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On a common compact solution domain, a right-hand side continuous in parameters and uniformly Lipschitz in the state yields continuous dependence of the solution on those parameters. The Gronwall inequality controls the difference of two solutions. A transverse zero of an event function persists under a small parameter change, which allows a simple stellar surface in the Newtonian limit to persist in relativistic models.

With the metric and stress-energy tensor fixed and zero cosmological constant, changing the sign convention for the Riemann curvature tensor changes the signs of the Ricci tensor, scalar curvature and Einstein tensor. The same physical field equation is therefore written with the opposite coupling sign. For signature $(+---)$, the convention $R^a{}_{bcd}={\partial }_d{\Gamma }_{bc}^a-{\partial }_c{\Gamma }_{bd}^a-{\Gamma }_{ec}^a{\Gamma }_{bd}^e+{\Gamma }_{ed}^a{\Gamma }_{bc}^e$ and $R_{ab}=R^c{}_{acb}$ gives $R_{00}\simeq -\Delta \Phi$ in a static Newtonian limit. The trace-reversed equation and $\Delta \Phi =4\pi G\rho$ then require $G_{ab}=-8\pi G{T}_{ab}$.

General relativity describes gravity through the geometry of spacetime. Its metric tensor specifies distances, proper time measured by clocks, and the causal structure that limits communication. Matter and radiation affect this geometry through the Einstein field equations; freely falling test particles follow geodesics of the resulting geometry. A planet's orbit and a falling apple therefore belong to the same geometric story. The Newtonian limit recovers familiar Newtonian gravity when fields are weak and speeds are small compared with the speed of light. See David Tong's introduction to general relativity.

The most direct way to understand spacetime curvature is to compare neighboring freely falling bodies. Even when each body feels weightless, their separation can change: this is geodesic deviation, observed as a tidal force. Curvature concerns these relative measurements, rather than the shape of a drawing in an imagined surrounding space. A second accessible measurement is gravitational time dilation: identical clocks maintained at different heights accumulate different proper times. Together, tides and clocks turn an abstract geometry into laboratory questions.

For metric signature $(-,+,+,+)$, $x^0=ct$ and small dimensionless potentials, the metric perturbation is $h_{00}=-2\Phi /c^2$, $h_{ij}=-2\Psi {\delta }_{ij}/c^2$. Substitution into the linearized Ricci tensor and scalar formula givesThe contraction is $R=(4{\nabla }^2\Psi -2{\nabla }^2\Phi -6{\partial }_0^2\Psi )/c^2$, so $G^0{}_0=-2{\nabla }^2\Psi /c^2$. These signs use $R^{\mu }{}_{\nu \alpha \beta }{V}^{\nu }=[{\nabla }_{\alpha },{\nabla }_{\beta }]V^{\mu }$. Static zero-anisotropic-stress boundary conditions give $\Phi =\Psi$ and the Newtonian limit; without those assumptions the two potentials and their time derivatives must be retained.

For this question use the Levi-Civita connection, its metric compatibility and the usual curvature symmetries. In the index ordering of Question 1, put $R_{\mu \nu \rho \sigma }=g_{\sigma \lambda }{R}_{\mu \nu \rho }{}^{\lambda }$ and $R_{\nu \rho }=g^{\mu \sigma }{R}_{\mu \nu \rho \sigma }$. Contract the second Bianchi identity in its derivative index $\mu$ and curvature index $\rho$. The curvature symmetries giveNow contract $\lambda ,\sigma$. The left-hand side becomes $-{\nabla }^{\rho }{R}_{\nu \rho }$, while the right-hand side becomes ${\nabla }^{\sigma }{R}_{\nu \sigma }-{\nabla }_{\nu }R$. ThusUsing $\nabla g=0$ yields the contracted Bianchi identityTaking the covariant divergence of the Einstein field equations, with constant $G$, then gives $8\pi G{\nabla }^{\mu }{T}_{\mu \nu }=0$. The matter equation is therefore consistent with the geometric identity rather than imposing an extra incompatible condition on the metric.

Stress-energy conservation expresses the local balance of energy and momentum, including their transport and the effects of the gravitational Levi-Civita connection. In a freely falling frame at a point it reduces there to the familiar ordinary divergence law. A global conserved energy requires additional geometric structure, such as a timelike Killing vector field; it does not follow in an arbitrary spacetime just by replacing a covariant derivative with an ordinary derivative.

