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Let $u$ be the tangent vector to a geodesic with affine parameter $\lambda$. Its geodesic equation is ${\nabla }_{u}u=0$. For a Killing vector field $\xi$, metric compatibility givesThusThis geodesic conserved quantity from a Killing vector is independent of whether the geodesic is timelike, spacelike or null. For a timelike geodesic one can use proper time; for a null geodesic choose an affine parameter. An arbitrary nonaffine tangent rescales the displayed quantity and need not make it constant.

Start with the four-dimensional Minkowski metric $d{s}^2=-d{t}^2+d{r}^2+r^{2}d{\Omega }^2$, where $r\ge 0$. Choose an arbitrary length $L>0$, and use the retarded and advanced null coordinates $u=t-r$, $v=t+r$. For the Minkowski conformal compactification, setBoth $p,q$ lie between $-\pi /2$ and $\pi /2$. Since $v\ge u$, we have $R\ge 0$; the remaining inequalities are $|T|+R<\pi$. Moreover,Multiply by the square of the conformal factor ${\Omega }_c=2\cos p\cos q/L$. The resulting metric is regular on the appropriate boundary pieces and preserves the null directions. Suppressing the angular two-spheres gives a triangular Penrose diagram with radial null geodesics at $45$ degrees.

The line $R=0$ is the ordinary timelike centre $r=0$. The upper sloping edge $T+R=\pi$ is future null infinity, reached with $v\to +\infty$ and finite $u$; the lower sloping edge $T-R=-\pi$ is past null infinity, reached with $u\to -\infty$ and finite $v$. The vertices $(T,R)=(\pi ,0)$ and $(-\pi ,0)$ are future and past timelike infinity, denoted $i^{+}$ and $i^{-}$. The vertex $(0,\pi )$ is spacelike infinity, $i^0$. These are limiting endpoints in the conformal completion, rather than ordinary physical events. In particular, finite diagram coordinates at null infinity do not imply finite physical affine parameter.

The four-dimensional radial diagram is the triangle $R\ge 0$, $|T|+R\le \pi$. If one instead draws two-dimensional Minkowski spacetime with a signed Cartesian spatial coordinate, the diagram is the full diamond. The centre is a boundary of the radial quotient, not a boundary of the physical four-dimensional Minkowski spacetime.

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 $B^a{}_b{U}^b=\kappa U^a$.

One standard form of the Penrose singularity theorem assumes a time-oriented globally hyperbolic spacetime with a noncompact Cauchy hypersurface, the null convergence condition $R_{ab}{k}^a{k}^b\ge 0$ for every null vector $k$, and a closed future trapped surface. It concludes future null geodesic incompleteness: some future-inextendible null geodesic has a finite upper endpoint of its affine parameter.

The focusing mechanism is the Null Raychaudhuri equation. The normal generators have zero null twist, soAn initial ${\theta }_0<0$ therefore gives a conjugate point to a spacelike surface within affine distance at most $2/|{\theta }_0|$, assuming the generator can be continued that far. Such a generator ceases to lie on the achronal boundary after its first focal point. Compactness of the trapped surface supplies a uniform bound for all normalized initial null normals. Future completeness would consequently make its future boundary compact. Projection along timelike curves to a connected Cauchy hypersurface is injective on the achronal boundary and has open image. Compactness makes the image closed as well; the nonempty image must therefore be the entire hypersurface, contradicting its noncompactness. This explains why the global assumptions supplement local focusing.

For Reissner-Nordstrom spacetime, the Maxwell stress-energy tensor satisfies the null energy condition; the Einstein field equations imply the required null convergence condition. The spheres in the band just found are closed and trapped. Apply the theorem to a maximal Cauchy development with a noncompact Cauchy hypersurface and containing one such sphere. That globally hyperbolic spacetime must be future null-geodesically incomplete.

The Penrose theorem at a Cauchy horizon needs care: the full maximal analytic Reissner-Nordstrom spacetime has inner Cauchy horizons and is not globally hyperbolic. It does not satisfy every hypothesis of the displayed theorem. In the exact solution some incomplete geodesics of the globally hyperbolic development reach a smoothly extendible Cauchy horizon in finite affine parameter. The theorem asserts incompleteness of the development, not that every such endpoint is a curvature singularity. The separate curvature singularity at $r=0$ does not justify silently dropping the theorem's global hypothesis.

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.

In the two null coordinates the inverse metric has $g^{UV}=g^{VU}=-f^{-1}$ and $g^{UU}=g^{VV}=0$. ThereforeBoth tangents are null and future-directed, since $f>0$ and both coordinate directions are future-directed. To prove the affine parameter property, use the null-gradient affine-geodesic identity. The one-form $k_a=-{\nabla }_{a}U$ is closed, so ${\nabla }_a{k}_b={\nabla }_b{k}_a$. ConsequentlyThe same argument applies to ${\ell }_a=-{\nabla }_{a}V$. Thus both vector fields satisfy the geodesic equation without a nonaffinity term, and generate affinely parametrized null geodesic congruences. The coordinate vectors alone need not be affine; the factors $f^{-1}$ are essential.

