Photon "escaping" from photon sphere in Schwarzchild space

[N00813]

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.

 

[mark726]

The condition for escape to infinity is that the photon's energy (or equivalently, its frequency) at infinity is greater than or equal to its energy at the emission point. This can be expressed mathematically as:

\frac{E_{\infty}}{E_{emission}} \geq 1

Using the equations of motion, we can express the photon's energy at infinity as:

E_{\infty} = c^2 (\frac{dt}{d\sigma})_{\infty}

where (\frac{dt}{d\sigma})_{\infty} is the time component of the photon's 4-momentum at infinity. Similarly, the energy at the emission point can be expressed as:

E_{emission} = c^2 (\frac{dt}{d\sigma})_{emission}

Using the equation for (\frac{dt}{d\sigma}), we can rewrite this as:

E_{emission} = c^2 \frac{k}{1-\frac{2\mu}{r}}

Now, we can use the condition for escape to infinity to solve for (alpha), the angle at which the photon is emitted. We can set E_{\infty} = E_{emission} and solve for (\frac{dt}{d\sigma})_{emission}. Then, we can use the equation for (\frac{dt}{d\sigma}) to solve for (alpha).

I hope this helps!