# 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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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.