= Solution
A <null geodesic congruence> is a smooth family of <null geodesics> filling a region without crossings there. Choose a smooth future-directed tangent $U^a$ and an <affine parameter> $\lambda$, so
$$
U^aU_a=0,\qquad U^b\nabla_bU^a=0.
$$
The space orthogonal to $U$ is degenerate because it contains $U$ itself. To measure genuine transverse separation, introduce an auxiliary <null vector> $N^a$ with $U\cdot N=-1$ and the <screen-space projector>
$$
q_{ab}=g_{ab}+U_aN_b+N_aU_b.
$$
In four dimensions $q$ has a positive-definite two-dimensional image and annihilates $U$ and $N$. The <optical tensor> is the transverse part of the velocity <derivative>,
$$
\widehat B_{ab}=q_a{}^cq_b{}^d\nabla_dU_c.
$$
Its irreducible decomposition defines the three optical quantities:
$$
\boxed{\theta=q^{ab}\widehat B_{ab},\qquad
\widehat\sigma_{ab}=\widehat B_{(ab)}-\frac12\theta q_{ab},\qquad
\widehat\omega_{ab}=\widehat B_{[ab]}.}
$$
The <null expansion> is the fractional rate of change of a small transverse area, $\theta=d\log\delta A/d\lambda$. The symmetric trace-free <null shear> distorts the shape of the transverse beam without changing its area at first order. The antisymmetric <null twist>, also called rotation of the congruence, describes the rotational part of its transverse deformation. These are properties of the screen quotient $U^\perp/\langle U\rangle$; the auxiliary vector chooses a convenient representative of that quotient. For an affinely parametrized <null geodesic congruence>, the contraction constraints also give $\theta=\nabla_aU^a$.
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