systematically multiply each term in the new octonion
by h.
Ah = (ah + bi)h = ah2 + bhi = −a + bhi = [B],Ah2 = [B]h = (−a + bhi)h = −ah − bi = D = −A,
Ah3 = [B]h2 = Dh = (−ah − bi)h = a − bhi = [E] = −[B],Ah4 = [B]h3 = Dh2 = [E]h = (a − bhi)h = ah + bi = A.
Thus, multiplication by h performs a line-symmetric translation along the coordinate axes. The point cyclically enters and exits the positive and negative worlds.
Like Theorem 22, Theorem 23 is realized only when the double structure of fourdimensional space–time is assumed.
Theorem 24
Multiplying by i/h = −hi is equivalent to a line-symmetric translation
along the world line of light in two-dimensional space-time.

146

Here, we consider two-dimensional space–time in the positive world, as shown in
Figure 13.14. If the coordinates of point A in the positive world are (ah, bi), thenew octonion is A = ah + bi. We systematically multiply each term in the newoctonion by i/h. We find that
A(i/h) = A(−hi) = (ah + bi)(−hi) = −bhi2 − ah2 i = bh + ai = B,
A(i/h)2 = A(−hi)2 = B(−hi) = (bh + ai)(−hi) = ah + bi = A.
Therefore, multiplication by i/h = −hi performs a line-symmetric translation alomg
the world line of light. In this case, the point moves only in the positive world.
Similarly, multiplying point A = a + bhi in the negative world by i/h = −hi isequivalent to a line-symmetrc translation along the world line of light in the negative
world.
By the same method, we can prove that multiplication by hi performs a linesymmetric translation along the world line x = −ct.
Theorem 25
Both sides of an equality cannot be divided by a new octonion whose
absolute value is zero.
Consider the following equality:(h + i)hi = h2 i + hi2 = −(h + i).

(13.10)Assuming that both sides of (13.10) are divisible by (h + i), we find thathi = −1.
This result contradicts Axiom 1, which implies that hi is a fundamental number
and cannot be replaced by other numbers. Therefore, both sides of the equality are
not divisible by (h + i).
We now discuss the reason for this result. In Axiom 4, the division A ÷ B of two
new octonions A and B is given by
A÷B =

AB
|B|

2.

That is, the new octonion cannot be divided by (h + i), but instead must be divided
2
by |h + i| . Then, the equation becomes2|h + i| = (h + i)(h − i)
= h2 − i2
= 0.

147

Since mathematics forbids the division of both sides of an equality by zero, bothsides of (13.10) cannot be divided by (h + i). Besides (h + i), there exist many newoctonions, for example (h + j), (h + k), and (h − i), whose absolute values are zero.
Theorem 26
The derivative world distance dl in two-dimensional space-time is√dl = chdt 1 − (dx)2 /(cdt)2 .

As shown in Figure 13.15, the equation of the world line of a point mass acceleratingin the x-direction of a rest frame in two-dimensional space–time is x = at2 /2, wherea is the mass’s acceleration. However, in relativity theory, a is treated as a constant
only at slow velocities. The case of varying a is treated in Section 20.2. In addition,
though (dl)2 is usually written as dl2 in calculus, we adopt the (dl)2 notation forthe benefit of readers who lack a calculus background.
Suppose that two points A and B approach extremely close on the world line
x = at2 /2. The infinitesimal time and distance between A and B are dt and dx,
respectively. Since the new octonion describing the infinitesimal line segment AB
is cdth + dxi, the square of the infinitesimal world distance dl is(dl)2 = (cdth + dxi)(cdth − dxi)
= c2 (dt)2 h2 − (dx)2 i2 .

(13.11)

When cdt > dx, i.e., c > dx/dt = v, the square root is rendered positive byextracting the term cdth, as explained in Axiom 9. Thus, we have√dl = c2 (dt)2 h2 − (dx)2 i2√= cdth 1 − (dxi)2 /(cdth)2√= chdt 1 − (dx)2 /(cdt)2 .
148(13.12)

This formula gives the derivative world distance in two-dimensional space–time. It
is important to note that, unlike flat space–time, (dl)2 cannot be
c2 (dt)2 h2 + (dx)2 i2orc2 (dt)2 + (dx)2in (13.11). From the equation x = at2 /2 of the world line, we find thatddx
= (at2 /2)
dt
dt
= at.
Substituting this expression into (13.12), we obtain√dl = chdt 1 − (at/c)2 .
When we assume that the coordinates of A are (ct0 , x0 ), the distance l of the world
line from the origin O to point A is
∫ t0
√
l=
chdt 1 − (at/c)20∫ t0 √= ch
1 − (at/c)2 dt.
(13.13)0Equation (13.13) gives the world distance of curvilinear x = at2 /2 in two-dimensional
space–time.
We now determine the infinitesimal distance in a timeless three-dimensional space.
If two points A and B approach extremely closely in this three-dimensional space,
the infinitesimal components in the x-, y-, and z-directions are dx, dy, and dz,
respectively. Since the new octonion describing the line segment AB is dxi + dyj +dzk, the square of the infinitesimal distance dl is(dl)2 = (dxi + dyj + dzk)(−dxi − dyj − dzk)
= −(dxi)2 − (dyj)2 − (dzk)2= (dx)2 + (dy)2 + (dz)2 ,√dl = (dx)2 + (dy)2 + (dz)2 .

(13.14)Equation (13.14) is exactly the infinitesimal distance between two points in flat
three-dimensional space, derived from calculus.

149

Theorem 27
If the new octonions of vectors A and B are specified as A and B, respectively, their inner product A·B and outer product A × B are givenbyBA = A·B + A × B.
This theorem will be proved in Section 14.2.

Theorem 28
If the new octonions of two vectors A and B are specified by A and B,
respectively, their inner product A·B and outer product A × B are givenby
A·B = (BA + AB)/2,
A × B = (BA − AB)/2.
This theorem will be proved in Section 14.4.

Theorem 29
If the new octonions of three vectors A, B, and C are specified by A,
B, and C, respectively, their triple scalar product A·(B × C) and triple
vector product A × (B × C) are given by(CB − BC)A/2 = A·(B × C) + A × (B × C).
This theorem will be proved in Section 14.4.

Theorem 30
New octonion geometry obeys non-Euclidean and non-Riemannian geometry.
This theorem does not appear in the revised Japanese second edition.
There is no bend in space and parallel lines do not cross in Euclidean geometry.
Space curves and parallel lines cross in Riemannian geometry. However, althoughspace curves, parallel lines do not cross in the new octonion geometry, i.e., the newoctonion geometry is the third geometry. We prove this here.
150Section 13.1 explained that in the process of proving the fifth axiom (postulate)of Euclid, i.e., parallel lines do not cross, we discovered Riemannian geometry whereparallel lines do cross. In addition, it was found that space curves in the Riemannspace where parallel lines cross. Einstein’s general relativity is based on Riemannian
geometry. Four-dimensional space–time is bent in the new octonion geometry as
proven in Section 8.2 and by Theorem 15. Do parallel lines cross in this curved
space–time? If parallel lines cross, the new octonion geometry is a Riemannian
geometry. If they do not cross, it is a new geometry.

As shown in Figure 13.16, we assume that the straight lines x = mct and x =mct + a lie in the cth-xi new complex plane. As explained in Section 5.2, because astraight line in the new complex plane can be calculated with the equation of a real
number, the following calculations are done with real numbers. Point P (ct0 , mct0 )is on the straight line x = mct and point Q(ct1 , mct1 + a) is on the straight line
x = mct + a. It was proved in Theorem 17 that when the Pythagorean theorem
is applied, two straight lines are line symmetric about the world line x = ct oflight. Therefore, if the straight line P Q and the straight line x = mct intersect
rectangularly, the gradient of another oblique axis of x = mct and the gradient of
the straight line P Q are the same. As explained in Section 4.3, the equation ofanother oblique axis of x = mct can be obtained by replacing x and ct by ct and x,
respectively. If we actually calculate, we havect = mx,1
x = (ct).
m
Since the units of the cth-axis are ct, the gradient of the oblique axis is 1/m.
Therefore, from the condition that the straight line passes point P , the equation of151

the straight line P Q is1
(ct − ct0 ).
m
Because this straight line passes point Q, we can write
x − mct0 =(mct1 + a) − mct0 =1(ct1 − ct0 ).
m

By solving this equation, we find that
(m −

1
1
)ct1 = (m − )ct0 − a,mm1
ct1 = ct0 − a/(m − ),mmat1 = t0 −.
2(m − 1)c(13.15)

We assume that the x-axial component of point Q is x1 . By substituting (13.15)
into x = mct + a, we havex1 = mc[t0 − ma/(m2 c − c)] + am2 a+am2 − 1−m2 a + m2 a − a= mct0 +m2 − 1a= mct0 − 2.
m −1= mct0 −If we assume that the new octonions showing the line segments OP, OQ, and P Q
are (OP ), (OQ), and (P Q), respectively, we find that(OP ) = ct0 h + mct0 i,(OQ) = ct1 h + x1 i
= [ct0 − ma/(m2 − 1)]h + [mct0 − a/(m2 − 1)]i,(P Q) = (OQ) − (OP )= [ct0 − ma/(m2 − 1) − ct0 ]h + [mct0 − a/(m2 − 1) − mct0 ]i= −mah/(m2 − 1) − ai/(m2 − 1).
Thus, we can write|P Q|2 = [mah/(m2 − 1) + ai/(m2 − 1)][mah/(m2 − 1) − ai/(m2 − 1)]= [ma/(m2 − 1)]2 h2 − [a/(m2 − 1)]2 i2152

−m2 a2a2+(m2 − 1)2(m2 − 1)2
2
2−(m − 1)a=(m2 − 1)2
−a2
= 2
.
m −1
=

(13.16)

Because m and a are constant, from (13.16), it is proved that regardless of thelocation of point P (ct0 , mct0 ), |P Q|2 is constant. From the above result, because
the new octonion space–time is bent but parallel lines do not cross there, the newoctonion geometry is a non-Euclidean geometry and non-Riemannian geometry,
respectively.

As shown in Table 13.1, four kinds of geometry can be considered from the intersection of parallel lines and the bend of space–time. Of these, the geometry of
the region (#) will not be found because Riemannian geometry with curved space
was found in the process of seeking the geometry where parallel lines cross in aflat plane, as explained in Section 13.1. If our universe has the structure indicatedby the new octonion geometry, the present cosmology obtained using Riemannian
geometry needs to be reconsidered.

153

154

14
New Octonion and Vectors

14.1 Basic properties of vectors
In this chapter, we prove that vector calculations can be reformulated as new octonions. To prepare the reader, we first review the basic properties of vectors.
Quantities with magnitude and no direction, such as length, time, and mass, arecalled scalars. Scalars are expressed only as numbers. On the other hand, quantities with both magnitude and direction, such as velocity, acceleration and force,are called vectors. Vectors are represented by arrows whose length and orientation
indicate their magnitude and direction, respectively. Vectors are expressed in bold
font; for example, A, B, a, b, · · ·, and their magnitudes are expressed as absolutevalues |A| , |B| , · · ·.
Vector can be multiplied using two methods; as an inner product and as an outer
product. Given two vectors A and B on a plane, the inner product is denoted by
A·B and the outer product is denoted by A × B. These multiplications are visuallyinterpreted in Figure 14.1.

An inner product is a scalar. If we assume that the angle between two vectors A
and B is θ (theta), the inner product is defined by
A·B = |A| |B| cos θ.

155(14.1)

Since this result is a scalar, it has no direction. The outer product is a vector of
magnitude |A × B|, defined by|A × B| = |A| |B| sin θ.

(14.2)The direction of the outer product is perpendicular to the plane described by thevectors A and B. In Figure 14.1, where A and B lie in the horizontal plane, theouter product is oriented both upward and downward. By convention, the positiveorientation of the vector A × B is defined as the direction of a right-handed screw
rotating from vector A to vector B. However, this convention is quite arbitrary;
we could also define the opposite orientation as positive. Regarding B × A, when aright-handed screw rotates from B to A, it follows a direction opposite to A × B.
Thus, we can state
B × A = −A × B.
Why the outer product is oriented perpendicular to the plane described by thevectors A and B is not explained in mathematics and physics texts. The definition isprovided while the rationale is not explained, probably because the calculations show
no contradictions; therefore, the definition is assumed true without mathematical
proof. For this reason, when the vector was proposed at the end of the 19th century,
it was regarded by some mathematicians as a non-mathematical construct. The
direction of the outer product will be discussed in Section 14.3.
In addition, since the positive and negative directions of A × B may be arbitrarily chosen, the right-hand screw rule that determines the positive direction of
A × B has no mathematical basis. If the positive direction were redefined as the
direction followed by a left-handed screw, the mathematics would remain consistent.
Therefore, vector operations are not admitted as mathematics by those who insist
on strict theorems based on minimum axioms. Euclid, who wrote Elements, will
not acknowledge the arbitrariness of the screw rule that a right-handed screw or aleft-handed screw can be used.

156

Since this concept is not discussed in conventional texts, we here provide a mathematical basis for the freedom of the positive and negative directions. We consider
the x-y coordinates of a two-dimensional plane (see Figure 14.2). The coordinate
axes can be decided in two ways. If upward is selected as the positive direction of
the y-axis, the positive direction of the x-axis can be considered as either leftward
or rightward. The two coordinate systems are mirror symmetric.
Hypothetical creatures inhabiting a two-dimensional world cannot add height tothese two coordinate planes. Because three-dimensional space does not exist in
their world, they can shift a geometric figure only within the coordinate plane.
In contrast, life forms inhabiting three-dimensional space can fold two coordinate
planes into a three-dimensional structure. That is, while our hypothetical twodimensional counterparts view the x-y coordinate planes as diﬀerent, we regard
them as the same.
This view is extendable to three-dimensional space and four-dimensional space–time. In three-dimensional space, the direction of A × B may be either up or
down. Since they are mirror symmetrc, we view them as diﬀerent. However, sinceboth directions of A × B can be piled up in four-dimensional space–time, they are
actually the same.
In Section 11.1, it was explained that only three kinds of space–times; onedimensional, two-dimensional, and four-dimensional, are mathematically possible.
Since three-dimensional space cannot independently exist, we must consider the vector in four-dimensional space–time. Relativity theory asides, contemporary mathematics and physics treat the vector in three-dimensional space. Thus, two apparent
directions for A × B exist. However, from the viewpoint of four-dimensional space–time, a single direction exists. That is, since A × B exists in four-dimensional
space–time, the mathematics is consistent if the positive and negative directions ofan outer product are reveresed in three-dimensional space.

157

Next, we explain how a vector is described by its coordinate components. The
vectors of unit length 1 along the x-, y-, z-axes in three-dimensional space are called
fundamental vectors, expressed as ex , ey , and ez . Then, the x-, y-, and z-axial
components of two vectors A and B are denoted Ax , Ay , Az and Bx , By , Bz ,respectively. In terms of the unit vectors, these are expressed as
A = Ax ex + Ay ey + Az ez , B = Bx ex + By ey + Bz ez .

(14.3)The unit vector representation of A is illustrated in Figure 14.3. From the definitionof the inner product, i.e.,
A·B = |A| |B| cos θ,(14.1)and given that cos 0◦ = 1, cos 90◦ = 0, the inner products of the fundamental
vectors ex , ey , and ez are
ex ·ex = 1, ey ·ey = 1, ez ·ez = 1,

(14.4)

ex ·ey = 0, ey ·ez = 0, ez ·ex = 0.

(14.5)

Therefore, from (14.3), (14.4), and (14.5), the inner product A·B is expressed interms of its unit vectors as
A·B = (Ax ex + Ay ey + Az ez )·(Bx ex + By ey + Bz ez )= Ax Bx + Ay By + Az Bz .

(14.6)

Similarly, from the definition of the outer product, i.e.,|A × B| = |A| |B| sin θ,(14.2)and given that sin 0◦ = 0, sin 90◦ = 1, and the right-hand screw rule gives the outer
products of the fundamental vectors ex , ey , and ez as
ex × ex = 0, ey × ey = 0, ez × ez = 0,(14.7)

ex × ey = ez , ey × ez = ex , ez × ex = ey ,(14.8)

ey × ex = −ez , ez × ey = −ex , ex × ez = −ey .

(14.9)

Therefore, from (14.3), (14.7), (14.8), and (14.9), the outer product A × B is expressed in terms of unit vectors as
A × B = (Ax ex + Ay ey + Az ez ) × (Bx ex + By ey + Bz ez )= Ax By ex × ey + Ax Bz ex × ez
+ Ay Bx ey × ex + Ay Bz ey × ez
158

+ Az Bx ez × ex + Az By ez × ey
= Ax By ez − Ax Bz ey− Ay Bx ez + Ay Bz ex+ Az Bx ey − Az By ex= (Ay Bz − Az By )ex + (Az Bx − Ax Bz )ey + (Ax By − Ay Bx )ez .
Summarizing this result, we obtain
A × B = (Ay Bz − Az By )ex + (Az Bx − Ax Bz )ey + (Ax By − Ay Bx )ez . (14.10)
As explained previously, the vector must be considered in four-dimensional space–time. However, since the angles made by the fundamental vector et on the temporalaxis and the fundamental vectors ex , ey , and ez in three-dimensional space areunknown, the formulae describing the inner and outer products in four-dimensional
space–time by their coordinate components cannot be verified as correct. In Section
14.6, we show that the inner and outer products in four-dimensional space–time canbe expressed in coordinate components using the new octonion.

14.2 Vectors and coordinate transformations
In this section, we prove that vector multiplications (inner and outer products)
are coordinate transformations. Multiplying vector A by vector B is equivalent tochanging vector B into the coordinate system of vector A. In addition, we provethat since vector multiplication is a coordinate transformation, both inner and outer
products can be rewritten as new octonions.

Again, we consider two vectors A and B on the x-y plane (Figure 14.4). Vector
A intercepts with a perpendicular line drawn from the tip E of the vector B at
node D. The lengths of the line segments OD and DE are written as |OD| and159

|DE|, respectively. From the definitions of the inner product and outer product ofthe vector, i.e.,
A·B = |A| |B| cos θ,(14.1)|A × B| = |A| |B| sin θ,(14.2)

we have
A·B = |A| |OD| ,|A × B| = |A| |DE| .
Rearranging these formulae, we obtain
A·B
,
|A|
|A × B||DE| =.
|A||OD| =(14.11)(14.12)The coordinate plane of Figure 14.4 is considered as a complex plane in which the xand y-axes are real and imaginary axes, respectively. We assume that the complex
numbers describing vectors A and B are A and B, respectively. As explained in
Section 2.2, the coordinate transformation BA/ |A| of the complex number indicates
how B is seen from A and we can write
BA= |OD| + |DE| i.
|A|
Substituting (14.11) and (14.12) into this equation, we obtain
A·B|A × B|
BA
=
+
i.
|A|
|A|
|A|
Since the magnitude |A| of vector A is the magnitude |A| of the complex number
A, it can be written as
A·B|A × B|
BA
=
+
i.
|A|
|A|
|A|
Multiplying both sides by |A|, this equation becomes
BA = A·B + |A × B| i.

(14.13)Equation (14.13) shows that the inner product A·B of vectors and magnitude|A × B| of an outer product are performed together when the coordinate transformation BA/ |A| of a complex number is multiplied by |A|. That is, the innerproduct A·B is obtained by multiplying the component of vector B parallel to160vector A by |A|. The magnitude |A × B| of an outer product is obtained by multiplying the component of vector B perpendicular to vector A by |A|. Thus, themultiplication of vectors A and B, involving both the inner and outer products, isthe multiplication of |A| and the coordinate transformation of B by A.
Since (14.13) is computed in a two-dimensional plane, the direction of |A × B| idoes not correspond to the direction of the outer product A×B in three-dimensional
space. Thus, we must extend the above theory to three-dimensional space. If thetime t of the new octonion cth + xi + yj + zk in the positive world of curved fourdimensional space–time is set to zero, the new octonion in three-dimensional spacebecomes xi + yj + zk. Thus, the new octonions A and B describing vectors A and
B in three-dimensional space are written as
A = Ax i + Ay j + Az k, B = Bx i + By j + Bz k.

(14.14)

In addition, the three-dimensional space component of Hamilton’s quaternion in flat
four-dimensional space–time is xi + yj + zk; thus, (14.14) is realized regardless ofthe curvature of space. From (14.14), we find
BA = (Bx i + By j + Bz k)(−Ax i − Ay j − Az k)= −Bx Ax i2 − Bx Ay ij − Bx Az ik− By Ax ji − By Ay j 2 − By Az jk− Bz Ax ki − Bz Ay kj − Bz Az k 2= Bx Ax − Bx Ay k + Bx Az j
+ By Ax k + By Ay − By Az i
− Bz Ax j + Bz Ay i + Bz Az= (Ax Bx + Ay By + Az Bz )+ (Ay Bz − Az By )i + (Az Bx − Ax Bz )j + (Ax By − Ay Bx )k.

(14.15)

We know that in terms of the fundamental vectors ex , ey , and ez the outer productis written as
A × B = (Ay Bz − Az By )ex + (Az Bx − Ax Bz )ey + (Ax By − Ay Bx )ez . (14.10)
Equivalently, in teams of the imaginary numbers i, j, and k, we can write
A × B = (Ay Bz − Az By )i + (Az Bx − Ax Bz )j + (Ax By − Ay Bx )k,(14.16)

because the fundamental vectors and the imaginary numbers have the same magnitudes and directions. Since the left side of (14.16) is a vector and the right side is a161scalar, an equal symbol cannot be used. However, since the semantic of both sides
is the same, we adopt this notation hereafter.
From (14.15), (14.16), and the coordinate components expression of the inner
vector product, i.e.,
A·B = Ax Bx + Ay By + Az Bz ,(14.6)BA = A·B + A × B.

(14.17)

we can write

Equation (14.17) extends the formula in the two-dimensional plane, i.e.,
BA = A·B + |A × B| i,(14.13)to three-dimensional space. From (14.17), we see that BA involves the simultaneous calculation of the inner and outer vector products in three-dimensional space.
Furthermore, the multiplication of vectors A and B containing both products is themultiplication of |A| and the coordinate transformation of B by A. In other words,
the multiplication of two vectors A and B indicates the multiplication of B viewed
from A by |A| in three-dimensional space.
We now present an alternative proof that the inner and outer products of two vectors are coordinate transformations. The inner product A·B multiplies |A| with thecomponent of B parallel to vector A. The magnitude |A × B| of an outer productdenotes the multiplication of |A| with the component of B perpendicular to vector
A. The magnitude |B| of B remains unchanged after a coordinate transformation.
Thus, by Pythagoras’ theorem, we can write
|A·B|
|A|

2

2

+

2

|A × B||A|2= |B| ,

2

2

2

2

2

|A·B| + |A × B| = |B| |A| .

(14.18)

If an inner and outer products, expressed in terms of their coordinate components,
can be proven to satisfy (14.18), then the inner and outer vector products arecoordinate transformations. Substituting the following two equations
A·B = Ax Bx + Ay By + Az Bz ,(14.9)

A × B = (Ay Bz − Az By )i + (Az Bx − Ax Bz )j + (Ax By − Ay Bx )k(14.16)

into the left-hand side of (14.18), we find that2

2

|A·B| + |A × B| = (Ax Bx + Ay By + Az Bz )2+ [(Ay Bz − Az By )i + (Az Bx − Ax Bz )j + (Ax By − Ay Bx )k]× [−(Ay Bz − Az By )i − (Az Bx − Ax Bz )j − (Ax By − Ay Bx )k] .
162

Since the calculations are lengthy, they are omitted here. If the algebra is followed
through, we obtain
2

2|A·B| + |A × B| = (Bx2 + By2 + Bz2 )(A2x + A2y + A2z )22= |B| |A| .

14.3 New octonions and direction of outer products of vectors
Section 14.1 explained that, in mathematics and physics texts, the outer productA × B of two vectors A and B is assumed perpendicular to the plane described by
the vectors without providing proof. The definition is provided without a rigorous
rationale. In this section, using
BA = (Ax Bx + Ay By + Az Bz )+ (Ay Bz − Az By )i + (Az Bx − Ax Bz )j + (Ax By − Ay Bx )k(14.15)BA = A·B + A × B,

(14.17)

and

we prove that if a vector is rewritten by the new octonion, the outer product isinevitably oriented perpendicular to the plane described by vectors A and B.

As shown in Figure 14.5, if the vectors A and B lie on the xi-yj plane, sinceAz = Bz = 0, (14.15) becomes
BA = (Ax Bx + Ay By ) + (Ax By − Ay Bx )k.
By this formula and (14.17), we find that
A·B + A × B = (Ax Bx + Ay By ) + (Ax By − Ay Bx )k.

163

Since the scalar A·B is directionless but the vector A × B has a direction, we canwrite
A·B = Ax Bx + Ay By ,(14.19)A × B = (Ax By − Ay Bx )k.

(14.20)Since in (14.20), the imaginary number k lies along the zk-axis, the outer product of
two vectors A and B on the xi-yj plane orients in the zk-direction. This statement
is a mathematical proof of the direction of the outer product A × B. It cannot befound in standard mathematics and physics texts. Vector mathematics is successful
only if the direction of the outer product A × B is set perpendicular to the plane
described by vectors A and B. Above, this universal fact has been mathematically
proven.

14.4 Calculation of three-vector products by the new octonion
The previous analyses were performed on two vectors A and B. Multiplications of
three vectors A, B, and C are called triple products. The important three-vector
products are the triple scalar product A·(B × C) and the triple vector productA × (B × C). This section shows that triple products can be rewritten in terms ofthe new octonion.
In Hamilton’s notation, the components (x, y, z) of the vectors A, B, and C
are written as(Ax , Ay , Az ), (Bx , By , Bz ), (Cx , Cy , Cz ).
In terms of the fundamental vectors of magnitude 1 along the x-, y-, and z-axes,
namely, ex , ey , and ez , the vectors become
A = Ax ex + Ay ey + Az ez ,B = Bx ex + By ey + Bz ez ,C = Cx ex + Cy ey + Cz ez .
The calculations can be found in any standard text on vectors, and are hence omitted. From the properties of the inner and outer products of the fundamental vectors
ex , ey , and ez , given by Equations (14.4), (14.5), (14.7), (14.8), and (14.9), thetriple scalar product is obtained as
A·(B × C) = Ax (By Cz − Bz Cy ) + Ay (Bz Cx − Bx Cz ) + Az (Bx Cy − By Cx ), (14.21)

164

and the triple vector product is
A × (B × C) = [(Ay Cy + Az Cz )Bx − (Ay By + Az Bz )Cx ]ex+ [(Az Cz + Ax Cx )By − (Az Bz + Ax Bx )Cy ]ey+ [(Ax Cx + Ay Cy )Bz − (Ax Bx + Ay By )Cz ]ez .

(14.22)

Next, we express the triple products as new octonions. The new octonions describing vectors A, B, and C are A, B, and C, respectively. From the expression
BA = A·B + A × B(14.17)

obtained in Section 14.2, we can write
CB = B·C + B × C,

(14.23)BC = C·B + C × B.

(14.24)

From the definition of the outer product, we have
C × B = −B × C.
Thus, (14.24) becomes
BC = C·B − B × C.

(14.25)Since B·C = C·B, subtracting (14.25) from (14.23) yields
CB − BC = 2(B × C).
Thus, we can write
B × C = (CB − BC)/2.

(14.26)

In addition, adding (14.26) to (14.23), we obtain
CB + B × C = B·C + B × C + (CB − BC)/2.
Thus, we can write
B·C = (CB + BC)/2.

(14.27)Equations (14.26) and (14.27) were incorporated into Theorem 28 in Section 13.3.
Since the purpose of this section is to formulate the triple products as new octonions, we verify the results in advance. For simplicity, we set D = B × C, so that
(14.26) becomes
D = (CB − BC)/2.

165

Since the new octonion D is considered identical to vector D, we also define
D = (CB − BC)/2.

(14.28)BA = A·B + A × B(14.17)

From the equation

obtained in Section 14.2, we can write
DA = A·D + A × D.
In terms of D = B × C and (14.28), this equation becomes(CB − BC)A/2 = A·(B × C) + A × (B × C).

