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The Einstein tensor and its contractions areTo match the supplied Ricci tensor components, use the Riemann curvature tensor conventionThis convention also fixes the sign used for geodesic deviation above. In vacuum, tracing the Einstein field equations gives $-R+4\Lambda =0$. Substitution therefore yields $R_{\mu \nu }=\Lambda g_{\mu \nu }$.

Differentiate the angular field equation to obtainConsequently $R_{00}=\Lambda A=\Lambda g_{00}$ and $R_{11}=-\Lambda /A=\Lambda g_{11}$. The other angular component follows from $R_{33}={\sin }^2\theta R_{22}$. Thus the single angular equation implies all the displayed vacuum equations wherever this coordinate chart is regular.

Integrating $(rA{)}^{\prime }=1-\Lambda r^2$ gives $A=1+C/r-\Lambda r^2/3$. Match $g_{tt}\simeq 1+2\Phi$ with the Newtonian gravitational potential $\Phi =-GM/r$ in the weak-field overlap region to identify $C=-2GM$. Hence the Schwarzschild-de Sitter spacetime hasNegative $\Lambda$ instead gives the Schwarzschild--anti-de Sitter black hole case. This matching fixes the normalization locally; a spacetime with nonzero $\Lambda$ is not asymptotically flat.

For timelike geodesics in the orbital plane, take proper time as parameter and use $L=(A{\dot{t}}^2-{\dot{r}}^2/A-r^2{\dot{\varphi }}^2)/2$. The cyclic coordinates give $A\dot{t}=k$ and $r^2\dot{\varphi }=h$; the angular canonical momentum is $-h$ in this signature. The four-velocity normalization $2L=1$ then givesHere $k$ is conserved Killing energy per unit rest mass in the chosen time normalization, and $h$ is conserved angular momentum per unit rest mass. There is no general interpretation of $k$ as energy measured at an asymptotically flat infinity when $\Lambda \ne 0$.

For radial motion $h=0$, differentiation gives $2\dot{r}\ddot{r}=-A^{\prime }\dot{r}$. The radial Euler-Lagrange equation extends the resulting expression through a turning point where division by $\dot{r}$ would fail. ThusThis is the proper-time second derivative of the areal radius, not a nonzero four-acceleration of the freely falling particle. Positive $\Lambda$ contributes an outward term; negative $\Lambda$ reinforces inward attraction. Restoring units in the slow-motion, weak-field Newtonian limit givesThe extra term describes the gravitational effect of a cosmological constant, not drag or a contact force. The approximation requires $GM/(r{c}^2)\ll 1$, $|\Lambda |r^2\ll 1$ and speed much less than $c$.

For geodesic motion of uncharged dust, use the pressureless stress-energy form, $T^{\mu \nu }=\rho u^{\mu }{u}^{\nu }$ and $u_{\nu }{u}^{\nu }=c^2$. stress-energy conservation expands toContracting with $u_{\nu }$ kills the second term, since $u_{\nu }{\nabla }_{\mu }{u}^{\nu }=\frac{1}{2}{\nabla }_{\mu }(c^2)=0$. Hence ${\nabla }_{\mu }(\rho u^{\mu })=0$, and wherever $\rho >0$,On a flow line $u^{\mu }=d{x}^{\mu }/d\tau$, so this becomes $d^2{x}^{\nu }/d{\tau }^2+{\Gamma }^{\nu }{}_{\alpha \beta }(d{x}^{\alpha }/d\tau )(d{x}^{\beta }/d\tau )=0$. Each relativistic dust element therefore follows a timelike geodesic with proper time as an affine parameter. pressure gradients or external nongravitational forces would add an acceleration term.

Throughout this solution $c=1$ and $a>0$.

