# 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?

• 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 May 31 '13 at 9:21

According to this source (which is also a good source for all fun things regarding Schwarzschild black holes), there is a critical emission angle for photons from a stationary source some radius, $R$, from the black hole. Note that this equation technically should work for radii less than the Schwarxschild radius (the event horizon radius, $r_s=\frac{2GM}{c^2}$), but it'll give you negative angles because photons can't escape. Also note that all angles are given relative to the radial direction. $\theta=0$ means directed radially outwards and $\theta=\pi$ is radially inwards.
Inside the photon sphere, $R\le{3\over2}r_s$, the angles at which photons can escape are given by: $$\theta\le\arcsin\left[\frac{\sqrt{27}r_s}{2R}\sqrt{1-\frac{r_s}{R}}\right]$$
Outside the photon sphere, $R\ge{3\over2}r_s$, the escape angles are: $$\theta\le\pi-\arcsin\left[\frac{\sqrt{27}r_s}{2R}\sqrt{1-\frac{r_s}{R}}\right]$$
To get correct angles, just assume that arcsin always results in values between $-\pi/2$ and $\pi/2$. You'll note that for $R=r_s$, you find that $\theta=0$, which means only photons directed radially outwards escape. For $R=\frac{3}{2}r_s$, $\theta=\frac{\pi}{2}$ as my first paragraph stated. And that a radially inward photon ($\theta=\pi$) is always absorbed (this angle is asymptotically approached as $R\to\infty$).