Define
$$
s=\sqrt{1-\frac{J^2}{M^4}}=\sqrt{1-\frac{a^2}{M^2}}.
$$
The <Bekenstein-Hawking entropy> and <Hawking temperature> become
$$
S_{BH}=\frac A4=2\pi Mr_+=2\pi M^2(1+s),
\qquad
T_H=\frac{s}{4\pi M(1+s)}.
$$
At fixed $J$,
$$
\frac{ds}{dM}=\frac{2(1-s^2)}{sM},
\qquad
\left.\frac{dT_H}{dM}\right|_J
=\frac{T_H}{M}\frac{2-2s-s^2}{s^2}.
$$
The fixed-$J$ first law gives $(\partial S_{BH}/\partial M)_J=1/T_H$, so
$$
C_J=T_H\left(\frac{\partial S_{BH}}{\partial T_H}\right)_J
=\left(\frac{\partial T_H}{\partial M}\right)_J^{-1}.
$$
Therefore the <Kerr black-hole heat capacity at fixed angular momentum> is
$$
\boxed{C_J=4\pi M^2
\frac{s(1+s)}{2-2s-s^2}}.
$$
At $a=0$, this correctly reduces to $C_J=-8\pi M^2$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 311 / 4 / c / iii / Solution

{c}
{title2=$C_J$}

Writing $s=\sqrt{1-J^2/M^4}=\sqrt{1-a^2/M^2}$, the heat capacity of a <Kerr black hole> at fixed angular momentum is
$$
C_J=4\pi M^2\frac{s(1+s)}{2-2s-s^2}.
$$
It is positive for $\sqrt{2\sqrt3-3}<|a|/M<1$, diverges at the lower endpoint, and tends to zero in the extremal limit.

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