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For $d{s}_5^2={\gamma }_{ij}d{x}^{i}d{x}^j+(dz+A_{i}d{x}^i{)}^2$ with $z$-independent coefficients, the geodesic Lagrangian makes $k=\dot{z}+A_i{\dot{x}}^i$ constant. Its remaining Euler-Lagrange equations give the displayed Lorentz force, where $F=dA$ and $D$ uses the Levi-Civita connection of $\gamma$. For a particle of five-dimensional mass $M$ with proper-time parameter $s$, the conserved momentum is $p_z=Mk$ and the four-dimensional rest mass is $m=M\sqrt{1+k^2}$. Reparametrizing by four-dimensional proper time gives $D{u}^i/d\tau =(p_z/m)F^i{}_j{u}^j$.

For the full unit-radius two-dimensional de Sitter spacetime, the coordinate ranges areOne can use $0\le \chi <2\pi$ with endpoints identified. If the identification is omitted, the same line element describes the universal covering spacetime instead. The periodicity follows from the hyperboloid embeddingThe induced ambient Minkowski metric is precisely the given Lorentzian metric.

Using an affine parameter $s$, the geodesic Lagrangian is $\frac{1}{2}({\dot{t}}^2-{\cosh }^{2}t{\dot{\chi }}^2)$. The nonzero Christoffel symbols areThus the geodesic equations and their first integrals areHere $\epsilon =1$ for unit-speed timelike geodesics, $0$ for null geodesics and $-1$ for unit-speed spacelike geodesics. The conserved $J$ is the geodesic conserved quantity from a Killing vector for the angular symmetry.

All geodesics of two-dimensional de Sitter spacetime can be written especially simply in the embedding. If $P$ is the initial point and $V$ the initial tangent, with $P^2=-1$, $P\cdot V=0$, $V^2=\epsilon$, the geodesic has normal ambient acceleration $X^{\prime \prime }=\epsilon X$. ThereforeThese curves stay in the two-plane spanned by $P$ and $V$, preserve both constraints and have no tangential acceleration. They consequently solve the geodesic equation and exhaust all initial directions.

An isometry can move the chosen point to $P=(0,1,0)$, namely $(t,\chi )=(0,0)$. For future-directed timelike geodesics take $V=(\cosh \alpha ,0,\sinh \alpha )$, obtainingThe continuous angular branch has $|\chi |<\pi /2$, and every real $\alpha$ gives a distinct timelike direction. The two null geodesics have $V=(1,0,\pm 1)$ andFor spacelike geodesics take $V=(\sinh \alpha ,0,\cosh \alpha )$, givingThey are closed with period $2\pi$ in affine arclength and $|t|\le |\alpha |$. In particular, $\alpha =0$ is the equatorial geodesic $t=0$. These three families, with reversals of parametrization allowed, are the requested complete set through the chosen point.

The conformal time substitution givesHenceMultiplying by ${\cos }^{2}T$ gives the regular conformal cylinder of two-dimensional de Sitter spacetime. Adding $T=\pm \pi /2$ completes the conformal compactification; both boundary circles are spacelike. Null geodesics are straight lines $\chi =\pm T$ through the origin. The other families have the useful diagram equationsThe image sketches representative directions from each family; the formulas allow all intermediate directions.

For the complete observer $\chi =0$, let $d(\chi ,0)\in [0,\pi ]$ be the shortest angular distance on the circle. Since light travels one unit of angular distance per unit conformal time, an event at $(T,\chi )$ can signal to the observer at some finite future time precisely whenEquality defines its future observer event horizon. Likewise an event can receive a signal emitted by the observer precisely whenand equality defines the past horizon. In the lift $-\pi \le \chi \le \pi$, their branches are $\chi =\pm (\pi /2-T)$ and $\chi =\pm (T+\pi /2)$. The common communication region is the diamond $|\chi |+|T|<\pi /2$, the observer's static patch of de Sitter spacetime. These are the observer horizons in two-dimensional de Sitter spacetime. Every complete timelike observer has analogous horizons after a de Sitter isometry; they depend on the observer, rather than enclosing an intrinsic black-hole region.

