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[EDITED] I'm a postdoc working in cond-mat/quant theory, and I've heard some explanations of Hawking radiation that strike me as inconsistent or silly (e.g., in terms of pair production). I'm hoping someone can confirm that my concerns below don't apply to the rigorous derivation of Hawking radiation. I'm also fine with QFT / technical details, and I've obviously read Wiki lol.

Apologies for the lengthy question, and a huge thank you to anyone who takes the time to respond!

I have been assured since posting my original question that the pair-production picture is cartoony and I need not worry about it. Thanks for that, and I'll skip ahead to other derivations.

To add to the original version, my real question pertains to why should we expect to see Hawking radiation? Basically, if the derivation comes from the Unruh effect, then I would like to know why the experience of an accelerating observer is relevant to our experience. After all, aren't we inertial with respect to any black hole (i.e., our solar system follows its geodesic)? Or in Hawking's derivation, in what sense does what happens in two extreme stationary limits have any bearing on what we experience at intermediate times? Note that I could easily be missing details about the intricacies of doing QFT on curved space time.

I'm told that one explanation derives from the Unruh effect. As I recall, what an inertial observer perceives as the QFT vacuum will appear otherwise to an accelerating observer, who will observe lots of excitations (perhaps of a different QFT). While the vacuum state is certainly in thermal equilibrium at $T=0$, I don't see why the noninertial observer perceives the system as being in thermal equilibrium at temperature $T>0$ (as opposed to out of equilibrium)? Additionally, assuming thermal equilibrium, I am curious as to the precise argument that the noninertial observer perceives "radiation"?

More importantly still, I fail to see what this has to do with inertial physics. We learn very early on that the physics of inertial frames need not apply in noninertial frames (and vice versa). Clearly, an inertial observer (who perceives the $T=0$ QFT vacuum) cannot perceive the Unruh radiation experienced by a noninertial observer (who experiences a $T>0$ thermal bath). Moreover, I have seen several papers disputing the existence of Unruh radiation. So it is not clear to me where the Unruh effect should enter into the derivation of Hawking radiation.

Basically, I'm confused as to how the required noninertial frame comes into play. After all, while "free fall" under gravity appears to us as acceleration, in GR, this results from how matter modifies the embedding of 3+1D spacetime in itself. There is a choice of frame in which "free fall" is simply inertial, geodesic motion, no? I'm going to ignore the fact that gravity often leads to rotating orbits...I don't know if this is important or not, but will assume it is inconsequential for the simplest cases.

Naïvely I would think our solar system is in an inertial frame, following geodesics in the same way an object in free fall does. To me this seems to imply that, from the perspective of GR, any observer accelerating under gravity alone is in the same inertial frame as all others, and should agree on physics. I also assume that what we identify as the QFT vacuum is defined in this same frame (where we live and eat). So I don't see how the Unruh effect applies.

That is, unless one identifies an accelerating observer—e.g., one falling toward the event horizon who accelerates away to forestall their own demise. However, that observer would cease to be in the same inertial frame as the rest of us (and where the QFT vacuum lives). While they would experience the Unruh effect, the physics they observe would not be relevant our own perception. Any "test particle" we use to probe physics near the event horizon should be in an inertial frame relative us, right? Basically, no conclusion of an accelerating observer has bearing on inertial physics.

Hopefully I've clearly explained my own confusion; if not, I'm happy to clarify!

[UPDATE] Thanks to your answers I am now aware of a much better derivation, which is on much stronger footing and makes physical sense. Thanks also for the references, I plan to read through these!

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    $\begingroup$ See the PBS SpaceTime video Hawking Radiation $\endgroup$
    – mmesser314
    Nov 29, 2022 at 3:36
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    $\begingroup$ There is a rather formal approach which involves — I think — analytically continuing the black hole metric to imaginary time and observing that it is periodic. In quantum field theory at finite temperature, temperature is “equivalent” to periodic imaginary time, because $\exp(iH\Delta t)$ becomes $\exp(-\beta H)$. Here is an approachable article about this. $\endgroup$
    – Ghoster
    Nov 29, 2022 at 4:28
  • $\begingroup$ My impression is that this is the fastest, “deepest”, and most elegant way to derive the Hawking temperature, and as far as I know it works correctly in all cases. I am by no means an expert in this area, however, and don’t want to write an answer because I might get something wrong. $\endgroup$
    – Ghoster
    Nov 29, 2022 at 4:30
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    $\begingroup$ Did you actually read Hawking's paper? $\endgroup$
    – Avantgarde
    Nov 29, 2022 at 5:42
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    $\begingroup$ @justaphase Here is Hawking's paper: projecteuclid.org/journals/…. I strongly recommend thoroughly reading this. $\endgroup$
    – Avantgarde
    Nov 30, 2022 at 9:05

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You seem to be looking for an answer that explains the main ideas behind the formal computations, without actually going through them (otherwise, the answer would be just as complex as any review of Hawking radiation). Hence, I'll try to explain the main ideas.

First and foremost, pair production is not a derivation of the Hawking effect. It is, at most, a pictorial explanation to make the effect more intuitive. Nevertheless, it is extremely misleading.

The core idea behind the Hawking, and Unruh, effects is that the notion of particle depends on the observer. "Particle" is not a fundamental concept in Physics, despite being useful in many situations. This is not a problem from a theoretical perspective, because we describe physics in terms of a quantum theory of fields, not of particles. Furthermore, it is not incompatible between different observers. It is not different than saying that what an observer calls an electric force might have a magnetic component for another observer. The physical result is the same, but the explanation for why it is so is very different. Similarly, two observers might see a particle detector going off, but disagree on whether it is because it detected a particle or because it interacted with the vacuum.