Null metric tensor and boost. Inverting the coordinate definitions gives $t=(p+q)/2$, $x=(q-p)/2$, and henceFor a frame moving at speed $v$ along positive $x$, use $t^{\prime }=\gamma (t-vx)$, $x^{\prime }=\gamma (x-vt)$. Their difference and sum giveThe reciprocal scale factors preserve $dpdq$. They are $e^{\pm {\eta }_v}$ in rapidity notation.

acceleration invariants. Proper-time normalization gives ${\dot{t}}^2-{\dot{x}}^2=1$, so $\dot{p}\dot{q}=1$. The four-acceleration has squared norm ${\ddot{t}}^2-{\ddot{x}}^2=-a^2$; thereforeFor future motion write $\dot{p}=e^{-\eta }$ and $\dot{q}=e^{\eta }$. Then the second equation is $-(\dot{\eta }{)}^2=-a^2$. acceleration toward positive $x$ selects $\eta ={\eta }_0+a\tau$.

hyperbolic motion from an initial event and rapidity. The stated initial event fixes $p(0)=-1/a$, $q(0)=1/a$, but not the initial rapidity ${\eta }_0$. Integrating with positive signed acceleration gives the general familyThe initial event alone does not fix the initial velocity or signed acceleration. The trajectory explicitly requested in the next paragraph selects the additional conditions ${\eta }_0=0$ and positive-$x$ acceleration. With that convention,and consequentlyWith arbitrary ${\eta }_0$ the equivalent expressions are $t=[\sinh ({\eta }_0+a\tau )-\sinh {\eta }_0]/a$ and $x=1/a+[\cosh ({\eta }_0+a\tau )-\cosh {\eta }_0]/a$. Thus position data alone do not select the displayed special hyperbola. The following calculations use that intended hyperbola, which is also explicitly given by the paper.

Diagram and horizon. The selected world line is the right-hand branch $x^2-t^2=a^{-2}$. Its future asymptote $x=t$ is a reception horizon: rightmoving signals have constant $p=t-x$, whereas the observer has $p<0$ and approaches zero only as $\tau \to \infty$. Events with $p\ge 0$ cannot send a future signal to this observer if the acceleration continues indefinitely. This is an observer-dependent causal restriction in flat spacetime, not a curvature singularity.

photon reception by a uniformly accelerated observer. A rightmoving photon from the stationary source has four-momentum $P^{\mu }=h{\nu }_0(1,1)$. Contracting with the observer's four-velocity givesThe same result follows by differentiating its emission-time relation. A photon emitted at $x=L$ and time $t_e$ has $p=t_e-L$, so intersection requiresThus $t_e=L$ is the supremum of receivable emission times, reached only in the limit of infinite receiver proper time. A photon emitted exactly then lies on the horizon and is never received at finite time; later photons cannot catch the ship. Restoring dimensions gives the limiting emission time $L/c$ for this chosen origin and trajectory.

Rindler coordinates. Substitution of the selected trajectory into the coordinate transformation givesDifferentiating the transformation also gives $d{s}^2=e^{2a\zeta }(d{\xi }^2-d{\zeta }^2)$. Curves of fixed $\zeta$ are hyperbolae, while curves of fixed $\xi$ are straight rays $t/x=\tanh (a\xi )$ within the right wedge. The coordinate $\xi$ is the selected ship's proper clock; at another fixed $\zeta$, proper time obeys $d{\tau }_{\zeta }=e^{a\zeta }d\xi$. The spatial coordinate is logarithmic in $\rho =\sqrt{x^2-t^2}=e^{a\zeta }/a$, rather than being proper distance itself. Fixed-$\zeta$ observers have proper acceleration $a{e}^{-a\zeta }$.

Boost and velocity transformation. For the moving frame,The inverse has $v\mapsto -v$. Taking differentials of that inverse and dividing by $dt$ gives

Constant proper acceleration. For collinear motion, the instantaneous rest-frame acceleration is $a={\gamma }^{3}dv/dt$. Since $dt/d\tau =\gamma$,Write $v=c\tanh \eta$; then $d\eta /d\tau =a/c$. Starting from rest gives $\eta =a\tau /c$, $dt/d\tau =\cosh \eta$ and $dx/d\tau =c\sinh \eta$. integration with the stated initial origin yields

photon rocket mass ratio. In the instantaneous rest frame, ejecting backward photon energy $d{E}_{\gamma }$ gives, to first order, $c^{2}dM=-d{E}_{\gamma }$ and $Mcd\eta =d{E}_{\gamma }/c$. Hence $dM/M=-d\eta$, andHere $M$ includes all stored fuel and energy. This assumes complete conversion into a perfectly collimated photon exhaust without other losses.

