At time $t_0 = 0$ someone on earth sends a (frequency modulated) signal to a spacecraft, which is at distance $R$ and moves with speed $v \geqslant 0$ along the $x$-axis. Suppose the signal is reflected instantaneously at the spacecraft and "travels" back to the transmitter. How to compute the signal received on earth using a Lorentz transformation?


In case of a linear frequency modulated signal, the frequency and the corresponding sinals are given as $$ \begin{align*} f(t) &= f_0 + \frac{B}{T}t\\ u(x, t) &= u_0 \cos\left( 2\pi \left(f_0\left(t - \frac{x}{c}\right) + \frac{B}{2T}\left(t - \frac{x}{c}\right)^2 \right) \right), \end{align*} $$ where $f_0>0$ denotes the carrier frequency, $B$ the bandwidth and $T$ the duration of the frequency ramp. With the help of a simple sketch and the already known formula for the Doppler shift, we can derive the received frequency (after the signal is reflected from the spacecraft and returned to the transmitter).

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Hence, the received frequency is given as $$ f(t) = \frac{c-v}{c+v} \left( f_0 + \frac{c-v}{c+v} \frac{B}{T}(t-\tau_0) \right). $$ Note that the domain of the received signal is $t \in [t_0 + \tau_0, t_1 + \tau_1]$ and therefore the slope/gradient differs from the transmitted one. Consequently, the received signal is given as $$ u(t) = u_0 \cos\left( 2\pi \frac{c-v}{c+v}\left(f_0(t-\tau_0) + \frac{c-v}{c+v}\frac{B}{2T}(t-\tau_0)^2 \right) \right). $$

Problme: According to Lorentz transformation of a frequency modulated signal, the signal at the spacecraft can be computed with the following transformation: $$ \begin{align*} t^\prime &= \gamma \left( t + \frac{v(x+R)}{c^2} \right)\\ x^\prime &= \gamma \left( x + R + vt \right). \end{align*} $$ If we apply this transformation twice (and set $x=0$), we will not end up with the exact same signal.

Question: How to obtain the exact same result using a Lorentz transformation?


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