How Does Relativistic Motion Affect Photon Reception and Luminosity Angles?

[Silviu]

Homework Statement

A body emits photons of frequency $\omega_*$ at equal rates in all direction in its rest frame. An observer is moving with speed V relative to the body in the x direction. Find the rate at which the photons are received per unit solid angle $dN/dt'd\Omega'$ a large distance away in the observer's frame as a function of angle $\alpha'$ from the x'-axis. Find the luminosity per unit solid angle $dL'd\Omega$ a large distance away as a function of $\alpha'$

Homework Equations

$cos\alpha'=\frac{cos\alpha + V}{1+Vcos \alpha}$

The Attempt at a Solution

If in the stationary S frame we have N photons per $dt$ per $d\Omega$ this means in the S' the same number of photons per $dt'=dt/\gamma$ per $d\Omega'$. $d\Omega = sin(\theta)d\theta d\phi$ so in S' $d\Omega' = sin(\theta') d\theta' d\phi'$. As they say long distance, I assume we can work in small angle approximation. Also as they want as a function of $\alpha'$ I can take both $\omega$ and $\phi$ to be $\alpha$? But I am not sure how to go from the original formula to something useful to plug in here.

 

[_coyote_]

The flux received per unit solid angle at an angle $\alpha'$ is proportional to the number of photons emitted per unit solid angle in the source frame, which is uniform in all directions. The transformation of the solid angle element $d\Omega$ from the source frame to the observer frame is given by $d\Omega' = sin(\theta') d\theta' d\phi' = \frac{sin(\theta)d\theta d\phi}{\gamma^2(1-Vcos\alpha)^2}$ where $\alpha$ is the angle between the direction of emission and the direction of motion of the observer. The rate of photons received per unit solid angle in the observer's frame is then given by $\frac{dN}{dt'd\Omega'} = \frac{N}{\gamma^2(1-Vcos\alpha)^2}$The luminosity per unit solid angle is given by the total power emitted per unit solid angle in the source frame divided by the speed of light. In the source frame, the power emitted per unit solid angle is $\omega_*N$, and the speed of light is $c$. Therefore, the luminosity per unit solid angle in the observer's frame is $\frac{dL'}{d\Omega'} = \frac{\omega_*N}{c\gamma^2(1-Vcos\alpha)^2}$