photon-rocket fuel for acceleration and braking. With effectively instantaneous changes of speed, the cruise proper time is $D/(\gamma v)=10\mathrm{yr}$, givingBraking to rest costs another factor $e^{-\eta }$, so $M_f/M_i=e^{-2\eta }=(\sqrt{101}-10{)}^2=0.00248758$. If equal rest masses of matter and antimatter annihilate, the consumed antimatter is half the total consumed fuel:This is the fraction of initial total mass, not the antimatter fraction of the consumed matter–antimatter fuel, which is one half.

finite-acceleration correction to a relativistic cruise. Each constant-proper-acceleration leg has proper duration ${\tau }_a=c\eta /a$ and Earth-frame length $x_a=(c^2/a)(\gamma -1)$. ThusSubtracting the original $D/(\gamma v)$ givesFor $a=10\mathrm{m}{\mathrm{s}}^{-2}$, each accelerating leg takes about $2.85$ proper years and covers $8.60$ light years in the Earth frame, leaving positive cruise distance. The new total is about 13.98 proper years. Adding $2{\tau }_a$ to ten years would overcount the cruise segment replaced by acceleration and braking.

geodesic integrals and affine normalization. Put $F(r)=1-2\mu /r$. Choose the affine parameter to be proper time, so $g_{ab}{\dot{x}}^a{\dot{x}}^b=c^2$. Cyclic $t$ and $\varphi$ in the geodesic Lagrangian give $F\dot{t}=k$ and $r^2\dot{\varphi }=h$. Substituting into the four-velocity normalization giveshenceAn arbitrary constant rescaling of affine parameter changes its normalization; the displayed unit constant assumes the proper-time choice.

circular orbit. Define $V(r)=h^2/r^2-2\mu h^2/r^3-2\mu c^2/r$. The radial geodesic equation is $\ddot{r}=-V^{\prime }/2$. A circular geodesic therefore needs $V^{\prime }(R)=0$, not merely $\dot{r}=0$ at an instant. Solving and substituting into the radial first integral givesExistence requires $R>3\mu$. circular orbits below $6\mu$ are radially unstable, although they still define the instantaneous comparison here.

orthogonal local velocities in Schwarzschild spacetime. Use Alice's static orthonormal frame at the meeting event. Her proper clock is $d{\tau }_A=\sqrt{F}dt$ and tangential proper distance is $Rd\varphi$. Bob's local tangential speed isCarol has $h_C=0$ and $k_C=1$, because she starts from rest at infinity. The radial integral gives ${\dot{r}}_C=-c\sqrt{2\mu /R}$. Her static-frame radial speed is $(dr/\sqrt{F})/(\sqrt{F}dt)=dr/(Fdt)$, and $F{\dot{t}}_C=1$, soThis is inward radial motion, perpendicular to Bob's tangential motion in Alice's frame.

Their Lorentz factors relative to Alice are ${\gamma }_{\mathrm{AB}}=\sqrt{(R-2\mu )/(R-3\mu )}$ and ${\gamma }_{\mathrm{AC}}=\sqrt{R/(R-2\mu )}$. The four-velocity contraction therefore gives

Special-relativistic consistency. Let Bob's tangential velocity in Alice's local frame be $u$ and Carol's inward radial speed be $w$. A boost to Bob's frame gives Carol components $v_{\mathrm{tang}}^{\prime }=-u$ and $v_{\mathrm{rad}}^{\prime }=-w/{\gamma }_u$. ConsequentlyThus the three measured relative speeds agree with local Lorentz velocity addition. Ordinary Euclidean subtraction of the two velocity vectors would omit the transverse-velocity transformation.

Put $f(r)=1-2\mu /r$, where $\mu =GM/c^2$, and work in the orbital plane. A dot denotes an affine-parameter derivative; for a material particle choose that parameter to be proper time.

