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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.

A future trapped surface is a smooth compact spacelike two-surface without boundary whose two future-directed orthogonal null expansions are strictly negative. In four-dimensional general relativity, the Penrose singularity theorem states: a time-oriented globally hyperbolic spacetime with a noncompact Cauchy hypersurface, a trapped surface, and the null convergence condition $R_{ab}{k}^a{k}^b\ge 0$ for every null vector is future null-geodesically incomplete. With the Einstein field equations, the null energy condition implies this null convergence condition; a cosmological constant drops out of the null contraction.

The Kruskal spacetime is an example. In its black hole interior, use future null normals in Ingoing Eddington-Finkelstein coordinates:For a round sphere of areal radius $r$, its area is $4\pi r^2$ and its null expansions areBoth are negative for $0<r<2M$. The vacuum Einstein field equations give $R_{ab}=0$, and a two-ended Kruskal Cauchy hypersurface is noncompact. The future radial null geodesics reaching $r=0$ in finite affine parameter provide precisely the incompleteness predicted by the Penrose singularity theorem. A trapped surface at $r<2M$ is strictly trapped; the horizon sphere has one zero null expansion instead.

The second law of black-hole mechanics says that the total area of future event horizon cross-sections cannot decrease toward the future, assuming the classical Einstein field equations, the null energy condition, suitable global predictability and regularity, and here the stipulated future completeness of its generators.

A future event horizon is an achronal boundary generated by null geodesics. If a smooth generator had $\theta <0$, the null focusing theorem would produce a conjugate point to a spacelike surface within finite affine parameter. Future completeness lets the generator reach it, but beyond such a point it could no longer remain on the achronal boundary, a contradiction. Hence $\theta \ge 0$ wherever the null expansion is defined. A transported area element obeysEvery generator present on an early cut continues to a later cut without decreasing its area contribution. New generators can join at past endpoints, adding area; future endpoints in a regular spacetime are excluded. ThereforeThis argument also covers the sum of initially disconnected event horizons of the black holes in a merger. Nonsmooth crease sets require the standard regularity treatment; they do not provide a classical area-decrease loophole under these hypotheses. Quantum negative-energy flux can violate the classical null energy condition, so this is a classical black-hole area theorem, not an unconditional law for evaporating black holes.

Use an affine parameter $\lambda$ along a generator of a null hypersurface. Its null twist vanishes by hypersurface orthogonality. The screen metric is positive definite, so ${\hat{\sigma }}^{ab}{\hat{\sigma }}_{ab}\ge 0$. The Einstein field equations and null energy condition imply $R_{ab}{U}^a{U}^b\ge 0$, giving from the Null Raychaudhuri equationStarting with ${\theta }_0<0$, the null expansion stays negative as long as the congruence remains regular. Set $z=-1/\theta >0$. Then $z^{\prime }={\theta }^{\prime }/{\theta }^2\le -1/2$, soA finite negative null expansion cannot persist to $\lambda -{\lambda }_0=2/|{\theta }_0|$. Thus the null focusing theorem givesThis is a focal breakdown of the transverse congruence, provided the generator and regular geometry extend that far; it is not by itself a curvature singularity or a statement that the individual null geodesic terminates there.

Use spatial indices $i,j=1,2,3$, set ${\varphi }_i={\partial }_i\varphi$, and assume $\varphi$ is smooth. The inverse metric has $g^{tt}=-c^{-2}{e}^{-2\varphi /c^2}$ and $g^{ij}={\delta }^{ij}$. The Christoffel symbols of the Levi-Civita connection follow from ${\Gamma }^{\alpha }{}_{\beta \gamma }=g^{\alpha \delta }({\partial }_{\beta }{g}_{\gamma \delta }+{\partial }_{\gamma }{g}_{\beta \delta }-{\partial }_{\delta }{g}_{\beta \gamma })/2$:All other components vanish. Although $g_{tt}=-c^2-2\varphi +O(c^{-2})$ diverges, its affine connection has a smooth limit on compact subsets:This is the Newtonian connection from an exponential lapse. It is torsion-free, and its geodesic equation, using $t$ as an affine parameter when $\dot{t}\ne 0$, is $d^2{x}^i/d{t}^2=-{\partial }_i\varphi$. Thus $\varphi$ has the role of a Newtonian gravitational potential. There is no finite limiting nondegenerate Lorentzian metric in these fixed coordinates; $-g(c)/c^2$ tends to $d{t}^2$, a rank-one tensor. The limiting connection preserves $dt$ and the contravariant spatial tensor with components $h^{ij}={\delta }^{ij}$, $h^{t\mu }=0$, which describe the degenerate temporal/spatial structures of this limit.

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 $\lambda$, soThe 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 projectorIn 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,Its irreducible decomposition defines the three optical quantities: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 $U^{\perp }/⟨U⟩$; the auxiliary vector chooses a convenient representative of that quotient. For an affinely parametrized null geodesic congruence, the contraction constraints also give $\theta ={\nabla }_a{U}^a$.

