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The inverse of the $(v,\lambda )$ metric tensor block isThe normal covector $d\lambda$ has squared norm ${\lambda }^2/m^2$. ThereforeMore explicitly, the induced metric tensor is $-{\lambda }_0^{2}d{v}^2+m^{2}d{\Omega }_2^2$; at zero its sole zero direction is ${\partial }_v$. The coefficients are independent of $v$, so $K={\partial }_v$ is a Killing vector field. On $\lambda =0$,Hence $K$ is both tangent and normal to this null hypersurface, proving it is a Killing horizon. The definition of surface gravity is ${\nabla }_{K}K=\kappa K$ on the Killing horizon, or equivalently $d(K^2)=-2\kappa K^{♭}$. Since $d(K^2)=-2\lambda d\lambda$ vanishes there while $K^{♭}\ne 0$,Thus this is a degenerate Killing horizon, and $v$ is an affine parameter on each of its null geodesics.

For a concrete horizon frequency comparison, fix the same angular position for a narrow high-frequency light beam and two freely falling radial observers. Work in geometric optics, with the beam's affinely transported wavevector tangent to the Killing horizon, $k=qK$. Here $q>0$ is constant because ${\nabla }_{K}K=0$. Choose the observers as translates of one timelike geodesic by the isometry generated by $K$, crossing at $v_1$ and $v_2$. Give them the same positive Killing energy per unit rest mass, $E=-K\cdot W$. At their crossings their normalized four-velocities havewhich satisfy $g(W,W)=2m{W}^v{W}^{\lambda }=-1$. They can be realized by local geodesics with precisely these initial data. The measured angular frequency isTo connect this measurement to surface gravity, if ${\nabla }_{K}K=\kappa K$ with constant $\kappa$, affine transport of $k=q(v)K$ would give $dq/dv=-\kappa q$. For observers related by the same Killing vector field translation, $-K\cdot W$ agrees at the crossings, giving ${\omega }_2/{\omega }_1=e^{-\kappa (v_2-v_1)}$. Here there is no accumulated redshift along the Killing horizon for these equally prepared observers. Arbitrary observers with different four-velocities can still measure different frequencies through their different Killing energies.

Before its second decay A moves uniformly with $x=-Vt$. Its Lorentz factor isThe proper time along its world line satisfies $d\tau =dt/{\gamma }_A$. Since the first decay occurs at the spacetime origin, integration givesEquivalently these are the components of A's constant four-velocity times $\tau$. The assumption $m>0$ is needed for a daughter rest frame and its proper-time clock; a massless daughter has neither.

Take constant kinematic viscosity $\nu$ and the inner specific angular momentum found above. The Angular velocity of a circular Schwarzschild geodesic gives $-{\Omega }^{\prime }=3c\sqrt{m}/(2r^{5/2})$. Hence the inward radial component of four-velocity in the stated viscous model isTo convert proper time to Schwarzschild time, neglect ${\dot{r}}^2$ in the normalization as prescribed and insert the circular value of $h$. One obtainsThe change in observed Schwarzschild time equals this coordinate-time change when differential light travel time is neglected. Since $dr/d\tau =-u$,This time is nonnegative: $r^2-12m(r-3m)=(r-6m{)}^2\ge 0$ on the integration interval.

Put $x=r/(3m)$ and $X=r_1/(3m)$. The integral becomesPolynomial division gives $x^2/(x-1)=x+1+1/(x-1)$. Separately, writing $y=x-1$ gives $\int 2x/\sqrt{x-1}dx=(4/3)(x-1{)}^{3/2}+4\sqrt{x-1}$. Therefore the antiderivative inside the integral isIts value at $x=2$ is $-4/3$. Subtracting this lower endpoint yields the formal viscous inspiral time in Schwarzschild spacetimeThe logarithm has a positive coefficient. The minus sign in the printed final expression is inconsistent with its preceding integral. Differentiating the corrected bracket gives $6X[X-2\sqrt{X-1}]/(X-1)$, precisely the positive integrand after rescaling. The bracket vanishes at $X=2$; replacing $+6\log (X-1)$ by its negative would give derivative $-12$ there and a negative elapsed time for $X$ just above $2$. This is a sign error, not a choice of potential or time convention.

