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The power emitted by a Schwarzschild black hole via Bekenstein-Hawking radiation is usually given for an observer at spatial infinity.

What is the emitted power for an observer hovering just above its horizon at a radial distance $r$? (Here, "emitted" is meant to describe only the flow of energy away from the black hole.)

An observer nearer to the horizon will see a higher temperature $T$, due to red shift. In all cases, the emitted power $P(r)$ at radius $r$ is expected to be given by the black hole horizon surface times $T^4$ times the Stefan-Boltzmann constant.

The emitted power should thus increase when getting closer to the horizon. What is the value of $P(r)$?

KlausK
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For the temperature, there is just a redshift factor. The temperature for an observer hovering at radial coordinate $r$ is $$T_r = \frac{T_H}{\chi},$$ where $T_r$ is the temperature at $r$, $T_H$ is the temperature at infinity, and $\chi = \sqrt{- \chi^a \chi_a}$, where $\chi^a$ is the Killing field. For Schwarzschild, $$\chi = \sqrt{1 - \frac{2M}{r}},$$ in units with $G = c = 1$.

This is discussed in books about QFT in curved spacetime and black hole thermodynamics. For example, see Wald's book, Eqs. (5.3.3) and (7.2.10). This ends up matching what one would expect by considering the proximities of the black hole to look like Minkowski spacetime from the point of view of an accelerated observer and then computing the temperature according to the Unruh effect.

As mentioned in the comments to the question, it doesn't make sense to speak of particles in the vicinity of the black hole. They are only defined at infinity. Still, a thermal state is a thermal state and one can still compute the stress tensor associated to the quantum field and see a flux of negative energy into the black hole. These notions are also mentioned in QFTCS books.

Knowing the temperature measured locally, I don't see any immediate problems with using the Stefan–Boltzmann formula to obtain the power from the temperature.