(14.29)Equation (14.29) relates the triple products to the new octonion, and is used asfollows.
(1) Construct the new octonions A, B, and C from the coordinate components ofvectors A, B, and C, respectively.
(2) Calculate (CB − BC)A/2.
(3) The real-number component of (CB − BC)A/2 is the triple scalar product
A·(B × C), while the imaginary component is the triple vector productA × (B × C).
The inner and outer products of vectors are computed together in (14.29), as established earlier for two vectors A and B. For convenience, areas and volumes arecalculated from outer products and from triple scalar products, respectively. However, because the vector multiplications are essentially coordinate transformations,the inner and outer products appear together in the new octonion.
We now demonstrate that (14.29) is correct. Substituting
A = Ax i + Ay j + Az k,
B = Bx i + By j + Bz k,
C = Cx i + Cy j + Cz kfor (CB − BC)A/2, we find(CB − BC)A/2= [(Cx i + Cy j + Cz k) (−Bx i − By j − Bz k) − (Bx i + By j + Bz k) (−Cx i − Cy j − Cz k)]× (−Ax i − Ay j − Az k) /2= Ax (By Cz − Bz Cy ) + Ay (Bz Cx − Bx Cz ) + Az (Bx Cy − By Cx )166

+ [(Ay Cy + Az Cz )Bx − (Ay By + Az Bz )Cx ]i+ [(Az Cz + Ax Cx )By − (Az Bz + Ax Bx )Cy ]j+ [(Ax Cx + Ay Cy )Bz − (Ax Bx + Ay By )Cz ]k.
The calculations are lengthy, and are therefore omitted. Since
A·(B × C) = Ax (By Cz − Bz Cy ) + Ay (Bz Cx − Bx Cz ) + Az (Bx Cy − By Cx ),(14.21)
A × (B × C) = [(Ay Cy + Az Cz )Bx − (Ay By + Az Bz )Cx ]ex+ [(Az Cz + Ax Cx )By − (Az Bz + Ax Bx )Cy ] ey
+ [(Ax Cx + Ay Cy )Bz − (Ax Bx + Ay By )Cz ]ez(14.22)

as previously discussed, the above equation is simplified to(CB − BC)A/2 = A·(B × C) + A × (B × C).
However, since we consider that the fundamental vectors, ex , ey , and ez , andimaginary numbers, i, j, and k, express the same content, we also regard them as
mutually interchangeable.

14.5 Calculation of four-vector products by the new octonion
By the method used to obtain the triple product, we can relate the quadruple
product to the new octonion in the case of four vectors, A, B, C, and D. The
sum of the quadruple scalar and vector products is(A × B)·(C × D) + (A × B) × (C × D).
From the formula
QP = P ·Q + P × Q,
we find{}{C × D} A × B = (A × B)·(C × D) + (A × B) × (C × D).

(14.30)}{Here, {C × D} is a new octonion describing the vector C × D, and A × B is thenew octonion conjugate of the new octonion {A × B} describing the vector A × B.
From the formula{B × C} = (CB − BC)/2,

167(14.26)

we can write{A × B} = (BA − AB)/2,{C × D} = (DC − CD)/2.

(14.31)

In addition, given that P Q = Q P , the new octonion conjugate of the new octonion
BA − AB is AB − BA. Thus, we obtain{}
A × B = (AB − BA)/2.
(14.32)
Substituting (14.31) and (14.32) into the left-hand side of (14.30), we have(DC − CD)(AB − BA)/4 = (A × B)·(C × D) + (A × B) × (C × D).
This equation relates the quadruple product to the new octonion.

14.6 New octonion and the four-dimensional vector
The previous analyses were conducted in three-dimensional space with the time
dimension set to zero. This section examines the inner and outer products of vectors in the whole curved four-dimensional space–time, including both positive andnegative worlds.
If we assume that the coordinates (cth, xi, yj, zk) of two four-dimensional vectors
A and B in the positive world are(At h, Ax i, Ay j, Az k), (Bt h, Bx i, By j, Bz k),
the new octonions A and B describing vectors A and B are
A = At h + Ax i + Ay j + Az k,
B = Bt h + Bx i + By j + Bz k.
If we assume that the formula
BA = A·B + A × B
is also realized in four-dimensional space–time, we find that
BA = (Bt h + Bx i + By j + Bz k)(At h − Ax i − Ay j − Az k)= Bt At h2 − Bt Ax hi − Bt Ay hj − Bt Az hk+ Bx At hi − Bx Ax i2 − Bx Ay ij − Bx Az ik+ By At hj − By Ax ji − By Ay j 2 − By Az jk+ Bz At hk − Bz Ax ki − Bz Ay kj − Bz Az k 2
168(14.17)

= −Bt At − Bt Ax hi − Bt Ay hj − Bt Az hk+ Bx At hi + Bx Ax − Bx Ay k + Bx Az j
+ By At hj + By Ax k + By Ay − By Az i
+ Bz At hk − Bz Ax j + Bz Ay i + Bz Az= (−Bt At + Bx Ax + By Ay + Bz Az )+ (Bx At − Bt Ax )hi + (By At − Bt Ay )hj + (Bz At − Bt Az )hk+ (Bz Ay − By Az )i + (Bx Az − Bz Ax )j + (By Ax − Bx Ay )k= (−At Bt + Ax Bx + Ay By + Az Bz )+ (At Bx − Ax Bt )hi + (At By − Ay Bt )hj + (At Bz − Az Bt )hk+ (Ay Bz − Az By )i + (Az Bx − Ax Bz )j + (Ax By − Ay Bx )k.
Thus, we obtain
A·B = −At Bt + Ax Bx + Ay By + Az Bz ,(14.33)A × B = (At Bx − Ax Bt )hi + (At By − Ay Bt )hj + (At Bz − Az Bt )hk+ (Ay Bz − Az By )i + (Az Bx − Ax Bz )j + (Ax By − Ay Bx )k.

(14.34)Equations (14.33) and (14.34) are expressed in terms of the coordinate componentsof the inner and outer products of the four-dimensional vectors. If the time components At and Bt in these formulae are set to zero, these equations reduce to
A·B = Ax Bx + Ay By + Az Bz ,A × B = (Ay Bz − Az By )i + (Az Bx − Ax Bz )j + (Ax By − Ay Bx )k.
These are identical to Equations (14.6) and (14.10), i.e.,
A·B = Ax Bx + Ay By + Az Bz ,(14.6)

A × B = (Ay Bz − Az By )ex + (Az Bx − Ax Bz )ey + (Ax By − Ay Bx )ez , (14.10)
describing the inner and outer products of two three-dimensional vectors. However,(14.10) is expressed in terms of the fundamental vectors, ex , ey , and ez , rather
than the imaginary numbers i, j, and k.
Since (14.34) is diﬃcult to memorize, we rewrite it as a determinant. The determinant of a matrix is found by multiplying the right-diagonal elements (from top
to bottom rows) and adding the components, then multiplying the left-diagonal elements (from top to bottom rows) and subtracting the components from the positive169

components. Below, this process is demonstrated by a pair of simple examples.
a1 a2= a1 a4 − a2 a3 ,

a3 a4
i j k
a1 a2 a3
b1 b2 b3

= (a2 b3 − a3 b2 )i + (a3 b1 − a1 b3 )j + (a1 b2 − a2 b1 )k.

In terms of the determinants of its components, (14.34) becomes
At Ax
Bt Bx

A×B =
+

Ay Az
By Bz

hi +

i+

At Ay
Bt By
Az Ax
Bz Bx

hj +

j+

At Az
Bt Bz
Ax Ay
Bx By

hk

k.

(14.35)

Similarly, the i, j, and k components of (14.35) are rewritten asi j k
Ax Ay Az.

Bx By Bz
In this form, the determinant does not show the relationships between the i, j, andk components and the hi, hj, and hk components; these become clarified when thedeterminant is rewritten as (14.35).
In new octonion algebra, the real number component expresses the time component in the negative world (see Section 11.4). Thus, from
A·B = −At Bt + Ax Bx + Ay By + Az Bz ,(14.33)the inner product of two vectors in four-dimensional space–time denotes the timecomponent in the negative world. Because this time component is inaccessible tous, we observe only the outer product when multiplying two vectors. This partly
explains why we observe the inner and outer vector products as diﬀerent.
In addition, because the outer productA × B = (At Bx − Ax Bt )hi + (At By − Ay Bt )hj + (At Bz − Az Bt )hk+ (Ay Bz − Az By )i + (Az Bx − Ax Bz )j + (Ax By − Ay Bx )k(14.34)contains hi, hj, and hk components, it denotes the spatial component in both
positive and negative worlds. Like the inner product, we cannot observe the spatial
component in the negative world. Thus, setting the hi, hj, and hk components of(14.34) to zero, we retrieve the definition of the outer product until now.
170

Equations (14.33) and (14.34) correctly describe the inner and outer products,
respectively, in curved four-dimensional space–time. Furthermore, replacing theimaginary number h of the new octonion by 1, we restore Hamilton’s quaternion,
which describes the mathematics of flat four-dimensional space–time. Thus, in flat
four-dimensional space–time, (14.33) and (14.34) can be rewritten as follows:
A·B = At Bt + Ax Bx + Ay By + Az Bz ,
At Ax
Bt Bx

A×B =
+

+

i+

At Ay
Bt By
At Az
Bt Bz

j+

k+

Ay Az
By Bz

(14.35)

i

Az Ax
Bz Bx
Ax Ay
Bx By

j

k.

(14.36)Since no negative world is recognized in flat four-dimensional space–time, Equations(14.35) and (14.36) express the temporal and spatial components, respectively, in thepositive world. While these formulae are easily understood, they do not acknowledge
relativity theory in their realized world. Assuming that relativity theory is correct,(14.33) and (14.34) correctly formulate the inner and outer vector products in fourdimensional space–time.
The correctness of Equations (14.33)−−(14.36) is easily proved by demonstratingthat the equations satisfy
2

2

2|A·B| + |A × B| = |B| |A|2

derived in Section 14.2. Calculations of this proof are tedious, and are hence omitted.
In this chapter, we have demonstrated that vector calculations can be rewritten
in terms of the new octonion. Since the concept of the vector is now established inphysics, rewriting physics in terms of the new octonion remains a major challenge.
Successors of Hamilton who discovered the quaternion failed in their attempts torewrite Maxwell’s electromagnetic equations, because their new mathematics is applicable only to flat space–time.
Since standard vector analysis is the mathematics of flat three-dimensional space,
they must be rewritten as the mathematics of curved four-dimensional space–timeusing Equations (14.33) and (14.34) which describe the inner and outer products of
the new octonion.

171

14.7 New octonion and rotation vector
This section shows how the rotation vector, which plays a central role in physics,
may be rewritten as a new octonion. This is an example of rewriting physical lawsby the new octonin. First, we explain what is meant by a rotation.

Suppose that a point mass A moves along the circumference of the circle of radiusr from point P to point Q through an angle θ (theta). The length l of the arc
described by the movement of the point mass is
l = rθ.

(14.37)

The reader should note that θ is expressed in radians, not degrees. A radian is thelength of arc equivalent to unit radius (1 radian ∼ 57.3◦ ). If a point mass travels
through an infinitesimal angle dθ during infinitesimal time dt, the infinitesimal distance traveled is the infinitesimal arc length dl. Therefore, the infinitesimal limit of(14.37) is
dl = rdθ.
Dividing both sides of this formula by the infinitesimal time dt, we obtain
dl
dθ
=r ,
dt
dt

(14.38)where dl/dt is the velocity v of the rotating point mass at point P . This direction
is tangential to the circumference. In addition, since dθ/dt describes the temporalchange in angle, it is defined as the angular velocity ω (omega). Therefore, (14.38)
becomesv = rω.
To ensure agreement with Equation (14.40), derived later, this expression is rewritten asv = ωr.

172(14.39)Japanese physics students are not introduced to vectors until they enter university.
It is the rotation vector that is frequently bewildering to students encounteringvectors for the first time. Below, we give a detailed description of the rotation
vector.

Suppose that point P travels with velocity v around the circumference of a circleof radius r centered at O. Then, vectors r and v are related to the angular velocity
vector ω, through the outer vector product:v = ω × r.
The direction of ω is perpendicular to the plane described by the vectors r andv. Since ω is perpendicular to the rotational plane, its direction is that of therotation axis. Although the rotation axis may be oriented up or down, its direction
is conventionally defined as that moved by a right-handed screw turning in the
direction of v. In Figure 14.7, this direction is upward.
The definition of the angular velocity vector ω is bewildering to students because,
although rotation is planar, the angular velocity vector is directed perpendicular,
and therefore contradictory to the direction of angular movement. By contrast,
force and velocity vectors are oriented parallel to the force and velocity. However,
the rotational and actual directions of the angular velocity vector ω are not thesame, for reasons that are mentioned, but not discussed, in standard physics texts.
This case is analogous to the outer product of two vectors A and B being assigned,
but not proved, perpendicularly oriented to the plane described by the vectors.
Establishing a rule because it is convenient is neither rigorous nor logical.
In Section 14.3, we used the new octonion to prove that the outer product of two
vectors A and B is perpendicular to the plane described by the vectors. Similarly,
the new octonion can prove that the direction of the angular velocity vector ω isperpendicular to the plane described by the vectors r and v. In (14.39), we assume
that the angular velocity ω is an unknown function of two vectors r and v. Since r

173

and v are vectors while ω is not, (14.39) becomesv = ωr.

(14.40)In terms of the coordinate components, the new octonions of r and v can be writtenasv = vx i + vy j + vz k,
r = xi + yj + zk,
by which (14.40) becomes(vx i + vy j + vz k) = ω(xi + yj + zk).
This formula relates two new octonions by a rotation. For ease of calculation, thisexpression can be reverted to the new octonion itself, giving
v = ωr.

(14.41)

Although (14.41) appears identical to (14.39), the latter is formulated as a real
number while (14.41) is formulated in new octonions. Furthermore, ω in (14.41) isan unknown function that relates vectors r and v. Multiplying both sides of (14.41)
by a right-handed r , we obtain
vr = ωrr2= ω |r| .
For r 6= 0, we find that
ω=

vr
|r|

2,

(14.42)

which expresses the rotation velocity ω in terms of the new octonion.
Using this equation, we now determine the direction of ω. From the relationshipbetween the new octonion and the vector presented in Section 14.2, i.e.,
BA = A·B + A × B,
we can write
vr = r·v + r × v.
Thus, (14.42) becomes
ω=

1
|r|

2 (r·v + r × v).

174(14.43)

When the radius r is constant, i.e., when the path is truly circular, the angle
made by r and v is a right angle (i.e., π/2 radian). Thus, from the definition of an
inner product, we haveπr·v = |r| |v| cos = 0.
2
Therefore, if the radius r is constant, (14.43) reduces to
ω=

1
|r|

2 (r × v).

Since ω is a new octonion, it expresses the same content as the vector ω. Hencer, v, and the angular velocity vector ω are related by
ω=

1
|r|

2 (r × v).

(14.44)

Together with the formula of the outer vector product, (14.44) shows that the angular velocity vector ω is perpendicular to the plane described by r and v. Equation(14.44) provides mathematical proof of the perpendicularity between v and angularvelocity vector ω. In addition, when point P travels an ellipsoid path, the innerproduct r·v is non-zero. Thus, the angular velocity vector ω possesses a temporalcomponent r·v in the negative world, as well as a spatial component perpendicular
to the rotational plane.
When r and v are parallel to the direction of motion, r × v = 0. In this case, theequation1
ω = 2 (r·v + r × v)
(14.43)|r|
reduces to
ω=

1
|r|
1

2 r·v

2 |r| |v|
|r|
|v|
=
|r|
|r| /t
=
|r|
1
= ,
t

=

where t denotes time. Then, we can writev = ωrr= .
t175

This equation relates the velocity to the distance traveled in linear motion.
From the above results, the equation
ω=

1
|r|

2 (r·v + r × v)

(14.43)expresses the general relationship between the position vector r, rate vector v, andangular velocity ω in three-dimensional space. Equation (14.43) is varid only whencalculated by the new octonion.

176

15
New Octonions and Tensors

15.1 Reasons why tensors were madeA student is not introduced to tensors until he/she enters the third year of the scienceand engineering degree at a University in Japan. Thus, many ordinary readers andcollege students will be unfamiliar with the tensor, although tensors are an essential
component of higher science and engineering learning. In the engineering field, they
are used to calculate tensions in objects. Einstein’s general relativity theory ofgravity is also written in a tensor form.
Despite its importance in present-day science and engineering, the tensor remains
obscure to most students, most likely for the following reasons:(1) Few books provide a rationale for developing the tensor.
(2) The tensor cannot be matched to a concrete image.
Most tensor texts begin by defining the tensor and then presenting examples
and tensor formulas. As a result, the necessity of the tensor is neglected. When theimaginary number is introduced at high school, its rationale is explained; the squareof the imaginary number i equals −1. We can intuitively understand that trigonometric functions describe the relationships between the angles and side lengths oftriangles. Quantities possessing both magnitude and direction are easily represented by vectors. However, by comparison, the tensor lacks an obvious rationale,
and therefore seems mysterious and obscure. This obscurity is partly responsible
for why tensors are so diﬃcult to understand.
Tensors are further complicated by their lack of connection to any concrete image.
The imaginary number described previously can be plotted as a point on a complex
plane, while a trigonometric function is easily visualized on a triangle. A vector isdenoted by an arrow. Every number can be visualized as an image.
Individuals who memorize sequences of hundreds of irregular numbers, such as thecircular constant π, do so by relating numbers to concrete images and memorable
tale. Since the tensor can neither be plotted on a figure nor be interpreted as an
177

image, it cannot be readily understood and memorized. To redress this issue, thissection explains why the tensor was postulated and why it cannot be expressedby a diagram. This explanation is essential for understanding the content of thefollowing section, 15.2.
The wise student who has studied vectors in high school might recognize that two
vectors cannot be divided. At least, nobody has proposed the division of a vector.
Real numbers, imaginary numbers, and trigonometric functions are all divisible. As
vectors are indivisible, they cannot be manipulated by a full set of mathematicaloperations. The tensor was proposed to accommodate division in vector quantities.
First, we consider real number division. If two real numbers x and y are relatedbyy = axand x 6= 0, then deviding both sides of the equation by x, we obtain
a=

y
.
x

Furthermore, if x = 1 and y = 2, we have
a=

2
= 2.
1

Thus, y = ax becomesy = 2x.
Inserting x = 3 into this equation gives
y = 6.
Therefore, in the equation y = 2x, numbers 1 and 2 and numbers 3 and 6 are related
in the same way (through the constant 2).
Similar relationships exist between two vectors. We assume the following relationship between vectors A and B, where T denotes a tensor.
B = T A.

(15.1)At present, the contents of T are unknown. As explained in Section 14.1, if vectors
A and B are expressed in terms of their coordinate components (Ax , Ay , Az ) and(Bx , By , Bz ), and their fundamental vectors ex , ey , and ez , (15.1) becomes
Bx ex + By ey + Bz ez = T (Ax ex + Ay ey + Az ez ).

(15.2)

Assuming that both sides of (15.2) are divisible by Ax ex + Ay ey + Az ez , we can
write
T =

Bx ex + By ey + Bz ez.
Ax ex + Ay ey + Az ez
178(15.3)

To eliminate the fundamental vectors ex , ey , and ez from the denominator of(15.3), we multiply the numerator and denominator by some vector. Since multiplication of two vectors yields an inner or an outer product, the explicit form of
the multiplication is unknown, and hence, we retain (15.3) in the above format.
Multiplying (15.3) by vector C(Cx , Cy , Cz ), vector D is obtained as
D = TC=Bx ex + By ey + Bz ez
× (Cx ex + Cy ey + Cz ez ).
Ax ex + Ay ey + Az ez

Even in this format, whether the result of multiplication of the fundamental vectors
ex , ey , and ez is an inner or outer product is indeterminable and further calculations are impossible. As the vector D is incalculable, it is considered that a vector
cannot be divided by another vector.
If vector D cannot be obtained from C through the relationship T between vectors
A and vector B, we can regard vector mathematics as incomplete. The tensor wasformulated to resolve this issue. Next, we explain the solution by the tensor. In(15.2), Bx , By , and Bz are assumed as primary functions of Ax , Ay , and Az , thatis, they do not contain secondary A2x or tertiary A3x . At this time, (15.2) is rewritten
as follows:
Bx ex + By ey + Bz ez = (a1 Ax + a2 Ay + a3 Az )ex+ (b1 Ax + b2 Ay + b3 Az )ey+ (c1 Ax + c2 Ay + c3 Az )ez .

(15.4)

The calculations can be much simplified by compacting (15.4) in a matrix form.
Using the matrix multiplication notation


 

a1 a2 a3Ax
a1 Ax + a2 Ay + a3 Az


 

 b1 b2 b3   Ay  =  b1 Ax + b2 Ay + b3 Az  ,c1 c2 c3Az

c1 Ax + c2 Ay + c3 Az

we rewrite (15.4) as follows:


Bx





a1 a2 a3



Ax




 


 By  =  b1 b2 b3   Ay  .
Bz
c1 c2 c3
Az

(15.5)

B = T A,

(15.1)

Comparing (15.5) and179

we can writea1 a2 a3T =  b1 b2 b3  ,c1 c2 c3

(15.6)

where T is a tensor. Using T , we can construct vector D from vector C. Since
D = T C, we observe that




Cx
Dx




 Dy  = T  Cy 
Cz
Dz



a1 a2 a3Cx=  b1 b2 b3   Cy c1 c2 c3Cz
a1 Cx + a2 Cy + a3 Cz=  b1 Cx + b2 Cy + b3 Cz  .
c1 Cx + c2 Cy + c3 Cz
The above result exemplifies how the tensor construct can express the relationshipbetween two vectors that are indivisible in their vector form.
Once we appreciate that tensors express relationships between two vectors, we
can understand that tensors cannot be drawn on a figure. If an equation exists thatchanges a triangle into a quadrangle, both the triangle and the quadrangle can be
drawn as figures but the conversion formula cannot be represented in this manner.
Similarly, we can draw two vectors on a figure, but not the tensor that relates them.
As mentioned above, the tensor construct overcomes the divisibility problem inherent in vector. However, we can also obtain vector D by division of new octonions.
In the new octonion notation, three vectors A, B, and C are expressed as
A = Ax i + Ay j + Az k, B = Bx i + By j + Bz k, C = Cx i + Cy j + Cz k,(15.7)and the equation
B = TA(15.1)

B = HA.

(15.8)

becomes

H is assumed as the function relating A and B. As this function completely diﬀers
from the tensor, it is denoted as H. H implies the inclusion of the fourth imaginary
number h.

180Right-multiplying both sides of (15.8) by A, we obtain
BA = HAA
2= H |A| .
Thus, if A 6= 0, we have
H=

BA

2.

|A|
Since D is the product of C and H, we write
D = HC
=

BA

2 C.

|A|

(15.9)

Equation (15.9) shows that vector division is possible if the vectors are rewritten as
new octonions. Thus, D can be algebraically calculated without invoking tensors.
Furthermore, this calculation is manageable by junior high school students, whereas
tensor calculations are diﬃcult even for college students.
As mentioned above, if a new octonion is used, a vector can be easily constructedfrom other vectors without using tensors. However, this discussion applies only tostraight coordinates and is not realized in curved coordinate axes. This limitation
occurs because the multiplication of the two new octonions in (15.8) is realized onlyin straight coordinates. Whether coordinate rotation by a new octonion is realizedin curvilinear coordinates is yet to be proven. In relativity theory, special andgeneral versions are formulated in straight and curvilinear coordinates, respectively.
Here we restrict our discussion to special relativity.

15.2 Diﬀerences between the tensor and the new octonion
The previous section demonstrated how, similar to the tensor, a vector can beconstructed from another vector by expressing the vectors as new octonions. In thissection, we explain the diﬀerences between the tensor and the new octonion. To
simplify the formulas, coordinate components are expressed in Hamilton’s notation.
That is, i, j, and k are not attached to their coordinate components.
Assume that four points A(1, 1, 1), B(−1, 1, 1), C(−1, −1, −1), and D(1, −1, −1)
are distributed in three-dimensional space, as shown in Figure 15.1. Clearly, points
A, B, C, and D exist in the same plane, leaning at 45◦ to the x-y plane. Denotethe vectors connecting the origin O and points A, B, C, and D as A, B, C, andD, respectively, and let vector B be related to vector A, through the tensor T :B = T A.
181

In a matrix notation, this equation becomes

 

−1
1


 
=
T
 1 
 1 .
1(15.10)1

We now obtain the tensor T that realizes (15.10). Substituting the tensor introducedin the previous section
a1 a2 a3
T =  b1 b2 b3 
(15.6)
c1 c2 c3
into (15.10) yields
 


−1
a1 a2 a3
1

 


 1  =  b1 b2 b3   1 1
c1 c2 c31
a1 + a2 + a3

=  b1 + b2 + b3  .


c1 + c2 + c3
This equation is equivalent to−1 = a1 + a2 + a3 ,1 = b1 + b2 + b3 ,1 = c1 + c2 + c3 .
As we have three equations in nine unknowns, this system cannot be solved. Thus,
the unknown tensor components are labeled a, b, and c, and all other elements are
assumed as 0. For convenient arrangement of a, b, and c, we also introduce a tensor
182

that is symmetric to the straight line drawn from the upper left to the lower right.
This tensor is written as follows:


abc


(15.11)
T =  b 0 0 .
c00
Substituting this tensor into (15.10), we obtain




 
abc
1
 
 
1  =  b 0 0  1 
1
c00
1


a+b+c


=
b
.

−1





c
Thus, we havea + b + c = −1, b = 1, c = 1,
from whicha = −3, b = c = 1.
Substituting these results into (15.11) yields


T =

−3 1 1
1 0 0 .
100

(15.12)This tensor constructs vector B from vector A.
We now use (15.12) to construct a vector from vector C. Since the components
of C are (−1, −1, −1), we can write−3 1 1−1
T C =  1 0 0   −1 
100
−1


3−1−1


=  −1



−1

1


=  −1  .


−1
183

This is vector D. That is, the tensor T that constructs vector B from vector A
also constructs vector D from vector C.
Comparing 4OAB described by A and B with 4OCD described by C and D, weobserve from Figure 15.1 that the triangles are congruent. In other words, ∠AOB
equals ∠COD and the lengths of the three sides are identical. If the coordinate
components of point B are doubled to (−2, 2, 2), the tensor (15.12) is also doubled.
Specifically, T becomes−6 2 2T =  2 0 0 .
200

If vector C is multiplied by this tensor, the coordinate components of point D arealso doubled to (2, −2, −2). That is, 4OCD and 4OAB remain similar undersize scaling operations. Moreover, when the size of vector C is d × A, 4OCD isconverted to a similar triangle of size d × 4OAB. These results are easily provedby calculations.
From the above results, if vector D is the product of C and T , (as vector B is
the product of A and T ), we find that 4OAB and 4OCD are similar. This is afundamental property of tensors. We now discuss the diﬀerences between the tensor
and the new octonion.
(1) The first diﬀerence between the tensor and the new octonion
The first diﬀerence between the tensor and the new octonion is that the new octonion oﬀers a unique solution. Equation (15.11) is one of multiple tensors T that
construct a vector B from vector A. Because (15.11) was temporarily assumed,
the existence of other tensors was not considered in our previous example. As an184

example, consider the following tensors:


a0b





T =  0 0 c .

(15.13)bc0
Substituting this tensor into (15.10) yields








 
a0b
1
 
 
1  =  0 0 c  1 
1
bc0
1


a+b


= c
,

−1

b+c
from which we obtaina + b = −1, c = 1, b + c = 1,
and therefore
a = −1, b = 0, c = 1.
Substituting these results into (15.13) gives


T =

−1 0 0
0 0 1 .
010

(15.14)Multiplying vector C by this tensor, we find that


 

−1 0 0
−1
1


 

 0 0 1   −1  =  −1  ,
−1

010

−1

which is again vector D. Therefore, this tensor also indicates a similarity between
4OAB and 4OCD.
Other tensors that relate vector B to A and vector D to C are












0bc
0bc
00b
00b
a00
0ab












 b a 0 ,  b 0 0 ,  0 a c ,  0 0 c ,  0 b 0 ,  a 0 c .
c00

c0a

bc0

bca

00c

bc0

That is, vector B can be constructed from vector A in multiple ways, all of which
are mathematically valid.
185

We now show that a unique new octonion constructs vector B from vector A. The
new octonions A, B, C, and D describing vectors A, B, C, and D, respectively,are
A = i + j + k, B = −i + j + k, C = −i − j − k, D = i − j − k.

