# How can one reconcile the temperature of a black hole with asymptotic flatness?

A stationary observer very close to the horizon of a black hole is immersed in a thermal bath of temperature that diverges as the horizon is approached. $$T^{-1} = 4\pi \sqrt{2M(r-2M)}$$ The temperature observed by a stationary observer at infinity can then be obtained through the gravitational redshift formula (see http://en.wikipedia.org/wiki/Hawking_radiation#Emission_process) to be $$T^{-1} = 8 \pi M$$ which is what is often quoted as the temperature of a black hole.

As QGR points out here in an answer to my related question here, the resulting non-zero stress-energy tensor at infinity is incompatible with the asymptotic flatness of the Schwarzschild spacetime. What exactly is going wrong here?

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May I suggest that a link to your earlier related question is probably in order? – dmckee May 29 '11 at 1:50
You can't. If you have a black hole which is in equilibrium with radiation, the spacetime is not asymptotically flat. – Ron Maimon Aug 28 '11 at 20:52

Dear D-brane, indeed, a uniform thermal radiation would curve the Universe. Even if one doesn't immerse the black hole in a thermal bath, the outgoing Hawking radiation may violate the asymptotically flat conditions at any finite time, although just mildly.

However, an evaporating black hole that is not surrounded in the thermal bath ultimately evaporates and the Hawking radiation dilutes arbitrarily, so that the Universe will be asymptotically flat.

And a black hole immersed in a thermal bath of the same temperature does curve the Universe, but the curvature is much smaller than the curvature near the black hole as long as the black hole is much greater than the Planck length (or Planck mass). There is a parametric gap here. In the Planck units, if the radius is $R$, then the mass is also $M=R$ (in four dimensions), but the temperature is $1/R$, the density of radiation is $1/R^d$ i.e. $1/R^4$ in four dimensions, and the amount of radiation (energy per unit time) above the horizon is $R^{d-2}/R^d = 1/R^2$, in any dimension. That's $R^3$ times smaller than $R=M$, in $d=4$, so the Hawking radiation will evaporate the black hole mass in time $R^3$ - more generally, $R^{d-1}$, which is still $R^{d-2}=R^2$ times longer than the characteristic time scale of the black hole (orbital time for light, for example).

The bigger a black hole is, the more you can neglect those things. The factors $R^2$ or $R^3$ are huge because, for example, the black hole at the center of the Milky Way has 3+ million solar masses which is almost $10^{37}$ kilograms or $M=10^{45}$ Planck masses. The energy carried by the Hawking radiation is smaller by a factor that is a positive power of $10^{45}$. It's small, indeed.

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It is simply not true that a non-vanishing stress tensor is incompatible with asymptotic flatness. The Schwarzschild spacetime is asymptotically flat, period. The semiclassical Hawking-calculation does not in any way change this background, unless you consider effects from backreactions.

Once you take into account backreactions the Hawking radiation does change the background, but it does so by essentially diminishing the Bondi mass at scri$^+$. (You can think of the Bondi mass as measuring the amount of energy stored in the black hole; as opposed to the ADM mass, which is defined at spatial infinity and which measures the total energy, including radiation to or from scri, it is defined on light-like infinity and can thus change with the advanced or retarded time, depending on whether you are on scr$^\pm$.) The spacetime remains asymptotically flat in this process.

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Once you start asking questions about the global backreaction effects (important roughly when M is of order 1 eg the Planck mass), its not clear to me how you define a concept like the Bondi mass. Intuitively what you say makes good sense, but I don't know of a way to see that precisely. Also its clear that thermal equilibrium no longer makes sense when you enter the regime where backreaction effects are important and its thus difficult to talk about a Hawking temperature. – Columbia May 28 '11 at 8:02
Thinking about it more. I think you are talking about using the quasistatic approximation here, before the hole becomes Planckian (and the approximation loses its power and quantum gravity becomes important) – Columbia May 28 '11 at 8:17
Yes, to talk about Hawking temperature in the first place we should assume the semi-classical approximation to be valid. Then backreaction effects will be small, the Hawking flux will be stationary to leading order, and a Bondi mass can be well-defined. Regarding thermal equilibrium there are the standard issues with negative specific heat, that can be resolved by putting the black hole inside a cavity, which provides a heat bath. If the cavity is sufficiently close to the horizon you have positive specific heat. However, then it is meaningless to talk about asymptotic observers. – Daniel Grumiller May 28 '11 at 10:27
-1: If the whole universe is filled with radiation at a given temperature, how can you expect it to be asymptotically flat? This answer is totally wrong. – Ron Maimon Aug 28 '11 at 20:51