# Lorentz invariance of the wave equation

I want to show that the 2-d wave equation is invariant under a boost, so, the starting point is the wave equation

$$\frac{\partial^2\phi}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2\phi}{\partial t^2}$$

and the Lorentz transformation:

$$t'=\gamma(t-\frac{v}{c^2}x) \\ x'=\gamma(x-vt)$$

My question is should I write $\displaystyle\frac{\partial}{\partial t}$ as a derivative with respect to $x'$ and $t'$ and then substitute?

Work done so far

$$\frac{\partial}{\partial t}=\frac{\partial }{\partial x'}\frac{\partial x'}{\partial t}+\frac{\partial}{\partial t'}\frac{\partial t'}{\partial t}=-\gamma v\frac{\partial}{\partial x'}+\gamma\frac{\partial}{\partial t'}$$

$$\frac{\partial^2}{\partial t^2}=\frac{\partial }{\partial t}\left(\frac{\partial}{\partial t} \right)=\frac{\partial}{\partial t}\left( -\gamma v\frac{\partial}{\partial x'}+\gamma \frac{\partial}{\partial t'}\right)= \\ =-\gamma v\frac{\partial}{\partial x'}\left( -\gamma v\frac{\partial}{\partial x'}+\gamma \frac{\partial}{\partial t'}\right)+\gamma\frac{\partial}{\partial t'}\left( -\gamma v\frac{\partial}{\partial x'}+\gamma \frac{\partial}{\partial t'}\right)=\\ = \gamma^2v^2\frac{\partial^2}{\partial x'^2}-2\gamma^2v\frac{\partial ^2}{\partial x'\partial t'}+\gamma^2\frac{\partial^2 }{\partial t'^2}$$

-Edit-

The same applies to $x$

$$\frac{\partial}{\partial x}=\frac{\partial }{\partial x'}\frac{\partial x'}{\partial x}+\frac{\partial }{\partial t'}\frac{\partial t'}{\partial x}=\gamma\frac{\partial}{\partial x'}-\frac{\gamma v}{c^2}\frac{\partial }{\partial t'}$$

$$\frac{\partial^2}{\partial x^2}=\frac{\partial}{\partial x}\left( \frac{\partial }{\partial x}\right)=\frac{\partial}{\partial x}\left(\gamma\frac{\partial}{\partial x'}-\frac{\gamma v}{c^2}\frac{\partial }{\partial t'} \right)= \\ = \gamma\frac{\partial}{\partial x'}\left(\gamma\frac{\partial}{\partial x'}-\frac{\gamma v}{c^2}\frac{\partial }{\partial t'} \right)-\frac{\gamma v}{c^2}\frac{\partial }{\partial t'}\left(\gamma\frac{\partial}{\partial x'}-\frac{\gamma v}{c^2}\frac{\partial }{\partial t'} \right)=\\= \gamma^2\frac{\partial^2 }{\partial x'^2}-2\frac{\gamma^2v}{c^2}\frac{\partial^2 }{\partial x'\partial t' }+\frac{\gamma^2v^2}{c^4}\frac{\partial^2}{\partial t'^2}$$

Edit 2 with the hints given by nervxxx

The wave equation becomes

$$\frac{\gamma^2v^2}{c^2}\frac{\partial^2 \phi}{\partial x'^2}-\frac{2\gamma^2v}{c^2}\frac{\partial ^2 \phi}{\partial x'\partial t'}+\frac{\gamma^2}{c^2}\frac{\partial^2 \phi}{\partial t'^2}=\gamma^2\frac{\partial^2 \phi}{\partial x'^2}-2\frac{\gamma^2v}{c^2}\frac{\partial^2 \phi}{\partial x'\partial t' }+\frac{\gamma^2v^2}{c^4}\frac{\partial^2\phi}{\partial t'^2}$$

$$\frac{\gamma^2 v^2}{c^2}\frac{\partial ^2 \phi}{\partial x'^2}+\frac{\gamma^2}{c^2}\frac{\partial^2 \phi}{\partial t'^2}=\gamma^2\frac{\partial^2 \phi}{\partial x'^2}+\frac{\gamma^2v^2}{c^4}\frac{\partial ^2\phi}{\partial t'^2}$$

But I still don't get... since all $\gamma^2$ cancel

Final edit. done!

$$\frac{\gamma^2 v^2}{c^2}\frac{\partial ^2 \phi}{\partial x'^2}-\gamma^2\frac{\partial^2 \phi}{\partial x'^2}=\frac{\gamma^2v^2}{c^4}\frac{\partial ^2\phi}{\partial t'^2}-\frac{\gamma^2}{c^2}\frac{\partial^2 \phi}{\partial t'^2}$$

$$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

$$\left( \frac{1}{1-\frac{v^2}{c^2}}\right)\frac{v^2}{c^2}\frac{\partial ^2 \phi}{\partial x'^2}-\left( \frac{1}{1-\frac{v^2}{c^2}}\right)\frac{\partial^2 \phi}{\partial x'^2}=\left( \frac{1}{1-\frac{v^2}{c^2}}\right)\frac{v^2}{c^4}\frac{\partial ^2\phi}{\partial t'^2}-\left( \frac{1}{1-\frac{v^2}{c^2}}\right)\frac{1}{c^2}\frac{\partial^2 \phi}{\partial t'^2}$$

$$\left( \frac{v^2}{c^2-v^2}\right)\frac{\partial ^2 \phi}{\partial x'^2}-\left( \frac{1}{1-\frac{v^2}{c^2}}\right)\frac{\partial^2 \phi}{\partial x'^2}=\left( \frac{v^2}{c^2-v^2}\right)\frac{\partial ^2\phi}{\partial t'^2}\frac{1}{c^2}-\left( \frac{1}{1-\frac{v^2}{c^2}}\right)\frac{1}{c^2}\frac{\partial^2 \phi}{\partial t'^2}$$

$$\frac{\partial \phi^2}{\partial x'^2}\left(\frac{v^2}{c^2-v^2}- \frac{1}{1-\frac{v^2}{c^2}}\right)=\frac{\partial^2 \phi}{\partial t'^2}\frac{1}{c^2}\left(\frac{v^2}{c^2-v^2}- \frac{1}{1-\frac{v^2}{c^2}}\right)$$

$$\frac{\partial^2\phi}{\partial x'^2}=\frac{1}{c^2}\frac{\partial^2\phi}{\partial t'^2}$$

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If you combine the two results you have and use the definition of $\gamma$ do you not get the right answer? – BebopButUnsteady Mar 17 '13 at 17:51
What you have done is correct. If you cancel $\gamma^2$ at this stage, you will get $(v^2/c^2 - 1)*(\text{wave equation})$. This is just the wave equation multiplied by a constant factor, which is still a wave equation, but this is what's confusing you because you're expecting the constant to be equal to $1$. Instead of cancelling $\gamma^2$ at this stage, substite for what is, and you will find that the wave equation emerges, with no pesky multiplicative constant outfront. – nervxxx Mar 17 '13 at 19:19

Second, you made a mistake in the cross terms for the $\partial^2 /\partial x^2$ term. The cross term should have the coefficient $-2\gamma^2 v/c^2$.
Third, use the fact that $\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$.