The Traveling Electromagnetic Wave, Qualitatively — Question 8

Question 8

A moving sampler crosses fewer wave crests when it follows the wave

A laboratory wave has phase $\phi=kx-\omega t$ with omega=ck. A sampler moves along x=ut, where $|u|<c$. Its readings are indexed by laboratory time t, not by its own proper time.

1. Find the angular oscillation frequency in the recorded laboratory-time trace.

2. Compare the observed cycle rates for u=+c/4 and u=-c/4. Explain why this is not the full relativistic Doppler formula.

Question 8 - Solution

A moving sampler crosses fewer wave crests when it follows the wave

(a) Method and calculation

$$\phi(t)=(ku-\omega)t=-\omega(1-u/c)t.$$$$\boxed{\omega_{\mathrm{trace}}=\omega(1-u/c)}\quad(|u|<c).$$

(b) Interpretation and further calculation

$$\boxed{\omega_+=3\omega/4,\quad\omega_-=5\omega/4,\quad\omega_-/\omega_+=5/3}.$$$$\text{Proper-time rate}=\gamma\omega_{\mathrm{trace}},\quad\gamma=(1-u^2/c^2)^{-1/2}.$$

Physical check

The defined laboratory clock avoids confusing crest-crossing kinematics with the time read by a moving observer.