# 3D Geometric Algebra and Special Relativity

   Without the belief that it is possible to grasp reality with our theoretical constructions, without the belief in the inner harmony of the world, there can be no science. This belief has and always will be the fundamental motive for all scientific creation.

## Project Summary

Currently this page is just notes that i'll clean up eventually.

### Project Goals

This learning project aims to:
* investigate the similarities and differences between certain rotations in 3D Geometric Algebra and the Lorentz transformation that forms the mathematical basis for Special Relativity

### Assumed background knowledge/skills

For this learning project, if you are familiar with:
* basic Geometric Algebra in 3 dimensions (see Investigating 3D Geometric Algebra).
* An understanding of Special Relativity will be useful to see the similarity and differences that exist between Special Relativity and 3D Geometric Algebra. Wikipedia has a separate, less-technical Introduction to special relativity, and Wikibooks has a Special Relativity text.

## Introduction

Most applications of Geometric Algebra to Special Relativity (from Hestenes' work) start by defining the space as 4D Minkowskian space. Eg. $e0^{2}=1=e1^{2}=e2^{2}=-e3^{2}$ (re-phrase using more familiar x,y,z and w. But instead of starting with this assumption, start by investigating some of the properties of rotations in normal 3D geometric algebra.

## Example Rotations

Setting up a rotation by ${\frac {\pi }{6}}$
$\mathbf {a} =\mathbf {x}$
$\mathbf {b} ={\frac {\sqrt {3}}{2}}\mathbf {x} +0.5\mathbf {y}$
Therefore, rotor $\mathbf {r} =\mathbf {a} \mathbf {b}$ (need diagrams to visualise... and lots of them!)
Imagine a unit cube (again, need to draw). If we apply normal 3D rotation as follows, for each point $\mathbf {p}$ of the cube:
$\mathbf {p'} =\mathbf {r} \mathbf {p}$$\mathbf {\tilde {r}}$
Answer:
can't display the inverse of r [Anonymous editor: Shouldn't $\mathbf {\tilde {r}}$ be "reverse" rather than "inverse" since $\mathbf {\tilde {r}} =\mathbf {b} \mathbf {a}$ ? Also,it looks as though the Geometric Algebra community is not united on the symbol to be used to denote the operation of reversion for the reverse geometric product. Some use tilde and others a small superscript dagger. The LaTeX dagger is too big and needs to be scaled down a bit and moved higher. $\mathbf {R} \dagger$]
https://tex.stackexchange.com/questions/524293/is-there-a-shorter-dag-dagger-or-how-could-we-move-it-a-bit-vertically
Rotations and Geometric algebra (uses tilde)
https://www.mrao.cam.ac.uk/~clifford/introduction/intro/node9.html
"We call $\mathbf {\tilde {R}} \equiv \mathbf {n} \mathbf {m}$ the 'reverse' of $\mathbf {R}$, because it is obtained by reversing the order of all geometric products."
(uses tilde, https://en.wikipedia.org/wiki/Geometric_algebra)
(also uses tilde on slide 25, REVERSION) https://www.ime.unicamp.br/~agacse2018/GA_Lecture1.pdf
(uses a long tilde in equation 7) http://geometricalgebratutorial.com/ga-basics/
Oersted Medal Lecture 2002: Reforming the Mathematical Language of Physics by David Hestenes
"To facilitate algebraic manipulations, it is convenient to introduce a special symbol for the operation (called reversion) of reversing the order of multiplication. The reverse of the geometric product is defined by
$(\mathbf {ab} )\dagger =\mathbf {ba}$ (32)
We noted in (20) that this is equivalent to complex conjugation in 2D."
https://geocalc.clas.asu.edu/pdf/OerstedMedalLecture.pdf
http://geocalc.clas.asu.edu/pdf/NFMPchapt1.pdf (also uses dagger for reversion on pages 7 and 8)
(uses asterisk on pg. 10 (equation 1.28) Geometric Algebra for Physicists by Chris Doran and Anthony Lasenby https://assets.cambridge.org/97805214/80222/sample/9780521480222ws.pdf
(Alan Macdonald seems to be using superscript -1 (inverse??) for the operation of reversion??) https://philpapers.org/archive/MACASO-16.pdf
http://www.faculty.luther.edu/~macdonal/
and the cube is rotated by ${\frac {\pi }{3}}$. Note that only the $\mathbf {x}$ and $\mathbf {y}$ components are affected by rotation. $\mathbf {z}$ component of all vectors remains unchanged.
Now taking the same unit cube, apply the following rotation for each point:
$\mathbf {p'} =\mathbf {r} \mathbf {p} \mathbf {r}$ (without the inverse of r this time)
we find that for all points of the cube the $\mathbf {x}$ and $\mathbf {y}$ components remain unchanged, but the $\mathbf {z}$ and, a new component of our multivector, $\mathbf {xyz}$ are modified. It's still a rotation, as $\mathbf {p'} \mathbf {~p'} =\mathbf {p} \mathbf {~p}$, but not one that we're used to.
Seems as those cube is compressed along z-axis (only 0.5 in length along z-axis), not dissimilar to Lorentz contraction, but the $\mathbf {xyz}$ component of the points p4 to p7 has been changed from zero to ${\frac {\sqrt {3}}{2}}\mathbf {xyz}$. If this component is a time-like component, could provide interesting interpretation of such rotations.
For eg, rotating one unit of $\mathbf {xyz}$ results in:
$\mathbf {r} \mathbf {xyz} \mathbf {r} =-{\frac {\sqrt {3}}{2}}\mathbf {z} +0.5\mathbf {xyz}$
That is, for each unit of time in the unrotated frame, only half a unit passes in the rotated frame. And, furthermore, for each unit of time in the unrotated frame, the rotated frame shifts a distance along the z-axis. Again, not dissimilar to time-dilation and velocity.
Finally, to calculate the shape of the actual cube in the rotated frame, want to get a snapshot for all the points when they have the same temporal value of $\mathbf {xyz}$. If we translate the original p4-p7 back in time by ${\sqrt {3}}\mathbf {xyz}$ (again, diagram very necessary here) then it turns out that the cube is elongated along the z-axis (in fact it's doubled in length) as it moves along.
Next, try rotation of ${\frac {\pi }{4}}$ as this corresponds to point of singularity in SR (c=1), but behaves nicely here...