(15.15)

Denoting the new octonion that constructs vector B from vector A as H, we have
B = HA.
Substituting (15.15) into this expression, we obtain−i + j + k = H(i + j + k).
Right-multiplying both sides of this equation by the new octonion conjugate A =−i − j − k, the equations become(−i + j + k)(−i − j − k) = H(i + j + k)(−i − j − k),(−i + j + k)(i + j + k) = H(i + j + k)(i + j + k),−i − ij − ik + ji + j 2 + jk + ki + kj + k 2 = H(i2 + ij + ik + ji + j 2 + jk + ki + kj + k 2 ),2

1 − k + j − k − 1 + i + j − i − 1 = H(−1 + k − j − k − 1 + i + j − i − 1),−1 + 2j − 2k = H(−3),
H = (1 − 2j + 2k)/3.

(15.16)Equation (15.16) is the new octonion that constructs vector B from vector A.
We now investigate whether vector D is obtained as the product of C and H.
From (15.15) and (15.16), we obtain
HC = (1 − 2j + 2k)(−i − j − k)/3
= (−i − j − k + 2ji + 2j 2 + 2jk − 2ki − 2kj − 2k 2 )/3= (−i − j − k − 2k − 2 + 2i − 2j + 2i + 2)/3= (3i − 3j − 3k)/3= i − j − k,which is precisely vector D from (15.15). That is, the new octonion
H = (1 − 2j + 2k)/3(15.16)constructs vector B from vector A as well as vector D from vector C. Moreover,
H is a unique solution. This calculation clarifies the first diﬀerence between thetensor and the new octonion.
186

(2) The second diﬀerence between the tensor and the new octonion
We now explain the second diﬀerence between the tensor and the new octonion. In
Figure 15.1, vectors A, B, C, and D lie on the same plane. Consider vector E
extending from the origin O to a point E(1, 2, 3) lying outside the plane.

We examine the vector transformed from vector E by the tensor
−3 1 1T =  1 0 0 ,
100

(15.12)

which constructs vector B from vector A. Explicit calculation gives the following
equations:TE = 

−3 1 1



1



 
1 0 0  2 
100
3


−3 + 2 + 3


=
1





1

2
 
=  1 .
1
That is, the tensor T transforms vector E into vector F , which is the vector connecting the origin O to point F (2, 1, 1). Moreover, as shown in Figure 15.1, point
F lies on the plane occupied by points A, B, C, and D. Indeed, regardless of its
location, if E is multiplied by the tensor T , the point F return to the plane occupied
by points A, B, C, and D. This is another fundamental property of tensors.
187We have established that 4OAB and 4OCD described by points A, B, C, andD on the same plane are similar. However, are 4OAB and 4OEF similar? By
Pythagoras’ theorem, the lengths of the sides OE and OF are calculated as
2|OE| = 12 + 22 + 32 = 1 + 4 + 9 = 14,
2|OF | = 22 + 12 + 12 = 4 + 1 + 1 = 6.
As |OE|2 6= |OF |2 , 4OAB and 4OEF are not similar. Thus, while tensor T
preserves congruency between 4OAB and 4OCD in the same plane, it does notconstruct congruent triangles from vectors lying outside the plane. In other words,
the tensor is restricted to vectors lying in the same plane. Vectors lying outside theplane cannot be synchronized with vectors occupying a common plane. However,
we emphasize that these discussions relate only to straight coordinates, i.e., thepremises of special relativity, and may not apply to curvilinear coordinates, i.e.,
general relativity.
Unlike the tensor, vectors transformed by the new octonion can be synchronizedthroughout the entire three-dimensional space. The new octonion H that constructs
vector B from vector A is uniquely specified as
H = (1 − 2j + 2k)/3.

(15.16)Since the new octonion describing vector E is E = i + 2j + 3k, the new octonion ofpoint G obtained as the product of E by H is
G = HE= (1 − 2j + 2k)(i + 2j + 3k)/3= (i + 2j + 3k − 2ji − 4j 2 − 6jk + 2ki + 4kj + 6k 2 )/3= (i + 2j + 3k + 2k + 4 − 6i + 2j − 4i − 6)/3
= (−2 − 9i + 4j + 5k)/3.

(15.17)Since G has four coordinate components, the vector E in three-dimensional spaceis transformed to a vector G in four-dimensional space–time.
For readers who cannot accept that a three-dimensional vector becomes a fourdimensional vector, we present a typical example. If a stationary observer A detectsthat a point mass D does not move in time (i.e., t = 0 always), the coordinates of
D are denoted by the three-dimensional vector (x, y, z). Substituting t = 0 into
the Lorentz transformations188

t − (v/c2 )xt0 = √,
1 − v 2 /c2
x − vtx0 = √,
1 − v 2 /c2
y 0 = y,

(3.3)
(3.4)
(3.5)

0

z = z,(3.6)

we have−(v/c2 )xt0 = √,
1 − v 2 /c2x,x0 = √
1 − v 2 /c2
y 0 = y,z 0 = z.
The coordinates of D observed from a linearly moving observer B are described by afour-dimensional vector (t0 , x0 , y 0 , z 0 ). Thus, a three-dimensional vector naturally
changes to a four-dimensional vector under a familiar coordinate transformation.
Algebraic theory dictates that possible worlds are restricted to one, two, four,or eight dimensions (concepts developed in this book permit overlapping fourdimensional worlds). Thus, independent three-dimensional space is forbidden bymathematics. In this context, a vector E in three-dimensional space naturally converts to a vector G in four-dimensional space–time.
Next, we investigate whether a semantic can be defined for a vector G in fourdimensional space–time constructed from vector E by
H = (1 − 2j + 2k)/3.
First, we find that2

|OE| = EE= (i + 2j + 3k)(−i − 2j − 3k)= −i2 − 2ij − 3ik − 2ji − 4j 2 − 6jk − 3ki − 6kj − 9k 2
= 1 − 2k + 3j + 2k + 4 − 6i − 3j + 6i + 9
= 14.

189

(15.16)

From
G = (−2 − 9i + 4j + 5k)/3,(15.17)

we obtain2|OG| = GG= (−2 − 9i + 4j + 5k)(−2 + 9i − 4j − 5k)/9= (4 − 18i + 8j + 10k + 18i − 81i2 + 36ij + 45ik
− 8j + 36ji − 16j 2 − 20jk − 10k + 45ki − 20kj − 25k 2 )/9= (4 − 18i + 8j + 10k + 18i + 81 + 36k − 45j− 8j − 36k + 16 − 20i − 10k + 45j + 20i + 25)/9= (4 + 81 + 16 + 25)/9= 126/9= 14.
Thus, we have2

2

|OE| = |OG| = 14.

(15.18)

These equations indicate that the four-dimensional space–time vector G is obtainedby rotating the three-dimensional vector E in four-dimensional space–time.
Next, we calculate |EG|2 . If the new octonion connecting points E and G isdenoted (EG), we have(EG) = G − E= (−2 − 9i + 4j + 5k)/3 − (i + 2j + 3k)2
4
5= − − 3i + j + k − i − 2j − 3k
3
3
3
2
4
2
= − − 4i − j − k.
333
Thus, as2

|EG| = (EG)(EG)
2
2
4
2
2
4
= (− − 4i − j − k)(− + 4i + j + k)
3
3
3
3
3
3
2 2
2 2
4 2
2
=( ) +4 +( ) +( )
3
3
3
4
4 16
= + 16 + +
9
9
9
24
+ 16
=
9
190

8
+ 16
3
56
=
,
3

=

√

we find that
|EG| =

√
56
2 14
= √ .
3
3

From (15.18) and (15.19), the ratios of the sides of 4OEG are√√√√
2 14 √|OE| : |OG| : |EG| = 14 : 14 : √ = 3 : 3 : 2.
3

(15.19)(15.20)From Figure 15.1, the side lengths of 4OAB are|OA| = |OB|√= 12 + 12 + 12√
= 3,
|AB| = 2.
Thus, we find that
|OA| : |OB| : |AB| =√√3 : 3 : 2.

(15.21)Equations (15.20) and (15.21) demonstrate that 4OEG and 4OAB are similar.
Here the reader must recognize that 4OEG exists in four-dimensional space–time,
while 4OAB is a three-dimensional spatial entity. That is, the new octonion transformation converts a vector in three-dimensional space into another vector in fourdimensional space–time, and realizes congruent triangles in four-dimensional space–time. As previously explained, the tensor is applicable only to vectors describing
two-dimensional planes in three-dimensional space. The new octonion transformation is more general, being applicable to all vectors regardless of their relationships.
191

In three-dimensional linear coordinate spaces, the tensor is made to convert athree-dimensional vector to another three-dimensional vector. However, vectors
occupying diﬀerent planes cannot be synchronously calculated using tensor mathematics. In other words, when the tensor operates on a vector in three-dimensional
space, it converts the vector to a two-dimensional vector. This is clear from Figure15.1. As demonstrated previously, if any point E extending beyond the plane occupied by points A, B, C, and D is multiplied by a tensor T , the resulting point F
returns to the plane occupied by points A, B, C, and D. As there are three coordinate components, we tend to consider that tensors manipulate three-dimensional
vectors. However, correctly speaking, since tensor mathematics is applicable only
to vectors in the same plane, the tensor confines the vectors to two dimensions andprevents the four-dimensional transformation of three-dimensional vectors.
Engineers would prefer that tensor transformation do not convert three-dimensional
vectors to four-dimensional vectors. However, tensor mathematics may lead researchers of space–time or fundamental particles to incorrect conclusions. Considering a tensor of three columns and three rows and a tensor of four columns andfour rows, the latter will convert a four-dimensional vector into a three-dimensional
vector, which exists in a cross-section of four-dimensional space–time. Regardlessof whether we intend to work with four-dimensional vectors, the result of tensor
mathematics is a three-dimensional vector. This is easily demonstrated by performing transformations by a tensor of four columns and four rows. Moreover, a tensor
of three columns and three rows does not elucidate the relationship between twodimensional planes, while a tensor of four columns and four rows precludes us fromunderstanding the relationships between cross-sections in four-dimensional space–time. If we use tensors to examine the structure of four-dimensional space–time, wewill obtain a sum of many cross-sections rather than appreciating the true natureof space–time.

15.3 Kronecker δ
The Kronecker δ (delta), which commonly appears in tensor and relativity texts, is{defined asδji =1 (i = j)
0 (i 6= j).

Some readers may regard δ to be somewhat incongruous. The reason for this discomfort is that variables, whose values change according to conditions, are typicallyabsent in mathematically elegant geometries and algebras. Given a function y = x,
192

if x > 0, then y > 0, and if x < 0, then y < 0. This causes no concern, since y is afunction. In pure geometry and algebra, the parameters are fixed, unlike δ, whichis conditional.
In three-dimensional space, i and j denote the natural numbers 1 − 3. The matrix
form of δji is


100


δji =  0 1 0  .
001
Multiplying this matrix by a vector A(1, 1, 1), we obtain


 
100
1

 
i
δj A =  0 1 0   1 
001
1
 
1
 
= 1 
1= A.
Since the δji transformation retrieves the vector, it is a tensor that constructs thesame vector.
Next, we calculate δji in terms of the new octonion. We denote the new octonion
that changes vector A(1, 1, 1) into vector A is assumed as H. Since
A = HA,
we can write
i + j + k = H(i + j + k).
Right-multiplying both sides of this equation by the new octonion conjugate A =−i − j − k of A = i + j + k, the equations become(i + j + k)(−i − j − k) = H(i + j + k)(−i − j − k),1 + 1 + 1 = H(1 + 1 + 1),
3 = H3,
H = 1.
That is, Kronecker δji is 1 by the new octonion and the number does not conditionallychange its value.

193

As explained in Section 15.2, the tensor collapses the three-dimensional vector
into a two-dimensional plane and prevents the construction of a four-dimensional
vector. Therefore, tensor mathematics requires the Kronecker δ, which conditionally changes its value, and is thus a cause for concern among some purists. Since
Kronecker δ is unconditionally 1 in the new octonion interpretation, it is not required in new octonion space–time investigations. However, since our discussion isrestricted to straight coordinates, we may draw diﬀerent conclusions in curvilinear
coordinates.

15.4 Metric tensor
An important tensor in Einstein’s relativity theory is the metric (distance) tensor,
also known as the fundamental tensor. This section relates the new octonion to themetric tensor.
General relativity theory generalizes the time ct and the distances x, y, z, which
are written as X 0 , X 1 , X 2 , and X 3 in special relativity. Thus, the square of theworld distance s
s2 = (ct)2 − x2 − y 2 − z 2(7.1)

is written in relativity textbooks as
s2 = (X 0 )2 − (X 1 )2 − (X 2 )2 − (X 3 )2 .

(15.22)

However, as explained in Section 7.2, the world distance in terms of the new octonioniss2 = −(ct)2 + x2 + y 2 + z 2 .
(7.6)
Note that Equation (7.1) formulates the world distance in the negative world.
Equation (15.22) is compactly rewritten in terms of η (eta), µ (mu), and ν (nu)ass2 = ηµν X µ X ν .

(15.23)

η00 = 1, η11 = η22 = η33 = −1, ηµν = 0. (: µ 6= ν)(15.24)

However,

In tensor form, (15.24) is written as



1
0 0 0


 0 −1 0 0 .
ηµν = 
 0 0 −1 0 
0 0 0 −1
194

(15.25)

Equation (15.25) defines a metric tensor in special relativity. Similar to (15.23), we
can rewrite the four-dimensional infinitesimal world distance ds as
ds2 = ηµν dX µ dX ν ,(15.26)

where ds2 is an alternative form of (ds)2 .
What is the result of applying the new octonion to the above calculation? In thenotation X 0 , X 1 , X 2 , and X 3 , the new octonion describing a world point is
A = X 0 h + X 1 i + X 2 j + X 3 k.
The square of the world distance s becomess2 = AA= (X 0 h + X 1 i + X 2 j + X 3 k)(X 0 h − X 1 i − X 2 j − X 3 k)
= −(X 0 )2 + (X 1 )2 + (X 2 )2 + (X 3 )2 .

(15.27)

As explained in Section 7.2, (15.27) is the mathematically calculated world distance.
Similarly, the square of the infinitesimal four-dimensional world distance ds becomes
ds2 = dAdA
= (dX 0 h + dX 1 i + dX 2 j + X 3 k)(dX 0 h − dX 1 i − dX 2 j − dX 3 k)
= −(dX 0 )2 + (dX 1 )2 + (dX 2 )2 + (dX 3 )2 .
The square of the world distance s can be calculated from the new octonion A andits conjugate A avoiding the need for metric tensors such as (15.23) and (15.26). As
explained in Section 15.2, tensor mathematics calculates physical quantities in fourdimensional space–time by confining them to a cross section of four-dimensional
space–time. In contrast, the new octonion mathematics calculates physical quantities throughout the entire four-dimensional space–time. Thus, tensors are notrequired in straight coordinate systems. However, this situation may alter in curvilinear coordinate systems.

195

196

16
Synchrotron Radiation

16.1 About synchrotron radiation
Consider an electron moving with uniform horizontal velocity v watched by a stationary observer. The situation is illustrated in Figure 16.1 (a) and (b). The coordinate axes of the static system are x, y, and z, while those of the moving electron
are x0 , y 0 , and z 0 . Moreover, the x- and x0 -axes overlap. The y- and y 0 -axes as wellas the z- and z 0 -axes point in the positive direction. When the electron moving at
near-light velocity emits light in the y 0 -axial direction, how does the light appear
to the stationary observer? The light seen by this observer is called synchrotron
radiation.
Because relativity theory and the new octonion space–time theory yield diﬀerent
Lorentz transformations of y 0 and z 0 , the two approaches may lead to diﬀerent
conclusions. This chapter focuses on the conclusions obtained from both approaches,and discusses their diﬀerences.

16.2 Proof under Lorentz transformationsFirst, we use the Lorentz transformation of special relativity to explain the directionof synchrotron radiation. This content is written in standard relativity texts. The
Lorentz-transformed velocity was discussed in Section 10.4.
197

Let the velocity of a point mass D be V (Vx , Vy , Vz ) with respect to a stationary
observer A, and V 0 (Vx0 , Vy0 , Vz0 ) with respect to an observer B moving along a
straight line with uniform velocity v in the x-direction of A. Then, we have
Vx − v,
1 − (v/c2 )Vx√
Vy 1 − v 2 /c20Vy =,1 − (v/c2 )Vx√
Vz 1 − v 2 /c2
Vz0 =.
1 − (v/c2 )Vx

Vx0 =(10.16)(10.17)(10.18)

Now, consider that D is the light emitted in the y 0 -direction by an electron withvelocity components Vx0 = 0, Vy0 = c, and Vz0 = 0. Inserting Vx0 = 0 into (10.16), we
find that0=

Vx − v,
1 − (v/c2 )Vx

Vx = v,(16.1)

while inserting Vy0 = c into (10.17) yields√
Vy 1 − v 2 /c2c=.
1 − (v/c2 )Vx
Substituting (16.1) into the above equation, we obtain√
Vy 1 − v 2 /c2c=
1 − (v/c2 )v√
Vy 1 − v 2 /c2=
1 − v 2 /c2Vy=√,
1 − v 2 /c2√
Vy = c 1 − v 2 /c2 .

(16.2)In addition, inserting Vz0 = 0 into (10.18) yields√
Vz 1 − v 2 /c20=,
1 − (v/c2 )Vx
Vz = 0.

(16.3)

Vx = v,√
Vy = c 1 − v 2 /c2 ,(16.1)(16.2)

Vz = 0.

(16.3)

Collectively, we have198

As the velocity of the electron v approaches the velocity of light c, (16.1), (16.2),and (16.3) become
Vx = c,Vy = 0,Vz = 0.
These equations indicate that, according to the stationary observer, light moves in
the forward x-direction with velocity c. That is, in special relativity, light, whichis emitted in the perpendicular direction by an electron moving with velocity v
approaching that of light, is measured as it concentrates ahead of the stationary
observer. The situation is shown in Figure 16.2.

16.3 Proof under the new Lorentz transformations
Next, we derive the direction of the synchrotron radiation from the new Lorentz
transformation. The velocities under the new Lorentz transformation are given by(for proof, see Section 10.5)
Vx − v,
1 − (v/c2 )Vx
Vy + (v/c)Vz h
Vy0 =,1 − (v/c2 )Vx
Vz − (v/c)Vy hVz0 =.
1 − (v/c2 )Vx
Vx0 =(10.25)(10.26)(10.27)

When D is the light emitted in the y 0 -direction by an electron, its velocity components are Vx0 = 0, Vy0 = c, and Vz0 = 0. Inserting Vx0 = 0 into (10.25), we obtain0=

Vx − v,
1 − (v/c2 )Vx

Vx = v.

(16.4)

199

Inserting Vy0 = c into (10.26) yields
Vy + (v/c)Vz h.
1 − (v/c2 )Vxc=

Substituting (16.4) into the above equation, we getc=Vy + (v/c)Vz h.
1 − v 2 /c2(16.5)In addition, inserting Vz0 = 0 into (10.27) yields
0=Vz − (v/c)Vy h,
1 − (v/c2 )Vx

Vz = (v/c)Vy h.

(16.6)

Substituting (16.6) into (16.5), we obtainc=

Vy (1 − v 2 /c2 ),1 − v 2 /c2

Vy = c.

(16.7)

Substituting (16.7) into (16.6) gives
Vz = vh.

(16.8)

Collectively, (16.4), (16.7), and (16.8) are given by
Vx = v,(16.4)

Vy = c,(16.7)

Vz = vh.

(16.8)

Comparing these equations to those obtained by the Lorentz transformation
Vx = v,√
Vy = c 1 − v 2 /c2 ,(16.1)

Vz = 0,(16.3)(16.2)

we see that in the new Lorentz transformation, Vy is always c and Vz is a [zhk]-axis
component vh in the negative world.

200

16.4 Synchrotron radiation and the constancy of the velocity
of light
We obtained (16.1), (16.2), and (16.3) from the Lorentz transformation and (16.4),(16.7), and (16.8) from the new Lorentz transformation. Which is correct? First, weinvestigate whether (16.1), (16.2), and (16.3) are consistent with a constant velocity
of light c. The square of the synthetic velocity V in terms of its components Vx , Vy ,and Vz isV 2 = Vx2 + Vy2 + Vz2 .
Substituting
Vx = v,√
Vy = c 1 − v 2 /c2 ,(16.1)(16.2)

Vz = 0(16.3)

into this equation, we have
V 2 = v 2 + c2 (1 − v 2 /c2 )= c2 ,
V = c.
This result fulfills the condition that c is constant.
Repeating the calculations with
Vx = v,(16.4)

Vy = c,(16.7)

Vz = vh(16.8)

obtained by the new Lorentz transformation, we find thatV 2 = v 2 + c2 − v 2= c2 .
Thus, (16.4), (16.7), and (16.8) also fulfill the condition that c is constant.

201

16.5 Slant of synchrotron radiation
We now investigate the slant of light emitted in the y 0 -direction by an electron, as
observed in the static system. By the Lorentz transformation, we have
Vx = v,√
Vy = c 1 − v 2 /c2 ,(16.1)

Vz = 0.

(16.3)(16.2)

Thus, the slant of light with respect to the x-axis, defined as Vy /Vx , becomes√c 1 − v 2 /c2
Vy
=
.
Vx
v
As the electronic velocity v approaches the velocity of light c, the limiting slant
becomes√c 1 − v 2 /c2
Vy
lim
= lim
v→c Vxv→cv√
2c 1 − c /c2
=
c
= 0.

That is, the light travels along the x-axis in a straight line of gradient 0. For thisreason, synchrotron radiation concentrates ahead of the stationary observer A.
If the calculations are repeated using the formulae (16.4), (16.7), and (16.8) obtained from the new Lorentz transformation, we require a diﬀerent approach. In
applying the Lorentz transformation to compute the slant, we only need to consider the x-y plane, since Vz = 0. However, in the new Lorentz transformation, thecalculations are complicated by the Vz = vh term.
To advance discussions, we propose a new axiom. In four-dimensional space–timewith a double structure, if the velocity has three components in the x-, y-, andz-directions, we must first calculate the combined velocity Vyz in the y-z plane,
and then calculate Vyz /Vx to obtain the gradient with respect to the x-axis. This
calculation is correct in the positive world. However, it has not been proven correctwhen the velocity has a component in the negative world, such as Vz = vh. Thus,
we declare this situation as an axiom.
First, we calculate the gradient with respect to the x-axis. From the equations
Vy = c,(16.7)

Vz = vh,(16.8)

202

we find that2
Vyz
= Vy2 + Vz2= c2 + (vh)2= c2 − v 2 .
Thus, conditional on c > v, we haveVyz =√c2 − v 2 .

(16.9)

From (16.9) and the x velocity component obtained from the new Lorentz transformation, i.e.,
Vx = v,(16.4)

the gradient with respect to the x-axis of the synchrotron radiation is√Vyz
c2 − v 2=.
Vxv
As the electron velocity v approaches the velocity of light c, the limiting slant isobtained as√Vyz
c2 − v 2lim= lim
v→c Vx
v→c
v
√
2
2
c −c
=
c
= 0.
(16.10)That is, the light travels along the x-axis in a straight line of gradient 0, consistent
with the result obtained by the Lorentz transformation.
Next, we calculate the gradient with respect to the y-axis. From the x and z
velocity components obtained under the new Lorentz transformation, i.e.,
Vx = v,(16.4)

Vz = vh,(16.8)

we find that2
Vxz
= Vx2 + Vz2= v2 − v2= 0,
Vxz = 0.
203

(16.11)

From (16.11) and the y velocity component obtained under the new Lorentz transformation, i.e.,
Vy = c,(16.7)
we have
Vxz
0
=
Vy
c
= 0.

(16.12)This result indicates that synchrotron radiation is observed as light in the y-direction,
regardless of the electron velocity v.
According to (16.10), which obtains the gradient with respect to the x-axis, thesynchrotron radiation must be directed along the x-axis. However, in (16.12), it
also becomes oriented in the y-direction. What is the cause of this contradiction?
Before pursuing this result, we calculate the gradient with respect to the z-axis.
From the x and y velocity components under the new Lorentz transformation, i.e.,
Vx = v,(16.4)

Vy = c,(16.7)

we obtain2
Vxy= Vx2 + Vy2= v 2 + c2 ,√Vxy = v 2 + c2 .

(16.13)

From (6.13) and the z velocity component obtained under the new Lorentz transformation, i.e.,
Vz = vh,
we find that

(16.8)√
v 2 + c2vh√
= −h 1 + c2 /v 2 .

Vxy
=
Vz

As the electron velocity v approaches the velocity of light c, the limiting slant
becomes√Vxylim= − lim h 1 + c2 /v 2
v→c Vzv→c√= −h 1 + c2 /c2√= − 2h.
204

According to this result, the gradient with respect to the z-axis extends into thenegative world, and is hence invisible to us.
From the above results, the light emitted in the y 0 -direction from an electron
moving in the x-direction at near-light velocity is observed by a stationary observer
as light traveling in both the x- and y-directions. In addition, a component of thelight exists in the negative world, but cannot be observed.
This result leads to a contradiction that the light emitted as a single world linefrom the moving electron is observed as the three world lines by a stationary observer. However, if we add the temporal component cth to the velocity components
obtained under the new Lorentz transformation, i.e.,
Vx = v,(16.4)

Vy = c,(16.7)

Vz = vh,(16.8)

we obtain the new octonion of the world line of light:
cth + Vx ti + Vy tj + Vz tk = cth + vti + ctj + vthk.
This equation implies a single world line observed in the static system; hence, theappearance of the three world lines is an artifact. The reasons for this anomaly arepresently unknown. However, since the hk component resides in the negative world,
the world line of the light emitted in the y 0 -direction ranges over both positive andnegative worlds.
The axiom proposed above, that is, the gradient with respect to the x-axis iscalculable if we first calculate Vyz /Vx , may be incorrect in four-dimensional space–time with double structure. Alternatively, the new octonion itself may be incorrect.
However, in our analysis of special relativity, the direction of synchrotron radiation
alone is inconsistent with the new octonion results. A correct interpretation of thisresult may be realized in future.

16.6 Observation of the perpendicular light of synchrotron
radiation
In the previous section, we mentioned that the light emitted in the y 0 -direction froman electron moving in the x-direction at near-light velocity apparently arrives at a
stationary observer from both the x- and y-directions. This raises the question:
how is light traveling at velocity c in the y-direction perceived by the observer?
205

Consider a light detector placed on a straight line parallel to the direction of amoving electron, as shown in Figure 16.3 (a) and (b). The distance between theparallel lines is l0 in the static system and l in the reference frame of the electron.
We investigate the relationship between l0 and l. Inserting z 0 = 0, y 0 = l, andy = l0 into the new Lorentz transformationsy + (v/c)zh
y0 = √,
1 − v 2 /c2z − (v/c)yhz0 = √,
1 − v 2 /c2(10.5)(10.6)

we havel0 + (v/c)zhl= √,1 − v 2 /c2z − (v/c)l0 h0= √.
1 − v 2 /c2(16.14)(16.15)

From (16.15), we havez = (v/c)l0 h.

(16.16)

Substituting this result into (16.14), we obtainl0 − (v/c)2 l0
l= √1 − v 2 /c2√= l0 1 − v 2 /c2 .
According to this formula, if the velocity of the electron v approachs the velocity oflight c, l reduces to 0 in the reference frame of the electron, and light reaches thedetector within a very short time.
Conversely, in the reference frame of the stationary observer A, the light velocity
in the y-direction is c and the distance to the detector is l0 . Thus, light reaches thedetector in the time given by
l0
.
c
206

However, as shown in (16.16), the light velocity has a z-axial component in thenegative world, whose detectability is unknown.

16.7 Light and inertial law
Does the locus of the light emitted in the y 0 -direction from the electron become astraight line in the reference frame of the electron in Figure 16.1 (a) and (b)?

Since light has dual wave–particle characteristics, it is called a photon in a particle
context. In Figure 16.1, we ask whether the initial and final emitted photons canbe traced from the same y 0 -axis. If the photon has mass, it will travel on a verticalline in the electron’s reference frame. The situation is analogous to a ball thrown
upwards in a moving train, which returns to the hand that supplied the motion.
Since the law of inertia acts on particles with mass and the speed of the ball inthe direction of the moving train matches that of the train, the ball returns to its
original location. Similarly, if a photon has mass and travels with velocity v in thex-direction, the light (as seen by the electron) should trace a vertical line. However,according to special relativity, light possesses a kinetic mass, but no rest mass. Here,
we investigate the locus of the light emitted from an electron without imposing thelaw of inertia.
As explained in Section 8.2, the world distance of light is always 0.

