Are black holes perfect heat sinks?

Imagine a closed system with a black body and a black hole, where the black hole has a lower temperature. The black body will radiate heat and the BH will absorb that heat, gaining mass in the process. The obvious catch is that the BH's temperature is actually sinking as it's absorbing heat energy. Which means it can do this endlessly. Both objects are getting colder.

This seems deeply wrong to me and my admittedly limited understanding of thermodynamics. Which assumption is wrong? Also, is there some sort of equilibrium temperature?

• Why do you say the black hole's temperature will decrease? – electronpusher Sep 4 at 22:40
• Indeed, a (Schwartzchild) black hole has negative heat capacity, precluding it coming to equilibrium with another object. Here's a quite interesting related question: physics.stackexchange.com/questions/194347/… – Rococo Sep 4 at 22:51
• Note also that gravitational objects in general have this property, not just black holes: physics.stackexchange.com/questions/142461/… – Rococo Sep 4 at 22:55
• @Rococo Thanks for the clarification, I'm a self-learner in most areas of physics and I never came across negative heat capacity. Appreciated. – electronpusher Sep 5 at 2:16
• @electronpusher of course! That's what the site is here for. – Rococo Sep 5 at 2:50

I have to amend my initial comment: the fact that the black hole has negative heat capacity means that it is not guaranteed that an equilibrium is possible, but in a closed system, as is the case you specified, it is possible that equilibrium exists. As you'll see, this is a pretty rich problem.

You can simplify your setting by considering a black hole in a closed box, without the initial blackbody. Because the black hole will emit blackbody radiation itself, and that radiation will fill the box and bring it to some temperature, it ends up being an equivalent situation. The equilibrium conditions of this setup have been studied by Hawking (1) and Page (2), and by the standards of scientific papers both are pretty readable.

Consider a box with some fixed total energy, $$E_0$$, that can be divided between a black hole and the radiation around it. The energy in the radiation is $$E$$ and the energy in the black hole is $$E_0-E$$. Then the temperatures of each are easily found:

$$k_BT_{BH}=\frac{\hbar c^5}{8\pi G(E_0-E)}$$ $$k_BT_{rad}=\left[\frac{15(\hbar c)^3}{\pi^2}\frac{E}{V}\right]^{1/4}$$

The first equation is the usual black hole temperature formula, and the second equation comes from the blackbody radiation formula one learns in undergraduate thermodynamics. Note that this depends on $$V$$, the volume of the box.

We can plot both of these to learn about the possible stable configurations. There are two possibilities, depending on $$V$$:

The way to read these diagrams is that when the black hole has higher temperature than the radiation, there will be a net energy flow from the black hole to the radiation, moving you towards the right, and vice versa when the radiation temperature is higher. That determines the direction of the arrows marking the flow.

In case 1, for a large box, there is no stable equilibrium with a black hole, and the system always moves towards the black hole evaporating and leaving only radiation.

In case 2, for a small box, there is a stable equilibrium with the black hole and radiation, and also an unstable equilibrium. This corresponds to the situation originally asked by the OP: for example, imagine starting at x=0.4. So we can see that while the OP was correct that both systems initially get colder, because they do so at different rates this does not go on indefinitely.

While the black hole is locally stable here, over an unfathomably long time there will occasionally be thermodynamic fluctuations, some of which push the system into the unstable region near the right of the plot. So, you would expect that the system sometimes fluctuates from the equilibrium with the black hole to the equilibrium without, and vice versa. The relative probability of these two equilibria are not equal; there is another critical volume at 0.256 $$V_c$$ that separates whether the system spends more time near the black hole equilibrium or the radiation equilibrium.

The value of $$V_c$$, as found in the above papers, corresponds to $$E=E_0/5$$: $$V_c=\frac{3*2^{20}\pi^2E_0^5}{125}\frac{G^4}{\hbar c^{17}}$$. This assumes that the black hole emits only photons; there is a slight modification when you allow it to emit more species of particles.

For completeness, there is one more final possibility that I've ignored until now. If the initial energy density is very large, or the volume very small, the Schwartzchild radius of the system will be larger than V, meaning that when you form a black hole it is larger than the box itself. If this is the case, then this whole notion of an isolated black hole breaks down. This happens at $$V_s=\frac{8 G^3 E_0^3}{c^{12}}$$.

Edit: In response to Rob's question, one may look at the ratio $$V_c/V_s$$:

$$\frac{V_c}{V_s}=\left( \frac{E_0}{E^*} \right)^2,$$ where $$E^*=\sqrt{\frac{125\hbar c^5}{3*2^{17} \pi^2 G}} \approx 0.0056m_pc^2$$. Here $$m_p$$ is the Planck mass of about 21 $$\mu$$g. In other words, for a black hole with an initial mass of greater than 0.0056 $$m_p,$$ around 0.1 $$\mu$$g, one can put a box around it such that the box is large enough to contain the black hole but small enough that the black hole is stable.

