How do we determine the temperature of a Black Hole? Since we cannot see a Black Hole, which I presume, is because it absorbs light, would it not also prevent radiation from escaping, making temperature determination difficult.

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    $\begingroup$ Probably via entropy.. IIRC, if you replace 'entropy' and 'temperature' with 'horizon surface area' and 'surface gravity' in all the ThermoD laws, they check out for a black hole. The radiation becomres Hawking radiation. Related to some fellow called Bekestein, IIRC. I would post an answer but I'm extremely unsure of this. $\endgroup$ Apr 1, 2012 at 2:13

3 Answers 3


Classically, you're exactly right.

Quantum mechanically, something strange happens. It turns out that the way that we define what a particle is and, therefore, what the vacuum is, is dependent on our notion of time--in particular, you are stuck with particles being objects that propagate information forward in time, rather than backward in time.

In general relativity, though, we know that the notion of time on a black hole horizon is inequivalent to that far from the black hole. What this means is that if you take a vacuum solution far from the black hole, and in the distant past of the black hole, and trace it forward in time, you will find that by the time you reach the black hole, it will appear that that state will contain particles! Furthermore, some of these particles will be travelling above escape velocity, and at late times, will be arbitrarily far from the black hole. So, Hawking radiation is inherently a quantum mechanical effect deriving from the fact that the definition of a 'particle' is ambiguous in general relativity.

Furthermore, when you analyze this gas of radiating particles, you will find that it is exactly equivalent to a gas having a temperature that is proportional to the gravitational force felt at the horizon by an observer falling into the black hole.

  • $\begingroup$ @RonMaimon Thank you all, for your answers. My main reason for the question, was the differences in temperature that have been given for different black holes. Generally black holes temperatures are given as very high, and then all of a sudden someone talks about a cold black hole. I think Wiki "Hawkins Radiation" explains this. I.E. Mass & CMB radiation balance. Are you aware of anyone developing a means to prove this? - NASA etc. $\endgroup$ Apr 3, 2012 at 21:50
  • $\begingroup$ @Clive Ballard: What do you mean "prove"? I gave you an outline of the proof. $\endgroup$ Apr 4, 2012 at 11:57
  • $\begingroup$ Sorry, I did take on board what you said. Ron Maimon said much the same thing and so does (Wiki) - Hawking radiation. There are so many different satellites out there now, measuring different forms of radiation etc. When I said developing, what I meant was developing a system to look directly into a black hole, at a distance, and take its temperature. Probably not feasible, since that sort of information, is not escaping. $\endgroup$ Apr 4, 2012 at 15:40
  • $\begingroup$ I am a layman, as far as much of this is concern, and I do struggle a little bit, to take it all in. That particularly applies to the quantum side of things. I have worked with logic all my life, and that probably explains why I do not get on, with the quantum theory. Though, I have no objection to it, if it proves a point. $\endgroup$ Apr 4, 2012 at 15:42
  • $\begingroup$ @CliveBallard Hawking radiation has never been directly observed, and likely will not unless you get black holes made in Earth-based particle accelerators. The reason is that astrophysical black holes are likely to be extremely low in temperature, and at that point, the effects of quantum gravity, such as Hawking radiation, are likely to be extremely small. $\endgroup$ Apr 4, 2012 at 16:07

The temperature is determined using the equivalence principle, and the temperature detected by accelerating detectors.

When a detector is accelerating forever in a certain fixed direction with acceleration $a$, it sees a thermal background temperature equal to $a/2\pi$. This thermal background is associated with the horizon that is formed by the accelerating particle, since a beam of light a certain distance away from the accelerated observer never catches up.

The reason is that an accelerated path is a hyperbola, because the acceleration is the curvature of the particle's path through space-time, and the curvature is constant, and in relativistic geometry, the constant curvature curve is a hyperbola, just as in Euclidean geometry it is a circle (this comes from the minus sign in the relativistic version of the pythagorean theorem).

The hyperbolic time coordinate is the analog of the angle in Euclidean polar coordinates, and if you formulate a quantum field theory for the accelerating coordinates, the imaginary time continuation is periodic with period $2\pi$, which means that the theory is thermal. This is a little technical, but you can just take Unruh's result on faith. For a free field theory interacting with a detector, like photons interacting with an accelerated atom, Unruh's result that the atom behaves as if it were in a thermal environment can be verified by direct calculations of the response of an accelerated detector, not just by fancy methods.

If you are near the horizon of a black hole, you must accelerate to keep from falling in. This means that you must see an Unruh temperature, by the equivalence principle. This Unruh temperature is in the "t" coordinate of the exterior, which turns into the Unruh angle-like coordinate near the horizon. This radiation must be self-consistent as you move away from the black hole, and the self-consistency condition is that the radiation is at a temperature which is smaller by an amount proportional to the rate at which clocks tick. the faster they tick, the colder the temperature.

When you redshift the near-horizon Unruh temperature to infinitely far from the black hole, you find a finite answer. This answer is $8\pi M$, or proportional to the mass of the black hole. This is equivalent to finding the analytic continuation of the black hole solution to imaginary time, and identifying the periodicity of this solution, and the reason is the physical argument above (since Unruh radiation is periodicity in imaginary time).

This argument has in the past been confusing, because most solutions do not have an analytic continuation. How does a physical temperature depend on analytic continuation? It doesn't really. The only thing that is used is the equivalence principle, to link the periodicity in imaginary time near the horizon (where it is just Minkowski space) to the periodicity far away. Mathematically, this is the same as the periodicity in imaginary time of the continued general relativistic solution.

These things are explained in a somewhat more complete and technical way on Wikipedia, under Unruh effect and Hawking radiation.


A black hole's temperature is given by:

$T = \frac{\hbar c^3}{8\pi kGM} \tag{1}$

So, if the mass of the black hole is known, normally by calculating the affect the black hole has in the surrounding space, you can calculate the temperature of the black hole.

  • $\begingroup$ why is this incorrect? $\endgroup$
    – psitae
    Jan 16, 2021 at 16:13

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