Except from the mathematical derivation, is there a physical reason of why do we call the event horizon of a black hole a null surface? Is the concept of event horizon being a null surface any how similar to a light cone being a null surface?
2 Answers
To precisely answer the question you have to understand the concept of future null infinity (sometimes called "scri-plus"). In flat spacetime, future null infinity is the place where all future-directed light rays end up asymptotically. In an asymptotically flat spacetime, future null infinity is similarly where light rays end up, at least if they are launched sufficiently near the flat region. But it could be that light launched further in just does not make it out to future null infinity. In this case we say there is a black hole.
We can imagine starting near the flat region and trying to send light to future null infinity. We move further and further away from the flat region, still sending out light. At some point the light stops making it out to future null infinity. The very first light ray that does not make it out is part of the boundary of the black hole. The collection of all such light rays -- those that barely don't make it out -- defines a null surface. (The null tangents to these light rays also serve as normal vectors to this surface -- null surfaces are counterintuitive.) This null surface is called the event horizon.
A light cone in flat spacetime is also a null surface, generated by rays emanating from a single point. But it is not an event horizon because the light does make it out to future null infinity. In fact, the light cone is a classic example to help understand why the event horizon is defined the way it is, with reference to future null infinity. If you just define a black hole as a region that nothing can escape from, then the interior of the light cone is a black hole! Obviously this is not what we want.
See Wald's GR book for a technical explanation of these concepts.
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$\begingroup$ Can you recommend me any book that has questions and nice explanation on solving geodesic paths in space-time as i am pretty weak in that? $\endgroup$ Jul 20, 2017 at 3:25
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$\begingroup$ Carroll --> Wald is the right progression for the graduate level IMO. For undergraduate level, I'm not too familiar with the books, but Schutz and Hartle seem to be popular. $\endgroup$ Jul 20, 2017 at 16:17
By definition, the event horizon of a black hole is the shell such that it is impossible for light inside the shell to escape. That is, events inside the event horizon cannot affect an outside observer, since information cannot escape from the inside.
Compare the scenario described above to a light cone in flat, Minkowski spacetime. Events inside the light cone are spacelike-separated from events outside the light cone, so events inside cannot causally influence events outside.
I believe it is in this sense both the event horizon and the light cone are termed null surfaces, but I admit I could be wrong about some of the exact definitions or terminology used.
Edit: More precisely, a null surface is a hypersurface whose normal vector at every point is a null vector (has zero length with respect to the local metric tensor). From Wikipedia: https://en.wikipedia.org/wiki/Null_hypersurface So, I think the characterization above based on timelike and spacelike separation still holds, but if someone can be more mathematically precise than me, that'd be better.
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$\begingroup$ nicely told friend. I just have one more thing to ask. Is null surface a surface on which all events are light like separated? If so, this is true for light cone. Is it also true for event horizon? Also, if i shoot a photon on the event horizon in a tangential direction, how and where would the photon move? $\endgroup$ Jul 19, 2017 at 8:27
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$\begingroup$ Also, if one tries to send some info inside the black hole or influence something inside the event horizon, one can as one can send anything inside. here events seem to be timelike related. So events being space or time like separated also dependent on the way the events are covered? $\endgroup$ Jul 19, 2017 at 8:31
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$\begingroup$ Again from Wikipedia, it says that for each point on a null hypersurface, there is exactly one lightlike worldline containing that point, which is also entirely contained in the null hypersurface. So that means if you shoot off a photon in the event horizon, its spacetime trajectory would be confined to the event horizon, and also be uniquely determined. $\endgroup$ Jul 19, 2017 at 8:31