The result assumes constant kinematic viscosity; if $\nu$ varies with radius it remains inside the time integral. There is also a physical limitation to the stipulated approximation: $h-h_0$ vanishes quadratically at $6m$, so its formal $u$ diverges and eventually violates the neglected-radial-motion condition. The finite integral is the formal extrapolation of the nearly circular viscous model. A real accretion disk must instead match to its plunging region of a black-hole accretion disk; this approximation cannot describe that transition arbitrarily closely.

The Pound-Rebka experiment measures a small gravitational redshift by comparing nuclear emission and resonant absorption at different heights in Earth's field. The $14.4\mathrm{keV}$ transition of iron-57 supplies the gamma-ray frequency standard. The Mössbauer effect permits recoil-free emission and absorption by nuclei bound in solids, giving a sufficiently narrow resonance. A source and absorber separated vertically by about $22.6\mathrm{m}$ have a predicted fractional frequency difference of order $2.5\times 10^{-15}$. A calibrated Doppler effect from slow source motion measures the shift through its effect on resonant absorption. Interchanging the upper and lower arrangements reverses the gravitational contribution and helps separate it from intrinsic source-absorber offsets; temperature-dependent frequency shifts must also be controlled. The observed change agrees with the weak-field prediction.

The prediction follows either from an accelerating laboratory and the equivalence principle, or directly from a static metric. With $x^0=ct$, write $g_{00}=N^2\simeq 1+2\Phi /c^2$. The time-translation Killing vector field gives a conserved photon energy $E_{\xi }=p_a{\xi }^a$, while a static observer has four-velocity $u^a={\xi }^a/N$ and measures $h\nu =E_{\xi }/N$. ThusFor an upward ray over height $H$, $\Delta \nu /\nu \simeq -gH/c^2$; downward propagation has the opposite sign. The equivalent compensating Doppler speed has magnitude $gH/c$, about $7.4\times 10^{-7}\mathrm{m}{\mathrm{s}}^{-1}$ for this baseline. Identical clocks at rest have $d\tau =Ndt$, so a clock at higher potential runs faster relative to the common static coordinate time. This does not mean that the nuclear transition changes its locally measured frequency with height: the comparison involves transporting a photon between different local proper-time standards.

The weak equivalence principle asserts universal free fall for test bodies. The Einstein equivalence principle adds local Lorentz invariance and local position invariance for nongravitational experiments in a freely falling laboratory. The redshift comparison tests the clock aspect of local position invariance and supports a metric description of gravity. The strong equivalence principle extends these requirements to self-gravitating bodies and local gravitational experiments, including the contribution of gravitational binding energy. Their outcomes should be independent of the external location and motion of a freely falling laboratory when tidal effects are negligible. A nuclear redshift experiment is therefore not a test of the whole strong equivalence principle, and its success does not uniquely establish the gravitational field equations.

The geometrical realization uses the Levi-Civita connection of a spacetime metric. At any event, normal coordinates make $g_{ab}={\eta }_{ab}$ and the Christoffel symbols vanish. Freely falling test particles follow geodesics; local nongravitational equations take their special-relativistic form at that event. The Riemann curvature tensor need not vanish, so finite laboratories can measure tidal geodesic deviation. The strong principle concerns removing a uniform external gravitational field locally, not removing curvature throughout an extended region.

For a perfect fluid, local rest-frame isotropy and the absence of heat flux or viscosity give $T^{\hat{a}\hat{b}}=\mathrm{diag}(\rho ,p,p,p)$, where $\rho$ is rest-frame energy density. With the PDF's signature $(+---)$, $u^a{u}_a=1$ and $h^{ab}=g^{ab}-u^a{u}^b$ projects spatially. Its covariant form isThe pressure sign follows from the negative spatial metric; it gives positive diagonal stresses in the rest frame. Local stress-energy conservation, ${\nabla }_a{T}^{ab}=0$, produces the two projectionsThe first is the energy equation, including pressure work; the second is the relativistic Euler equation. Its rest-frame spatial pressure force is $-\nabla p$.

