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Original source: physical PDF pages 203–206; paper pages 1–4. All three optional notes and both longer alternatives are answered below. The metric signature is $+---$; Question 3 uses $c=1$.

Use metric signature $(+---)$ and four-velocities $U_i^a={\gamma }_{i}c(1,{\mathbf{\beta }}_i)$, with ${\gamma }_i=(1-{\beta }_i^2{)}^{-1/2}$. Their Lorentz invariant inner product isIn the rest frame of either particle this same scalar is the relative Lorentz factor ${\gamma }_{\mathrm{rel}}$. HenceLet $d={\mathbf{\beta }}_1\cdot {\mathbf{\beta }}_2$. The numerator after combining terms isUse $|{\mathbf{\beta }}_1\times {\mathbf{\beta }}_2{|}^2={\beta }_1^2{\beta }_2^2-d^2$ to obtainThe two rest frames agree on the relative speed, although their velocity vectors are expressed in different spatial frames. Checks include equal velocities giving zero, collinear velocities giving $|{\beta }_1-{\beta }_2|/|1-{\beta }_1{\beta }_2|$, and low speeds reducing to the Galilean relative speed. For timelike particles the four-velocity inner product also ensures $0\le {\beta }_{\mathrm{rel}}<1$.

Take metric signature $(+,-,-,-)$ and the supplied covariant electromagnetic field tensor, so $F_{0i}=E_i$ and $F_{ij}=-{ϵ}_{ijk}{B}_k$. Raising both indices reverses the time-space components but leaves the spatial components unchanged. Its complete contraction is thereforeDefine the dual electromagnetic field tensor by ${\tilde{F}}^{\mu \nu }=\frac{1}{2}{\epsilon }^{\mu \nu \rho \sigma }{F}_{\rho \sigma }$, with ${\epsilon }^{0123}=+1$. Direct contraction givesThese electromagnetic field invariants are complete tensor contractions. The second is a pseudoscalar: it is invariant under proper Lorentz transformations and changes sign under spatial inversion.

For $T^{0i}$ the metric trace term vanishes. Since $F^0{}_j=E_j$,Consequently the Poynting vector is exactly the time-space row in the stated $c=1$ units:

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, with the time component of four-momentum equal to $E/c$. The incident particle has energy $\gamma m{c}^2$ and momentum of magnitude $mc\sqrt{{\gamma }^2-1}$. Adding the stationary target's four-momentum gives total laboratory quantitiesThe centre-of-momentum energy $W$ follows from the invariant Minkowski norm of the total four-momentum:At the relativistic collision mass threshold, the two final particles have zero relative velocity in their centre-of-momentum frame. Their total energy there is therefore $2\alpha m{c}^2$. Four-momentum conservation requires $W\ge 2\alpha m{c}^2$, soThe condition $\alpha >3/2$ ensures ${\gamma }_{\min }>1$. This is the minimum required Lorentz factor; larger values allow final kinetic energy in the centre-of-momentum frame. Equivalently, the incident laboratory kinetic energy at threshold is $({\alpha }^2-9/4)m{c}^2$.

Let $\mathbf{a}=d\mathbf{u}/dt$ denote coordinate acceleration and use $A^{\mu }$ for four-acceleration, to distinguish these two objects. Applying the chain rule to the Lorentz factor givesSince $dt/d\tau ={\gamma }_u$, differentiation of the four-velocity $U^{\mu }={\gamma }_u(c,\mathbf{u})$ with respect to proper time yieldsThis is also ${\gamma }_u^2[{\gamma }_u^2(\mathbf{u}\cdot \mathbf{a})/c,\mathbf{a}+{\gamma }_u^2(\mathbf{u}\cdot \mathbf{a})\mathbf{u}/c^2]$, as required.

