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The phase $Φ$ of wave is defined as $kx-wt$. It should be the case that all observers moving relative to each other in the non relativistic case will agree on this.

So given the transforms $x'=x-vt$ and $t=t'$,

$Φ'=kx'-wt'$

$=k(x-vt)-wt$

$=kx-wt-kvt$

Seeing as this is wrong, how does one properly show that the phase of a wave is Galilean invariant.

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The ordinary wave equation is not galilean invariant. It is invariant only under Lorentz transformations with "$c$" being the wave velocity. This is not unreasonable as the usual wave equation refers to motion in a medium, and if you are moving with respect to the medium things will seem different. Your algebra shows that in the moving frame $$ kx-\omega t \mapsto kx+\omega' t $$ where $\omega'= \omega+kv$ is the Doppler-shifted frequency.

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  • $\begingroup$ Does this also imply that everyone agrees on the wavelength of the harmonic wave, while they may disagree on its speed and frequency? That seems interesting. $\endgroup$ – Vishal Jain Mar 13 at 17:11
  • $\begingroup$ @Vishal. Yes. But then it is obvious that if you ride past a wave, then you will pass the wavecrests at a different rate than if you were stationary... $\endgroup$ – mike stone Mar 13 at 17:54

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