Thus,

if the world points of light in the static and moving systems are expressed as(cth, xi, yj, zk) and (ct0 h, x0 i, y 0 j, z 0 k), respectively, we can write−(ct0 )2 + x02 + y 02 + z 02 = −(ct)2 + x2 + y 2 + z 2 = 0.

(16.17)

If the photons emitted in the y 0 -direction from an electron follow a straight line, aworld point on the locus is expressed as A(ct0 h, 0, ct0 j, 0). This point fulfills thecondition of the world line of light if its locus satisfies (16.17). Substituting A into

207

the left-hand side of (16.17), we have−(ct0 )2 + x02 + y 02 + z 02 = −(ct0 )2 + 0 + (ct0 )2 + 0= 0.
Thus, it is proved that the light emitted in the y 0 direction from an electron follows
a straight line in the electron’s reference frame
Consider a point B on the locus of light observed in the static system. In termsof the velocity components obtained from the new Lorentz transformation, i.e.,
Vx = v,(16.4)

Vy = c,(16.7)

Vz = vh,(16.8)

B is expressed as B(cth, vti, ctj, vthk). Substituting these coordinate components
into the right-hand side of (16.17), we obtain−(ct)2 + x2 + y 2 + z 2 = −(ct)2 + (vt)2 + (ct)2 + (vth)2= −(ct)2 + (vt)2 + (ct)2 − (vt)2= 0.
Therefore, the world line traced by B fulfills the condition of the world line of light.
This result shows that the new Lorentz transformation correctly interprets lightemitted perpendicularly from an electron.
Although light is massless in the above verification, we have realized the same
phenomenon as the law of inertia. Some readers may argue that the law of inertia
applies only to objects with mass. However, as explained in Section 9.3, the lawof inertia can be paraphrased as follows: if no force is applied, the world line of apoint mass is linear. In this context, mass is not required. In fact, Newton’s law ofinertia is precluded for massless entities, and should also apply to the kinetic massof light. However, the kinetic mass is not required in the above verification.

16.8 Upthrow of a ball
In Section 16.5, we mentioned that the light emitted in the y 0 -direction from an electron moving in the x-direction at near-light velocity is observed in the stationaryreference frame as lights emitted to both x- and y-directions. If a ball is perpendicularly thrown with velocity w in a train moving at velocity v, does the ball appear
as two entities to a stationary observer, as in the case of light?
208

The upthrow of a ball in a moving train is expressed by inserting Vx0 = 0, Vy0 = w,
and Vz0 = 0 into the equations used in Section 16.3, i.e.,
Vx − v,
1 − (v/c2 )Vx
Vy + (v/c)Vz h
Vy0 =,1 − (v/c2 )Vx
Vz − (v/c)Vy hVz0 =.
1 − (v/c2 )Vx

Vx0 =(10.25)(10.26)(10.27)

Substituting Vx0 = 0 into (10.25), we have
Vx = v,(16.18)

while substituting Vz0 = 0 into (10.27) yields
Vz = (v/c)Vy h.

(16.19)From (10.26), (16.18), (16.19), and Vy0 = w, we obtainw=

Vy − (v/c)2 Vy.
1 − (v/c2 )v

Thus,
Vy = w.

(16.20)In addition, from (16.19) and (16.20), we can write
Vz = (v/c)wh.

(16.21)

Collectively, the equations governing the upthrow of a ball in a moving train are
Vx = v,(16.18)

Vy = w,(16.20)Vz = (v/c)wh.

(16.21)

Using these formulae, we can investigate the gradient with respect to the coordinate
axis of the ball locus, as seen by the static observer. From (16.18) and (16.20), we
have
2
Vxy
= Vx2 + Vy2= v 2 + w2 ,√Vxy = v 2 + w2 .
209

(16.22)

From (16.20) and (16.21), we also have2
Vyz
= Vy2 + Vz2= w2 − (v/c)2 w2 ,√Vyz = w 1 − v 2 /c2 .

(16.23)

From (16.18) and (16.20), we obtain2
Vzx= Vx2 + Vz2= v 2 − (v/c)2 w2 ,√Vzx = v 1 − (w/c)2 .

(16.24)

The gradients with respect to the x-, y-, z-axes are obtained from (16.18), (16.20),(16.21), (16.22), (16.23), and (16.24) as follows:√√
w 1 − v 2 /c2Vyz= w 1/v 2 − 1/c2 ,(16.25)=Vxv√√
v 1 − (w/c)2
Vzx(16.26)== v 1/w2 − 1/c2 ,
Vy
w
√
√
Vxy
v 2 + w2
(16.27)== −ch 1/v 2 + 1/w2 .
Vz(v/c)whIf the ball is to appear as a single object in the stationary reference frame, (16.25)
and (16.26) must multiply to unity. In other words, (16.25) and (16.26) must be
the reciprocals of each other. Since this is not the case, a ball tossed upward in amoving train must be visible as two pieces from the static reference frame, besides
the (non-observable) component that appears in the negative world.
The ball appears in one piece because v and w are very much less than the velocity
of light c. When 1/c2 = 0, (16.25), (16.26), and (16.27) reduce to√Vyz= w 1/v 2 = w/v,
Vx√
Vzx
= v 1/w2 = v/w,
Vy√
Vxy
= −ch 1/v 2 + 1/w2 = −∞.
Vz
Since,(Vyz /Vx )(Vzx /Vy ) = (w/v)(v/w) = 1,
the ball converges to the same location along the x- and y-axes and is visible as one
object from a static perspective. The reciprocal of Vxy /Vz = −∞ is Vz /Vxy = 0,
implying that the ball resides on the x-y plane, and no contradiction occurs.
210

It is hard to believe that a ball thrown upwards at velocity w in a train moving at
velocity v moves perpendicularly in three directions, when observed by a stationary
observer. This idea could be experimentally tested by a suﬃciently precise detector.

211

212

17
Dynamics by the New Octonion

17.1 Formulae of the physical quantity that is constant by
coordinate transformation
In Section 3.5, we obtained the formulae of the new Lorentz transformation of timeand distance using the coordinate transformation B/|B| that makes the absolutevalue of world distance constant. Since we consider velocity, momentum, and acceleration in this chapter, we first obtain the conditions that should be fulfilled byphysical quantities whose absolute values become constant, even if they are changed
into the coordinates of the observer with linear uniform motion.
It is assumed that physical quantity P seen by stationary observer A has four
components (Eh, F i, Gj, Hk) in four-dimensional space–time . E is a temporalcomponent and F, G, and H are space components. Suppose that the coordinate
components of the physical quantity P are P 0 (E 0 h, F 0 i, G0 j, H 0 k), as seen from
B moving in the x-direction of A with uniform velocity v. The formulae of thecoordinate transformation that makes the absolute value of P constant, i.e., |P | =|P 0 |, can be obtained by multiplying P by B/ |B| when we assume the new octonionof observer B to be B, as explained in Section 3.5. Since B = cth + vti, we canwrite

P (cth − vti)
P0 = √.
(cth + vti)(cth − vti)

If c > v, substituting P = Eh + F i + Gj + Hk into this equation, we have
(Eh + F i + Gj + Hk)(cth − vti)√
(cth + vti)(cth − vti)1√=
cth 1 − (vti)2 /(cth)2()
× Ecth2 + F cthi + Gcthj + Hcthk − Evthi − F vti2 − Gvtji − Hvtki

P0 =

213

=

1
√
cth 1 − v 2 /c2
× (−Ect + F cthi + Gcthj + Hcthk − Evthi + F vt + Gvtk − Hvtj)

=

ct

√

−h
1 − v 2 /c2× [(−Ect + F vt) + (F ct − Evt)hi + (Gcth − Hvt)j + (Hcth + Gvt)k]1=√
1 − v 2 /c2
× [(E − F v/c)h + (F − Ev/c)i + (G + Hvh/c)j + (H − Gvh/c)k] .
Since P 0 = E 0 h + F 0 i + G0 j + H 0 k, by comparing the coeﬃcients, we find thatE − (v/c)F
E0 = √,
1 − v 2 /c2
F − (v/c)E
F0 = √,
1 − v 2 /c2
G + (v/c)Hh
G0 = √,
1 − v 2 /c2
H − (v/c)Gh
H0 = √,
1 − v 2 /c2(17.1)(17.2)(17.3)(17.4)

where (17.1), (17.2), (17.3), and (17.4) are formulae in which the physical quantitythat does not change in absolute value should fulfill, even if it is seen by observer
B undergoing linear uniform motion.
We can confirm |P | = |P 0 | by substituting (17.1), (17.2), (17.3), and (17.4) into
|P 0 |2 = (E 0 h + F 0 i + G0 j + H 0 k)(E 0 h − F 0 i − G0 j − H 0 k),
and having|P 0 |2 = −E 02 + F 02 + G02 + H 02
[
]2 [
]2 [
]2 [
]2
E − (v/c)F
F − (v/c)E
G + (v/c)Hh
H − (v/c)Gh
=− √
+ √
+ √
+ √
1 − v 2 /c2
1 − v 2 /c2
1 − v 2 /c2
1 − v 2 /c2= −E 2 + F 2 + G2 + H 2= (Eh + F j + Gj + Hk)(Eh − F j − Gj − Hk)= |P |2 .
Intermediate calculations are omitted.
The following three points are important regarding (17.1), (17.2), (17.3), and(17.4).
214

(1) The denominator is the constant√1 − v 2 /c2 , which does not contain the vari-ables t or x.
(2) The equation |P | = |P 0 | is realized even if we multiply P and P 0 by the same
constant.
(3) If the forms of (17.1), (17.2), (17.3), and (17.4) are used, any physical quantity
becomes constant under coordinate transformation. Thus, a certain physicalquantity must be written as (17.1), (17.2), (17.3), and (17.4) after proving byanother method that it is constant by coordinate transformation. We must notclaim that it is constant under coordinate transformation by using the previously
written forms of (17.1), (17.2), (17.3), and (17.4).
In special relativity, velocity, acceleration, momentum, and force, which do notchange their absolute values under coordinate transformation in four-dimensional
space–time, are called four-velocity, four-acceleration, four-momentum, and fourforce, respectively. After coordinate transformations, their denominators contain√1 − v 2 /c2 to make their absolute value constant, for reason (1). Moreover, forreason (2), their numerators are multiplied by the mass m0 , which is constant, and√their denominators contain several factors of 1 − v 2 /c2 several times.
About (3), an explanation is required. For instance, it is assumed that the quantity of heat Q has the coordinate components (Qt h, Qx i, Qy j, Qz k), and we write
the formulae as (17.1), (17.2), (17.3), and (17.4), and using these formulae, we claimthat the absolute value |Q| is constant under coordinate transformation. The causeof the mistake in this theory is having written the formulae of (17.1), (17.2), (17.3),
and (17.4) without proving the invariance of the absolute value |Q| of the quantityof heat by another method. In this example, since Q is a scalar and it is clearthat it does not have coordinate components, we recognize a mistake immediately.
However, in vector quantities such as velocity or acceleration, it is easy to commit
the above-mentioned mistake. A note is required.

17.2 Mass and coordinate transformations of velocity andmomentum of the x-axial motion
In this section, using the new octonion, we examine whether velocity is constantunder coordinate transformation. Suppose that observer B moves along the xdirection of stationary observer A with uniform velocity v. Moreover, the coordinates of the point mass D observed by A are (cth, xi, yj, zk) and those by B are(ct0 h, x0 i, y 0 j, z 0 k). At this time, the four-dimensional space–time diagram becomes that in Figure 17.1. Since the z-axis is not visible, it is drawn with a dashed215

line. Moreover, the velocity of D as observed by A is V (Vt h, Vx i, Vy j, Vz k) andthat by B is V 0 (Vt0 h, Vx0 i, Vy0 j, Vz0 k). However, V and V 0 are new octonions andnot vectors.

As explained in Section 10.5, the conversion equations of the velocities by the new
Lorentz transformation are
Vx − v,
1 − (v/c2 )Vx
Vy + (v/c)Vz h
Vy0 =,1 − (v/c2 )Vx
Vz − (v/c)Vy hVz0 =,1 − (v/c2 )Vx
Vx0 =(10.25)(10.26)(10.27)and the temporal component of the velocity Vt0 is
Vt0 =

cdt0
= c.
dt0

(17.5)The square of the absolute value of V 0 , including Vt0 , is|V 0 |2 = (Vt0 h + Vx0 i + Vy0 j + Vz0 k)(Vt0 h − Vx0 i − Vy0 j − Vz0 k)= −Vt02 + Vx02 + Vy02 + Vz02 .
As explained in Section 17.1, since the denominators of (10.25), (10.26), (10.27),√
and (17.5) do not contain 1 − v 2 /c2 , it is obvious that |V 0 |2 is not constant undercoordinate transformation, i.e., |V 0 |2 6= |V |2 . Citing only the calculation result, theproof is as follows:|V 0 |2 = −Vt02 + Vx02 + Vy02 + Vz02=

1 − v 2 /c2(1 − Vx2
v/c2 )()−Vt2 + Vx2 + Vy2 + Vz2 .

216

To solve this issue, we assume that observer B approaches the point mass D,
infinitesimally and Vx0 along the x-direction of D with respect to B is 0. At thattime, in the system of observer A, Vx along the x-direction of D and the velocity v
of B are the same. That is,Vx = v.

(17.6)

However, the velocities of B and D along the y- and z-axes need not be the same.
Moreover, (17.6) can also be obtained by substituting Vx0 = 0 into (10.25).
Substituting (17.6) into (10.26) and (10.27) yields
Vy + (v/c)Vz h,
1 − v 2 /c2
Vz − (v/c)Vy hVz0 =.
1 − v 2 /c2
Vy0 =(17.7)(17.8)

From (17.5), (17.6), (17.7), (17.8), Vx0 = 0, and Vt = cdt/dt = c, we obtain|V 0 |2 = (Vt0 h + Vx0 i + Vy0 j + Vz0 k)(Vt0 h − Vx0 i − Vy0 j − Vz0 k)= −Vt02 + Vx02 + Vy02 + Vz02
[
]2 [
]2
Vy + (v/c)Vz h
Vz − (v/c)Vy h22= −c + 0 ++1 − v 2 /c2
1 − v 2 /c2
1= −c2 +
(1 − v 2 /c2 )2][ 2
× Vy + 2(v/c)Vy Vz h + (v/c)2 Vz2 h2 + Vz2 − 2(v/c)Vy Vz h + (v/c)2 Vy2 h2[ 2]
1= −c2 +
Vy − (v/c)2 Vz2 + Vz2 − (v/c)2 Vy2222(1 − v /c )1
(1 − v 2 /c2 )(Vy2 + Vz2 )= −c2 +
(1 − v 2 /c2 )2
Vy2 + Vz2
= −c2 +
1 − v 2 /c2
1=(−c2 + v 2 + Vy2 + Vz2 )1 − v 2 /c2
1=(−Vt2 + Vx2 + Vy2 + Vz2 )1 − v 2 /c2
1=(Vt h + Vx i + Vy j + Vz k)(Vt h − Vx i − Vy j − Vz k)
1 − v 2 /c2
1|V |2 .
=
1 − v 2 /c2
The square root is1|V 0 | = √|V | .
1 − v 2 /c2
217(17.9)

From (17.9), we can see that, in four-dimensional space–time, the absolute valueof velocity is not constant under a coordinate transformation. Moreover, even if
the time components Vt = Vt0 = c of velocity are not included in a calculation of
the absolute value of velocity, |V | does not become constant under a coordinatetransformation. This can be easily proven by setting Vt and Vt0 to 0 in the above
calculations.
What can we understand from (17.9)? If both sides of (17.9) are multiplied by
the rest mass m0 of the point mass D, the equation becomesm0
m0 |V 0 | = √|V | .
1 − v 2 /c2(17.10)

The rest mass m0 is not the mass with respect to stationary observer A, but themass with respect to the point mass D itself, or the moving observer B whose
velocity along the x-direction is the same as that of D. That is, it is the mass inthe coordinate system in which D is observed at rest. On the other hand, the mass
with respect to observer A, who is moving with respcet to the point mass D, is the
kinetic mass m. Substitutingm0
m= √1 − v 2 /c2(17.11)m0 |V 0 | = m |V | .

(17.12)

into (17.10), we have

The left-hand-side of (17.12) is the product of mass and velocity with respect tothe observer B and the point mass D, who assume that the point mass D is at
rest along the x-direction. The right-hand-side is the product of mass and velocity
with respect to the observer A who is assumed to be moving with respect to thepoint mass D. Thus, (17.12) shows that the momentum of D is constant under acoordinate transformation. Moreover, from (17.11), we see that if the velocity v
increases, the kinetic mass m increases, and if v approaches the velocity of light c,
the mass becomes infinite.
The following two conclusions are obtained from the above results:(1) In three-dimensional space or four-dimensional space–time, the absolute valueof velocity is not constant under coordinate transformation.
(2) When the relation between the rest mass m0 and the kinetic mass m is
m0
m= √
,
1 − v 2 /c2
the momentum m |V |, including the temporal component, is constant under218

coordinate transformation.
Conclusion (2) can also be rewritten as follows:(3) If the momentum m |V |, including the temporal component, is constant undercoordinate transformation, the relation between the rest mass m0 and the kinetic
mass m is
m0
m= √.
1 − v 2 /c2This result indicates that if we do not have mass variance through movement orinvariance of momentum in axiom, we cannot obtain either law. Axiom is a lawthat is considered correct but cannot be proven. It is unknown which is an axiom.
Moreover, there is a possibility that the two above laws are obtained from anotheraxiom. At that time, they become theorems. We will discuss this possibility in
Section 18.1.
Next, we examine whether
m0 |V 0 | = m |V |(17.12)is correct. If we rewrite the right-hand-side of (17.12) with the new octonion, from
(17.6) and (17.11), we havem0√
(ch + vi + Vy j + Vz k) .
1 − v 2 /c2(17.13)

From (17.5), (17.7), (17.8), and Vx0 = 0, the left-hand-side of (17.12) can de rewrittenwith the new octonion as
{
[
]
[
] }
Vy + (v/c)Vz h
Vz − (v/c)Vy h
m0 ch + 0i +j+k .
1 − v 2 /c2
1 − v 2 /c2(17.14)

If (17.13) becomes (17.14) under coordinate transformation, it is a proof of thecorrectness of (17.12). As explained in Section 17.1, we can perform the coordinatetransformation by multiplying (17.13) by B/ |B|. The equations becomem0(cth − vti)√
(ch + vi + Vy j + Vz k) √1 − v 2 /c2(cth + vti)(cth − vti)
m0(cth − vti)√=√
(ch + vi + Vy j + Vz k)22
1 − v /ccth 1 − (v/c)2m0=(ch + vi + Vy j + Vz k) (1 − vi/ch)
1 − v 2 /c2
m0=(ch + vi + Vy j + Vz k) (1 + vhi/c)
1 − v 2 /c2()
m0=
ch + vi + Vy j + Vz k + vh2 i + v 2 hi2 /c + Vy vhji/c + Vz vhki/c2
2
1 − v /c()
m0=
ch + vi + Vy j + Vz k − vi − v 2 h/c − Vy vhk/c + Vz vhj/c2
2
1 − v /c
219

(
)
m0
ch − v 2 h/c + Vy j + Vz vhj/c + Vz k − Vy vhk/c
2
2
1 − v /c{}m0=c(1 − v 2 /c2 )h + [Vy + (v/c)Vz h] j + [Vz − (v/c)Vy h] k2
2
1 − v /c
]
] }
{
[
[
Vy + (v/c)Vz h
Vz − (v/c)Vy h= m0 ch + 0i +j+k .
1 − v 2 /c2
1 − v 2 /c2=

Since this result is identical to (17.14), we have confirmed that the momentum()
m00
m0 |V | = m |V |: m= √(17.15)1 − v 2 /c2
is constant under coordinate transformation using the new octonion. In (17.15),
velocity v of observer B is the same as x-axial component Vx of the velocity of thepoint mass D. However, the y-axial component Vy and the z-axial component Vz
have no limit. Thus, when the velocity v of B is in the x-direction of A, the changeof the kinetic mass m of D is independent of Vy and Vz .

17.3 Mass and coordinate transformations of velocity andmomentum of arbitrary axial motion
In this section, we investigate conservation of momentum and change of mass inthe case where the velocity v of observer B is along an arbitrary direction in threedimensional space. As explained in Section 10.8, when the velocity of observer B,
who is moving along a straight line with uniform velocity as seen from static observer
A, is v(vx , vy , vz ), the general formulae of the new Lorentz transformation are
t0 = √x0 = √
y0 = √
z0 = √

1
1 − v 2 /c2
1
1 − v 2 /c2
1
1 − v 2 /c2
1
1 − v 2 /c2(t − vx x/c2 − vy y/c2 − vz z/c2 ),(10.38)(x − vx t + vz yh/c − vy zh/c),(10.39)(y − vy t + vx zh/c − vz xh/c),(10.40)(z − vz t + vy xh/c − vx yh/c).

(10.41)Infinitesimal components are
dt0 = √
dx0 = √

1
1 − v 2 /c2
1
1 − v 2 /c2(dt − vx dx/c2 − vy dy/c2 − vz dz/c2 ),(17.16)(dx − vx dt + vz dyh/c − vy dzh/c),(17.17)220

1(dy − vy dt + vx dzh/c − vz dxh/c),
dy 0 = √1 − v 2 /c2
1
dz 0 = √(dz − vz dt + vy dxh/c − vx dyh/c).
1 − v 2 /c2(17.18)(17.19)By dividing (17.17), (17.18), and (17.19) by (17.16), we find thatdx0
dt0
dx − vx dt + vz dyh/c − vy dzh/c=
dt − vx dx/c2 − vy dy/c2 − vz dz/c2
dx/dt − vx + (vz h/c)dy/dt − (vy h/c)dz/dt=
1 − (vx /c2 )dx/dt − (vy /c2 )dy/dt − (vz /c2 )dz/dt
Vx − vx + (vz h/c)Vy − (vy h/c)Vz=,
1 − (vx /c2 )Vx − (vy /c2 )Vy − (vz /c2 )Vz
dy 0
Vy0 = 0dt
Vy − vy + (vx h/c)Vz − (vz h/c)Vx,=
1 − (vx /c2 )Vx − (vy /c2 )Vy − (vz /c2 )Vz
dz 0
Vz0 = 0dt
Vz − vz + (vy h/c)Vx − (vx h/c)Vy=.
1 − (vx /c2 )Vx − (vy /c2 )Vy − (vz /c2 )Vz

Vx0 =(17.20)(17.21)(17.22)

Intermediate calculations were omitted. In addition, the temporal component of
the velocity of D as seen from B is
Vt0 =

cdt0
= c.
dt0

(17.23)

The absolute square of V 0 , including Vt0 , is|V 0 |2 = (Vt0 h + Vx0 i + Vy0 j + Vz0 k)(Vt0 h − Vx0 i − Vy0 j − Vz0 k)= −Vt02 + Vx02 + Vy02 + Vz02 .
As explained in Section 17.1, since the denominators of (17.20), (17.21), (17.22), and√(17.23) do not contain 1 − v 2 /c2 , it is obvious that |V 0 |2 is not constant undercoordinate transformation, i.e., |V 0 |2 6= |V |2 .
We now consider the case where B coincides with the point mass D. It is thesame as that of the case where the velocity v of observer B is along the x-directionof the static observer A. Since the velocity V (Vx , Vy , Vz ) of D, as seen from A
becomes the same as the velocity v(vx , vy , vz ) of B, we can write
Vx = vx , Vy = vy , Vz = vz .
221

(17.24)

Since the velocity V 0 (Vx0 , Vy0 , Vz0 ) of D as seen from B is 0, we have
Vx0 = Vy0 = Vz0 = 0.

(17.25)

From (17.23) and (17.25), the equation becomes|V 0 |2 = −Vt02 + Vx02 + Vy02 + Vz02
= −c2 .

(17.26)

From Vt = cdt/dt = c, v 2 = vx2 + vy2 + vz2 , and (17.24), we find that|V |2 = −Vt2 + Vx2 + Vy2 + Vz2= −c2 + vx2 + vy2 + vz2= −c2 + v 2= −c2 (1 − v 2 /c2 ).

(17.27)

From (17.26) and (17.27), we have
|V |2 = |V 0 |2 (1 − v 2 /c2 ),1|V 0 |2 =|V |2 .
1 − v 2 /c2
Taking the square root, the equation becomes1|V 0 | = √|V | .
1 − v 2 /c2Thus, even in four-dimensional space–time, the velocity is not constant under coordinate transformation. Thus, if both sides of this formula are multiplied by therest mass m0 , we find thatm0
|V | .
m0 |V 0 | = √1 − v 2 /c2
If we assume the kinetic mass m is
m0
m= √
,
1 − v 2 /c2
we havem0 |V 0 | = m |V | .

(17.28)The point mass D moves with velocity v along an arbitrary direction as seen fromresting observer A. From (17.28), if the rest mass of D is assumed to be m0 andthe kinetic mass as m as seen from observer A, the kinetic mass is dependent on the222uniform velocity v in an arbitrary direction, and the momentum is constant undercoordinate transformation using the new octonion. The equation()
m00
m0 |V | = m |V |: m= √(17.15)1 − v 2 /c2
explained in Section 17.2, and (17.28) are the same. However, the velocity v of
(17.15) is the velocity in the x-direction and the velocity v of (17.28) is the velocity
that is composed of the x-, y-, and z-axial components.

17.4 Four-velocity and four-momentum
In the last section, it was proven that velocity is not constant under coordinatetransformation in four-dimensional space–time. In special relativity, the velocity
constant under coordinate transformation in four-dimensional space–time is called
four-velocity. And when explaining four-velocity, it is written as follows. If the√velocity is obtained using the proper time τ = t 1 − v 2 /c2 instead of the time tof each coordinate system, the absolute value of velocity becomes constant undercoordinate transformation. Thus, a static system and a kinetic system use thecommon time τ .
In this section, we prove that we can obtain the four-velocity of special relativity
using the new octonion. As explained in Section 17.2, the velocity of point mass D,
as seen from stationary observer A, is V (Vt h, Vx i, Vy j, Vz k) and the velocity ofD, as seen from observer B, who moves along a straight line in the x-direction of A
with uniform velocity v, is V 0 (Vt0 h, Vx0 i, Vy0 j, Vz0 k). When the velocity v of B andthe velocity in the x-direction of D, i.e., Vx , are the same as seen from A, we have1
|V 0 | = √|V | .
1 − v 2 /c2(17.9)

Thus, we cannot obtain a four-velocity that is constant under coordinate transformation. To solve this problem, although there is no rationale, Vt0 , Vx0 , Vy0 , and Vz0
are made using the proper time τ . At this time, to distinguish from Vt0 , Vx0 , Vy0 , andVz0 calculated using dt0 , four-velocities obtained using dτ are written as Ut0 , Ux0 , Uy0 ,and Uz0 . From the new Lorentz transformationst − (v/c2 )xt0 = √,
1 − v 2 /c2
x − vt,x0 = √
1 − v 2 /c2
223(10.3)(10.4)

y + (v/c)zh
y0 = √,
1 − v 2 /c2(10.5)z − (v/c)yhz0 = √,
1 − v 2 /c2(10.6)and Vt = cdt/dt, we have
Ut0 =

cdt0
dτ

cdt − c(v/c2 )dx√=√
1 − v 2 /c2 dt 1 − v 2 /c2
cdt/dt − (v/c)dx/dt=
1 − v 2 /c2
Vt − (v/c)Vx=,
1 − v 2 /c2
dx0
Ux0 =
dτ
dx − vdt√=√
2
1 − v /c2 dt 1 − v 2 /c2
dx/dt − (v/c)cdt/dt=
1 − v 2 /c2
Vx − (v/c)Vt=,
1 − v 2 /c2
dy 0
Uy0 =
dτ
dy + (v/c)dzh√=√
1 − v 2 /c2 dt 1 − v 2 /c2
dy/dt + (v/c)(dz/dt)h=
1 − v 2 /c2
Vy + (v/c)Vz h=,
1 − v 2 /c2
dz 0
Uz0 =
dτ
dz − (v/c)dyh√=√
1 − v 2 /c2 dt 1 − v 2 /c2
dz/dt − (v/c)(dy/dt)h=
1 − v 2 /c2
Vz − (v/c)Vy h.
=
1 − v 2 /c2(17.29)(17.30)(17.31)(17.32)

If the absolute value |U 0 | of velocity U 0 of the point mass D is calculated using224

(17.29), (17.30), (17.31), and (17.32), we find that
|U 0 |2 = (Ut0 h + Ux0 i + Uy0 j + Uz0 k)(Ut0 h − Ux0 i − Uy0 j − Uz0 k)= −Ut02 + Ux02 + Uy02 + Uz02
(
)2 (
)2
Vt − (v/c)Vx
Vx − (v/c)Vt=−+
1 − v 2 /c2
1 − v 2 /c2
(
)2 (
)2
Vy + (v/c)Vz h
Vz − (v/c)Vy h++
1 − (v/c)2
1 − v 2 /c2
1=
2(1 − v 2 /c2 )[× −Vt2 + 2(v/c)Vt Vx − (v/c)2 Vx2 + Vx2 − 2(v/c)Vt Vx + (v/c)2 Vt2=+ Vy2 + 2(v/c)Vy Vz h − (v/c)2 Vz2 + Vz2 − 2(v/c)Vz Vy h − (v/c)2 Vy21]2(1 − v 2 /c2 )
[ (
)
(
)
(
)
(
) ]
× − 1 − v 2 /c2 Vt2 + 1 − v 2 /c2 Vx2 + 1 − v 2 /c2 Vy2 + 1 − v 2 /c2 Vz2)(
1=−Vt2 + Vx2 + Vy2 + Vz222
1 − v /c1=
|V |2 .
1 − v 2 /c2Thus, if c > v, we can write|U 0 | = √1
1 − v 2 /c2|V | .

