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Consider the following scenario, very similar to the one proposed in this question, but this time, the pure quantum radiation used for the black hole collapse, is now being split with down-converter crystals in pairs of entangled photons. So we have two groups of photons, but the overall state is still pure (it is described by an eigenvector, density matrix satisfies $\rho^2 = \rho$)

Brave and Courageous assertion No.1 So the overall quantum state has Von Neumann entropy zero or nearly zero (right? wrong? why?)

If i were to create a single black hole with the two groups of photons, by the argument explained in the previous question, the entropy of the black hole would be zero or nearly zero, and hence the black hole would be mysteriously extremal, with no measurable charge or angular momentum.

But if i take the two groups separately and have them collapse in individual black holes, it would seem to external observers that the Von Neumann entropy of the parts is non-zero, since the partial traces of a pure entangled state will be density matrices of mixed states.

Which leads me to my Brave and Courageous assertion No.2: So each black hole in the pair would seem to have a non-zero temperature to far-away observers, and they would look classical (right? wrong? why?)

Assuming both brave and courageous assertions hold true, the question i have is this:

If the two black holes are made to coalesce in a single black hole, what will be the thermodynamic behaviour of this merged black hole: Would it be a nearly-zero entropy black hole? or would it radiate as a classical black hole?

if the answer turns out to be: it radiates as a classical black hole, then what happened to the Von Neumann information that is inside the merged black hole and why the formation and merge of black holes changed it so much?

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