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If a null vector tangent satisfies ${\nabla }_{K}K=\kappa K$, then $U=hK$ is affine when $K(\log h)=-\kappa$. Choose $h=1$ on the initial transverse surface. The screen-space projector kills derivative terms proportional to $K$, so the initial optical tensor and its null expansion agree with those of $K$. In ingoing spherical coordinates, $K={\partial }_v+(f/2){\partial }_r$ has $\kappa =f^{\prime }/2$.

A nonzero Lorentz four-vector orthogonal to a timelike vector is spacelike: in the latter's rest frame the former has zero temporal component. Orthogonality alone does not ensure one timelike and one spacelike vector. Two orthogonal spatial coordinate vectors are both spacelike, and parallel null vectors are also orthogonal.

For metric tensors related by $\tilde{g}=e^{2\omega }g$, differentiating the rescaled metric tensor in the Christoffel symbol formula gives the displayed difference, where ${\omega }_a={\partial }_a\omega$ and the index on ${\omega }^a$ is raised with $g$. Consequently ${\tilde{\nabla }}_{V}V={\nabla }_{V}V+2V(\omega )V-g(V,V){\mathrm{grad}}_g\omega$. The last term vanishes on a null vector, proving conformal preservation of null geodesic paths; for a non-null geodesic its orthogonal component generally changes the path.

Contract the Einstein field equations with a null vector. Both the trace term and the cosmological constant term multiply $g_{ab}{\ell }^a{\ell }^b=0$ and disappear. The null energy condition therefore still implies the null convergence condition, independently of the sign of the cosmological constant. A black-hole area theorem still needs the appropriate global horizon assumptions as well as this local focusing condition.

A nonzero Hermitian rank-one positive semidefinite matrix has zero determinant and positive trace, representing a future-directed null vector. It factors as $\chi {\chi }^{†}$, with an irrelevant spinor phase. A past-directed null vector instead requires a minus sign; a positive outer product alone does not cover both time directions.

A smooth manifold carrying a nondegenerate bilinear form on each tangent space, with one negative and all remaining positive eigenvalues, or the overall reversed-sign convention. Its nonzero null vectors have vanishing metric quadratic form.

With the mostly-plus metric signature $(-,+,+,+)$ and coordinates $X^{\mu }=(ct,x,y,z)$, the Minkowski metric is the nondegenerate bilinear formIt is a symmetric bilinear form, but not a positive-definite bilinear form. For example, $U=(1,1,0,0)$ is nonzero yet $U\cdot U=0$, making it a null vector. This does not make the Minkowski metric degenerate: some other vectors have nonzero pairing with $U$.

Applied to infinitesimal displacements, the Minkowski metric givesFor a timelike curve, a clock records $d\tau =\sqrt{-d{s}^2}/c$. For a null curve, $d{s}^2=0$. For spacelike separation, the positive Minkowski interval measures a different kind of separation. A Lorentz transformation preserves this form, expressed in matrix notation by ${\Lambda }^T\eta \Lambda =\eta$, although it need not preserve ordinary Euclidean distance in the four-component coordinate list.

The null convergence condition requires $R_{ab}{k}^a{k}^b\ge 0$ for every null vector. Through the Einstein field equations, it follows from the null energy condition, since all terms proportional to the metric tensor vanish on a null vector.

A nonzero covector $\ell$ is null when its squared dual Lorentzian inner product is zero: $g^{\alpha \beta }{\ell }_{\alpha }{\ell }_{\beta }=0$. Raising its index gives a null vector. Null covectors describe normals to characteristic light cones and are the test directions in the classical null condition for wave equations.

A null curve has tangent $u$ satisfying $g(u,u)=0$. It represents the possible spacetime path of a light signal.

Its nonzero tangent is a null vector at every regular point; this distinguishes the curve from a single null direction.

Null directions are tangent directions of the light cone; their Minkowski squared norm is zero.