For radial null geodesics, the metric gives $dt/dr=\pm f^{-1}$ in the exterior. Define the tortoise coordinate byAn outgoing future radial ray has $dt/dr=f^{-1}$, so $u=t-r_{\star }$ is constant; an ingoing ray has $dt/dr=-f^{-1}$, so $v=t+r_{\star }$ is constant. These describe the null trajectories independently of the choice of affine parameter.

For the tortoise coordinate of a five-dimensional charged black hole, with $a=r_{+}$, $b=r_{-}$ and $\Delta =a^2-b^2$, partial fractions giveThus a useful explicit tortoise coordinate isThe logarithms signal failure of the static coordinates at the horizons, rather than a curvature singularity there.

The product of the magnetic field with the velocity in this question is the three-dimensional cross product. Fix ${ϵ}_{123}=1$ and write $K^i{}_j={ϵ}^i{}_{kj}{B}^k$, so $K\mathbf{v}=\mathbf{B}\times \mathbf{v}$. In coordinates $y^0=t$, $y^i=x^i$ on ${\mathbb{R}}^3\times \mathbb{R}$, prescribe an affine connection by the following Christoffel symbols:with all other Christoffel symbols zero. These are smooth globally in the given Cartesian coordinates. The lower-index symmetry makes this a torsion-free connection. It is the geometrization of the Lorentz force by an affine connection; no condition on the curl of $\mathbf{E}$ or the divergence of $\mathbf{B}$ is required for this construction.

An affinely parametrized geodesic, with primes denoting differentiation with respect to $\lambda$, satisfiesOn the branch $t^{\prime }=c\ne 0$, use $t$ itself as an affine parameter. Dividing the spatial equation by $c^2$ givesConversely, each solution of this equation makes $\lambda =t$, $y(t)=(t,\mathbf{x}(t))$ an affinely parametrized geodesic of the constructed affine connection. Therefore its image is also an unparametrized geodesic. A general change of parameter adds a term proportional to the tangent in the geodesic equation, leaving the curve unchanged. The branch $t^{\prime }=0$ is not a trajectory with time as parameter.

For the metric realization, assume $\mathbf{E}=-\nabla U$ and $\nabla \cdot \mathbf{B}=0$. Introduce the differential formThen $dF=(\nabla \cdot \mathbf{B})d{x}^1\wedge d{x}^2\wedge d{x}^3=0$. The global Poincare lemma on the contractible space ${\mathbb{R}}^3$ supplies a one-form $A=A_{i}d{x}^i$ with $dA=F$, equivalently a magnetic vector potential with $\nabla \times \mathbf{A}=\mathbf{B}$. An explicit radial-gauge potential for a divergence-free magnetic field isThis is the radial homotopy formula for a closed two-form and is smooth even at the origin. For a constant magnetic field it gives $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{x}$.

Consider the Eisenhart-Duval lift with the Lorentzian metricThe independent one-forms $d{x}^1,d{x}^2,d{x}^3,dt,du-2A_{i}d{x}^i-Udt$ exhibit three positive directions and a two-dimensional block with one positive and one negative direction. Thus the metric is nondegenerate, of signature $(4,1)$. Its coefficients do not depend on $u$, and $g_{uu}=0$, so $\xi ={\partial }_u$ is a null Killing vector field. Indeed $g_{Au}$ are constant, so ${\Gamma }^A{}_{Bu}=0$ for the Levi-Civita connection and $\xi$ is parallel.

The geodesic Lagrangian for an affine parameter $\lambda$ isThe cyclic coordinate $u$ gives the conserved quantity $p_u=\partial L/\partial u^{\prime }=t^{\prime }$. Work at a nonzero value of $p_u$ and rescale the affine parameter to set $t^{\prime }=1$. This is the essential step in the null Kaluza-Klein reduction of a stationary force: one fixes the momentum along the null isometry and projects its geodesics, rather than dividing by $g_{uu}$.

The spatial Euler-Lagrange equations, before setting $t^{\prime }=1$, areSince $t^{\prime \prime }=0$ and $F_{ij}{v}^j=-(\mathbf{B}\times \mathbf{v}{)}_i$, their reduction isEquivalently, after taking $t$ as the affine parameter, the term $\dot{u}$ is a total derivative and the reduced Lagrangian is $L_{\mathrm{red}}=\frac{1}{2}|\dot{\mathbf{x}}{|}^2-2\mathbf{A}\cdot \dot{\mathbf{x}}-U$. Its Euler-Lagrange equations give the same sign and factor two.

There is also an explicit converse using null geodesics. For any physical trajectory, setThis makes its five-dimensional tangent null. The conserved momentum of the cyclic coordinate $t$ becomesThe quantity in parentheses is conserved, because its derivative is $\dot{\mathbf{x}}\cdot (2\mathbf{B}\times \dot{\mathbf{x}})=0$. Hence the $t$ equation, as well as the spatial and $u$ Euler-Lagrange equations, is satisfied. Every trajectory admits a null geodesic lift with $p_u=1$, and every such lift projects to the required trajectory.

