Proof of double Doppler shift

Question

We know that the Doppler effect for EM radiation is $\Delta$$f$ = $\frac{v}{c}$$f_0$

$\Delta$$f$ = the change in frequency

v = the relative speed of the source and observer

c = the speed of light in a vacuum

$f_0$ = the original frequency

I have been told simply the change in frequency, $\Delta$$f$, of an EM wave is doubled when the source sends a wave to a moving observer and then it reflects off a moving observer and it gets received by the source.

But I don't understand why this happens. I tried multiple times to use the above equation to prove the total $\Delta$$f$ is twice of that of the first wave. But every time I fail. So I would like to see a mathematical proof rather than word explanations. Thank you!

Answer 1

The correct formula for the Doppler shift is $$f_r=f_e\sqrt{\frac{c+v}{c-v}}$$ where $f_r$ is the received frequency and $f_e$ is the emitted frequency. Now, for small velocities we can perform a first-order Taylor series expansion in $v$ and write $$f_r\approx f_e \left( 1+\frac{v}{c} \right)$$ but remember that this is an approximation.

In this case a reflection counts as a reception followed by an immediate emission at the same frequency. So there are two Doppler shifts, one for the shift of the received frequency and the other for the shift of the emitted frequency. So then we have two Doppler shifts $$f_r = f_e\sqrt{\frac{c+v}{c-v}}\sqrt{\frac{c+v}{c-v}} \approx f_e \left(1+\frac{2v}{c} \right)$$

Answer 2

Here's a variation of the answers by @Dale and @JEB using the Bondi $k$-calculus.

Using a variant of my diagram from https://www.physicsforums.com/insights/relativity-using-bondi-k-calculus/ to handle two approaching astronauts, we draw

$k$ (the Bondi-$k$-factor) is introduced as a proportionality constant that depends only on the relative-velocity (and not on the period $T$). By radar measurement definitions, we get $k=\sqrt{\displaystyle \frac{c\Delta t+\Delta x}{c\Delta t- \Delta x}}=\sqrt{\displaystyle \frac{c+v}{c-v}}$, the formula for the Doppler factor.

We have \begin{align*} \Delta f &=f_r -f_e\\ &=\frac{k^2}{T}-\frac{1}{T}\\ &=\frac{\displaystyle\left( \frac{c+v}{c-v} \right)}{T}-\frac{1}{T}\\ &=\left( \displaystyle\left( \frac{c+v}{c-v} \right)-1\right)\frac{1}{T}\\ &= \displaystyle\left( \frac{2v}{c-v} \right)f_e\\ \end{align*} For $v\ll c$, we have $\Delta f\approx \displaystyle\frac{2v}{c}f_e$.

To connect with @Dale's answer, write instead \begin{align*} f_r &= \frac{k^2}{T}\\ &= k^2 f_e \end{align*} in agreement with @Dale.

Answer 3

The double Doppler shift can be derived kinematically: the 4-acceleration ($a^{\mu}$) imparted by the light must be orthogonal to mirror's 4-velocity ($v_{\mu}$). (This is required to keep the mirror "on-shell").

The initial photon has 4-momentum:

$$ p_{\mu} = (p, 0, 0, p) $$

The reflected photon has 4-momentum:

$$ p'_{\mu} = (p', 0, 0, -p') $$

So the momentum transfer, which is proportional to $a_{\mu}$, is:

$$ q_{\mu} = p'_{\mu} -p_{\mu}=(p'-p,0,0, -p'-p) $$

Meanwhile, the mirror moving away along the $z$-axis has 4-velocity:

$$ v^{\mu} = \gamma(c, 0, 0, v) $$

The condition is then:

$$q_{\mu}v^{\mu} = \gamma\big((p'-p)c - (p'+p)v\big)= 0$$

so that:

$$(p-p')c = (p+p')v$$

Subbing in $ pc = E= hf $:

$$ f' = f \frac{c-v}{c+v} $$

which agrees with the checked answer.