# Black Hole Geodesics

1. Dec 10, 2003

### ObsessiveMathsFreak

Light cannot escape from black holes, hence their name.

But Since light has the same speed everywhere does that mean that the space/time geodesics in and around black holes are closed loops?

2. Dec 10, 2003

### Ambitwistor

Not generally (although there is one location at which light can (unstably) orbit a black hole in a closed loop.) Why should the constancy of the local speed of light imply that light should travel in closed loops, around a black hole or anywhere else?

3. Dec 10, 2003

### quantum

My understanding of black holes is that photons travellining in the right trajectory would be caught by the massive gravity of the black hole and orbit the black hole perpetually. This is known as a photon sphere, but I'm sure is just theory.

4. Dec 16, 2003

### franznietzsche

well the light should just fall into the singularity, but then again i may be confused on the actual scenario.

5. Dec 16, 2003

### chroot

Staff Emeritus
At a distance of

$$\frac{3 G M}{c^2}$$

light can orbit a black hole, though unstably, as Ambi said.

- Warren

6. Dec 17, 2003

### franznietzsche

unstably because if its distance changes at all, then the force of gravity changes with it and the orbit either becomes a fall or an escape?

7. Dec 17, 2003

### chroot

Staff Emeritus
You got it. You'd have to have pretty good aim with your laser!

- Warren

8. Dec 17, 2003

### franznietzsche

lol

edit: just realized something, the point at which it orbits exactly is in fact the event horizon is it not?

9. Dec 17, 2003

### chroot

Staff Emeritus
No, light can orbit at exactly 1.5 times the event horizon radius.

The event horizon exists at the "Schwarzschild radius,"

$$r_s = \frac{2 GM}{c^2}$$

- Warren

10. Dec 18, 2003

### franznietzsche

oh, ok.

11. Nov 28, 2006

### Chris Hillman

Closed null curves and two reference books

Hi, OMF,

I'd add to what the others told you a possibly confusing and distracting comment: deep inside the Kerr solution, well hidden from outside viewers by the event horizon, lie closed null curves. These are generally thought to be unphysical and to represent a mathematical artifact of the symmetry of the Kerr vacuum (while the exterior field on the other hand is thought to be in some sense the "preferred state" of the exterior; according to gtr, an isolated black hole will radiate away any deviations from the Kerr geometry in the form of gravitational radiation).

These CNCs are truly closed curves; don't confuse them with the "unstable circular orbits" in the exterior region, which are spiral-shaped null geodesics.

Two excellent books which offer extensive discussions of geodesics in the Kerr solution are The Geometry of Kerr Holes, by Barrett O'Neill, and The Mathematical Theory of Black Holes by Subrahmanyan Chandrasekhar.

Chris Hillman

12. Nov 28, 2006

### Hurkyl

Staff Emeritus
There's an easy heuristic reason, I think: if you inserted a mirror into the orbit, then the light could be deflected, and thus escape the black hole. So, the orbit has to be outside of the event horizon.

13. Jan 1, 2010

### Favicon

Sort of - it depends on whether your photon gets within the event horizon or not.

According to general relativity, massive objects cause space-time to warp in such a way that other objects (including photons) are pulled towards them, so a photon passing close to a black hole will always have its path deflected towards the hole. If you send a photon past with just the right trajectory at just the right distance from the hole, then its path should curve around the black hole and into an orbit (just like getting a satellite to orbit the earth, except that with a satellite we have the advantage of being able to use boosters to continuously fine-tune its position and keep it in orbit).

Inside the event horizon of a black hole the curvature is so strong that the escape velocity is higher than the speed of light and therefore any photon caught in there will spiral inwards towards the singularity.