The invariant Minkowski norm of the four-velocity is $U^{\mu }{U}_{\mu }=c^2$. Differentiating this identity gives $U^{\mu }{A}_{\mu }=0$. In the instantaneous rest frame, $U^{\mu }=(c,0)$, so $A^0=0$ and the spatial components equal the physically experienced proper acceleration $\mathbf{\alpha }$. Therefore ${\alpha }^2=-A^{\mu }{A}_{\mu }$ in the $+---$ metric signature.

For a direct calculation, put $q=\mathbf{u}\cdot \mathbf{a}$ and expand the temporal and spatial squares:The last step uses $1-u^2/c^2={\gamma }_u^{-2}$. This proper acceleration from coordinate velocity and acceleration also reads ${\alpha }^2={\gamma }_u^6{a}_{\parallel }^2+{\gamma }_u^4{a}_{\perp }^2$, providing the familiar one-dimensional limit $\alpha ={\gamma }_u^3|a_{\parallel }|$. Each expression has units ${\mathrm{m}}^2{\mathrm{s}}^{-4}$.

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.

For a massive particle choose $\lambda =\tau$, its proper time, and obtain $\kappa =c^2$. For a photon, the tangent is null and $\kappa =0$. A general rescaling of a massive particle's affine parameter rescales a positive $\kappa$; the numerical value $c^2$ requires the proper time choice. With that choice, $h=L_z/m$ is signed specific angular momentum, of units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$, and $k=E/(m{c}^2)$ is dimensionless conserved specific Killing energy. The time coordinate is normalized to proper time at infinity, so this $E$ is the particle's conserved energy referred to infinity. A photon's constants scale with the normalization of its null tangent; it has no rest frame or proper time parametrization.

Substitution of $\dot{t}=k/f$ and $\dot{\varphi }=h/r^2$ into the norm gives the general radial first integralFor a massive particle this becomesAdopting the usual half-normalization for the effective potential for timelike Schwarzschild geodesics, defineAll terms have units ${\mathrm{m}}^2{\mathrm{s}}^{-2}$, or energy per unit mass. The first two terms are the Newtonian gravitational and centrifugal contributions; the last is the relativistic correction.

A circular orbit requires both $\dot{r}=0$ and $V_{\mathrm{eff}}^{\prime }(R)=0$. Differentiating at fixed $h$ givesThe radial first integral also givesFor the future-directed circular orbit at $R=4\mu$, these constants are $|h|=4\mu c$, $k=1$, and $f=1/2$. ThereforeSince Schwarzschild time agrees with the clock time of a static observer at infinity, the intended infinity-referred tangential orbital speed isThis uses the circumferential areal radius $R$ with the clock at infinity; it is not a local measurement made at the orbit. A local static observer's clock instead has $d{\tau }_{\mathrm{stat}}=\sqrt{f}dt$ and measures $v_{\mathrm{local}}=v_{\infty }/\sqrt{f}=\boxed{c/\sqrt{2}}$. These two operationally different speeds explain the significance of the phrase “at infinity”.

The effective-potential stability of a circular orbit follows fromConsequently the $R=4\mu$ orbit exists but is unstable to radial perturbations. The original plot below shows its position at a maximum, for $h=4\mu c$.

For a weak-field elliptic orbit, the relativistic correction makes the direction of closest approach advance after each radial cycle. The derivations under i and ii below giveper orbit to first order, where $a_{\mathrm{orb}}$ is the semi-major axis and $e$ the orbital eccentricity. For Mercury, the additional Schwarzschild perihelion precession is approximately 43 arcseconds per century, over and above the much larger Newtonian precession produced primarily by other planets. It is the unexplained residual after those effects have been included that provides the classic quantitative test of general relativity; one should not identify the whole observed precession with this relativistic correction.^[ref]

The labels i and ii in the original paper introduce optional orbit-equation aids, rather than separate mandatory subquestions. They are retained here for derivations of both aids and of the perihelion result.

