# Black hole no-hair theorems vs. entropy and surface area

I was revisiting some old popular science books a while ago and two statements struck me as incompatible.

• No-hair theorems: a black hole is fully-described by just a few numbers (mass, spin etc) irrespective of the type or configuration of the matter/energy within the event horizon.

• Surface area measures entropy: you can't reduce the total entropy of the universe by throwing a box of hot gas into a black hole, its size/entropy will increase by the necessary amount.

Suppose I have two black holes, A and B. They are identical and interchangeable, same mass, spin, charge...

I have two boxes, Box A and Box B. Box A contains a kilogram of salt in the form of a single crystal and Box B contains a kilogram of salt as loose powder. BoxA has less entropy than Box B but is otherwise identical.

I throw Box A into A and Box B into B.

The no-hair theorems seem to imply that A and B will increase in size by the same amount. The entropy theorems seem to imply that B will end up larger than A.

What am I missing here?

• The entropy difference between the boxes is negligible--- the black hole is maximal entropy for a given mass-energy, so by throwing the boxes in, they become part of the ultimate maximum entropy state, their entropy increases to the maximum possible as the black hole settles down, and the result has the same entropy regardless of which box you throw in. By the way, your example of a single crystal vs. a powder has only the tiniest immeasurable entropy difference, you should use a block of salt and a slightly warmer infiniesimally smaller block of salt with the same total mass-energy. – Ron Maimon May 7 '12 at 17:13

Here is my answer. I should preface it by warning that this is a subject that can provoke intense discussion, and I'm sure there are physicists would would disagree. You should be aware that I'm an expert on thermodynamics but not on general relativity.

But basically, as far as I understand it, the process of converting matter into black-hole-stuff is an irreversible one, in the usual macroscopic sense. Throwing your boxes of salt into identical black holes is somewhat analogous to what would happen if you emptied them into two identical vats of water. You would end up with two identical vats of salty water, with the same mass, temperature, and salt concentration, and the same entropy.

The no hair theorem for black holes is an asymptotic one. It says that if you throw some stuff into a black hole and wait long enough, the black hole will become an arbitrarily good approximation to an "ideal" black hole (which is to say, a black hole solution of Einstein's equations), which can can be completely described by its mass, charge and spin. It also says (I believe) that this convergence happens rather rapidly. But converging towards something is not the same as ever actually reaching it. In reality, nothing can cross the event horizon as seen from an outside perspective (see my answer to this question), it just gets very hard to detect because its light is red-shifted to extremely long wavelengths.

So in my view the apparent loss of information comes from assuming that the black hole actually becomes an ideal one rather than just closely approximating it. It's very similar to the question of how the entropy of an isolated vat of salt+water can increase as the salt dissolves, even though on the microscopic level, the laws of physics seem to preserve information. The resolution is that when you switch to a macroscopic description (in terms of temperature, pressure etc.), you throw away some information about the microscopic state. After the salt has dissolved, the information about its previous state (crystal or powder) is still there, but it's hidden in fine correlations between the molecules' motions. When you choose to describe the final state as an equilibrium ensemble you're basically admitting that those fine correlations can never practically be measured, and therefore choosing to ignore them. Similarly, when you choose to approximate a real black hole as an ideal one, you're basically choosing to ignore any information about what kind of salt was thrown into it in the past, on the basis that there's no longer any practical way to recover it. In both cases, the fundamental reason for the increase in entropy is the same.

Note that I'm not saying the box's entropy increases as it passes the event horizon. I'm actually saying that the box never crosses the event horizon, as seen from an outside point of view. That would take an infinite amount of time. However, the outside observer would very rapidly find the box very hard to see due to the red-shifting. At some point you, as the observer, might decide as an approximation that the box might as well have crossed the event horizon, since you basically can't detect it anymore. When you do this, your approximation has a higher entropy than the "real" black hole, and that's where the entropy increase comes from.

That might seem like a weird concept. But in fact all increases in entropy are due to approximations of one kind or another. In principle you could always reverse the velocities of every particle making up a system and watch it "run backwards in time" to its initial state (unscrambling an egg or whatever). So the information about the initial conditions is always still there. We just treat things as irreversible (i.e. information-destroying or entropy-producing) because it's a very useful approximation that helps us make predictions about macroscopic systems.

Of course, the observer falling in with the box of salt would not want to make the same approximation as the outside observer. It would be a bad approximation from the infalling observer's point of view, because she can still see the box perfectly clearly. (If it's a big enough black hole it won't even get torn apart.) But that's ok - although we often treat it as an observer-independent physical quantity, entropy is actually observer-dependent, even for everyday things like gases. See this rather wonderful paper by Edwin Jaynes. (Jaynes, E. T., 1992, `The Gibbs Paradox, ' in Maximum-Entropy and Bayesian Methods, G. Erickson, P. Neudorfer, and C. R. Smith (eds.), Kluwer, Dordrecht).

• +1: I don't think anyone would argue with this. The analogy to dissolving the salt is spot on. – Ron Maimon May 7 '12 at 17:11
• @RonMaimon well, Stephen Hawking might have argued with it a few years ago ;) – Nathaniel May 7 '12 at 18:29
• You seem to be saying that the box of salt reaches a state of maximal entropy as soon as it crosses the event horizon. An observer going across with it might disagree. An outside observer, even if he could witness the tearing apart of the box, would see it take an infinite amount of time, no? – spraff May 7 '12 at 19:27
• Of course, the observer falling in with the box of salt would not want to make the same approximation as the outside observer. It would be a bad approximation from the infalling observer's point of view, because she can still see the box perfectly clearly. (If it's a big enough black hole it won't even get torn apart.) But that's ok - although we often treat it as an observer-independent physical quantity, entropy is actually observer-dependent, even for everyday things like gases. See bayes.wustl.edu/etj/articles/gibbs.paradox.pdf – Nathaniel May 7 '12 at 20:42
• @Robotbugs fixed. (I'm not sure what happened there.) – Nathaniel Oct 27 '15 at 8:45

I'm not really sure of this, I seem to have read it somewhere. Comments appreciated.

Entropy of the box-black hole system need not stay the same. It can increase (remember, it isn't an isolated system--gravitational waves can exit).

IIRC, the entropy of a black hole is basically the "number of ways it can be formed to give the same mass/charge/L". A black hole that's just swallowed the box contains both box cases in its "number of ways". So, no-hair is preserved.

Just that entropy of the system increases by a different amount during both processes.

I think.

• yes, this is right too, but the difference is miniscule for this example, because powdering is a negligible entropy increase, and in some special cases where the surface would like to be in contact with other surface for a certain grain size it can be a decrease of entropy. The notion of entropy is atomic, and macroscopically visible disorder is negligible compared to atomic scale disorder. – Ron Maimon May 7 '12 at 18:41
• @Ron hmm, yep.. Didn't occur to me (though the question still holds no matter how small the difference is) – Manishearth May 7 '12 at 18:44

Nathaniel's answer is excellent, but I'll just mention an even more fundamental qualitative difference between the no-hair theorem and the statement that the surface area measures entropy. The former is a purely classical result of general relativity; roughly speaking, it refers to a situation where you've zoomed out enough that you can't see any quantum effects. The latter result applies to quantum (or really semiclassical) gravity; the black hole entropy formula contains a factor of $$\hbar$$.

So they're fundamentally different beasts: the Bekenstein-Hawking entropy of a black hole roughly counts the number of quantum microstates, while the no-hair theorem describes classical macrostates.