# Photon Spheres

Ok, now that I have finished the preamble over here, I can finally start talking about what I originally intended to blog about: Photon spheres.

The defining feature of a black hole is that the gravitational attraction beneath the event horizon is so strong that even light can’t escape. That is, the curvature of spacetime in the vicinity of the black hole is so intense that there are no geodesics which are able to leave. This is what makes black holes so interesting, since anything (including light) which is dropped into a black hole is lost forever (we’re talking classical black holes at the moment, so no Hawking radiation). On the other hand, light (or matter) travelling near a black hole can escape, provided it stays outside of the event horizon.

Light from the blue object is deflected towards the grey massive object

General relativity predicts that light passing near a heavy object will be deflected by the gravitational field of that object. Equivalently, light will follow a geodesic in the curved spacetime around the heavy object.

This situation strongly resembles the situation we have in orbital mechanics, where a small object like a satellite or asteroid is moving near a large object, like a planet. Depending on the velocity of the object, it either escapes the planet, falls into the planet, or starts orbiting the planet. It seems like a natural question, then, to ask:

Can light orbit a black hole?

The answer is yes! General relativity predicts that light can orbit a black hole! For a Schwarzchild black hole, the black hole is spherically symmetric, and the possible orbits of light form what is called the Photon sphere of the black hole. We’re now going to derive the radius of the photon sphere directly from the Schwarzchild metric and geodesic equation.

We begin with the Schwarzchild metric in spherical polar coordinates $(ct, r, \theta, \phi)$ given by

$ds^2 = -(1-\frac{R_s}{r})c^2dt^2 + (1-\frac{R_s}{r})^{-1}dr^2 + r^2 d\theta^2 + r^2 \sin^2 \theta d \phi^2$

where $R_s = \frac{2GM}{c^2}$ is the Schwarzchild radius of the black hole. Recall that this metric describes the spacetime around an uncharged, non-rotating, spherically symmetric massive body. Since we want to consider the motion of light around this body, we are looking for null geodesics, that is, geodesics for which $ds^2 = 0$. We also want our paths of light to orbit the black hole, so the radial coordinate should be constant, that is, $dr^2 = 0$. To simplify things, we may as well consider orbits which are confined in a plane, say $\theta = \frac{\pi}{2}$, so $d\theta^2 = 0$. Substituting this into our metric and rearranging gives us the relation

$\frac{\text{d}\phi}{\text{d}t} = \frac{c}{r}\sqrt{1 - \frac{R_s}{r}}$

We now look at the radial geodesic equation

$\frac{\text{d}^2 r}{\text{d}t^2} + \Gamma^{r}_{\alpha \beta} \frac{\text{d} x^{\alpha}}{\text{d} t} \frac{\text{d} x^{\beta}}{\text{d} t} = 0$

For the Schwarzchild metric, the only non-vanishing Christoffel symbols are

$\Gamma^r_{t t} = \frac{B B'}{2}$

and

$\Gamma^r_{r r} = \Gamma^r_{\theta \theta} = \Gamma^r_{\phi \phi} = -Br$

where $B = (1- \frac{R_s}{r})$ and $B' = \frac{\text{d}B}{\text{d}r}$. Substituting these into the radial geodesic equation we get

$c^2 \frac{B B'}{2} - Br \dot{\phi}\dot{\phi} = 0$

which rearranges to give us the following relation

$\frac{\text{d}\phi}{\text{d}t} = \frac{c}{r}\sqrt{\frac{R_s}{2r}}$

But we already have an expression for $\frac{\text{d}\phi}{\text{d}t}$! By equating these two expressions, it follows that

$\frac{R_s}{2r} = 1- \frac{R_s}{r}$

or equivalently

$r = \frac{3R_s}{2}$

That is, in order for light to orbit a black hole, it must do so at precisely 1.5 times the Schwarzchild radius. There are a couple of things to note about this. First of all, since the Schwarzchild metric is the metric for ANY spherically symmetric (non-rotating, uncharged) mass distribution, our analysis would also apply to any compact object with a radius less than 1.5 times the Schwarzchild radius. Thus a neutron star with, say, a radius of $1.2 R_s$ would still have a photon sphere. Secondly, it should be noted that photon spheres are unstable. What I mean by this is that if light in a photon sphere is perturbed slightly, it will either decay into the black hole or escape it entirely.

So what makes a photon sphere special? Aside from the fact that light is orbiting a black hole and that’s totally badass? Well, consider what would happen if I stuck my head inside the photon sphere. What would I see?

The back of my head! Light reflected from the back of my head would travel in a circular orbit around the black hole and into my eyes. So next time someone says they have eyes in the back of their head, chances are they have their head in a photon sphere.

Next post I’m going to be writing about Photon spheres around rotating black holes!