Local equivalence and tides. In Newtonian mechanics, inertial mass $m_{\mathrm{i}}$ measures the response to a force, whereas passive gravitational mass $m_{\mathrm{g}}$ couples to an imposed Newtonian gravitational field. A freely falling body obeysThe observed composition independence of free fall says that this ratio is universal. With a common choice of units it is one. This is the Newtonian content of the weak equivalence principle for test particles; it does not by itself specify all nongravitational physics.

The Einstein equivalence principle further says that local nongravitational experiments in a sufficiently small freely falling laboratory have the form predicted by special relativity, independently of the laboratory's location and motion. In such a local inertial frame, the metric tensor is locally the Minkowski metric and the Christoffel symbols vanish at the selected event. The strong equivalence principle extends the universality claim to local gravitational experiments and bodies with significant gravitational self-energy. None of these statements removes measurable tidal forces across a finite apparatus.

In an ideal small freely falling lift, a released ball initially floats relative to the cabin, because both undergo the same free fall. Two separated balls, however, need not retain their separation: they sample different parts of the gravitational field. For a point source, linearizing $\mathbf{g}=-GM\mathbf{r}/r^3$ across a small separation $\mathbf{\xi }$ gives the tidal tensor predictionA radial separation stretches with coefficient $2GM/r^3$, whereas a transverse separation is compressed with coefficient $-GM/r^3$. Initially parallel velocities therefore develop relative motion even though the individual bodies experience zero proper acceleration.

The relativistic statement is geodesic deviation. With the Riemann curvature tensor conventiona connecting vector between neighboring freely falling worldlines satisfiesHere $D/d\tau$ is the covariant derivative along the reference geodesic, and $U^{\mu }$ its four-velocity. Curvature can be nonzero where the Christoffel symbols have been made zero. In the weak, static, slow-motion limit this equation reduces to ${\ddot{\xi }}^i=-{\partial }_i{\partial }_j\Phi {\xi }^j$, recovering the tidal tensor above. A finite falling lift can detect gravity through these relative accelerations; local equivalence does not imply a globally uniform field.

Weak-field connection and potential. Take $x^0=ct$ and the $+---$ metric signature. There is an index error in the printed formula for the Christoffel symbols: its first differentiated metric component must be $g_{\nu 0}$, not $g_{\mu 0}$. The correctly contracted expression isThe offending term vanishes under the stipulated time independence, so this correction does not change the intended calculation. Write $g_{\mu \nu }={\eta }_{\mu \nu }+h_{\mu \nu }$ and treat both $h$ and its derivatives as first order. To that order $g^{00}=1-h_{00}$, $g^{0i}=O(h)$, and $g^{ij}=-{\delta }^{ij}+O(h)$. A stationary metric tensor givesFor the temporal component the inverse-metric coefficient is already first order, soFor a spatial component the zeroth-order inverse metric contributes:The time component of the approximate geodesic equation says that $dt/d\tau$ is constant to the retained order. In its spatial component, convert the proper time derivatives to coordinate-time derivatives and cancel $(dt/d\tau {)}^2$:Comparison with the Newtonian equation $d^2\mathbf{x}/d{t}^2=-\nabla \Phi$ identifies $h_{00}=2\Phi /c^2+C$. For an isolated source, choose the gravitational potential to tend to zero at infinity and normalize the time coordinate there, giving $C=0$. ThusThe quantity $\Phi$ is the Newtonian gravitational potential, or gravitational potential energy per unit mass, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}={\mathrm{m}}^2{\mathrm{s}}^{-2}$. For a point source, $\Phi =-GM/r$. A static clock has $d\tau =\sqrt{g_{00}}dt\approx (1+\Phi /c^2)dt$, connecting the Newtonian limit with gravitational time dilation.

Frequency of the orbiting source. Treat the gas blob as a test particle on a timelike circular Schwarzschild geodesic, with pressure and other nongravitational forces negligible. Treat our very distant location as infinity, so the observer's own gravitational potential is negligible. These are the assumptions that make a unique answer in terms of $R_S/r$ possible.

