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How can Hawking radiation with a finite (greather than zero) temperature come from the event horizon of a black hole? A redshifted thermal radiation still has Planck spectrum but with the lower temperature (remember CMB with temperature redshifted by expansion of the universe). Now, redshift at the event horizon is infinite (time is frozen for a distant observer) so the temperature of the radiation would be zero for him/her, that is no radiation is detected

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The radiation comes from the region just outside the horizon. If you search the site I'm sure something like this has been asked before. –  John Rennie May 15 '13 at 17:53
    
related: physics.stackexchange.com/q/22498/4552 –  Ben Crowell May 15 '13 at 19:01
    
@JohnRennie I see - so the radiation comes from the just outside the horizon and the redshift is quite large but finite, reducing the temperature to low values observed but still greater than zero. –  Leos Ondra May 16 '13 at 6:53
    
It's a bit more complicated than that because the radiation isn't really emitted in the sense that a hot filament emits EM. A distant observer calculates the QFT vacuum to be different to an observer nearer the event horizon, and as a result the distant observer finds the "vacuum" near the horizon to contain a finite particle density - this constitites the Hawking radiation. I'm pretty certain (but wouldn't swear to it) that the Hawking calculation calculates the temperature at infinity. An observer falling freely into the black hole would not see any Hawking radiation even at the horizon. –  John Rennie May 16 '13 at 7:46
    
physics.stackexchange.com/questions/30597/… is related, though not a duplicate. I'm not sure I understand this area well enough to hazard an answer rather than just this comment :-) –  John Rennie May 16 '13 at 7:49
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