I just watched this video from MIT explaining Galilean transformation of ordinary waves.
Early on, the professor goes on to say that the phase is exactly the same in all inertial frames of reference, and I can't really see why. I intuitively have a feeling that the angular frequency of the wave changes if we are in a translating frame of reference relative to a stationary one, however, is there any way to understand why the phase is the same.
Maybe I'm just misunderstanding the definition of phase. But I'd be glad if anyone could clarify as to why this is the case (meaning without any proof of that $\omega' = \omega(1-\frac{v}{c})$ ) which he already shows in the video. More of a intuitive explanation is searched for, maybe some real life example.
Thanks.
Best Answer
Galilean transformation (GT) means $$ \newcommand{\dd}[1]{\text{d}{#1}} \begin{align} x&\rightarrow x-ut \\ t&\rightarrow t \\ v&\rightarrow v-u \end{align} $$
The condition on the wave speed is that, $$ v = \omega/k $$
For wave speed $v$, $\phi$'s definition can be rewritten by introducing $0$ in the following way and using the expressions above. $$ \begin{align} \phi &\equiv kx-\omega t \\ &= k(x-vt) \\ &= k((x-ut)-(v-u)\,t) \\ &= k(x'-v't') \\ &= kx'-\omega't' \end{align} $$
All that remains is to show how $k$ transforms under GT.
We know that $k$ is a length between nodes, so in either frame, the observer could choose two adjacent nodes and observe the space between them, $$ L = x_b-x_a = 2\pi/k $$
and in the primed frame, the same approach $$ L' = x_b'-x_a' = x_b-x_a + (ut-ut) = L $$
So $L'=L$ implies $k'=k$, therefore $k\rightarrow k$ and $\phi\rightarrow\phi$
Deriving the short result you ask for, use that $v' = v-u$
$$ \begin{align} \frac{\omega'}{k'} &= v - u \\ v\frac{\omega'}{\omega}\frac{k}{k'} &= v - u \\ \frac{\omega'}{\omega} &= \frac{k'}{k}(1-u/v) \end{align} $$
again relying on $k'=k$
I hope someone comes up with an insightful statement motivating this result.