Consider the discrete mapIt is an isometry: $\cos \theta$ and $d\varphi$ both change sign, leaving $\cos \theta d\varphi$ unchanged, while $d{\theta }^2$ and ${\sin }^2\theta d{\varphi }^2$ are also unchanged. A local component of its fixed-point set is $\theta =\pi /2$, $\varphi ={\varphi }_0$.

If a geodesic starts tangent to that component, the isometry fixes both its initial position and its initial tangent. Applying the isometry therefore gives a geodesic with the same initial data. Uniqueness of the geodesic equation makes it the same curve, so it remains in the fixed component. This proves that the selected submanifold is a totally geodesic submanifold; it has dimension three, not two. The proof continues through a horizon when expressed in regular coordinates, so coordinate singularities of the original chart do not invalidate it.

The geodesic conserved quantities from Killing vectors associated with ${\partial }_{\psi }$ and ${\partial }_{\varphi }$ areDots denote derivatives with respect to an affine parameter. On the chosen submanifold $\dot{\varphi }=0$ and $\cos \theta =0$, so $L_{\varphi }=0$. Requiring also $L_{\psi }=0$ givesMore generally, away from the polar-coordinate axes, both displayed angular charges being zero imply $\dot{\varphi }=0$ and the same relation. In particular, zero conserved angular momentum does not imply constant $\psi$: the motion follows the local dragging of the angular coordinate by the rotating geometry.

Put $h(r)=1-\mu /r^2$. For the equatorial geodesic, the geodesic Lagrangian is $\mathcal{L}=\frac{1}{2}[-h{\dot{t}}^2+h^{-1}{\dot{r}}^2+r^2{\dot{\varphi }}^2]$, where dots denote derivatives with respect to the affine parameter. The cyclic coordinates $t,\varphi$ give constants $p_t=-h\dot{t}=-E$ and $p_{\varphi }=r^2\dot{\varphi }=L$. A third constant is the tangent norm $g_{ab}{\dot{x}}^a{\dot{x}}^b=-\kappa$, preserved by the geodesic equation and metric compatibility. For a timelike geodesic rescale the affine parameter to proper time so $\kappa =1$; for a null geodesic $\kappa =0$.

Substitute $\dot{t}=E/h$, $\dot{\varphi }=L/r^2$ into the norm identity and multiply by $h$. It gives the effective potential equationOutside the horizon $r=\sqrt{\mu }$, its derivative is $V^{\prime }=-(L^2-\kappa \mu )/r^3+2\mu L^2/r^5$. A nontrivial circular null geodesic requires $L\ne 0$, givingFor a circular timelike geodesic,At it $V^{\prime \prime }=-4\mu L^2/r^6<0$, and its radius exceeds $\sqrt{2\mu }$. If $L^2\le \mu$ no finite circular timelike radius exists. Thus all finite circular null and timelike orbits in this five-dimensional exterior are radially unstable. The energy of a circular orbit is fixed by $E^2=2V(r)$, not independently arbitrary.

For comparison, with four-dimensional Schwarzschild horizon radius $r_s$, the photon circle is at $3r_s/2$ and is unstable. Circular timelike orbits exist for $r>3r_s/2$; they are unstable below $3r_s$, marginal at $3r_s$ and stable above $3r_s$. The five-dimensional potential has no corresponding stable outer circular timelike branch.

Let $U$ be the smooth nonzero tangent along a regular null curve, and $a={\nabla }_{U}U$. Metric compatibility gives $g(U,a)=\frac{1}{2}U[g(U,U)]=0$. In a two-dimensional Lorentzian manifold, the orthogonal complement of a nonzero null vector is its own span; this follows directly in the preceding null basis. Therefore $a=\kappa (t)U$.

Consequently every regular null curve is a geodesic up to reparametrization. To obtain an affine parameter $\lambda$, write $q=dt/d\lambda >0$ and solve $dq/dt+\kappa (t)q=0$. Then ${\nabla }_{qU}(qU)=q^2(\kappa +q^{-1}dq/dt)U=0$. A solution $q=C\exp (-\int \kappa dt)$ exists locally and along any nonsingular parameter interval. The original parameter need not be affine; the literal assertion of zero covariant acceleration for every parameter would be false.

The Euler-Lagrange equations for the kinetic Lagrangian areExpanding and multiplying by the inverse metric gives the geodesic equationThe kinetic variational principle uses an affine parameter; the energy $g_{ab}{\dot{x}}^a{\dot{x}}^b/2$ is constant along its solutions. Constant curves are also geodesics.

For the Poincare half-plane model, translation in $x$ gives the conserved quantity $p=\dot{x}/y^2$, and the energy is $E=({\dot{x}}^2+{\dot{y}}^2)/(2y^2)$. If $p=0$, the nonconstant geodesics are vertical lines, with $y=y_0{e}^{\pm \sqrt{2E}t}$. If $p\ne 0$,Writing $R^2=2E/p^2$ and integrating $dx/dy=\pm y/\sqrt{R^2-y^2}$ yields $(x-x_0{)}^2+y^2=R^2$. Thus nonconstant geodesic images are vertical lines or upper semicircles meeting the real axis orthogonally.