Further assumptions are needed to derive the Einstein field equations. Take a local generally covariant metric equation, linear in curvature and of second differential order, and demand compatibility with stress-energy conservation. The available rank-two tensors are $R_{ab}$, $R{g}_{ab}$ and a constant multiple of $g_{ab}$. Defining $R_{ab}=R^c{}_{acb}$, the contracted Bianchi identity gives ${\nabla }^a{R}_{ab}=\frac{1}{2}{\nabla }_{b}R$. Hence the combination $R_{ab}+\alpha R{g}_{ab}$ is identically divergence-free only when $\alpha =-\frac{1}{2}$. This determines the Einstein tensor $G_{ab}=R_{ab}-R{g}_{ab}/2$; a cosmological constant term remains allowed.

The coupling sign must be matched to the actual curvature convention. Here $R^a{}_{bcd}$ represents $[{\nabla }_d,{\nabla }_c]$, and a static weak field has $R_{00}\simeq -{\nabla }^2\Phi /c^2$. With negligible pressure and cosmological term, a trial equation $G_{ab}=C{T}_{ab}$ implies $R_{00}\simeq C{\rho }_{\mathrm{mass}}{c}^2/2$. Comparing with the Poisson equation ${\nabla }^2\Phi =4\pi G{\rho }_{\mathrm{mass}}$ fixes $C=-8\pi G/c^4$. Thus the Einstein equation in the sequential-derivative curvature convention isEquivalently, $R_{ab}=\Lambda g_{ab}-(8\pi G/c^4)(T_{ab}-T{g}_{ab}/2)$, with $T=\rho -3p$. Pressure therefore gravitates as well as supplying a material force. The cosmological constant is not fixed by the local Newtonian matching.

Setting $c=1$ for the action normalization, the same metric equation follows from the Einstein-Hilbert action $S_g=(2\kappa {)}^{-1}\int \sqrt{-g}(R-2\Lambda )d^{4}x$, where $\kappa =8\pi G$, together with a matter action obeying $\delta S_m=\frac{1}{2}\int \sqrt{-g}{T}_{ab}\delta g^{ab}{d}^{4}x$. After the metric boundary term is removed or supplied separately, the bulk variation is proportional to $G_{ab}+\Lambda g_{ab}+\kappa T_{ab}$. This also makes the sign convention explicit. General covariance, the field equation and the contracted Bianchi identity are mutually consistent with the fluid conservation laws. The experiment motivates the local metric description; the curvature and conservation assumptions, plus the Newtonian limit, supply the additional route to the field equation.

Let $Q=g^{ab}{\chi }_{,a}{\chi }_{,b}$ and $\mathcal{L}=p(ϵ)+n(Q-{ϵ}^2)/(2ϵ)$. Vary $n,ϵ,\chi$ independently, taking compactly supported variations. The multiplier variation is $\partial \mathcal{L}/\partial n=(Q-{ϵ}^2)/(2ϵ)=0$. With $u_a={\chi }_{,a}/ϵ$, it gives the unit four-velocity condition $u^a{u}_a=1$. The enthalpy variation isUsing $Q={ϵ}^2$ yields $p^{\prime }(ϵ)=n$. Finally,The integration by parts boundary term vanishes. Thus the three matter equations areThe last is particle number conservation in a relativistic fluid. Off shell $n,ϵ$ are independent variational fields; on shell the thermodynamic relation ties them to the specified function $p(ϵ)$.