Despite these horizons, each constant-$T$ circle is a Cauchy hypersurface of the full spacetime: any inextendible causal curve crosses it exactly once. The global spacetime is therefore globally hyperbolic and has no Cauchy horizons. The conformal boundaries lie at infinite proper time for complete timelike geodesics and infinite affine distance for complete null geodesics, so their finite height in the diagram is not a physical incompleteness.

Let $f(r)=1-2M/r$, with $M>0$, and let a dot denote differentiation with respect to an affine parameter $\lambda$. The geodesic Lagrangian is one half the squared tangent norm. Its Euler–Lagrange equations givewhere $f^{\prime }=df/dr$ in these formulas only. The constants from time translation and axial rotation areThey are the Killing energy and axial angular momentum per chosen affine normalization. A rescaling of $\lambda$ rescales both, while their ratio is invariant. More generally the rotational Killing fields conserve the full angular-momentum vector: differentiating any Killing contraction along an affine geodesic gives $k^a{k}^b{\nabla }_{(a}{\xi }_{b)}=0$. Thus its direction can be chosen as the polar axis.

Equivalently rotate the initial position and angular tangent into one plane through the centre. The angular geodesic equation then has initial data $\theta =\pi /2$, $\dot{\theta }=0$, for which $\theta =\pi /2$ solves the equation identically. Uniqueness keeps the geodesic in that plane. A radial geodesic has no distinguished orbital plane but can be placed in any equatorial plane. Hence no physical restriction is imposed by setting $\theta =\pi /2$.

Write $L=L_z$ in that plane. For a null geodesic the tangent norm vanishes, givingThe full radial equation also gives $\ddot{r}=L^2(r-3M)/r^4$. These equations are used away from the coordinate singularity $r=2M$; the geodesics themselves extend across the horizon in regular coordinates. For radial rays $L=0$, $\dot{r}=\pm E$ and azimuth is not an orbit parameter.

For the nonradial case $L\ne 0$, set $u=M/r$ and now use a prime for $d/d\varphi$. Since $u^{\prime }=-M\dot{r}/L$, the null first integral becomesHere $b$ is the signed impact parameter; its magnitude is the ordinary positive impact parameter for an exterior future-directed ray. The second-order orbit equation is $u^{\prime \prime }+u=3u^2$. It follows from the radial geodesic equation and therefore holds even where $u^{\prime }=0$; differentiation of the squared first integral alone would not establish a circular solution at a turning point.

A finite circular photon orbit has $u^{\prime }=u^{\prime \prime }=0$ and nonzero $u$. Hence $u=1/3$, while the first integral fixes ${\alpha }^2=1/27$:The null radial potential $L^{2}f/r^2$ has second derivative $-2L^2/(81M^4)$ there, so the photon sphere is unstable. The formal root $u=0$ lies at infinite radius, not a finite orbit. Radial null generators of the event horizon have fixed angles and are not circular photon orbits.

At the critical impact parameter the first integral factors asChoose the oriented branch $u^{\prime }=(u-1/3)\sqrt{1+6u}/\sqrt{3}$. Put $w=\sqrt{1+6u}$; then $w^{\prime }=(w^2-3)/(2\sqrt{3})$. The functionsatisfies $dQ/d\varphi =Q$ by direct differentiation. Integrating givesThis establishes the displayed family as actual critical-impact-parameter geodesics. Reversing the azimuthal orientation gives the complementary $A{e}^{-\varphi }$ parametrization.

To expose the geometry, put $x=A{e}^{\varphi }$. Solving for the positive square root givesFor physical $r>0$ this branch uses $-1/2<x<(2-\sqrt{3})/2$, with the upper endpoint corresponding to infinite radius. As $\varphi \to -\infty$,Thus the critical photon orbits of Schwarzschild spacetime wind around the photon circle infinitely many times with an exponentially shrinking radial displacement. For $A>0$ the approach is from outside $r=3M$; for $A<0$ it is from inside; for $A=0$ the orbit is the circle itself. Increasing $\varphi$ along the chosen orientation sends the positive branch outward toward infinity and the negative branch inward, through the horizon toward the singularity. The limiting description at $\varphi \to -\infty$ concerns the asymptotic photon sphere rather than a finite-angle radial turning point.

