Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I know the Schwarzschild event horizon is a null surface generated by null geodesics. But what does that actually mean in terms of the path of a light ray that reaches it? Does that mean the geodesic trajectory of light on the surface will be along the surface? Which way will the light ray go? And what equation shows this?

share|cite|improve this question
Related: – Qmechanic Jun 5 '13 at 22:10

The incoming light ray can pass through the horizon. The easiest way to see this is to imagine light cones in a 2d (1 space + 1 time) spacetime. As you get near the horizon, the light cones are more and more tipped up. A light cone in 1+1 dimensions consists of two lines. At the horizon, imagine one of these null lines being the null geodesic horizon generator through that point. The other one represents the path of an incoming light ray. This is nicely illustrated on the Eddington Finkelstein diagram in this wikipedia article.

The Eddington Finkelstein coordinates are a system that's non singular on the horizon (Kruskal Szekeres is another). In this coordinate system, the incoming light ray is just represented by the equation $v=const$.

share|cite|improve this answer
Thank you, I know this - but how does the fact that the horizon is a null surface and null geodesics are tangent to it affect the movement of light? Will it move tangent to the surface? What happens if I, falling into a black hole, turn on a flashlight tangent to the horizon just as I reach the surface? – Judy Jan 23 '13 at 17:45
@Judy: exactly. You can think of the light ray just sitting at the surface at a constant $r$, neither escaping nor falling into the horizon, until it intersects with future timelike infinity. At least to the extent where you can take these sorts of metaphors seriously. – Jerry Schirmer Jan 23 '13 at 17:54

Provided you're willing to consider only radial light rays, i.e. normal to the event horizon, I calculated the behaviour at the event horizon in my answer to Why is a black hole black?.

You have to give some thought as to what you mean by the velocity of the light ray at the event horizon. The calculation I did uses Gullstrand-Painlevé co-ordinates and these don't correspond to physical co-ordinates measured by any observer. Nevertheless the result is quite convincing i.e. that for a light ray directed outward the velocity at the event horizon is zero.

I may be misinterpreting your question (if I am ignore this paragraph) but you seem to be suggesting that the null geodesics run tangential to the surface, i.e. constant $r = r_s$, so a light ray tangential to the event horizon would circle the event horizon. That doesn't happen. There are no null geodesics tangential to the event horizon. The closest distance that a null geodesic at constant $r$ exists is as $r = 3M$ i.e. the last stable orbit.

share|cite|improve this answer
If there are no null geodesics tangent to the event horizon than why is it a null surface? – Judy Jan 24 '13 at 8:41
From Wikipedia ( In the theory of relativity, a null surface is a 3-surface whose normal vector is everywhere null (zero length with respect to the local Lorentz metric), but the vector is not identically zero. For example, light cones are null surfaces. The event horizon is a null surface because light rays normal to the surface have a velocity that is zero at the surface. See the link in my article for the proof that the velocity of a radial light ray is zero at the event horizon. – John Rennie Jan 24 '13 at 10:55

Before you reach the horizon region you can find an interesting region, called photon sphere that will be full of unstable photon orbits, that spire down the black hole or escape to infinity.

On the other hand, to an in-falling observerat black hole, nothing special happens at the horizon. At least classically! The same is true for light. This is a result of equivalence principle. In particular, an in-falling observer is unable to determine exactly when he crosses the event horizon, as it is impossible to determine the location of the boundary from local observations. The only reason it is special is that it is a causal boundary i.e. once you cross it there is no way to go back.

In the quantum case, this question is subject to a recent heated debate, to what exactly happens at the horizon. This debate started with the information loss paradox. One of the logical possibilites, could be that our current local description fails near the horizon and we are sensitive to non-local quantum gravity. For example look at Black Holes: Complementarity or Firewalls?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.