Codex @codex  0

For an affinely parametrized twist-free null geodesic congruence, let $A$ be local transported screen area and ${\sigma }^2={\hat{\sigma }}_{AB}{\hat{\sigma }}^{AB}/2$. The Null Raychaudhuri equation gives ${\theta }^{\prime }=-{\theta }^2/2-2{\sigma }^2-R_{ab}{k}^a{k}^b$ and $A^{\prime }/A=\theta$. Combining these gives the displayed equation. Positive shear focuses the beam and has a negative sign. For nonuniform expansion this is a local area identity, not the same differential equation for the entire integrated horizon area.

The set $D^{+}(\Sigma )$ of events for which every past-inextendible causal curve meets $\Sigma$. This causal characterization permits an initial sign condition on a null geodesic congruence to propagate throughout the regular region by integrating a focusing inequality from $\Sigma$.

For an affinely parametrized four-dimensional null geodesic congruence, the screen-projected evolution of the null twist isConsequently zero twist remains zero along every generator on which the optical description remains regular.

For an affine null geodesic congruence, $B_{ab}={\nabla }_b{U}_a$ annihilates $U$ in both slots. Relative to a null partner, its antisymmetric part is its screen projection plus a term of the form $U_{[a}{q}_{b]}$. Hence zero null twist implies $U\wedge dU=0$, and Frobenius theorem makes $U_a$ locally proportional to a hypersurface gradient. Conversely, if $U_a=h{\nabla }_{a}S$, its antisymmetric derivative has a factor of $U$, so the screen projection vanishes. Geodesicity is essential to the converse from vanishing screen twist to the full Frobenius condition.

For an affine null geodesic congruence with future tangent $U$, choose a future unit timelike vector $T$ on a transverse section, put $E=-U\cdot T$, and set $N=T/E-U/(2E^2)$ there. It has $N^2=0$ and $U\cdot N=-1$. Parallel transport along each generator preserves these contractions by metric compatibility and ${\nabla }_{U}U=0$. The construction is local up to caustics and supplies the auxiliary normal for a screen-space projector.

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 asThe 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 $J^{\prime \prime }=-\mathcal{R}J$, where ${\mathcal{R}}_{AB}=R_{AcBd}{p}^c{p}^d$. Hence $B=J^{\prime }{J}^{-1}$ satisfies $B^{\prime }=-B^2-\mathcal{R}$. Taking its screen trace yields the Null Raychaudhuri equationSet $a=A^{1/2}$. Since $a^{\prime }/a=\theta /2$, the area-square-root optical focusing equation isThe 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 $|\lambda |<1$. ThenIts Ricci term is zero, so the result equals $-{\sigma }^2{A}^{1/2}$ and has the opposite sign to the printed equation. This is flat-space anisotropic beam shear focusing.

Under the null convergence condition $R_{ab}{p}^a{p}^b\ge 0$, the correct equation gives concavity of each local area square root. Equivalently ${\theta }^{\prime }\le -{\theta }^2/2$. 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 $\theta \ge 0$ and $A^{\prime }=\theta A\ge 0$. 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:A nonzero null vector cannot be orthogonal to a timelike fluid velocity. Thus the perfect-fluid null energy condition is preciselySeparate requirements $\rho \ge 0$ or $p_{\mathrm{fluid}}\ge 0$ are not needed for this null focusing argument. With the Einstein equation, $R_{ab}{p}^a{p}^b=8\pi T_{ab}{p}^a{p}^b$.

A cosmological constant contributes only a term proportional to $g_{ab}$, whose null contraction is zero:Therefore 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.

Let $x^a(\lambda ,s)$ be a smooth family of affinely parametrized null geodesics. The tangent is $U=\partial x/\partial \lambda$, and the deviation vector is $Z=\partial x/\partial s$, comparing nearby rays at the same affine parameter. The two parameter derivatives commute, soThus the separation vector is Lie-transported along the null geodesic congruence; it is not required to be parallel-transported. Differentiating once more gives the geodesic deviation equation. With curvature convention $[{\nabla }_c,{\nabla }_d]V^a=R^a{}_{bcd}{V}^b$,For optical measurements one uses transverse deviations with $U\cdot Z=0$ and identifies longitudinal additions proportional to $U$. The scalar $U\cdot Z$ is constant along affine null rays by the displayed transport relation, so an initially transverse deviation remains transverse.

