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LORENTZ TRANSFORMATION WITHOUT
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Alain Mailliat

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Alain Mailliat. LORENTZ TRANSFORMATION WITHOUT ASSUMING LIGHT OR GROUP PROPERTIES Revision 1. 2024. hal-04588480

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<figure><img src="image_1.png" /></figure>

HAL
open science

LORENTZ TRANSFORMATION WITHOUT
ASSUMING LIGHT OR GROUP PROPERTIES

Revision 1

Alain Mailliat

► To cite this version:

Alain Mailliat. LORENTZ TRANSFORMATION WITHOUT ASSUMING LIGHT OR GROUP PROPERTIES Revision 1. 2024. hal-04588480

HAL Id: hal-04588480

https://hal.science/hal-04588480v1

Preprint submitted on 27 May 2024

HAL is a multi-disciplinary open access
archive for the deposit and dissemination of sci-
entific research documents, whether they are pub-
lished or not. The documents may come from
teaching and research institutions in France or
abroad, or from public or private research centers.

L'archive ouverte pluridisciplinaire HAL, est
destinée au dépôt et à la diffusion de documents
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émanant des établissements d'enseignement et de
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LORENTZ TRANSFORMATION
WITHOUT ASSUMING LIGHT OR GROUP PROPERTIES

Revision 1

Alain Mailliat

Commission Locale d'Information de Cadarache

# 1 INTRODUCTION

This paper objective is the derivation of Lorentz transformation between reference frames having uniform motion of translation, the so-called Galileans frames, without assuming neither light speed constancy cf. A. Einstein [1] APPENDIX I nor group structure properties for space-time cf. Levy Leblond [2].

After having assumed the homogeneity of space and time, the unique hypothesis taken into account for this derivation is the principle of relativity as stated by A. Einstein [1] chapter V.

"If, relative to R, R' is a uniformly moving co-ordinate system devoid of rotation, then natural phenomena run their course with respect to R' according to exactly the same general laws as with respect to R. This statement is called the principle of relativity (in the restricted sense)"

Hereafter, this principle of relativity is used in a practical way based on the following. If an observer in a reference frame R obtains a given result for a frame R' then, an observer in R' gets the same result for R. With such an understanding, this principle of relativity can be viewed as a principle of symmetry between Galilean reference frames.

The principle is applied three times successively for Lorentz transformation derivation.

Transformation formulas are of the same forms for coordinate changes from R' to R and from R to R'.

The equality of the measure from R of a segment with a given length lying in R' and the measure from R' of an identical segment in R.

Equality of frame velocities through speed addition formulas for three Galilean frames.
LORENTZ TRANSFORMATION
WITHOUT ASSUMING LIGHT OR GROUP PROPERTIES

Revision 1

Alain Mailliat

Commission Locale d'Information de Cadarache

# 1 INTRODUCTION

This paper objective is the derivation of Lorentz transformation between reference frames having uniform motion of translation, the so-called Galileans frames, without assuming neither light speed constancy cf. A. Einstein [1] APPENDIX I nor group structure properties for space-time cf. Levy Leblond [2].

After having assumed the homogeneity of space and time, the unique hypothesis taken into account for this derivation is the principle of relativity as stated by A. Einstein [1] chapter V.

"If, relative to R, R' is a uniformly moving co-ordinate system devoid of rotation, then natural phenomena run their course with respect to R' according to exactly the same general laws as with respect to R. This statement is called the principle of relativity (in the restricted sense)"

Hereafter, this principle of relativity is used in a practical way based on the following. If an observer in a reference frame R obtains a given result for a frame R' then, an observer in R' gets the same result for R. With such an understanding, this principle of relativity can be viewed as a principle of symmetry between Galilean reference frames.

The principle is applied three times successively for Lorentz transformation derivation.

Transformation formulas are of the same forms for coordinate changes from R' to R and from R to R'.

The equality of the measure from R of a segment with a given length lying in R' and the measure from R' of an identical segment in R.

Equality of frame velocities through speed addition formulas for three Galilean frames.
## 2 NO REFERENCE FRAME ABSOLUTELY AT REST

Let us recall several points from A. Einstein [1] chapter V.

* If the Galilean frames are not equivalent there is a special one offering the advantage of simplified natural laws. “We should then be justified (because of its merits for the description of natural phenomena) in calling this system “absolutely at rest,” cf. [1] “and all other Galilean systems “in motion.”

* Physics laws in any reference frame in uniform motion have to include its velocity relative to the absolutely at rest frame because two Galileans frames are only identified through their relative velocity.

* The terrestrial globe runs in opposite directions at two dates having a difference of 6 months and then cannot be regarded as motionless reference frame. So physical laws include the globe velocity relative to a frame absolutely at rest.

* But observations on earth do not reveal any dependence to neither its velocity nor to its direction. Therefore, physics laws in Galilean reference frames do not involve its own velocity relative to any other reference frame.

* So, A, Einstein [1] assumed that there is no reference frame absolutely at rest and “If, relative to R, R' is a uniformly moving coordinate system devoid of rotation, then natural phenomena run their course with respect to R' according to exactly the same general laws as with respect to R. This statement is called the principle of relativity (in the restricted sense)”

Then, there is no way to discriminate between two Galilean frames. Their velocities are relative and frames can be regarded at rest or in motion in turn.

Any observations of R from R' are identical to those of R' from R. Two Galilean reference frames being indistinguishable we have here a kind of principle of symmetry which will be used here after for Lorentz transformation derivation.

## 3 FORMULAS FOR FRAME CHANGES

The 3 coordinates for space and time $\vec{X}$ for a point E or event are observed from a first Galilean frame R of dimension 4. If an other frame R' is used, the same event is seen with coordinates $\vec{X}'$.

To exchange coordinates from a frame to another one, we need a relationship involving some parameters $\{a\}_{1,n}$. A general form for such a relationship is proposed by Levy-Leblond [2]:

$$ (1) \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \vec{X}' = \vec{F}(\vec{X}, \{a\}_{1,n}). $$

This formula has to provide only one $\vec{X}'$ for a given $\vec{X}$. If (1) includes 4 parameters or more, in general, we can associate any $\vec{X}'$ arbitrary chosen to our given $\vec{X}$. So the $\{a\}_{1,n}$ have only 3 parameters $\{a\}_{1,3}$,
## 2 NO REFERENCE FRAME ABSOLUTELY AT REST

Let us recall several points from A. Einstein [1] chapter V.

* If the Galilean frames are not equivalent there is a special one offering the advantage of simplified natural laws. “We should then be justified (because of its merits for the description of natural phenomena) in calling this system “absolutely at rest,” cf. [1] “and all other Galilean systems “in motion.”

* Physics laws in any reference frame in uniform motion have to include its velocity relative to the absolutely at rest frame because two Galileans frames are only identified through their relative velocity.

* The terrestrial globe runs in opposite directions at two dates having a difference of 6 months and then cannot be regarded as motionless reference frame. So physical laws include the globe velocity relative to a frame absolutely at rest.

* But observations on earth do not reveal any dependence to neither its velocity nor to its direction. Therefore, physics laws in Galilean reference frames do not involve its own velocity relative to any other reference frame.

* So, A, Einstein [1] assumed that there is no reference frame absolutely at rest and “If, relative to R, R' is a uniformly moving coordinate system devoid of rotation, then natural phenomena run their course with respect to R' according to exactly the same general laws as with respect to R. This statement is called the principle of relativity (in the restricted sense)”

Then, there is no way to discriminate between two Galilean frames. Their velocities are relative and frames can be regarded at rest or in motion in turn.