Put $f=1-R_S/r$ and use a local static observer at the emission point as an intermediate measuring frame. The remote observer lies on the symmetry axis of the orbit. A ray arriving on that axis has zero conserved axial angular momentum, so its local direction is perpendicular to the emitter's azimuthal velocity. Bending within a meridional plane does not alter this orthogonality. The transverse relativistic Doppler effect therefore giveswhere $v$ is the orbital speed measured by that local static observer. The gravitational redshift between static Schwarzschild observers then givesA gravitational circular orbit has $v^2=GM/r+O((GM/r{)}^2/c^2)$ in the weak-field limit, so $v^2/c^2=R_S/(2r)+O((R_S/r{)}^2)$. Expanding both factors,Solving for the emitted frequency givesThe gravitational part is $R_S/(2r)$ and the transverse kinematic time dilation contributes another $R_S/(4r)$; the absence of line-of-sight motion does not remove this second effect.

As an exact check, the local speed on a timelike circular Schwarzschild geodesic satisfies $v^2/c^2=\mu /(r-2\mu )$, where $\mu =GM/c^2=R_S/2$. Therefore the face-on frequency shift of a circular Schwarzschild source iswhose expansion agrees with the boxed answer. If the gas is supported or accelerated by pressure or magnetic stresses rather than following a geodesic, its orbital speed is not fixed by the mass and radius alone; the appropriate first-order result is instead ${\nu }_{\mathrm{em}}\approx {\nu }_{\mathrm{ob}}[1+GM/(r{c}^2)+v^2/(2c^2)]$, with $v$ supplied independently.

Use the metric signature $\eta =\mathrm{diag}(1,-1,-1,-1)$ and $c=1$ throughout this question. The trace-reversed metric perturbation satisfies $\varphi ={\eta }_{\mu \nu }{\varphi }^{\mu \nu }=-h$, so its inverse relation isSet $C_{\nu }={\partial }^{\sigma }{\varphi }_{\nu \sigma }$. Substituting this inverse relation into the linearized field operator cancels the second derivatives of the trace, leavingThe imposed Lorenz gauge in linearized gravity is $C_{\nu }=0$. Therefore the Linearized Einstein equations in vacuum reduce toThe overall sign of the field operator is immaterial for this homogeneous equation.

A fundamental observer follows the coarse-grained cosmological matter, with fixed comoving coordinates. The cosmological principle asserts spatial homogeneity and cosmological isotropy on large scales for this family of observers. A moving observer generally sees a preferred direction, for example a Doppler dipole in the cosmic microwave background, even when the underlying cosmology is isotropic.

In an FLRW metric, these worldlines are orthogonal to homogeneous spatial slices. Synchronize their clocks to define cosmic time $t$, which is their proper time. Along a fixed spatial label, $d{s}^2=c^{2}d{t}^2$, giving $g_{tt}=c^2$. Orthogonality of the spatial coordinate directions to the observers' four-velocity gives $g_{ti}=0$. Thus, in the $+---$ metric signature,where the spatial matrix $g_{ij}$ is negative definite. Spatial homogeneity and cosmological isotropy further restrict it to $-a(t{)}^2{\gamma }_{ij}(\mathbf{x})$, with $\gamma$ a constant-curvature spatial metric.

At any event on a massive observer's worldline, choose an orthonormal tetrad with timelike basis vector $e_{(0)}=U/c$. Its components obey $g(e_{(a)},e_{(b)})={\eta }_{ab}=\mathrm{diag}(1,-1,-1,-1)$, so locally $d{s}^2={\eta }_{ab}d{X}^{(a)}d{X}^{(b)}$: the Minkowski metric. A local inertial frame can also remove the Christoffel symbols at that event. Curvature cannot be removed on a finite region, and a frame following an accelerated observer need not remain inertial in its neighbourhood.

