# Black hole complementarity

Black hole complementarity states that two observers, one falling into a black hole, and one observing outside, experience two different histories but since they can not communicate there is no inherent contradiction..imagine an observer far from a black hole watching some matter fall into a black hole. The observer sees the matter freeze at the horizon, but the matter itself does fall in. Now, if you measure the mass of the black hole before and after the matter falls in, would you not measure a higher mass? And wouldn't this violate the principal of complementarity? You don't see the matter fall in, yet you see the black holes mass increase. Help?

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I don't think your description of black hole complementarity is correct. Black hole complementarity is a phenomenon in quantum gravity, but your description is purely classical. What you're asking about is a FAQ about the classical theory of black holes, but it has nothing to do with black hole complementarity. –  Ben Crowell Nov 13 '11 at 23:00
The black home complementarity is nontrivial even in the classical limit. The question makes sense as it is. –  Ron Maimon Dec 15 '11 at 20:31

I'm pretty sure that the general relativistic equivalent of this Newtonian answer is qualitatively the same (I'd love to hear someone expand it) but:

for a spherically symmetric system like a black hole, it wouldn't matter whether all the mass forms a shell on the horizon or whether it's formed some other spherical (possibly singular) distribution at the centre, the Shell Theorem will imply that you wouldn't be able to tell the difference by making any local measurements, like say, measuring the acceleration due to gravity to find the $M$ in $GM/r^2$.

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Of course, this answer does not explain what happens if mass is dumped into the black hole asymmetrically, which in the case of a single object dropped into the black hole would be pertinent. –  Niel de Beaudrap Oct 16 '11 at 11:25
The asymmetric mass, from the point of view of the exterior observer, smears itself symmetrically on the surface when it gets close. You can think of this as each string-theory string making up the dumped object getting longer by being boosted to insane speeds, and wrapping around the black hole many times. –  Ron Maimon Oct 16 '11 at 11:41
Yes, I assumed the OP was aware of Ron Maimon's point. It is explained nicely in Susskind and Lindesay's very nice book. –  dbrane Oct 17 '11 at 2:33