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I just watched this video from MIT explaining Galilean transformation of ordinary waves.

https://www.youtube.com/watch?v=YdtHAIh-kas

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.

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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.

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  • $\begingroup$ Thank you Orion. I appreciate the help. Very clear :) $\endgroup$
    – Tanamas
    Apr 5, 2022 at 14:46
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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.
[...] is there any way to understand why the phase is the same.

There is nothing mysterious about the invariance of the phase. To understand this consider 2-dimensional water waves on the sea and a swimmer at its surface.

Let an observer at the beach watch the waves and the swimmer. From his reference frame (with coordinates $x,y,t$) the height of the waves is described by $$h=A\cos(\underbrace{k_x x + k_y y - \omega t}_{=\ \phi}).$$ Suppose at a certain point in time ($t$) the swimmer is at the peak of a wave. Because $\cos(\phi)$ has a maximum at $\phi=0$, that means the wave phase at the swimmer is $\phi=0$.

Let a second observer from an airplane flying above the sea also watch the waves and the swimmer. From his reference frame (with coordinates $x',y',t$) the height of the waves is described by $$h=A\cos(\underbrace{k'_x x' + k'_y y' - \omega' t}_{=\ \phi'}).$$ As seen from the airplane, at the same certain point in time ($t$) the swimmer has other spatial coordinates ($x',y'$). But of course he is still at the peak of a wave, meaning his phase is $\phi'=0$.

You can repeat the reasoning above for the swimmer being at the bottom of a wave (meaning $\phi=\phi'=180°$) or at any other intermediate place/time of a wave ($\phi=\phi'$). Hence it is always $$\phi =\phi'$$ or $$k_x x + k_y y - \omega t = k'_x x' + k'_y y' - \omega' t$$

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  • $\begingroup$ Thanks Thomas. Your example made it very clear to me as to why the phase is invariant. $\endgroup$
    – Tanamas
    Apr 5, 2022 at 14:47

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