For relativistic dust,With signature $(+,-,-,-)$, $u^{\nu }{u}_{\nu }=1$ implies $u_{\nu }{u}^{\mu }{\nabla }_{\mu }{u}^{\nu }=0$. Contracting gives ${\nabla }_{\mu }(\rho u^{\mu })=0$; where $\rho >0$, substitution then gives $u^{\mu }{\nabla }_{\mu }{u}^{\nu }=0$. Thus the continuity law combines rest-mass conservation and free geodesic motion for this matter model.

For the Newtonian limit, take $c=1$ and $\eta =\mathrm{diag}(1,-1,-1,-1)$, consistently with Question 3. Use the static weak field sourced by the time-independent, slowly moving matter, retaining first order in $ϵ$ and leading order in the spatial speed. Write spatial gradients with ${\partial }^i={\delta }^{ij}{\partial }_j$ here; these are Euclidean spatial indices in the Newtonian equation. The Levi-Civita connection formula givesThe exact change from proper time to $t$ in the spatial geodesic equation isTerms containing spatial velocities are negligible at leading order, leavingAn additive constant in $\Phi$ is immaterial to this acceleration and is normally fixed by the asymptotic time normalization. With units restored, $g_{00}=1+2{\Phi }_{\mathrm{physical}}/c^2$.

At the same order, staticity and the curvature definition giveThe trace of the four-dimensional Einstein equations is $R=-8\pi GT$, so their trace-reversed form isFor nonrelativistic dust, $u^0\simeq 1$, $u^i\simeq 0$, so $T_{00}\simeq \rho$ and $T\simeq \rho$. Keeping leading matter terms therefore gives $R_{00}=4\pi G\rho$. Combining this with the geometric expression yieldsThis trace reversal is essential: equating $R_{00}$ directly to $8\pi G\rho$ would miss the factor of two. The approximation is the quasistatic Newtonian sector; arbitrary independently imposed gravitational radiation is not part of that sector.

Approximate Earth by an isolated, nonrotating, spherically symmetric body. By Birkhoff's theorem, the exterior vacuum metric tensor is the Schwarzschild metric. With $x^0=ct$ it isHere $r$ is the areal radius, $\theta$ and $\varphi$ are spherical angles, $t$ is Schwarzschild time normalized to the proper time of a static observer at infinity, $M$ is Earth's mass, $G$ the gravitational constant, and $c$ the speed of light. The nonzero diagonal metric tensor components are $g_{00}=-f$, $g_{rr}=f^{-1}$, $g_{\theta \theta }=r^2$, $g_{\varphi \varphi }=r^2{\sin }^2\theta$.

The apple is a test particle; neglect its effect on the metric tensor, air resistance, Earth's rotation and nonspherical structure, and external tidal forces. Near Earth's surface, $GM/(R_E{c}^2)\simeq 7\times 10^{-10}$, $h/R_E\ll 1$, and the falling speed remains much smaller than $c$. Thus the Newtonian limit of this metric tensor is appropriate, while the exact radial geodesic is available as a check. A static observer under the tree has proper time $T=\sqrt{f(R_E)}t$, up to an additive constant. Such an observer is supported by the ground and has nonzero proper acceleration; the released apple instead has zero proper acceleration.

Use $x^0=ct$ and metric signature $(-,+,+,+)$. With the source's curvature sign convention, the Ricci tensor isIts time-time component is thereforeFor a slowly moving, pressure-negligible source, the stress-energy tensor is approximated by relativistic dust: $T^{\mu \nu }=\rho U^{\mu }{U}^{\nu }$, where $\rho$ is mass density and $U^{\mu }$ is the four-velocity. In its approximate rest local inertial frame, $U^{\mu }\simeq (c,0,0,0)$ and $U_0\simeq -c$. Thusand the trace-reversed stress-energy tensor satisfies

Independently, introduce the metric perturbation $g_{\mu \nu }={\eta }_{\mu \nu }+h_{\mu \nu }$, with $|h_{\mu \nu }|\ll 1$ and constant Minkowski metric ${\eta }_{\mu \nu }$. The Christoffel symbols are first order in $h$, so their products are discarded in the weak-field approximation. With a comma denoting a partial derivative, the linearized Riemann curvature tensor isTaking the tensor contraction gives, to this order,In the static Newtonian limit, all time partial derivatives vanish. The constant ${\eta }_{00}$ also has zero partial derivatives, leavingHere ${\nabla }^2={\sum }_i{\partial }_i^2$ is the spatial Laplace operator in the approximately Cartesian background coordinates. This sign follows the paper's curvature sign convention.