Any observations of R from R' are identical to those of R' from R. Two Galilean reference frames being indistinguishable we have here a kind of principle of symmetry which will be used here after for Lorentz transformation derivation.

## 3 FORMULAS FOR FRAME CHANGES

The 3 coordinates for space and time $\vec{X}$ for a point E or event are observed from a first Galilean frame R of dimension 4. If an other frame R' is used, the same event is seen with coordinates $\vec{X}'$.

To exchange coordinates from a frame to another one, we need a relationship involving some parameters $\{a\}_{1,n}$. A general form for such a relationship is proposed by Levy-Leblond [2]:

$$ (1) \qquad \vec{X}' = \vec{F}(\vec{X}, \{a\}_{1,n}). $$

This formula has to provide only one $\vec{X}'$ for a given $\vec{X}$. If (1) includes 4 parameters or more, in general, we can associate any $\vec{X}'$ arbitrary chosen to our given $\vec{X}$. So the $\{a\}_{1,n}$ have only 3 parameters $\{a\}_{1,3}$,
The ${a}_{1,3}$ can be regarded as 3 components of a 4 dimension vector with one of them being arbitrary $a_4=1$ for example. The ${a}_{1,3}$ are functions of the relationship between the two frames and (1) can rewritten as:

$$ \vec{X}' = \vec{F}(\vec{X}, \vec{A}). $$

## 4 SPACE TIME HOMOGENEITY.

Space time is homogeneous if its properties do not depend of location and time translations. Let us consider a small movement of the event E. Its previous location $\vec{X}$ in frame R is modified by $\delta \vec{X}$ and its location $\vec{X}'$ in R' is increased of $\delta \vec{X}'$. Applying (1) to the event movement provides

$$ \vec{X}' + \delta \vec{X}' = \vec{F}(\vec{X} + \delta \vec{X}, \vec{A}), $$

where, using $\vec{X}'=\vec{F}(\vec{X},\vec{A})$ we have:

$$ \delta \vec{X}' = \vec{F}(\vec{X} + \delta \vec{X}, \vec{A}) - \vec{F}(\vec{X}, \vec{A}). $$

This equation can be approximated using a first order development

$$ \delta X'_{k} = \sum_{i=1}^{4} \frac{\partial F_{k}(\vec{X}, \vec{A})}{\partial X_{i}} \delta X_{i}. $$

In order to have $\delta \vec{X}'$ independent of $\vec{X}$, we need that all the partial derivatives $\frac{\partial F_k(\vec{X}, \vec{A})}{\partial X_i}$ be constant with respect to space and time. So, they are functions of $\vec{A}$ parameter only and the $F_k$ are linear functions of $\vec{X}$. Writing $T_{ki}(\vec{A})=\frac{\partial F_k(\vec{X}, \vec{A})}{\partial X_i}$, we have:

$$ F_k(\vec{X}, \vec{A}) = \sum_{i=1}^{4} T_{ki}(\vec{A}) X_i + C_k \quad \forall k = \{1, \dots, 4\}. $$

Within this previous relation, $C_k$ are the coordinates of event E in the frame R' when it is observed at $\vec{X}=0$ from frame R, so to say at the origin of this last one. If frame R' is chosen in such a way all the $C_k$s are zero, then the relation between the two frames is:

$$ (2) \qquad \vec{X}' = T \vec{X} \quad \text{where T is 4x4 matrix.} $$

Now, our intention is to derive the parameters ${a}_{1,3}$ which define the transformations between the Galileans frames.
The ${a}_{1,3}$ can be regarded as 3 components of a 4 dimension vector with one of them being arbitrary $a_4=1$ for example. The ${a}_{1,3}$ are functions of the relationship between the two frames and (1) can rewritten as:

$$ \vec{X}' = \vec{F}(\vec{X}, \vec{A}). $$

# 4 SPACE TIME HOMOGENEITY.

Space time is homogeneous if its properties do not depend of location and time translations. Let us consider a small movement of the event E. Its previous location $\vec{X}$ in frame R is modified by $\delta \vec{X}$ and its location $\vec{X}'$ in R' is increased of $\delta \vec{X}'$. Applying (1) to the event movement provides

$$ \vec{X}' + \delta \vec{X}' = \vec{F}(\vec{X} + \delta \vec{X}, \vec{A}), $$

where, using $\vec{X}'=\vec{F}(\vec{X},\vec{A})$ we have:

$$ \delta \vec{X}' = \vec{F}(\vec{X} + \delta \vec{X}, \vec{A}) - \vec{F}(\vec{X}, \vec{A}). $$

This equation can be approximated using a first order development

$$ \delta X'_{k} = \sum_{i=1}^{4} \frac{\partial F_{k}(\vec{X}, \vec{A})}{\partial X_{i}} \delta X_{i}. $$

In order to have $\delta \vec{X}'$ independent of $\vec{X}$, we need that all the partial derivatives $\frac{\partial F_k(\vec{X}, \vec{A})}{\partial X_i}$ be constant with respect to space and time. So, they are functions of $\vec{A}$

parameter only and the $F_k$ are linear functions of $\vec{X}$. Writing $T_{ki}(\vec{A})=\frac{\partial F_k(\vec{X}, \vec{A})}{\partial X_i}$, we have:

$$ F_k(\vec{X}, \vec{A}) = \sum_{i=1}^{4} T_{ki}(\vec{A}) X_i + C_k \quad \forall k = \{1, \dots, 4\}. $$

Within this previous relation, $C_k$ are the coordinates of event E in the frame R' when it is observed at $\vec{X}=0$ from frame R, so to say at the origin of this last one. If frame R' is chosen in such a way all the $C_k$s are zero, then the relation between the two frames is:

$$ (2) \qquad \vec{X}' = T \vec{X} \quad \text{where T is 4x4 matrix.} $$

Now, our intention is to derive the parameters ${a}_{1,3}$ which define the transformations between the Galileans frames.
## 5 RELATIONS BETWEEN FRAMES.

Hereafter we consider the usual case with 2 dimensions, the first one for space, the second one for time. The two frames $R$ and $R'$ are two parallel axes having uniform motions. Locations on the axes are called $x$ et $x'$ and the frame times are measured by quantities $t$ and $t'$. $R'$ origin is $x'=0$ the $R$ one is $x=0$.

According to the conclusion of chapter 4, $R$ et $R'$ having 2 dimensions the transformation between $R$ et $R'$ is a linear relationship (2) with coefficients function of a unique parameter $a$:

$$ (3) \qquad \begin{aligned} x' &= M(a)x - N(a)t \\ t' &= O(a)x - P(a)t \end{aligned} $$

Let us assume that $R$ postulates that $R'$ is in motion compared with him with a velocity $v$. With $R$ et $R'$ origins corresponding at time $t=0$, the location $x$ of $R'$ in $R$ is $x=vt$. Using this in the first equality of (3) we have:

$$ 0 = [M(a)v - N(a)]t. $$

This relation will be verified for any $t$ only if:

$$ (4) \qquad \frac{N(a)}{M(a)} = v. $$

This last condition (4) provides $a$ as a function $a(v)$ of $v$ and we set down $\gamma(v)=M(a(v))$.

