What's the color of Hawking radiation?

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SUMMARY

The discussion centers on the characteristics of Hawking radiation, specifically its "color" or wavelength. It is established that Hawking radiation arises from virtual particle pairs near a black hole's event horizon, where one particle falls into the black hole, reducing its mass, while the other escapes. The escaping particle is severely red-shifted, leading to the conclusion that Hawking radiation emits "black light" with an almost zero frequency. Additionally, Hawking radiation is thermal and follows a Planck spectrum, resulting in an average temperature that is extremely low, making direct observation nearly impossible.

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What's the "color" of Hawking radiation?

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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I think you've got it about right except for the quantum tunneling effect, which I don't think has anything to do with Hawking radiation. For reasons I can no longer remember (I THINK it "got" it at one point) the particle that falls in always has negative energy, thus dissipates a tiny bit of mass of the BH, and the other one has positive energy and moves away from the BH, although I don't think it has to be absolutely vertical, just very close. And yes, it has to be severely red shifted as it leaves the BH.
 


It is said that Hawking radiation is thermal, so doesn't have a specific color but follows a Planck spectrum. All the same, it has a temperature, and that means it has an average "color." There is a formula for the temperature of Hawking radiation, but for any black hole that is thought to exist at present, that temperature is incredibly low. So low, in fact, that it should win some kind of award for most widely believed phenomenon that is most difficult to ever have any hope of directly observing!
 

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