So I have this long standing problem. I know that the wave equation (with or without source term) changes form when one makes a Galilean transformation of coordinates. My question is about the physical meaning of this fact: is it correct to say that the phenomenon which appeared as a wave in the original coordinates is not a wave anymore? Since it satisfies a different equation I have been tempted to say so, but I am still not convinced Since the everyday Experience does not adapt well to this view. I know there's something I'm missing, Maybe is stupid but I would like to have somebody else's opinion on this!

  • $\begingroup$ It is not a wave anymore only if you move with it at the same speed. $\endgroup$ – safesphere Jul 3 '18 at 14:35

This should possibly be a comment, but the speed of light will no longer be invariant after the transformation, it's as straightforward as that.

Try the transform yourself and see.

  • $\begingroup$ Thank you for your answer but that is not my problem...I am talking about the General wave equation, not necessarily concerning EM waves $\endgroup$ – user199710 Jul 3 '18 at 12:27
  • $\begingroup$ Force of habit on my part sorry, $\endgroup$ – user198207 Jul 3 '18 at 12:51

The thing is just as Countto pointed out. The wave equation in one dimension without sources for speed $c$ is $$\frac{1}{c^2}\frac{\partial^2\phi}{\partial t^2}-\frac{\partial^2\phi}{\partial x^2}=0.$$

This has solutions of the form


The first term is a wave propagating with velocity $+c$, and the second one with $-c$. The Galilean transformation is $$t'=t$$$$x'=x-vt$$ Which transforms our solution to $$\phi(x',t')=\psi_+(x'-(c-v)t')+\psi_-(x'+(c+v)t'),$$ Which is still a superposition of two waves. These waves however have different velocities, $c-v$ and $-(c+v)$.

  • $\begingroup$ You said "is still a superposition of Two waves" but having them different velpcities, would Φ still be a wave? A superposition of waves of different velpcities does not satisfy any kind of wave equation am I right? $\endgroup$ – user199710 Jul 3 '18 at 13:51
  • $\begingroup$ Which is the meaning of the fact that the D'Alambert operator is not invariant under galilean transformation i Think, that is what I am trying to understand $\endgroup$ – user199710 Jul 3 '18 at 13:52
  • $\begingroup$ There is a loop in which I Think I'm stuck, and it could be the definition of what a wave is... If you take by definition "function which satisfies the wave equation" this implies that waves of different velpcities cannot be summed in A meaningful way, since the wave equation has only one parameter in it! $\endgroup$ – user199710 Jul 3 '18 at 13:57
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    $\begingroup$ Hello, sorry for not paying attention for two months! A "wave" can definitely be defined as you've mentioned, but in an intuitive sense, it's a perturbation that transfers energy over some distance without an average displacement of the medium. $\endgroup$ – Gabriel Golfetti Sep 13 '18 at 11:02
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    $\begingroup$ So yeah, in both the senses we've discussed here, that final superposition is not a wave in a rigorous sense. It doesn't satisfy a wave equation, and the medium has an average motion (with speed $v$). $\endgroup$ – Gabriel Golfetti Sep 13 '18 at 11:04

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