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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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  • $\begingroup$ let us continue this discussion in chat $\endgroup$ – Alan Rominger Nov 29 '12 at 19:05
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    $\begingroup$ Thermodynamics aside, it's classically impossible for a black hole to split into two, as a consequence of the causal structure of a black hole region. See theorem 12.2.1 on page 308 of Wald's 'General Relativity'. $\endgroup$ – gj255 Oct 26 '16 at 10:35
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I) Let us choose units where $c=1=G$ for simplicity. Recall that a Kerr-Newman black-hole with mass $M > 0$, charge $Q\in [-M,M]$, and angular momentum $J\geq 0$, has surface area given by

$$\frac{A}{4\pi}~:=~ r^2_+ +a^2~=~ M^2+ \delta + 2M \sqrt{\Delta}, \tag{1}$$

where

$$ r_+~:=~M+\sqrt{\Delta}, \qquad \Delta~:=~ \delta -a^2~\geq~0,\qquad \delta~:=~M^2-Q^2~\geq~0, \qquad a~:=~\frac{J}{M}.\tag{2}$$

The entropy

$$S~=~\frac{k_B}{\ell_P^2}\frac{A}{4}\tag{3}$$

is proportional to the area $A$.

II) An interesting question asks the following.

If we merge $n$ Kerr-Newman black-holes
$$(M_i>0, Q_i, J_i),\qquad i\in\{1,\ldots,n\},\tag{4}$$ into one Kerr-Newman black-hole $(M>0,Q,J)$, such that mass and charge are conserved$^1$ $$ M~=~\sum_i M_i , \qquad Q~=~\sum_i Q_i , \qquad J~\leq ~\sum_i J_i, \tag{5}$$ and the angular momentum satisfies the triangle inequality; would the discriminant $$ \Delta~\geq~0 \tag{6}$$ for the merged black hole be non-negative, and would the Kerr-Newman area formula (1) respect the second law of thermodynamics $$ A~>~ \sum_i A_i~? \tag{7}$$

The answer is in both cases Yes! The ineq. (7) in turn shows that the opposite splitting process is impossible, cf. OP's question.

Proof of ineqs. (6) & (7): First note that

$$ \delta~\stackrel{(2)}{=}~(M+Q)(M-Q)~ \stackrel{(5)}{=}~\sum_i(M_i+Q_i)(M_i-Q_i) +\sum_{i\neq j}(M_i+Q_i)(M_j-Q_j)$$ $$~\stackrel{(2)}{\geq}~\sum_i(M_i+Q_i)(M_i-Q_i) ~\stackrel{(2)}{=}~ \sum_i \delta_i , \tag{8}$$

and hence

$$ \frac{\delta}{2}~\stackrel{(8)}{\geq}~\frac{\delta_i +\delta_j}{2}~\geq~ \sqrt{\delta_i \delta_j}, \tag{9}$$

due to the ineq. of arithmetic & geometric means. Next consider

$$ M^2\Delta - \left(\sum_i M_i\sqrt{\Delta_i}\right)^2 ~\stackrel{(2)}{=}~(M^2\delta -J^2) - \sum_i M_i^2\Delta_i - \sum_{i\neq j}M_i\sqrt{\Delta_i}M_j\sqrt{\Delta_j} $$ $$~\stackrel{(2)+(5)}{\geq}~\left(\delta \sum_i M_i^2 + \delta\sum_{i\neq j} M_iM_j -J^2\right) - \sum_i (M_i^2\delta_i -J^2_i) - \sum_{i\neq j}M_i\sqrt{\delta_i}M_j\sqrt{\delta_j} $$ $$~\stackrel{(8)+(9)}{\geq}~ \sum_{i\neq j}M_i\sqrt{\delta_i}M_j\sqrt{\delta_j} +\sum_i J^2_i -J^2 ~\stackrel{(2)}{\geq}~ \sum_{i\neq j}J_iJ_j +\sum_i J^2_i -J^2 $$ $$~=~\left(\sum_i J_i\right)^2 -J^2 ~\stackrel{(5)}{\geq}~0. \tag{10}$$

