Hawking Radiation without a horizon?

Question

I’m reading this article for a straightforward derivation of the Hawking effect https://www.researchgate.net/publication/356129307_A_Pedagogical_Review_of_Black_Holes_Hawking_Radiation_and_the_Information_Paradox

The derivation looks at the quantum field in a Schwarzschild spacetime. Using the tortoise coordinates, it expands the field in a mode expansion. It then compares this to the mode expansion using Kruskal coordinates. This yields two different vacua $|0_{T}\rangle$ and $|0_{K}\rangle$. Writing out the tortoise creation/annihilating operators in terms of the Kruskal coordinates, we arrive at the bogoliubov transformation and taking $|\beta|^2$ yields the average number of particles the tortoise observer will see. My problem with this derivation is that the Schwarzschild spacetime is valid for any static spherical body. It is not excluded to just a black hole. And therefore, the entire derivation would apply to all spherical bodies. But of course, spherical bodies don’t have a thermal temperature for the mean number of particles, so what am I missing?

Answer 1

You are correct: there is no Hawking radiation for a general spherical body. The reason is subtle, although the reference you gave comes near it on p. 17: the field's state.

In the reference you provided, the author decides that the physical choice of vacuum is $|0_K\rangle$ because $|0_T\rangle$ has an unphysical build up of modes near the event horizon. This is not seem, so $|0_T\rangle$ is not physical and must be abandoned.

However, suppose that we have a planet instead of a black hole. Then we have two remarks:

1. $|0_T\rangle$ doesn't have an unphysical build up of energy near the horizon, for the horizon literally does not exist (the solution near the "would-be-horizon" is an interior solution, not Schwarzschild)
2. $|0_K\rangle$ has unphysical modes that go into the horizon, which literally does not exist. Hence $|0_K\rangle$ is not an acceptable state anymore.

This is the point in which the derivation fails. The physical states in a black hole spacetime and in a planet/star spacetime are different. If the state is different, the behavior is different, and hence we end up with no thermal effects.

To give a few more nomenclature, here are three of the main states considered in Schwarzschild spacetime:

1. Hartle–Hawking vacuum: this is the vacuum associated with the Unruh effect in curved spacetime. This vacuum is compatible with the symmetries of Schwarzschild spacetime and it has no unphysical build ups of energy. It predicts a static observer hovering above the black hole will see particles coming from the black hole and from infinity with a thermal spectrum.
2. Unruh vacuum: this is the vacuum associated with the Hawking effect (yes, the Unruh vacuum is associated with the Hawking effect, and the Hartle–Hawking vacuum with the Unruh effect). This vacuum is compatible with the symmetries of Schwarzschild spacetime and it an unphysical build up of energy near the white hole horizon. This is not an issue if you consider a black hole that formed out of a stellar collapse, in which case the white hole itself is unphysical. It predicts a static observer hovering above the black hole will see particles coming from the black hole but not from infinity with a thermal spectrum.
3. Boulware vacuum: this is the vacuum in which static observers hovering at a constant distance don't see any particles. It has unphysical build ups of energy near the event horizon (I believe both in the black and white holes, if my memory is not failing me). This, of course, is not an issue if the spacetime has a planet or star that renders the horizons unphysical. This vacuum does not predict a static observer will see particles with a thermal spectrum coming from either the spherical object or from infinity.

Notice that the Unruh effect and the Hawking effect are physically different phenomena (as pointed out in Wald's QFTCS book). It is very common, however, to derive the Unruh effect in Schwarzschild spacetime and call it the "Hawking effect", although Hawking's original derivation was concerned with a completely different field state and hence has different predictions concerning what is observed.

For more information, I suggest checking out Chap. 5 of Wald's book. Everything I said in this answer is probably there somewhere.

Answer 2

You might want to look up the Unruh effect instead. They're basically the same but the Unruh effect/temperature is usually used in the context of a free-falling observer in Minkowski spacetime vs an accelerating observer in Minkowski spacetime. Because of their acceleration through spacetime, they only have access to a portion of Minkowski called the Rindler wedge. This Rindler wedge is bounded by a future and past horizon. Of course, these horizons are not present for the free-falling observer.

The main cause of this effect that a free-falling observer and an accelerating one disagree on the definition of a particle, and hence on their vacuum. A Minkowski observer doesn't see particles, a Rindler one sees the Minkowski vacuum as a thermal bath at the Unruh temperature. The converse is also true, the Rindler vacuum is empty according to a Rindler observer but contains particles with respect to a Minkowski observer.

The Hawking effect is basically the Unruh effect in which the acceleration is provided by the Schwarzschild metric. So in theory this is true for all spherical bodies, but the temperature is so low because of the weak acceleration that you won't notice it.

The temperature is a statement about the particle bath, not about the spherical body.