Locally write the <null hypersurface> as a level set $u=0$ and its normal as $n_a=\nabla_a u$. Since a null normal is also tangent, $n^a$ lies within the hypersurface. The symmetry of the <Levi-Civita connection> gives
$$
n^b\nabla_bn_a=n^b\nabla_an_b
=\frac12\nabla_a(n^bn_b).
$$
The scalar $n^2$ vanishes on the hypersurface, so its gradient there is normal and hence proportional to $n_a$:
$$
n^b\nabla_bn_a=\kappa n_a.
$$
This is the nonaffine <geodesic equation>. A rescaling of $n$ removes $\kappa$, proving that the <null hypersurface normal generates null geodesics> lying in the hypersurface.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 311 / 2 / a / Solution

If $n_a=\nabla_a u$ is normal to a null level set, then on the hypersurface
$$
n^b\nabla_bn_a=\frac12\nabla_a(n^2)=\kappa n_a.
$$
Thus the normal is tangent to null pregeodesics and can be rescaled to give an affine null tangent.


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