Let $k$ generate the <Killing horizon> and let $\ell=fk$ be an affinely parametrized generator wherever $k\ne0$. Choose an auxiliary null vector $n$ with $n\cdot\ell=-1$ and the <screen-space projector> $q_{ab}=g_{ab}+\ell_an_b+n_a\ell_b$. The <optical tensor> is the screen projection $B_{ab}=q_a{}^cq_b{}^d\nabla_d\ell_c$.

Since the projector annihilates $k$, the symmetric part is
$$
B_{(ab)}=f q_a{}^cq_b{}^d\nabla_{(d}k_{c)}=0
$$
by the <Killing equation>. Its trace is the <null expansion> and its trace-free symmetric part is the <null shear>, so both vanish. Also $\ell$ is normal to the null hypersurface. <Frobenius theorem> gives $\ell_{[a}\nabla_b\ell_{c]}=0$, which implies $q_a{}^cq_b{}^d\nabla_{[d}\ell_{c]}=0$: the <null twist> vanishes. These arguments use derivatives tangent to the horizon and hence apply to any smooth <null geodesic congruence> containing its generators, independently of its extension off the horizon. At a regular <bifurcation surface>, use the smooth affine generator and continuity. Thus
$$
\boxed{\theta=0,\qquad\hat\sigma_{ab}=0,\qquad\hat\omega_{ab}=0\quad\text{on the horizon}.}
$$
This is the vanishing of the <optical scalars on a Killing horizon>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 311 / 2 / a / Solution

{title2=$\theta=\hat\sigma=\hat\omega=0$}

The <null expansion>, <null shear> and <null twist> of a congruence tangent to the generators of a <Killing horizon> all vanish there. If $k$ is its Killing generator and $\ell=fk$ is an affine generator, the <screen-space projector> kills every term proportional to $k$. The symmetric part of the projected derivative of $\ell$ therefore vanishes by the <Killing equation>. The antisymmetric part vanishes by <Frobenius theorem>, since $\ell$ is hypersurface-orthogonal. These conclusions extend through a regular <bifurcation surface> by continuity.

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 Incoming links: Optical scalars on a Killing horizon