Can the Lorentz Transformation Matrix be Derived using the Change of Basis Formula?

Question

So I watched a MinutePhysics video explaining the idea of Lorentz transforms geometrically and the way he described it sounded very similar to how 3Blue1Brown explained the idea of change of basis. I know special relativity can be modeled using linear algebra and I found a derivation of the Lorentz transform that was based on first defining a 4d vector space to model spacetime, but it made no mention of change of basis. I've looked online and I couldn't find anything specifically talking about Lorentz transforms and change of basis together, but the description of what the Lorentz transform does sounds so incredibly similar to what change of basis transformations do that it would shock me if it's not a change of basis transformation or at least something similar. So is it a change of basis matrix? And if so, can we derive it by using the general matrix formula for a change of basis matrix?

Answer 1

Lorentz transformations can be viewed in two equivalent ways. First as active transformations that take one vector in Minkowski vector space $\mathbb{R}^{1,3}$ and give a new vector, preserving the Minkowski inner product $\eta : \mathbb{R}^{1,3}\times \mathbb{R}^{1,3}\to \mathbb{R}$. Second as passive transformations that are induced on the coordinates of a vector in one orthonormal basis when you perform a change of basis to another orthonormal basis. I believe here you want to understand the second viewpoint.

In basic treatments it is common to see the Lorentz transformations be viewed as changes of coordinates. In this scenario what is really going on is the following. You have one orthonormal basis $\{e_\mu\}$ on $\mathbb{R}^{1,3}$. Therefore any vector $x\in \mathbb{R}^{1,3}$ can be written uniquely as $$x=x^\mu e_\mu.\tag{1}$$

If $\{e_\mu'\}$ is a second basis on $\mathbb{R}^{1,3}$ the same vector can be written also uniquely as $$x = {x'}^\mu e'_\mu \tag{2}.$$

Of course (1) and (2) must be equal. Since $e_\mu$ is a basis it can itself be written in terms of $e'_\mu$ as $$e_\mu = \Lambda^\nu_{\phantom{\nu}\mu}e'_\nu\tag{3}.$$

Using this to equate (1) and (2) gives $${x'}^\nu e_\nu' = \Lambda^\nu_{\phantom{\nu}\mu}x^\mu e'_\nu\tag{4},$$ and now uniqueness of the expression of a vector in a basis gives you ${x'}^\mu$ in terms of $x^\nu$ as $${x'}^\mu = \Lambda^\mu_{\phantom{\mu}\nu} x^\nu\tag{5}.$$

This is one general analysis which can be done in any vector space, really. Summarizing what is written above is basically this:

1. In a vector space, expansion in a particular basis, defines coordinates on that space: the coordinates of a vector are just the expansion coefficients;

2. If you change the basis, you have new coordinates which can be written in terms of the old ones following the above derivation.

Now remember when I said the first basis was orthonormal? By that I meant that if $\eta : \mathbb{R}^{1,3}\times \mathbb{R}^{1,3}\to \mathbb{R}$ is the Minkowski inner product, we have $$\eta(e_\mu,e_\nu)=\eta_{\mu\nu},\quad \eta_{\mu\nu}=\begin{pmatrix}-1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix}\tag{6}.$$

Now if you restrict attention to the class of bases which are orthonormal, and you demand that $\{e_\mu'\}$ be orthonormal as well, then we must also have $\eta(e_\mu',e_\nu')=\eta_{\mu\nu}$. Write down this condition using (3)

$$\eta(e_\mu,e_\nu)=\eta(\Lambda^\alpha_{\phantom{\alpha}\mu}e'_\alpha,\Lambda^\beta_{\phantom{\beta}\nu}e'_\beta)=\Lambda^\alpha_{\phantom \alpha \mu}\Lambda^\beta_{\phantom \beta \nu}\eta(e'_\alpha,e'_\beta)= \eta_{\alpha\beta}\Lambda^\alpha_{\phantom \alpha \mu}\Lambda^\beta_{\phantom \beta \nu}\tag{7}.$$

This is the same as $$\eta_{\alpha\beta}\Lambda^\alpha_{\phantom \alpha \mu}\Lambda^\beta_{\phantom \beta \nu}=\eta_{\mu\nu},\tag{8}$$

which is just the standard condition that Lorentz transformations must obey.

Answer 2

Allow me to show how to derive Lorentz transforms in the standard configuration using change of basis. Let $X=(t, x, y, z)$ be a position 4-vector as measured in some inertial frame. By default, it is expressed in orthonormal basis: $X=t\hat{t}+x\hat{x}+y\hat{y}+z\hat{z}$ where

$$\hat{t}= \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} \qquad \hat{x}= \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} \qquad \hat{y}= \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} \qquad \hat{z}= \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} $$

Suppose we want to write $X$ as $X=t'\hat{t}'+x'\hat{x}'+y'\hat{y}'+z'\hat{z}'$ where

$$\hat{t}'= \begin{bmatrix} \cosh\phi \\ \sinh\phi \\ 0 \\ 0 \end{bmatrix} \qquad \hat{x}'= \begin{bmatrix} \sinh\phi \\ \cosh\phi \\ 0 \\ 0 \end{bmatrix} \qquad \hat{y}'=\hat{y} \qquad \hat{z}'=\hat{z} $$

is another set of orthonormal basis and $\phi=\frac{v}{c}$ is some rapidity. These are the basis vectors for an observer moving along positive x-axis with speed $v$. We can find $t', x', y', z'$ by projecting $X$ on each $\hat{t}', \hat{x}', \hat{y}', \hat{z}'$:

$$t'=-X\cdot\hat{t}'\qquad x'=X\cdot\hat{x}'\qquad y'=X\cdot\hat{y}'\qquad z'=X\cdot\hat{z}'\qquad$$

So,

$$ \begin{align*} t'&=-(-t\cosh\phi+x\sinh\phi)=t\cosh\phi-x\sinh\phi \\ x'&=-t\sinh\phi+x\cosh\phi \\ y'&=y \\ z'&=z \end{align*} $$

We are done!

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Addendum

To see how to obtain the first two basis vectors $\hat{t}'$ and $\hat{x}'$, we can do a bit of hyperbolic trigonometry on spacetime diagram:

Source

In this diagram, the basis vectors for primed axis are related to basis vectors for unprimed axis by

$$\hat{t}'=\cosh\alpha\hat{t}+\sinh\alpha\hat{x}$$ $$\hat{x}'=\sinh\alpha\hat{t}+\cosh\alpha\hat{x}$$

This is similar to the relationship between the basis vectors of Cartesian system and its rotated counterpart.

Source

$$\hat{x}'=\cos\theta\hat{x}+\sin\theta\hat{y}$$ $$\hat{y}'=-\sin\theta\hat{x}+\cos\theta\hat{y}$$