Ineq. (10) implies ineq. (6) and

$$ M\sqrt{\Delta} ~\stackrel{(10)}{\geq}~ \sum_i M_i\sqrt{\Delta_i}. \tag{11}$$

Together with

$$ M^2 ~>~ \sum_i M^2_i,\tag{12} $$

eqs. (8) & (11) yield ineq. (7). $\Box$

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$^1$ We assume that the system can be treated as isolated. In particular, we ignore outgoing gravitational radiation. As we know from recent gravitational wave detections, this assumption is violated in practice for black hole merges. However, for the opposite hypothetical splitting process, which OP asks about, this is a reasonable assumption.

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  • $\begingroup$ Best answer. Precise and correct. $\endgroup$ – Physics Guy Oct 28 '16 at 16:40
  • $\begingroup$ I do not think that this answers the question. Is there a way to split? implies a sequence of actions performed by experimenter, not a spontaneous disintegration. There may be a sequence of reasonably simple actions: 1) charge BH to $Q=M/2$ 2) Irradiate BH with EM wave of wavelength, intensity … which would split the BH but that are not covered in your argument, $\endgroup$ – A.V.S. May 21 '18 at 17:06
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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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  • $\begingroup$ Which is probably why repeated black mergers (early in the universe's lifetime) end up up forming ever larger black holes - makes one wonder if this was how intermediate mass black holes or even super-massive black holes formed. $\endgroup$ – dualredlaugh Oct 27 '16 at 22:26
  • $\begingroup$ In the above answer should be In the below answer. $\endgroup$ – Ruslan Oct 28 '16 at 16:28
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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.

[edit]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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  • $\begingroup$ 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' ? $\endgroup$ – Chris Gerig Dec 1 '12 at 6:47
  • $\begingroup$ 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). $\endgroup$ – Johannes Dec 1 '12 at 9:39
  • $\begingroup$ The problem is: nobody knows thermodynamics holds below event horizon or not. $\endgroup$ – Schrödinger's Cat Dec 7 '12 at 5:55
  • $\begingroup$ 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". $\endgroup$ – Johannes Dec 7 '12 at 9:11
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Setting the principle of entropy aside for the moment, I believe it should be logical for a black hole to be able to split apart. The energy added to the system would simply have to be larger than the total sum energy released by all that matter which entered the black hole.

Actually, the inside of a black hole must be very hot. Think about it. All that mass that accelerated inward as the black hole coalesced, with nowhere for that energy to escape. The logical end conclusion is that it doesn't take much energy to skim off matter from the surface of a black hole, relatively speaking of course. (Of course, as matter is skimmed off there would be adiabatic cooling of the inside, analogous to what happens with liquid nitrogen boiling off and keeping the remaining liquid cold)

Here's a thought experiment. Imagine two extremely heavy spherical objects. Their sum total mass is enough to create a black hole as soon as they meet. As they approach each other they begin moving faster and faster, finally at close contact approaching light speed. Well there is an unusual effect here. As mass is transferred to energy, the rest masses of each individual object decrease, even though the net mass of the two-body system does not change. So rest mass is decreasing while velocity is increasing. You can see that it approaches light speed exponentially. However, here's the thing: You have all that energy, yet if some of the mass were to move away from the center of the black hole it would gain more mass. In other words, the traditional escape velocity equation does not apply.

There's a saying from Newton, "What goes up must come down". Well, when it comes to a black hole one could say, in a sense, that "What goes down must come up". All the energy converted from mass when the object fell in is also enough to bring the object back out.

It's commonly repeated, and assumed as truth, that light cannot escape a black hole, but that's not entirely true. It just becomes red-shifted "out of existence" (but not literally out of existence), or the ray of light does not leave at a narrow enough angle to being perpendicular to the the black hole's tangent, so it gets bent back.

Now this is all on the most basic theoretical level and doesn't take into account energy loss through gravitational waves, which could very well make the temperature of a black hole much cooler than classical mechanics would predict if it had resulted from the collision of two smaller black holes.

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  • $\begingroup$ There are numerous errors in this answer. $\endgroup$ – PM 2Ring Aug 29 '18 at 17:40

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