Black Hole Geodesics

[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?

[Ambitwistor]

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

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?

[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.

[franznietzsche]

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

[chroot]

At a distance of

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

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

- Warren

[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?

[chroot]

Originally posted by 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?
You got it. You'd have to have pretty good aim with your laser! [:)]

- Warren

[franznietzsche]

Originally posted by chroot
You got it. You'd have to have pretty good aim with your laser! [:)]

- Warren

lol

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

[chroot]

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

[franznietzsche]

oh, ok.

[Chris Hillman]

Hi, OMF,


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


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

[Hurkyl]

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