# Photon Escape Angle From Black Hole

Consider a photon source emitting photons near the surface of a Schwarzschild black hole. What angle, as a function of the source's radius from the event horizon, must the photons be emitted at such that they can escape to an observer at infinity?

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when you say the source's radius, are you asking at what angle a photon can be emitted and still escape as a function of distance from the event horizon? – Jim May 27 '13 at 13:39
Yes, with the angle being measured from outward radial direction. – Andyb May 27 '13 at 13:44
Comment to the question (v4): The tag escape-velocity is usually only for massive particles. – Qmechanic May 27 '13 at 20:07
This is one of the more interesting and challenging homework like question, so I think it should not get closed even though there are some closevotes. – Dilaton May 31 '13 at 9:12
@Dilaton Agreed, though the OP should show what they've tried and where they are getting stuck. – Michael Brown May 31 '13 at 9:21

Within the photon sphere, the geometry is complicated and most of what I've read indicates that computer simulation is needed. However, outside the photon sphere (from about $2r_s$), the following formula should give you the maximum emission angle that could escape:
$$\theta(R)=cos^{-1}(2{r_s\over R}-1)$$
This formula also works for at the event horizon, it gives $\theta=0^\circ$.
Summary: $$R=r_s,~~\theta=0^\circ$$ $$r_s<R<1.5r_s,~~0\le\theta<90$$ $$\lim_{R\rightarrow(1.5r_s)^+}\theta=90^\circ$$ $$R\ge2r_s,~~\theta=cos^{-1}(2{r_s\over R}-1)$$