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It is often said that the event horizon of a Schwarzschild black hole is lightlike. Is this correct and, if so, what exactly does this mean?

Intuitively, this may mean that any two points on the horizon surface are lightlike separated, but I am unable to see this from the metric. The temporal part of the Schwarzschild metric is zero at the horizon. The radial part approaches zero in the limit. However, the angular part remains well defined positive thus apparently making the interval spacelike.

What am I missing? I’ve searched the web, but couldn’t find anything relevant. I would appreciate if someone points me in the right direction.

safesphere
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2 Answers2

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It is often said that the event horizon of a Schwarzschild black hole is lightlike. Is this correct and, if so, what exactly does this mean?

It means that only lightlike geodesics can stay stationary at the horizon, all other geodesics must fall in (or escape, assuming there are space-like geodesics as well) and the proper time intervall they can spend at the horizon is infinitesimal. If an observer on a timelike path emitts a radially outwards directed photon when he crosses the horizon, it will stay there forever. So the only stationary probes you can have at the horizon are light-like (and inside space-like, which means that only Tachyons, if they existed, could remain stationary inside the horizon). Outside of the horizon the stationary (with respect to the black hole) probes (called Fidos for fiducional observers) are on time-like paths.

Yukterez
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Lightlike simply means that your interval $ds^2=dx^2+dy^2+dz^2-c^2dt^2$ evaluates to zero.

It does not mean that real black hole would have light encircled forever in eternal loop over its event horizon. In fact, any non-zero energy object cannot do that. Real particles would lose energy because they would have accelerated orbit and gradually fall down. Even virtual particles are split and sucked under event horizon (BH is a heating engine).

sanaris
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