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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.

With $u=t-z$, the quadratic geodesic Lagrangian isIts transverse Euler-Lagrange equations areThe longitudinal equations areafter using the difference of those same equations. In particular,This conserved quantity also follows from the Killing vector field ${\partial }_t+{\partial }_z$ and the geodesic conserved quantity from a Killing vector.

Affinely parametrized geodesics are the critical curves of the geodesic LagrangianFor $w=1+r^2$ and $w^{\prime }=2r$, the nonzero Christoffel symbols, up to symmetry in the lower indices, areThey follow either from the Euler-Lagrange equations or directly from the Levi-Civita connection formula.

The geodesics of the product-weighted diagonal plane metric follow from the geodesic Lagrangian $L=(y^2{\dot{x}}^2+x^2{\dot{y}}^2)/2$. With dots denoting an affine parameter derivative, its Euler-Lagrange equations giveSubstituting $y=mx$ with constant nonzero $m$ reduces both equations to $x\ddot{x}+{\dot{x}}^2=0$. Thus $x\dot{x}$ is constant, and a suitable affine parametrization exists; explicitly, $x^2=C_1\lambda +C_2$ on a segment of fixed sign.

These straight rays are geodesics only in the regular domain $xy\ne 0$. The metric tensor determinant is $x^2{y}^2$. On the axes, including the origin, this is a degenerate metric tensor: the calculation does not establish a regular geodesic continuing through the origin. The slope-zero ray is outside the regular domain. Geometrically, on each quadrant the change $\rho =|xy|/\sqrt{2}$, $\vartheta =\log |y/x|$ gives $d{s}^2=d{\rho }^2+{\rho }^{2}d{\vartheta }^2$. A fixed slope has constant $\vartheta$ and is therefore locally a radial geodesic of a flat metric tensor.

Put $f=1-2\mu /r$ and let dots now denote an affine parameter derivative. The null geodesic lagrangian in the orbital plane is $L=(f{c}^2{\dot{t}}^2-{\dot{r}}^2/f-r^2{\dot{\varphi }}^2)/2$. The time and azimuthal Euler-Lagrange equations integrate once; the null normalization givesFor completeness, the differential radial geodesic equation isthe remaining differential equations are $\ddot{t}+(f^{\prime }/f)\dot{t}\dot{r}=0$ and $\ddot{\varphi }+2\dot{r}\dot{\varphi }/r=0$. Spherical symmetry permits the equatorial choice.

Eliminating $\dot{t},\dot{\varphi }$ yields ${\dot{r}}^2+h^{2}f/r^2=c^2{k}^2$. For $h\ne 0$, divide by $h^2$:The impact parameter is unchanged by a common rescaling of the affine parameter. At infinity it is $c$ times angular momentum divided by energy. Radial photons have $h=0$ and obey the undivided equation ${\dot{r}}^2=c^2{k}^2$.

The effective potential is zero at $r=2\mu$, positive outside, and tends to zero as $r\to \infty$. Its derivative is $V_{\mathrm{eff}}^{\prime }=2(3\mu -r)/r^4$. ThusThe circular null geodesic lies at a maximum, since $V_{\mathrm{eff}}^{\prime \prime }(3\mu )=-2/(81{\mu }^4)<0$. It is unstable: a small radial displacement away from the maximum permits motion away from the orbit. For incoming photons from infinity in the exterior of a black hole, an outer turning point exists when $b>b_{\mathrm{crit}}$; for $b<b_{\mathrm{crit}}$ the photon can cross the event horizon. For a material central body, the vacuum circular orbit is relevant only if $3\mu$ lies outside its surface.

For the new coordinates, set $q=2\mu /r$, so $f=1-q$. The geodesic Lagrangian is $L^{\prime }=[f{c}^2{\dot{t}}^{\prime 2}-2qc{\dot{t}}^{\prime }\dot{r}-(1+q){\dot{r}}^2-r^2{\dot{\varphi }}^2]/2$. Cyclic-coordinate conservation and the null condition give the required analogous equations:Here $k$ denotes the same conserved time-translation quantity as before. The transformation is $t^{\prime }=t+(2\mu /c)\log |r/(2\mu )-1|$. In particular, the displayed time coordinate belongs to Ingoing Kerr-Schild coordinates for Schwarzschild spacetime; the usual advanced null coordinate of ingoing Eddington-Finkelstein coordinates is $v=t^{\prime }+r/c$. Using a metric tensor with a vanishing $d{r}^2$ coefficient while retaining the name $t^{\prime }$ would therefore solve a different coordinate representation.