For the Newtonian gravitational potential $\Phi$, the weak static metric tensor has $g_{00}\simeq -1-2\Phi /c^2$. The Newtonian limit derived above consequently givesMatching the Poisson equation for Newtonian gravity, ${\nabla }^2\Phi =4\pi G\rho$, fixesHere $G$ is the gravitational constant and $c$ is the speed of light. This negative coupling agrees with the source formula sheet. The familiar positive coupling uses the opposite curvature sign convention for the Ricci tensor; the physical Poisson equation for Newtonian gravity is the same.

Use metric signature $(+---)$ and the curvature sign fixed by the stated Newtonian limit. In flat spacetime the source-free Maxwell equations are ${\partial }_a{F}^{ab}=0$ and ${\partial }_{[a}{F}_{bc]}=0$. The stress-energy tensor in mixed form is $T^a{}_b=F^{ac}{F}_{cb}+{\delta }_b^a{F}_{cd}{F}^{cd}/4$. Its divergence isContracting the cyclic Maxwell equation with $F^{ac}$ gives $F^{ac}{\partial }_a{F}_{cb}=-{\partial }_b(F_{cd}{F}^{cd})/4$. The first term also vanishes, proving $\boxed{{\partial }_a{T}^{ab}=0}$. The component $b=0$ expresses local electromagnetic energy conservation, and the spatial components express momentum conservation. Integrating converts them into the usual energy and momentum flux balances.

In curved spacetime the tensor equations areThey agree with the flat equations in coordinates of a local inertial frame at a point, where the connection vanishes, and covariance then extends them to arbitrary coordinates. Antisymmetry makes the homogeneous equation equivalent to $dF=0$. Metric compatibility permits the same contraction proof, now giving $\boxed{{\nabla }_a{T}^{ab}=0}$. In general,which need not vanish. Covariant local balance remains valid; ordinary coordinate components do not provide a universal conserved global energy-momentum vector in a curved geometry. Additional spacetime symmetries can furnish conserved currents. A locally inertial frame removes the connection at a point, not the curvature or the global obstruction.

The alternating array with fixed component ${ϵ}_{0123}=1$ is appropriate in oriented inertial special-relativistic coordinates, whose proper Lorentz transformations have determinant one. It is not itself a covariant tensor under general coordinate changes. For an oriented Lorentzian metric, replace it by the metric metric volume tensor ${\epsilon }_{abcd}=\sqrt{-\det g}[abcd]$. The genuine dual isThis is the Hodge star; the determinant factor, orientation and raising conventions are essential. Even in flat spacetime arbitrary curvilinear coordinates require the metric volume tensor, so the distinction is about coordinate covariance rather than curvature alone.

The isolated equation for $R_{00}$ singles out a coordinate component and cannot be a generally covariant law. Replacing it by a proportionality $R_{ab}\propto T_{ab}$ still fails: the contracted Bianchi identity gives ${\nabla }^a{R}_{ab}={\nabla }_{b}R/2$, whereas stress-energy conservation requires ${\nabla }^a{T}_{ab}=0$. Such a law would force $R$ constant and hence impose an unphysical restriction on a varying relativistic dust density. Missing units or a coupling constant in the proposed proportionality would not repair this structural problem.

The Einstein tensor $G_{ab}=R_{ab}-R{g}_{ab}/2$ is divergence-free. Thereforeis consistent with covariant stress-energy conservation in the source's curvature convention. Its trace gives $R=8\pi GT$ and hence $R_{ab}=-8\pi G(T_{ab}-T{g}_{ab}/2)$. For slowly moving relativistic dust with $V^a{V}_a=1$, $T_{00}\simeq \rho$, $T\simeq \rho$, $g_{00}\simeq 1$, so $R_{00}\simeq -4\pi G\rho$, exactly matching $-\Delta \varphi$ and Poisson's equation.

