Deriving Lorentz Transformations
An axiomatic approach
Steve Trettel
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I’m running an independent study in Lorentzian geometry right now, and was thinking this morning about how to introduce spacetime (for context, we have already spent some time learning mathematically about manifolds that have Lorentzian metrics, but we haven’t connected any of this abstract stuff to what it models in physics: the world around us).
In putting together a narrative, I found a rather elegant axiomatic way to proceed which shows (essentially) if you assume Euclidean geometry is true of space, and Galileo’s principle holds (that uniform motion is the indistinguishable from being at rest) then there are only two possible ways for spacetime to behave: it can be the Galilean spacetime of Newton, or the relativistic spacetime of Einstein.
Without input from an experiment (or another axiom like ‘the speed of light is constant for all observers’) one can’t get further than this. But this is already so constraining from such simple axioms that I figured I would write it down for future reference, and for other interested students. I am by no means claiming this is original; the Lorentz transformations have been derived hunderds of different ways over the past century and a half, and I am sure this, and many very similar approaches are among them.
The Axioms
We start with the spacetime topologically $\mathbb{R}^4=\mathbb{R}^3\times \mathbb{R}$, where we write $(p,t)$ for the point $p\in\mathbb{R}^3$ at the time $t$, and assume the following properties:
- Symmetries of space are symmetries of spaceitme: If $A\colon\mathbb{R}^3\to\mathbb{R}^3$ is a Euclidean isometry, then $\mathcal{A}(p,t)=(A(p),t)$ is a symmetry of spacetime.
- Symmetries of time are symmetries of spacetime: If $B$ is any isometry of $\mathbb{R}$ (a translation, reflection or combination) then $\mathcal{B}(p,t):=(p,B(t))$ is a symmetry of spacetime.
- Galileo’s Principle: If $\ell_1$ and $\ell_2$ are the worldlines of any two constant speed observers, there is a symmetry of spacetime taking $\ell_1$ to $\ell_2$.
- Space is different than time: There is no symmetry of spacetime which takes the vertical axis $(0,t)$ to a line in space $(\ell,0)$.
Reduction to 2 Dimensions
Defining Boosts
A boost is an orientation preserving transformation which fixes the origin, takes a constant-speed trajectory through the origin to the time axis. From your perspective (ie sitting still, so moving only along the time axis) this means that an object you used to see moving away at speed $v$ is now staying stationary right next to you: you’ve boosted your speed by $v$. Let’s write such a transformation as $B(v)$.
You might ask, why we are insisting on orientation preserving: are we missing out on anything by doing so? But we are not: PROOF INVOLVING EUCLIDEAN REFLECTIONS
The Derivation
Step I
Let $B(v)$ be the boost taking taking $x=tv$ to the time axis. Then $$\begin{pmatrix}0\ \star\end{pmatrix}=\begin{pmatrix}a(v) & b(v)\ c(v) & d(v)\end{pmatrix}\begin{pmatrix}vt\ t\end{pmatrix}$$
Looking to the first entry (in space) we see $0=a(v) vt + b(v)t$: simplifying yields $b(v)=-a(v)v$, which reduces our list of unknown functions by one!
$$B(v)=$$