# Is there an explanation for why Hawking radiation causes a black hole to shrink *without* referring to particle/anti-particle pairs?

An observer far from a black hole sees thermal radiation at the Hawking temperature. As a consequence, we are told that the black hole loses energy over time: that is, it evaporates. This logical step from radiation to evaporation is usually justified by considering particle pairs near the horizon, and showing that the infalling particle has negative energy. This is ok, but it would be nice if there were an explanation not involving particles at all, i.e. only working at the level of fields moving on curved spacetime.

What I have in mind is some sort of conservation law which says that "Energy of spacetime + expected energy of quantum fields is the same for early- and late-time observers". Then, if the late-time observer sees more energy in the field, the energy of spacetime must decrease to compensate, i.e. the BH must shrink.

Does such a conservation law hold? And can it indeed be used to show the BH shrinks?

• Related: See physics.stackexchange.com/q/251385/50583 for the explanation of Hawking radiation itself. Commented Dec 10, 2021 at 12:11
• For starters, Hawking's original argument never mentions pair creation. It is, as you want, a semiclassical computation using only fields in curved spacetime. Commented Dec 10, 2021 at 12:58
• @AccidentalFourierTransform It is the implication "hawking radiation => black hole loses mass" which I'm asking about. This implication is not derived in the paper you mention. Commented Dec 10, 2021 at 22:09

You pretty much solved the problem yourself. I'll sketch the argument given on Wald's Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics, Sec. 7.3. The idea is exactly what you said, so I think I pretty much just need to write down a couple of equations to make it clearer.

The energy properties of the quantum field that gives origin to Hawking radiation on the spacetime are governed by the expectation value of its energy momentum tensor, $$\langle \hat{T}_{ab} \rangle$$. Since outside the black hole spacetime is stationary (for simplicity, let us assume a Schwarzschild black hole, so that it really is static), we know the four-current $$J_a = \langle \hat{T}_{ab} \rangle \xi^b$$ is conserved, where $$\xi^a$$ is the spacetime's Killing vector. This gives us the notion of conservation of energy you mentioned.

On dimensional grounds, the energy flux seen by an observer at infinity should be $$F = \frac{\alpha}{M^2}$$, where $$M$$ is the black hole's mass and $$\alpha$$ is a constant. This equation should hold for any field whose mass is much smaller than the black hole's, so that we can consider the black hole mass as being the relevant one.

At each point of the evaporation process we can approximate the geometry of spacetime as a Schwarzschild black hole with varying mass $$M(t)$$, at least while we can assume $$M \gg \sqrt{\frac{c \hbar}{G}}$$, i.e., at least while the black hole's mass is much larger than the Planck mass and hence the local backreaction effects are small. In this case, we can write $$\frac{\textrm{d} M}{\textrm{d} t} = - F = - \frac{\alpha}{M^2}, \tag{7.3.4}$$ where the tag refers to the equation on Wald's text. Solving this ODE, we get $$M(t) = \left[M_0^3 - 3 \alpha t\right]^{\frac{1}{3}}, \tag{7.3.5}$$ which implies the black hole evaporates within time $$t = \frac{M_0^3}{3 \alpha}$$.

Notice, of course, that the semiclassical approximations hold throughout the entire process. It could be that Quantum Gravity effects come into play when the black hole's mass reaches the Planck scale and stops the process, hence stopping the evaporation process. arXiv: 1703.02140 [hep-th], for example, discusses steps on the derivation where things could go wrong and stop the evaporation process.

• I agree that $J_a$ is conserved, but I can't quite see how this implies $\frac{\textrm{d} M}{\textrm{d} t} = - F$. It seems you are invoking some sort of conservation law which includes both the energy of the classical metric and the (expected) energy of the quantum field. Could you be precise about the statement of this conversation law? Commented Dec 10, 2021 at 22:05
• The idea is to get the flux of $J_a$ on spatial infinity and on the black hole horizon. Notice that in between these two regions (i.e., in $M - B$, where $M$ is spacetime and $B$ is the black hole) we only have energy associated with the quantum field (apart from it, the spacetime is in vacuum). Since there is an outgoing flux $F$ from $M - B$at infinity, there must be an incoming energy flux due to conservation of energy. This incoming flux must come from the horizon, which is the other boundary of $M - B$. Hence, there is an outgoing flux of energy $F$ at the black hole. However, all + Commented Dec 10, 2021 at 22:25
• energy at the black hole region is associated with the black hole's mass. Hence, we get a loss of mass at rate $\frac{\textrm{d} M}{\textrm{d} t} = - F$. Commented Dec 10, 2021 at 22:26
• Just to clarify: how are you defining the flux $F$ in terms of $J_a$? Presumably we want to integrate $\ast J$ over some 3-surface? Commented Dec 10, 2021 at 23:59
• @nodumbquestions Hmm I must admit I never stopped to think about the details of the calculation hahaha, but I'll sketch my guess. Nevertheless, since we want to pick a flux we want to think in three dimensions and then take a surface integral at infinity. I'd say first project $J_a$ on the submanifold of constant $t$. Now we have a "three-vector" and what remains to compute the flux is to compute its surface integral on a sphere of large radius, i.e., compute $n_a J^a$ and integrate it on the large $2$-sphere I mentioned, $n^a$ being the normal 3-vector to the sphere Commented Dec 11, 2021 at 2:15