For the Hilbert stress-energy tensor, keep these matter fields fixed while varying the inverse metric. The volume variation is $\delta \sqrt{-g}=-\frac{1}{2}\sqrt{-g}{g}_{ab}\delta g^{ab}$, and $\delta Q={\chi }_{,a}{\chi }_{,b}\delta g^{ab}$. ThereforeOn shell, $\mathcal{L}=p$ and $nϵ=\rho +p$. With signature $(+---)$,For variation of the covariant metric the same definition reads ${\delta }_{g}S=-\frac{1}{2}\int \sqrt{-g}{T}^{ab}\delta g_{ab}{d}^{4}x$. In a local rest frame the tensor has entries $\mathrm{diag}(\rho ,p,p,p)$, which checks both the energy density and pressure sign.

Define a derivative along the flow by $\dot{F}=u^a{\nabla }_{a}F$, and set $\theta ={\nabla }_a{u}^a$, $a^b=u^a{\nabla }_a{u}^b$. The number equation immediately gives $\dot{n}+n\theta =0$. The specific enthalpy per particle relation and $dp=ndϵ$ giveConsequently the energy equation follows from the number equation: $\dot{\rho }=ϵ\dot{n}=-(\rho +p)\theta$.

The velocity equation follows directly from the potential, rather than requiring an independent assumption about the fluid force. Since $ϵu_a={\nabla }_a\chi$ and the connection has zero torsion, the canonical vorticity of a relativistic fluid vanishes:Contract with $u^a$ and use $u^a{u}_a=1$ and $u^a{\nabla }_b{u}_a=0$. This gives $ϵa_b={\nabla }_bϵ-u_b\dot{ϵ}$. Multiplication by $n$ and use of ${\nabla }_{b}p=n{\nabla }_bϵ$ produces the relativistic Euler equation. The evolution system is thereforeIt is equivalent to stress-energy conservation together with number conservation, subject to the potential-flow restriction. The action describes an isentropic barotropic sector of a perfect fluid in general relativity; it does not provide general vortical or entropy-carrying fluid motions. The constraint on canonical vorticity is stronger than the Euler equation alone.

For the weak static metric, use $c=1$. To the retained order, $\sqrt{-g}=1+\varphi$, $u^0=1-\varphi +O({\varphi }^2,|\mathbf{v}{|}^2)$ and $u^i=v^i$. The correction to $u^0$ is required by normalization, even though its zeroth-order value is one. The relevant Christoffel symbols arewith no spatial connection at this order. The cancellation $\sqrt{-g}{u}^0=1+O({\varphi }^2,|\mathbf{v}{|}^2)$ makes the leading number equation particularly simple:The corresponding leading energy transport equation is ${\partial }_t\rho +\mathbf{v}\cdot \nabla \rho +(\rho +p)\nabla \cdot \mathbf{v}=0$. The spatial Euler equation contains the gravitational acceleration $-\nabla \varphi$ and pressure force $-\nabla p/(\rho +p)$. In a strict first-order expansion, convective velocity acceleration and products of the small quantities are discarded.

To exhibit a fully linear system, perturb a homogeneous rest state: $n=n_0+\delta n$, $\rho ={\rho }_0+\delta \rho$, $p=p_0+\delta p$, and $ϵ={ϵ}_0+\delta ϵ$, with the perturbations, $\mathbf{v}$ and $\varphi$ all first order. Let $w_0={\rho }_0+p_0=n_0{ϵ}_0$. The weak-field slow-motion equations for a relativistic fluid areThe equation of state gives $\delta \rho ={ϵ}_0\delta n$ and $\delta p=n_0\delta ϵ$. Where $p^{\prime \prime }({ϵ}_0)\ne 0$, it also gives the barotropic relativistic sound speedWithout this invertibility, the primitive conservation equations remain valid and this sound-speed parametrization must not be used.

The potential furnishes a further check on the pressure and gravity signs. Write $\chi ={ϵ}_{0}t+\delta \chi$. Its time and spatial components implyTheir compatibility reproduces the linear Euler equation. Pressure gradients oppose gravitational acceleration, and hydrostatic balance is $\nabla \delta p=-w_0\nabla \varphi$. Combining the energy and Euler equations gives the forced acoustic equationFor positive $c_s^2$, disturbances propagate as sound waves; the prescribed potential supplies a static forcing. A nonrelativistic equation of state gives $p\ll \rho \simeq mn$, $ϵ\simeq m$, and reduces the inertia $\rho +p$ to mass density, recovering the ordinary linearized continuity and Euler equations. Slow velocity alone does not justify dropping $p$ from the inertia of a relativistically hot fluid. The given metric supplies an external gravitational potential; the matter action by itself supplies no Poisson equation for it.

