# Relativistic Transverse Doppler Effect

In Minkowski spacetime, two observers, $$A$$ and $$B$$, are moving at uniform speeds $$u$$ and $$v$$, respectively, along different trajectories, each parallel to the y-axis of some inertial frame $$S$$. Observer $$A$$ emits a photon with frequency $$\nu_{A}$$ that travels in the x-direction in $$S$$ and is received by observer $$B$$ with frequency $$\nu_B$$. Show that the Doppler shift $$\frac{\nu_B}{\nu_A}$$ in the photon frequency is independent of whether $$A$$ and $$B$$ are travelling in the same direction or opposite directions.

Relevant equations: $$\frac{\lambda}{\lambda'} = \frac{\nu_B}{\nu_A} = \gamma(1-\beta\cos\theta)$$

Aberration formula: $$\cos\theta' = \frac{\cos\theta - \beta}{1-\beta\cos\theta} = -\beta$$
(for transverse case)
The answer is apparently that the Doppler shift is independent of the relative direction of motion. I have tried to transform to the frame $$S'$$ where $$B$$ is stationary, finding the velocity of $$A$$ using the addition of velocities formula - to then get gamma. I have used the abberation formula to insert $$\cos\theta'$$ into the Doppler shift formula above to get $$\frac{\nu_B}{\nu_A} = \gamma(1+\beta^2)$$. Plugging in the velocity of the emitter $$A$$ in frame $$S'$$ doesn't seem to get the required result.

• Dec 9, 2019 at 16:20

$$\nu_B=\nu_A \frac{1+\frac{w}{c}\cos\theta_A}{\sqrt{1-\frac{w^2}{c^2}}}\space ,$$
where $$w=\frac{u+v}{1+{uv}/{c^2}}$$ is the relativistic velocity of $$A$$ as measured by $$B$$. (Assume that $$A$$ and $$B$$ recede from each other along $$y_A$$ and $$y_B$$.) However, remember that using the aberration formula, we should find the original angle of emission in $$A$$'s frame of reference, while the emission angle is calculated as $$\theta_S=\pi/2$$ in $$\boldsymbol S$$. Therefore, we must consider $$A$$'s velocity of $$u$$ WRT $$S$$, rather than $$w$$:
$$\cos \theta_A=\frac{\cos \theta_S-\frac{u}{c}}{1-\frac{u}{c}\cos \theta_S}=-\frac{u}{c} \space.$$
Subtituting the second equation in the first one and using $$w=\frac{u+v}{1+{uv}/{c^2}}$$ implies:
$$\nu_B=\nu_A \frac{1-\frac{wu}{c^2}}{\sqrt{1-\frac{w^2}{c^2}}}=\nu_A × \frac{c^2-\frac{u(u+v)}{1+uv/c^2}}{c\sqrt{c^2-\frac{u^2+v^2+2uv}{(1+uv/c^2)^2}}}=\nu_A \frac{c^2-u^2}{c\sqrt{c^2+\frac{u^2v^2}{c^2}-u^2-v^2}}=\nu_A \sqrt{\frac{c^2-u^2}{c^2-v^2}}\space .$$
As you see, if you change $$u$$ into $$-u$$, or $$v$$ into $$-v$$, the result remains the same.