First integrals and normalization. The geodesic Lagrangian isTime translation and axial rotation make $t$ and $\varphi$ cyclic, givingThe tangent norm is constant along an affine geodesic because the Levi-Civita connection preserves the metric. ThereforeFor proper-time parametrization of a timelike worldline this norm is $c^2$. For a null geodesic it is zero, for every affine normalization. Thus $\mathcal{E}=c^2$ for the chosen material-particle normalization and $\mathcal{E}=0$ for light. Proper time cannot itself parametrize a null ray, despite the same printed symbol $\tau$ being used for its affine parameter.

Null normalization and impact parameter. Under $\tau \mapsto A\tau +B$ with $A>0$, both $k$ and $h$ scale as $1/A$, while the null path is unchanged. Their ratio is invariant. At large radius the spatial speed with respect to asymptotic time is $c$, and orbital angular momentum per affine tangent givesFor an orientation chosen with $h>0$, this is $h=bc$. Setting $k=1$ is a choice of affine scale, not a claim that all photons have equal measured energy. Once a tangent is identified with a physically normalized four-momentum, its energy fixes that scale.

Orbit shape. Substituting the constants in the norm and multiplying by $f$ givesFor $u=1/r$ and $h\ne 0$, $du/d\varphi =-\dot{r}/h$. HenceThe $h=0$ radial orbit must be treated with the radial equation, because azimuth is then constant.

Circular photons and stability. For a null ray define $W(r)=f(r)/r^2$. Its radial equation becomesThe circular orbit requires $\dot{r}=0$ and $W^{\prime }=0$, soIt is at a maximum of the null effective potential and is unstable. Equivalently the Binet equation $u^{\prime \prime }+u=3\mu u^2$ linearizes about $u_0=1/(3\mu )$ to $\delta u^{\prime \prime }=\delta u$, admitting exponentially growing radial deviations in azimuth.

Capture from infinity. Divide the null radial equation by $k^2{c}^2$. The quantity opposing inward motion is $b^{2}W(r)$, whose exterior maximum is $b^2/(27{\mu }^2)$. There is no outer radial turning point precisely whenAt equality an exactly critical ray from infinity asymptotes to the photon sphere; it does not cross the horizon at a finite affine parameter. Above the threshold an initially inward ray from infinity reaches an outer turning point and scatters. The radial case $b=0$ is also captured. These statements refer to inward rays launched from the asymptotically flat exterior, not arbitrary initial data inside the angular-momentum barrier.

Approach to the singularity. A captured nonradial ray bends inward, crosses $r=3\mu$ and then the event horizon $r=2\mu$, and continues toward the curvature singularity. A plot in $(r\cos \varphi ,r\sin \varphi )$ is only a coordinate projection, not a Euclidean spatial slice inside the horizon. For large $u$ the cubic term in the shape equation dominates:for the inward branch with $h>0$. Its integral isThus the azimuth tends to a finite limit at the singularity, rather than winding infinitely often there. Near-critical rays can accumulate many turns near $3\mu$ before capture, which is a separate phenomenon. A strictly radial ray has constant azimuth and does not admit this azimuthal asymptotic form. The radial equation also gives $r\propto ({\tau }_{\ast }-\tau {)}^{2/5}$ for a nonradial null ray, confirming finite remaining affine parameter.

Material capture and the condition at infinity. To recover the quoted threshold, take the usual marginally bound orbit, initially at rest at infinity: $k=1$. With $\mathcal{E}=c^2$,The turning-point quadratic has discriminant $h^2(h^2-16{\mu }^2{c}^2)$. It is positive for large $r$ and has no real positive root when $0<|h|<4\mu c$, while the radial case also falls in. ThereforeAt equality the quadratic is proportional to $(r-4\mu {)}^2$ and the orbit asymptotes to the unstable marginally bound circular orbit at $r=4\mu$. For larger $|h|$, an incoming marginally bound particle encounters the outer turning point and scatters.