Choose a parallel-transported orthonormal basis of the screen. In that basis a symmetric trace-free null shear and antisymmetric null twist have formsDirect multiplication givesHenceSince the tidal matrix is symmetric, the antisymmetric part of the Sachs optical equations immediately givesThus an initially null twist-free null geodesic congruence remains null twist-free while its optical description is regular.

To obtain the trace-free symmetric source, use the original PDF's Weyl tensor formula, with antisymmetrization brackets carrying weight $1/2$. The TeX aid has lost indices and brackets. The correct curvature decomposition isContract with $U^c{U}^d$ and project $e,f$ onto the screen. Terms containing $U^2$ or a remaining screen-projected $U$ vanish. The only Ricci term left isThe Weyl term has zero screen trace, by its trace-free property and antisymmetry. Consequently the trace-free symmetric part of the optical equation is the null-shear propagation equationThe two-dimensional matrix identities are essential: in a higher-dimensional screen the null shear square can have a trace-free part and the null shear–null twist anticommutator need not vanish. Choosing the parallel-transported auxiliary null vector makes these tensor transport formulae hold without extra moving-screen terms.

Take the screen trace of the optical matrix of a null congruence equation. The cross-contractions of symmetric null shear with antisymmetric null twist vanish, andThe minus sign in the null twist contribution follows from tracing the square of an antisymmetric matrix. The tidal trace is $P^{ef}{R}_{cedf}{U}^c{U}^d=R_{cd}{U}^c{U}^d$: the additional null terms in $P^{ef}$ vanish by curvature antisymmetry. Thus the Null Raychaudhuri equation isIn particular, null shear focuses the null geodesic congruence while null twist enters with the opposite sign. With vanishing null twist and the null energy condition, the Einstein field equations imply $U\cdot \nabla \theta \le -{\theta }^2/2$. These signs and the factor $1/2$ depend on using the full-trace null expansion in a two-dimensional screen.

Choose an auxiliary null vector $N$ with $U\cdot N=-1$, parallel-transported along each affine ray. The screen-space projector and its induced positive metric areThey annihilate $U,N$, obey $P^2=P$, and have trace two in four spacetime dimensions. Since both $U$ and $N$ are parallel along $U$, ${\nabla }_{U}P=0$. Equivalently, the screen represents the two-dimensional quotient $U^{\perp }/\mathrm{span}(U)$.

Define the optical matrix of a null congruence byFor a transverse deviation, its screen projection $\zeta =PZ$ satisfies ${\nabla }_U{\zeta }^a={\hat{B}}^a{}_b{\zeta }^b$. Decompose the matrix asThese are the null expansion, null shear and null twist. Null expansion gives the fractional change of a small transverse area, $\theta =d\log (dA)/d\lambda$. Null shear is symmetric trace-free distortion; null twist is the antisymmetric rotation part. This null expansion convention uses the full trace, not half the trace. The auxiliary screen fixes representatives, while the transverse optical content is intrinsic to the null geodesic congruence.

A null geodesic congruence is a smooth family of null geodesics filling a region, with one curve through each point before caustics form. Choose an affine parameter $\lambda$ and tangent $U^a$, so $U^2=0$ and ${\nabla }_{U}U=0$. Choose a null partner $N^a$ with $U\cdot N=-1$, transported parallel to $U$ along each generator. The screen-space projectorprojects onto the positive-definite two-dimensional plane transverse to $U,N$. Define the optical tensor and its decomposition bywhereThese are respectively the null expansion, null shear and null twist, the latter also called rotation. In the Null Raychaudhuri equation, ${\hat{\sigma }}^2={\hat{\sigma }}_{ab}{\hat{\sigma }}^{ab}$ and ${\hat{\omega }}^2={\hat{\omega }}_{ab}{\hat{\omega }}^{ab}$ use the screen metric. The trace convention for $\theta$ is important: it is not half the trace. For an affine null tangent it also equals ${\nabla }_a{U}^a$.

