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The equation for the relativistic doppler effect in a medium where waves travel at speed $c_m$ is $$\frac{f_r}{f_s} = \frac{c_m - v_r}{c_m-v_s}\frac{\gamma_r}{\gamma_s}$$

My question is whether this value stays invariant under velocity shifts as it does in the galilean case. That is we apply transformation, $v' = \frac{v-u}{1-\frac{uv}{c^2}}$ to the speed of waves, velocity of source and receiver and see if the equation stays the same. This doesn't seem at first glance to be true. If not, why?

Moreover let us consider the case of the doppler effect for light. The equation simplifies to $$\frac{f_r}{f_s} = \sqrt{\frac{(c-v_r)(c+v_s)}{(c+v_r)(c-v_s)}}$$ If we look carefully, we can see that this is equivalent to the standard 'correct' relativistic doppler equation 1 cited on this wikipedia page, compounded twice. To clarify, this is equivalent to a receiver in an intermediate frame transmitting another light beam of the frequency observed by them towards the receiver. Because of the constancy of $c$ this should give us the same value (if the equation were correct), right? But still it doesn't seem like the equation is actually invariant with velocity shifts. Which then implies a contradiction to the postulates of relativity.

Just to make sure I have everything in check, I will post my derivation of the equation below.

I have tried to derive the doppler effect equation the following way. Let there be 3 frames. The source frame $S$, the medium frame $S'$ and the receiver frame $S''$. The time period in $S$ is $\Delta t$ and the velocity of the source and receiver w.r.t $S'$ is $v_s,v_r$ respectively. Applying the inverse Lorentz transforms we go from $S$ to $S'$ to find

$\Delta t'=\gamma_s\Delta t$

$\Delta x'=\gamma_sv_s\Delta t$

The path of the two periodic disturbances in $S'$ is

$x=c_mt$ and

$x-\gamma_sv_s\Delta t = c_m(t-\gamma_s\Delta t)$

and the path of the receiver is $x = v_rt+\alpha$ where $\alpha$ is the displacement of the reciever from the source. Solving for the difference between the two solutions for the two times as the wave intersects the receiver in $S'$ gives us $\tilde{\Delta t'}= \frac{c_m-v_s}{c_m-v_r}\gamma_s\Delta t$, $\tilde{\Delta x'}= v_r\tilde{\Delta t'}$. Applying Lorentz transforms gives us $\Delta t'' = \tilde{\Delta t'}(1-\frac{v_r^2}{c^2})\gamma_r$ so $\Delta t'' = \frac{c_m-v_s}{c_m-v_r}\frac{\gamma_s}{\gamma_r}\Delta t$

So

$$\frac{f_r}{f_s} = \frac{c_m - v_r}{c_m-v_s}\frac{\gamma_r}{\gamma_s}$$

I believe this derivation should apply for light waves in any frame where we have the special case $c_m = c$. If this is the case, then it is necessary that the equation should be invariant under velocity shifts. That is, $v' = \frac{v-u}{1-\frac{uv}{c^2}}$

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Some algebraic busy work shows that the equation for light is in fact invariant under velocity shifts. One way of doing this is to see that when we plug in $$u = \frac{v_r-v_s}{1-\frac{v_rv_s}{c^2}}$$, it is possible to show that the factor becomes $$\sqrt{\frac{1-\beta}{1+\beta}}$$ where $\beta = \frac{u}{c}$.

Now it remains to show that $u$, the velocity of the receiver in the source frame is independent of frame of reference. Indeed, this is an underlying symmetry of Lorentz transformations, that stems from the fact that the spacetime vector in frame $A'$ is the same no matter how many intermediate steps you take to get from $A$ to $A'$. To check this, you can plug in velocity transformations in $u$ and see that it is invariant.

It's also invariant for waves in a medium, just the algebra is extremely tedious.

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It's also invariant for waves in a medium, just the algebra is extremely tedious.

I think the equation for waves in a medium can be proved invariant in the following fairly simple way. You can rearrange it to give

$$\frac{f_r/\gamma_r}{f_s/\gamma_s}=\frac{c_m-v_r}{c_m-v_s}.$$

If $f_r$ refers to the proper frequency, then $\gamma_r/f_r$ is the period referred to the frame of the medium, and similarly for $\gamma_s/f_s$. Therefore the entire equation is invariant because every quantity is explicitly referred to a particular frame, so they don't transform at all.

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  • $\begingroup$ but the proper frequencies could change if $v_r$,$v_s$ changes? Your last statement is obviously necessary for relativity. But it is not known to me a priori. See my next question: physics.stackexchange.com/q/447469 $\endgroup$
    – user86425
    Dec 15, 2018 at 18:14
  • $\begingroup$ Otherwise, I would have stopped just at getting to the point where I showed that the ratio was $\sqrt{\frac{1-\beta}{1+\beta}}$ for the first case $\endgroup$
    – user86425
    Dec 15, 2018 at 18:18
  • $\begingroup$ what does proper frequency refer to here anyway? $\endgroup$
    – user86425
    Dec 15, 2018 at 18:20
  • $\begingroup$ @lucky-guess: By proper frequency, I mean the frequency as measured by the observer, i.e., it's the inverse of the proper time that that observer measures for the period of the wave. So by definition the proper frequency doesn't change, because it's explicitly referred to a particular observer. $\endgroup$
    – user4552
    Dec 16, 2018 at 15:12

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