(17.33)

However, U 0 on the left side of (17.33) is the velocity divided by the infinitesimal
proper time dτ of the point mass D. V on the right-hand-side is the velocity divided
by the infinitesimal time dt of static observer A.
As explained previously, in special relativity, the proper time τ of the point mass
D is used instead of each observer’s time t. All velocities are calculated using theproper time τ and are called four-velocities. Thus, if the velocity V of D, as seenfrom static observer A, is recalculated using the infinitesimal proper time dτ andthe obtained four-velocity is written as U , the relation between Ux and Vx is
Ux =

dx
dτ

dx
√
dt 1 − v 2 /c2
1
=√
Vx .
1 − v 2 /c2=

225(17.34)

From the same method, Ut , Uy , and Uz become
11
1
Ut = √
Vt , Uy = √Vy , Uz = √
Vz .
2
2
2
2
1 − v /c
1 − v /c
1 − v 2 /c2(17.35)Thus, from (17.34) and (17.35), Equation (17.33) becomes|U 0 | = |U | .

(17.36)This indicates that four-velocity is constant under coordinate transformation. Moreover, if both sides of (17.36) are multiplied by the rest mass m0 of D, we havem0 |U 0 | = m0 |U | .

(17.37)Here, if m0 |U | is defined as four-momentum, it is constant under coordinate transformation. In addition, if (17.34) and (17.35) are applied toVt − (v/c)Vx,
1 − v 2 /c2
Vx − (v/c)Vt,
Ux0 =1 − v 2 /c2
Vy + (v/c)Vz hUy0 =,1 − v 2 /c2
Vz − (v/c)Vy hUz0 =,1 − v 2 /c2
Ut0 =(17.29)(17.30)(17.31)(17.32)

we can have the formulae of the coordinate transformation of four-velocity, i.e.,
Ut − (v/c)Ux,
Ut0 = √
1 − v 2 /c2
Ux − (v/c)Ut
Ux0 = √,
1 − v 2 /c2
Uy + (v/c)Uz h
Uy0 = √,
1 − v 2 /c2
Uz − (v/c)Uy h
Uz0 = √.
1 − v 2 /c2
The above verification in special relativity looks logical. However, it is not mathematical. The reason is that since the result is not contradictory, the proper time τ isused. Using the theory that is assumed correct since the desired result is obtained,
although the reason is unknown, we understand only a part of the reality.
As explained in Section 17.1, if the formulaeE − (v/c)F,
E0 = √
1 − v 2 /c2
226(17.1)

F − (v/c)E
F0 = √,
1 − v 2 /c2
G + (v/c)Hh
G0 = √,
1 − v 2 /c2
H − (v/c)Gh
H0 = √
1 − v 2 /c2(17.2)(17.3)(17.4)

are used, any physical quantity becomes constant under coordinate transformation.
Thus, after proving that a certain physical quantity is constant under coordinatetransformation using another method, we must put it into the form of (17.1), (17.2),(17.3), and (17.4). It is not logical that after making the transformation formulaeof four-velocity, i.e., (17.1), (17.2), (17.3), and (17.4), using the proper time τ , theinvariance under coordinate transformation is asserted based on these formulae.
Thus, in the next section, we explain the defects of four-velocity.

17.5 Defects of four-velocity
In Section 17.2, we proved that in four-dimensional space–time, velocity is notconstant under coordinate transformation. However, in the last section, we provedthat four-velocity is constant under coordinate transformation. Comparing the two
conclusions, we find that the theory of four-velocity and four-momentum has thefollowing four defects.
(1) In the theory of four-momentum, mass does not change by movement.
(2) In order to make four-velocity constant under coordinate transformation, velocity is calculated using the proper time τ . Thus, the coordinate transformation
invariance of four-velocity is not proven.
(3) The direction of the velocity v of observer B is in the x-direction of observer A,
and the direction of the velocity of the point mass D is in an arbitrary direction
in four-dimensional space–time. However, the case where B and D are in agreement is considered.
(4) For observer B moving together with the point mass D, the velocity of D must
be 0. However, velocity U (Ut0 , Ux0 , Uy0 , Uz0 ) of D observed by B is considered.
Next, we examine the above-mentioned defects of four-velocity and four-momentum.
Defect (1)From
m0 |U 0 | = m0 |U | ,227(17.37)

mass is always constant. It is proven in special relativity that mass changes according to velocity. Thus, it is considered that four-velocity is an expedient concept that
does not express the property of space–time.
Defect (2)As explained in Section 17.1, the coordinate transformation invariance of fourvelocity must be proved by another method besides by the formulae of coordinatetransformation. In order to prove that there is no mathematical basis for using theproper time τ of the point mass D in calculations of all velocities, we prove that byusing time t of stationary observer A instead of the proper time τ , four-velocity canbe made constant under coordinate transformation. As explained in Section 10.5,
the infinitesimal quantities of the new Lorentz transformations are
dt − (v/c2 )dx
dt0 = √,
1 − v 2 /c2dx − vdt
dx0 = √,
1 − v 2 /c2
dy + (v/c)dzh
dy 0 = √,1 − v 2 /c2dz − (v/c)dyh.
dz 0 = √1 − v 2 /c2(10.21)(10.22)(10.23)(10.24)

If (10.21), (10.22), (10.23), and (10.24) are divided by the infinitesimal time dt ofobserver A, we have
cdt0
dt
cdt − (v/c)dx= √dt 1 − v 2 /c2c − (v/c)dx/dt= √1 − v 2 /c2c − (v/c)Vx=√,
1 − v 2 /c2
dx0
Vx0 =
dt
dx − vdt= √dt 1 − v 2 /c2
dx/dt − v=√
1 − v 2 /c2
Vx − v,=√
1 − v 2 /c2
Vt0 =

228

(17.38)(17.39)dy 0
dt
dy + (v/c)dzh= √dt 1 − v 2 /c2
dy/dt + (v/c) (dz/dt)h√=
1 − v 2 /c2
Vy + (v/c)Vz h,= √1 − v 2 /c2
dz 0
Vz0 =
dt
dz − (v/c)dyh= √dt 1 − v 2 /c2
dz/dt − (v/c)(dy/dt)h√=
1 − v 2 /c2 t
Vz − (v/c)Vy h= √.
1 − v 2 /c2
Vy0 =(17.40)(17.41)

Next, if we calculate |V 0 |2 using (17.38), (17.39), (17.40), and (17.41), we findthat
|V 0 |2 = (Vt0 h + Vx0 i + Vy0 j + Vz0 k)(Vt0 h − Vx0 i − Vy0 j − Vz0 k)= −Vt02 + Vx02 + Vy02 + Vz02
[
]2 [
]2
c − (v/c)Vx
Vx − v=− √+ √1 − v 2 /c2
1 − v 2 /c2
[
]2 [
]2
Vy + (v/c)Vz h
Vz − (v/c)Vy h+ √+ √1 − v 2 /c2
1 − v 2 /c2
1=
1 − v 2 /c2[
× −c2 + 2c(v/c)Vx − (v/c)2 Vx2 + Vx2 − 2vVx + v 2+ Vy2 + 2(v/c)Vy Vz h − (v/c)2 Vz2 + Vz2 − 2(v/c)Vz Vy h − (v/c)2 Vy2 ]
]
[ 2
1
=
−c − (v/c)2 Vx2 + Vx2 + v 2 + Vy2 − (v/c)2 Vz + Vz2 − (v/c)2 Vy222
1 − v /c][ 2
1=
−c (1 − v 2 /c2 ) + Vx2 (1 − v 2 /c2 ) + (Vy2 + Vz2 )(1 − v 2 /c2 )1 − v 2 /c2= −c2 + Vx2 + Vy2 + Vz2 .
Since Vt = cdt/dt = c, the above equation becomes|V 0 |2 = −Vt2 + Vx2 + Vy2 + Vz2229

= (Vt h + Vx i + Vy j + Vz k)(Vt h − Vx i − Vy j − Vz k)= |V |2 .
Thus, we have
|V 0 | = |V | .
From this formula, we see that even if we calculate all velocities using the time t ofstationary observer A instead of the proper time τ of the point mass D, velocities
become constant under coordinate transformation. That is, there is no reason forhaving to use the proper time τ when we obtain the four-velocity.
As explained in Section 17.1, if the denominator of the formula of a certain phys√
ical quantity has a factor of 1 − v 2 /c2 , the physical quantity becomes constantunder coordinate transformation. Since the formula for the velocity obtained using
the proper time τ of the point mass D, and the formula for the velocity obtained√using the time t of stationary observer A, both have 1 − v 2 /c2 in the denominator, they become constant under coordinate transformation. Without proving thecoordinate transformation invariance using another method, the invariance of thecoordinate transformation of velocity and four-momentum are asserted by using the√formula that has 1 − v 2 /c2 in the denominator. Thus, we have the contradictory
result that when using four-velocity, mass does not change by movement.
Defect (3)The Lorentz transformation of special relativity does not have general formulae forthe case of the velocity of observer B being in an arbitrary direction. The directionof velocity of B is always in the x-direction of stationary observer A. The y- and zaxial components of the velocity of observer B do not exist. However, the velocity ofthe point mass D has y- and z-axial components. In the process in which the Lorentz
transformation is applied to four-velocity, the direction of the velocity v changes toan arbitrary direction without verification. On the other hand, in Section 17.3, we
proved correct the formulae of four-velocity, in the case of velocity v of observer B
being in an arbitrary direction, using the new Lorentz transformation.
Defect (4)In special relativity, it is assumed that although observer B is approaching the point
mass D, infinitesimally, B is not in accordance. Since they are not in agreement, it
is assumed that four coordinate components of the velocity of D can be observed by
B. However, since B is in agreement with D, the proper time τ of D is used whenfinding the velocity. Thus, this theory is not mathematical. Moreover, velocity is230

obtained by assuming that according to observer B moving with the point massD, D moves an infinitesimal distance ds(cdt, dx, dy, dz) at a certain moment.
This theory does not have a clear basis either.
As shown in Section 17.3, from a purely mathematical standpoint, velocity is
not constant under coordinate transformation but momentum is. At that time,
mass varies with velocity. The reason for the existence of coordinate transformation
invariance of momentum, the change of mass according to velocity, and the meaningof the temporal component of velocity, will be discussed in Section 18.1. From now
on, in this book, when velocity and acceleration are calculated, we use proper timeτ for point masses and time t for the observer. Only in the case of four-velocity,
four-acceleration, four-momentum, and four-force of special relativity, the proper
time τ is used for calculations of all velocities.

17.6 Conservation of momentum
In Section 17.3, we explained the coordinate transformation invariance of the momentum. In this section, we explain a law of conservation of momentum. The
coordinate transformation invariance of momentum means that the absolute valueof momentum is constant in any coordinate system. On the other hand, the law ofconservation of momentum means that the sum of each coordinate component of
momentum is constant before and after movement. Although two terms are alike,semantics diﬀer. The law of conservation of momentum is proven by special relativity. In this section, the law of conservation of momentum is proved using thenew octonion. Moreover, although the case where a point mass and an observer
move together in three-dimensional space is treated in special relativity, using thenew octonion, we consider the case where the velocity in the x-direction of a point
mass and an observer’s velocity are in agreement. However, in order to simplify, theimaginary numbers i, j, and k are not used, but the imaginary number h is used
when the component in the negative world is expressed.
As shown in Figure 17.2, it is assumed that a point mass A with rest mass M0
is at rest at origin O, which is the origin of the x-, y-, and z-coordinates of restingobserver O. The z-axis is omitted. Next suppose that A is divided into point
masses B and D with rest masses m0 by an internal explosion. It is assumed thatthe fission takes place in the x-y plane. If the velocity of B is v(vx , vy , 0) at thattime, the velocity of D is −v(−vx , −vy , 0) by the symmetry property. In addition,
suppose that observer O0 who moves along the x-axis at the same velocity as vx in231

x-direction of B. The coordinates axes of O0 are the x0 -, y 0 -, z 0 -axes. However, thez 0 -axis is omitted in Figure 17.2.

Since M0 and m0 are rest masses, we assume the kinetic masses to be M and m,
respectively. As explaned in Section 17.2, the relation between kinetic mass andrest mass isM0M=√,
1 − v 2 /c2
m0
m= √
.
1 − v 2 /c2(17.11)

However, v is a three-dimensional space velocity and is not four-velocity.
First, we prove the law of conservation of momentum as seen by stationary observer O. The velocity of A before the fission is (0, 0, 0). The velocity of B after
the fission is v(vx , vy , 0) and the velocity of D is −v(−vx , −vy , −0). Therefore,
the sum of the momenta of the x-, y-, and z-axes before and after the fission arecalculated by
M0 × 0 = √M0 × 0 = √M0 × 0 = √m0
1 − (vx2 + vy2 )/c2

vx + √m0
1 − (vx2 + vy2 )/c2(−vx ),m0

m0
vy + √(−vy ),
1 − (vx2 + vy2 )/c2
1 − (vx2 + vy2 )/c2
m0

m0
×0+ √
× 0.
1 − (vx2 + vy2 )/c2
1 − (vx2 + vy2 )/c2

Thus, the law of conservation of momentum is realized about the x-, y-, and zcomponents.
Next, we investigate whether or not the law of conservation of momentum isrealized as seen by observer O0 who moves in the x-direction with velocity vx . As
232

explained in Section 10.5, if the velocity of a certain point mass is observed as
V (Vx , Vy , Vz ) by stationary observer O, and as V 0 (Vx0 , Vy0 , Vz0 ) by observer O0 ,who is moving in the x-direction with velocity vx , the relation between V and V 0 is
Vx − vx,
1 − (vx /c2 )Vx
Vy + (vx /c)Vz h
Vy0 =,1 − (vx /c2 )Vx
Vz − (vx /c)Vy hVz0 =.
1 − (vx /c2 )Vx

Vx0 =(10.25)(10.26)(10.27)In Figure 17.2, the velocity of D as seen by O0 is v 00 (vx00 , vy00 , vz00 ). Since the velocity
of D as seen by O is −v(−vx , −vy , 0), by replacing Vx → −vx , Vy → −vy , Vz →0, Vx0 → vx00 , Vy0 → vy00 , and Vz0 → vz00 in (10.25), (10.26), and (10.27), we find that−vx − vx
1 + (vx /c2 )vx−2vx=,
1 + vx2 /c2−vyvy00 =,
1 + vx2 /c2(vx vy /c)hvz00 =.
1 + vx2 /c2
vx00 =(17.42)(17.43)(17.44)

From (17.42), (17.43), and (17.44), we havev 002 = vx002 + vy002 + vz002=

4vx2 + vy2 − vx2 vy2 /c2.
(1 + vx2 /c2 )2(17.45)

Then, the velocity of B as seen by O0 is v 0 (vx0 , vy0 , vz0 ). Since the velocity of B as
seen by O is v(vx , vy , 0), by replacing Vx → vx , Vy → vy , Vz → 0, Vx0 → vx0 , Vy0 →vy0 , and Vz0 → vz0 in (10.25), (10.26), and (17.27), we find thatvx0 =vx − vx
1 − vx2 /c2= 0,(17.46)vy,
1 − vx2 /c2−(vx vy /c)h.
vz0 =1 − vx2 /c2

vy0 =From (17.46), (17.47), and (17.48), we have233

(17.47)(17.48)

v 02 = vx02 + vy02 + vz02
vy2 − vx2 vy2 /c2(1 − vx2 /c2 )2vy2 (1 − vx2 /c2 )=(1 − vx2 /c2 )2vy2=.
1 − vx2 /c2=(17.49)In addition, as shown in Figure 17.3, we assume that the velocity of A from O0
before the fission is v 000 (vx000 , vy000 , vz000 ). Since the velocity of A as seen by O isv(0, 0, 0), by replacing Vx → 0, Vy → 0, Vz → 0, Vx0 → vx000 , Vy0 → vy000 , and
Vz0 → vz000 in (10.25), (10.26), and (10.27), we find that
vx000 = −vx ,(17.50)vy000 = 0,(17.51)vz000 = 0.

(17.52)

From (17.50), (17.51), and (17.52), we havev 0002 = vx0002 + vy0002 + vz0002= vx2 .

(17.53)

If (17.45), (17.49), and (17.53) are written collectively, we have4vx2 + vy2 − vx2 vy2 /c2,(1 + vx2 /c2 )2
vy2
v 02 =,
1 − vx2 /c2v 002 =

v 0002 = vx2 .

(17.45)(17.49)(17.53)234

Using these velocities, we can calculate the kinetic mass of the law of conservationof momentum below.
Before examining the momentum conservation law in the x-direction as seen by
O0 , we examine the momentum conservation law in the y- and z-directions. The
momentum in the y-direction of A before the fission as seen by O0 is p000
y , and themomenta in the y-direction of B and D after the fission are p0y and p00y , respectively.
If we replace v 2 in the general formula of momentum, i.e.,py = √m0 vy
1 − v 2 /c2,

byv 02 =vy2
1 − vx2 /c2(17.49)and replace vy with the previously obtained expression
vy0 =

vy
,
1 − vx2 /c2(17.47)

we find thatvym0×p0y = √[]
1 − vx2 /c2
1 − vy2 /(1 − vx2 /c2 ) /c2
m 0 vy=√[](1 − vx2 /c2 )2 − (1 − vx2 /c2 )2 vy2 /(1 − vx2 /c2 ) /c2
m0 vy=√(1 − vx2 /c2 )2 − (1 − vx2 /c2 )vy2 /c2m 0 vy=√[](1 − vx2 /c2 ) (1 − vx2 /c2 ) − vy2 /c2m 0 vy=√.
(1 − vx2 /c2 )(1 − vx2 /c2 − vy2 /c2 )(17.54)

Similarly, from the previously obtained expression−vy,
1 + vx2 /c2
4vx2 + vy2 − vx2 vy2 /c2v 002 =,
(1 + vx2 /c2 )2vy00 =(17.43)(17.45)

we find that
p00y = √1−[

m0(4vx2 + vy2 − vx2 vy2 /c2 )/(1 + vx2 /c2 )2
235

]

/c2

×

−vy
1 + vx2 /c2

−m0 vy=√2
2
2(1 + vx /c ) − (4vx2 /c2 + vy2 /c2 − vx2 vy2 /c4 )−m0 vy=√1 + 2vx2 /c2 + vx4 /c4 − 4vx2 /c2 − vy2 /c2 + vx2 vy2 /c4−m0 vy=√22(1 − vx /c )2 − vy2 /c2 (1 − vx2 /c2 )−m0 vy.
=√(1 − vx2 /c2 )(1 − vx2 /c2 − vy2 /c2 )(17.55)

Similarly, from the previously obtained expression
vy000 = 0,v

0002

(17.51)= vx2 ,

(17.53)

we find that
p000
y = √

M0
1 − vx2 /c2

×0

= 0.

(17.56)

From (17.54), (17.55), and (17.56), we have
p0y + p00y = 0 = p000y .
Thus, the momentum conservation law in the y-direction is realized. Since theequation of the momentum in the z-direction becomes the formula obtained byreplacing vy of the numerator of (17.54), and (17.55) by −vx vy h/c, we can write
p0z + p00z = 0 = p000z .
Thus, the momentum conservation law in the z-direction is also realized.
Next, we investigate the momentum conservation law in the x-direction. The
momentum in the x-direction of A before the fission as seen by O0 is p000
x , and themomenta in the x-direction of B and D after the fission are p0x and p00x , respectively.
From the previously obtained expression
vx0 = 0,(17.46)vy2
v 02 =,
1 − vx2 /c2(17.49)236

we find thatp0x = √m0× vx0
1 − v 02 /c2
m0
=√
×0
[
]
2
1 − vy /(1 − vx2 /c2 ) /c2= 0.

(17.57)

Similarly, from the previously obtained expression−2vx,
1 + vx2 /c2
4vx2 + vy2 − vx2 vy2 /c2,v 002 =
(1 + vx2 /c2 )2
vx00 =(17.42)(17.45)

we find that
p00x = √
=√
=√
=√
=√

m0

[

]

1 − (4vx2 + vy2 − vx2 vy2 /c2 )/(1 + vx2 /c2 )2 /c2

×−2vx
1 + vx2 /c2−2m0 vx(1 + vx2 /c2 )2 − (4vx2 /c2 + vy2 /c2 − vx2 vy2 /c4 )−2m0 vx
1 + 2vx2 /c2 + vx4 /c4 − 4vx2 /c2 − vy2 /c2 + vx2 vy2 /c4−2m0 vx(1 − vx2 /c2 )2 − vy2 /c2 (1 − vx2 /c2 )−2m0 vx(1 − vx2 /c2 )(1 − vx2 /c2 − vy2 /c2 ).

(17.58)Similarly, from the previously obtained expression
vx000 = −vx ,(17.50)v 0002 = vx2 ,(17.53)

we find that
p000
x = √

M0

× (−vx )1 − vx2 /c2−M0 vx=√.
1 − vx2 /c2(17.59)From (17.57), (17.58), and (17.59), we have
−2m0 vx−M0 vxp0x + p00x − p000
−√
.
x = √
2
2
2
2
2
2
1 − vx2 /c2(1 − vx /c )(1 − vx /c − vy /c )237

(17.60)We now consider what is necessary for (17.60) to become 0. If we assume
p0x + p00x − p000
x = 0,
from (17.60), we have√

2m0 vx(1 − vx2 /c2 )(1 − vx2 /c2 − vy2 /c2 )

=√

M0 vx
1 − vx2 /c2

,

2m0
√
= M0 .
2
1 − vx /c2 − vy2 /c2(17.61)Since the velocity v of point mass B is v = vx2 + vy2 + 0, (17.61) becomes√

2m0
1 − v 2 /c2= M0 .

(17.62)

When v is infinitesimal compared to c, from1√
+ 1 − (v 2 /c2 )(−1/2)
1 − v 2 /c2= 1 + (v 2 /c2 )/2,
(17.62) becomes
2m0 + 2m0 (v 2 /c2 )/2 + M0 .
By multiplying both side by c2 , the equation becomesM0 c2 + 2m0 c2 + 2m0 v 2 /2.

(17.63)

As proven in Einstein’s special relativity, if the rest mass m0 is changed into energy
E, we have
E = m0 c2 .
The left-hand-side of (17.63) expresses the rest energy of A before fission. The firstpart of the right-hand-side expresses the sum of the rest energies of B and D. The
second part expresses the sum of the kinetic energies of the Newtonian mechanicsof B and D. That is, (17.63) indicates that the rest energy of the point mass Achanges to sum of the rest energies and kinetic energies of the point masses B and
D. The conclusion obtained from this verification is the same as the conclusion of
special relativity. This shows that a contradiction does not occur in calculations
using the new octonion.
In the above discussion, velocities are not the four-velocities using the proper
time τ but the velocities using the time t of each observer. The reason is that if
238four-velocity is used, mass does not change according to the velocity. At that time,
we cannot conclude that the rest mass M0 becomes two times the rest mass m0 ,and that the energy becomes two times the kinetic energy m0 v 2 /2 after fission.
Also from this, it is concluded that four-velocity does not represent a real physicalphenomenon and is the quantity invented for the convenience of calculations. In
addition, there may be some readers who assume that the energy added before
fission converts into kinetic energy after fission. However, even if we use gunpowder
to trigger fission, the mass of the gunpowder is contained in the rest mass M0 , andwe can consider that mass changed to energy. The energy used for fission is not
added from external sources.

17.7 Acceleration and four-acceleration
In this section, the acceleration as seen by the observer who is undergoing uniform
motion is considered. Calculations are performed by the new octonion.