Use Minkowski spacetime coordinates $x^{\mu }=(ct,x,y,z)$ and the Minkowski metric of signature $(+,-,-,-)$. For a massive particle with three-velocity $\mathbf{v}=d\mathbf{x}/dt$ and $|\mathbf{v}|<c$, proper time satisfiesThus the four-velocity components areHere upper indices indicate components in the chosen coordinates; lowering with the Minkowski metric changes the signs of the spatial components. Its invariant product defined by the Minkowski metric is

The four-momentum is $p^{\mu }=(E/c,\mathbf{p})=m{U}^{\mu }$, so $E=\gamma m{c}^2$ and $\mathbf{p}=\gamma m\mathbf{v}$. The Taylor expansion of the Lorentz factor at small $|\mathbf{v}|/c$ isConsequently $\mathbf{p}=m\mathbf{v}+O(m|\mathbf{v}{|}^3/c^2)$, recovering ordinary momentum. At rest the four-momentum has energy $E_0=m{c}^2$, so subtracting this rest energy gives the kinetic energyThus the nonrelativistic limit preserves the rest-energy offset as well as the usual kinetic energy.

The comoving observer's four-velocity has $u_0=-a$, so its measured photon energy is $E=-u_{\mu }{p}^{\mu }=a{p}^0$, and the comoving momentum is $q=aE=a^2{p}^0$. Let $P^i=p^i/p^0=d{x}^i/d\tau$. The null geodesic condition isDivide the temporal geodesic equation by $p^0=d\tau /d\lambda$, and use the supplied Christoffel symbols:Differentiating $q=a^2{p}^0$ cancels the homogeneous redshift terms. In the remaining first-order term, replace $P^i$ by the unperturbed unit propagation direction $n^i$:This is the synchronous photon momentum redshift. Background physical momentum decreases as $a^{-1}$, while the background comoving momentum stays constant.

The spatial geodesic equation gives $d{p}^i/d\tau =-2\mathcal{H}{p}^i$ in the unperturbed universe. Since the temporal equation gives the identical factor for $p^0$, their ratio $P^i$ is constant at zeroth order. More explicitly, at first order,Every displayed term contains a metric perturbation or its derivative. The orthonormal-frame unit direction differs from $P^i$ by a first-order metric correction, $n^i=P^i+h^i{}_j{P}^j/2+O(h^2)$, whose derivative is also first order. Hence $d{n}^i/d\tau =O(h)$, with no background change of direction. Direction corrections multiplied by an already first-order source contribute only at second order.

Use the position four-vector $X^a=(ct,\mathbf{x})$ and Minkowski signature $(+,-,-,-)$. Along a massive particle's timelike path,Thus its four-velocity is $U^a=d{X}^a/d\tau =\gamma (c,\mathbf{v})$, and multiplying by the invariant rest mass givesIts temporal component is $E/c$, and the invariant relation is $(E/c{)}^2-|\mathbf{p}{|}^2=M^2{c}^2$.

For a massless particle, use the null energy–momentum relation rather than substituting $M=0$ into $M\gamma$: a massless particle has no proper-time parametrization of a timelike rest frame. With positive energy, $p^a{p}_a=0$ implies $E/c=|\mathbf{p}|$. HenceThe original PDF retains the spatial vector in this expression; the TeX conversion incorrectly replaced it by its magnitude.

In the instantaneous rest frame the total radiation has zero net three-momentum, so its emitted four-momentum is proportional to the four-velocity. A Lorentz boost multiplies its energy by $\gamma$, while the elapsed coordinate time for the source is also multiplied by $\gamma$. Their ratio, total instantaneous radiated power, is therefore invariant. This concerns total emitted power, not the directional flux or an observer's photon arrival rate.