The source says only “from infinity”, without explicitly fixing the asymptotic speed. For a general incoming speed $v_{\infty }$, $k=(1-v_{\infty }^2/c^2{)}^{-1/2}\ge 1$ andThe same bound $|h|<4\mu c$ remains sufficient for capture, since the added constant is nonnegative. It is the exact threshold only for $k=1$. At fixed $k>1$, the critical unstable orbit satisfiesThen $h_c>4\mu c$. This states both the printed sufficient condition and the additional assumption needed to call it a sharp capture boundary.

Consider an ideal spherical collapse with a vacuum exterior of constant mass $M$. Once pressure support is insufficient, a surface can continue inward and cross the Schwarzschild event horizon $r_s=2GM/c^2$ in finite comoving proper time. An event horizon is a causal boundary: light emitted from inside cannot reach distant observers. It is not a material shell or, in the classical Schwarzschild solution, a local curvature singularity at $r=r_s$.

The distant view of gravitational collapse differs from the surface clock's history. Exterior Schwarzschild time has $dt/d\tau \propto (1-r_s/r{)}^{-1}$ for regular infall and diverges as the surface approaches the horizon. Outgoing radial light also takes an increasingly long coordinate time to escape: $dt/dr=[c(1-r_s/r){]}^{-1}$ has a logarithmically divergent integral at the horizon. Signals received far away are progressively delayed and redshifted; both gravity and the inward emitter's Doppler shift contribute. photons carry less received energy and arrive at a lower rate, while the available escape directions also narrow.

Thus a distant observer does not receive a light signal announcing the crossing; the image fades and reddens rather than remaining a permanently bright frozen surface. A finite-band detector with finite sensitivity loses it in finite observation time. The divergence of one coordinate chart does not prevent the material crossing in finite proper time; horizon-regular coordinates describe that continuation. The pressureless Oppenheimer-Snyder model makes this distinction explicit, without representing all the pressure, rotation and radiation of a real stellar collapse.^[ref]

The plotted example is a test-particle geodesic falling from rest at infinity, used to isolate the coordinate and signal effects; it is not a fitted stellar-collapse light curve. No actual observer can hover on the horizon to supply a static local frame there.

Take $\tau$ as particle proper time, dots as $d/d\tau$, and $f(r)=1-2\mu /r$ with $\mu =GM/c^2$. In the equatorial plane the geodesic Lagrangian isThe cyclic coordinates give $d(c^{2}f\dot{t})/d\tau =0$ and $d(-r^2\dot{\varphi })/d\tau =0$. Set the resulting constants to $c^{2}k$ and $-h$ respectively. The explicit second-order equations areThe polar equation is satisfied by $\theta =\pi /2$, $\dot{\theta }=0$, consistently with spherical symmetry. The proper-time normalization $d{s}^2=c^{2}d{\tau }^2$ supplies the third first integral:The normalization is $2\mathcal{L}=c^2$; it is distinct from choosing an arbitrary affine tangent's norm.

Stationarity and axial symmetry make the covariant time and azimuthal momenta constant. For physical momentum and coordinate $t$ in seconds, $P_t=m{c}^{2}f\dot{t}=m{c}^{2}k$ and $P_{\varphi }=-m{r}^2\dot{\varphi }=-mh$. Thus $m{c}^{2}k$ is the conserved energy normalized at infinity and $mh$ is the signed axial angular momentum. This is Killing energy with positive time signature; a convention with negative timelike norm uses the opposite contraction sign. The supplied covariant-momentum form of the geodesic equation makes these cyclic constants immediate, but its parametrization must be understood: for $p^{\mu }=d{x}^{\mu }/d\lambda$ it is exactly as written. For physical $P^{\mu }=md{x}^{\mu }/d\tau$, proper-time differentiation instead gives $d{P}_{\mu }/d\tau =({\partial }_{\mu }{g}_{\nu \sigma })P^{\nu }{P}^{\sigma }/(2m)$; equivalently, the displayed mass-free form uses $\lambda =\tau /m$. This affine momentum normalization does not change either conservation law.

