- #1
thepopasmurf
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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.
Show that ratio between emitted and received frequencies (for two observers E and R moving with velocities u and v respectively) is given by
[itex] \frac{\nu_E}{\nu_R} =\frac{\textbf{u} \bullet\textbf{k}}{\textbf{v}\bullet\textbf{k}}[/itex]
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
[itex] k^a = \left( \omega / c, \vec{k}\right) [/itex]
Then I also proceed with the knowledge that scalar products of four-vectors are invariant under frame transformations
Transform to emitter's frame (E). In this frame
[itex][u^{a}]=\left(c,0,0,0\right)[/itex]
and
[itex][k^{a}_E]=\left(\omega_E , \vec{k_E}\right)[/itex]
Taking the scalar product gives [itex]\textbf{u} \bullet \textbf{k}_{E} = \omega_{E}[/itex]
Performing similar steps gives
[itex]\textbf{u} \bullet \textbf{k}_{R} = \omega_{R}[/itex]
This bit I'm unclear on. [itex]\textbf{k}_E = \textbf{k}_R[/itex]
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 [itex]\textbf{p}_0[/itex] 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
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.
Homework Statement
Show that ratio between emitted and received frequencies (for two observers E and R moving with velocities u and v respectively) is given by
[itex] \frac{\nu_E}{\nu_R} =\frac{\textbf{u} \bullet\textbf{k}}{\textbf{v}\bullet\textbf{k}}[/itex]
Homework 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
[itex] k^a = \left( \omega / c, \vec{k}\right) [/itex]
Then I also proceed with the knowledge that scalar products of four-vectors are invariant under frame transformations
The Attempt at a Solution
Transform to emitter's frame (E). In this frame
[itex][u^{a}]=\left(c,0,0,0\right)[/itex]
and
[itex][k^{a}_E]=\left(\omega_E , \vec{k_E}\right)[/itex]
Taking the scalar product gives [itex]\textbf{u} \bullet \textbf{k}_{E} = \omega_{E}[/itex]
Performing similar steps gives
[itex]\textbf{u} \bullet \textbf{k}_{R} = \omega_{R}[/itex]
This bit I'm unclear on. [itex]\textbf{k}_E = \textbf{k}_R[/itex]
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 [itex]\textbf{p}_0[/itex] 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