Use the metric of signature $(+---)$ and dots for proper-time derivatives. From $\gamma =(1-v^2{)}^{-1/2}$ obtain $\dot{\gamma }={\gamma }^3\alpha$, where $\alpha =\mathbf{v}\cdot \dot{\mathbf{v}}$. Differentiating $U=(\gamma ,\gamma \mathbf{v})$ givesSquaring in the Minkowski metric, using $1-v^2={\gamma }^{-2}$, gives $\dot{U}\cdot \dot{U}={\gamma }^6{\alpha }^2(1-v^2)-2{\gamma }^4{\alpha }^2-{\gamma }^2|\dot{\mathbf{v}}{|}^2$, henceIn the rest frame, $-{\dot{U}}^2$ is the squared ordinary acceleration, so invariance of power and the low-speed Larmor formula give $P=-({\mu }_0{q}^2/6\pi ){\dot{U}}^2$. In terms of coordinate-time acceleration $\mathbf{a}=d\mathbf{v}/dt$, this is the Lienard formulaFor circular motion perpendicular to $\mathbf{B}$, $\mathbf{a}\perp \mathbf{v}$ and the Lorentz force gives $\gamma m|\mathbf{a}|=|q|v|\mathbf{B}|$. ThusAt low speed ${\gamma }^2-1\sim v^2$, reproducing ordinary Larmor radiation with magnetic acceleration $|q|vB/m$.

Let $N=\sqrt{K^a{K}_a}>0$ on the fluid region and choose the Killing vector field $K$ to be future-directed. The normalized four-velocity isThe Killing equation implies both $K^b{\nabla }_{b}N=0$ andConsequently its Killing-flow four-acceleration isFor a scalar, the Lie derivative is ${\mathcal{L}}_{K}p=K^b{\nabla }_{b}p$. Thus the assumed stationarity of the pressure gives $U^b{\nabla }_{b}p=0$. These are the ingredients needed for hydrostatic equilibrium along a Killing vector; hypersurface orthogonality of $K$ is not needed.

Use metric signature $(+---)$ throughout this solution. Stress-energy conservation for the perfect fluid is ${\nabla }_b{T}^{ab}=0$. Expanding it and using metric compatibility givesThe normalization of the four-velocity implies $U_a{U}^b{\nabla }_b{U}^a=0$. Contracting with $U_a$ therefore gives the relativistic perfect-fluid energy equation,while subtracting the component parallel to $U$ gives the relativistic Euler equation,Equivalently, apply the orthogonal projection ${\delta }_a{}^c-U_a{U}^c$ to the conservation equation. This form makes clear that a pressure gradient orthogonal to the four-velocity supplies the fluid's four-acceleration.

For $|\mathbf{v}|<c$, define the Lorentz factor $\gamma =(1-|\mathbf{v}{|}^2/c^2{)}^{-1/2}$. A particle of rest mass $m$ hasThese expressions give the relativistic energy-momentum relation directly:so $\boxed{E^2=c^2\mathbf{p}\cdot \mathbf{p}+m^2{c}^4}$.

The four-momentum iswhere $U^{\mu }$ is the four-velocity, the proper-time derivative of $(ct,\mathbf{x})$. It is a four-vector, so its Minkowski metric norm is invariant under Lorentz transformations. Choose metric signature $(+,-,-,-)$; in the particle's rest frame, $P^{\mu }=(mc,0)$ and $P_{\mu }{P}^{\mu }=m^2{c}^2$. In any frame the same norm is $E^2/c^2-|\mathbf{p}{|}^2$. Equating the two yields the same relativistic energy-momentum relation. With the opposite metric convention both norm expressions change sign, leaving the energy relation unchanged.

Use $x^{\mu }=(ct,x,y,z)$ and metric signature $(+,-,-,-)$. Since $d\tau =dt/\gamma$, the four-velocity and four-momentum areThese use the proper time and Lorentz factor; the mass shell identity is $E^2-c^2|\mathbf{p}{|}^2=M^2{c}^4$.