An elimination that stays finite at the horizon is especially simple:The middle equality uses $q^2+f(1+q)=1$. Hence the radial energy equation and $V_{\mathrm{eff}}(r)$ are unchanged, including the horizon limit. The radial metric tensor block has determinant $-c^2$ and is regular at $r=2\mu$. Its radial null directions are $dr/(cd{t}^{\prime })=-1$ and $dr/(cd{t}^{\prime })=f/(1+q)$, as shown in the original schematic.

This coordinate invariance of the Schwarzschild photon radial potential follows because the transformation changes the time slicing but preserves the areal radius and the stationary Killing vector field ${\partial }_{t^{\prime }}={\partial }_t$. Consequently the conserved Killing energy, angular momentum and impact parameter are the same. photon capture, the circular orbit and the radial turning points cannot depend on coordinate labels; coordinate velocities can.

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.

Use arc length $s$ as the affine parameter, and put a dot over derivatives with respect to $s$. The geodesic Lagrangian on the sphere isThe cyclic coordinate $\varphi$ gives $\partial L/\partial \dot{\varphi }=a^2{\sin }^2\theta \dot{\varphi }=\text{constant}$. With the constant normalized as in the paper,Here $h$ has units of inverse length, because $s$ is physical arc length. Unit-speed normalization givesFor $h\ne 0$, division by ${\dot{\varphi }}^2$ yields the spherical-geodesic first integralThe turning latitudes satisfy $\sin \theta =|ah|$. In the usual polar-coordinate range $0\le \theta \le \pi$,The maximum departure from the equator, often the geometric quantity intended by an angular amplitude, is insteadThus polar angle and signed latitude must not be interchanged. At $|ah|=1$ the geodesic is the equator. At $h=0$ it is a meridian through the poles; $\varphi$ cannot serve as a global parameter and the divided equation is inapplicable. The limiting polar range is then $[0,\pi ]$.

The original PDF prints an ambiguous derivative with the square on $\theta$ rather than a conventional derivative notation. If the intended equation was second order, differentiating the first integral, or eliminating $\dot{\theta }$ from the Euler-Lagrange equation, gives the equally useful formIt has the same normalized geodesics and the same turning angles. Giving both forms resolves the typographical ambiguity without pretending that a second derivative is the square of a first derivative. Geometrically these trajectories are great circles, with inclination $\arccos |ah|$ to the equator.

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.

Use an affine parameter $\lambda$, with dots denoting its derivatives, and take $\alpha \ne 0$ so the metric is nondegenerate for $r>0$. The geodesic Lagrangian is $\mathcal{L}=({\dot{r}}^2+{\alpha }^2{r}^4{\dot{\varphi }}^2)/2$. The Euler-Lagrange equations giveor $\ddot{\varphi }+4\dot{r}\dot{\varphi }/r=0$. The conserved cyclic angular momentum isThis is the cyclic angular momentum in a rotational metric, associated with the rotational Killing vector ${\partial }_{\varphi }$. Explicitly, $\dot{J}={\alpha }^2(4r^3\dot{r}\dot{\varphi }+r^4\ddot{\varphi })=0$ by the angular equation.

Differentiating the Lagrangian and substituting both equations givesThus the squared tangent norm $2\mathcal{L}$ is constant. If $\lambda$ is arc length, it equals one; its numerical value otherwise depends on affine normalization. An arbitrary nonaffine parametrization would not preserve this constant-speed statement.

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.

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.

Use the $+---$ metric signature and abbreviate $f(r)=1-2\mu /r$, with $\mu =GM/c^2$ a length. Dots initially mean differentiation with respect to an affine parameter $\lambda$. The geodesic Lagrangian in the equatorial plane isThe planarity of geodesics in spherical symmetry permits this plane to be chosen. More explicitly, the full angular Euler-Lagrange equation is $\ddot{\theta }+2\dot{r}\dot{\theta }/r-\sin \theta \cos \theta {\dot{\varphi }}^2=0$, so initial data $\theta =\pi /2$, $\dot{\theta }=0$ preserve it.

Both $t$ and $\varphi$ are cyclic coordinates. Their Euler-Lagrange equations give conserved conjugate momenta $p_t=c^{2}f\dot{t}$ and $p_{\varphi }=-r^2\dot{\varphi }$. Defining constants $k,h$ therefore givesThe corresponding second-order geodesic equations areand variation of $r$ givesThe geodesic Lagrangian has no explicit $\lambda$ dependence. Its conserved Lagrangian energy function is ${\dot{x}}^{\alpha }\partial \mathcal{L}/\partial {\dot{x}}^{\alpha }-\mathcal{L}=\mathcal{L}$, since it is quadratic in the tangent components. Thus twice this function is a constant:This is the constant squared tangent norm of an affinely parametrized geodesic.