In Figure 17.4, observer B is moving in the x-direction of static observer E with aconstant velocity v. The point mass D as seen by B is exerting the force that causes
the acceleration movement. When B approaches and meets with D, B cannot measure the acceleration of D. This is because acceleration cannot be measured byoneself. The famous Gedanken experiment of general relativity has a freely falling
elevator. The idea is that in the elevator, whose tie is broken and is undergoing
free fall, a person inside cannot feel any gravitational acceleration because the position between the elevator and the person does not change. This idea is termed
equivalence principle in general relativity. However, in special relativity, an observer
measures its own acceleration, which is called proper acceleration. To solve this contradiction, in special relativity, the case where B infinitesimaly aproaches D is notconsidered, whereas the case in which B, moving together with D, stands still at a239

certain moment and measures the acceleration of D is considered. This is called an
instantaneous static system. However, in this book, the case where B infinitesimaly
aproaches D is considered. At that time, the time t0 infinitesimaly approaches the√√proper time τ = t 1 − v 2 /c2 of D. Thus, we use dτ = dt 1 − v 2 /c2 instead of dt0 .
Since the new Lorentz transformations aret − (v/c2 )xt0 = √,
1 − v 2 /c2
x − vtx0 = √,
1 − v 2 /c2
y + (v/c)zh
y0 = √,
1 − v 2 /c2z − (v/c)yhz0 = √,
1 − v 2 /c2(10.3)(10.4)(10.5)(10.6)

the infinitesimal quantities are
dt − (v/c2 )dx
dt0 = √,
1 − v 2 /c2dx − vdt
dx0 = √,
1 − v 2 /c2
dy + (v/c)dzh,
dy 0 = √1 − v 2 /c2dz − (v/c)dyh
dz 0 = √.
1 − v 2 /c2(10.21)(10.22)(10.23)(10.24)

To obtain the proper velocities, we divide (10.21), (10.22), (10.23), and (10.24) by√infinitesimal proper time dτ = dt 1 − v 2 /c2 , and find that
Vt0 =

cdt0
dτ

cdt − c(v/c2 )dx√=√
1 − v 2 /c2 dt 1 − v 2 /c2
cdt/dt − (v/c)dx/dt=
1 − v 2 /c2c − (v/c)Vx=,
1 − v 2 /c2
dx0
Vx0 =
dτ
dx − vdt√=√
2
1 − v /c2 dt 1 − v 2 /c2

240(17.64)

dx/dt − (v/c)cdt/dt
1 − v 2 /c2
Vx − v=,
1 − v 2 /c2
dy 0
Vy0 =
dτ
dy + (v/c)dzh√=√
1 − v 2 /c2 dt 1 − v 2 /c2
dy/dt + (v/c) hdz/dt=
1 − v 2 /c2
Vy + (v/c)Vz h=,
1 − v 2 /c2
dz 0
Vz0 =
dτ
dz − (v/c)dyh√=√
1 − v 2 /c2 dt 1 − v 2 /c2
dz/dt − (v/c) hdy/dt=
1 − v 2 /c2
Vz − (v/c)Vy h=.
1 − v 2 /c2=(17.65)(17.66)(17.67)

However, Vx , Vy , and Vz are the velocities found using the infinitesimal time dt ofobserver E.
In the case where observer B approaches point mass D, infinitesimally, the acceleration a0 of D is also calculated by dividing V 0 by the proper time τ . However,
the acceleration of D as seen by static observer E is obtained by dividing V by thetime t, and is written as a. From (17.64), (17.65), (17.66), and (17.67), we find that
a0t =

dVt0
dτ

0 − (v/c)dVx√(1 − v 2 /c2 )dt 1 − v 2 /c2−(v/c)dVx /dt=(1 − v 2 /c2 )3/2−(v/c)ax=,(1 − v 2 /c2 )3/2
dVx0
a0x =
dτ
dVx − 0√=2
(1 − v /c2 )dt 1 − v 2 /c2
dVx /dt=(1 − v 2 /c2 )3/2=

241(17.68)ax,(1 − v 2 /c2 )3/2
dVy0
a0y =
dτ
dVy + (v/c)hdVz√=(1 − v 2 /c2 )dt 1 − v 2 /c2
dVy /dt + (v/c) hdVz /dt=(1 − v 2 /c2 )3/2
ay + (v/c)az h=,(1 − v 2 /c2 )3/2
dVz0
a0z =
dτ
dVz − (v/c)hdVy√=
1 − v 2 /c2 dt 1 − v 2 /c2
az − (v/c) ay h=.
(1 − v 2 /c2 )3/2=

(17.69)(17.70)(17.71)

Next, using the new octonion a0 = a0t h + a0x i + a0y j + a0z k and a = at h + ax i + ay j +az k to express accelerations, we examine whether acceleration is constant undercoordinate transformation. From (17.68), (17.69), (17.70), and (17.71), we find that|a0 |2 = (a0t h + a0x i + a0y j + a0z k)(a0t h − a0x i − a0y j − a0z k)
02
02
02= −a02t + ax + ay + az
{
}
1
2
2
2 2
2
=
−(v/c)a+a+[a+(v/c)a
h]
+
[a
−
(v/c)
a
h]
y
z
z
y
x
x
(1 − v 2 /c2 )3[]1=−(v/c)2 a2x + a2x + a2y − (v/c)2 a2z + a2z − (v/c)2 a2y2
23(1 − v /c )1
(1 − v 2 /c2 )(a2x + a2y + a2z )=(1 − v 2 /c2 )31=(a2 + a2y + a2z ).
(1 − v 2 /c2 )2 x

Since at = dVt /dt = 0 from Vt = cdt/dt = c, the above equation becomes
1(0 + a2x + a2y + a2z )(1 − v 2 /c2 )2
1(−a2t + a2x + a2y + a2z )=(1 − v 2 /c2 )21=(at h + ax i + ay j + az k)(at h − ax i − ay j − az k)(1 − v 2 /c2 )2
1|a|2 .
=(1 − v 2 /c2 )2|a0 |2 =242

Thus, we have|a|,
(17.72)
1 − v 2 /c2
which indicates that an acceleration is not invariant under coordinate transformation.
|a0 | =In special relativity, the acceleration as seen by the static system is also calculatedusing the proper time τ of a moving point mass. This is called four-acceleration, andis written as A. Four-acceleration is invariant under coordinate transformation likefour-velocity. We explain this below. As explained in Section 17.4, four-velocity Ux ,which is obtained by calculating all velocities using proper time τ , is Ux = dx/dτ .
Thus, the acceleration A, as seen from the static system isdUxdτd(dx/dτ )=dτ √d(dx/dt 1 − v 2 /c2 )√=dt 1 − v 2 /c2
1d(dx/dt)=
×2
2
1 − v /c
dt
1
dVx
=
×
1 − v 2 /c2
dt
ax
=
.
1 − v 2 /c2

Ax =Similary, the other proper accelerations At , Ay , and Az become
at
,
At =
1 − v 2 /c2
ay
Ay =
,
1 − v 2 /c2
az
Az =
.
1 − v 2 /c2
In addition, since a0t , a0x , a0y , and a0z are obtained using the proper time τ , we canwrite
A0t = a0t , A0x = a0x , A0y = a0y , A0z = a0z .
Thus, (17.72) becomes|A0 | = |A| .

(17.73)

However, A and A0 are not vectors, but new octonions expressing four-acceleration.
Therefore, four-acceleration is constant under coordinate transformation. The above
result is in agreement with the result of special relativity. It indicates that a contradiction does not occur in calculations using the new octonion.
243

244

18
New Octonion and Mass

18.1 Mass is a time component of the unit world line
Newtonian mechanics is underpinned by two main assumptions: existence of stationary space and homogeneity of time. Movement is conceptualized in stationary
space as a diﬀerence between points in the space. Moreover, time passes uniformly
everywhere in space, and is identical for two or more observers. However, as Euclid
wrote in Elements, if the rules of nature are divided into axioms and theorems, stationary space and homogeneous time are imprecise concepts. Although consistent
with our experience, both premises are mathematical axioms rather than theorems.
Because an axiom is experientially considered as correct and cannot be proven, it
may be denied anytime.
Diﬀering from Newton, Einstein wondered how light would appear to an observer
traveling at velocity of light. Subsequently, he derived the Lorentz transformation,in which the time and distance of a point mass depend on the observer’s velocity. Einstein questioned the status of stationary space and homogeneity of time asaxioms. As explained in Section 13.1, questioning axioms leads to novel scientificconcepts.
Einstein’s relativity theory was seeded by his denial of stationary space, i.e.,
absolute space, and homogeneity of time, i.e., absolute time, assumed in Newtonianmechanics. However, he embraced the concept of absolute mass. He presumed
the existence of mass itself, without querying its source. In Einstein’s formulation,
mass is treated as another physical quantity that depends on time t and distance x.
Before mass and energy were equated in Einstein’s famous formula, E = mc2 , they
were related in the classical formula E = mv 2 /2. Both formulations associate massm, time t, and distance x through an equivalent relationship. Similarly, Einstein’s
formulam0,(17.11)m= √1 − v 2 /c2245which expresses the velocity-dependent change in kinetic mass, does not elucidate
the fundamental meaning of mass. In general relativity, mass is the pre-existententity that bends space and creates gravity. In other words, relativity theory adopts
the axiom that mass exists without questioning its origin. Thus, similar to Newton’s
acceptance of absolute space and absolute time, Einstein accepted absolute mass.
However, axioms tend to be improved throughout their history. In this section,
we posit an axiom that mass is not an absolute quantity; rather, it is a time component of a unit world line. Subsequently, we investigate whether this axiom admits
the velocity-dependent mass formulation of (17.11). In special relativity, since nocontradiction is found, the energy is related to the time component pt of the fourdimensional momentum multiplied by the velocity of light c. We test our axiom on
this theory also.
Consider an elementary particle B (constituting substance) moving along thepositive x-direction of a stationary observer A with uniform velocity v as shown in
Figure 18.1. The four-dimensional space–time diagram in this situation is shown in
Figure 18.2, where the y- and z-axes are omitted.

The line parallel to the cth-axis drawn from B intercepts the xi-axis at node D.
The equation of the world line of B is x = vt. The new octonions of points A, B,
and D are A = cth, B = cth + vti, and D = vti, respectively. The distance between
the origin O and B is |OB|. If c > v, we find that√
|OB| = BB√= (cth + vti)(cth − vti)√
= cth 1 − v 2 /c2 .
(18.1)The length of a unit world line is denoted by η (eta). However, it is a real number.
When |OB| is the length of a unit world line, the coordinates of B are assumed as
B0 (ct0 h, vt0 i). Since |OB0 | in (18.1) is an imaginary number, |OB0 | is η × h = ηh.
From (18.1), we obtain√ηh = ct0 h 1 − v 2 /c2 .
246

Thus, we find thatη.
t0 = √
c 1 − v 2 /c2(18.2)Let A0 be the coordinates of A when |OB| is the length of a unit world line. From
(18.2), we find that
A0 = ct0 h=√ηh
1 − v 2 /c2.

(18.3)

If the rest mass m0 comprises N elementary particles B (see Figure 18.3), B traces
N world lines. Hereafter, if B is associated with a single world line, it is referred
to as an elementary particle, and if it is associated with N world lines, it is calleda substance. From the axiom that mass is a time component of a unit world line,
the rest mass m0 can be obtained by multiplying N times the length |OB0 | = ηhof the unit time component of B (as seen from B) by a conversion factor δ (delta).
Thus, we have
m0 = δN ηh.

(18.4)

The conversion factor δ achieves unit consistency between N ηh and the weight measured in grams (g). An analogous conversion factor normalizes the weight measured
in UK (pounds) and the weight measured in USA (grams). The kinetic mass m is
the product of δN and the time component A0 of |OB0 |, which is observed by A.
From (18.3), we find thatm = δN A0
δN ηh
=√
.
1 − v 2 /c2
From (18.4) and (18.5), we have
m0
m= √
,
1 − v 2 /c2
247(18.5)

consistent with the formula of special relativity
m0
m= √
.
1 − v 2 /c2(17.11)

If the elementary particle B in Figure 18.4 travels at higher velocity v, the slope ofthe straight line x = vt increases, and the denominator of the previously calculated
formulaη
t0 = √(18.2)
c 1 − v 2 /c2
reduces. Thus, at higher velocities, time t0 becomes greater and point B0 (ct0 h, vt0 i),
whose distance from the origin O is a unit world line, recedes from the origin O.
This is a feature of four-dimensional space–time and beyond our common sense.
Then, the time component |OA0 | of |OB0 | becomes greater. Since |OA0 | expresses
the kinetic mass m as seen from static observer A, increasing the velocity v increases
the kinetic mass and intrinsically accounts for the increase of kinetic mass at higher
velocities.
From (18.4) and (18.5), we find that mass is expressed by an imaginary number
in our familiar positive world. Next, we reformulate the previous equations√|OB| = BB√= (cth + vti)(cth − vti)√
= cth 1 − v 2 /c2(18.1)
in terms of the new octonion B = ct + vthi in the negative world. If c > v, we find
that
√
BB
√
= (ct + vthi)(ct − vthi)√= c2 t2 − v 2 t2√
= ct 1 − v 2 /c2 .

|OB| =

248(18.6)

Thus, mass is expressed by a real number in the negative world. Since time anddistance in our positive world are expressed by imaginary numbers (as explained in
Section 11.4), mass as the time component of a world line is also expressed by animaginary number.
To express mass in our world by a real number, consider the positive and negative
world points ct + xhi + yhj + zhk and cth + xi + yj + zk, respectively. Assuming
the former as our world point, the square of the distance in three-dimensional spacebecomes negative:(xhi + yhj + zhk)(−xhi − yhj − zhk) = −x2 − y 2 − z 2 .
Since this result contradicts everyday observation, the world point in the positive
world can reasonably be assumed as cth + xi + yj + zk and it yields an imaginarymass.
If mass is the time component of a unit world line, η is related to the mass ofan elementary particle, but is considered so short as to be unobservable. Indeed, ηis comparable to the length of a fundamental string in string theory, as will be explained in Section 20.3. Since η is extremely small, it may be related to the Planck
constant, an important basic quantity in quantum mechanics. Moreover, since elementary particles of diﬀerent types extensively diﬀer in mass, various categories of
elementary particles should trace diﬀerent world lines. As explained in Section 8.3,
we posit an axiom that the world line of the particle is a wave as light, and thatparticles and light diﬀer only by their paths in four-dimensional space–time. Fromthat axiom, the wave amplitude Ψ becomes equivalent to the magnitude of a world
line, and induces diﬀerences among particle masses. However, in the simple proof
provided, Ψ is omitted from the calculations.

18.2 Energy and momentum
In Figure 18.3, δN A0 expresses the kinetic mass m of the substance B.

249

Applying the famous formula E = mc2 of special relativity, the energy E of thesubstance is δN A0 c2 . That is, we recover the theorem that energy is c2 times thetime component of a unit world line; however, we have also factored in the number
N of world lines, the conversion factor δ, and the amplitude Ψ of a world line (tosimplify the expression, Ψ has been omitted). This rule is a theorem because it isderived from the theorem E = mc2 and the axiom that mass is a time componentof a unit world line. The requirement for c2 is examined in Section 18.4.
With no rationale except consistency, special relativity equates the product pt c,of the time component of the four-dimensional momentum by c to the energy E.
In special relativity, this is a vector quantity; however, it can be re-expressed interms of the new octonion. Denoting the momentum components in the x-, y-, andz-directions as px , py , and pz , respectively, the new octonion p of the momentum
in four-dimensional space–time isp = pt h + px i + py j + pz k
= (E/c)h + px i + py j + pz k,(18.7)

where E, px , py , and pz are real numbers. However, since we examine their imaginary number status in the following verifications, we transform (18.7) intop = pt + px + py + pz
= (E/c) + px + py + pz .
In this equation, E, px , py , and pz are not real numbers. Moreover, the momentum
quantities pt , px , py , and pz involve the time t of observer A, diﬀerent from thefour-momentum calculated in proper time τ of the substance B.
Discerning readers of the book of special relativity may be perplexed by describing
the energy E as the product of c and the time component pt of the momentum infour-dimensional space–time. However, he/she could conceptualize mass as the timecomponent of the world line in Figure 18.3 and energy as the same time component
through the relationship E = mc2 . Such readers will also notice that the x-axial
component vt0 i in Figure 18.3, if multiplied by δN , is equivalent to the x-componentof the momentum px . However, the energy in a substance is mc2 and the momentum
px is 1/c × E from (18.7). Thus we expect to obtainpx = δN vt0 i × c2 × 1/c = δN vt0 ci.
This theorem states that the x-, y-, and z-components of the momentum are theproducts of c and the x-, y-, and z-axial components of a unit world line, respectively.
250

We prove this theorem as follows. Denoting the coordinates of D at the time when|OB0 | = η as D0 , we have
D0 = vt0 i.
Thus, x-axial component summed over N world lines is
N D0 = N vt0 i.
Then, we can rewrite
N D0 = N vt0 i= N ct0 hvi/(ch).
Multiplying both sides of this expression by the conversion factor δ, we find thatδN D0 = δN ct0 hvi/(ch).
Since mass is a time component of a unit world line, δN ct0 h is the kinetic mass mof substance B, and the equation becomesδN D0 = mvi/(ch).

(18.8)Let us focus on a pertinent issue. Previously, we expressed the mass of a substance
as

δN ηh
m= √
,
1 − v 2 /c2(18.5)

which contains the imaginary number h. Thus, if we apply the Newtonian definitionof momentum, i.e., px = mv, we find that px is imaginary and therefore becomesa time component. Defining px = mvi/h, the imaginary number h in the mass mcancels and px contains the imaginary number i of the x-coordinate. Substituting
px = mvi/h in (18.8), we obtainδN D0 = px /c.
This formula transforms intopx = δN D0 c.
The above expression is consistent with the theorem that momentum is c times
the x-, x-, and x-axial components of a unit world line. However, to acquire themomentum of a substance, we must multiply by the number N of world lines andthe conversion factor δ (omitting the amplitude Ψ of the world line).

25118.3 The sum of mass or energy in four-dimensional
space-time
The Universe is believed to have expanded from nothing during the Big Bang. According to this theory, the mass of the universe was initially zero. Although thenascent substance contained both positive and negative mass during the explosion,
the positive and negative masses collided and canceled; however, the excess positive
mass remained to form the Universe we know today. However, as explained in Section 18.1, the mass in our inhabited positive world is a positive imaginary quantity,
while the mass in the overlapping negative world is a positive real quantity. The newoctonion space–time theory admits only a positive imaginary mass and a positive
real mass. Thus, whether this mass theory is compatible with the Big Bang theory
is explored in this section.

As shown in Figure 18.3, the kinetic mass m (denoted by |OA0 |) is a function of
the velocity v of the substance B, for the following reason. Since the slope of thestraight line x = vt alters with velocity, the cth-axial component of a unit world
line, i.e., the time component, changes. In contrast, the rest mass m0 is the unit
world line |OB0 |, which is fixed for all slopes of the straight line x = vt. That is,
the mass residing in the Universe is the invariant rest mass m0 , while the kinetic
mass m is the apparent mass, which may diﬀer among observers.
In the reference frame of substance B, the coordinates of a point located at unit
length ηh from the origin O along the world line of B are (ηh, 0, 0, 0) in thepositive world and (η, 0, 0, 0) in the negative world. Denoting the rest mass of B
in the positive and negative worlds as m0 and m00 respectively, we have
m0 = δN ηh,(18.4)

m00 = δN η.

(18.9)

Recall that δ is a conversion factor. From (18.4) and (18.9), we find that

252

2
2
m20 + m02
0 = (δN ηh) + (δN η)= −δ 2 N 2 η 2 + δ 2 N 2 η 2= 0.
That is, the squares of the rest masses in the positive and negative worlds sum tozero. Since mass and energy are equivalent, the squares of the energies also sum tozero. From this, we see that when summing quantities in the entire Universe, wemust first square the quantity. If we sum the positive and negative energies, thecalculation is easy. However, the Universe is not described by such simple values.
Since this conclusion is obtained by the new octonion and the axiom that mass isa time component of a unit world line, we cannot determine whether the predictedvalue is consistent with that of the actual universe.

18.4 Meaning of c in E = mc2
Some readers may seek a special semantic to c2 in the mass/energy equivalence
relation E = mc2 . Here, we prove that c2 is required in calculations and we can
express the equivalence relation as E = m. Unlike Newtonian mechanics, relativity
theory allows interchange of time and distance units by changing t into ct. The
momentum obtained by changing dt into cdt in the definitionpx = mdx
dtof Newtonian mechanics, defined as pc , is called the new octonion momentum.
Expressing the x-axial component of pc as pcx , we have
pcx = m

dx
px
=
.
cdt
c

From (18.10) and the equationpx = δN D0 c
which is proved in Section 18.2, we find thatpxc= δN D0 c/c

pcx == δN D0 .
253(18.10)From this equation, we see that D0 in Figure 18.3 is the new octonion momentumof the elementary particle B (however, we must multiply by the number N of worldlines and the conversion factor δ). Since we calculate in terms of time t, we need thelight velocity c in the theorem that the x-, y-, and z-components of the momentum
are the products of c and the x-, y-, and z-axial components of a unit world line.
If the velocities are calculated from ct, the theorem describes the new octonion
momentum as the x-, y-, and z-axial components of a unit world line.

Similarly, we express the kinetic energy obtained by changing dt into cdt in the
Newtonian kinetic energy formulaE=m dx 2( )2 dt

as Ec , called the new octonion kinetic energy. Now we haveEc =m dx 2E() = 2.
2 cdt
c(18.11)

From (18.11) and E = mc2 of special relativity, (18.11) becomes
Ec =

mc2
= m.
c2

That is, replacing by ct, the theorem that energy is c2 times the time component ofa unit world line is equivalent to stating that the new octonion energy is the timecomponent of a unit world line, and is identical to mass.
The c2 in E = mc2 has no special semantic. Although the velocity could be
calculated in terms of ct, it is habitually calculated in terms of t, with the factor
c2 explicitly stated in E = mc2 . If ct were to become an accepted unit, E = mc2
would be rewritten as E = m and mass, energy, and momentum could be illustrated
on a four-dimensional space–time diagram.
The unit of a physical quantity is derived by a process called dimensional analysis.
Applying dimensional analysis to ct, we have

254

ct =

[distance][time][time]= [distance].
If we accept that mass is a time component of a unit world line, the mass unit also
converts to a unit [distance]. In above scenario, the energy E and momentum px
would have the same unit [distance], rather than being related by 1/c. Expressing
quantities of time, distance, mass, energy, and momentum in the same unit [distance]
would remove the idea of unit.

18.5 Mass of light and the Higgs boson
In special relativity, light possesses no rest mass; however, it has kinetic mass andmomentum. The idea of light acquiring mass only while it is moving cannot be
visualized. However, if we use the axiom that mass is the time component of a unit
world line, and construct a four-dimensional space–time diagram, the concept canbe easily understood.

Figure 18.5 is the four-dimensional space–time diagram of light emitted in thepositive x-direction of a stationary observer A. The y- and z-axes are omitted. The
equation of the world line of light is x = ct. Substituting x = ct into cth + xi,
which expresses the coordinates of a point mass, the new octonion of light is given
as cth + cti. Thus, the coordinates of point B on the world line of light are (cth, cti)
and light passes through all points that are equidistant from the cth- and xi-axes.
The point located a unit world line ηh from the origin O is denoted B0 (ct0 h, ct0 i).
Assuming that mass is the time component of a unit world line, the rest mass m0
and the kinetic mass m of light are given as δ |OB0 | and δA0 , respectively. The
conversion factor δ changes the mass unit in four-dimensional space–time into thestandard mass unit.
255

We now calculate the rest mass m0 of light. From B0 = ct0 h + ct0 i, we have
m0 = δ |OB0 |√= δ (ct0 h + ct0 i)(ct0 h − ct0 i)√= δct0 h 1 − (ct0 i)2 /(ct0 h)2√= δct0 h 1 − 1= 0.
That is, wherever point B0 exists on the line x = ct, the length of light from theorigin O is always zero. This is a feature of four-dimensional space–time and cannot
be realized in three-dimensional space. Therefore, since light lacks a unit world lineof distance ηh, its rest mass becomes zero.
What about the kinetic mass m? Figure 18.5 clearly shows the existence of anonzero cth component of light; therefore, light possesses kinetic mass. However,since the point B0 located at a distance ηh from the origin O does not settle, ct0 h
is unfixed. Any particle other than light with finite mass and constant velocity
possesses a fixed kinetic energy, given by E = mv 2 /2 at nonrelativistic velocities.
However, although light travels at constant velocity c, its energy is proportional toits frequency. More specifically, the energy of light E is related to its frequency ν(nu) through the Planck constant h as
E = hν.

(18.12)

Although Planck’s constant is correctly written as h, we here express it as h todistinguish it from h in the new octonion. As explained above, the kinetic mass
m of light is the variable quantity δct0 h; thus, all kinetic masses (or energies) arepermitted. This explains the proportionality between the energy of light and its
frequency, given by (18.12), which admits all values, although the velocity of lightis constant.
Assuming that energy is c2 times the time component of a unit world line, theenergy of light is,
E = δct0 × c2
= δc3 t0 .

(18.13)Here, the imaginary number h is omitted to avoid potential ambiguity with Planck’s
constant h. From (18.12) and (18.13), we find thatδc3 t0 = hν,
256

hν
.
(18.14)
δc2
From (18.14), the cth-axial component A0 of the unit world line of light of the
ct0 =frequency ν is hν/(δc2 ). Moreover, from the theorem that momentum is c times thex-, y-, and z-axial components of a unit world line, if the space imaginary number
i is omitted, the light of frequency ν has a momentum of
p = δct0 × c= δc2 t0 .
Substituting (18.14) into this equation, we havep = δc ×

hν
hν
=
.
2
δc
c

This result is consistent with the result of special relativity.
We can also prove by calculation alone (without constructing a four-dimensional
space–time diagram) that light has zero rest mass but unfixed kinetic mass, i.e.,
unfixed energy hν. Inserting v = c in the relationship between rest mass m0 andthe kinetic mass m, i.e.,
m0
m= √
,
(17.11)1 − v 2 /c2
we have

m0
m0
m= √
,
=
2
2
0
1 − c /cwhich is meaningless unless m0 = 0. That is, the rest mass of light is zero. Then,
we have0m= .
(18.15)0
Rearranging (18.15) gives
m × 0 = 0,
which admits any value of the kinetic mass m. The energy mc2 of light is similarly
unfixed. If mc2 = hν, the energy of light depends only on its frequency.
We now consider the Higgs boson. The Higgs boson, which supposedly gives
particles their mass, has long been sought by elementary particle researchers. Immediately following the Big Bang, all particles are considered to have been masslessand free-moving. Entities other than light gained mass only after their velocities
were reduced by interaction with Higgs bosons.
If we admit the axiom explained in Section 8.3 that the world line of light diﬀers
from that of a substance only in its path in four-dimensional space–time, a point257

mass, which passes a point equidistant from the cth- and xi-axes in a coordinate
plane, has velocity c, since its slope is c. Moreover, a point mass, which passesa point not equidistant from these axes, has velocity v, since its slope is not c.
Because the world distance of a point on x = ct is zero, light possesses no rest
mass. The world distance of the point on x = vt, on the other hand, is non-zero;hence, substance has mass. That is, mass is not given; however, it is finite or zerodepending on the locus in four-dimensional space–time. In this picture, no Higgs
boson is required. However, if the world distance of all world lines was zero at thetime of the Big Bang, all entities would have been massless light. If at that time
the world lines of light and a Higgs boson collided, shifting the world distance oflight from zero, a Higgs boson could be born. However, this scenario implies thepre-existence of the Higgs boson at the time of the Big Bang, which complicates thetheory. Rather, we consider that the world lines of light and a substance diﬀer onlyin their locus in the four-dimensional space–time, and that light possesses no rest
mass because its world distance is zero. In contrast, a substance is characterized bya non-zero world distance, and hence has finite mass. This theory is simpler than
the theory of the Higgs boson.

18.6 Constancy of the coordinate transformation of energy
conservation
The energy conservation law states that the total energy of colliding bodies is unaltered by the collision. As proven in Section 17.6, the total momenta in the x-,y-, and z-directions are the same before and after a collision. Since this law holds
true under a coordinate change, it is called a coordinate transformation invariance.
Although standard relativity texts verify the coordinate transformation invarianceof momentum, the same treatment for energy conservation is not found. We may
naturally consider that total energy E is also invariant under coordinate transformation, since the energy E is the product of c and the time component pt of the
momentum, and the total momenta in the x-, y-, and z-directions are invariant
under coordinate transformation. However, a mathematical demonstration of thisphenomenon is lacking. This section proves that energy conservation is invariant
under coordinate transformation using the new octonion.
Assume that two spheres A and B of rest masses m0 and M0 respectively collidewith respective velocities v0 and V0 , and move at respective velocities v and V after
the collision. The situation is illustrated in Figure 18.6.

258

From the mass–energy relationship E = mc2 , and the relationship between rest
mass and kinetic mass
m0
m= √
,
(17.11)1 − v 2 /c2
the energy conservation law is written as,
m c2
M c2
m c2
M c2
√ 0
√ 0
√ 0
√ 0
+
=
+
.
1 − v 2 /c2
1 − V 2 /c2
1 − v02 /c2
1 − V02 /c2(18.16)Equation (18.16) is formulated in the reference frame of a stationary observer. If
we can prove (18.16) in a moving reference frame; for example, in the coordinates ofsphere A moving at velocity v after the collision, then we can prove the coordinatetransformation invariance of energy conservation.

Expressing the collision as world lines, we obtain Figure 18.7. If the time of thecollision is assumed as the origin O, the equations of the world lines of A and B
before the collision are x = v0 t and x = V0 t, respectively, and x = vt and x = V t,
respectively, after the collision. Now, we implement a coordinate transformation
to the coordinates of sphere A moving at velocity v after the collision. The point
D(cth, vti) on the world line x = vt of A moves to point D0 (cth, 0) on the world line
x = 0, i.e., the cth-axis. If this transformation can be expressed as a new octonion259

H, given D = cth + vti and D0 = cth, we can write(cth + vti)H = cth.
Rearranging this expression, we obtaincth
cth + vti
cth(cth − vti)=
(cth + vti)(cth − vti)
cth(cth − vti)=(cth)2 (1 − v 2 /c2 )1 − vti/(cth)=1 − v 2 /c2
1 + vhi/c=.
1 − v 2 /c2

H=(18.17)

Next, the coordinates of the world line x = V t of sphere B are transformed
through H. If A observes B moving at velocity V 0 after the coordinate transformation, the equation of B’s world line (according to A) is x0 = V 0 t0 . Since thecoordinates of the point on the world line are (ct0 h, V 0 t0 i), we can write
ct0 h + V 0 t0 i = (cth + V ti)H.
Substituting (18.17) into this equation, we find that
1 + vhi/c
1 − v 2 /c2
cth − vti + V ti − V vth/c=
1 − v 2 /c2(1 − V v/c2 )cth + (V − v)ti=.
1 − v 2 /c2

ct0 h + V 0 t0 i = (cth + V ti)

Comparing the coeﬃcients, we find that
(1 − V v/c2 )t,
1 − v 2 /c2(V − v)t
V 0 t0 =.
1 − v 2 /c2
t0 =Eliminating t0 from these equations gives
V0 =

V −v
.
1 − V v/c2

260

That is, the equation x0 = V 0 t0 of the world line of B isx0 =(V − v)t0.
1 − V v/c2

(18.18)Since V 0 is given by the Lorentz transformation of the velocity
Vx − v
1 − (v/c2 )Vx(10.16)

as explained in Section 10.4, the coordinate transformation by the new octonion isverified.

