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Describe the geodesic variation by a smooth map $F(s,\tau )$, with each curve $\tau \mapsto F(s,\tau )$ a timelike geodesic parametrized by proper time. Its tangent vector is $T=F_{\ast }{\partial }_{\tau }$ and its deviation vector is $S=F_{\ast }{\partial }_s$. Thus $S$ measures the first-order separation of neighboring members at the same value of proper time.

Mixed coordinate derivatives on the parameter surface commute. Consequently, along the variation,For an embedded two-dimensional variation this is the usual Lie bracket of vector fields on its image; more generally the same identity is read along the map $F$ using its commuting parameter derivatives. Since the Levi-Civita connection is torsion-free, it follows that ${\nabla }_{T}S={\nabla }_{S}T$.

Use precisely the stated convention for the Riemann curvature tensor, and let $D={\nabla }_T$. The geodesic equation says ${\nabla }_{T}T=0$ because proper time is an affine parameter. The commuting variation and the torsion-free connection giveThus the geodesic deviation equation isHere $(R(X,Y)Z{)}^a=R^a{}_{bcd}{Z}^b{X}^c{Y}^d$. Declaring this index convention fixes the sign: writing $R(S,T)T$ instead would require a minus sign. A deviation vector satisfying this equation is a Jacobi field.

For every member of the geodesic variation, proper time normalization gives $g(T,T)=-1$. By metric compatibility, the geodesic equation, and ${\nabla }_{T}S={\nabla }_{S}T$,The last equality uses normalization across the whole variation, not merely constancy along one curve. Hence $g(S,T)$ is constant along each timelike geodesic, provingThis is orthogonality of proper-time deviation vectors.

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.

The unimodular spatial metric separates volume from shape: writing $h_{ij}=a^2{\tilde{h}}_{ij}$ and $a^6=h=\det h_{ij}$ gives $\det \tilde{h}=1$. Contract the extrinsic curvature with a negative shift with $h^{ij}$. Metric compatibility and the Jacobi determinant derivative formula giveConsequentlyFor zero shift vector, $\dot{h}/h=6\dot{a}/a$, soThe infinitesimal spatial volume is $\sqrt{h}{d}^{3}x=a^3{d}^{3}x$; its logarithmic change per unit normal proper time is $3H$. Thus this $H$ is the local volume Hubble parameter, reducing to the usual Hubble parameter in a homogeneous universe. The determinant-defined $a$ refers to fixed spatial coordinates, while this normal volume expansion has a geometric interpretation.

A static observer has four-velocity $u=F^{-1/2}{\partial }_t$. Its only nonzero acceleration component is $a^r={\Gamma }^r{}_{tt}/F=F^{\prime }/2=-r/R^2$. Using $g_{rr}=F^{-1}$, the magnitude isIt vanishes only at $r=0$, where $F=1$ and $d\tau =\sqrt{F}dt=dt$. Thus $t$ is proper time for the central inertial static observer. Other static observers need inward acceleration to resist de Sitter separation and have a redshifted proper time.

For the static patch of de Sitter spacetime, $r_{\ast }=\int dr/F=(R/2)\log |(R+r)/(R-r)|$. With $v=t+r_{\ast }$ or $u=t-r_{\ast }$, respectively,Both radial blocks have determinant $-1$ and smooth coefficients at $r=R$. The normal has norm $g^{rr}=F$, so that surface is null. There $k={\partial }_v$ has covector $dr$, or $k={\partial }_u$ has covector $-dr$, making it normal as well as tangent: it is a Killing horizon.

For the signed surface gravity of a de Sitter horizon, since ${\partial }_r{k}^2=-F^{\prime }(R)=2/R$, part a gives $\kappa =-1/R$ in the ingoing chart on the past branch, and $\kappa =+1/R$ in the outgoing chart on the future branch, with $k$ future-directed in the static patch. The physical magnitude is $\boxed{|\kappa |=1/R}$, normalized by central proper time. A negative signed $F^{\prime }(R)/2$ must not become a negative temperature.

Choose spin orientation so that $0<a<M$; the printed $M/a$ threshold assumes nonzero positive $a$. Put $H=2Mr/(r^2+a^2)$ and $F=1-H$. The axial metric has components $(-F,H,1+H)$ and determinant $-1$. Conserved energy and timelike normalization giveEliminating $\dot{\tilde{t}}$ gives the Kerr axial timelike effective potentialIt extends through the zeros of $F$ in the regular original coordinates. Since $V_{\mathrm{eff}}(0)=1$, reaching zero requires ${\epsilon }^2\ge 1$. At equality the infalling orbit reaches zero with zero radial velocity and turns back, because the nearby negative sheet is forbidden. Along the axis zero is regular, unlike the off-axis ring.

