In discussions of Black hole information paradox, people usually argue that the falling observer shall not see radiation at the horizon or near the horizon due to the equivalence principle. That is, for a falling observer, what around him is just vacuum in a locally Minkowski spacetime. This is because, for a falling observer, the Christoffel connection vanishes. The problem is, if the equivalence principle holds, the asymptotic static observer at infinity also has the same status of a "free-falling" observer. Because, spacetime is asymptotically flat and the static observer is a free-moving observer in locally Minkowski spacetime. Then why shall the static observer at infinity observe Hawing radiation?


Consider a static observer far away from the black hole (but not at infinity) and the usual free falling observer crossing the horizon. The latter follows a geodesic, while the former must accelerate to not fall in.

In analogy to the Unruh effect, the accelerated observer will see particles. All of this can be made less naive by computing the Bogoliubov transformations relating the regular vacuum at the horizon with the singular one in schwarzschild coordinates.

Now, since the observer far away from the hole will see a flux of radiation, the one at infinity will see it too.

  • $\begingroup$ Thank you for your answer. But I wonder, at infinity, the static observer coincide with a free falling observer up to a boost, right? $\endgroup$ – Wein Eld Oct 15 '17 at 11:33
  • $\begingroup$ For the Unruh effect, the accelerated observer at infinity approaches to an inertial observer which should be the fastest way to see there is no radiation at infinity. Now the static observer at infinity in Schwartzschild spacetime also approaches to an inertial observer, why should he see radiation? $\endgroup$ – Wein Eld Oct 15 '17 at 11:38
  • $\begingroup$ No, a static observer doesn't change his spatial position, while a free falling observer falls toward the center of the hole. Regarding the radiation, infinity is a tricky concept, so as I answered it's better to think at a closer observer, and then conclude that since the radiation is there, it must eventually go to infinity. $\endgroup$ – Rexcirus Oct 15 '17 at 17:20
  • $\begingroup$ This is not satisfactory. In the contex of Unruh effect, the "static obsever" (there we call the accelerated obsever) at infinity can not observe radiation since its acceleration is essentially zero. I know the difference of Schwarzschild spacetime and Rindler spacetime. But once people use equivalence principle to argue falling obsever see nothing, they assume Hawking radiation is a local effect. Then the static observer at infinity has no acceleration and should see nothing too. $\endgroup$ – Wein Eld Oct 15 '17 at 19:58
  • $\begingroup$ To be honest, I've never seen an author making the claim "not see radiation at the horizon or near the horizon due to the equivalence principle", and I think that's were confusion arise. People usually say that the horizon is no special place since an observer there would not experience anything singular, like the naive (t,r) coordinates would suggest. $\endgroup$ – Rexcirus Oct 15 '17 at 22:42

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