How to derive Lorentz Transforms from the wave equation?

Question

It is well known that the wave equation agrees with special relativity. So, my question is: it is possible to derive the Lorentz transforms from the equation?

First, I started by assuming a linear transformation of the form: $$ t'= \alpha x + \beta t \\ \; \\ x' = \gamma x + \delta t $$ The wave equation in the S frame is: $$ \frac{1}{c^2}\frac{\partial^2 \psi}{ \partial t^2} = \frac{\partial^2 \psi}{ \partial x^2} $$ The wave equation in the S' frame should be: $$ \frac{1}{c^2}\frac{\partial^2 \psi}{ \partial t'^2} = \frac{\partial^2 \psi}{ \partial x'^2} $$ Applying the chain rule and a change of variables, we get the following: $$ \frac{\partial^2 \psi}{ \partial t^2} \left [ \left ( \frac{1}{c} \frac{\partial t}{ \partial t'} \right ) ^2 - \left (\frac{ \partial t}{ \partial x'} \right ) ^2 \right ] + \frac{\partial^2 \psi}{ \partial t \partial x} \left [ \frac{1}{c^2} \frac{\partial t}{ \partial t'} \frac{\partial x}{ \partial t'} - \frac{ \partial t}{ \partial x'}\frac{ \partial x}{ \partial x'} \right ] + \frac{\partial^2 \psi}{ \partial x^2} \left [ \left ( \frac{1}{c} \frac{\partial x}{ \partial t'} \right ) ^2 - \left (\frac{ \partial x}{ \partial x'} \right ) ^2 \right ] = 0 $$ So we conclude that: $$ \left ( \frac{1}{c} \frac{\partial t}{ \partial t'} \right ) ^2 - \left (\frac{ \partial t}{ \partial x'} \right )^2 = \frac{1}{c^2} \\ $$ $$ \frac{1}{c^2} \frac{\partial t}{ \partial t'} \frac{\partial x}{ \partial t'} - \frac{ \partial t}{ \partial x'}\frac{ \partial x}{ \partial x'} = 0 \\ $$ $$ \left ( \frac{1}{c} \frac{\partial x}{ \partial t'} \right ) ^2 - \left (\frac{ \partial x}{ \partial x'} \right )^2 = -1 \\ $$ But now I'm stuck, how should I continue? Is this line of reasoning right? Any mistake that I made?

Answer 1

$$\begin{align} \left ( \frac{1}{c} \frac{\partial t}{ \partial t'} \right ) ^2 - \left (\frac{ \partial t}{ \partial x'} \right )^2 &= \frac{1}{c^2} \\ \frac{1}{c^2} \frac{\partial t}{ \partial t'} \frac{\partial x}{ \partial t'} - \frac{ \partial t}{ \partial x'}\frac{ \partial x}{ \partial x'} &= 0 \\ \left ( \frac{1}{c} \frac{\partial x}{ \partial t'} \right ) ^2 - \left (\frac{ \partial x}{ \partial x'} \right )^2 &= -1 \end{align} \tag{1}$$ But now I'm stuck, how should I continue? Is this line of reasoning right? Any mistake that I made?

So far everything is correct. So let's continue here.

For solving (1) you are looking for transformations $t(t',x')$ and $x(t',x')$. Note that it is advantageous to write it this way around, instead of the opposite way with $t'(t,x)$ and $x'(t,x)$. Assume the transformation is linear. $$\begin{align} t &= \alpha't'+\beta'x' \\ x &= \gamma't'+\delta'x' \end{align} \tag{2}$$ Plugging the transformation (2) into the $3$ equations (1) above gives:

$$\begin{align} \frac{1}{c^2}\alpha'^2-\beta'^2 &= \frac{1}{c^2} \\ \frac{1}{c^2}\alpha'\gamma'-\beta'\delta' &= 0 \\ \frac{1}{c^2}\gamma'^2-\delta'^2 &= -1 \end{align} \tag{3}$$

