Define <Ingoing BTZ coordinates> by
$$
dv=dt+\frac{dr}{f(r)},
\qquad
d\chi=d\phi+\frac{\Omega(r)}{f(r)}\,dr.
$$
Thus $dt=dv-dr/f$ and $d\phi=d\chi-\Omega\,dr/f$, so the potentially singular terms cancel:
$$
d\phi-\Omega dt=d\chi-\Omega dv,
$$
and
$$
-f\left(dv-\frac{dr}{f}\right)^2+\frac{dr^2}{f}
=-f\,dv^2+2\,dv\,dr.
$$
The metric becomes
$$
\boxed{ds^2=-f(r)dv^2+2\,dv\,dr
+r^2[d\chi-\Omega(r)dv]^2}.
$$
For $r_+>r_-$, antiderivatives may be chosen as
$$
v=t+\frac{L^2}{2(r_+^2-r_-^2)}
\left[r_+\log\left|\frac{r-r_+}{r+r_+}\right|
-r_-\log\left|\frac{r-r_-}{r+r_-}\right|\right]
$$
and
$$
\chi=\phi+\frac{L}{2(r_+^2-r_-^2)}
\left[r_-\log\left|\frac{r-r_+}{r+r_+}\right|
-r_+\log\left|\frac{r-r_-}{r+r_-}\right|\right],
$$
with the extremal case obtained by taking the limit. The transformed metric contains neither $1/f$ nor any other singular coefficient at $r=r_+$; since $f$, $\Omega$, and $r^2$ are analytic there, it gives an analytic extension across the outer horizon.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 311 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-311/1/b/solution