Doppler Effect for Light — Question 9

Question 9

Opposite sightlines provide a consistency test for Doppler data

A source with known rest frequency $f_0=500\,\mathrm{MHz}$ is reportedly observed at $f_1=800\,\mathrm{MHz}$ and $f_2=400\,\mathrm{MHz}$ along opposite laboratory directions $\theta$ and $\pi-\theta$. For a source of rest frequency $f_0$ moving at +v, use $f=f_0/[\gamma(1-\beta\cos\theta)]$, where $\beta=v/c$ and $\theta$ is the photon propagation angle from +x in the laboratory.

1. Eliminate the unknown angle to infer $\gamma$.

2. Decide whether any physical speed and angle can produce the data under this model.

Question 9 - Solution

Opposite sightlines provide a consistency test for Doppler data

(a) Method and calculation

$$\frac1{f_1}=\frac\gamma{f_0}(1-\beta\cos\theta),\quad\frac1{f_2}=\frac\gamma{f_0}(1+\beta\cos\theta).$$$$\boxed{\gamma=\frac{f_0}2\left(\frac1{f_1}+\frac1{f_2}\right)}.$$

(b) Interpretation and further calculation

$$\gamma=250\left(\frac1{800}+\frac1{400}\right)=\boxed{0.9375}.$$$$\gamma\ge1\quad\Rightarrow\quad\boxed{\text{no physical solution}}.$$

Physical check

An algebraic fit that demands gamma below one fails the physical admissibility test; at least one measurement or model assumption is inconsistent.