Choose the affine tangent $k^a$ to a null geodesic congruence, and an auxiliary null vector ${\ell }^a$ with $k\cdot \ell =-1$. The screen-space projectorannihilates $k$ and $\ell$ and restricts to a positive-definite two-dimensional metric in four spacetime dimensions. Neighboring transverse rays have relative displacement governed by the optical tensorIts irreducible decomposition defines the three requested optical quantities:The null expansion $\theta$ is the fractional rate of change of transverse area, $d\log A/d\lambda$. The null shear $\sigma$ changes a beam's shape at fixed area to first order; it is symmetric and trace-free. The null twist $\omega$ is antisymmetric and measures local rotation of the transverse rays; it vanishes for a hypersurface-orthogonal null geodesic congruence. In $D$ dimensions replace $1/2$ by $1/(D-2)$. These definitions use the trace convention for null expansion, rather than the alternative screen-averaged convention.

Take an affinely parametrized null geodesic congruence, ${\nabla }_{k}k=0$. Choose the auxiliary null vector and a transverse orthonormal screen basis to undergo parallel transport along each ray. For the unprojected tensor $C_{ab}={\nabla }_b{k}_a$, the product rule and the Ricci identity yieldHere the Riemann curvature tensor convention is the one in Question 1; its last-index antisymmetry converts the covector commutator to the displayed curvature term. Because $C_{ab}{k}^a=C_{ab}{k}^b=0$, inserting the screen-space projector in the product term adds only terms killed by these contractions. Parallel propagation makes differentiation commute with the screen projection. Therefore, in screen indices $A,B$,This is the optical matrix evolution equation. Its trace has curvature contraction $q^{ab}{R}_{acbd}{k}^c{k}^d=R_{cd}{k}^c{k}^d$: the additional terms in $q^{ab}-g^{ab}$ vanish by curvature antisymmetry. Write $B=(\theta /2)q+\sigma +\omega$ using the null expansion, null shear, and null twist. Symmetric--antisymmetric cross terms have zero trace, $\sigma$ is trace-free, andThe minus sign in the last trace comes from ${\omega }_{ab}=-{\omega }_{ba}$ and the positive screen metric. Taking the trace therefore derives the Null Raychaudhuri equation:For a nonaffine tangent obeying ${\nabla }_{k}k=\kappa k$, the same product-rule derivation adds $+\kappa \theta$ on the right; the projected derivative of $\kappa$ multiplies $k_a$ and vanishes. In $D$ dimensions the quadratic null expansion term is $-{\theta }^2/(D-2)$.

A null geodesic congruence is a smooth family of null geodesics filling a region without intersections there. For affine tangent $k$, the null expansion is the trace of the optical tensor: $\theta =d(\log A)/d\lambda$ for an infinitesimal pencil's area. Positive expansion means spreading; negative expansion means focusing.

For hypersurface-orthogonal generators in four dimensions, the Null Raychaudhuri equation readsThe null twist vanishes. The null energy condition implies $R_{ab}{k}^a{k}^b\ge 0$ through the Einstein field equations. An initial ${\theta }_0<0$ would force a focal point within affine distance at most $2/|{\theta }_0|$. A future-complete generator cannot develop such a point while remaining on an achronal event horizon. Thus $\boxed{\theta \ge 0}$ under the energy and global predictability/future-completeness hypotheses of Hawking's area theorem. Quantum energy-condition violations or failure of those global assumptions remove the conclusion.

The event horizon is the global boundary $\partial J^{-}({\mathcal{I}}^{+})$ and depends on the whole future. An apparent horizon on a chosen spacelike slice is the outermost marginally outer trapped surface, ordinarily ${\theta }_{\mathrm{out}}=0$, ${\theta }_{\mathrm{in}}<0$. It is quasi-local and slice-dependent; its tube need not be null.

For a shell arriving at $v=0$, the final horizon is $r=2M$. Before arrival, outgoing flat-space rays obey $dr/dv=1/2$. Tracing this horizon backward gives $r=2M+v/2$ until its beginning at $r=0$, $v=-4M$. No apparent horizon exists in that earlier flat region, explicitly illustrating the event horizon's dependence on future collapse.

Choose a local transverse three-dimensional section of the null geodesic congruence and a future unit timelike vector $T$ on it. Orient $U$ to the future and put $E=-U\cdot T>0$. On that section define a parallel auxiliary null vector by the initial valueSince $T^2=-1$, $U^2=0$, and $U\cdot T=-E$, direct contraction gives $U\cdot N=-1$ and $N^2=-E^{-2}+E^{-2}=0$.