Multiply the normalization equation by $f$ and substitute the two cyclic integrals. Rearrangement gives the radial energy equationFor radial marginal escape $h=0$ and the future-directed constant is $k=1$, because the speed at infinity vanishes. Choose the outward root:Hence the marginal escape geodesic in Schwarzschild spacetime hasThe constant $a$ has units of length to the power $3/2$ divided by time. This radius is an areal coordinate, not accumulated local proper distance.

A Schwarzschild static observer at exterior radius $r$ uses local radial rulers $d\ell =dr/\sqrt{f}$ and a local clock $dT=\sqrt{f}dt$. Therefore the local radial speed of a Schwarzschild geodesic isFor the marginal trajectory $k=1$ this givesIts equality to $\dot{r}$ is particular to $k=1$; the local definition, rather than that accidental equality, is the required observer measurement. The Schwarzschild-coordinate velocity is instead $dr/dt=f\sqrt{2GM/r}$.

For nonzero asymptotic speed $0\le u<c$, the same energy constant is $k={\gamma }_{\infty }=(1-u^2/c^2{)}^{-1/2}$. Combining the radial equation with the local-ruler relation gives radial Schwarzschild escape with asymptotic speedThe positive root describes the outward particle, at radii on its trajectory with $r\ge r_0$. Independently, a static observer has four-velocity $U=(1/\sqrt{f},0,0,0)$ in these coordinates, giving relative Lorentz factor $g(U,u_{\mathrm{particle}})/c^2=\sqrt{f}\dot{t}=k/\sqrt{f}$ and the same speed. As $r\to \infty$ it tends to $u$; for $u=0$ it gives the preceding result. Its limiting value approaches $c$ as $r\downarrow 2GM/c^2$, but that is a limit through exterior observers: no static timelike observer exists on or inside the horizon.

Write $r_s=2GM/c^2$ and $f=1-r_s/r$. The Schwarzschild tortoise coordinate satisfies $d{r}_{\ast }/dr=1/f$ and can be chosen as $r_{\ast }=r+r_s\ln |r/r_s-1|$. Define the advanced time $v=t+r_{\ast }/c$. Substituting $dt=dv-dr/(cf)$ into the Schwarzschild metric gives the ingoing Eddington-Finkelstein coordinates formThe coefficients are regular at $r=r_s$ and the radial-block determinant is $-c^2$, so the exterior Schwarzschild-time divergence is a coordinate singularity at the future event horizon. The physical curvature singularity at $r=0$ remains.

The two families of radial null trajectories in ingoing Eddington-Finkelstein coordinates obeyOutside the event horizon, the outgoing family increases $r$; at the horizon it generates the horizon itself; inside, even the outgoing future-directed family decreases $r$. An infaller crosses the future horizon at finite $v$ and finite proper time. The chart makes this causal geometry visible without mistaking Schwarzschild coordinate time for the infaller's clock.

Let $x^{\mu }(\tau )$ denote the spacetime coordinates, with dots meaning differentiation with respect to an affine parameter; for a massive particle take proper time $\tau$. The metric tensor determines the quadratic geodesic Lagrangian $L=g_{\mu \nu }{\dot{x}}^{\mu }{\dot{x}}^{\nu }$. The Euler-Lagrange equations areSymmetry of the metric tensor gives $\partial L/\partial {\dot{x}}^{\mu }=2g_{\mu \nu }{\dot{x}}^{\nu }$. Hence the covariant four-momentum is $p_{\mu }=m{g}_{\mu \nu }{\dot{x}}^{\nu }=(m/2)\partial L/\partial {\dot{x}}^{\mu }$. If $x^{\mu }$ is a cyclic coordinate, its Euler-Lagrange equation immediately gives ${\dot{p}}_{\mu }=0$.

