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I recently read about the Hawking effect and I noticed that formally, in its general lines, the derivation is very similar to that of the Unruh eff. In the latter case we write the Fock states in a Rindler metric and find a relationship with the Fock states in a Minkoski metric, finding the famous result. Since a Rindler space is essentially the region of space-time that can be explored by a uniformly accelerated body, the interpretation of the effect becomes quite intuitive: if an inertial observer is immersed in the void, an accelerated observer will believe instead of being immersed in a thermal bath. In the case of Hawking, what one does is to find the Fock states in a Schwarschild metric and compare them with those defined in a minkoskian metric, so it seems to me that the result is a bit different: whereas before the inertial and Rindler observers can also be in the same place, now we should have an observer away from any gravitational field and one near a black hole; moreover, while the latter would see the black hole radiate as if it were a black body at Hawking's temperature, Minkoski's observer would still be immersed in the void and therefore, according to him, the evaporation of black holes does not exist.

I'm pretty sure there is something wrong with my reasoning but I cannot say what, would anyone like to give me some elucidation?

Thank you in advance

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  • $\begingroup$ I don't know enough about this to risk an answer, but as I understand it the reason we see Hawking radiation at infinity is because of the horizon. A massive body that is not a black hole does not produce Hawking radiation. That's the difference between the Hawking and Unruh calculation. $\endgroup$ – John Rennie Nov 6 '19 at 12:19
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    $\begingroup$ Just a note, the usual policy is to restrict to one question only, so I only tried to answer the main body. If you'd like feel free to open a separate question regarding cosmic creation of particles in RW metric. $\endgroup$ – cesaruliana Nov 7 '19 at 1:14
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You're certainly on the right track. I'll review the situations as you stated, so that the comparison is clear.

1) In Minkowski spacetime we are interested in comparing two families of observers, the inertial ones and the uniformly accelerated ones. As you clearly stated what the inertial observers consider to be vacuum, i.e. absence of particles, the accelerated ones measure as a thermal bath, i.e. a thermal distribution of particles.

2) In Schwarzschild spacetime it is exactly the same. What inertial observers measure as vacuum the uniformly accelerated ones identify as thermal bath. The difference here is what constitutes inertial and accelerated motion. As per usual reasoning, in general relativity inertial movement is free-falling movement in the gravitational field. Therefore in Schwarzschild spacetimes the family of inertial observers are the ones falling towards the black hole. The uniformly accelerated observers here are the ones who manage to keep a constant distance from the event horizon, the acceleration being there to counter the gravitational pull.

What seems to be the source of confusion is that you are comparing one observer near the black hole with one far away. If both are in stationary orbit then both are accelerating and would see thermal bath, the only difference is that the one near the event horizon needs to accelerate more, and therefore sees a higher temperature than the one far away, just as in Minkowski spacetime. So what you are calling Minkowski observers in Schwarzschild, the inertial observers, should be the ones in free fall. Those really do not see radiation. You can check that the comparison holds because accelerated observers in Minkowski and stationary observers in Schwarzschild are confined to outside the horizon, while inertial observers in both cases have access to the whole spacetime.

3) As a side note, the previous case is not really the Hawking radiation, and in the literature it is usually called Unruh effect in Schwarzschild spacetime. What is called Hawking radiation is slightly different. What Hawking did was to consider an initial time where space is essentially Minkowski, but with some dust scattered. After enough time all the dust collapses into a black hole. What he found out was that one unique family of observers, who manage to stay outside the horizon, will eventually see thermal emission from the black hole.

In essence, the Unruh effect is about a single stationary spacetime and you are comparing two different family of observers. The Hawking radiation is about an evolving spacetime and what a single family of observers measure. The result are indeed connected, but different. In the stationary Schwarzschild spacetime we are in a equilibrium state, the stationary observers in orbit see radiation being emitted from the horizon, but also radiation going in. For Hawking's case the spacetime is evolving and the state is not in equilibrium. Those observers far away see the black hole emitting radiation but none going in (which should be the case, as there was no radiation outside to begin with).

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  • $\begingroup$ Thank you, but I have still some doubt. The fact that the Unruh effect on Schwarzschild spacetime reproduce the same result of Hawking's collapse, was it to be expected or is it just a coincidence? Moreover, I known for sure that the event horizon is necessary for the effect, but I haven't understood why. I opened a thread about this last question here: physics.stackexchange.com/questions/514920/…. I hope you can still help me. $\endgroup$ – silviozzo Nov 19 '19 at 18:47
  • $\begingroup$ inertial observer can remain in a semi-stable orbit around the BH without falling. Do orbiting inertial observers see a heat bath or vacuum? $\endgroup$ – lurscher Nov 19 '19 at 19:25
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    $\begingroup$ @silviozzo, regarding the similarity between the Unruh effect and Hawking radiation, I do think it is more of a coincidence than to be expected, since they both have horizons and thermal bath, but a horizon does not imply thermality. $\endgroup$ – cesaruliana Nov 26 '19 at 16:39
  • $\begingroup$ @lurscher, I apologize but I'm not familiar with inertial orbits that remain semi-stably outside the horizon in Schwarzschild. Could you point me some reference? What I can say in Schwarzschild is that any family of observers not in free-fall nor in stable orbit would observe particles but not in a thermal spectrum. In essence the distribution of particles observed is dependent on the trajectory of observers. $\endgroup$ – cesaruliana Nov 26 '19 at 16:44
  • $\begingroup$ well, perfectly stable orbits do not exist, since energy is emitted through gravitational waves, but the farther the orbit is from the EH, the more stable becomes. Those orbits are locally inertial frames $\endgroup$ – lurscher Nov 27 '19 at 13:48
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whereas before the inertial and Rindler observers can also be in the same place, now we should have an observer away from any gravitational field and one near a black hole; moreover, while the latter would see the black hole radiate as if it were a black body at Hawking's temperature, Minkoski's observer would still be immersed in the void and therefore, according to him, the evaporation of black holes does not exist.

This not really true, you actually have two Minkowskian observers, one in the infinite past and one in the inifinite future. What is a vacuum state (no particles) for the minkowskian observer in the past, is a thermal state for the observer in the infinite future, this observer would see the thermal radiation coming from the black hole. You don't have any observer near the black hole, there you have curved spacetime and the notion of particle in the proximity of the horizon is not well defined, that's why you choose as your second observer a Minkowskian observer in the infinite future. The point I want to make clear is the following: Hawking radiation is something an observer in the infinite future, far from the black hole sees.

Not only the observer can be in the same place, they can actually be the same observer: Suppose the only existing things are you and a collapsing star, at a certain moment, say $t=-\infty$, you are far from the star in what could be considered Minkowski spacetime, furthermore suppose you are in the vacuum that is you can't see any particles. If you wait long enough, say at at $t=+\infty$, after the star has collapsed and formed a black hole and therefore an event horizon, you will see that you are not in vacuum anymore, but you see thermal radiation coming from the BlackHole, so in this case the two observer can be in the same place (space coordinates), just not at the same time.

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  • $\begingroup$ Thanks to you too, but I have still some doubt. The fact that the Unruh effect on Schwarzschild spacetime reproduce the same result of Hawking's collapse, was it to be expected or is it just a coincidence? Moreover, I known for sure that the event horizon is necessary for the effect, but I haven't understood why. I opened a thread about this last question here: physics.stackexchange.com/questions/514920/…. I hope you can still help me. $\endgroup$ – silviozzo Nov 20 '19 at 11:51

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