Let a dot denote differentiation with respect to an affine geodesic parameter $s$. The five-dimensional geodesic Lagrangian isBecause $z$ is cyclic, its Euler-Lagrange equation gives the conserved quantityThis is the conserved momentum per unit five-dimensional mass associated with the Killing vector field ${\partial }_z$. For the other coordinates, differentiate $\partial L/\partial {\dot{x}}^i={\gamma }_{ij}{\dot{x}}^j+k{A}_i$. Using $\dot{k}=0$ givesThe first two terms combine into ${\gamma }_{ij}({\ddot{x}}^j+{\Gamma }^j{}_{lm}{\dot{x}}^l{\dot{x}}^m)$, with the Christoffel symbols of $\gamma$. Define the four-dimensional electromagnetic field tensor $F_{ij}={\partial }_i{A}_j-{\partial }_j{A}_i$ and raise its index with $\gamma$. The complete reduced geodesic equations areThis is the Kaluza-Klein geodesic Lorentz force. The Christoffel symbols describe the gravitational motion in general relativity in the four-dimensional metric tensor, while the right-hand side is antisymmetric in its lowered indices and gives the relativistic Lorentz force. In particular,Thus the force preserves the four-dimensional velocity norm, as a Lorentz force must. A particle with $k=0$ is neutral and follows an ordinary four-dimensional geodesic. Nonzero compact momentum gives electric charge, with its sign determined by the direction of motion around the circle.

To identify the actual charge-to-mass ratio, suppose the five-dimensional particle has mass $M>0$ and $s$ is its five-dimensional proper time, so $g_5(\dot{X},\dot{X})=-1$. Then $\gamma (\dot{x},\dot{x})=-(1+k^2)$. Four-dimensional proper time obeys $d\tau /ds=\sqrt{1+k^2}$; since $k$ is constant, $u^i=d{x}^i/d\tau$ satisfiesHere $q$ is in the geometric normalization of the supplied potential $A$. The relation $m^2=M^2+p_z^2$ also follows directly from the five-dimensional mass shell, and continues to identify the effective mass when the higher-dimensional particle is massless. The distinction between the two proper times prevents mistaking $k$ itself for the physical four-dimensional charge-to-mass ratio. A rescaling of $A$ to a canonically normalized electromagnetic potential correspondingly rescales $q$.

For the wave calculation, take a scalar field of five-dimensional mass $M$ in units $\hslash =c=1$, obeying $({\square }_5-M^2)\Psi =0$. In the coordinate order $(x^i,z)$, direct inversion of the metric tensor givesThe determinant follows by the triangular change of coframe from $(d{x}^i,dz)$ to $(d{x}^i,dz+A_{i}d{x}^i)$, whose determinant is one. Write $J=\sqrt{|\det \gamma |}$. Since all coefficients are $z$-independent, the scalar covariant wave operator isExpanding the right-hand side verifies both cross terms $-2A^i{\partial }_i{\partial }_z\Psi$, the term $-J^{-1}{\partial }_i(J{A}^i){\partial }_z\Psi$, and the coefficient $1+A_i{A}^i$ of ${\partial }_z^2\Psi$.

Single-valuedness under the compact coordinate period requires the Fourier seriesFor the $n$th Kaluza-Klein mode, put $q_n=n/R$ and $D_i={\partial }_i-i{q}_n{A}_i$. The five-dimensional Klein-Gordon equation reduces toThis is the charged four-dimensional Klein-Gordon equation: the divergence expression includes both the spacetime connection and the gauge connection. Its interpretation as electric charge can also be checked directly. Under the fibre coordinate change $z^{\prime }=z-\chi (x)$, the metric retains its form with $A^{\prime }=A+d\chi$, and the scalar Fourier coefficient changes by ${\psi }_n^{\prime }=e^{i{q}_n\chi }{\psi }_n$. Hence $D_i^{\prime }{\psi }_n^{\prime }=e^{i{q}_n\chi }{D}_i{\psi }_n$, exactly the transformation law for a field of charge $q_n$.

The Kaluza-Klein charge quantization is thereforePositive and negative integers give opposite charges, and the zero mode is neutral. Restoring $\hslash$ gives compact momenta $p_z=n\hslash /R$, so their fundamental spacing is $\hslash /R$. These are charges in the metric's geometric normalization. If $A_i=\kappa {\mathcal{A}}_i$ for a physically normalized electromagnetic potential ${\mathcal{A}}_i$, the physical charge unit is $\kappa /R$ in natural units, or $\kappa \hslash /R$ with $\hslash$ restored. Determining $\kappa$ requires an electromagnetic action normalization not specified by the metric ansatz alone.

