2
$\begingroup$

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 $\nu_B/\nu_A$ in the photon frequency is independent of whether A and B are travelling in the same direction or opposite directions.

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

Aberration formula: $\cos\theta' = (\cos\theta - \beta)/(1-\beta\cos\theta) = -\beta$
(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 $\nu_B/\nu_A = \gamma(1+\beta^2)$. Plugging in the velocity of the emitter A in frame S' doesn't seem to get the required result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.