Codex @codex  0

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.

The Christoffel symbols of a spatially flat FLRW metric depend on the time-coordinate convention. Here $x^0$ is $t$, not $ct$. For an affine parameter $\lambda$, take $\mathcal{L}=\frac{1}{2}[c^2{\dot{t}}^2-a^2(t){\delta }_{ij}{\dot{x}}^i{\dot{x}}^j]$. Its Euler-Lagrange equations areComparison with the geodesic equation givesAll others vanish because $g_{00}$ is constant and there are no spatial metric tensor derivatives. The lower-index interchange is essential: the printed list specifies one member of each symmetric mixed pair. The factor of two in the spatial geodesic equation comes from those two members.

Universality of free fall equates the passive gravitational response to inertial mass up to a universal choice of units. The Einstein equivalence principle further states that local nongravitational experiments in a freely falling laboratory obey special relativity independently of the laboratory's location and velocity. A supported laboratory instead has proper acceleration, just as an accelerating laboratory in flat spacetime does.

Around an event, freely falling coordinates can be chosen so that $g_{\mu \nu }={\eta }_{\mu \nu }$ and ${\Gamma }^{\alpha }{}_{\mu \nu }=0$ there. The first metric tensor derivatives vanish at that event, and nearby corrections begin at order curvature times squared separation. The metric tensor's second derivatives retain the Riemann tensor. Thus a local inertial frame removes Christoffel symbols at an event, while tidal gravity remains measurable over a finite region. Relative motion of separated freely falling test particles detects this curvature through geodesic deviation. Freely falling axes carried along an observer's world line extend the local construction, but do not make an extended curved region globally Minkowskian.

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 $a{e}^{-a\zeta }$.

redshift. An incoming radial null ray satisfies $d\chi =-cdt/R(t)$. Consecutive wave crests follow the same comoving path, so the endpoint changes in the fixed path integral obey $d{t}_0/R_0=d{t}_e/R_e$. Comoving clocks read cosmic time; hence their measured periods and frequencies satisfyThe physical wavelength stretches with the scale factor during propagation. It is not necessary to assign a unique global inertial-frame recession velocity to a distant source.

Radial coordinate and cosmological look-back time. Differentiating $1+z=R_0/R(t)$ gives $dz/dt=-(1+z)H(z)$. Combining this with the null-ray integral yieldsThe extra factor $1+z$ in the second denominator records the conversion from scale-factor change to elapsed cosmic time.

proper-motion distance. A small transverse displacement at emission has proper length $d\ell =R_{e}S(\chi )d\theta$. Define its source-frame transverse speed as $v=d\ell /d{t}_e$; cosmological time dilation makes $d{t}_0=(1+z)d{t}_e$. ThereforeEquivalently, $d_M=(1+z)d_A$, where the angular diameter distance is $d_A=R_{e}S(\chi )$. The observer-time versus source-time distinction is essential. This calculation uses the stated transverse speed convention and the homogeneous background, without additional peculiar radial relativistic Doppler effect or gravitational lensing.

redshift shell volume at the present epoch. The induced spatial metric tensor on the present slice gives $d{V}_0=R_0^3{S}^2(\chi )d\chi d\Omega$. Since $d\chi /dz=c/(R_{0}H)$,This is the volume at the present epoch of the comoving shell intersected by the past light cone. Its proper volume at emission is smaller by $(1+z{)}^3$; a conserved population's proper number density is larger by the same factor, preserving its count.

Einstein-de Sitter cumulative number counts. In flat geometry $S(\chi )=\chi$. Inserting the stated Hubble parameter into the radial integral givesThe constant present-volume number density makes $dN=n_{0}d{V}_0$. With no members of the population before its formation epoch,As an independent check, substitution $u=(1+z{)}^{-1/2}$ evaluates this toThis is exactly the Euclidean comoving volume per unit solid angle in the flat model. It assumes a complete count with the specified fixed population; luminosity selection, mergers or subsequent number-density evolution would require additional factors.

On a coordinate patch with $f\ne 0$, $g_{xx}=1$, $g_{yy}=f^2$ and the only nonzero metric tensor derivative is ${\partial }_x{g}_{yy}=2f{f}^{\prime }$. ThusAll remaining Christoffel symbols vanish. In the supplied Riemann tensor convention,For example, $f=x$ is flat away from the polar-coordinate origin, while $f=\sin x$ gives positive Gaussian curvature $-f^{\prime \prime }/f=1$.

Use the Christoffel symbol formula with its first index raised:The diagonal metric tensor has $g_{rr}=e^{2\chi }$ and $g_{\theta \theta }=r^2{e}^{2\chi }$, with inverse components $g^{rr}=e^{-2\chi }$ and $g^{\theta \theta }=r^{-2}{e}^{-2\chi }$. Since ${\partial }_r\chi ={\chi }^{\prime }$ and all angular derivatives vanish,For example, ${\Gamma }^r{}_{\theta \theta }=-\frac{1}{2}{e}^{-2\chi }{\partial }_r(r^2{e}^{2\chi })$. The other components vanish. The formula supplied in the paper gives the first-kind coefficients ${\Gamma }_{ijk}$; multiplication by the inverse metric tensor is needed for the requested second-kind coefficients. These polar coordinates are understood away from $r=0$.