For the two specified points, the circle centre is $(6,0)$ and the radius is five:For example, an affine unit-speed parametrization is $x=6+5\tanh t$, $y=5\mathrm{sech}t$. The endpoints on the real axis are $(1,0)$ and $(11,0)$, outside the hyperbolic surface.

Let $D_{abc}={\nabla }_a{\nabla }_b{K}_c$. Differentiating the Killing equation gives $D_{abc}=-D_{acb}$. For a covector, the Ricci identity isCall this difference $C_{abc}$. The differentiated Killing equation then givesTaking the first minus the second plus the third yieldsThe pair symmetries of the Riemann curvature tensor and the first Bianchi identity reduce the coefficient to $2R_{cbad}$. Therefore the second covariant derivative of a Killing vector is

For an affine parameter $\lambda$ on a geodesic, its tangent obeys $V^b{\nabla }_b{V}^a=0$. ThusThe second term vanishes by the geodesic equation; the first contracts a symmetric product $V^a{V}^b$ with the antisymmetric derivative from the Killing equation. This establishes the geodesic conserved quantity from a Killing vector:

By spherical symmetry, rotate a geodesic into the equatorial plane $\theta =\pi /2$. The geodesic conserved quantities from Killing vectors arewhere the dot denotes a suitable affine parameter. Normalize the tangent by $g(\dot{x},\dot{x})=-ϵ$, with $ϵ=1$ for timelike proper time, $ϵ=-1$ for spacelike proper length, and $ϵ=0$ for null geodesics. ThenTherefore the geodesic potential of the magnetic dilaton black hole isTogether with the two conserved first integrals, this reduces the radial motion to ${\dot{r}}^2/2+\tilde{V}=0$. To cover turning points as well, the radial geodesic equation itself givesThus a constant-radius geodesic must satisfy both $\tilde{V}=0$ and ${\tilde{V}}^{\prime }=0$. The intermediate static-coordinate formulas apply where $V\ne 0$; the radial equation extends through a regular horizon in regular coordinates.

A twice differentiable curve $\gamma :I\to S$ is a geodesic when its velocity is parallel along the curve:For a surface embedded in ${\mathbb{R}}^3$, the ambient acceleration criterion for a surface geodesic gives the equivalent conditionThe parameter $t$ is then an affine parameter, and $|\dot{\gamma }|$ is constant.

Suppose the closed trapped surface $T$ were not wholly in the black hole. The stated consequence of strong asymptotic predictability then supplies a point $p\in {\dot{J}}^{+}(T)\cap {\mathcal{I}}^{+}$ lying on a future null generator $\gamma$ orthogonal to $T$, with no point conjugate to $T$ before $p$. Both future null expansions at $T$ are negative. The null focusing theorem therefore forces the expansion along $\gamma$ to diverge after finite affine parameter, producing a conjugate point before the generator reaches future null infinity. A null geodesic with such a point cannot continue to generate the achronal boundary ${\dot{J}}^{+}(T)$, contradicting the property of $p$. Hence no point of $T$ can communicate with future null infinity, and

Forthe nonzero Christoffel symbols needed here areOn a line of constant longitude, $\dot{\varphi }=0$, and the geodesic equation reduces to $\ddot{\theta }=0$ after choosing an affine parameter. Thus every meridian is a geodesic.

For a nonconstant line of constant latitude, $\dot{\theta }=0$ and the $\theta$ equation requiresAway from the coordinate-degenerate poles, this holds only whenso only the equator is a constant-latitude geodesic.

For an affine parameter $\lambda$, the geodesic equation and the connection coefficients from part (a) givewhere dots denote derivatives with respect to $\lambda$.

On a line of constant $\varphi$, one has $\dot{\varphi }=0$. The equations reduce to $\ddot{r}=0$, so $r$ itself can be chosen as an affine parameter. Every such radial line is therefore a geodesic.

For a circle $r=r_0>0$, the first equation instead requiresBoth hyperbolic factors are positive when $r_0>0$, so $\dot{\varphi }=0$. This gives only a constant point, not a parametrized circle. Thus no circle of constant positive $r$ is a geodesic, in agreement with the coordinate geodesics of the hyperbolic polar metric.

For the quadratic geodesic Lagrangianthe Euler-Lagrange equation isRaising $\mu$ and symmetrizing the coefficient of the two velocities produces exactlyThe quadratic action fixes an affine parameter; proper time is an affine parameter for a timelike geodesic.

Write $f(r)=1-r_{+}^2/r^2$ and let a dot denote differentiation with respect to an affine parameter $\tau$. The time-translation and $z$-translation Killing vector fields, together with rotational symmetry of the unit $S^3$, give the conserved quantitiesDefine $\epsilon =-g_{ab}{\dot{x}}^a{\dot{x}}^b$, so $\epsilon =1,0,-1$ for timelike, null and spacelike geodesics respectively. The normalization equation becomesIt therefore has the effective potential form