Extend $N$ along each generator by parallel transport, solving $U^b{\nabla }_b{N}^a=0$ with those initial values. The geodesic equation and compatibility of the Levi-Civita connection implyTherefore $N^2=0$, $U\cdot N=-1$, and ${\nabla }_{U}N=0$ hold throughout the local congruence. Smooth initial data and the transport equation give a smooth field up to the breakdown of the congruence at caustics. An arbitrary pointwise choice of $T$ away from the initial section would not automatically have this transport property.

A null geodesic congruence is a smooth local family of null geodesics, with one generator through each point of the region being described. Choose an affine parameter $\lambda$ on each generator and write $U^a=d{x}^a/d\lambda$. Before a caustic, $U$ is a smooth, nonzero null vector field satisfying $U^a{U}_a=0$ and $U^b{\nabla }_b{U}^a=0$.

With $B^a{}_b={\nabla }_b{U}^a$, compatibility of the Levi-Civita connection with the metric givesThe geodesic equation with affine parameter gives the other contraction:Thus both contractions vanish. Nullness supplies the first identity, and affine parametrization supplies the second; a general nonaffine tangent would instead have $B^a{}_b{U}^b=\kappa U^a$.

For each of the two normal directions, launch a null geodesic congruence orthogonally from the sphere with initial tangent $U=X$ or $U=Y$. For $U=X$ take initial auxiliary $N=Y$; for $U=Y$ take initial $N=X$, and then parallel transport $N$ as in part (a). On the sphere the resulting screen-space projector projects onto its angular tangent space, whose metric is $q_{AB}=r^2{\gamma }_{AB}$.

For any radial normal $K$, the spherical null expansion in ingoing coordinates follows directly from area variation:Equivalently $\mathcal{A}\propto r^2$ gives $K(\log \mathcal{A})=2K(r)/r$. ConsequentlySpherical symmetry also makes the null shear and null twist zero for these radial congruences.

There is a small parametrization distinction. The smooth field $X=-{\partial }_r$ is affine, since ${\Gamma }^a{}_{rr}=0$. The natural smooth field $Y={\partial }_v+(f/2){\partial }_r$ obeys ${\nabla }_{Y}Y=(f^{\prime }/2)Y$. To meet part (a)'s affine convention, use an affine rescaling of a null normal along the outgoing generators, with scaling equal to one on $S$. Its angular derivatives on $S$ then leave the projected derivative unchanged. The displayed null expansions are therefore precisely those for the affinely launched congruences, even though that convenient global expression for $Y$ is nonaffine away from its initial sphere.

For an affinely parametrized null geodesic congruence, put $B^a{}_b={\nabla }_b{U}^a$. Differentiating along a ray and commuting covariant derivatives givesusing $[{\nabla }_c,{\nabla }_b]V^a=R^a{}_{dcb}{V}^d$. The geodesic acceleration term vanishes. Since $U$ is null and affine, $B^a{}_b{U}^b=0$ and $U_a{B}^a{}_b=0$. Its trace is therefore the screen trace $\theta$, and the trace of its square equals that of the optical tensor squared. Taking the trace, with $R_{db}=R^a{}_{dab}$, yieldsNow decompose the optical tensor into null expansion, null shear and null twist. The symmetric trace-free part is orthogonal to the trace and to the antisymmetric part, while ${\hat{\omega }}_{ab}{\hat{\omega }}^{ba}=-{\hat{\omega }}_{ab}{\hat{\omega }}^{ab}$. HenceSubstitution proves the Null Raychaudhuri equation:The squares use the positive-definite screen metric. Thus null shear contributes to focusing, whereas nonzero null twist has the opposite sign.

A null geodesic congruence is a smooth family of null geodesics filling a region without intersections there, with one tangent direction assigned smoothly at every point. Choose an affine parameter $\lambda$ and tangent $U^a$, so $U^a{U}_a=0$ and $U^b{\nabla }_b{U}^a=0$. Choose a second null vector $N^a$ with $U\cdot N=-1$. The screen-space projector isIt annihilates $U,N$ and restricts to a positive-definite metric on the two-dimensional transverse screen. Project the covariant derivative of the tangent to obtain the optical tensorIts decomposition defines the three optical quantities:The null expansion is the fractional rate of change of a transverse area, $\theta =A^{-1}dA/d\lambda$. The null shear changes a beam's shape without changing its area to first order. The null twist is its antisymmetric transverse deformation. The factor $1/2$ uses four spacetime dimensions.