These are $3$ equations for $4$ unknowns. So we have one free parameter, and we can choose it to be $\delta'$. Solving (3) for $\alpha'$, $\beta'$ and $\gamma'$ we get: $$\begin{align} \alpha' &= \delta' \\ \beta' &= \frac{1}{c}\sqrt{\delta'^2-1} \\ \gamma' &= c\sqrt{\delta'^2-1} \end{align} \tag{4}$$ This doesn't yet look very much like the expected Lorentz transform. But without gain or loss of generality we can choose $\delta'=\frac{1}{\sqrt{1-b^2}}$ with an arbitrary $b$. Inserting this into (4) we get: $$\begin{align} \alpha' &= \frac{1}{\sqrt{1-b^2}} \\ \beta' &= \frac{b}{c}\frac{1}{\sqrt{1-b^2}} \\ \gamma' &= cb\frac{1}{\sqrt{1-b^2}} \\ \delta' &= \frac{1}{\sqrt{1-b^2}} \end{align} \tag{5}$$

Plugging this into (2) we get the Lorentz transform in a more familiar form: $$\begin{align} t &= \frac{1}{\sqrt{1-b^2}} \left(t'+\frac{b}{c}x'\right) \\ x &= \frac{1}{\sqrt{1-b^2}} (bct'+x') \end{align} \tag{6}$$

For getting the physical meaning of $b$ consider an object at rest in frame $S$ at position $x=0$. From the second equation of (6) you get $0=bct'+x'$ or $x'=-bct'$. That means in frame $S'$ this object moves with velocity $v=-bc$. Or in shorter words: $-bc$ is the relative velocity between frames $S$ and $S'$.

We can rewrite (6) in terms of this relative velocity $v$ instead of $b$ to get the Lorentz transform in its most familiar form. $$\begin{align} t &= \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \left(t'-\frac{v}{c^2}x'\right) \\ x &= \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} (-vt'+x') \end{align} \tag{7}$$

Answer 2

You need to solve $t'=\alpha x +\beta t$ and $x' = \gamma x + \delta t$ for $t$ and $x$ and then plug into the last three equations. This should fix the coefficients

Answer 3

1. The d'Alembertian in an arbitrary coordinate system $x^{\mu}$ of Minkowski spacetime $(M,\mathbb{g})$ is given by the 4D version of the Laplace-Beltrami operator$^1$ $$\Box~=~\frac{1}{\sqrt{|g|}}\partial_{\mu}\sqrt{|g|}g^{\mu\nu}\partial_{\nu}.\tag{1}$$ One may show that the formula (1) is invariant under arbitrary coordinate transformations $$x^{\mu}~~ \longrightarrow~~ x^{\prime\nu}~=~f^{\nu}(x).\tag{2}$$

2. Now OP wants to consider coordinates $x^{\mu}$ where the metric tensor is of the diagonal form $$\begin{align} \mathbb{g}~=~&\eta_{\mu\nu} \mathrm{d}x^{\mu} \odot \mathrm{d}x^{\nu} \cr ~=~&-\mathrm{d}x^0 \odot \mathrm{d}x^0 +\mathrm{d}x^1 \odot \mathrm{d}x^1 +\mathrm{d}x^2 \odot \mathrm{d}x^2 +\mathrm{d}x^3 \odot \mathrm{d}x^3.\tag{3}\end{align} $$ Then the operator (1) simplifies to $$ \Box~=~\eta^{\mu\nu}\partial_{\mu}\partial_{\nu}. \tag{4}$$

3. OP is interested in the coordinate transformations (2) between coordinates systems of type (3) that preserves (4). Well, that's any coordinate transformation (2) that preserves (3), which is one way of characterizing the Lorentz group of Lorentz transformations. In particular, it is shown in this Phys.SE post that such transformations necessarily are affine.

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$^1$ We put the speed of light $c=1$ to one for simplicity.