 Solution

Let $t$ label the reception event in the source reference frame. Its distance from the source and the relevant retarded time areThe received phase is $2\pi f_0{t}_e$. Because the observer's proper time obeys $d\tau =dt/\gamma$, its measured frequency isEquivalently, write $\beta =v/c$ and $s(t)=vt/r(t)$. The incoming Doppler factor is $D_{\mathrm{in}}=\gamma (1-\beta s)$, obtained also by contracting the photon four-momentum with the observer's four-velocity. The frequency decreases monotonically, withIn particular, the instant of closest approach is not the zero-shift point: a moving receiver looking perpendicular to its motion sees a factor $\gamma$, not $1/\gamma$. The latter factor applies to a different moving-emitter geometry. The graph uses source-frame reception time, with observer proper time $\tau =t/\gamma$.