That being said, Hawking's original derivation of his effect consists in the following setup: you have a collapsing star that, in the distant past, was stationary, but it eventually collapses to a black hole that, in the distant future, becomes stationary. Being stationary means there is a time-translation symmetry in the spacetime, so from Noether's theorem we can get a notion of energy in the far past and in the far future. Since the symmetry doesn't hold at finite times, these notions of energy might (and usually do) differ. Nevertheless, if we have a notion of energy, we can get a notion of particle: the vacuum is the state of minimum energy and states with more energy quanta have particles.

Notice, however, that our notion of particle in the far past and in the far future might be different. This is because our notion of energy is different. This isn't usually an issue in QFT in flat spacetime because "positive energy" means the same thing for all inertial observers, but in this context it doesn't need to. What Hawking then did was to consider a wavepacket at the far future (i.e., a "particle" as judged by a stationary observer at the far future) and then propagate it backwards in time until the far past, where he could decompose this "particle" in terms of the notion of particle of an observer in the far past. It turns out these two notions are not equivalent. Hence, if you are a stationary observer in the far past and see no particles, after the star collapses you'll eventually start to see particles in the spacetime. The reason is because the very notion of what is a particle has changed. If you compute the particle distribution you see at late times, you find out it is thermal at the Hawking temperature.

I should mention that the relation between particles at the far past and at the far future is given by a Bogoliubov transformation.

Notice this analysis considers what is happening very far away from the black hole. At no point do we talk about particles near the horizon: in fact, it is not clear which time-translation symmetry we could use over there to even define what we mean by particle, so the word doesn't even make sense. The pair production analogy was introduced by Hawking because you do find that, in the final state, you get particle modes at the far future entangled with modes inside the black hole. In this sense, you can interpret this entanglement as if it was due to pair production near the horizon. Notice, however, that this is just pictorial.

While this is Hawking's original derivation, there are others. As people mentioned in the comments, you can analytically continue Schwarzschild spacetime to imaginary time and find that you end up with a thermal state at the Hawking temperature. You can try to understand Hawking radiation in terms of the Unruh effect, although I dislike this approach because I feel like it is ignoring the field's quantum state (which is relevant to determine, for example, whether there is radiation incoming from infinity as well). There is also a more rigorous derivation in the context of algebraic quantum field theory (I believe it is due to Fredenhagen and Haag).

In case it interests you, I have also written this answer concerning the general idea behind how quantum field theory in curved spacetime works and this answer giving a more "scicomm-like" discussion of how the Hawking effect works. The latter is similar to a watered-down version of this answer.

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  • $\begingroup$ Thanks for the careful answer! It seems like an accelerating observer / noninertial frame isn't required to formulate Hawking radiation? I definitely find that reassuring. BTW, I'm fine with QFT and technical details, I just don't know all the HEP/gravity jargon. It seems to me the steady states are extreme limits, and everything we actually observe corresponds to intermediate times? Is this not a problem for the argument for Hawking radiation based on a wave packet? I'll follow up with more questions shortly. $\endgroup$ Nov 29, 2022 at 19:37
  • $\begingroup$ Some technical questions: (1) Does Bogoliubov transformation mean the same thing as in cond-mat (i.e., a rotation between creation/annihilation operators / $\Psi$ and $\Psi^{\dagger}$)? (2) How do you know how to evolve in time between the distant past and distant future? What generates dynamics? (3) Presumably the times relevant to our observation are intermediate? (4) Why do you assume the distant-past observer sees no particles? (5) In what sense can we be sure that the particles were emitted by the collapsed star (which is already a black hole? Sorry if these are naïve. Thanks! $\endgroup$ Nov 29, 2022 at 19:53
  • $\begingroup$ @justaphase "n. It seems to me the steady states are extreme limits, and everything we actually observe corresponds to intermediate times? Is this not a problem for the argument for Hawking radiation based on a wave packet?" typically, one starts talking about QFT by defininig "in" and "out" states for your QFT, which amount to a choice of vacuum. Initial conditions are usually set on the "in" states, and observables are equated with "out" states. Hawkings analysis, shown in general terms, above, basically says that the "out" vacuum has to be different than the "in" vacuum $\endgroup$ Nov 29, 2022 at 19:59
  • $\begingroup$ just a quibble: I believe that it is possible to derive this by doing a similar sort of computation, but comparing asymptotic future infinity to the schwarzschild horizon, using the fact that the outgoing null vector is a killing vector of the horizon, which removes the need to appeal to collapse, and makes the importance of the horizon manifest. it's been a very long time, but I think this is done in that Wald "QFT in curved spacetime" booklet $\endgroup$ Nov 29, 2022 at 20:10
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    $\begingroup$ @justaphase QFTs are generally computed perturbatively, with the base states being the exactly solved "free" theories, whose solutions are just superposed plane waves, and the fields are considered non-overlapping enough that the interaction terms are all zero. All interactions and interesting physics is hidden in an "S matrix" that maps "in" states to "out" states, so those spaces can be considered to be the (related to) the upper and lower limits of the path integral, ultimately, though computations are nearly always done in momentum domain $\endgroup$ Nov 29, 2022 at 21:38

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