Four-Momentum of Photon / Doppler shift in General Relativity

In summary, the ratio between emitted and received frequencies for two observers moving with velocities u and v is given by the scalar product of their velocities and four-momentum, divided by the four-momentum in both frames. This is possible because the metric is independent of time, allowing us to equate the four-momentum at the time of emission and reception.
  • #1
thepopasmurf
76
0
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.

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
 
Physics news on Phys.org
  • #2
for your help.The Attempt at a Solution (cont'd)From the given four-wavevector of the photon, we can calculate the four-momentum by p^a = hbar * k^aTransform to emitter's frame (E). In this frame,[u^{a}]=\left(c,0,0,0\right)and[p^{a}_E]=hbar * \left(\omega_E , \vec{k_E}\right)Taking the scalar product gives \textbf{u} \bullet \textbf{p}_{E} = hbar*\omega_{E}Performing similar steps in the receiver's frame gives\textbf{v} \bullet \textbf{p}_{R} = hbar*\omega_{R}Since the metric is independent of 't', then \textbf{p}_0 at time of emission is the same as at the time of reception, i.e. \textbf{p}_E = \textbf{p}_R.Therefore we can divide both scalar products giving the answer.\frac{\nu_E}{\nu_R} =\frac{\textbf{u} \bullet\textbf{p}}{\textbf{v}\bullet\textbf{p}}
 

1. What is the four-momentum of a photon in general relativity?

The four-momentum of a photon in general relativity is a mathematical representation of the photon's energy and momentum in the context of curved spacetime. It is a four-dimensional vector that includes the photon's energy and three components of momentum.

2. How is the four-momentum of a photon related to the Doppler shift in general relativity?

In general relativity, the Doppler shift is a result of the photon's energy and momentum being affected by the curvature of spacetime. The four-momentum of a photon is used to calculate the change in energy and momentum due to this curvature, resulting in the observed Doppler shift.

3. Can the four-momentum of a photon be used to explain gravitational redshift?

Yes, the four-momentum of a photon is crucial in understanding gravitational redshift. In general relativity, gravitational redshift occurs when a photon loses energy as it travels through a curved spacetime, resulting in a longer wavelength and a shift towards the red end of the electromagnetic spectrum.

4. How does the four-momentum of a photon differ from that of a massive particle in general relativity?

The four-momentum of a photon differs from that of a massive particle in general relativity mainly in terms of the magnitude of energy and momentum. While a massive particle's four-momentum is dependent on its rest mass, a photon's four-momentum is solely determined by its frequency and wavelength.

5. Can the four-momentum of a photon be used to study the effects of gravitational lensing?

Yes, the four-momentum of a photon is used in the study of gravitational lensing, which is the bending of light by massive objects. By analyzing the changes in the photon's energy and momentum as it travels through a gravitational lens, scientists can gain insights into the distribution of matter and the curvature of spacetime in the lensing object.

Similar threads

  • Advanced Physics Homework Help
Replies
7
Views
1K
Replies
1
Views
811
  • Advanced Physics Homework Help
Replies
4
Views
944
  • Special and General Relativity
Replies
17
Views
2K
  • Advanced Physics Homework Help
Replies
3
Views
1K
Replies
4
Views
962
  • Advanced Physics Homework Help
Replies
6
Views
975
  • Advanced Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
575
  • Introductory Physics Homework Help
Replies
14
Views
1K
Back
Top