# 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

At the Schwarzschild radius, a photon must be emitted exactly normal to the surface in order to escape. As you travel outwards, the angle of emission decreases such that just above 1.5 times the Schwarzschild radius (i.e. the Photon Sphere) the photon can be emitted parallel to the tangent of the horizon and still escape.

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:

$$\theta(R)=cos^{-1}(2{r_s\over R}-1)$$

The theta produced is the emission angle relative to the radial direction.
This formula also works for at the event horizon, it gives $\theta=0^\circ$.

Summary: $$R=r_s,~~\theta=0^\circ$$ $$r_s<R<1.5r_s,~~0\le\theta<90$$ $$\lim_{R\rightarrow(1.5r_s)^+}\theta=90^\circ$$ $$R\ge2r_s,~~\theta=cos^{-1}(2{r_s\over R}-1)$$

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