# Doppler effect and Lorentz transformation

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?

Example:

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).

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?