Next, we use (18.18) to derive the mass and energy of sphere B after the coordinatetransformation. Denote the point located a unit distance ηh from the origin O on
the world line (18.18) as E 0 (ct0 h, x0 i). In Figure 18.7, point E moves to E 0 after
the coordinate transformation. From|ηh|2 = |OE 0 |2
= (ct0 h + x0 i)(ct0 h − x0 i),
we have−η 2 = −c2 t02 + x02 .
Sustituting (18.18) into (18.19), we find that(V − v)2 t02(1 − V v/c2 )2−c2 t02 (1 − V v/c2 )2 + (V − v)2 t02=(1 − V v/c2 )2−c2 t02 (1 − 2V v/c2 + V 2 v 2 /c4 ) + (V 2 − 2V v + v 2 )t02=
(1 − V v/c2 )2−c2 t02 + 2V vt02 − V 2 v 2 t02 /c2 + V 2 t02 − 2V vt02 + v 2 t02=(1 − V v/c2 )2−η 2 = −c2 t02 +261(18.19)−c2 t02 − V 2 v 2 t02 /c2 + V 2 t02 + v 2 t02(1 − V v/c2 )22 02−c t (1 − v 2 /c2 ) + V 2 t02 (1 − v 2 /c2 )=(1 − V v/c2 )22 02−c t (1 − v 2 /c2 )(1 − V 2 /c2 )=.
(1 − V v/c2 )2=Conditional on V < c and v < c, this equation becomesη 2 (1 − V v/c2 )2,(1 − v 2 /c2 )(1 − V 2 /c2 )η(1 − V v/c2 )√
ct0 = √.
1 − V 2 /c2 1 − v 2 /c2

c2 t02 =(18.20)

Therefore, from (18.18) and (18.20), the new octonion expressing the point E 0 located at unit distance ηh from the origin O is
E 0 = ct0 h + x0 i
(V − v)t0 i
1 − V v/c2
η(1 − V v/c2 )hη(V − v)(1 − V v/c2 )i
√
√
√
=√
+
1 − V 2 /c2 1 − v 2 /c2(1 − V v/c2 )c 1 − V 2 /c2 1 − v 2 /c2= ct0 h +=√η(V − v)iη(1 − V v/c2 )h
√
√
+ √
.
1 − V 2 /c2 1 − v 2 /c2
c 1 − V 2 /c2 1 − v 2 /c2This is a purely imaginary number whose first term (involving h) expresses kinetic
mass and energy. The second term (involving i) expresses momentum as explainedin Section 18.2. Thus, multiplying by the number N of world lines and the massconversion factor δ, the kinetic mass of B observed by A after the coordinate transformation isδN η(1 − V v/c2 )h√√.
(18.21)
1 − V 2 /c2 1 − v 2 /c2
Note that the amplitude Ψ of the world lines is not considered.
Suppose that sphere A traces n world lines. Since the rest mass of A before thecollision is m0 = δnηh, replacing N by n and V by v0 in (18.21), the kinetic massof A before the collision and after the coordinate transformation becomesδnη(1 − v0 v/c2 )hm0 (1 − v0 v/c2 )√√A(bef ore) : √=√.
2
1 − v0 /c2 1 − v 2 /c2
1 − v02 /c2 1 − v 2 /c2
Similarly, since the rest mass of B before the collision is M0 = δN ηh, replacing
V by V0 in (18.21) gives the kinetic mass of B before the collision and after the262

coordinate transformation asB(bef ore) : √M0 (1 − V0 v/c2 )δN η(1 − V0 v/c2 )h
√
√
√
=
.
1 − V02 /c2 1 − v 2 /c2
1 − V02 /c2 1 − v 2 /c2

Now, replacing N by n and V by v in (18.21), the coordinate-transformed kinetic
mass of A after the collision isA(af ter) : δnηh = m0 .
Since the coordinate-transformed kinetic mass of B after the collision is given by(18.21), we haveB(af ter) : √δN η(1 − V v/c2 )h
M0 (1 − V v/c2 )√√=√.
1 − V 2 /c2 1 − v 2 /c2
1 − V 2 /c2 1 − v 2 /c2

Multiplying each rest mass by c2 , the energy conservation law after the coordinatetransformation is√

m0 c2 (1 − v0 v/c2 )M c2 (1 − V0 v/c2 )
√ 0
√
√
+
1 − v02 /c2 1 − v 2 /c2
1 − V02 /c2 1 − v 2 /c2

M0 c2 (1 − V v/c2 )√= m0 c2 + √.
1 − V 2 /c2 1 − v 2 /c2(18.22)

If we can obtain (18.22) from the energy conservation law before the coordinatetransformation, i.e.,
M c2
m c2
M c2
m c2
√ 0
√ 0
√ 0
√ 0
+
=
+
,
1 − v 2 /c2
1 − V 2 /c2
1 − v02 /c2
1 − V02 /c2(18.16)and the momentum conservation law proven in Section 17.6, i.e.,
m v
M0 V0
m0 v
M0 V
√ 0 0
+√
=√
+√
,
2
2
2
2
2
2
1 − v /c
1 − V 2 /c2
1 − v0 /c
1 − V0 /c(18.23)then we can prove that energy conservation is invariant under a coordinate transformation. To this end, we attempt to derive (18.22) from (18.16) and (18.23).
Dividing (18.16) by√
1 − v 2 /c2 ,we obtain
√

m0 c2
M0 c2
√
√
√
+
1 − v02 /c2 1 − v 2 /c2
1 − V02 /c2 1 − v 2 /c2

=

m0 c2
M0 c2
√
+√
.
2
2
1 − v /c
1 − V 2 /c2 1 − v 2 /c2
263(18.24)

Multiplying (18.23) byv√
1 − v 2 /c2gives
m 0 v0 v
M0 V0 v
√
√
+√
2
2
2
2
2
1 − v0 /c 1 − v /c
1 − V0 /c2 1 − v 2 /c2
m0 v 2
M0 V v
√
=
+√
.
2 2
2 2
1 − v 2 /c2√

1 − V /c

1 − v /c

(18.25)

Subtracting (18.25) from (18.24), we find that√

M c2 (1 − V0 v/c2 )m0 c2 (1 − v0 v/c2 )
√ 0
√
√
+
1 − v02 /c2 1 − v 2 /c2
1 − V02 /c2 1 − v 2 /c2=

m0 c2 (1 − v 2 /c2 )M0 c2 (1 − V v/c2 )√√.
+1 − v 2 /c2
1 − V 2 /c2 1 − v 2 /c2This equation becomes√

M c2 (1 − V0 v/c2 )m0 c2 (1 − v0 v/c2 )
√ 0
√
√
+
1 − v02 /c2 1 − v 2 /c2
1 − V02 /c2 1 − v 2 /c2

M0 c2 (1 − V v/c2 )√= m0 c2 + √.
1 − V 2 /c2 1 − v 2 /c2This formula is identical to the energy conservation law obtained from the unit
length ηh of the world line after a coordinate transformation, i.e.,√

m0 c2 (1 − v0 v/c2 )M c2 (1 − V0 v/c2 )
√ 0
√
√
+
1 − v02 /c2 1 − v 2 /c2
1 − V02 /c2 1 − v 2 /c2

M0 c2 (1 − V v/c2 )√= m0 c2 + √.
1 − V 2 /c2 1 − v 2 /c2(18.22)

The above derivations prove that the energy conservation law, given by
M c2
m c2
m c2
M c2
√ 0
√ 0
√ 0
√ 0
+
=
+
,
1 − v 2 /c2
1 − V 2 /c2
1 − v02 /c2
1 − V02 /c2(18.16)is invariant after a coordinate transformation. That is, Equation (18.16) holds inany reference frame.

18.7 Force and a reflection of the world line
One of the first equations encountered by the high school student of mechanics is
Newton’s equation of motion
F = ma,264(18.26)

where F, m, and a denote force, mass, and acceleration, respectively. The acceptedsemantic of (18.26) is that a force F applied to a body of mass m induces anacceleration a. However, to what extent is this interpretation actually correct? Since
F and ma are equated in (18.26), an applied force gives rise to an instantaneous
acceleration (that is, the object accelerates in zero time). In special relativity, asignal cannot be instantaneously transmitted. In other words, a signal cannot betransmitted at infinite velocity. Thus, specifying the force F as the quantity that
imposes an acceleration a on mass m contradicts the premise of special relativity.
Equation (18.26) is more properly interpreted in the following form:
ma = F.

(18.27)This formula means that the status of a body of mass m having an acceleration a
is a force F . In this interpretation, the force F , which imposes acceleration, does
not exist but the status of ma is force. Force is merely a word which expresses thestatus of movement. In this way, the above contradiction is avoided, despite theequality of ma and F in (18.27).
To explain that force does not exist in four-dimensional space–time, we show thefollowing example, in which the velocity changes in the absence of an applied force.
Two steel balls of the same size and mass, traveling at v and −v collide. After thecollision, the balls reverse their direction, traveling at −v and v, respectively. This
collision is both symmetric and perfectly elastic (no energy loss). Now consider that,on collision, the velocity of the gravitational center of the steel ball instantaneously
reverses from v to −v. Here, we need not consider the force exerted at the time of thecollision. Rather, we may consider that the slope of the world line changes from v
to −v. The force imparted during a collision is frequently called an action-reaction.
However, no one can explain what it is.
From the above speculation, we here establish a new axiom that force does notexist, but is a status that world lines bend and reflect. If a world line continues
bending after a collision, we say that force continues to act. If a body accelerates, its
world line continues turning in the four-dimensional space–time diagram. Moreover,linear world line implies that no force is acting. In the world-line interpretation offorce, we must establish the reflection law of world lines. When light is reflected by aplane mirror, the angle of reflection equals the angle of incidence. Recall our axiom
that light and substance diﬀer only by their locus in four-dimensional space–time.
An analogous reflection law should therefore apply to the world lines of substances.

265

In Figure 18.6, the velocities v and V after the collision must be collectively determined from the velocities v0 and V0 before the collision. For this purpose, we
require simultaneous equations in v0 , V0 , v, and V . The required equations are theenergy conservation law
√

m0 c2

M c2
m c2
M c2
√ 0
√ 0
√ 0
+
=
+
1 − v 2 /c2
1 − V 2 /c2
1 − v02 /c2
1 − V02 /c2

and the momentum conservation law
M0 V0
m0 v
m v
M0 V
√ 0 0
+√
=√
+√
,
1 − v 2 /c2
1 − V 2 /c2
1 − v02 /c2
1 − V02 /c2(18.16)(18.23)

which yield v and V from v0 and V0 . Moreover, the reflection law of world lines
must be invariant in any coordinates frame. In Sections 17.6 and 18.6, we provedthat (18.16) and (18.23) are invariant under coordinate transformation. That is,since (18.16) and (18.23) satisfy two conditions required by the reflection law, they
are considered as reflection laws of world lines.
Energy and momentum conservation are fundamental physical laws that are ubiquitously applied in physics and engineering problems. However, why these two laws
exist is unresolved. The upper verification shows that energy and momentum conservation are merely reflection laws of world lines. Moreover, since many otherphysical laws can be proved by energy and momentum conservation, the reflectionof world lines might explain many physical phenomena.
To conclude, mass and energy are both time components of a unit world line,
while momentum is the space component of a unit world line. Force is a word
which expresses the status of a world line. Furthermore, energy and momentum
conservation laws are reflection laws of world lines. That is, only new octonions andworld lines exist in four-dimensional space–time. All physical phenomena may be
explained by reflections of world lines. Einstein interpreted gravity as the bendingof space by mass. However, in the theory of the new octonion and world lines, mass
is a time component of a unit world line, and does not bend space. As explained in
Section 13.3, four-dimensional space–time is intrinsically bent, regardless of mass.
266

18.8 Extinction of the world line
In Section 17.6, we internally exploded a substance of rest mass M0 into two substances, each of rest mass m0 and velocities v and −v. Subsequently, we used
momentum conservation to show that
2m0
M0 = √.
1 − v 2 /c2(17.62)In this section, (17.62) is obtained from energy conservation and the axiom thatmass is the time component of a unit world line. We also prove that a world line
disappears and is transformed into energy.

Figure 18.8 shows a substance A of rest mass M0 located at the origin O. An
interior explosion splits A into two substances B and D, each of rest mass m0 . The
division occurs along the x-axis and substances B and D travel horizontally withvelocities v and −v, respectively. The corresponding four-dimensional space–timediagram is shown in Figure 18.9.

Since the substance A remains at rest, its distance x is zero. As time t passes, theworld line x = 0 is generated for substance A. That is, substance A shifts along thecth-axis from the left to the origin O. If the fission occurs at the origin O in time,267

the world line of substance B after the fission satisfies x = vt while that of substance
D is given by x = −vt. On the straight lines x = 0, x = vt, and x = −vt, points
E, F , and G are located one unit world line ηh from the origin O. The straight line
F G intersects the cth-axis at node H. The coordinates of H are (ct0 h, 0). Since
the coordinates of F are (ct0 h, vt0 i), it follows thatηh = |OF |√= (ct0 h + vt0 i)(ct0 h − vt0 i)√= ct0 h 1 − (vt0 i)2 /(ct0 h)2√= ct0 h 1 − v 2 /c2 .
Thus, we find that
ct0 = √

η1 − v 2 /c2.

(18.28)Assume that A traces N world lines before the fission, while B and D each trace nworld lines after the fission. From E = mc2 and the axiom that mass is the timecomponent of a unit world line, the energy before and after the fission is δN |OE| c2
and 2δn |OH| c2 respectively. Recall that δ is a mass conversion factor. Energy
conservation gives
δN |OE| c2 = 2δn |OH| c2 .
Thus, we have
N |OE| = 2n |OH| .
Substituting |OE| = ηh and |OH| = ct0 h into this equation, we find thatN η = 2nct0 .
Substituting (18.28) into this equation, we have2nN=√.
1 − v 2 /c2(18.29)Equation (18.29) relates the number N of world lines before the fission to the total
number of world lines 2n after the fission. Rearranging (18.29), we obtain thenumber of world lines after fission as
2n = NSince√
1 − v 2 /c2 .

√1 − v 2 /c2 < 1, the number of the world lines is reduced after the fission.

This analysis reconfirms the proof in Section 17.6 that some of the world lines havebeen converted to the kinetic energy of the substances B and D.
268

Given that E = mc2 , (18.29), and accepting that mass is the time component ofa unit world line, the rest energy before and after the fission diﬀers by
M0 c2 − 2m0 c2 = δN ηhc2 − 2δnηhc22δnηhc2=√− 2δnηhc2
1 − v 2 /c2+ 2δnηhc2 (1 +v2
) − 2δnηhc2
2c2

v2
h
2
v2
= 2m0 h.
2

= 2δnη

(18.30)The quantity m0 v 2 /2 is only the Newtonian kinetic energy of substances B and D.
Therefore, we have reconfirmed that the missing world lines after the fission areconverted to kinetic energy. When a launched firework divides by explosion of theinternal gunpowder, the world lines of the whole firework become fewer as someare converted to energy. When mitotic division occurs at near-light velocities, theresult is nuclear fission, which releases vast amounts of energy.

18.9 Energy-momentum equation and Dirac’s γ matrix
The energy–momentum equation is discussed in all special relativity texts. A stationary observer A views a point mass B of rest mass m0 moving along a straightline with uniform velocity v. In A’s reference frame, B carries momentum p and its
energy E is
E 2 = (m0 c2 )2 + c2 p2 ,(18.31)

where m0 c2 is the rest energy of point mass B. However, p is not the four-momentumof special relativity but the momentum in the three-dimensional space of Newtonianmechanics. In this section, the energy–momentum equation (18.31) is derived interms of the new octonion. We also derive a relationship between the new octonion
and Dirac’s γ (gamma) matrix in quantum mechanics.
As proved in Section 17.2, the momentum
mc

dt
dx
dy
dz
h+m i+m j+m k
dt
dt
dt
dt

(18.32)of point mass B in four-dimensional space–time is invariant under a coordinatetransformation. In other words, the absolute values of (18.32) are identical in thereference frames of both A and B. However, since the momenta observed by A
269

and B are calculated in terms of time t and the proper time τ , respectively, theydiﬀer from the four-momentum of special relativity, expressed in terms of τ only.
Although the mass is invariant in the four-momentum formulation, it depends onthe velocity in (18.32).
First, we calculate the absolute value of the momentum in the coordinates of point
mass B. Since B observes the proper time τ and is stationary in its own reference
frame, dx = dy = dz = 0. Moreover, B’s mass is the rest mass m0 . Therefore, themomentum pB and the square of its absolute value |pB |2 observed by B aredτ000h + m0 i + m0 j + m0 k
dτ
dτ
dτ
dτ
= m0 ch,pB = m0 c|pB |2 = m20 c2 h2= −m20 c2 .

(18.33)The momentum pA of B and the square of its absolute value |pA |2 as observed by
A are similarly calculated. In this reference frame, the time is t and the mass is the
kinetic mass
m0
m= √
.
1 − v 2 /c2Moreover, the momenta of the x-, y-, z-directions are px , py , and pz , respectively.
Thus, (18.32) becomes
dt
dx
dy
dz
h+m i+m j+m k
dt
dt
dt
dt
= mch + px i + py j + pz k,

pA = mc|pA |2 = (mch + px i + py j + pz k)(mch − px i − py j − pz k)= −m2 c2 + p2x + p2y + p2z .

(18.34)Since momentum is invariant under coordinate transformation, we have|pB |2 = |pA |2 .
Substituting (18.33) and (18.34) into this equation, we find that
−m20 c2 = −m2 c2 + p2x + p2y + p2z .
Substituing E = mc2 and the Newtonian momentum p2 = p2x + p2y + p2z into theabove expression yields−m20 c2 = −E 2 /c2 + p2 ,E 2 /c2 = m20 c2 + c2 p2 .
270

Multiplying both sides of this equation by c2 , we obtain
E 2 = (m0 c2 )2 + c2 p2 .
That is, the energy–momentum equation (18.31) is recovered by the new octonion
formulation.
Finally, we relate the new octonion to Dirac’s γ matrix. Paul Dirac, who largely
contributed to the development of quantum mechanics, factored the energy-momentum
equation
E 2 = (m0 c2 )2 + c2 p2(18.31)

as follows:
(

E
c

)2
= p2 + (m0 c)2= p2x + p2y + p2z + (m0 c)2= (α1 px + α2 py + α3 pz + βm0 c)2 ,E= α1 px + α2 py + α3 pz + βm0 c.
c
In this formulation, αn and β are matrixes given by





0 0 0 10 0 0 −i
0 0





 0 0 1 0 
 0 0 i 0 
 0 0




α1 = 
 0 1 0 0  , α2 =  0 − i 0 0  , α3 =  1 0





1 0 0 0i 0 0 00 −1


1 0 0 0
 0 1 0 0 

.
β=

 0 0 −1 0 
0 0 0 −1

(18.35)

1 0

0 −1 
,
0 0 

0 0

αn and β are related through the γ matrix as
iβαn = −γn .
Dirac incorporated the γ matrix into his classical wave equation for a free electron
(
)
∂Ψ
∂Ψ
∂Ψ
m0 c
iβ ∂Ψ
− γ1
+ γ2
+ γ3
−
Ψ = 0.
c ∂t
∂x1
∂x2
∂x3
~
Here, Ψ is a wave function and ~ is obtained by dividing Planck’s constant h by 2π.

271

When deriving the energy–momentum equation using the new octonion, we specified|m0 ch| = |mch + px i + py j + pz k| .
Since mc = E/c (rearranging E = mc2 ), the above formula can be rewritten as|m0 ch| =E
h + px i + py j + pz k .
cComparing this equation toE= α1 px + α2 py + α3 pz + βm0 c,c(18.35)

we must question whether the new octonion relates to the γ matrix. In fact, Hiroyuki
Kamada (Kyushu Institute of Technology) demonstrated that Dirac’s γ matrix andthe new octonion are mathematically equivalent (personal communication).

18.10 An easy method for obtaining E = mc2
Here, we derive Einstein’s famous special relativity formula E = mc2 in terms ofthe new octonion. The derivation is very simple. In Newtonian mechanics, force,
distance, and time are written as F, x, and t, respectively, and the energy E andmomentum p are given as
E = F x,(18.36)

p = F t.

(18.37)Equations (18.36) and (18.37) can be added or subtracted, provided that their units
are the same. To this end, we multiply both sides of (18.37) by the velocity of lightc, and rewrite the equation as
cp = F ct.

(18.38)The right-hand sides of (18.36) and (18.38) have identical units. Therefore, E and cp(or E/c and p), have identical units. Expressing the momentum in four-dimensional
space–time in terms of the new octonion, we have
m

d
dt
dx
dy
dz
(cth + xi + yj + zk) = mc h + m i + m j + m k
dt
dt
dt
dt
dt
= mch + px i + py j + pz k.

(18.39)As explained in Section 18.2, the time component mc of the momentum (18.39) infour-dimensional space–time is equivalent to energy. Thus, assigning the same unit272

to the time component mc of the momentum and the energy E through E/c, wefind thatmc = E/c,
E = mc2 .

273

274

19
Special Relativity at
the Superluminal Velocity

19.1 New Lorentz transformations at the superluminal velocity
When deriving the new Lorentz transformation in Section 3.5, it was explained that
if we assume v > c, we obtain a new Lorentz transformation at the superluminal
velocity. Here we calculate this transformation. Consider a stationary observer A atorigin O and an observer B moving at uniform velocity v in the positive x-directionof A. In addition, a stationary observed point mass D is positioned at distance x
from the origin. B and A coincide at the origin O at time t = 0. After t seconds,
the coordinates of the three entities are A(cth, 0), B(cth, vti), and D(cth, xi), andthe corresponding complex numbers are A = cth, B = cth + vti, and D = cth + xi.
The space–time diagram at the superluminal velocity is illustrated in Figure 19.1.
Since v > c, the world line x = vt of B leans toward the xi-axis from the world lineof light x = ct.

Calculating the coordinate transformation DB/ |B| to obtain the new Lorentz
transformation, we find that275

DB(cth + xi)(cth − vti)=√
|B|
(cth + vti)(cth − vti)=(cth)2 − cvt2 hi + xcthi − xvti2√.
(cth)2 − (vti)2

From v > c and ct > 0, the equation becomes(cth)2 − cvt2 hi + xcthi − xvti2
DB
√
=
|B|
(cth)2 − (vti)2−(ct)2 − cvt2 hi + xcthi + xvt√−c2 t2 + v 2 t22−(ct) + xvt − cvt2 hi + xcthi√=
c2 t2 (v 2 /c2 − 1)=−(ct)2 + xvt − cvt2 hi + xcthi√
ct v 2 /c2 − 1−ct + xv/c − vthi + xhi√=
v 2 /c2 − 1==c(v/c2 )x − ct + (x − vt)hi√.
v 2 /c2 − 1(19.1)

Since hi is a real number, (19.1) is a purely real number. Therefore, the coordinatetransformation transfers the world point of D to the negative world. Thus, if the newoctonion of D as seen from B after the coordinate transformation is D0 = ct0 + x0 hi,
by coeﬃcient comparison, we find thatc(v/c2 )x − ct
ct0 = √,
v 2 /c2 − 1(v/c2 )x − t,t0 = √
v 2 /c2 − 1
x − vtx0 = √.
v 2 /c2 − 1(19.2)(19.3)

Equations (19.2) and (19.3) specify the new Lorentz-transformed time and distance,
respectively, at superluminal velocity.
We now examine the signs of (19.2) and (19.3). In Figure 19.1, when the point
mass D is located above observer B, we have x > vt. At this time, sincex0 = √x − vtv 2 /c2 − 1

vt − vt>√=0
v 2 /c2 − 1276

from (19.3), we find that x0 > 0. On the other hand, when the point mass D lies
below observer B, i.e., x < vt, we havex − vtvt − vtx0 = √<√= 0.
2
2
v /c − 1v 2 /c2 − 1That is, x0 < 0. Since the value of (19.2) is undefined when x < vt, we examine thecase of x > ct. We have√t v/c − 1
(v/c2 )x − t
xv/c2 − tvt/c − tt0 = √
=√
>√
= √
> 0.
v 2 /c2 − 1v 2 /c2 − 1v 2 /c2 − 1
v/c + 1Thus, when the point mass D is located above the world line x = ct of light, i.e.,x > ct, t0 is positive. Summarizing the above results, we obtain
x0 > 0, t0 > 0 (: x > vt),
x0 < 0, t0 > 0 (: vt > x > ct).

19.2 Proper time at the superluminal velocity
The equations(v/c2 )x − tt0 = √,
v 2 /c2 − 1
x − vtx0 = √
v 2 /c2 − 1(19.2)(19.3)

obtained in the last section specify the time and distance of the point mass D as
seen from observer B whose velocity exceeds the luminal velocity. Thus, we must
examine the proper time of D moving at superluminal velocity with respect to B.
For this purpose, we consider that observer B approaches D along the line AD, andspecify x0 = 0 when D coincides with B. Setting x0 = 0 in (19.3), we obtainx − vt0= √v 2 /c2 − 1and x = vt. Since the time t obtained by substituting x = vt into (19.2) is theproper time τ of D, we find that(v/c2 )vt − t
τ= √v 2 /c2 − 1
t(v 2 /c2 − 1)= √v 2 /c2 − 1√= t v 2 /c2 − 1.
277

(19.4)

Equation (19.4) is the proper time of the point mass moving at the superluminal
velocity. Since t > 0, τ > 0 and is therefore irreversible; a point mass D moving
at superluminal velocity cannot return to the past. Hypothetical particles calledtachyons, which move at superluminal velocities, have been suggested as the basisof time travel. However, the above calculations suggest that backward time travel is
impossible. Moreover, since the world point of the point mass at the superluminal
velocity is ct0 +x0 hi, which is a number in the negative world, the point mass exists inthe negative world, and is therefore unobservable from the positive world. Whether
a positive world substance can even exist in the negative world is unknown.
Furthermore, from (19.4), D’s proper time τ passes more rapidly as the velocity v
increases. This result is opposite to the time slowing eﬀect as subluminal velocities
approach c.
In determining Equation (19.4), we used the following two axioms, introduced in
Section 13.2:
Axiom 5The new octonion BA/ |A| is the coordinate transformation of B by A in
four-dimensional space-time.
Axiom 7
The new octonion, describing the world point in four-dimensional spacetime, is
A = ct0 h + x0 i + y0 j + z0 k + ct1 + x1 hi + y1 hj + z1 hk.
However, ct0 and ct1 are both positive.

The space–time between the cth-axis and the straight line x = ct in Figure 19.1 isour familiar subluminal space–time. The theorems and conclusions obtained fromthe above axioms in this space–time are entirely consistent. However, whether these278

axioms are realized in the space–time between the straight line x = ct and the xiaxis is unknown. If Axioms 5 and 7 do not hold in space–time at superluminal
velocities, the formulation of (19.4) may alter at these velocities.