It can be pointed out that, except $x$ and $t$, the only physical quantity available for deriving a relationship between the frames $R$ et $R'$ is their relative velocity $v$. Therefore, it is almost obvious to look for $a$ as a velocity function and write down immediately:

$$ \begin{aligned} x' &= M(a(v))x - N(a(v))t \\ t' &= O(a(v))x - P(a(v))t \end{aligned} $$

Relation (3) between $R$ and $R'$ is rewritten:

$$ (5) \qquad \begin{aligned} x' &= \gamma(v)[x - vt] \\ t' &= \gamma(v)[-\mu(v)x + \lambda(v)t] \end{aligned} $$

with:

$$ O(a(v)) = -\gamma(v)\mu(v) \qquad P(a(v)) = -\gamma(v)\lambda(v). $$

Applying the relativity principle, to observe $R'$ from $R$ is identical to observe $R$ from $R'$. So, the corresponding transformations have to present identical forms. If $v'$ is the uniform velocity assigned to $R$ by $R'$, the $R$ origin $x=0$ location seen from $R'$ is $x'=v't'$. Rewriting (5) we have:
## 5 RELATIONS BETWEEN FRAMES.

Hereafter we consider the usual case with 2 dimensions, the first one for space, the second one for time. The two frames $R$ and $R'$ are two parallel axes having uniform motions. Locations on the axes are called $x$ et $x'$ and the frame times are measured by quantities $t$ and $t'$. $R'$ origin is $x'=0$ the $R$ one is $x=0$.

According to the conclusion of chapter 4, $R$ et $R'$ having 2 dimensions the transformation between $R$ et $R'$ is a linear relationship (2) with coefficients function of a unique parameter $a$:

$$ (3) \qquad \begin{aligned} x' &= M(a)x - N(a)t \\ t' &= O(a)x - P(a)t \end{aligned} $$

Let us assume that $R$ postulates that $R'$ is in motion compared with him with a velocity $v$. With $R$ et $R'$ origins corresponding at time $t=0$, the location $x$ of $R'$ in $R$ is $x=vt$. Using this in the first equality of (3) we have:

$$ 0 = [M(a)v - N(a)]t. $$

This relation will be verified for any $t$ only if:

$$ (4) \qquad \frac{N(a)}{M(a)} = v. $$

This last condition (4) provides $a$ as a function $a(v)$ of $v$ and we set down $\gamma(v)=M(a(v))$.

It can be pointed out that, except $x$ and $t$, the only physical quantity available for deriving a relationship between the frames $R$ et $R'$ is their relative velocity $v$. Therefore, it is almost obvious to look for $a$ as a velocity function and write down immediately:

$$ \begin{aligned} x' &= M(a(v))x - N(a(v))t \\ t' &= O(a(v))x - P(a(v))t \end{aligned} $$

Relation (3) between $R$ and $R'$ is rewritten:

$$ (5) \qquad \begin{aligned} x' &= \gamma(v)[x - vt] \\ t' &= \gamma(v)[-\mu(v)x + \lambda(v)t] \end{aligned} $$

with:

$$ O(a(v)) = -\gamma(v)\mu(v) \qquad P(a(v)) = -\gamma(v)\lambda(v). $$

Applying the relativity principle, to observe $R'$ from $R$ is identical to observe $R$ from $R'$. So, the corresponding transformations have to present identical forms. If $v'$ is the uniform velocity assigned to $R$ by $R'$, the $R$ origin $x=0$ location seen from $R'$ is $x'=v't'$. Rewriting (5) we have:
$$ (6) \qquad \left. \begin{array}{l} x = \gamma(\nu')[x' - \nu't'] \\ t = \gamma(\nu')[-\mu(\nu')x + \lambda(\nu')t'] \end{array} \right\} $$

The relativity principle does not allow identifying which one of the reference frames is in motion. If $R$ see $R'$ moving forward with velocity $v$ then, $R'$ see $R$ moving as fast but backward so with velocity $v'=-v$. We carry on the velocity $v'=-v$ into (6).

$$ (7) \qquad \left. \begin{array}{l} x = \gamma(-v)[x' + vt'] \\ t = \gamma(-v)[-\mu(-v)x' + \lambda(-v)t'] \end{array} \right\} $$

It can be pointed out that the relations (7) can be derived directly from the following argument based on the principle of relativity.

The $R$ observer postulates being motionless and frame $R'$ in motion from its left to its right sides. He assigns a velocity $v$ to $R'$ and applies (5). Then, as the $R$ observer can not decide which one of the frames is motionless, he assumes that $R'$ is motionless and himself moving now from its right side to its left one, in the reverse direction than previously, and consequently, its velocity is now $-v$ in comparison to $R'$. After what, the $R$ observer assumes that observer $R'$ will apply the same transformation (5) than he did motionless but using $R'$ space and time coordinates, so switching the $x, t$ by the $x', t'$ and changing velocity $v$ by $-v$.

The coordinates $x$ and $t$ in (7) can be derived by solving (5):

$$ (8) \qquad \left. \begin{array}{l} x = \frac{1}{D(\nu)}[\lambda(\nu)x' + v t'] \\ t = \frac{1}{D(\nu)}[\mu(\nu)x' + t'] \end{array} \right\} , \qquad D(\nu) = \gamma(\nu)[\lambda(\nu) - v\mu(\nu)] $$

For any $x'$ and $t'$ the coordinates $x$ et $t$ have to be identical obtained from (7) or from (8). Such conditions require equality of their coefficients in the two relations.

We get $\gamma(-v)=\frac{\lambda(\nu)}{D(\nu)}$ and $v\gamma(-v)=\frac{v}{D(\nu)}$ providing $\lambda(\nu)=1 \ \forall v$ and there: $\lambda(-v)=1$. In brief:

$$ (9) \qquad \gamma(-\nu) = \frac{1}{\gamma(\nu)[1 - \nu\mu(\nu)]} \qquad \mu(-\nu) = -\mu(\nu). $$

$R$ et $R'$ transformations can be rewritten:

$$ (10) \qquad \left. \begin{array}{l} x = \gamma(-\nu)[x' + v t'] \\ t = \gamma(-\nu)[\mu(\nu)x' + t'] \end{array} \right\} \qquad \begin{array}{l} x' = \gamma(\nu)[x - v t'] \\ t' = \gamma(\nu)[-\mu(\nu)x + t'] \end{array} $$

When the frame $R'$ is motionless in comparison to $R$, so to say when $v=0$ then $x'=x$ and $t'=t$ are needed. So, from (10) we obtain $\gamma(0)=1$ et $\mu(0)=0$.
$$ (6) \qquad \left. \begin{aligned} x &= \gamma(\nu') [x' - \nu' t'] \\ t &= \gamma(\nu') [-\mu(\nu')x + \lambda(\nu')t'] \end{aligned} \right\} $$

The relativity principle does not allow identifying which one of the reference frames is in motion. If R see R' moving forward with velocity $v$ then, R' see R moving as fast but backward so with velocity $v'=-v$. We carry on the velocity $v'=-v$ into (6).

$$ (7) \qquad \left. \begin{aligned} x &= \gamma(-\nu) [x' + \nu t'] \\ t &= \gamma(-\nu) [-\mu(-\nu)x' + \lambda(-\nu)t'] \end{aligned} \right\} $$

It can be pointed out that the relations (7) can be derived directly from the following argument based on the principle of relativity.

The R observer postulates being motionless and frame R' in motion from its left to its right sides. He assigns a velocity $v$ to R' and applies (5). Then, as the R observer can not decide which one of the frames is motionless, he assumes that R' is motionless and himself moving now from its right side to its left one, in the reverse direction than previously, and consequently, its velocity is now $-v$ in comparison to R'. After what, the R observer assumes that observer R' will apply the same transformation (5) than he did motionless but using R' space and time coordinates, so switching the $x, t$ by the $x', t'$ and changing velocity $v$ by $-v$.

