Does Hawking radiation lead to black hole evaporation / reduction of black hole mass?

Physics Asked by emacs drives me nuts on December 9, 2020

As I understand, Hawking radiation leads to black hole evaporation, resp. a black hole would lose mass due to that effect.

Now Hawking radiation is very similar to Unruh radiation, i.e. some (apparent) horizon leads to a thermal bath:

Unruh Radiation:

  1. An inertial observer in Minkowski space does not see radiation.
  2. A Rindler observer sees Unruh radiation.

Hawking Radiation:

  1. A free falling observer does not observe Hawking radiation from a black hole.
  2. An observer hovering somewhere over the event horizon of a black hole does see Hawking radiation.

Hence in either case, the inertial observer (1.) sees no radiation whilst the accelerated observer (2.) sees thermal radiation.

Of course, the case U.2 is stationary, i.e. for a Rindler observer the spacetime does not change and the Rindler horizon does not disappear, evaporate or change its distance due to Unruh radiation.

Doesn’t this also apply to H.2, i.e. there is just some thermal bath due to acceleration (or due to some horizon), and the black hole does not change in mass?

Moreover, if the black hole did evaporate due to Hawking radiation, wouldn’t that lead to conflicting observations from a free falling observer (black hole does not evaporate because no loss of energy / mass because no radiation is emitted) vs. hovering observer (black hole does evaporate because it loses mass / energy due to Hawking radiation)?

Hawking’s derivation predicts named radiation, but does that derivation also show that the black hole’s mass is changing?

2 Answers

Unruh and Hawking effects are similar but not quite the same. In particular, a freely falling observer far from the black hole does detect Hawking radiation. An inertial observer far from the black hole could be momentarily at rest relative to say Schwarzschild coordinates (for a simple black hole) and they would detect the radiation. They would still detect it as they begin to gently fall in the direction of the black hole. The black hole has interacted with the electromagnetic field and emitted photons for all to see, just like a star (well the process is not like a star, but the end result is, for observers far away).

The observer freely falling near to the black hole may or may not be able to detect the radiation, depending on the size of his detector and the length of time he has to register the effect. He hasn't got much time to look before he enters the horizon, so none of his measurements are going to be precise. In fact his energy measurements will be imprecise by an amount of the order of the Hawking temperature. And if he tries to measure for a time long enough to get that sort of precision, then he can't help but notice that spacetime is not flat. That is, his location will have moved through a distance of the order of the Schwarzschild radius. That is one way to see why such an observer's observations are not the same as those of an inertial observer in flat spacetime. (The equivalence principle only applies in the limit of small regions of spacetime of course). The other way of seeing it is to note that the electromagnetic field tensor in the Hawking case is not the same as in the Unruh case.

Answered by Andrew Steane on December 9, 2020

I am going to try to explain this without using any equations. I hope I don't make any mistakes along the way.

A freely falling observer "close" to the black hole horizon will not detect any Hawking radiation, like a non-accelerating observer in Minkowski space. This does not mean that any freely falling observer cannot detect Hawking radiation. If that was the case then even the observers which are at rest at spatial infinity (and hence freely falling to the gravity of the black hole) were not able to detect Hawking radiation. Even though freely falling observers which are close to the horizon will locally see it flat, if they are asked to extend their reach to feel the curvature of the horizon, they will feel the geodesic deviation as an acceleration. (This means that we cannot have a solid extended freely falling observer near the horizon.) Therefore one might say that if they try to look at the black hole as a whole then they will indeed see that it radiates.

To put in slightly different words, if near the horizon observer uses a detector vast enough to feel the curvature of the horizon, then that detector will feel acceleration. This means vast enough detectors can see the radiation. Also, the detector should be a non-accelerating object in the reference frame of the observer. Otherwise it will not work correctly, for example, in the case of flat space-time.

Answered by Alphy on December 9, 2020

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