Initially the parent four-momentum is $(Mc,0)$. By four-momentum conservation, the remaining single particle has energy $M{c}^2-E$ and three-momentum $-\mathbf{p}$. Applying its mass shell identity gives $m^2{c}^4=(M{c}^2-E{)}^2-c^2|\mathbf{p}{|}^2$, henceThe positive root is selected for the rest mass; physically admissible decay data also give nonnegative remaining energy.

Use geometric units $G=c=1$, metric signature $(-+++)$, and a four-velocity normalized by the physical metric. Write $\varphi =1+\Phi$, where $|\Phi |\ll 1$. For a pressureless perfect fluid, the dust stress-energy tensor is $T_{ij}=\rho U_i{U}_j$ and $g^{ij}{U}_i{U}_j=-1$. Since ${\eta }^{ij}={\varphi }^2{g}^{ij}$, its background trace isConsequently the field equation is $(-{\partial }_t^2+\Delta )\varphi =4\pi \rho {\varphi }^3$. Neglecting time derivatives and higher weak-field orders gives the Newtonian limit of Nordstrom gravity:The constant shift between $\varphi$ and $\Phi$ does not affect the Poisson equation or the force. Here $\rho$ is the physical rest density; if fluid indices are instead defined using the background metric, the intermediate powers of $\varphi$ change but the same leading-order equation results.

For the motion, the Levi-Civita connection of the conformally flat metric isIn particular ${\Gamma }^{\alpha }{}_{00}={\partial }_{\alpha }\log \varphi$. Replacing the affine parameter in the geodesic equation by coordinate time gives, with $w^i=(1,\mathbf{v})$ and $\mathbf{v}=d\mathbf{x}/dt$,At small speeds and for a nearly static field, this becomesThus the Newtonian gravitational potential and freely falling trajectories agree to the stated weak-field, slow-motion order. The comparison is an approximation, not an exact equivalence of relativistic trajectories.

Assume $M>0$ and use geometric units. The stationary Killing vector field $K={\partial }_t$ gives the conserved Killing energy $E=-g(K,U)=f\dot{t}$. For radial motion, normalization of the four-velocity gives $-f{\dot{t}}^2+f^{-1}{\dot{r}}^2=-1$. Combining these identities yieldsAt rest at infinity, $f\to 1$ and $\dot{r}\to 0$, so $E=1$. Choose the inward branch $\dot{r}=-\sqrt{2M/r}$. Its proper time from the Schwarzschild event horizon to the curvature singularity isIn ordinary units this is $4GM/(3c^3)$. The endpoint $r=0$ is not a regular centre of the spacetime. This is the Schwarzschild horizon-to-singularity proper time for fall from infinity, rather than a universal time for all infall energies.

For this geodesic congruence, set $w(r)=\sqrt{2M/r}$. Its four-velocity and metric dual areThe right side is an exact one-form because $w/f$ depends only on $r$. Thus $U_a=-{\partial }_a\tau$, with free-fall proper time in Schwarzschild spacetime determined byIndeed $U^a{\partial }_a\tau =-U^a{U}_a=1$, so this scalar advances by the actual proper time on every member of the infalling geodesic congruence, with a synchronized choice of additive offsets.

Substitution of $dt=d\tau -w{f}^{-1}dr$ in the Schwarzschild metric givesUsing $f=1-w^2$, the Painlevé–Gullstrand coordinates therefore haveAll coefficients are finite at $r=2M$; the determinant of the $\tau ,r$ block is $-f-w^2=-1$. Hence the horizon is a coordinate singularity of Schwarzschild time, not a degeneracy of the metric. The new spatial slices have $d{\ell }^2=d{r}^2+r^{2}d{\Sigma }^2$, the flat Euclidean metric, and $g^{ab}{\partial }_a\tau {\partial }_b\tau =-1$ makes their normals the freely falling four-velocities. By contrast, the curvature scalar $R_{abcd}{R}^{abcd}=48M^2/r^6$ diverges at $r=0$; this genuine singularity is not removed.