alt Hacker NewsNot sure I follow your comment exactly, perhaps I am misunderstanding it but Hawking radiation doesn't involve anything escaping a black hole.
The radiation never originates from within the event horizon. Furthermore the radiation itself does not encode any information about what fell into the black hole apart from the electric charge, angular momentum, and mass (all of which must remain conserved).
Replies
The only caveat that I'd ask about is the death of a black hole after it evaporates away too much mass. If that explosion does occur, is that matter/energy that was from inside the singularity itself?
Indeed All Hawking radiation takes from a black hole is some mass. It's a bit like radiation resulting from annihilation of matter and antimatter: it does not "escape", it "gets born" in the process.
I don't know the slightest about this but... That list looks like it is supposed to be limited but that sounds like infinitely more information than zero?
The entire blackhole will eventually evaporate, so yes it is literally escaping from the black hole.
It's been ages since I read it and I am 99% sure I'm remember horribly wrong, but I was thinking it's something like this? Also - your "the only information" is precisely what we can measure of quantum states, so information is still preserved. That's more a limit of what we know?
I think that with Hawking radiation you have quantum entanglement of two pairs. You have a matter-antimatter pair production near the horizon. When the antimatter falls into the blackhole without being destroyed via mutual annihilation, the matter particle at the boundary now "radiates" away. HOWEVER.
The matter inside boundary which the anti-matter DOES interact with is destroyed, and the information of that interaction is is "passed" to the outside via the the anti-matter's entanglement with its co-created pair.
Something like that, isn't it? Please any physicists clarify :)