In general coordinates, $X^{\mu }(\tau )$ labels the trajectory, while the four-velocity and four-acceleration areCoordinate values $X^{\mu }$ are not themselves a vector under arbitrary nonlinear coordinate changes; their tangent $U$ is. For fixed rest mass $m$, the four-momentum is $P^{\mu }=m{U}^{\mu }$ and the four-force is $f^{\mu }=D{P}^{\mu }/d\tau =m{A}^{\mu }$. With the $+---$ metric signature,where $a$ is the proper acceleration. The middle identity follows by differentiating the constant velocity norm; in the instantaneous rest frame, $U=(c,0)$ and $A=(0,\mathbf{a})$, proving the last identity.

Source: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/9/Physics%20Papers%202020.pdf?cookie_passthrough=1

Source bundle pages 70–73; the cover identifies Trinity Term 2021 despite the attachment filename. Use coordinates $x^{\mu }=(ct,x,y,z)$, metric signature $(-,+,+,+)$, and $A^{\mu }=(\varphi /c,\mathbf{A})$. For the electromagnetic field tensor use $F^{0i}=E_i/c$ and $F^{ij}={ϵ}_{ijk}{B}_k$. Prescribed external fields are treated classically, neglecting radiation reaction in the trajectories; radiation losses are computed separately where requested.

Source: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/10/Physics%20Papers%202021.pdf?cookie_passthrough=1

Source bundle pages 71–75. The cover establishes 2022 despite the attachment filename. Throughout these solutions the metric signature is $(-,+,+,+)$, as fixed by the source's lowered time coordinate. Question 2 uses $c=1$. Signs and additional assumptions needed by particular printed conclusions are explained where they arise.

Metric-sign qualification. With this paper's $(-,+,+,+)$ metric signature, b givesThe minus sign in the printed Larmor formula would therefore make a positive radiative loss negative. A concrete counterexample is an instant of rest with nonzero acceleration: $F^{\alpha }=(0,m\mathbf{a})$, so the printed expression is $-q^2{a}^2/(6\pi {ϵ}_0{c}^3)$. That sign is appropriate for the opposite metric signature. With the convention already fixed in question 1, positive emitted power instead isAlternatively the printed negative quantity can be interpreted as the particle's signed radiative energy change, rather than the positive loss magnitude.

For fixed rest mass, resolve the acceleration parallel and perpendicular to the velocity, as in b. This givesWriting $C_q=q^2/(6\pi {ϵ}_0{c}^3)$, the Liénard radiated-power formula becomes $P_{\mathrm{rad}}=C_q({\gamma }^6{a}_{\parallel }^2+{\gamma }^4{a}_{\perp }^2)$. For uniform circular motion, the acceleration is perpendicular to the velocity; for straight-line motion it is parallel. Comparing the same particle at equal relativistic energy and equal laboratory acceleration magnitude $a$ therefore givesThis particular comparison makes the linear loss larger. The familiar larger loss from bending applies to a different comparison: at equal applied laboratory force magnitude $F$, $a_{\perp }=F/(m\gamma )$ and $a_{\parallel }=F/(m{\gamma }^3)$, giving $P_{\mathrm{circular}}=C_q{\gamma }^2{F}^2/m^2$ and $P_{\mathrm{linear}}=C_q{F}^2/m^2$. Stating which quantity is held fixed is essential.

Use metric signature $(+,-,-,-)$ and units with propagation speed one. The scalar Euler-Lagrange equation givesWith the stated spatial Fourier convention, $-{\nabla }^2$ becomes $k^2$. Each transformed mode obeysHere $F$ and $V$ are the initial transformed field and velocity. Reality requires $F(-\mathbf{k})=F(\mathbf{k}{)}^{\ast }$ and the same for $V$; the Fourier convention places no extra $(2\pi {)}^3$ in these evolution formulas.