For the rotating metric, $t,\varphi ,z$ are all cyclic. ThusCombining the first two equations cancels their $r$ dependence:The radial Euler-Lagrange equation is $-2\ddot{r}=-2r(\dot{\varphi }+\omega \dot{t}{)}^2$, soThis is the geodesic equation in a uniformly rotating Minkowski chart. With inertial angle $\Phi =\varphi +\omega t$, $-p_{\varphi }$ is the inertial axial angular momentum. The outward radial-coordinate acceleration expresses the familiar centrifugal term associated with angular motion: the underlying free trajectory is a straight line in Minkowski spacetime. It does not imply an external radial force. Holding a particle at rest in the rotating reference frame instead requires a real inward force.

Curvature contractions. Use the index contraction stipulated in the question. In three dimensions,The same contraction in four dimensions givesThen the Ricci scalar is $R=4\Lambda$ and the Einstein tensor is $G_{\mu \nu }=-\Lambda g_{\mu \nu }$. Thus $R_{\mu \nu }=\Lambda g_{\mu \nu }$ is precisely the vacuum Einstein field equations $G_{\mu \nu }+\Lambda g_{\mu \nu }=0$.

The static solution. For $A=1+{\lambda }^2{r}^2$, $A^{\prime }=2{\lambda }^{2}r$ and $A^{\prime \prime }=2{\lambda }^2$. The given Ricci tensor components becomeEach equals $-3{\lambda }^2$ times the corresponding metric tensor component. ThereforeThis is Anti-de Sitter spacetime in static coordinates on anti-de Sitter spacetime, with curvature radius $1/\lambda$. The overall metric signature is opposite to the convention used in some descriptions of those coordinates; the geodesic constants below use the question's $(+,-,-,-)$ convention.

Conserved quantities and radial motion. In the equatorial plane, the geodesic Lagrangian is $L=A{\dot{t}}^2-{\dot{r}}^2/A-r^2{\dot{\varphi }}^2$. Its cyclic coordinates giveFor rest mass $m$, $mk=p_t$ is the conserved Killing energy and $mh=-p_{\varphi }$ is the axial angular momentum. Timelike normalization $L=1$ givesFor purely radial motion $h=0$, differentiating away from a turning point gives $\ddot{r}=-{\lambda }^{2}r$, and continuity extends that equation through the turning point. An outward launch from the origin at $\tau =0$ hasThis is the radial timelike geodesic in anti-de Sitter spacetime. Its first return to the origin isindependent of the outward launch speed for every nontrivial launch $k>1$. The coordinate $r$ is nonnegative, so this sine expression describes the excursion up to that return; subsequent passage through the origin requires changing the radial direction or using a signed diameter coordinate. The speed changes the amplitude, not the harmonic frequency.

Circular periods. The effective radial potential is $V_{\mathrm{eff}}=A(1+h^2/r^2)$. For a circle at $r_0$, $V_{\mathrm{eff}}^{\prime }=0$ gives $h^2={\lambda }^2{r}_0^4$, and normalization gives $k=A(r_0)$. Thus $\dot{t}=1$ and $\dot{\varphi }=\pm \lambda$. A stable circular timelike geodesic in anti-de Sitter spacetime therefore hasindependently of radius. This is the same isochronous behavior as the central harmonic radial equation. A static clock placed at the orbit's radius instead measures $\sqrt{1+{\lambda }^2{r}_0^2}\Delta t$, which does depend on radius; it is a different clock from the orbiting particle's proper time.

The cosmological principle gives spatial homogeneity and cosmological isotropy for the large-scale spatial slices. About any chosen comoving observer, spherical coordinates therefore separate a radial direction from the angular directions. In particular,is the metric tensor on the unit sphere. Its appearance expresses cosmological isotropy, rather than a privileged centre of the universe. The factor $(1-K{r}^2{)}^{-1}$ permits the radial geometry to have nonzero spatial curvature of an FLRW universe.

The scale factor $a(t)$ changes every spatial separation between fixed comoving coordinates by the same multiplicative factor. For example, the radial proper distance in cosmology between fixed radial labels at a fixed cosmic time isExpansion or contraction through this single function preserves spatial homogeneity and cosmological isotropy; direction-dependent scale factors would generally spoil the latter. The Friedmann-Lemaître-Robertson-Walker metric specifies this possible evolution geometrically; a dynamical theory is needed to determine $a(t)$.