When $r_-=0$, $\Omega=0$ and $f=(r^2-r_+^2)/L^2$. A circle $S$ at fixed $v,r$ has circumference $2\pi r$. Choose its future null normals as
$$
\ell=\partial_v+\frac f2\partial_r,
\qquad
n=-\partial_r,
\qquad
\ell\mathbin\cdot n=-1.
$$
For a one-dimensional transverse surface, each <null expansion> is the logarithmic derivative of its length:
$$
\theta_{(\ell)}=\frac1r\ell(r)=\frac{f}{2r},
\qquad
\theta_{(n)}=\frac1r n(r)=-\frac1r.
$$
Inside the horizon, $0<r<r_+$ implies $f<0$, so both expansions are negative. Every such circle is therefore a <trapped surface>, specifically a <Trapped circle inside a nonrotating BTZ black hole>.

Solution

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

{c}

For the nonrotating BTZ metric in ingoing coordinates, future null normals to a circle of fixed $v,r$ may be chosen as $\ell=\partial_v+(f/2)\partial_r$ and $n=-\partial_r$. Their expansions are $\theta_{(\ell)}=f/(2r)$ and $\theta_{(n)}=-1/r$, so both are negative inside the outer horizon.


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 Incoming links: Trapped circle inside a nonrotating BTZ black hole