A direction represented by a nonzero null vector has zero Minkowski norm. Such directions form the generators of the light cone.

Null dust is matter with a stress-energy tensor $T_{ab}=\epsilon k_a{k}_b$ for a future-directed null vector $k$. For every future-directed timelike vector $t$, the energy current is $-T^a{}_b{t}^b=\epsilon (-k\cdot t)k^a$. Therefore its dominant energy condition holds exactly when $\epsilon \ge 0$, allowing the zero current. Rescaling $k$ rescales $\epsilon$ inversely by the square and leaves the stress-energy tensor unchanged. Ingoing Vaidya metric matter has $k=-{\partial }_r$, $k_a=-(dv{)}_a$, and $\epsilon =\dot{m}/(4\pi r^2)$.

The null energy condition requires $T_{\mu \nu }{k}^{\mu }{k}^{\nu }\ge 0$ for every null vector $k$. Through the Einstein field equations this is equivalent to $R_{\mu \nu }{k}^{\mu }{k}^{\nu }\ge 0$ when the cosmological term is absent.

The classical stress-energy tensor of a canonical scalar field has a gradient-product term and a metric-proportional term. Contracting with a real null vector removes the latter, leaving the displayed square for a real field. This proves the null energy condition for a minimally coupled canonical scalar, including the massless case. It is not a statement about arbitrary curvature couplings, wrong-sign kinetic terms or a renormalized quantum stress-energy tensor.

In a planar warped spacetime, contraction of the Ricci tensor with a null vector in an orthonormal frame gives $R_{ab}{k}^a{k}^b=-2(\log A{)}^{\prime \prime }(k^3{)}^2$. The Einstein field equations identify this with $8\pi T_{ab}{k}^a{k}^b$; metric terms, including a cosmological constant, vanish in a null contraction. Thus the null energy condition is equivalent to concavity of $\log A$ in the proper transverse coordinate $z$. Null directions confined to a planar slice saturate the condition.

In a flat two-dimensional Penrose diagram, a straight line tangent to a null vector; with radial metric tensor $-d{T}^2+d{X}^2$ it is $T-X=\mathrm{constant}$ or $T+X=\mathrm{constant}$. Such lines represent radial null geodesics after a conformal transformation.

Distinct spacetime events have null separation when their displacement has zero Minkowski interval, equivalently $c|\Delta t|=|\Delta \mathbf{x}|$. They can be joined by a straight vacuum light ray. Their nonzero displacement is a null vector, so the zero squared interval does not mean that the spacetime events coincide. For example, $(t,x)=(3\mathrm{s},3\mathrm{ls})$ is null separated from the origin.

A worldline whose tangent is a null vector. In flat Minkowski spacetime, a freely propagating light signal follows a straight null geodesic. Its coordinate speed is $c$ in an inertial frame, even when another observer moves relative to that frame.

In the extremal case, the squared norm of the Killing vector field isFor $M=|Q|>0$, it is negative for every regular $r\ne M$ and vanishes precisely at $r=M$ in the extended spacetime. ThusThe static chart cannot itself test the geometry at its horizon. In the regular Eddington-Finkelstein coordinates, the normal to $h=r-M=0$ is $n=dr$, andAt $r=M$ this gives $n^{♯}={\partial }_v=K$, a nonzero null vector tangent to the surface. Therefore $r=M$ is a null hypersurface, even though $F$ has the same sign on its two sides.

The coordinate basis vector $K={\partial }_v$ has constant components $K^a={\delta }_v^a$. Its norm isso it is a nonzero null vector everywhere. The Lie derivative of the metric tensor isThus $\boxed{K={\partial }_v\text{ is a null Killing vector field}.}$ In fact it is a parallel vector field: $K_b=g_{bv}={\delta }_b^u$ and the Christoffel symbols found below give ${\nabla }_a{K}_b=-{\Gamma }^u{}_{ab}=0$. This stronger property is characteristic of this plane-fronted gravitational wave.