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Photon "escaping" from photon sphere in Schwarzchild space

  • #1
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Homework Statement


Close to a Schwarzchild black hole, a photon is emitted between r = 2(mu) and 3(mu), where [itex] \mu = \frac{GM}{c^2} [/itex]. The photon is emitted at an angle (alpha) to the radial direction. At r = 2(mu), the highest angle that the photon can escape at is (alpha) = 0; at r = 3(mu), the highest angle is (pi/2). For a general r between 2(mu) and 3(mu), what is the highest angle (alpha) in which the photon can escape to infinity?

Homework Equations


Eqns of motion (EQM):
[itex]
(1-\frac{2\mu}{r}) \frac{dt}{d\sigma} = k [/itex]

[itex] c^2 (1-\frac{2\mu}{r}) (\frac{dt}{d\sigma})^2 - (1-\frac{2\mu}{r})^(-1) (\frac{dr}{d\sigma})^2 -r^2 ((\frac{d\phi}{d\sigma})^2 =0 [/itex]
[itex]
r^2 (\frac{d\phi}{d\sigma}) =h
[/itex]

Impact parameter:
[itex] b = \frac{h}{ck} = r sin (\phi) [/itex]

The Attempt at a Solution


I guessed that the condition to escape was that [itex] \frac{dr}{d\phi} =0 [/itex] at r = 3 mu to give b, and then using that to give (phi), but that didn't get the right result for r = 2(mu).
I think that I don't really know what the condition for escape to infinity is.
 

Answers and Replies

  • #2
18,087
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Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 

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