- #1

N00813

- 32

- 0

## Homework Statement

Close to a Schwarzschild 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.