# Relativistic Doppler effect for parallel-moving observers

Suppose two observers (A, B) are moving along trajectories parallel to the x-axis in some inertial frame. Now A emits a photon with $$\nu_a$$ in the positive y-direction, which is recieved by B with frequency $$\nu_B$$. I've been asked to prove that the Doppler shift $$\frac{\nu_B}{\nu_A}$$ is the same regardless A and B traveling in the same or the opposite direction.

I've tried obtaining the formula from scratch, and I got the usual $$\frac{\nu_B}{\nu_A}=\gamma(1+\beta cos(\theta))$$, where $$\theta$$ is the angle which B sees between its x-axis and the photon trajectory. It makes sense for me that the Doppler effect is the same since they are moving perpendicular to the photon, so the cosine cancels and it just deppends on gamma, evaluated over the relative speed of A with respect to B in the rest frame of this last one... But I'm not completely satisfied with the answer, it would be of much help if someone could confirm or debunk my reply.

Yeah, you are alright, this is because $$\frac{\nu_B}{\nu_A}=\frac{1}{\sqrt{1-\beta^2}}$$ that is the lorentz factor this is because $$\beta cos\pi/2=0$$ that is the angle between the trayectory of the light and the observer, so then the lorentz factor is the same por $$-v$$ or $$v$$ because $$\beta^2=(-\beta)^2$$, $$\frac{v^2}{c^2}=\frac{(-v)^2}{c^2}$$ so it no depend on the direction between observers in fact just in the velocity between them