The two initial <Kerr black holes> have total area
$$
A_i=16\pi\left(M^2+\sqrt{M^4-J^2}\right),
$$
whereas the final <Schwarzschild black hole> has $A_f=16\pi M'^2$. <Hawking's area theorem> implies
$$
M'^2\geq M^2+\sqrt{M^4-J^2}.
$$
Since $\eta=1-M'/(2M)$, the radiated fraction is bounded by
$$
\boxed{\eta\leq
1-\frac12\sqrt{1+\sqrt{1-\left(\frac{J}{M^2}\right)^2}}}.
$$
This upper bound increases with $|J|/M^2$ and reaches
$$
\boxed{\eta_{\rm max}=\frac12}
$$
for two initially extremal holes, $|J|=M^2$. For the final hole to be Schwarzschild, their spins must be oppositely directed so the total angular momentum vanishes. The bound assumes an idealized merger saturating the area law.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 311 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-311/3/c/solution