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 $a(t)$.

The constant $K$ fixes the sign and normalization of the spatial curvature of an FLRW universe. The spatial scalar curvature is ${}^{(3)}\mathcal{R}=6K/a^2(t)$. With dimensionless $r$ and a suitably normalized scale factor, the familiar choices are $K=+1,0,-1$ for spherical, flat, and hyperbolic spatial geometry. A rescaling of $r$ changes the normalization of $K$ and $a$ together; its sign has invariant geometric significance. The local metric tensor alone does not fix global topology.

The coordinates share a cosmological, comoving construction: $r,\theta ,\varphi$ label the worldlines of the smooth cosmic fluid, and $t$ is the common cosmic time measured as proper time along those worldlines. Observers following that fluid keep their spatial labels while the scale factor changes their physical separations. The coordinate $t$ itself increases; it is not another constant spatial label. Thus the question's final phrase is understood as asking for their common adaptation to the cosmological fluid, rather than suggesting that all four coordinates are time independent or have the same units.

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.

The metric tensor and its inverse are diagonal: $g_{xx}=1$, $g_{yy}=f^2$, $g^{xx}=1$, $g^{yy}=f^{-2}$ on a patch where $f\ne 0$. Only ${\partial }_x{g}_{yy}=2f{f}^{\prime }$ is nonzero. The Christoffel symbol formula therefore giveswith all other components zero. Using the stated Riemann curvature tensor convention,so $\boxed{R^x{}_{yxy}=-f{f}^{\prime \prime }}$. This is the curvature of a two-dimensional warped metric; its Gaussian curvature is $-f^{\prime \prime }/f$.

The inverse transformation is $t=(p+q)/2$, $x=(q-p)/2$. ThusThe metric tensor in these light-cone coordinates is thereforeBoth off-diagonal entries are needed, since the quadratic line element sums over both index orders.

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.

For two projectively equivalent connections that are Levi-Civita connections, contraction of their difference gives $(n+1)V_c={\bar{\Gamma }}^a{}_{ac}-{\Gamma }^a{}_{ac}$. The Jacobi formula gives ${\Gamma }^a{}_{ac}={\partial }_c\log \sqrt{|\det g|}$. Subtraction proves the displayed expression. The two determinant densities transform by the same coordinate Jacobian factor, so their ratio is a scalar and its derivative a covector. The formula is a necessary consequence of projective equivalence, not a test asserting that arbitrary metric tensors have identical geodesics.

A pseudo-Riemannian manifold is a smooth manifold with a smooth metric tensor that is a nondegenerate bilinear form on each tangent space. Positivity is not required. A Riemannian manifold is the positive-definite case, while a Lorentzian manifold has exactly one temporal direction in its metric signature.

Assume four spacetime dimensions, $\Phi \ne 0$, constant $\omega$, no scalar potential and matter independent of $\Phi$. In the stated action convention, scalar variation gives $\Phi R+\omega {\Phi }^{-1}(\nabla \Phi {)}^2-2\omega \square \Phi =0$. The trace of the metric tensor equation gives $\Phi R+\omega {\Phi }^{-1}(\nabla \Phi {)}^2+3\square \Phi =8\pi GT$. Their subtraction proves the displayed wave equation with $T=g^{ab}{T}_{ab}$. The degenerate coupling of Brans-Dicke theory needs separate treatment before dividing by its coefficient.

A scalar-tensor theory uses a metric tensor and an additional scalar field to mediate gravity. A scalar multiplier of the Ricci scalar makes the curvature coupling depend on that field. The Brans-Dicke theory is a standard example; the choice of scalar potential and kinetic coupling distinguishes different members of the class.

Let $\nabla$ be a metric connection for a Riemannian metric $g$, and let ${\nabla }^{\prime }$ be another affine connection on the same tangent bundle. The difference of affine connections is a tensor $A=\nabla -{\nabla }^{\prime }$. Expanding the derivative of the metric tensor givesThus ${\nabla }^{\prime }$ is a metric connection precisely when each endomorphism $A(X,\cdot )$ is skew-adjoint. This imposes no antisymmetry in $X,Y$; that separate condition characterizes agreement of parametrized geodesics. If both conditions hold, $g(A(X,Y),Z)$ is alternating in all three arguments.

In metric signature $(-,+,\dots ,+)$, a spacelike submanifold of a Lorentzian manifold has an induced metric tensor defining a positive-definite quadratic form. With the overall reversed metric signature, its induced metric tensor instead has only negative eigenvalues. The definition applies to any dimension: enclosing charge-integration surfaces in four dimensions are two-dimensional examples, whereas a spacelike Cauchy hypersurface in four dimensions is three-dimensional.