Is there a way to split a black hole?

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.

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.

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]