The coordinates $x$ and $t$ in (7) can be derived by solving (5):

$$ (8) \qquad \left. \begin{aligned} x &= \frac{1}{D(\nu)} [\lambda(\nu)x' + \nu t'] \\ t &= \frac{1}{D(\nu)} [\mu(\nu)x' + t'] \end{aligned} \right\} , \qquad D(\nu) = \gamma(\nu)[\lambda(\nu) - \nu\mu(\nu)] $$

For any $x'$ and $t'$ the coordinates $x$ et $t$ have to be identical obtained from (7) or from (8). Such conditions require equality of their coefficients in the two relations.

We get $\gamma(-\nu) = \frac{\lambda(\nu)}{D(\nu)}$ and $\nu\gamma(-\nu) = \frac{\nu}{D(\nu)}$ providing $\lambda(\nu)=1 \ \forall \nu$ and there: $\lambda(-\nu)=1$. In brief:

$$ (9) \qquad \gamma(-\nu) = \frac{1}{\gamma(\nu)[1 - \nu\mu(\nu)]} \qquad \mu(-\nu) = -\mu(\nu). $$

R et R' transformations can be rewritten:

$$ (10) \qquad \left. \begin{aligned} x &= \gamma(-\nu) [x' + \nu t'] \\ t &= \gamma(-\nu) [\mu(\nu)x' + t'] \end{aligned} \right\} \qquad \begin{aligned} x' &= \gamma(\nu) [x - \nu t'] \\ t' &= \gamma(\nu) [-\mu(\nu)x + t] \end{aligned} $$

When the frame R' is motionless in comparison to R, so to say when $v=0$ then $x'=x$ and $t'=t$ are needed. So, from (10) we obtain $\gamma(0)=1$ et $\mu(0)=0$.
To find the two functions $\gamma$ and $\mu$ we consider a segment of length $l$ motionless in $R'$. Observed from $R$, this segment has a measure $m(l)$. Symmetrically, a segment with length $l$ motionless in $R$ has a measure $m'(l)$ seen from $R'$. Results in $R$ obtained from $R'$ can not differ from results in $R'$ obtained from $R$. According to the relativity principle, we can not distinguish which one of the frames is in uniform motion in comparison to the other. So, the two observer have measures of length to be identical:

$$ (11) \qquad m(l)=m'(l). $$

Two points $x'_1$ et $x'_2=x'_1+l$ in $R'$ are observed at the same time<fnref n="1" /> $t'_2=t'_1$. Seen from $R$ the corresponding point locations in $R$ verify $x_2-x_1=m(l)$ and their time difference is $\Delta t=(t_2-t_1)$. From (10) we have:

$$ \left| \begin{array}{lcl} x'_{2}-x'_{1} & = & l \qquad \gamma(\upsilon)[m(l)-\upsilon\Delta t] \\ 0 & = & \gamma(\upsilon)[-\mu(\upsilon)m(l)+\Delta t] \end{array} \right. \\ \text{Then, eliminating } \Delta t \text{ in the 2 relations, we get:} \\ (12) \qquad l=\gamma(\upsilon)m(l)[1-\upsilon\mu(\upsilon)]. $$

In the same way, two points $l$ apart in $R$ considered at same time $t_2=t_1$ are measured with a length $m'(l)$ from $R'$ and, according to (10), verify

$$ \left| \begin{array}{lcl} x_2-x_1 & = & \gamma(-\upsilon)[m'(l)+\upsilon\Delta t'] \\ 0 & = & \gamma(-\upsilon)[\mu(\upsilon)m'(l)+\Delta t'] \end{array} \right. \\ \text{and we arrive to:} \\ (13) \qquad l=\gamma(-\upsilon)m'(l)[1-\upsilon\mu(\upsilon)]. $$

Finally, relations (11) (12) (13) provide the condition:

$$ (14) \qquad \gamma(-\upsilon)=\gamma(\upsilon), $$ which is carried into the first equality of (9) and gives the function $\gamma(\upsilon)$ we were looking for:

$$ (15) \qquad \gamma^2(\upsilon)=\frac{1}{(1-\upsilon\mu(\upsilon))}. $$ As $\gamma$ is squared in the previous relation, we have an additional condition for $\mu(\upsilon)$ namely:

According to the previous results, the transformations (10) between the Galileans frames can be rewritten as following:

<footnote n="1">¹ The observer and its segment are both motionless, so there is no difficulty for observing segment ends simultaneously.</footnote>
To find the two functions $\gamma$ and $\mu$ we consider a segment of length $l$ motionless in $R'$. Observed from $R$, this segment has a measure $m(l)$. Symmetrically, a segment with length $l$ motionless in $R$ has a measure $m'(l)$ seen from $R'$. Results in $R$ obtained from $R'$ can not differ from results in $R'$ obtained from $R$. According to the relativity principle, we can not distinguish which one of the frames is in uniform motion in comparison to the other. So, the two observer have measures of length to be identical:

$$ (11) \qquad m(l)=m'(l). $$

Two points $x'_1$ et $x'_2=x'_1+l$ in $R'$ are observed at the same time<fnref n="1" /> $t'_2=t'_1$. Seen from $R$ the corresponding point locations in $R$ verify $x_2-x_1=m(l)$ and their time difference is $\Delta t=(t_2-t_1)$. From (10) we have:

$$ \left| \begin{array}{lcl} x'_{2}-x'_{1} & = & l \qquad \gamma(\upsilon)[m(l)-\upsilon\Delta t] \\ & & 0 \qquad \gamma(\upsilon)[-\mu(\upsilon)m(l)+\Delta t] \end{array} \right. \\ \text{Then, eliminating } \Delta t \text{ in the 2 relations, we get:} \\ (12) \qquad l=\gamma(\upsilon)m(l)[1-\upsilon\mu(\upsilon)]. $$

In the same way, two points $l$ apart in $R$ considered at same time $t_2=t_1$ are measured with a length $m'(l)$ from $R'$ and, according to (10), verify

$$ \left| \begin{array}{lcl} x_2-x_1 & = & \gamma(-\upsilon)[m'(l)+\upsilon\Delta t'] \\ & & 0 \qquad \gamma(-\upsilon)[\mu(\upsilon)m'(l)+\Delta t'] \end{array} \right. \\ \text{and we arrive to:} \\ (13) \qquad l=\gamma(-\upsilon)m'(l)[1-\upsilon\mu(\upsilon)]. $$

Finally, relations (11) (12) (13) provide the condition:

$$ (14) \qquad \gamma(-\upsilon)=\gamma(\upsilon), $$ which is carried into the first equality of (9) and gives the function $\gamma(\upsilon)$ we were looking for:

$$ (15) \qquad \gamma^2(\upsilon)=\frac{1}{(1-\upsilon\mu(\upsilon))}. $$ As $\gamma$ is squared in the previous relation, we have an additional condition for $\mu(\upsilon)$ namely:

According to the previous results, the transformations (10) between the Galileans frames can be rewritten as following:

<footnote n="1">¹ The observer and its segment are both motionless, so there is no difficulty for observing segment ends simultaneously.</footnote>
$$ (17) \quad \begin{aligned} x &= \gamma(\nu)[x' + \nu t'] \\ t &= \gamma(\nu)[\mu(\nu)x' + t'] \end{aligned} \qquad \begin{aligned} x' &= \gamma(\nu)[x - \nu t'] \\ t' &= \gamma(\nu)[-\mu(\nu)x + t'] \end{aligned} $$

To finalize the above transformations we have to find $\mu(\nu)$. It will be done now through the relationship between frame velocities.

