Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

What is the basis for black hole evaporation?

I understand that Hawking-radiation is emitted at the event horizon, a theoretical result originating in General Relativity and Quantum Field Theory, but it seems to me that additionaly one has to assert an integral conservation law for mass/energy, ie. for a sphere surrounding the black hole.

Does such a conservation law hold for the simplest case of a Schwarzschild metric?

I am grateful for any related classic paper references.

EDIT: The usual heuristic for understanding Hawking-radiation is: virtual pair, one falls in, one goes out; the ones going out are called Hawking-radiation. But what about the ones going in? Naively, it seems there should also be Hawking-radiation going inward, which would actually increase the black hole's mass.

share|improve this question
Classical GR does not have a conserved, scalar measure of mass-energy that applies to all spacetimes. However, it does have such measures in certain special cases, such as stationary and asymptotically flat spacetimes. Since the Schwarzschild metric is both stationary and asymptotically flat, these things are conserved (Komar mass, ADM and Bondi masses). There are various heuristic ways of understanding Hawking radiation; here is a skeptical discussion of those heuristics: math.ucr.edu/home/baez/physics/Relativity/BlackHoles/… –  Ben Crowell Nov 13 '11 at 19:35
@Ben, that link discusses the heuristic understanding of Hawking-radiation: virtual pair, one falls in, one goes out. This picture is exactly what lead to my question: the ones going out are called Hawking-radiation, but what about the ones going in? Naively, it seems to me that we should also have Hawking-radiation going inward, which would actually increase the black hole's mass. –  mtrencseni Nov 13 '11 at 20:12
I see. That makes the motivation for your question much clearer. Maybe you could edit the question a little to clarify that. I don't know the answer, but it might be helpful to compare with Unruh radiation. If an observer is accelerating in a rocket ship, he has a horizon that is similar to a black hole's event horizon, and that horizon has a temperature and emits blackbody radiation. The source of the energy for that radiation is the rocket engines. This is discussed in Three Roads to Quantum Gravity by Lee Smolin, pp. 82-83. –  Ben Crowell Nov 13 '11 at 21:07
Here's a possible answer to your question based on the Unruh-effect analogy. This could be totally wrong, though :-) The accelerating astronaut A has a horizon and observes Unruh radiation, but an inertial observer B on the other side of A's horizon does not have a horizon and does not observe radiation. Similarly, I don't think an observer inside a black hole's event horizon sees a horizon himself, and I don't think he can observe any infalling Hawking radiation. –  Ben Crowell Nov 13 '11 at 21:15
Possibly related: physics.stackexchange.com/questions/9359/… –  Ben Crowell Nov 13 '11 at 21:16
show 6 more comments

2 Answers

up vote 2 down vote accepted

The best calculation of Hawking radiation is the one given on Wikipedia, in the article for Hawking radiation. This is the method implicitly described by Unruh in his famous paper on accelerated observers in QFT, and it is mathematically equivalent to the method Hawking uses in papers of the 1977-78 period, which use periodicity of the analytically continued solution in imaginary time.

Hawking's original calculation assumes that the emitted field quanta are non-interacting. The result is clearly true for interacting theories as well, but this is not completely clear in Hawking's original calculation. Further, Hawking's 1976 method uses a pure-black hole solution (not an eternal solution, no white hole), meaning that the horizon was created at a finite affine parameter in the past. The outgoing radiation, when traced back in time, is all coming from a sub-microscopic infinitesimal region right at the moment the horizon first formed. During the sojourn next to the even horizon, the peeling-off property of the horizon (the fact that outgoing light rays peel away from stationary light rays right on the horizon) lead to a big magnification of the tiny region where the black hole is first formed. Any outgoing photon mode, extended back, comes from a place where its wavelength is trans-trans-trans Planckian, this is right the moment of formation of the black hole horizon. The trans-Planckian property led many people to be skeptical of Hawking's result.

Analytic continuation is impossible for general metrics, and it seems strange to use an analytic continuation to find a physical temperature. The calculation below is equivalent to the analytic continuation to imaginary time, but explains why this method works.

