# is there a way to split a black hole?

Classically, black holes can merge, becoming a single black hole with an horizon area greater than the sum of both merged components.

Is it thermodynamically / statistically possible to split a black hole in multiple black holes? If the sum of the areas of the product black holes would exceed the area of the original black hole, it seems to be a statistically favorable transition by the fact alone that would be a state with larger entropy than the initial state

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But the radius is proportional to mass, and surface area is proportional to radius squared. That means that entropy is proportional to mass squared. $(M_1+M_2)^2>M_1^2+M_2^2$ so splitting the black hole results in less area, not more. – Alan Rominger Nov 29 '12 at 18:24
right, which means that splitting a black hole requires at least an increase of entropy of $2 M_1 M_2$, or adding $\sqrt{2 M_1 M_2}$ energy – lurscher Nov 29 '12 at 18:45
We're still off by a sign difference. It requires an entropy decrease of $2 M_1 M_2$, since the equation I wrote was really saying $S^2 > S_1^2 + S_2^2$ if we take variables with no subscript to represent the black hole before splitting. – Alan Rominger Nov 29 '12 at 18:58
what i meant is that the missing entropy needs to be accounted by adding it to the splitted black holes, and as you noticed, the mass-energy-entropy strict relation implies it won't happen unless energy is added to the system – lurscher Nov 29 '12 at 19:00
Oh, gotcha. I see how the wording of the question allows that now. – Alan Rominger Nov 29 '12 at 19:03

## 2 Answers

The question asks for a black hole splitting such that "the product black holes would exceed the area of the original black hole".

In the above answer I have argued that to do so requires at least two black holes colliding.

However, the question continuous with the remark that such a splitting into black holes with larger horizon area "seems to be a statistically favorable transition by the fact alone that would be a state with larger entropy than the initial state". The edit in my answer above suggests this assertion to be correct.

However, this is not the case. To determine what is a statistically favorable transition requires a comparison between alternative results. If there is an outcome that can be realized in overwhelmingly more ways than any of the alternatives, that is the statistically favorable outcome.

Let's see how this works out for two colliding black holes. As an example we take two black holes of 4N Planck masses each. Let's consider two alternative scenarios:

A) 'splitting': 4N + 4N --> 6N + N + N

B) 'merging': 4N + 4N --> 8N

A black hole containing N Planck masses has entropy $S = 4\pi N^2$. Therefore, the initial state has total entropy $S = 128\pi N^2$ and can be realized in $e^S = e^{128\pi N^2}$ ways.

The end products from scenario A) has larger entropy ($S = 152\pi N^2$) and can be realized in $e^{152\pi N^2}$ ways. For large N this number is way larger than the number of realizations for the initial state. Yet, scenario A) does not represent the statistically favorable transition.

This is because scenario B) leads to entropy $S = 256\pi N^2$ encompassing overwhelmingly more microscopic states: $e^{256\pi N^2}$.

The conclusion is that although entropy-increasing black hole splitting reactions can be defined, these are not realizable from a statistical physics perspective.

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Thermodynamics forbids the splitting of a black hole in multiple smaller black holes. Reason being that the result of such a splitting would violate the first law of thermodynamics (energy conservation) and/or the second law of thermodynamics (entropy non-decrease).

If energy conservation is honored, the product would be multiple black holes with a sum of circumferences (sum of energy contents) equal to the circumference (energy content) of the original black hole. As a consequence, the sum of surface areas would be less than the surface area of the original black hole. As surface area corresponds to entropy, this would violate the second law of thermodynamics.

However, it is possible to split a black hole in multiple non black hole components. This is in fact easy: just sit back and let Hawking radiation do its thing. Key is that the resulting long-wavelength non black hole components are not localized enough to form small black holes.

What can happen though, is that this Hawking radiation gets captured by other black holes. This would effectively give a simple scenario for splitting a small black hole into components that feed multiple larger black holes (with much longer evaporation times). This is thermodynamically feasible, but probably not what OP has in mind.

If you interpret 'splitting' broadly and classify the latter scenario as 'black hole splitting', then lots of 'splitting processes' are thermodynamically allowed. For instance, you can in theory have two colliding black holes of three solar masses each, yielding three black holes, two of one solar mass and one of four solar masses: 3M + 3M --> 4M + M + M. Key is that the splitting of one black hole into two is not possible. You need an additional black hole participating in the process to ensure energy conservation and at the same time avoid entropy decrease.[/edit]

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That would be an explanation for why black holes won't spontaneously split, but the "lower entropy" could be compensated by something else (see the comments under the OP's question). In other words, can we get a reaction where: entropy of original hole = entropy of smaller holes PLUS entropy of 'something else' ? – Chris Gerig Dec 1 '12 at 6:47
Sure, but that 'something else' needs to be a localized source of entropy. Lots of it. In other words: it needs to be one (or more) black holes. For instance: starting with three equal mass black holes, there is no thermodynamics law that would forbid a reaction leading to two black holes of half the mass and one of double the mass. If you see that as a 'splitting process', yes it is possible (at least in principle). – Johannes Dec 1 '12 at 9:39
The problem is: nobody knows thermodynamics holds below event horizon or not. – Evil Angel Dec 7 '12 at 5:55
Thermodynamics holds for the whole observable universe. What is your observable universe depends on your position and your state of motion. But for each observer his/her whole observable universe stretches out over everything that he/she can observe at finite redshift. The boundaries where the redshifts diverge, we call horizons. Whatever might be behind these horizons is causally disconnected from the observer and utterly irrelevant to the physics he/she observes. Therefore it makes no sense to talk about "thermodynamics behind horizons". – Johannes Dec 7 '12 at 9:11