## 6 RELATIONSHIP FOR VELOCITIES.

Let us consider a point in a frame $R'$ having an arbitrary motion given by an implicit equation $f(x',t')=0$. This frame has a uniform motion with velocity $v$ in comparison to a Galilean frame $R$.

In $R'$ the point velocity $u'(t)$ is given by:

$$ u'(t') = - \frac{\frac{\partial f}{\partial t'}}{\frac{\partial f}{\partial x'}} $$

The movement of this point, observed from $R$, is obtained by carrying the coordinate transformations (17) into $f(x',t')=0 : g(x,t)=f(x'(x,t),t'(x,t))=0$. The velocity of the point seen from $R$ is given by:

$$ u(t, \nu) = - \frac{\frac{\partial f}{\partial t}}{\frac{\partial f}{\partial x}} = - \frac{\frac{\partial f}{\partial x'} \frac{\partial x'}{\partial t} + \frac{\partial f}{\partial t'} \frac{\partial t'}{\partial t}}{\frac{\partial f}{\partial x'} \frac{\partial x'}{\partial x} + \frac{\partial f}{\partial t'} \frac{\partial t'}{\partial x}} $$

Dividing both sides of the fraction by $\frac{\partial f}{\partial x'}$ for highlighting $u'$ velocity we arrive to:

$$ u(t, \nu) = - \frac{\frac{\partial x'}{\partial t} - u' \frac{\partial t'}{\partial t}}{\frac{\partial x'}{\partial x} - u' \frac{\partial t'}{\partial x}} $$

where we replace derivatives $\frac{\partial x'}{\partial t}$ and $\frac{\partial t'}{\partial t}$ deduced from (17).

Finally, the relationship for velocity addition is:

$$ (18) \qquad u(t, \nu) = \frac{\nu + u'(t)}{1 + \mu(\nu)u'(t)}. $$

We see that the velocity of a point in motion is not just a sum of its own velocity and the velocity of the frame where it is located. The usual classic sum has to be corrected by a quantity depending on both point and frame velocities. According to the result $\mu(0)=0$ obtained above, (18) reduces to the classic addition of speeds only when the frame velocity is zero.
$$ (17) \quad \begin{aligned} x &= \gamma(\nu)[x' + \nu t'] \\ t &= \gamma(\nu)[\mu(\nu)x' + t'] \end{aligned} \qquad \begin{aligned} x' &= \gamma(\nu)[x - \nu t'] \\ t' &= \gamma(\nu)[-\mu(\nu)x + t'] \end{aligned} $$

To finalize the above transformations we have to find $\mu(\nu)$. It will be done now through the relationship between frame velocities.

## 6 RELATIONSHIP FOR VELOCITIES.

Let us consider a point in a frame $R'$ having an arbitrary motion given by an implicit equation $f(x',t')=0$. This frame has a uniform motion with velocity $v$ in comparison to a Galilean frame $R$.

In $R'$ the point velocity $u'(t)$ is given by:

$$ u'(t') = - \frac{\frac{\partial f}{\partial t'}}{\frac{\partial f}{\partial x'}} $$

The movement of this point, observed from $R$, is obtained by carrying the coordinate transformations (17) into $f(x',t')=0 : g(x,t)=f(x'(x,t),t'(x,t))=0$.

The velocity of the point seen from $R$ is given by:

$$ u(t, \nu) = - \frac{\frac{\partial f}{\partial t}}{\frac{\partial f}{\partial x}} = - \frac{\frac{\partial f}{\partial x'} \frac{\partial x'}{\partial t} + \frac{\partial f}{\partial t'} \frac{\partial t'}{\partial t}}{\frac{\partial f}{\partial x'} \frac{\partial x'}{\partial x} + \frac{\partial f}{\partial t'} \frac{\partial t'}{\partial x}} $$

Dividing both sides of the fraction by $\frac{\partial f}{\partial x'}$ for highlighting $u'$ velocity we arrive to:

$$ u(t, \nu) = - \frac{\frac{\partial x'}{\partial t} - u' \frac{\partial t'}{\partial t}}{\frac{\partial x'}{\partial x} - u' \frac{\partial t'}{\partial x}} $$

where we replace derivatives $\frac{\partial x'}{\partial t}$ and $\frac{\partial t'}{\partial t}$ deduced from (17).

Finally, the relationship for velocity addition is:

$$ (18) \qquad u(t, \nu) = \frac{\nu + u'(t)}{1 + \mu(\nu)u'(t)}. $$

We see that the velocity of a point in motion is not just a sum of its own velocity and the velocity of the frame where it is located. The usual classic sum has to be corrected by a quantity depending on both point and frame velocities. According to the result $\mu(0)=0$ obtained above, (18) reduces to the classic addition of speeds only when the frame velocity is zero.
# 7 RELATION TIME VERSUS SPACE

Now we have to find the quantity $\mu(\nu)$ which provides the relation between time and space in equations (17). Two methods are available based on frame symmetries or dimensional analysis.

## 7.1 FRAME SYMMETRIES

Now we will find $\mu(\nu)$ using once more the relativity principle applied to three Galilean frames $R_1, R_2, R_3$ in uniform motions. $R_2$ has a velocity $v_{2/1}$ in comparison to $R_1, R_3$ with a velocity $v_{3/2}$ in comparison to $R_2$ and $R_3$ with a velocity $v_{3/1}$ in comparison to $R_1$.

The relationship for velocities (18) provides $v_{3/1}$

$$ (19) \qquad | \qquad v_{3/1} = \frac{v_{3/2} + v_{2/1}}{1 + v_{3/2} \mu(v_{2/1})}. $$

Now, we consider from $R_3$ the movements of the two others frames. The velocity of $R_2$ seen from $R_3$ is $v_{2/3}$, the $R_1$ one seen from $R_2$ is $v_{1/2}$ and finally the $R_1$ one seen from $R_3$ is $v_{1/3}$. Applying (18) from $R_3$ to $R_1$ via $R_2$ we obtain.

$$ (20) \qquad | \qquad v_{1/3} = \frac{v_{1/2} + v_{2/3}}{1 + v_{1/2} \mu(v_{2/3})}. $$

According to relativity principle we have the following equality $v_{1/2}=-v_{2/1}$, $v_{2/3}=-v_{3/2}$ and $v_{1/3}=-v_{3/1}$. Then, taking into account $\mu(-\nu)=-\mu(\nu)$ from (9), and keeping in mind that (19) and (20) has to be equal, we arrive to $v_{3/2}\mu(v_{2/1})=v_{2/1}\mu(v_{3/2})$. This condition of form $x\mu(y)=y\mu(x)$, is possible only if:

$$ (21) \qquad | \qquad \mu(\nu) = \kappa \nu. $$

The parameter $\kappa$ has the dimension of an inverse squared velocity. Calling $c$ this velocity with $c>0$, we write down the three possibilities for $\mu(\nu)$ as:

$$ (22) \qquad | \qquad \mu(\nu) = \epsilon \frac{\nu}{c^2}. \quad \epsilon = -1, 0, 1 $$

## 7.2 DIMENSIONAL ANALYSIS

This second method results of the dimensional analysis of equations (17) and (18) from which we deduce that function $\mu(\nu)$ needs to have the dimension of an inverse velocity. The available quantities to find such a parameter are time, space and the frame velocity $\nu$ itself.
# 7 RELATION TIME VERSUS SPACE

Now we have to find the quantity $\mu(\nu)$ which provides the relation between time and space in equations (17). Two methods are available based on frame symmetries or dimensional analysis.

