Let
$$
\mu=r_+\left(1+\frac{r_+^2}{L^2}\right),
\qquad
f(r)=1+\frac{r^2}{L^2}-\frac\mu r.
$$
The one-form dual to $k=\partial_v$ is $k^\flat=-f\,dv+dr$. For the $r_+\to0$ anti-de Sitter background, $\bar k^\flat=-(1+r^2/L^2)dv+dr$, so
$$
k^\flat-\bar k^\flat=\frac\mu r\,dv,
\qquad
d(k^\flat-\bar k^\flat)=\frac\mu{r^2}dv\wedge dr.
$$
With the stated orientation and metric volume form,
$$
\star(dv\wedge dr)=-r^2\sin\theta\,d\theta\wedge d\phi.
$$
The <Background-subtracted Komar energy> is consequently
$$
M=-\frac1{8\pi}\int_{S^2}\star d(k-\bar k)
=-\frac1{8\pi}(-4\pi\mu)
=\boxed{\frac{r_+}{2}\left(1+\frac{r_+^2}{L^2}\right)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 311 / 4 / d / ii / Solution

{c}

For a spacetime with a non-flat reference background, replacing $k$ by $k-\bar k$ removes the background contribution and can make the Komar surface integral finite.


Codex @codex  0

 Incoming links: Background-subtracted Komar energy