Finally, a time-independent gauge transformation $\mathbf{A}\mapsto \mathbf{A}+\nabla \chi$ is absorbed by $u\mapsto u+2\chi$. The one-form $du-2A_{i}d{x}^i-Udt$ and the five-dimensional metric are unchanged. These gauge transformations of an Eisenhart-Duval lift alter the reduced Lagrangian only by the total derivative $-2d\chi /dt$.

For radial motion, take an affine parameter $\lambda$ and use dots for $d/d\lambda$. The radial null geodesics of the Vaidya metric satisfy the two radial geodesic equationsThe radial squared norm is $-f{\dot{v}}^2+2\dot{v}\dot{r}$. If $v$ is constant it vanishes, and the geodesic equation reduces to $\ddot{r}=0$. Thus constant-$v$ radial curves are null geodesics, with $r=a\lambda +b$ an affine parametrization. Future ingoing motion has $a<0$.

For the other family let $dr/dv=f/2$. Its tangent $k={\partial }_v+(f/2){\partial }_r$ is a null vector. Using the Christoffel symbols of the Vaidya metric, direct differentiation givesHence these are also null geodesics, but $v$ is generally not affine. For completeness, put $p=dv/d\lambda$ and chooseThis makes $\ddot{v}=-(M/r^2){\dot{v}}^2$. Differentiating $\dot{r}=(f/2)\dot{v}$ then gives $\ddot{r}=(f_v/2){\dot{v}}^2=-M^{\prime }{\dot{v}}^2/r$; substitution verifies the radial $r$ equation as well. Thus the null condition and both radial geodesic equations hold, not just the null condition alone.

Let $\lambda$ be an affine parameter and $X=d{x}^{\mu }/d\lambda {\partial }_{\mu }$ the tangent to the geodesic. Since ${\nabla }_{{\partial }_{\mu }}{\partial }_{\nu }={\Gamma }^{\rho }{}_{\mu \nu }{\partial }_{\rho }$, the product rule for the covariant derivative givesIndependence of the coordinate basis vectors yields the coordinate geodesic equationThe affine parameter condition matters: an arbitrary reparametrization instead allows a multiple of the tangent on the right, as in the unparametrized geodesic equation.

Use $t$ as an affine parameter on each member of the geodesic variation, so ${\nabla }_{T}T=0$. The parameter directions commute, giving $[T,S]=0$ along the variation. The torsion-free connection then givesConsequently the covariant derivative and curvature definition giveThis is the geodesic-deviation equation with the stated curvature sign:The component expression defines the slot order: $\beta$ is the vector acted on, and $\gamma ,\delta$ are the two derivative directions of the Riemann curvature tensor. Both the commuting parameter directions and affine parametrization are essential; a general connecting vector field or a nonaffine parameter would add terms.

A spacetime is geodesically complete if every maximal geodesic has an affine parameter ranging over all of $\mathbb{R}$. For timelike geodesics this is equivalent to unbounded proper time in both directions. An extendible geodesic segment can be prolonged in the same spacetime; an inextendible geodesic cannot. A finite coordinate endpoint need not imply finite affine parameter.

For the Kruskal spacetime, use $U,V$ withIn the right exterior $U<0<V$. A truncated ray $U=U_0<0$, $V_1<V<V_2$ is an extendible geodesic of radial null type: neither artificial endpoint is a spacetime boundary. A future ray $U=U_0>0$ in the black hole reaches $UV=1$, hence $r=0$, and is inextendible geodesic and future null-geodesically incomplete. Its Killing energy $E\ne 0$ gives $dr/d\lambda =-|E|$, so the Schwarzschild singularity occurs at finite affine parameter. The maximal continuation toward the past supplies the other half of this same null geodesic.

There is no inextendible, complete radial timelike geodesic in positive-mass Kruskal spacetime. This requested example is impossible as printed. For a radial timelike geodesic, the conserved Killing energy and normalization giveIf $E^2<1$, $r$ has at most one turning point, a maximum $r_{\max }=2M/(1-E^2)$; a maximal trajectory runs from the white hole Schwarzschild singularity to the Schwarzschild singularity. If $E^2\ge 1$, there is no finite turning point; one end can lie at infinity but the other reaches $r=0$. The exceptional $E=0$ trajectory through the bifurcation surface also reaches $r=0$ in both time directions. Near $r=0$,whose integral is finite. Constant-$r$ radial timelike curves are accelerated, not geodesics.

Two plausible repairs have different meanings. Removing “radial” permits a complete circular timelike geodesic at $r=R>6M$, with nonzero angular momentum and proper time ranging over $\mathbb{R}$. Replacing “timelike” by “null” permits a complete horizon null geodesic: $U=0$, $V\in \mathbb{R}$, with $V$ an affine parameter. The Penrose diagram shows both repairs explicitly, together with the two valid requested examples; the circular trajectory is only a radial projection and is labelled as nonradial.

The horizontal boundaries are the Schwarzschild singularities, diagonal dashed lines are the Killing horizons, and outer diagonal edges are null infinity.