## 7.1 FRAME SYMMETRIES

Now we will find $\mu(\nu)$ using once more the relativity principle applied to three Galilean frames $R_1, R_2, R_3$ in uniform motions. $R_2$ has a velocity $v_{2/1}$ in comparison to $R_1, R_3$ with a velocity $v_{3/2}$ in comparison to $R_2$ and $R_3$ with a velocity $v_{3/1}$ in comparison to $R_1$.

The relationship for velocities (18) provides $v_{3/1}$

$$ (19) \qquad \qquad v_{3/1} = \frac{v_{3/2} + v_{2/1}}{1 + v_{3/2} \mu(v_{2/1})}. $$

Now, we consider from $R_3$ the movements of the two others frames. The velocity of $R_2$ seen from $R_3$ is $v_{2/3}$, the $R_1$ one seen from $R_2$ is $v_{1/2}$ and finally the $R_1$ one seen from $R_3$ is $v_{1/3}$. Applying (18) from $R_3$ to $R_1$ via $R_2$ we obtain.

$$ (20) \qquad \qquad v_{1/3} = \frac{v_{1/2} + v_{2/3}}{1 + v_{1/2} \mu(v_{2/3})}. $$

According to relativity principle we have the following equality $v_{1/2}=-v_{2/1}$, $v_{2/3}=-v_{3/2}$ and $v_{1/3}=-v_{3/1}$. Then, taking into account $\mu(-\nu)=-\mu(\nu)$ from (9), and keeping in mind that (19) and (20) has to be equal, we arrive to $v_{3/2}\mu(v_{2/1})=v_{2/1}\mu(v_{3/2})$. This condition of form $x\mu(y)=y\mu(x)$, is possible only if:

$$ (21) \qquad \qquad \qquad \qquad \mu(\nu) = \kappa \nu. $$

The parameter $\kappa$ has the dimension of an inverse squared velocity. Calling $c$ this velocity with $c>0$, we write down the three possibilities for $\mu(\nu)$ as:

$$ (22) \qquad \qquad \mu(\nu) = \epsilon \frac{\nu}{c^2}. \quad \epsilon = -1, 0, 1 $$

## 7.2 DIMENSIONAL ANALYSIS

This second method results of the dimensional analysis of equations (17) and (18) from which we deduce that function $\mu(\nu)$ needs to have the dimension of an inverse velocity. The available quantities to find such a parameter are time, space and the frame velocity $\nu$ itself.
On the one hand, according to the hypothesis of homogeneity, space and time cannot be used. On the other hand, we cannot assume $\mu(\nu)$ being proportional to $1/\nu$ because (17) and (18) would be undetermined when $\nu=0$.

So, we are constrained to define a new physical quantity c with the dimension of a velocity and to look for $\mu(\nu)$ with form :

$$ (23) \qquad \mu(\nu) = \frac{1}{c} \mu'(\frac{\nu}{c}). $$

Speed c cannot be a function of the frame velocity $v$, not even change with $v$ intervals, because the association between one or several frame velocities and values of c would provide a way to differentiate a Galilean frame from another one. Such a possibility is not acceptable due to the relativity principle which constrains frames to be indistinguishable, apart from their relative velocity.

Therefore, $c>0$ must have the same value what ever is the Galilean frame and has to be observed with this same value from any of them. So, function $\mu'(\frac{\nu}{c})$ in the

velocity addition (18) has to verify the condition

$$ c = \frac{v + c}{1 + \frac{1}{c} \mu'(\frac{v}{c})c}, $$

from where we find:

$$ (24) \qquad \mu'(\frac{v}{c}) = \frac{v}{c} \qquad \mu(\nu) = \frac{v}{c^2}. $$

In the same way, it can be pointed out that a negative value $-c<0$ will provide relation

$$ (25) \qquad \mu(\nu) = \frac{v}{c^2}, $$

having in mind that $\mu(-v)=-\mu(v)$.

Relations (24) and (25) can be completed for the trivial case of a classical transformation (17) for which $\mu(\nu)=0$ for all $\nu$:

$$ (26) \qquad \mu(\nu) = \epsilon \frac{v}{c^2}. \qquad \epsilon = -1, 0, 1 $$

# 8 THREE POSSIBILITIES

According to (22), (24) we have three possibilities for frame transformations.

## 8.1 GALILEI RELATIONS

When $\epsilon=0$ the condition $v\mu(\nu)<1$ is verified for any value of the velocity $v$. Equation (15) gives $\gamma(\nu)=1$ and the relations between frames (17) are the classic Galilean transformations for which the time measures of an event are the same in all the frames:
On the one hand, according to the hypothesis of homogeneity, space and time cannot be used. On the other hand, we cannot assume $\mu(\upsilon)$ being proportional to $1/\upsilon$ because (17) and (18) would be undetermined when $\upsilon=0$.

So, we are constrained to define a new physical quantity c with the dimension of a velocity and to look for $\mu(\upsilon)$ with form :

$$ (23) \qquad \mu(\upsilon)=\frac{1}{c} \mu'(\frac{\upsilon}{c}). $$

Speed c cannot be a function of the frame velocity $\upsilon$, not even change with $\upsilon$ intervals, because the association between one or several frame velocities and values of c would provide a way to differentiate a Galilean frame from another one. Such a possibility is not acceptable due to the relativity principle which constrains frames to be indistinguishable, apart from their relative velocity.

Therefore, $c>0$ must have the same value what ever is the Galilean frame and has to be observed with this same value from any of them. So, function $\mu'(\frac{\upsilon}{c})$ in the velocity addition (18) has to verify the condition

$$ c=\frac{\upsilon+c}{1+\frac{1}{c}\mu'(\frac{\upsilon}{c})c}, $$

from where we find:

$$ (24) \qquad \mu'(\frac{\upsilon}{c})=\frac{\upsilon}{c} \qquad \mu(\upsilon)=\frac{\upsilon}{c^2}. $$

In the same way, it can be pointed out that a negative value $-c<0$ will provide relation

$$ (25) \qquad \mu(\upsilon)=-\frac{\upsilon}{c^2}, $$

having in mind that $\mu(-\upsilon)=-\mu(\upsilon)$.

Relations (24) and (25) can be completed for the trivial case of a classical transformation (17) for which $\mu(\upsilon)=0$ for all $\upsilon$:

$$ (26) \qquad \mu(\upsilon)=\epsilon\frac{\upsilon}{c^2}. \qquad \epsilon=-1,0,1 $$

# 8 THREE POSSIBILITIES

According to (22), (24) we have three possibilities for frame transformations.

## 8.1 GALILEI RELATIONS

When $\epsilon=0$ the condition $\upsilon\mu(\upsilon)<1$ is verified for any value of the velocity $\upsilon$. Equation (15) gives $\gamma(\upsilon)=1$ and the relations between frames (17) are the classic Galilean transformations for which the time measures of an event are the same in all the frames:
$$ (27) \quad \begin{array}{c} x=x'+vt' \\ t=t' \end{array} \qquad \begin{array}{c} x'=x-vt \\ t'=t \end{array} $$

The second solution $\gamma(v)=-1$ provides the same relations with an opposite direction for $x$ and $t$.

