A Schwarzchild black hole has a "thermal atmosphere" in the region $r_{s} < r < (3/2)r_{s}$ where $r_{s}$ is the Schwarzchild radius [1, 2]. The thermal atmosphere is comprised of radiation quanta that are entangled with quanta behind the horizon [1, 3]. The radiation quanta that slowly leak out of the atmosphere constitute the radiation that Hawking found (Hawking radiation) [1, 4]. Far from the black hole, the radiation has a temperature $T_{hawking} = \frac{1}{8 \pi M}$. The temperature of the thermal atmosphere is $T_{atm} = \frac{1}{2 \pi \rho}$ where $\rho$ is the proper distance to the horizon. This shows that the temperature of the thermal atmosphere diverges as one approaches the horizon [1, 2, 3]

Hawking showed that a Schwarzchild black hole can be in thermal equilibrium with blackbody radiation at a temperature $T_{eq} = \frac{5}{32 \pi M} = \frac{5}{4}T_{hawking}$, if the "size" (radius) of the total (black hole + radiation) system does not exceed $\frac{3}{2}r_{s}$ [1, 2]. This implies that a black hole in thermal equilibrium with blackbody radiation has the same size as a black hole surrounded by its thermal atmosphere.

Given the fact that the radiation in the thermal atmosphere leaks away very slowly, can we conclude black hole is approximately in thermal equilibrium with its thermal atmosphere?


  1. Susskind, Leonard and Lindesay, James, An Introduction to Black Holes, Information and the String Theory Revolution. The Holographic Universe}, World Scientific, 1st Edition, 2005.

    1. Jacobson, Ted Introductory Lectures on Black Hole Thermodynamics http://www.physics.umd.edu/grt/taj/776b/lectures.pdf

    2. Harlow, Daniel Jerusalem Lectures on Black Holes and Quantum Information arxiv:1409.1231v4

    3. Preskill, John Black holes and information: A crisis in quantum physics http://www.theory.caltech.edu/people/preskill/talks/blackholes.pdf


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.