thermodynamics - 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

Answer

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). $\Box$

--

$^1$ 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.