Including the interaction gives the two coupled real Klein-Gordon fields,Adding and subtracting diagonalizes them:Both squared masses are positive for $0<g<m^2$. For arbitrary initial data $F_i={\tilde{\varphi }}_i(0)$ and $V_i={\dot{\tilde{\varphi }}}_i(0)$, defineThen the general solution is ${\tilde{\varphi }}_1=(U_{+}+U_{-})/2$, ${\tilde{\varphi }}_2=(U_{+}-U_{-})/2$, with the corresponding reality conditions on initial data.

For the specified sinusoidal initial displacement, put ${\Omega }_{\pm }=\sqrt{q^2+m^2\mp g}$. Both normal fields initially have amplitude $A\sin (\mathbf{q}\cdot \mathbf{x})$ and zero time derivative. ConsequentlyWriting $\bar{\Omega }=({\Omega }_{-}+{\Omega }_{+})/2$ and $\delta \Omega =({\Omega }_{-}-{\Omega }_{+})/2$ gives ${\varphi }_1=A\sin (\mathbf{q}\cdot \mathbf{x})\cos \bar{\Omega }t\cos \delta \Omega t$ and ${\varphi }_2=A\sin (\mathbf{q}\cdot \mathbf{x})\sin \bar{\Omega }t\sin \delta \Omega t$. This is field-amplitude beating of coupled Klein-Gordon modes. The conserved energy includes the interaction term $-g{\varphi }_1{\varphi }_2$; separate uncoupled-field energies need not be conserved, and an envelope zero is not by itself exact complete energy transfer.

The Penrose singularity theorem states that a globally hyperbolic spacetime with a noncompact Cauchy hypersurface, a closed trapped surface, and the null convergence condition is future null-geodesically incomplete. Time reversal replaces the trapped surface by an anti-trapped surface and gives past incompleteness.

In his 1965 paper Gravitational Collapse and Space-Time Singularities, Roger Penrose used trapped surfaces and global causal structure to address whether nonspherical perturbations could avert a breakdown during gravitational collapse. Quasars supplied the opening astrophysical motivation. The paper used metric signature $(+,-,-,-)$ and a stronger timelike local-energy assumption; the statement above uses the modern null convergence condition.^[ref]

For this version, take a smooth connected four-dimensional spacetime with a time orientation. A Cauchy hypersurface meets every inextendible causal curve exactly once; its existence expresses global hyperbolicity. The hypersurface must be noncompact, meaning not compact; this is a global spatial condition, not an assumption about the size of the collapsing body. The null convergence condition is $R_{ab}{k}^a{k}^b\ge 0$ for every null vector $k$. Through the Einstein field equations it follows from the null energy condition; a cosmological constant drops out of the null contraction.^[ref]

The conclusion is that at least one future-inextendible null geodesic has a finite upper endpoint of its affine parameter. It neither asserts infinite mass density or divergent curvature at every endpoint, nor says that such an endpoint is visible. Locating an event horizon requires additional global information; the theorem is not a horizon-finding rule for arbitrary spacetimes. Weak cosmic censorship conjecture is an additional issue.^[ref]

The supermassive black hole account describes the motivating observational evidence from stellar orbits at the Galactic centre. Such measurements infer a compact gravitating mass. They do not observe a trapped surface, measure a geodesic's complete future, or experimentally prove the Penrose singularity theorem.

For metric signature $(+---)$, a perfect fluid has $T_{ab}=(\rho +p)u_a{u}_b-p{g}_{ab}$. The Einstein field equations and static Killing field is a Ricci eigenvector imply $(\rho +p)(k\cdot u)u_{[a}{k}_{b]}=0$. Two nonzero timelike vectors cannot be orthogonal, so $k\cdot u\ne 0$. If $\rho +p\ne 0$, the four-velocity is parallel to the static Killing vector. Choosing both future-directed gives the displayed normalization. For $\rho +p=0$, the stress-energy tensor is proportional to the metric tensor and does not determine a preferred fluid velocity; the strict enthalpy hypothesis is therefore material.