Before quantum theory we knew accelerating electrons radiated electric fields. This is modelled classically (even though we know it is a quantum process emitting photons)

Similarly is there a possible classical theory of Hawking radiation? In this analogy the electron becomes the black hole radiating hawking radiation.

From a distance an astronomer may just observe a black hole as an object emitting classical electromagnetic waves and not feel the need to model this quantum mechanically.

The radiation coming out of a black hole would contribute to the stress energy tensor $T$ which is almost exclusively photons. Classically we could consider this just release of electromagnetic waves(?)

In GR we have:

$$G_{\mu\nu} =\kappa T_{\mu\nu}$$

which tells gravity how to evolve.

We have

$$\nabla_\mu T^{\mu\nu} = 0$$

which is the conservation of energy.

For any particular $T$ one could always solve it to find a $g$. But this $T$ might not correspond to any classically moving fields.

i.e. it may not satisfy:

$$\frac{\delta \int \sqrt{-g}T dx^4}{\delta \phi}=0$$

Of these 3 equations, which ones would we keep and which ones would we modify in order to model a semi-classical evaporating black hole. By semi-classical I mean modelling the radiation as smooth fields instead of particles (just as in the classical electron case).

Since we think of the electromagnetic radiation a "tunnelling" out of the black hole. Would this break one of the equations above?

Basically, if QM had not been discovered how would a classical physicists model an evaporating large black hole with classical field equations? (including the shrinking of the black hole as it evaporates)

Or perhaps this is not possibly classically due to the radiation being in a black-body spectrum which is only derivable using Plank's constant?

  • $\begingroup$ That variational equation doesn’t give the classical equation of motion for $\phi$. And what is $T$ supposed to be? The trace of $T^{\mu\nu}$? Whatever it is, it’s not the Lagrangian density. $\endgroup$
    – Ghoster
    Feb 17 at 7:08
  • $\begingroup$ For a massless scalar field? Are you sure? alves-nickolas.github.io/pdf/… $\endgroup$
    – user84158
    Feb 17 at 8:00
  • $\begingroup$ "if QM had not ... classical field equations?" - No. Its a consequence of Area theorem in classical GR. Quantum fields can allow violation of the required energy conditions $\endgroup$
    – KP99
    Feb 17 at 11:09
  • $\begingroup$ What is the variation $\frac{\delta \int \sqrt{-g} T d^4x}{\delta \phi}=0$ even supposed to mean? $\endgroup$
    – KP99
    Feb 17 at 11:10
  • $\begingroup$ @KPP99 It is called the variation of T. For a scalar field $T=\nabla^\mu \phi \nabla_\mu \phi$. It is equivalent to the Euler-Lagrange equations. It gives the wave equation in curved space-time $\endgroup$
    – user84158
    Feb 17 at 16:52

1 Answer 1


I can't see how one could use classical physics to model evaporating black holes for two main reasons:

1. The Area Theorem

The Area Theorem is a result in General Relativity that gives a few basic conditions under which the area of a black hole can't decrease. I've discussed it in this and this post, for example. It holds for matter satisfying the null energy condition, which is a very mild assumption for classical matter. If you assume classical matter always to have a positive energy density, then the null energy condition holds, and hence so does the area theorem. Therefore, black holes cannot evaporate.

Black hole evaporation can happen in semiclassical contexts because quantum matter doesn't have this sort of restriction. One actually can come up with situations in which there are regions of negative energy density. Hence, the null energy condition is violated and the black hole can evaporate.

Intuitively, the reason is that you can't have energy coming out of a black hole: nothing can come out of a black hole. Nevertheless, quantum matter allows you to send negative energy into the black hole, which has the same net effect.

Without previous knowledge of QM, it seems difficult to see how (or why) one would consider fields with negative energy density.

2. The Hawking Temperature is proportional to $\hbar$

The temperature of an evaporating Schwarzschild black hole is given by (see Wikipedia) $$T = \frac{\hbar c^3}{8 \pi G M k_B}.$$ Hence, the temperature vanishes in the classical limit $\hbar \to 0$. While $\hbar$ also occurs on the classical Klein–Gordon equation, the black hole temperature makes no reference to the field's mass, which should accompany $\hbar$ if that's where it came from.

I do admit this argument has a gap, though: one can actually relate classical radiation from accelerated charges to the Unruh effect, which is "just as quantum as the Hawking effect" (see arXiv: 1701.03446 [gr-qc], for example).

Final Comments

I might add that historically one would expect based on classical arguments that Kerr black holes should emit radiation (I think this paper discusses the history behind this). Shortly, one can extract rotation energy from black holes by means of something known as the Penrose process. It is possible to exploit this by sending in a wave near the black hole and having it come back with more energy, slowing down the black hole in the process. Since this is similar to stimulated emission, one could expect that a Kerr black hole might also have some form of spontaneous emission and radiate. It was particularly surprising, though, that even non-rotating black holes could radiate too.

  • $\begingroup$ Thanks. Nice answer $\endgroup$
    – user84158
    Feb 17 at 17:21
  • $\begingroup$ @zooby You're welcome! If the answer solves your problem, please consider accepting it. $\endgroup$ Feb 17 at 23:14

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