For $r<0$, $V_{\mathrm{eff}}^{\prime }=2M(r^2-a^2)/(r^2+a^2{)}^2$ and the global maximum is $V_{\mathrm{eff}}(-a)=1+M/a$. If ${\epsilon }^2>1+M/a$, an inward particle has no negative-sheet turning point. It passes through zero and continues toward $\boxed{r\to -\infty }$, another asymptotically flat end, with $\dot{r}\to -\sqrt{{\epsilon }^2-1}$ and infinite proper time to infinity. It first crosses the outer and inner horizons. Perturbative Cauchy-horizon instability can invalidate this ideal Kerr axial analytic extension physically.

This is the two-dimensional axis restriction. The dashed $r=0$ curve is regular; the four-dimensional ring lies off-axis and is not an axial singular boundary.

The future domain of dependence $D^{+}(\Sigma )$ consists of points through which every past-inextendible causal curve meets the partial Cauchy hypersurface $\Sigma$. Data on $\Sigma$ determine evolution there for suitable hyperbolic equations. Its future boundary is the Cauchy horizon $H^{+}(\Sigma )$; one precise definition is $\overline{D^{+}(\Sigma )}\setminus I^{-}(D^{+}(\Sigma ))$.

For a nonextremal Reissner-Nordstrom spacetime, $0<|Q|<M$, the metric function is $F=1-2M/r+Q^2/r^2$, with $r_{\pm }=M\pm \sqrt{M^2-Q^2}$. The outer event horizon lies at $r_{+}$. The future trapped region leads toward the inner null surface $r_{-}$, a future Cauchy horizon for bridge data on the illustrated $\Sigma$. Beyond it some past causal curves arrive from other analytic blocks or timelike singularities without meeting $\Sigma$. The CP diagram repeats these analytic blocks; $r=0$ is a timelike curvature singularity.

Near the inner horizon, arbitrarily late exterior radiation is compressed into finite infaller proper time. Its frequency is exponentially blueshifted, with scale $e^{|{\kappa }_{-}|v}$, so even a decaying flux can have diverging measured stress. Generic ingoing and outgoing perturbations produce mass inflation, growing internal mass and curvature. The smooth RN Cauchy horizon is expected to be unstable to back-reaction. This is a generic instability argument, not a universal theorem from every specially tuned test trajectory; the approximation of a harmless test particle becomes inconsistent.

Use metric signature $(-,+,+,+)$ and a future-oriented worldline. Variation of the worldline einbein gives the mass-shell condition; variation of momentum givesEliminating $p$ first leaves $L={\dot{x}}^2/(2e)-e{m}^2/2$. Its $e$ equation is ${\dot{x}}^2=-e^2{m}^2$. Choose the positive lapse branch $e=\sqrt{-{\dot{x}}^2}/m$; substituting gives $L=-m\sqrt{-{\dot{x}}^2}$. Thus $\boxed{S=-m\int \sqrt{-g_{\mu \nu }d{x}^{\mu }d{x}^{\nu }}=-m\int d\tau }$, the proper time action. The lapse branch fixes the sign convention. Eliminating auxiliary variables on their algebraic equations preserves the worldline equations, which are timelike geodesics up to parametrization.

Assume positive fluid density, ${\rho }_0>0$, and write $m=1+3w>0$. A perfect fluid with constant equation of state obeys the cosmological perfect-fluid continuity equation, givingIn the stated units the Friedmann equation is $H^2=\rho /3-k/a^2+\Lambda /3$. Multiplication by $a^2/2$ yields the Friedmann effective potential for a constant-equation-of-state fluid,The allowed region has $V\le 0$. The Friedmann acceleration equation is equivalently $\ddot{a}=-V^{\prime }(a)$, including turning points by continuity. Thus a zero-energy mechanical trajectory reproduces the cosmological evolution, with $a>0$.

For $k=0$, $\Lambda <0$, the potential increases strictly from negative infinity to positive infinity. Its unique zero givesExpansion from the Big Bang stops there, with negative acceleration, then reverses into a Big Crunch. Both the turning point and the final singularity occur in finite proper time: $dt=da/\sqrt{-2V}$ is integrable near a simple turning point and behaves as a constant times $a^{m/2}da$ near zero.