19.3 Zeno’s paradox and a discontinuity axiom
In the above calculations, we proved the possible existence of particles moving at
superluminal velocities, because we can evaluate a proper time for these particles.
However, c is generally regarded as an asymptotic speed; that is, the velocity v canapproach that of light, but cannot exceed it. This issue can be resolved throughZeno’s paradox.
Zeno was a Greek philosopher who lived in the 4th century B.C. Zeno’s paradox
takes various forms, but is typified by Achilles and the tortoise, which represent theswiftest running human being and one of the slowest moving animals, respectively.
To present an intelligible argument, we place Achilles at the origin O and the tortoiseslightly ahead of Achilles at distance x0 along the x-axis. Achilles and the tortoisesynchronously start to move. When Achilles arrives at the tortoise’s original position
x0 , the tortoise has moved to position x1 , ahead of its x0 . Later, when Achilles
arrives at x1 , the tortoise has arrived at position x2 , ahead of position x1 . After
repeating these movements several times, Achilles remains behind the tortoise. In
reality, however, Achilles will rapidly catch up with the tortoise. Since theory andreality contradict, this situation is a paradox. Mathematically, such contradictions
are treated by the theory of limits in calculus, expressed as lim (limit) or ε − δ(epsilon–delta). Intuitively, how an infinitesimal amount diﬀers from zero is better
explained by discussion rather than complicated mathematical theory.
As explained in Section 13.1, all theories have underlying axioms that can be
examined to solve contradiction of the theory. The axiom of Zeno’s paradox isthat space and time are continuous quantities. Zeno considered that space andtime are continuous with no gaps. In a continuous space–time, a quantity can be
infinitesimal but not zero. If this axiom is redefined in discontinuous space–time,
Zeno’s paradox is no longer a paradox. We call this axiom of discontinuous space–time a discontinuity axiom.
To facilitate understanding, we consider Zeno’s paradox in a diﬀerent setting.
Consider a river of width l that must be crossed, either by walking on a bridge orcrossing on stepping stones. When a person uses the bridge to reach the opposite
bank, he/she arrives halfway across the bridge (at l/2) at a certain time. Sometime
later, the walker has moved a further l/4, half of the remaining distance to the279

opposite bank. If the walker continues to reduce his/her distance to the opposite
bank by half, the distance to the opposite bank becomes infinitesimally small but
never zero, and the person cannot cross the river. In calculus, this situation is
resolved by taking the absolute limit
1
lim ( )n l = 0.
2n→∞Intuitively, however, an infinitesimal amount cannot be exactly zero.
What happens if we use the stepping stones to reach the opposite bank? Atsome time, we reach a stone that is halfway between the two banks. Sometime
later, we reach the last stepping stone that separates us from the opposite bank.
Since the position halfway between this last stone and the opposite bank is in theriver, we cannot move to this position. Instead, we jump to the opposite bank. If
Achilles competes with the tortoise on stepping stones rather than on a continuous
surface, he will eventually reach the tortoise’s position and surpass it. Therefore,
Zeno’s paradox is not realized in discontinuous space. From elementary particle
theory and quantum mechanics, space–time is known to be discontinuous at thevery smallest scales. By the discontinuity axiom, each coordinate axis of our fourdimensional space–time is discontinuous. The continuity axiom has been adopted
to comply with our everyday experience. Considering the discontinuity axiom, thevelocity v of a particle may approach and reach the velocity of light c and eventually
become superluminal. The axiom that space and time are discontinuous quantities
is necessary for a complete understanding of four-dimensional space–time.
Here, the size of the discrete quantity becomes important. In Section 18.1, we
specified the length of a unit world line as η. Since η expresses the mass of thesmallest fundamental particle, from the axiom that mass is the time component ofa unit world line, smaller lengths are impermissible. Thus, η may represent the unit
length of discretized space–time.

280

20
Future Problems

20.1 Electromagnetism and biquaternion
Up to and including Chapter 19, we proved that many physical phenomena can besolved in terms of the new octonion. This chapter is devoted to currently unsolved
problems, beginning with electromagnetism.
The basic laws of electromagnetism are the famous Maxwell field equations (named
after James Maxwell.) Maxwell described the then-known rules of electromagnetism
as twelve equations. Later, they were summarized to four fundamental equations.
Specifically, the electric field E, electric displacement D, magnetic field H, magnetic induction B, current density J , and volume density ρ (rho) of an electriccharge are related by∂D+ J = ∇ × H,∂t∇·D = ρ,∂B= ∇ × E,∂t∇·B = 0.

−Here, ∂ denotes a partial diﬀerence and ∇ (nabla) is specified in Cartesian coordinates as
∂
∂
∂
∇=i
+j
+k .
∂x
∂y
∂z
In the above equations, B, D, E, H, and J are vectors and i, j, and k are
not imaginary numbers of the new octonion, but unit vectors on the x-, y-, zaxes. The four Maxwell field equations are known to be invariant under the Lorentz
transformation.
The Lorentz transformation derived from special relativity and the new Lorentz
transformation derived from the new octonion yield the same formulae for t0 and281

x0 , but diﬀerent formulae for y 0 and z 0 . In the traditional Lorentz transformation,y 0 = y and z 0 = z. In Section 10.1, the new Lorentz transformation formulated by
the new octonion was given as
y + (v/c)zh
y0 = √,
1 − v 2 /c2(10.5)z − (v/c)yhz0 = √.
1 − v 2 /c2(10.6)As explained in previous chapters, these new y 0 and z 0 do not violate the constancyof the velocity of light, the world distance, momentum conservation law, or the
dependence of mass on velocity. However, as explained in Section 16.5, if lightis emitted perpendicular to the direction of an observer’s movement, the result
yielded by the new octonion diﬀers from that of the Lorentz transformation. The
correct formulation is indeterminable at present. Maxwell’s electromagnetic field
equations are certainly invariant under the standard Lorentz transformation. Their
possibility of being invariant under the new Lorentz transformation remains anunanswered question. If invariance is proven, what are the y 0 and z 0 components ofMaxwell’s equations in the new octonion formulation? Applying the new Lorentz
transformation to Maxwell field, equations may yield new solutions to phenomena
such as synchrotron radiation. Proving invariance is the first obstacle in applying
the new octonion to electromagnetism.
The identified problem mentioned above cannot be solved until Maxwell’s field
equations are appropriately expressed by the new octonion. As explained in Chapter
14, the vectors A and B are related to their corresponding new octonions A and B
byBA = A·B + A × B,

(14.17)

A × B = (BA − AB)/2,(14.26’)

A·B = (BA + AB)/2.

(14.27’)In terms of these expressions, Maxwell’s field equations might be reducible to simpler
forms that are easier to understand. Moreover, when a vector is rewritten by a newoctonion, its negative world component is easily identified. In the new octonion
formulation, the negative world components are the real number components hi, hj,
and hk. Thus, a portion of electromagnetic waves may reside in the negative world.
Since the time of Hamilton, many researchers have attempted and failed to rewrite
Maxwell field equations in terms of the quaternion. Thus, rewriting them by the
new octonion is challenging.
282

We now turn to the third problem. As explained in Section 8.3, this book has
accepted the axiom that the world line of a particle (constituting a substance) has
the same properties as the world line of light. If the world line of an entity follows thepath of zero world distance, where s2 = −c2 t2 + x2 + y 2 + z 2 = 0, its velocity equalsc, and it can be treated as light. Otherwise, the entity is called a substance because
its velocity diﬀers from c. Light is radiated as electromagnetic waves. If the worldlines of particles and light diﬀer only by their velocities, the world line of a substance
may also satisfy Maxwell’s field equations. Formulated in terms of the new octonion,
Maxwell’s field equations may convert v of the new octonion equations into c. In
addition, the force calculated by the new octonion equations may be intrinsic to theforce of the substance; that is, express its gravitational and nuclear force. Here lies
the third problem in relating the new octonion to electromagnetism; can we finda new octonion equation that generalizes Maxwell’s field equations? If such a newoctonion equation exists, it may unify all known forces. Moreover, if the world linesof substances and waves can be considered equivalent, the substance world line
is characterized by an amplitude Ψ (psi). The corresponding time component is∂ψ/∂t, the partial time diﬀerential of the wave function Ψ, which may be consistent
with quantum mechanics. To establish this relationship, Maxwell’s field equations
must be rewritten in terms of the new octonion.
As the third edition of this book was being translated into English, it was revealed that Koen J. van Vlaenderen and Andre Waser reformulated Maxwell’s field
equations using biquaternions (double quaternions) in 2001. The biquaternion algorithm is identical to that of the new octonion. In other words, the new octonion
and biquaternion are equivalent. Recall the algorithm presented in Section 3.4:
h2 = i2 = j 2 = k 2 = −1,
ij = −ji = k, jk = −kj = i, ki = −ik = j,hi = ih, hj = jh, hk = kh.
Moreover, the biquaternion A is obtained by re-expressing the coeﬃcients of thequaternion a + bi + cj + dk in complex number format as
A = (a + ph) + (b + qh)i + (c + rh)j + (d + sh)k.
Hamilton discovered the quaternion in 1843 and the biquaternion in 1844. The
following 22 years of his research career were devoted to applying the quaternion
to physics. However, he appears not to have appreciated the importance of thebiquaternion. Hamilton published proof of the biquaternion in his Lectures on283

Quaternions in 1853, nine years after its discovery, and discussed its applicability in
his famous Elements of Quaternions. However, the biquaternion has been neglected
in almost all quaternion texts to date.
Though the biquaternion and the new octonion are mathematically equivalent,
their interpretation and application methods markedly diﬀer in the following ways.
(1) In the biquaternion, h is treated as an attached imaginary number rather thana fourth imaginary number. Because the imaginary numbers i, j, and k form
coordinate axes, they have both magnitude and direction, whereas h is considered
as scalar. Thus, the algorithms of the biquaternion are
hi = ih, hj = jh, hk = kh.
The above relationships, where h is scalar, are assumed in all discussions of thebiquaternion. On the other hand, in the new octonion, h is the fourth imaginary
number and is more important than i, j, and k. As explained in Section 6.1, theimportant physical quantity of special relativity (proper time) is denoted by theimaginary number h. Mass and energy are also related through h, as explained in
Chapter 18. Moreover, in Section 3.4, we found that inserting hi = −ih in the newquaternion (or new octonion) does not yield the Lorentz transformation. Thus, inthis book, we have assumed that hi = ih. On the other hand, the biquaternion
allows hi = ih because h is a scalar. Furthermore, hi = ih indicates an important
structure of four-dimensional space–time. Although the imaginary numbers i, j, andk are mutually related by ij = k, h is not mutually related to i, j, and k throughhi = ih. Here, h expresses time and i, j, and k express space. Mathematically,
therefore, we can move from one space to another, but not from space to time. Thatis, time travel is mathematically impossible.
(2) In the biquaternion, a number may be rewritten as follows:
A = (a + ph) + (b + qh)i + (c + rh)j + (d + sh)k= (a + bi + cj + dk) + (p + qi + rj + sk)h.
However, as explained in Section 11.4, the new octonion expresses the world point ofa point mass in four-dimensional space–time. Thus, in the new octonion, a number
is expressed as
A = (ah + p) + (b + qh)i + (c + rh)j + (d + sh)k= (ah + bi + cj + dk) + (p + qhi + rhj + shk).

284

The left (imaginary) and right (real) halves of the right-hand term express the world
point in the positive and negative worlds, respectively. The new octonion, rather
than the biquaternion, appears to be an ideal way of expressing world points infour-dimensional space–time.
(3) In the biquaternion formulation, h is not a formal imaginary number but an
auxiliary number, and the biquaternion is a combination of quaternions. Thus, thebiquaternion is considered as a member of the quaternion family. However, h is
the most important component of the new octonion. In Section 11.8, hi, hj, andhk were identified as independent real numbers. Thus, the new octonion comprises
four real and four imaginary numbers and strictly belongs to the octonion family.
Since algebraic theory precludes more than eight dimensions, and the new octonion
algebra describes curved four-dimensional space–time, it presents as an ultimate
algebra. From this perspective, the new octonion is a more suitable mathematicalformulation than the biquaternion.
(4) Biquaternion calculations are performed on vector quantities. Denoting thescalar components of two quaternions p and q by p0 and q0 , respectively, and theirrespective vector components by p and q, we have p = p0 + p and q = q0 + q.
Furthermore, if the inner product is p.q and the outer product is p × q, the twoquaternions multiply as
pq = p0 q0 − p.q + p0 q + q0 p + p × q.
This formula is widely used in quaternion and biquaternion calculations. As explained in Chapter 14, the vectors can be rewritten in terms of the new octonions
as follows:
BA = A·B + A × B,

(14.17)

A × B = (BA − AB)/2,(14.26’)

A·B = (BA + AB)/2.

(14.27’)Moreover, as explained in Chapter 15, new octonions admit a wider range of calculations than tensors. Since complicated vector, tensor, and matrix calculations are
simplified in the new octonion formulation, physical research may be undertaken
purely by algebraic calculations in these formulations.

285

20.2 General relativity
In special relativity, an observer B moves along a straight line with uniform velocity
v, relative to a stationary observer A. If observer B accelerates relative to observer
A, his movement is described by general relativity. Succinctly, general relativity
postulates that
(1) The velocity of light c depends on the acceleration a.
(2) Gravity results from the bending of space by mass.

We now consider these postulates in terms of the new octonion. In general relativity texts, the first postulate is always exemplified by the case of a free-falling
elevator. The pulley of an elevator breaks, sending the elevator into free-fall asshown in Figure 20.1. The falling direction is assumed as the positive x-axis andthe horizontal direction is the y-axis. The coordinates of a stationary observer A
are denoted (x, y), while those of observer B inside the elevator are denoted by(x0 , y 0 ). A and B are located at origins O and O0 , respectively. Moreover, at any
given moment, the velocity of the elevator is v and B emits light in the y 0 -direction.
B observes the width of the elevator as lB .
First, we compute the Lorentz transformation of special relativity. The x-, y-,and z-axial components of the velocity of light, as seen by the stationary observer
A, are denoted Vx , Vy , and Vz , respectively. These components are given by (seeSection 16.2)
Vx = v,√
Vy = c 1 − v 2 /c2 ,(16.1)

Vz = 0.

(16.3)(16.2)

Since the length along the y-axis is unaltered in the Lorentz transformation, thewidth lB of the elevator observed by B is the same as that observed by A. Therefore,286

if light reaches the opposite wall of the elevator in time tA as observed by A, wefind that
tA =

lB
Vy

lB
= √
.
c 1 − v 2 /c2(20.1)On the other hand, since Vy0 = c, the time tB taken for light to traverse the elevatorin B’s reference frame is given by
tB =

lB
.
c

(20.2)

Although the velocity v alters under acceleration, we consider its value at a particular instant. From (16.2), (20.1), and (20.2), the velocity of light emitted in thehorizontal direction in the falling elevator is seen to be slower for a stationary observer A than for a moving observer B. The time at which light strikes the opposite
wall is also longer from A’s perspective. Because the infall velocity v increases under
acceleration a, by (16.1) and (16.2), Vx increases while Vy decreases. As shown in
Figure 20.2, A observes the path of light as downward parabolic.

However, under the new Lorentz transformation, observer A records the y component of the velocity of light emitted in the y 0 -direction in the elevator as c. This
result has been proven in Section 16.3. Under the new Lorentz transformation, we
have
Vx = v,(16.4)

Vy = c,(16.7)

Vz = vh.

(16.8)

287

We now determine the times tA and tB at which light reaches the opposite wallof the elevator. Diﬀerent from the traditional Lorentz transformation, the elevator
width changes with velocity v in the new Lorentz transformation. We seek therelationship between the width lB as seen from B and the width lA as seen from A.
The velocity of A relative to B is assumed as −v. At z = 0, setting y = lB , y 0 = lA ,and vx = −v in the new Lorentz transformationy + vx zh/cy0 = √,
1 − v 2 /c2
we find thatlA = √lB
1 − v 2 /c2(10.5).

Thus, from (16.7), we have
tA =

lA
Vy

lB
.
= √
c 1 − v 2 /c2(20.3)

Moreover,lB.
(20.4)
c
Since (20.3) and (20.4) are identical to (20.1) and (20.2), the locus of light observedby A is that of Figure 20.2. However, in (20.1) and (20.2), the y component of c is√
c 1 − v 2 /c2 , whereas in (20.3) and (20.4), it is exactly c. Although the y-axial comtB =ponents of the velocities are diﬀerent in the two formulations, the travel times are
identical because y 0 and y diﬀer in the new and traditional Lorentz transformations.
From the above discussion, we find that, even when one reference frame is accelerating, the velocity of light along the y-direction is c in the new Lorentz transformation, and no contradiction occurs. Denoting the position by x and acceleration
by a, general relativity gives the velocity of light c0 in the moving reference frameasax
c0 = (1 + 2 )c.
(20.5)c
The first problem in applying the new octonion to general relativity is to investigate
its eﬀect on (20.5).
We now examine the second postulate of general relativity; in that mass bends
space, and thereby producing gravity. Einstein’s theory of gravitation is wonderful
because it correctly predicted the existence of black holes. However, mass being
an independent physical quantity that bends space is debatable. As explained in
Section 6.1, when special relativity (formulated in non-accelerating reference frames)288

is rewritten in terms of the new octonion, the space–time relationship intrinsicallybends in the absence of mass. Moreover, in Chapter 18, we proved the special
relativity result, according to which energy equals c times the time component of
the momentum in four-dimensional space–time. We also demonstrated that, if massis regarded as the time component of a unit world line, we recover the energy–massequation E = mc2 . Einstein denied Newton’s concept of absolute space and timeand discovered that both quantities are relative. However, he accepted the conceptof absolute mass. This book introduces an axiom that mass is the time componentof a unit world line. If this axiom is accepted, absolute mass is denied and massacquires a distance unit. Therefore, under the new octonion formulation, whatis gravity? This constitutes the second problem in applying the new octonion to
general relativity.
The third problem is whether the new octonion yields the Lorentz transformationin an accelerating system. The Lorentz transformation of special relativity has noequivalent transformation in an accelerating system in general relativity texts forthe following reason. The Lorentz transformation is obtained using the constancyof the velocity of light, and is therefore inapplicable when the direction of thevelocity v of a moving observer is arbitrary or when the reference frame accelerates.
On the other hand, since we obtained the new Lorentz transformation using thecoordinate transformation by the new octonion, the new Lorentz transformation
may be obtainable in an accelerating system. The intermediate calculations areshown below.
Consider an observer B moving with proper acceleration a0 on the cth-xi plane.
A stationary observer A observes B accelerating with a. As explained in Section
17.7, a0 and a are related bya0 =a(1 − v 2 /c2 )3/2.

Since v is the velocity of B observed by A and a = dv/dt, we have
a0 =

dv/dt
.
(1 − v 2 /c2 )3/2

Rearranging and integrating over dt, we find that∫∫dv
a0 dt =,
(1 − v 2 /c2 )3/2v
+ C,
a0 t = √1 − v 2 /c2

289(17.69)where C is the constant of integration. If the initial velocity is zero, i.e., v = 0 at
t = 0, then C = 0, and we getv
a0 t = √.
1 − v 2 /c2(20.6)

Expressing (20.6) in terms of v, we obtain√a0 t 1 − v 2 /c2 = v,()
a20 t2 1 − v 2 /c2 = v 2 ,a20 t2 = v 2 + a20 t2 v 2 /c2 ,a20 t2 = v 2 (1 + a20 t2 /c2 ),
a0 t
v=√
.
1 + a20 t2 /c2
Note that v > 0 and t > 0. Since v = dx/dt, we can write
a0 t
dx
=√
.
dt
1 + a20 t2 /c2
Again rearranging and integrating over dt, we have
∫
∫
a tdt
√ 0
,
dx =
1 + a20 t2 /c2
which yieldsx=c2 √
1 + (a0 t/c)2 + C 0 ,a0where C 0 is a constant of integration. If x = 0 at t = 0, from (20.7), we havec2
+ C 0,a0c2
C0 = − .
a00=Substituting this result into (20.7), we find that
x=

]
c2 [√
1 + (a0 t/c)2 − 1 .
a0After some algebra, we obtain
√
a0 x
+
1
=
1 + (a0 t/c)2 ,
c2
(
)2
(a x
)2
a0 t
0
+
1
=
1
+
,
c2
c
290

(20.7)

(a x0c2)2 ( a t )2
0
+1 −
= 1.
c

(20.8)

Equation (20.8) describes the world line of observer B moving with proper acceleration a0 , and specifies the ct0 h-axis.
According to (20.8), a curve that is linearly symmetric to the line x = ct becomesthe x0 i-axis. The equation is solved by the method presented in Section 4.3. Interchanging x with ct; that is, replacing t by x/c in (20.8), we obtain the equation of
the x0 i-axis. Specifically, we have)2 (a 0 x )2
a0 ct
−
+
1
= 1,
2
c
c2
(
)2 (
a0 t
a 0 x )2
+1 −
= 1.
c
c2

(

(20.9)

Equation (20.9) is the equation of the x0 i-axis. (20.8) and (20.9) are hyperbolic
curves.

If the coordinates of a point mass D observed by A and B are (cth, xi) and(ct0 h, x0 i), respectively, the new complex plane becomes that of Figure 20.3. Allowing origin O to move along the curved x0 i-axis, we perform a parallel translationof curve OB. The position of O when the curve passes point D is denoted by F .
Similarly, if O moves along the curved ct0 h-axis, we perform a parallel translation of
curve OF . The position of O when the curve passes point D is E. At this instant,
we have
ct0 h + x0 i = |OE| h + |OF | i.
Here, |OE| and |OF | are the lengths of the curves OE and OF , respectively. Together with (20.8), and (20.9), the above formula describes a coordinate transformation in an accelerating reference frame in terms of the new octonion. Integration

291

must be used to obtain |OE| and |OF |. We also require the theorem proved in
Section 13.3, i.e.,
Theorem 26
The derivative world distance dl in two-dimensional space-time is√dl = chdt 1 − (dx)2 /(cdt)2 ,and (most likely) the hyperbolic functions (ex − e−x )/2 and (ex + e−x )/2. However, before performing the calculation, we must demonstrate that the exponentialfunction e is realized in the cth-xi plane.
As shown in Figure 20.3, since the ct0 h- and x0 i-axes traveled by B in an accelerating system are curved, B observes that space–time is curved. Thus, even in theabsence of mass, space–time is bent in an accelerating system. Importantly, thebending of space–time observed by B is relative. Absolute bending of space–time isrestricted to the cth-xi and yj-[zhk] planes in the new octonion formulation, as explained by Theorem 10 and Theorem 15 in Section 13.3. Obtaining the new Lorentz
transformation in an accelerating system is the third problem in applying the newoctonion to general relativity.

20.3 Five-dimensional space-time and string theory
String theory is one of the latest theories of modern physics. In this section, we
consider the relationship between string theory and the new octonion space–timetheory.
String theory evolved from the Kaluza–Klein theory. In the 1920s, physicists
were seeking to combine Maxwell’s field equations with the gravity theory of general relativity, into a so-called unified field theory. In 1919, Theodor Kaluza showedthat Einstein’s four-dimensional gravitational equation could be reformulated in
five-dimensions. The resulting formula was found to embody both general relativity
and electromagnetism. Moreover, in considering the physical meaning of this fifth
dimension, Oskar Klein suggested that it exists as a very small round coil. The
Kaluza-Klein theory was thought to have successfully united the electromagnetic
and gravitational fields. However, the theory was abandoned by around 1940, because the fifth dimension was smaller than the size of a fundamental particle, andtherefore unobservable.
Elementary particle physics studies the fundamental particles that constitute a
substance. This complicated theory has successfully explained the properties of292

particles. On the other hand, in 1970, Yoichiro Nambu and his colleagues mathematically demonstrated that nuclear particles behave as oscillating one-dimensional
strings of finite size, and the string theory was born. The first string theory required
26 dimensional space–time. However, in the supersymmetry theory of Pierre Ramond, every particle is coupled to a supersymmetric partner, reducing the numberof space–time dimensions to 10. Furthermore, the extra dimensions are considered
to be curled into small round structures, reminiscent of Kaluza–Klein theory.
The concepts of string theory are potentially consistent in the new octonion space–time theory. In the new octonion formulation, all particles are characterized by fivevariables; four variables (ct, x, y, z) describing their location in four-dimensional
space–time, and a variable Ψ indicating the amplitude of their world line. Moreover,as explained in Section 18.1, the size of the world line Ψ is smaller than a particle.
In Kaluza–Klein theory, the predecessor of string theory, the fifth dimension was
coiled into a structure that was too small to be observed. Brian Greene likened
the curled up fifth dimension to the circumference of the cross section of a garden
hose on the yard. The space–time theory of the new octonion similarly predicts aminiscule world line of an elementary particle with size Ψ. Thus, the fifth dimensionof Kaluza-Klein theory may be identical to the amplitude Ψ of a world line.
Since all particles are described by five variables, namely, ct, x, y, z, and Ψ,
five-dimensional space–time may appeal to physicists and mathematicians. However, since Ψ is the amplitude in four-dimensional space–time, space–time can only
be realized in four dimensions. According to algebraic theory, a numerical systemallowing the algebraic operations of addition, subtraction, multiplication, and division comprises only real numbers, complex numbers, quaternions, and octonions.
Therefore, five-dimensional space–time is precluded on algebraic grounds.
Moreover, since the new octonion space–time theory predicts a doubly-structured
four-dimensional space–time with overlapping negative and positive worlds, substances in the positive and negative worlds are each described by five variables,(ct, x, y, z, Ψ) and (ct0 , x0 , y 0 , z 0 , Ψ0 ), respectively. Since supersymmetric string
theory is realized only in ten-dimensional space–time, the ten variables of a substance in the new octonion space–time theory may be equivalent to ten-dimensional
space–time. Although ten variables are specified, only doubly-structured fourdimensional space–time is recognized by the new octonion space–time theory.
String theory postulates that all substances are ultimately composed of extremely
short vibrating strings. The new octonion space–time theory also accepts an axiom
that the world line of a substance vibrates in a light-like manner. As explained in
Section 19.3, space–time is discretized in the new octonion theory, and the length η293

of a unit world line is smaller than a particle. Thus, the new octonion space–timetheory shares many features of string theory, and the two theories may be fundamentally identical. However, whether string theory can be rewritten in terms of the
new octonion is uncertain. Kaluza–Klein theory is a tensor theory. As explainedin Section 15.2, tensor mathematics is limited to cross-sections of four-dimensional
space–time, whereas the new octonion admits the entire four-dimensional space–time. However, the diﬀerence between tensor and the new octonion formulations
is proven only in straight line coordinates. Whether the new octonion space–timetheory is consistent with the curvilinear coordinates of general relativity has notbeen examined in this text.

294

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295

296

Indexabsolute mass58, 245

absolute space 245
absolute time 245
associative law 125
axiom of the new octonion

energy–momentum equation

269

Euclid 117
extinction of the world line

267

five-dimensional space–timefour-acceleration 239

118

Big Bang 252
biquaternion 283

four-dimensional line 16
four-momentum 223

Cayley number

four real numbers 100
four-vector product 167
four-velocity 223

88

coeﬃcients comparison 128
complex number 1
conservation of momentum 231

free-falling elevator

constancy of the velocity of light13, 57
coordinate transformation 7, 14, 213
coordinate transformation of
momentum 215
curvature of space–time 136
curvature of the world line 67
defect of four-velocity
Dirac, P. 271

292

67, 263γ matrix 271
garden hose 293
general relativity 286
Graves, J. T. 88
gravity 288
Greene, B.

293

Hamilton, W. R. 18, 51, 87, 283
Higgs boson 255
hyperbolic curve 291

227

discontinuity axiom 279
double quaternion 283

imaginary number hdouble structure of space–timeinfinitesimal distance 148
inverse transformation 11, 82

E = mc2

90

253, 269

isotropy of space

72

Einstein, A. 67, 245, 289
electromagnetism 281

Kaluza–Klein theory

energy 249
energy conservation law

Kamada, H.
kinetic mass

258

297

18, 38, 80

272
218, 247

292

Kronecker δrectangular frame 141
reflection of the world line
rest mass 218

192

least-squares theory

54

265

length contraction 44, 45
Lorentz transformation 13, 16, 29

Riemannian geometry 118
rotation of coordinate axes 25

magnitude

rotation vector 172
rotational translation

119, 126

mass 245
mass of light 255
Maxwell field equations
metric tensor 194
Minkowski space–timemomentum 249
Nambu, Y.

scalar 155
special relativity

281

synchrotron radiation

292

69, 84

new quaternion 19
Newton’s equation of motion
oblique coordinate axes
octonion 88

197

tachyon 278
tensor 177

19

new Lorentz transformation
111
new octonion 89

parallel translation49

string theory 292
superluminal velocity 275
supersymmetry theory 293

8

negative world 91
new complex number

129, 134

theorem of the new octonion
three-vector product 164
time dilation 41
time travel 278
triple scalar product264

26, 103

133

twin paradox

59

unit world line

245

121

164

van Vlaenderen, K. J.

283

Planck constant 249, 256
positive world 91
proper length 44

vector 155
velocity transformation

75, 78, 215

proper time 41
Pythagoras’ theorem

Waser, A. 283
world distance 49

140

quadruple scalar productquantum mechanics
quaternion 18

world distance of lightworld line 16

167

58

55

world of imaginary numbers 92, 106
world of real number 92, 106

Ramond, P. 293
rectangular cross 137

Zeno’s paradox

298

279