• Interesting. Two questions: (1) You have two volume expressions near the end of the answer, one with $V\sim E^5$ and one with $V\sim E^3$. I was expecting $V\sim E^3$ in both places for dimensional reasons? (2) How does the length scale $R\sim V^{-1/3}$ compare to the Schwarzchild radius for the critical volume? – rob Sep 6 at 22:18
• Hi Rob: I agree that the scalings in this problem are quite unusual! That's what you get when you combine normal physics with something holographic, I suppose. 1. The $E^5$ dependence can be seen from the first two (dimensionful) equations that I give. If you equate the temperatures, and rearrange for V, you see that $V \sim E^5$. 2. Shouldn't it be $R \sim V^{1/3}?$ The final expression that I give is precisely the Schwartzchild volume, so this ratio is: $[(3*2^{20}\pi^2E_0^5/250)/(64\pi E_0^3/3)]^{1/3}\sim E_0^{2/3}$. – Rococo Sep 7 at 14:52
• So, it is not guaranteed that the critical volume is larger than the Schwartzchild volume- it only happens for sufficiently large $E_0$. – Rococo Sep 7 at 14:58
• Okay, this question gave me the nudge needed to put all the units in the equations into SI form- hopefully I got all the weird numerical factors right. Anyway, the reader can now easily check that the dimensions of both formulae work out. – Rococo Sep 7 at 16:17
• Just for fun, I've also added a slight elaboration on the critical energy scale needed for $V_c>V_s$. Inevitably it is set by the Planck mass, and therefore is very small by astrophysical standards. – Rococo Sep 8 at 16:58

As a commenter points out, black holes (and all systems whose primary interaction is gravitational) have negative heat capacity. For classical gravitational systems you can tell that the heat capacity is negative from the virial theorem: the average kinetic energy of a gravitationally-bound particle has half the magnitude of its potential energy. Adding heat to a gravitationally-bound system sends its particles into higher orbits, where their speeds are slower; to the extent that temperature is related to mean particle speed, the object cools off. (A consequence of this is that when a dying star shifts from hydrogen burning to helium burning, its overall brightness increases, but its average temperature gets cooler: our yellow sun will turn into a brighter red giant, not a brighter blue giant.)

To observe that black holes also have negative heat capacity, simply note that adding energy to a black hole increases its mass, which decreases its Hawking temperature.

Negative heat capacities screw around with temperature-based ideas about thermal equilibrium. But the reason we talk about temperature is because temperature relates energy exchange and entropy exchange. If you're not sure how a system will evolve, you can always ask what pathway gives the largest increase in entropy. And there are several proofs that a black hole is a maximum-entropy state: if you have a system that contains a black hole and some other stuff (including other black holes), that system has less entropy than if everything ended up in the one black hole. That's what will happen, if you wait long enough.

• I agree with everything said here but the last statement: in a situation in which a black hole does have a higher temperature than the surrounding space and radiates into it, the transmutation of the BH into diffuse radiation increases entropy: physics.stackexchange.com/questions/106864/… – Rococo Sep 5 at 2:55
• Of course you are correct that the disequilibrium goes the other way if the black hole begins as the hotter object. But no stellar-scale black hole has ever had a temperature warmer than the CMB, so in our universe only the smallest of primordial black holes has ever experienced any net mass loss due to evaporation. – rob Sep 5 at 5:11

A black hole is only a good heat sink if its horizon temperature is much lower than the background. Even then it is not perfect because it still emits Hawking radiation.

Suppose you have a black hole hole in a space or spacetime with a constant background temperature $$T$$. The temperature of the black hole is $$T~=~\frac{\kappa}{8\pi M}$$, where $$\kappa$$ contains constants. We set this black hole in this space with the same temperature as the background. This black hole has equiprobability of absorbing a photon from the background, in which case $$M~\rightarrow$$ $$M~+~\delta M$$ where the temperature will decrease. Similarly, the black hole may emit a Hawking photonswhere $$M~\rightarrow$$ $$M~-~\delta M$$ where the temperature will increase. This means the equal temperature situation is not stable, for in a random walk setting the black hole will either drift towards greater mass or smaller mass and quantum evaporate.

• Note that the black home mass corresponding to this unstable equilibrium is quite small; all stellar-mass black holes are colder than the cosmic microwave background. – rob Sep 5 at 0:08
• If I remember right a black hole with the mass of the moon would have the same temperature as the CMB. Of course the CMB is not static either, that will be half its current value in 10 billion years or so. The rule is that equilibrium appears not to hold in general relativity. – Lawrence B. Crowell Sep 6 at 1:34

Case 1: The black hole cools, the black body cools faster. After some time both objects have the same temperature, which is the equilibrium temperature.

Case 2: The black hole cools, the black body is so large that it cools slower. After a long time the black hole has become so massive that it cools slower than the black body, and we have the situation described in Case 1.