Using the affine parameter and the normalization requested for the geodesic Lagrangian,The coordinate $v$ is cyclic. Its Euler-Lagrange equation gives $\ddot{u}=0$. The transverse coordinate equations give $\ddot{x}+H_x{\dot{u}}^2=0$ and $\ddot{y}+H_y{\dot{u}}^2=0$. For the remaining coordinate,Its Euler-Lagrange equation, after using $\ddot{u}=0$, isCompare with ${\ddot{x}}^a+{\Gamma }_{bc}^a{\dot{x}}^b{\dot{x}}^c=0$. With indices $(1,2,3,4)=(u,v,x,y)$, the complete list of nonzero Christoffel symbols, including symmetry of the lower indices, isAll contracted symbols ${\Gamma }_{ad}^d$ vanish. For $R_{11}$, the first derivative term in the supplied Ricci tensor formula is ${\partial }_v{H}_u+{\partial }_x{H}_x+{\partial }_y{H}_y=H_{xx}+H_{yy}$, since $H$ is independent of $v$. The second derivative term vanishes because the trace symbols vanish. The third term vanishes for the same reason, and every possible product in the last term requires a zero symbol with upper index $1$ or a lower $2$, so it also vanishes. ThusThe Vacuum Einstein equations with zero cosmological constant require $R_{ab}=0$, and therefore $\boxed{H_{xx}+H_{yy}=0}$. The wave profile of this plane-fronted gravitational wave is harmonic in the transverse plane for each fixed $u$.

Vary the geodesic Lagrangian before imposing the normalization of proper time. With $u^{\mu }=d{x}^{\mu }/d\tau$,The Euler-Lagrange equation isequivalently the affinely parametrized geodesic equation. For the static weak metric, its leading spatial connection coefficient isAll terms with two spatial velocities are suppressed by $v^2$. Therefore $d^2{x}^i/d{\tau }^2\simeq -{\partial }_i\Phi (dt/d\tau {)}^2$.

Because $t$ is cyclic, the same Euler-Lagrange equation gives $(1+2\Phi )dt/d\tau =\text{constant}$. On converting to coordinate time, the additional term $(d^{2}t/d{\tau }^2)(d{x}^i/dt)$ is of order $v^2\nabla \Phi$ and can be dropped. Henceto leading order, the Newtonian limit of general relativity. Restoring SI units gives $\Phi ={\Phi }_{\mathrm{SI}}/c^2$, so the acceleration is $-\nabla (c^2\Phi )$. Substituting $L=1$ from proper-time normalization before varying would incorrectly erase the dynamics.

Set $f=1-2M(v)/r$, so $f_r=2M/r^2$ and $f_v=-2M^{\prime }/r$, where $M^{\prime }=dM/dv$. Dots in this solution mean derivatives with respect to proper time $\tau$. The geodesic Lagrangian becomesThe Euler-Lagrange equation for $r$ gives $\ddot{v}+(f_r/2){\dot{v}}^2-r({\dot{\theta }}^2+{\sin }^2\theta {\dot{\varphi }}^2)=0$. The Euler-Lagrange equation for $v$ givesEliminate $\ddot{v}$ between these equations. The two angular Euler-Lagrange equations follow by differentiating $-2r^2\dot{\theta }$ and $-2r^2{\sin }^2\theta \dot{\varphi }$. Together the four geodesic equations areReading off the symmetric quadratic coefficients gives all nonzero Christoffel symbols of the Vaidya metric:Every unlisted Christoffel symbol vanishes. The lower-index symmetry is that of the Levi-Civita connection. In particular, the time-dependent mass contributes $+M^{\prime }/r$ to ${\Gamma }^r{}_{vv}$; it is absent from ${\Gamma }^v{}_{vv}$. There are eleven independent nonzero entries, or fifteen when both lower-index orders are counted.

