# Four-Momentum of Photon / Doppler shift in General Relativity

1. Dec 6, 2011

### thepopasmurf

Hi,
I'm trying to understand a derivation / question for my revision. It has to do with the four momentum of a photon and doppler/redshifts.

1. The problem statement, all variables and given/known data

Show that ratio between emitted and received frequencies (for two observers E and R moving with velocities u and v respectively) is given by
$\frac{\nu_E}{\nu_R} =\frac{\textbf{u} \bullet\textbf{k}}{\textbf{v}\bullet\textbf{k}}$

2. Relevant equations
My first issue is deriving the four-wavevector for a photon. I know this is related to the four-momentum by p = hbar * k. Which one do I derive first? Assuming I'm given the four-wavevector
$k^a = \left( \omega / c, \vec{k}\right)$
Then I also proceed with the knowledge that scalar products of four-vectors are invariant under frame transformations

3. The attempt at a solution

Transform to emitter's frame (E). In this frame
$[u^{a}]=\left(c,0,0,0\right)$
and
$[k^{a}_E]=\left(\omega_E , \vec{k_E}\right)$

Taking the scalar product gives $\textbf{u} \bullet \textbf{k}_{E} = \omega_{E}$

Performing similar steps gives
$\textbf{u} \bullet \textbf{k}_{R} = \omega_{R}$

This bit I'm unclear on. $\textbf{k}_E = \textbf{k}_R$

With this, we divide both scalar products giving the answer.

Elsewhere in my notes I have a comment saying that "metric is independent of 't', therefore $\textbf{p}_0$ at time of emission is the same as at the time of reception"
This is related so could someone comment on how this fits into the overall picture.

Thanks