For $\Lambda =0$, $k=+1$, $V$ increases from negative infinity to $1/2$ and crosses zero atThis is again expansion followed by finite-time recollapse. For $k=-1$, the same increasing curve has asymptote $-1/2$ and never meets zero. Expansion continues without a finite maximum; asymptotically $\dot{a}\to 1$ and $a\sim t$, as spatial curvature dominates the diluted fluid.

For $k=0$, $\Lambda >0$, $V$ tends to negative infinity at both ends. It has a maximum atThere is no turning point. Expansion initially decelerates, then accelerates once $a>a_{\ast }$, and approaches de Sitter spacetime expansion $a\propto e^{\sqrt{\Lambda /3}t}$. Thus positive flat dark-energy expansion is unbounded; negative flat dark energy and positive curvature without dark energy recollapse.

Use conformal time, $dt=ad\tau$, and let primes mean $d/d\tau$. Then $a^{\prime }=a\dot{a}$ and $a^{\prime \prime }=a{\dot{a}}^2+a^2\ddot{a}$. For $k=1$, $\Lambda =0$, the Friedmann equation and acceleration equation giveSubstitution yieldsUse the supplied solution without deriving it. Its first maximum has sine equal to one, so its amplitude must equal the physical turnaround scale in part (a):Choose the expanding branch with $B=0$ and the first zero at $\tau =0$.

For pressureless matter, $w=0$, the closed matter-dominated Friedmann solution isThe Big Crunch is at ${\tau }_c=2\pi$, or proper time $t_c=\pi A=\pi {\rho }_0/3$, measured from the Big Bang.

For radiation, $w=1/3$, the closed radiation-dominated Friedmann solution isThe Big Crunch is at ${\tau }_c=\pi$, or $t_c=2A$.

A radial null geodesic in FLRW spacetime obeys $d\chi /d\tau =\pm 1$. The spatial slice is a unit three-sphere times $a$, so a great circle has comoving circumference $2\pi$. The dust lifetime supplies conformal distance $2\pi$: one circumference, with the return occurring only in the crunch limit. Strictly before the singular endpoint no full return is completed. The radiation lifetime supplies conformal distance $\pi$: the photon reaches the antipode, half a great circle, in the crunch limit. These are limiting null rays from near the initial singularity, not photons at a regular event on the singular surface.

Change independent variable from proper time to the scale factor. Then $\dot{\delta }=\dot{a}{\delta }_a$ and $\ddot{\delta }={\dot{a}}^2{\delta }_{aa}+\ddot{a}{\delta }_a$. Under radiation domination, $\ddot{a}=-{\dot{a}}^2/a$ and $H^2\simeq 8\pi G{\bar{\rho }}_r/3$. Since ${\bar{\rho }}_m/{\bar{\rho }}_r=a/a_{\mathrm{eq}}$, the linear matter perturbation growth equation becomesPutting $\delta =au$ givesThis keeps the matter self-gravity term in a radiation-dominated background. It is not an exact background equation through radiation-matter equality.

At $a/a_{\mathrm{eq}}\ll 1$, neglecting that small term gives ${\delta }_{aa}+{\delta }_a/a=0$. Integration yieldsThus matter has at most logarithmic growth during this leading radiation approximation, together with a constant independent mode. The constant is non-growing, not a mode that literally falls as $a^{-1}$; that power belongs to $u$ rather than ${\delta }_m$.

For an increasing/decreasing basis of the displayed equation with matter self-gravity retained, put $x=\sqrt{6a/a_{\mathrm{eq}}}$. Its equation becomes $x^2{\delta }_{xx}+x{\delta }_x-x^2\delta =0$. HenceThe Modified Bessel function of the first kind gives $I_0(x)=1+3a/(2a_{\mathrm{eq}})+O((a/a_{\mathrm{eq}}{)}^2)$, which increases slowly. The Modified Bessel function of the second kind gives $K_0(x)=-\frac{1}{2}\ln (a/a_{\mathrm{eq}})+\frac{1}{2}\ln (2/3)-{\gamma }_E+O((a/a_{\mathrm{eq}})|\ln (a/a_{\mathrm{eq}})|)$, where ${\gamma }_E$ is Euler's constant; this mode decreases as $a$ increases. Their leading span is precisely the constant/logarithmic pair above. Mode labels depend on the chosen basis and normalization; no rapid matter-era growth $\delta \propto a$ occurs here. Neglected background corrections can change subleading terms, so the Bessel basis should not be extrapolated through equality.

