= Solution
Interpret $A$ first as an infinitesimal screen area transported along generators of a future <event horizon>, not as the total area of an arbitrary trapped surface. Horizon generators form an affinely parametrized, hypersurface-orthogonal <null geodesic congruence>. The two-dimensional screen metric is positive definite. Its optical tensor has zero twist and decomposes as
$$
B_{AB}=\frac12\theta h_{AB}+\widehat\sigma_{AB},\qquad
\theta=\frac{A'}A,\qquad
\sigma^2=\frac12\widehat\sigma_{AB}\widehat\sigma^{AB}\geq0.
$$
The scalar here agrees with the supplied derivative definition: nullness and the affine geodesic equation remove the longitudinal terms in the contracted norm, leaving $B_{AB}B^{AB}-\theta^2/2$.
To derive the focusing equation, vary neighbouring geodesics and let $J$ be their screen Jacobi map. The <geodesic deviation> equation is $J''=-\mathcal R J$, where $\mathcal R_{AB}=R_{AcBd}p^cp^d$. Hence $B=J'J^{-1}$ satisfies $B'=-B^2-\mathcal R$. Taking its screen trace yields the <Null Raychaudhuri equation>
$$
\theta'=-\frac12\theta^2-2\sigma^2-R_{ab}p^ap^b.
$$
Set $a=A^{1/2}$. Since $a'/a=\theta/2$, the <area-square-root optical focusing equation> is
$$
\boxed{\frac{a''}a=\frac12\theta'+\frac14\theta^2
=-\frac12R_{ab}p^ap^b-\sigma^2.}
$$
The PDF's plus sign before $\sigma^2$ is inconsistent with its own positive shear definition. This is an actual source error, not an antisymmetrization or curvature-sign convention.
For an explicit check, take a flat-space twist-free beam with screen Jacobi factors $1+\lambda$ and $1-\lambda$, for $|\lambda|<1$. Then
$$
A=1-\lambda^2,\qquad
\sigma^2=(1-\lambda^2)^{-2},\qquad
(A^{1/2})''=-(1-\lambda^2)^{-3/2}.
$$
Its Ricci term is zero, so the result equals $-\sigma^2A^{1/2}$ and has the opposite sign to the printed equation. This is <flat-space anisotropic beam shear focusing>.
Under the <null convergence condition> $R_{ab}p^ap^b\geq0$, the correct equation gives concavity of each local area square root. Equivalently $\theta'\leq-\theta^2/2$. If a horizon generator had $\theta(\lambda_0)=\theta_0<0$, integration gives a focal point within affine distance at most $2/|\theta_0|$. Such a generator could not remain on an <achronal boundary> beyond the focal point. Under the usual global hypotheses that horizon generators remain regular and future complete, this contradicts their being generators of the <event horizon>. Hence $\theta\geq0$ and $A'=\theta A\geq0$. Integrating over the horizon patches, and including newly joining generators which can add area, proves <Hawking's area theorem>. Global predictability/completeness assumptions are essential; the local differential inequality alone is not a theorem about arbitrary trapped-surface areas.
For the <perfect fluid> stress tensor, null contraction eliminates the pressure-metric term:
$$
T_{ab}p^ap^b=(\rho+p_{\rm fluid})(u_ap^a)^2.
$$
A nonzero null vector cannot be orthogonal to a timelike fluid velocity. Thus the <perfect-fluid null energy condition> is precisely
$$
\boxed{\rho+p_{\rm fluid}\geq0.}
$$
Separate requirements $\rho\geq0$ or $p_{\rm fluid}\geq0$ are not needed for this null focusing argument. With the Einstein equation, $R_{ab}p^ap^b=8\pi T_{ab}p^ap^b$.
A <cosmological constant> contributes only a term proportional to $g_{ab}$, whose null contraction is zero:
$$
(G_{ab}+\Lambda g_{ab})p^ap^b=R_{ab}p^ap^b.
$$
Therefore \b[a nonzero <cosmological constant> does not change the fluid condition or the local area-focusing argument], though it can change the global asymptotics and which horizons and completeness hypotheses are appropriate.
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