Velocity addition relationship is deduced from (18) and is just a simple sum

$$ u(t)=v+u'(t). $$

## 8.2 IMPROPER RELATIONS

For $\epsilon=-1$ the condition $v\mu(v)<1$ is satisfied fore any value of the velocity $v$. Nevertheless, the relationship for the velocities (18) with $\epsilon=-1$

$$ (28) \qquad u(t) = \frac{v+u'(t)}{1-\frac{vu'(t)}{c^2}} $$

imposes the velocity $c$ as an upper limit for frame velocities. Otherwise, for higher values, the addition of positive speeds with (28) can bring to a negative velocity. Frame transformations are given by:

$$ \begin{array}{l@{\quad}l} (29) & \gamma(v) = \frac{1}{\sqrt{(1+\frac{v^2}{c^2})}} \\ (30) & \begin{aligned}[t] x &= \gamma(v)[x'+vt'] & x' &= \gamma(v)[x-vt] \\ t &= \gamma(v)[\frac{-vx'}{c^2}+t'] & t' &= \gamma(v)[\frac{+vx}{c^2}x+t'] \end{aligned} \end{array} $$

Note that these transformations are not the special relativity ones.

It has to be noted that with (28) a point having a velocity $c$ is observed with various speeds higher than $c$ according to the velocity of the Galilean frame in which it is. So, the transformations derived from $\epsilon=-1$ do not allow to attribute a feature of physical quantity to $c$ because the principle of relativity imposes that a physical quantity has the same value in any Galilean reference frame.

Furthermore, a measure of $u$ by an observer of a point having the speed $c$ would allow using (28) to find its own velocity as a fraction of $c$, so to say, as an absolute velocity and not as a relative velocity in comparison to another frame:

$$ \frac{v}{c} = \frac{\frac{u}{c} - 1}{\frac{u}{c} + 1}. $$

For $u=c$ an observer would conclude that its own velocity is zero and he is in an absolutely at rest frame from which he can assign absolute velocity to any other
$$ (27) \quad \begin{array}{c} x=x'+vt' \\ t=t' \end{array} \qquad \begin{array}{c} x'=x-vt \\ t'=t \end{array} $$

The second solution $\gamma(v)=-1$ provides the same relations with an opposite direction for $x$ and $t$.

Velocity addition relationship is deduced from (18) and is just a simple sum

$$ u(t)=v+u'(t). $$

## 8.2 IMPROPER RELATIONS

For $\epsilon=-1$ the condition $v\mu(v)<1$ is satisfied fore any value of the velocity $v$. Nevertheless, the relationship for the velocities (18) with $\epsilon=-1$

$$ (28) \qquad u(t) = \frac{v+u'(t)}{1-\frac{vu'(t)}{c^2}} $$

imposes the velocity $c$ as an upper limit for frame velocities. Otherwise, for higher values, the addition of positive speeds with (28) can bring to a negative velocity. Frame transformations are given by:

$$ \begin{array}{l@{\quad}l} (29) & \gamma(v) = \frac{1}{\sqrt{(1+\frac{v^2}{c^2})}} \\ (30) & \begin{array}{l@{\quad}l} x = \gamma(v)[x'+vt'] & x' = \gamma(v)[x-vt] \\ t = \gamma(v)[\frac{-vx'}{c^2}+t'] & t' = \gamma(v)[\frac{+vx}{c^2}x+t'] \end{array} \end{array} $$

Note that these transformations are not the special relativity ones.

It has to be noted that with (28) a point having a velocity $c$ is observed with various speeds higher than $c$ according to the velocity of the Galilean frame in which it is. So, the transformations derived from $\epsilon=-1$ do not allow to attribute a feature of physical quantity to $c$ because the principle of relativity imposes that a physical quantity has the same value in any Galilean reference frame.

Furthermore, a measure of $u$ by an observer of a point having the speed $c$ would allow using (28) to find its own velocity as a fraction of $c$, so to say, as an absolute velocity and not as a relative velocity in comparison to another frame:

$$ \frac{v}{c} = \frac{\frac{u}{c} - 1}{\frac{u}{c} + 1}. $$

For $u=c$ an observer would conclude that its own velocity is zero and he is in an absolutely at rest frame from which he can assign absolute velocity to any other
Galilean frame. This result is not acceptable according to the relativity principle which excludes the possibility of an absolute velocity for a Galilean frame. So, the case $\epsilon=-1$ has to be disregarded.

## 8.3 LORENTZ TRANSFORMATIONS

For $\epsilon=1$ the condition $v\mu(v)<1$ is satisfied only if frame velocities are lower than the speed $c$ arbitrary chosen as the constant parameter of (21) (22). Using equality (15), $\gamma(v)$ is determined and the relation known as the Lorentz transformation of Galilean reference frames is:

$$ \begin{aligned} (31) \qquad \gamma(v) &= \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \\ x &= \gamma(v) [x' + v t'] \qquad x' = \gamma(v) [x - v t] \\ (32) \qquad t &= \gamma(v) \left[ \frac{v x'}{c^2} + t' \right] \qquad t' = \gamma(v) \left[ \frac{-v x}{c^2} x + t \right] \end{aligned} $$

Relation (18) provides the sum of velocities:

$$ (33) \qquad \qquad \qquad \qquad u(t) = \frac{v + u'(t)}{1 + \frac{vu'(t)}{c^2}} $$

Relations (31) (32) (33) are the well known relations of special relativity cf. [1] Appendix I.

Two points can be underlined regarding the relationship (33).

First, if a point is in motion with a speed $c$ in a Galilean frame with velocity $v$, its velocity deduced from (33) is always equal to $c$ what ever is the velocity $v$ of frame where is the point. Namely, carrying on $u'=c$ in (33) we get $u=c$. This result allows to attribute to $c$ a property of physical quantity as explain in previous chapter,

Two, mutually, we look for a point velocity $u'$ requested to be constant what ever is the speed of the reference frame where is this point, so, which is not function of $v$. We take the derivative of (33) which provides the condition:

$$ \begin{aligned} & | \qquad 0 = 1 - \frac{u'^2}{c^2}, \\ \text{so:} \qquad & | \qquad u' = \pm c. \end{aligned} $$

Having stated previously $c>0$ in (22) we disregard the negative value. The conclusion is: if a point is such that its observed velocity does not depend of its reference frame speed, then this point velocity is $c$.
Galilean frame. This result is not acceptable according to the relativity principle which excludes the possibility of an absolute velocity for a Galilean frame. So, the case $\epsilon=-1$ has to be disregarded.

## 8.3 LORENTZ TRANSFORMATIONS

For $\epsilon=1$ the condition $v\mu(v)<1$ is satisfied only if frame velocities are lower than the speed $c$ arbitrary chosen as the constant parameter of (21) (22). Using equality (15), $\gamma(v)$ is determined and the relation known as the Lorentz transformation of Galilean reference frames is:

$$ \begin{aligned} (31) \qquad \gamma(v) &= \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \\ x &= \gamma(v)[x' + v t'] \qquad x' = \gamma(v)[x - v t'] \\ (32) \qquad t &= \gamma(v)[\frac{v x'}{c^2} + t'] \qquad t' = \gamma(v)[\frac{-v x}{c^2} x + t] \end{aligned} $$

Relation (18) provides the sum of velocities:

$$ (33) \qquad u(t) = \frac{v + u'(t)}{1 + \frac{vu'(t)}{c^2}} $$

Relations (31) (32) (33) are the well known relations of special relativity cf. [1] Appendix I.

Two points can be underlined regarding the relationship (33).