Unruh radiation

When an observer is accelerating in Minkowski space, the trajectory in space time is a hyperbola, which is the relativistic analog of a circle. The natural coordinates for an accelerating observer are the relativistic analog of polar coordinates. Choosing the x coordinate to be the direction of the acceleration, these coordinates are given by

$$ x = r\cosh(\tau)$$ $$ t= r\sinh(\tau)$$

the y,z coordinates transverse to the acceleration are unchanged, and the metric in these coordinates is the relativistic polar metric:

$$ ds^2 = dr^2 - r^2 d\tau^2 + dy^2 + dz^2 $$

And it is important to compare this the standard Euclidean polar metric $dr^2 + r^2 d\theta^2$. It is obvious then that the time coordinate is like the angle, while the r coordinate is radial.

A Green's function for a quantum field in the $x,t$ coordinates analytically continues to imaginary time, this is Wick rotation, and it is the whole point of the Feynman formalism. The natural Hamiltonian for the $\tau$ coordinate analytically continues to a generator of rotations in the x-t plane, so that the $\tau$ coordinate is periodic with period $2\pi$. This means that the accelerated observer measures a thermal Green's function for the field with a temperature that is diverging as $1\over r$ (the $\tau$ period is $2\pi$, but $\tau$ is dimensionless-- the physical local time coordinate is $r\tau$, so that the physical energies of high frequency quanta (those that can be localized) are $1/r$ bigger).

The divergence of the Unruh temperature at the Rindler acceleration horizon is obvious--- it is just the statement that a family of observers at different r in the $r,\tau$ coordinates have an acceleration that diverges as $1\r$.

The Unruh radiation, when back-traced, also has a trans-Planckian wavelength right at the acceleration horizon. The reason is just because any outgoing mode is redshifted an infinite amount from its value near the horizon, and the constant magnification of the near-horizon region is the analytic continuation of the constant rotation in imaginary time (rotation is by cosines and sines, which continue to cosh's and sinh's, giving the magnifying at the acceleration horizon. The Bogoliubov coefficients are given by these near-horizon blow-up factors. The Bogoliubov formalism is not particularly enlightening.). Since the Unruh radiation is in flat space, there is no debate about it's correctness. You can model an accelerating detector in ordinary Minkowski coordinates to see that the detector goes off as if it is immersed in thermal radiation.

Another important point about Unruh radiation is that the only reason you see it all around you is because the radiation going out of the acceleration horizon free-falls back in, to hit you going the other way on its way back to the horizon. None of this thermal background can escape, because the Rindler (pseudo) gravitational potential well is infinitely steep--- you need to do an infinite amount of energy to get away from an accelerating observer in the direction of acceleration.

Equivalence principle

By the equivalence principle, a near-horizon observer by a black hole can't tell the difference between the black hole and nothing at all. The black hole horizon is identified with the Rindler horizon for a nearby stationary observer, which must be accelerating quickly to keep from falling in.

This is a powerful statement, because knowing that the black-hole metric is locally Rindler (using the coordinate $s=\sqrt{r-2M}$) means that one can continue into the black hole just by assuming that the horizon is not a special location. This continuation past the horizon is standard, it is how the black hole interior is derived from the exterior solution. It is sometimes called "analytic continuation", but this is a misnomer, as it does not require analyticity to work. It just needs that the metric is asymptotically Rindler, so that the spacetime is really still locally Minkowksi space in different coordinates, there are no curvatures or singularities, and so continues into the interior.

There are not many different stationary parametrizations of Minkowski space, if you want the space to look stationary you need to use a Rindler coordinate of some type. This means that any locally flat horizon is locally Rindler. Extremal black holes are not locally flat, and their near horizon limit is AdS, not Minkowksi, reflecting the diverging curvature at extremality, due to the pinching off of the interior.

The local temperature seen by a near horizon observer diverges as $1/s$, which is proportional to the radial distance to the horizon. This local temperature is equal to the local periodicity of the solution in imaginary time, because this is true in Rindler space--- the $\tau$ periodicity translates into a $t$ periodicity near the horizon.

