Frequency of Photon in Schwarzschild Metric?

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Homework Help Overview

The problem involves the Schwarzschild metric and the behavior of photons in a gravitational field. The original poster seeks to determine the frequency of a photon at infinity in relation to its frequency at a specific radius, as well as the range of frequencies.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster has derived the geodesic equations and the Christoffel symbols relevant to the Schwarzschild metric. They are attempting to relate the energy of the photon to its frequency but are uncertain about how to find the four-momentum and four-velocity components.

Discussion Status

The discussion is ongoing, with participants seeking clarification on the relationships between the variables involved. There are indications of exploration into the implications of redshift, but no consensus has been reached on the specific methods to solve the problem.

Contextual Notes

Participants are working within the constraints of the problem statement and the mathematical framework of general relativity, focusing on the implications of the Schwarzschild metric for photon behavior.

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Homework Statement



The schwarzschild metric is given by ##ds^2 = -Ac^2 dt^2 + \frac{1}{A} dr^2 + r^2\left( d\theta^2 + sin^2\theta d\phi^2 \right)##. A particle is orbiting in circular motion at radius ##r##.

(a) Find the frequency of photon at infinity ##\omega_{\infty}## in terms of when it is at ##r##, ##\omega_r##.
(b) Find the range of frequencies.

2007_B5_Q3.png

Homework Equations

The Attempt at a Solution


I have found the geodesic equations:
\frac{d}{ds}\left[A \dot t \right] = 0
0 = -Ac^2 \dot {t}^2 + \frac{1}{A} \dot {r}^2 + r^2 \dot {\phi}^2
\frac{d}{ds} \left[ r^2 \dot \phi \right] = 0

I have also found the christoffel symbols: ##\Gamma^r_{rr}=-\frac{A'}{2A}## and ##\Gamma^r_{tt}=-\frac{1}{2}A A'c^2## and ##\Gamma^r_{\theta \theta} = -Ar## and ##\Gamma^r_{\phi\phi}=-Ar## and ##\Gamma^t_{rt}=\Gamma^r_{tr}=\frac{A'}{2A}##.

Part(a)
The differential equation is simply the geodesic equation for a photon:
\dot p^\mu + \frac{1}{\hbar} \Gamma^\mu_{\alpha \beta} p^\alpha p^\beta = 0
For the temporal component we have exactly what the question wants:
\dot p^0 = - \frac{A'}{A}p^0 \dot r

The energy of the photon is given by ##E = p_\mu U^\mu =p_0 U^0+p_1 U^1+p_2 U^2+p_3 U^3 = \hbar \omega##.

How do I find ##p^\mu## and ##U^\mu##?
 
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