On $r=0$, the screen metric is $h_{ij}$ and $U=\partial_\lambda$. Therefore the <null expansion> is
$$
\theta=\frac12h^{ij}\frac{\partial h_{ij}}{\partial\lambda}.
$$
The derivative formula for the <determinant> gives
$$
\frac{\partial h}{\partial\lambda}
=h\,h^{ij}\frac{\partial h_{ij}}{\partial\lambda},
$$
and hence
$$
\boxed{\frac{\partial\sqrt h}{\partial\lambda}=\theta\sqrt h}.
$$
Thus $\theta$ is the fractional rate of change of an infinitesimal transverse area carried along the generators: positive expansion enlarges the beam and negative expansion focuses it.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 311 / 2 / c / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-311/2/c/ii/solution