First, if a point is in motion with a speed $c$ in a Galilean frame with velocity $v$, its velocity deduced from (33) is always equal to $c$ what ever is the velocity $v$ of frame where is the point. Namely, carrying on $u'=c$ in (33) we get $u=c$. This result allows to attribute to $c$ a property of physical quantity as explain in previous chapter,

Two, mutually, we look for a point velocity $u'$ requested to be constant what ever is the speed of the reference frame where is this point, so, which is not function of $v$. We take the derivative of (33) which provides the condition:

$$ \begin{aligned} & | \qquad 0 = 1 - \frac{u'^2}{c^2}, \\ \text{so:} \qquad & | \qquad u' = \pm c. \end{aligned} $$

Having stated previously $c>0$ in (22) we disregard the negative value. The conclusion is: if a point is such that its observed velocity does not depend of its reference frame speed, then this point velocity is $c$.
## 8.4 COMMENTS

In order to apply special relativity transformations (31) (32) (33) to real world, a constant velocity $c$ has to be measured identical in all the Galilean frames. It is well known that a velocity with such property is the light one or zero mass particles, gravitational or electromagnetic velocity.

The following can be highlighted. Setting special relativity transformations do not require unicity of the constant velocity $c$. The transformations are valid what ever is $c$, greater or lower than the light velocity, if such observations happen in reality.

Using a reduced coordinate for time $y=ct$ allows suppressing any specific value for the velocity $c$ in transformations (31) (32) (33) where we only find $\beta$, the relative magnitudes of frame velocity $v$ and $c$.

$$ (34) \qquad \beta = \frac{v}{c}, \qquad -1 < \beta < 1 $$ $$ (35) \qquad \begin{aligned} x &= y(\beta)[ \quad x' + \beta y' ] \\ y &= y(\beta)[-\beta x' + \quad y'] \end{aligned} \qquad \begin{aligned} x' &= y(\beta)[ \quad x - \beta y' ] \\ y' &= y(\beta)[ \beta x + \quad y ] \end{aligned} $$

## 9 CONCLUSIONS

The relations (31) (32) (33) are the well known relations of special relativity cf, [1] Appendix I also call Lorentz Transformation. We derived them without assuming a priori neither group structure nor light properties. After having made the hypothesis of space and time homogeneity, we only applied the principle of relativity which can be also regarded as a principle of symmetry between two indistinguishable Galilean reference frames. So to say, if an observer in a reference frame $R$ obtains a given result for a frame $R'$ then, an observer in $R'$ get the same result for $R$,

In addition to the Lorentz transformation two other results were underlined. First: the frame speed limit is always the same what ever is the frame from where it is measured. This result allows to attribute to this speed limit a property of physical quantity. This speed limit is usually taken as the light velocity.

Two: mutually, we showed that if a point is such that its observed velocity does not depend of its reference frame speed, then the point velocity is the speed limit.
## 8.4 COMMENTS

In order to apply special relativity transformations (31) (32) (33) to real world, a constant velocity $c$ has to be measured identical in all the Galilean frames. It is well known that a velocity with such property is the light one or zero mass particles, gravitational or electromagnetic velocity.

The following can be highlighted. Setting special relativity transformations do not require unicity of the constant velocity $c$. The transformations are valid what ever is $c$, greater or lower than the light velocity, if such observations happen in reality.

Using a reduced coordinate for time $y=ct$ allows suppressing any specific value for the velocity $c$ in transformations (31) (32) (33) where we only find $\beta$, the relative magnitudes of frame velocity $v$ and $c$.

$$ (34) \qquad \beta = \frac{v}{c}, \qquad -1 < \beta < 1 $$ $$ (35) \qquad \begin{aligned} x &= y(\beta)[ \quad x' + \beta y' ] \\ y &= y(\beta)[-\beta x' + \quad y'] \end{aligned} \qquad \begin{aligned} x' &= y(\beta)[ \quad x - \beta y ] \\ y' &= y(\beta)[ \beta x + \quad y ] \end{aligned} $$

## 9 CONCLUSIONS

The relations (31) (32) (33) are the well known relations of special relativity cf, [1] Appendix I also call Lorentz Transformation. We derived them without assuming a priori neither group structure nor light properties. After having made the hypothesis of space and time homogeneity, we only applied the principle of relativity which can be also regarded as a principle of symmetry between two indistinguishable Galilean reference frames. So to say, if an observer in a reference frame $R$ obtains a given result for a frame $R'$ then, an observer in $R'$ get the same result for $R$,

In addition to the Lorentz transformation two other results were underlined. First: the frame speed limit is always the same what ever is the frame from where it is measured. This result allows to attribute to this speed limit a property of physical quantity. This speed limit is usually taken as the light velocity.

Two: mutually, we showed that if a point is such that its observed velocity does not depend of its reference frame speed, then the point velocity is the speed limit.
REFERENCES

[1] Albert Einstein, Relativity The Special and General Theory, Authorised Translation by Robert W. Lawson, University of Sheffield, 1916

[2] Jean-Marc Levy-Leblond, One more derivation of the Lorentz transformation, American Journal of Physics Vol. 44, No. 3, March 1976.

CONTENTS

<table><tr><td>1 INTRODUCTION</td><td>0</td></tr><tr><td>2 NO REFERENCE FRAME ABSOLUTELY AT REST</td><td>1</td></tr><tr><td>3 FORMULAS FOR FRAME CHANGES</td><td>1</td></tr><tr><td>4 SPACE TIME HOMOGENEITY</td><td>2</td></tr><tr><td>5 RELATIONS BETWEEN FRAMES</td><td>3</td></tr><tr><td>6 RELATIONSHIP FOR VELOCITIES</td><td>6</td></tr><tr><td>7 RELATION TIME VERSUS SPACE</td><td>7</td></tr><tr><td>7.1 FRAME SYMMETRIES</td><td>7</td></tr><tr><td>7.2 DIMENSIONAL ANALYSIS</td><td>7</td></tr><tr><td>8 THREE POSSIBILITIES</td><td>8</td></tr><tr><td>8.1 GALILEI RELATIONS</td><td>8</td></tr><tr><td>8.2 IMPROPER RELATIONS</td><td>9</td></tr><tr><td>8.3 LORENTZ TRANSFORMATIONS</td><td>10</td></tr><tr><td>8.4 COMMENTS</td><td>11</td></tr><tr><td>9 CONCLUSIONS</td><td>11</td></tr></table>
REFERENCES

[1] Albert Einstein, Relativity The Special and General Theory, Authorised Translation by Robert W. Lawson, University of Sheffield, 1916

[2] Jean-Marc Levy-Leblond, One more derivation of the Lorentz transformation, American Journal of Physics Vol. 44, No. 3, March 1976.

CONTENTS

<table><tr><td>1 INTRODUCTION</td><td>0</td></tr><tr><td>2 NO REFERENCE FRAME ABSOLUTELY AT REST</td><td>1</td></tr><tr><td>3 FORMULAS FOR FRAME CHANGES</td><td>1</td></tr><tr><td>4 SPACE TIME HOMOGENEITY</td><td>2</td></tr><tr><td>5 RELATIONS BETWEEN FRAMES</td><td>3</td></tr><tr><td>6 RELATIONSHIP FOR VELOCITIES</td><td>6</td></tr><tr><td>7 RELATION TIME VERSUS SPACE</td><td>7</td></tr><tr><td>7.1 FRAME SYMMETRIES</td><td>7</td></tr><tr><td>7.2 DIMENSIONAL ANALYSIS</td><td>7</td></tr><tr><td>8 THREE POSSIBILITIES</td><td>8</td></tr><tr><td>8.1 GALILEI RELATIONS</td><td>8</td></tr><tr><td>8.2 IMPROPER RELATIONS</td><td>9</td></tr><tr><td>8.3 LORENTZ TRANSFORMATIONS</td><td>10</td></tr><tr><td>8.4 COMMENTS</td><td>11</td></tr><tr><td>9 CONCLUSIONS</td><td>11</td></tr></table>