# Issue showing that the phase of a harmonic wave is invariant under a Galilean transform

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.

## 1 Answer

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.

• 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. – Vishal Jain Mar 13 at 17:11
• @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... – mike stone Mar 13 at 17:54