# Wave equation invariance under Lorentz transform

1. Aug 24, 2013

### fluidistic

1. The problem statement, all variables and given/known data
I must show that the one dimensional wave equation $\frac{1}{c^2} \frac{\partial u}{\partial t^2}-\frac{\partial ^2 u}{\partial x^2}=0$ is invariant under the Lorentz transformation $t'=\gamma \left ( t-\frac{xv}{c^2} \right )$ , $x'=\gamma (x-vt)$

2. Relevant equations
Already given. Chain rule for a function of several variables.

3. The attempt at a solution
Rather than posting pages of latex litterally, I'm going to explain what I did.
Hypothesis: u(t,x) is a solution to the wave equation $\frac{1}{c^2} \frac{\partial u}{\partial t^2}-\frac{\partial ^2 u}{\partial x^2}=0$. With this assumption, I want to see whether $u _L(t,x)$ is a solution to the wave equation $\frac{1}{c^2} \frac{\partial u _L (t,x)}{\partial t^2}-\frac{\partial ^2 u _L (t,x)}{\partial x^2}=0$ where $u_L(t,x)=u(t',x')$.
My goal is to express, first, $\frac{\partial ^2 u_L(t,x)}{\partial t^2}$ in terms of $\frac{\partial ^2 u (t',x')}{\partial t'^2}$ (and $\frac{\partial ^2 u_L(t,x)}{\partial x^2}$ in terms of $\frac{\partial ^2 u (t',x')}{\partial x'^2}$) in order to later use the hypothesis and simplify the math to reach the goal.
I've done it and I'm left to show that $\frac{\gamma v }{c} \{ \frac{\partial ^2 u(t',x')}{\partial x \partial t'} - \frac{\partial ^2 u(t',x')}{\partial t \partial x'} +\gamma \left [ \frac{\partial ^2 u(t',x')}{\partial t' \partial x'} - \frac{\partial ^2 u(t',x')}{\partial x' \partial t'} \right ] \}=0$. If the left hand side is worth 0 then $u _L(t,x)$ is a solution to the wave equation $\frac{1}{c^2} \frac{\partial u _L (t,x)}{\partial t^2}-\frac{\partial ^2 u _L (t,x)}{\partial x^2}=0$ and the wave equation is invariant under the Lorentz transform.
However I haven't been able to show that this last equation holds. Any ideas are welcome. At first glance it doesn't look obvious to me.

2. Aug 25, 2013

### TSny

In the first term is the x meant to be x' in the denominator and in the second term is it t or t' in the denominator?

Don't forget equality of mixed partials.

3. Aug 25, 2013

### fluidistic

Hmm I must redo the algebra then, because I didn't make any typo by writing the latex here, I mean this matches what I have on my draft.
With respect to mixed partial derivatives, thanks. I didn't know I could do that, for some reason I forgot that continuity of those derivatives was enough to ensure that the mixed derivatives are equal.

4. Aug 25, 2013

### WannabeNewton

Just out of curiosity, are you forced to do it explicitly? You can write the wave equation as $\partial_{\mu}\partial^{\mu}u = 0$ where $\partial_{\mu}$ is the 4-gradient. This equation is manifestly Lorentz covariant so there's nothing to really do.

5. Aug 25, 2013

### fluidistic

Yeah I'm almost sure I am supposed to do it explicitly for now at least. The exercises lag slightly behind the theory we learn and we haven't seen the equation you wrote under this form.
I'm not even sure what "Lorentz covariant" mean.

6. Aug 25, 2013

### WannabeNewton

It's the same thing as "Lorentz invariant" in the way you've used it above, it's just that the term "Lorentz covariant" is more appropriate for equations. "Lorentz invariant" is usually reserved for actual quantities that remain the same under Lorentz transformations, not for equations (although many times it is used for equations as well but it's an abuse of terminology).

7. Aug 25, 2013

### fluidistic

I see, thanks a lot.

8. Aug 25, 2013

### fluidistic

WOW GUYS I GOT IT! I've rechecked my whole arithmetics/algebra and I can't believe I made no mistake!!! I always mess up in these lenthy calculations but this time I didn't (unless I reach the result by accident).
In order to show that $\frac{\partial ^2 u(t',x')}{\partial x \partial t'} - \frac{\partial ^2 u(t',x')}{\partial t \partial x'}$ is worth 0, I rewrite $\frac{\partial ^2 u(t',x')}{\partial x \partial t'}$ as $\frac{\partial t'}{\partial x} \cdot \frac{\partial ^2 u(t',x')}{\partial t'^2} + \frac{\partial x'}{\partial x} \cdot \frac{\partial ^2u}{\partial x' \partial t'}$ and a similar expression for the other term. After this I have to use the hypothesis twice and I'm done.

9. Aug 25, 2013

### WannabeNewton

Nice! It's great that you got it all sorted out :)

By the way, this kind of problem will be way neater and easier if you use 4-vectors and their associated transformations under Lorentz boosts. For example, for a boost along the $+x$ direction of a given global inertial frame, we have $\partial_{t'} = \Lambda^{\mu}_{t'}\partial_{\mu} = \gamma \partial_{t} - \gamma\beta \partial_{x}$ and similarly $\partial_{x'} = -\gamma\beta \partial_{t} + \gamma\partial_{x}$ so $\partial^2_{t'}u - \partial^{2}_{x'}u = \gamma^{2}(\partial_{t}^{2}u - \partial_{x}^{2}u) - \gamma^{2}\beta^{2}(\partial^{2}_{t}u - \partial^{2}_{x}u) = 0$ hence the wave equation is Lorentz covariant.