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If an observer falls into a black hole then gradually he will encounter higher and higher energy Hawking radiation. Assuming that Hawking radiation is created just outside the horizon (say at $r=2M+ \varepsilon$, $\varepsilon$ infinitesimal) the temperature that the observer measures (wrt. his four velocity) will diverge at the horizon and at some point before entering the black hole he will be grilled. In the versions of the derivation of Hawking radiation known to me (a modern one I know can be found in arXiv:2012.05770) one assumes the horizon to be a smooth lightlike surface. However according to the line of thought before the horizon should be a very wild and non-smooth place, so the derivation of Hawking radiation would not be self consistent. In other words: The backreaction of Hawking radiation at the horizon would not be negligible.

Just to get a feeling for the scales: The temperature of a solar mass BH as measured at infinity is about $6*10^{-8}K$ and its radius is about $3\; km$. To measure a temperature of $1\;K$ one would need to get within a distance of $6*10^{-13}\;m$ to its horizon, so indeed very close.

What is wrong with this line of argumentation? Can one have a self consistent derivation of Hawking radiation within QFT on curved (but classical) spacetime or does one already need a theory of quantum gravity to properly describe the horizon?

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    $\begingroup$ A freely falling observer does not observe any Hawking radiation. The radiation is only observed by observers with a non-zero proper acceleration. What you describe would apply to an observer hovering at some fixed distance $r$. As $r \to r_s$ the radiation intensity diverges, but then the proper acceleration required to hover goes to infinity so by analogy with the Unruh effect we shouldn't be surprised that the intensity diverges. $\endgroup$ Commented Oct 7, 2021 at 17:29
  • $\begingroup$ I know that this happens for a uniformly accelerated observer in flat space (Unruh effect) who has no contact with degrees of freedom on the other side of a lightlike surface. And if I remember correctly one can expand the metric at the horizon and an observer staying at a fixed $r$ has a uniform acceleration. So the Hawking radiation that this observer feels can be explained in this coordinate system as Unruh radiation. But I still have difficulties with this. Locally an infalling observer and one at constant $r$ are related by a Lorentz transformation so if one sees particles the other... $\endgroup$ Commented Oct 7, 2021 at 19:01
  • $\begingroup$ ...should too? Do you have a good reference that treats these sort of topics thoroughly? Independently from observers and what they measure, is the backreaction of Hawking radiation negligible near the horizon? $\endgroup$ Commented Oct 7, 2021 at 19:02
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    $\begingroup$ @JonasMager The Hawking radiation relates to, bit is not equivalent to the Unruh radiation. The former is directional coming from the horizon. The latter is not directional, but a thermal bath experienced equally from all sides. $\endgroup$
    – safesphere
    Commented Oct 8, 2021 at 5:51
  • $\begingroup$ @JonasMager The flaw in your argument is using different coordinates while measuring the energy of the same Hawking particle. Energy is frame dependent, but when measured in the same frame of reference, energy is conserved in the static Schwarzschild spacetime. In the coordinates of the same remote observer, the total energy of a Hawking particle when emitted at the horizon is the same as its energy when absorbed by this observer. $\endgroup$
    – safesphere
    Commented Oct 8, 2021 at 6:01

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To answer your question quickly: there are problems with the derivation of the Hawking effect and it does involve assuming validity of results that could depend on quantum gravity, but it is not due to the reasons you pointed out and we have strong belief that the result holds nevertheless (see pp. 12–13 of arXiv: gr-qc/9912119 and references therein, DOI: 10.1038/s41586-019-1241-0 should also be mentioned). See also Wald's 1994 book Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics, especially pp. 162–163.

As mentioned in the comments, a freely falling observer does not observe Hawking radiation (see, e.g., Wald's book on p.171). The thermal effects are observer dependent and they assume explicitly the observer to be following the orbits of the static Killing field. By studying the problem from the point of view of such an observer, no issue should occur.

To further explain why this argument holds, notice that while the static and the freely falling observer are locally related by a Lorentz transformation, these effects are not local. Locally, we could also relate an inertial observer and an accelerated observer in Minkowski spacetime by a Lorentz transformation. However, one of them perceives the Minkowski vacuum as a freezing void, while the other one perceives it as a hot mess of particles. It is not sufficient to use the local structure to determine the quantum effects. This can also be seen, for example, from the fact that the derivation of the Unruh effect employs a bifurcate Killing horizon, which is a global notion.

To avoid misinterpretation of the previous paragraph, I should also recall, as was done in the comments, that Hawking radiation is not the same as the Unruh effect. The Unruh effect can be generalized to curved spacetimes, but in the case of a black hole, it does not correspond to the Hawking effect. As pointed out, the Unruh radiation comes from all around a static observer in the presence of an eternal black hole. Hawking radiation, on the other hand, is seen by an static observer as coming from the black hole, which is assumed to be formed by gravitational collapse, instead of being eternal. Both effects are very well discussed on Wald's book, and the fact that they are quite different is pointed out, for example, on p. 129.

As for an observer-independent argument for the backreaction, I believe you would agree with me that the backreaction effects would only be divergent if the expectation value of the stress energy tensor would diverge as well. However, the Hawking effect is derived in the Unruh vacuum, which is well-behaved (more specifically, it is a Hadamard state) near the (future) event horizon, and hence no such an issue occurs. This vacuum is characterized by the fact that static observers see particles coming from the black hole with a thermal distribution at the Hawking temperature and see no particles coming from infinity. If one analyses the problem in this observer-independent manner, no issue could possibly arise, and the problem would only happen in the frame of a static observer, where it is natural since, as pointed in the comments, one already needs infinite acceleration to stay there. More information on this can also be found at Wald's book.

As you might guess, I quite like Wald's book hahaha. I believe all of these issues are treated there in clear way, especially on Chap. 7, which concerns the Hawking effect. While it is not an easy read, it is definitely worth the effort.

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