Doppler Effect for Light — Question 10

Question 10

Shortened arrival intervals can produce apparent superluminal transverse motion

A moving emitter has speed $v=\beta c$ and approaches a distant stationary observer at angle $\alpha$ between velocity and the line of sight toward that observer. Across a short interval the direction is effectively constant. Its transverse displacement is $v\sin\alpha\,\Delta t$.

1. Derive the apparent transverse speed using reception-time separation.

2. For $\beta=0.80$ and $\cos\alpha=0.80$, find that speed and the received-to-proper pulse-frequency ratio.

Question 10 - Solution

Shortened arrival intervals can produce apparent superluminal transverse motion

(a) Method and calculation

$$\Delta t_R=\Delta t-v\cos\alpha\,\Delta t/c=\Delta t(1-\beta\cos\alpha).$$$$\boxed{v_{\mathrm{app}}/c=\frac{\beta\sin\alpha}{1-\beta\cos\alpha}}.$$

(b) Interpretation and further calculation

$$\boxed{v_{\mathrm{app}}/c=\frac{.80(.60)}{1-.80(.80)}=4/3}.$$$$\boxed{f_R/f_0=\frac1{\gamma(1-\beta\cos\alpha)}=5/3}.$$

Physical check

The apparent speed uses reception times, not simultaneous source positions. The actual emitter speed remains 0.80c.