Relativistic Doppler Shift vs Classical Doppler Shift Contradiction

Question

In classical/Newtonian mechanics, the doppler shift (for light) can be expressed as:

$$ \frac{f_r-f_s}{f_s}=\frac{1+\beta}{1-\beta}-1 $$

In relativity the doppler shift can be expressed as:

$$ \frac{f_r-f_s}{f_s}=\sqrt{\frac{1+\beta}{1-\beta}}-1 $$

In the Newtonian limit where $ \beta\ $is small, we should be able to approximate both of the above expressions in terms of first order Taylor expansions. If we do so, we get the following approximations (obtained from WolframAlpha):

$$ \frac{f_r-f_s}{f_s}=2\beta\ (Newtonian) $$ $$ \frac{f_r-f_s}{f_s}=\beta\ (Relativitic) $$

However, in the Newtonian limit these approximations should be the same (i.e. we expect the relativistic doppler shift equation to give the same results as the Newtonian doppler shift equation when $ \beta\ $is small). Why do these last two equations not match? Where is the flaw in my logic?

Definitions:

$ f_s $: Source frequency

$ f_r $: Received frequency

$ \beta=v/c $

Answer 1

In the relativistic formula $v$ is the relative velocity between source and observer.

For the Newtonian case you have the source moving towards the observer at a speed $v$ relative to the medium and the observer moving towards the source at a speed $v$ relative to the medium.
Thus, the relative velocity between the source and the observer is $2v$.

Answer 2

They should not be the same, and the post-Minkowski (i.e., Taylor expansion in $v/c$) has no reason to work because it's not about the velocity, it's about the medium of propagation.

In relativity, that is the electromagnetic field, which is inherently relativistic and has no absolute rest frame. For that reason along, the Doppler shift has to satisfy:

$$ f(-v) = 1/f(v) $$

that is, it's symmetric in transmit and receive to all orders.

Meanwhile, the Doppler shift in a medium has an absolute rest frame, and is as such defined relative to its rest frame.

Another way to look at it is: a wave in medium has a well defined frequency, it's the frequency at $v=0$. Meanwhile, in vacuum, a plane wave doesn't have a frequency. None of $E, \vec p, \omega, \vec k, \lambda$ are Lorentz scalars.

The physics is just different situations.