Finally, electromagnetic stress-energy is trace-free: $T^a{}_a=F^{ac}{F}_{ca}+F_{cd}{F}^{cd}=0$ by antisymmetry. The traced field equation then gives $\boxed{R=0}$ for a purely electromagnetic source. This does not say $R_{ab}=0$ or that spacetime is flat; its trace-free Ricci curvature can still be nonzero.

General relativity models spacetime by a smooth four-dimensional Lorentzian manifold whose metric tensor governs clock readings, causal cones and free-particle motion. The Einstein equivalence principle motivates local special relativity and universal geodesic free fall for test bodies; it does not by itself select a unique gravitational field equation. We assume a Levi-Civita connection, a metric tensor as the gravitational field with no additional dynamical gravitational variables, and matter with a covariantly conserved stress-energy tensor. Units have $c=1$.

The existence of local inertial frames is a local statement. At any regular event choose an orthonormal frame and Riemann normal coordinates, so that $g_{ab}={\eta }_{ab}$ and ${\Gamma }^a{}_{bc}=0$ at the event, with metric signature $(+---)$. Metric compatibility then gives ${\partial }_c{g}_{ab}=0$ there. The geodesic equation at that point is the special-relativistic inertial equation. Along a timelike geodesic, a parallel-transported orthonormal frame gives Fermi normal coordinates, eliminating the connection along the central worldline. These constructions do not in general eliminate second derivatives of the metric tensor in a neighborhood. The Riemann curvature tensor is built from the derivatives of the connection coefficients as well as their quadratic terms; nonzero Riemann curvature tensor components can remain when all connection coefficients vanish at a chosen point. Conversely, Polar coordinates in Minkowski spacetime can have nonzero connection coefficients but zero Riemann curvature tensor. Connection coefficients alone are therefore not an invariant measure of gravity.

For the tensorial content, consider a two-parameter family of geodesics parametrized by an affine parameter, with tangent $U$ and connecting field $J$. Their coordinate construction gives $[U,J]=0$, and torsion-freeness gives ${\nabla }_{U}J={\nabla }_{J}U$. Since ${\nabla }_{U}U=0$, commuting derivatives in the convention of Question 1 yields geodesic deviation:The left side is covariant relative acceleration, rather than the second coordinate derivative of an arbitrary separation. In an orthonormal frame in free fall with $U=(1,0,0,0)$, its spatial components are $D^2{J}^i/d{\tau }^2=-R^i{}_{00j}{J}^j$. This is a measurable tidal force, and a coordinate transformation cannot make a nonzero Riemann curvature tensor disappear. It is the distinction between gravity at one event and the behavior of an extended freely falling laboratory.

In a static Newtonian limit, $g_{00}=1+2\Phi$ and ${\Gamma }^i{}_{00}={\partial }_i\Phi$, so the test-body acceleration is $-\nabla \Phi$. The curvature-sign convention in Killing derivative identities gives $R^i{}_{00j}\simeq {\partial }_i{\partial }_j\Phi$. Hence relative acceleration is the negative Hessian of the gravitational potential. For the Newtonian potential of a point mass $\Phi =-GM/r$, the Hessian has radial eigenvalue $-2GM/r^3$ and two tangential eigenvalues $GM/r^3$: radial separations stretch and tangential ones compress. Outside the source the Ricci tensor can vanish while the Riemann curvature tensor remains nonzero. Its vacuum tidal part is encoded by the Weyl tensor; vacuum does not mean flatness. Gravitational waves likewise carry Riemann curvature tensor disturbances without a local material source.

To obtain the field equations, make the additional dynamical assumptions explicit: locality and diffeomorphism invariance of general relativity; a symmetric metric tensor field equation using at most second derivatives of the metric tensor; a geometrical left-hand side linear in the Riemann curvature tensor with constant coefficients; and compatibility with local matter conservation. These restrictions exclude higher-curvature or extra-field alternatives. Under them the available rank-two curvature contractions give the ansatzThe contracted Bianchi identity and metric compatibility giveFor this to vanish identically for arbitrary metric tensors, $b=-a/2$. Assuming $a\ne 0$ and absorbing its normalization into the coupling givesThe permitted constant term is the cosmological constant. Its value is not fixed by the equivalence principle. Conservation of $T$ follows for generally covariant matter on its equations of motion: varying the matter action under a compactly supported infinitesimal diffeomorphism, then integrating its term $T^{ab}{\nabla }_a{\xi }_b$ by parts, gives ${\nabla }_a{T}^{ab}=0$. The divergence-free Einstein tensor is consistent with this identity.

