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If I understand correctly, the Hawking radiation hypothesis predicts that if a pair of virtual particles appears very close to the event horizon of a black hole, it may happen that one of them falls into the black hole and the other escapes, becoming a real particle (which diminishes the mass of the black hole, perhaps through some kind of quantum tunneling effect).

Having a rough notion of the geometry of space close to the event horizon of a black hole, I'm assuming that for this to happen one of the virtual particles has to go straight "down" and the other straight "up" (or else it would just arc back and also fall into the hole). Straight up is the only possible direction a photon can take to escape the extreme vicinity of the event horizon.

I have been thinking: What is the "color" (ie. the wavelength) of these particles?

If I understand correctly, if an object falls towards a black hole, from an external point of view it will look like the object red-shifts more and more until it becomes completely black (ie. zero frequency) at the event horizon. In other words, light that this object emits gets more and more red-shifted as it approaches the horizon.

Thus, I deduce, Hawking radiation would be very extremely red-shifted, to almost zero frequency (perhaps an infinitesimally small frequency above zero, given that the virtual particles probably appeared infinitesimally close to the event horizon).

If this deduction is correct, it would mean that black holes emit "black light" (ie "light" that has an almost zero frequency).

Is this even close to correct?

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# What's the color of Hawking radiation?

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