For a timelike tangent $v=dx/du$, put $L=\sqrt{-g(v,v)}>0$ and $\mathcal{L}=L^2/2=-g(v,v)/2$, and use the Euler-Lagrange equations. For the quadratic geodesic Lagrangian, they reduce toHere the first equation has been multiplied by an irrelevant minus sign. For the length Lagrangian, its derivatives are $L^{-1}$ times those of $\mathcal{L}$, and differentiating that factor yields insteadThis is the unparametrized geodesic equation, the key to why timelike length and energy have the same geodesic images. Define proper time locally by $d\tau /du=L>0$ and put $w=dx/d\tau =v/L$. The product rule then gives ${\nabla }_{w}w=L^{-2}[{\nabla }_{v}v-(\dot{L}/L)v]=0$ and $g(w,w)=-1$. Conversely an affinely parametrized timelike geodesic has constant $L$, by metric compatibility, and therefore satisfies both equations.

Thus the two variational problems have the same oriented timelike geodesic images, after reparametrization. They do not have exactly the same parametrized solutions with an arbitrary fixed $u$: in Minkowski spacetime, $t(u)=u^2$, $\mathbf{x}=0$, $u>0$, satisfies the length equation but not the quadratic equation. The timelike assumption excludes $L=0$, where this argument and the length derivative would fail.

For the Expanding flat slicing of de Sitter spacetime, take the geodesic LagrangianThe Euler-Lagrange equation for $t$ iswhile each spatial coordinate $x^i$ obeysComparing these equations with the geodesic equation gives the complete list of nonzero Christoffel symbols:

For an affine parameter $\lambda$, the geodesic Lagrangian isVarying $S=\int \mathcal{L}d\lambda$ gives the geodesic equations

The rotational Killing vector fields of the spherical symmetry conserve the angular-momentum vector. Its direction is fixed, so every nonradial orbit lies in the plane through the symmetry centre orthogonal to that vector. A spatial rotation can make this the equatorial plane. Equivalently, the initial conditions $\theta =\pi /2$ and $\dot{\theta }=0$ solve the $\theta$ equation uniquely. This is the planarity of geodesics in spherical symmetry; a radial geodesic has zero angular momentum and may be assigned any plane.

On $\theta =\pi /2$, the cyclic coordinates $t$ and $\varphi$ produce geodesic conserved quantities from Killing vectors,For a timelike geodesic parametrized by proper time, the proper-time normalization is $g_{ab}{\dot{x}}^a{\dot{x}}^b=-1$. Substitution of $E$ and $L$ gives the first-order radial equationThis is the timelike geodesic effective potential for static Anti-de Sitter spacetime with curvature radius one.

A geodesic passing through $r=0$ must have $L=0$. If it moves away from the origin, $E>1$, andTaking proper time $s=0$ at departure givesIt turns around at $s=\pi /2$ and first returns to the origin atThis energy-independent refocusing is the characteristic radial timelike geodesic in anti-de Sitter spacetime result; for curvature radius $a$, the answer is $\pi a$.

This is the Schwarzschild-de Sitter spacetime. Its geodesic Lagrangian may be taken asTime-translation and rotational symmetry give three independent integrals of motion, which may be chosen asThe constant norm of the tangent is $g_{ab}{\dot{x}}^a{\dot{x}}^b=-\kappa$, where $\kappa =1$ for a timelike geodesic parametrized by proper time and $\kappa =0$ for a null geodesic. Substitution givesand hence

The first fundamental form of a surface of revolution isThe geodesic Lagrangian is consequentlyThe rotation angle $u$ is a cyclic coordinate, so its conjugate momentum is conserved by the Euler-Lagrange equation:Thus $f(v(s){)}^2{u}^{\prime }(s)$ is constant.

The positively oriented unit tangent to the parallel of a surface of revolution through $\gamma (s)$ is $e_u={\varphi }_u/f(v)$. Since $\gamma$ has unit speed and $\theta$ is its angle with that parallel,Multiplication by $f(v(s))$ yieldswhich is Clairaut's relation.

Use the geodesic Lagrangianwhere here the overdot on a coordinate means differentiation with respect to the affine parameter. Applying the Euler-Lagrange equation to each coordinate and comparing with the geodesic equationdetermines the Christoffel symbols. To display derivatives of the metric coefficients unambiguously, write ${\lambda }_t={\partial }_t\lambda$, ${\lambda }_r={\partial }_r\lambda$, and similarly for $\mu$. The nonzero symbols, together with those obtained by symmetry in the two lower indices, areThis is the Christoffel symbols of a diagonal spherical spacetime metric.