But the black hole solution is stationary, which means that once you know the period in imaginary time near the horizon, if you extend the thermal radiation using the gravitational redshift factor, it stays in equilibrium. Any outgoing radiation is redshifted to a lower temperature, while any incoming radiation is blueshifted to a higher temperature. The operator which moves $t$ to $t+a$ is not quite a Hamiltonian, since it generates rotations near the horizon (just like the Rindler $\tau$), so you find that for equilibrium, the thermal radiation is the Unruh radiation plus a bath at a temperature that goes as the inverse redshift.

For Unruh radiation, the redshift factor, the square root of $g_{00}$, is $r$, it goes to infinity at $r=\infty$. In the Schwartzschild solution, the square root of $g_{00}$ is $\sqrt{1-{2M\over r}}$, and it asymptotes to a finite limit at infinite r. The analytic continuation matching the Unruh temperature gives the Hawking temperature at infinity to be (see Wikipedia for the simple calculation, it is an exercise):

$$T_H = { 1\over 8\pi M}$$

the argument from matching to the Unruh limit shows why analytic continuation to imaginary time is correct--- the Unruh periodicity continues to the long distance periodicity. The generator of time translations is a symmetry, and if it has a period at some position, imposing the condition that the period is constant gives a consistent background for a path integral, and this path integral is necessarily thermal, with asymptotic temperature given by Hawking's formula.

Physical emission

The argument above only shows that the black hole is in equilibrium with a thermal gas whose temperature at infinity is $1\over 8\pi M$. It also shows that the correct thermal ensemble for the exterior observer living with an interacting quantum field theory is the one from the path integral in an imaginary-time background which is periodic in T with period $8\pi M$ everwhere, with a background metric equal to the analytic continuation of the Schwartschild metric.

But what if there is no radiation at infinity? Then there is no infalling radiation, and the thermal emissions from the black hole are not compensated, so the black hole glows thermally. The actual radiation from a black hole is given by detailed balance--- the black hole is in equilibrium with a thermal gas at infinity at the Hawking temperature, so the emission rate for a free field theory is related to the absorption by the condition that they balance out in thermal equilibrium, and, since the quanta are free, they are independent processes. This gives the graybody factors for black hole emission from a calculation of the absorption at the same wavenumber.

share|improve this answer
add comment

The Hawking radiation is calculated by the methods of "quantum field theory on curved backgrounds". One may show that the final state contains the thermal radiation (corresponding to the right, Hawking temperature) for any field that propagates on the black hole background.

Quantum field theory on a curved background tells us many more microscopic details than just the existence of the radiation, its temperature, and the fact that the black hole is shrinking. But yes, of course, the total mass/energy conservation law holds. By Noether's theorem, it holds for any physical system whose laws are invariant under translations in time. Emmy Noether proved it in general, for any system: but of course, if one works with a specific field theory such as "the Klein-Gordon quantum field theory on a curved background", one may see what the mass/energy conservation means in much more detail.

The key paper for the Hawking radiation is obviously Hawking's 1975 paper, Particle creation by black holes:


I think it contains everything needed to answer basic questions such as yours.

share|improve this answer
My understanding is that QFT on curved backgrounds is shaky foundationally, so if one had to believe in it in order to believe in Hawking radiation, I wouldn't necessarily believe in Hawking radiation. Aren't there more fundamental reasons to expect Hawking radiation to exist than a calculation performed in this particular framework? –  Ben Crowell Nov 13 '11 at 19:38
Ben, I may be wrong but my impression was that QFT on curved background is shaky in general, on conformally flat backgrounds it is as solid as usual QFT. –  MBN Nov 13 '11 at 20:28
@MBN: Thanks for your reply. I think I may have not been making sufficiently careful distinctions. What I think is clearly shaky is is semiclassical gravity a la Barcelo and Liberati, in which the background is not fixed but is made consistent with the vacuum energy. It requires renormalization, and also makes predictions that I'm skeptical about (e.g., black stars). I guess Hawking radiation can be calculated on a fixed background, so maybe there aren't such foundational issues. I'm not an expert on quantum gravity, so these are just my general impressions. –  Ben Crowell Nov 13 '11 at 20:42
Oh, I see, sorry, I thought by QFT on curved background you meant on a fixed background. I was unaware of background indpendent semiclassical gravity. Sounds interesting. –  MBN Nov 13 '11 at 22:17
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.