Use geometrized units $G=c=1$ and metric signature $(-+++)$; $\tau$ is proper time. The Killing vectors ${\partial }_t$ and ${\partial }_{\varphi }$ of the Schwarzschild metric give the conserved specific Killing energy and specific angular momentumPut $f=1-2M/r$. In the equatorial plane $\dot{\theta }=0$, and normalization of the four-velocity givesChoosing the inward branch therefore yieldsThe divergence of the time component at the Schwarzschild event horizon is a coordinate effect; the inward radial component tends to $-1$.

Capture from infinity. To reach the event horizon from infinity, the radial square must remain nonnegative throughout $r>2M$. Equivalently,Differentiating the right side gives $2Mr(r-4M)/(r-2M{)}^2$, so its minimum occurs at $r=4M$ and equals $16M^2$. Thusis the necessary capture bound. At equality the radial numerator is $2M(r-4M{)}^2$. An inward particle arriving from larger radii approaches the unstable orbit $r=4M$ only after infinite proper time, because $\dot{r}$ is proportional to $-(r-4M)$ near that orbit. Actual plunges from infinity require $|h|<4M$. This is Schwarzschild marginally bound capture.

The origin-at-infinity hypothesis is important and is not explicit in the PDF. A particle already inside the angular-momentum barrier can plunge with larger $|h|$. For example, $h=10M$ and initial radius $r_0=2.01M$ give positive radial numerator $7.0802M^3$, remaining positive as $r$ decreases to $2M$. This trajectory has $e=1$ and reaches the event horizon. Thus an unrestricted claim about every inward particle would be false; the bound is the intended capture-from-infinity statement.

Invariant collision energy. At a collision, the total four-momentum is $P^{\mu }=m(u_1^{\mu }+u_2^{\mu })$. The invariant center-of-mass energy uses the covariant metric:The PDF instead prints a raised metric multiplying raised velocities. That contraction is not a tensor scalar; the corrected expression above, or a raised metric with lowered momenta, is required. Since each four-velocity has norm $-1$,Let $D_i=2M{r}^2-h_i^2(r-2M)$. For two inward trajectories the radial product is positive, and direct substitution into the Schwarzschild metric givesPutting these terms over one denominator proves

Horizon limit and the upper bound. A cancellation-free way to take the limit is to write $a_i=1+h_i^2/r^2$ and ${\dot{r}}_i=-\sqrt{1-f{a}_i}$. As $f\to 0^{+}$,Henceand thereforeFor particles captured from infinity, $|h_1-h_2|\le 8M$, giving $E_{\mathrm{com}}\le m\sqrt{20}$ in the horizon limit. For actual captured trajectories the inequality is strict, but the supremum is approached by $h_1\to 4M^{-}$ and $h_2\to -4M^{+}$. Their azimuthal starting positions can be chosen so that the trajectories meet. If particles may instead be prepared near the event horizon, the counterexample $h_1=10M$, $h_2=-10M$ has limiting energy $m\sqrt{104}$; no universal $m\sqrt{20}$ bound then follows.

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.

For a Lorentz transformation to an inertial frame moving at speed $v_B$ in the positive $x$ direction, write ${\beta }_B=v_B/c$ and ${\Gamma }_B=(1-{\beta }_B^2{)}^{-1/2}$. Any four-vector transforms asThis convention applies to $X=(ct,x,0,0)$ and uses the same $+---$ Minkowski metric as the four-momentum calculation.

The definition of proper time gives $d\tau =dt\sqrt{1-u^2/c^2}=dt/\gamma$ for $|u|<c$. Hence the four-velocity isSince $\dot{\gamma }={\gamma }^{3}u\dot{u}/c^2$, differentiating with respect to proper time givesHere $1+{\gamma }^2{u}^2/c^2={\gamma }^2$ simplifies the spatial component of the four-acceleration.

At each event, boost with $v_B=u$ to the instantaneous rest inertial frame. Then $A^{\prime 0}=0$ and $A^{\prime 1}=\gamma {\gamma }^4\dot{u}(1-u^2/c^2)={\gamma }^3\dot{u}$. Thus the signed proper acceleration for one-dimensional motion isThe last equality follows by differentiating $\gamma u$; its magnitude is the usual scalar proper acceleration.

If this signed proper acceleration is constant and the particle starts from rest, integration gives $\gamma u=at$, soIntegrating once more and imposing $x(0)=0$ yields the hyperbolic motion worldlineFor $a=0$ the continuous limiting solution is $x=0$. Finally, the Taylor series of the square root givesThis recovers constant-acceleration Newtonian mechanics at low speed while the exact relativistic speed remains less than $c$.