The coupling sign must be fixed in the actual curvature-sign convention in Killing derivative identities, with $R_{ab}=R^c{}_{acb}$. For the static weak-field approximation, $R_{00}\simeq -\Delta \Phi$. Set $\Lambda =0$ for the local asymptotically flat Newtonian comparison. In four dimensions the trace equation gives $R=-\kappa T$, so $R_{00}=\kappa (T_{00}-\frac{1}{2}{g}_{00}T)\simeq \kappa \rho /2$ for slowly moving matter. Matching $\Delta \Phi =4\pi G\rho$ forces $\kappa =-8\pi G$. Thus, with this convention,Under the opposite curvature-sign convention in Killing derivative identities the coupling sign reverses, as in Einstein-equation coupling under reversed curvature. This sign statement is compatible with the supplied linearized handout. Local inertial frames remove the connection at an event; tidal Riemann curvature tensor effects remain tensorial, and the field equations require dynamical assumptions beyond the equivalence principle.

The strong equivalence principle asserts universality of free fall even for bodies with gravitational self-binding, together with independence of local experimental outcomes, including gravitational experiments, from the location and velocity of a freely falling laboratory. “Local” means that external tidal variation across the apparatus can be neglected. It does not mean that all curvature, or a laboratory's own gravitational field, can be removed. The weak principle concerns freely falling test bodies without appreciable self-gravity; laboratory composition tests alone therefore do not establish the strong version.

Lunar laser ranging supplies evidence specifically sensitive to the strong version. The Earth and Moon have different fractions of gravitational self-energy. If those fractions affected gravitational and inertial mass differently, their accelerations toward the Sun would differ and the lunar orbit would acquire a periodically varying polarization, the Nordtvedt effect. Combining lunar ranging with laboratory bounds on composition-dependent differential acceleration separates this self-energy effect. A determination available before this examination found the self-energy contribution to the Earth–Moon difference in $m_G/m_I$ to be $(-2.0\pm 2.0)\times 10^{-13}$, consistent with zero; see the 2004 experimental analysis. Universality-of-free-fall experiments, gravitational redshift measurements and tests of local Lorentz invariance support the weaker constituent principles, but should not be confused with a direct self-gravity test.

In general relativity, all matter couples to one metric tensor, and freely falling test particles follow its geodesics. At any event, Riemann normal coordinates give $g_{ab}={\eta }_{ab}$ and ${\Gamma }^a{}_{bc}=0$. Nongravitational local equations reduce to their special-relativistic forms in that frame. Gravitational binding is governed by the same metric field equations rather than a separate composition-dependent force, and the local gravitational coupling is universal. These features implement the strong equivalence principle. Curvature still governs geodesic deviation, so this construction removes uniform acceleration, not tidal gravity.

To geometrize Newtonian mechanics, use a four-dimensional manifold with coordinates $(t,x^1,x^2,x^3)$, absolute time ${\tau }_a=(dt{)}_a$ and a degenerate spatial inverse metric $h^{ij}={\delta }^{ij}$, $h^{0a}=0$. For a gravitational potential $\Phi (t,x)$ define the torsion-free Newton–Cartan gravitational connection byIt satisfies ${\nabla }_a{\tau }_b=0$ and ${\nabla }_a{h}^{bc}=0$. An affine geodesic has $d^{2}t/d{\lambda }^2=0$; choosing $\lambda =t$ makes its spatial equationsThis is Newton's free-fall law. The geometry is Newton–Cartan theory, with an affine connection and degenerate metric data, rather than a nondegenerate four-dimensional Lorentzian metric. The spatial slices themselves remain Euclidean.

Use the curvature conventionIn these coordinates the quadratic connection terms vanish, and the only nonzero curvature components, apart from antisymmetry in $c,d$, areIn particular $R_{00}={\nabla }^2\Phi$ and all other components of the Ricci tensor vanish. This Newtonian tidal curvature gives relative acceleration $\delta {\ddot{x}}^i=-{\partial }_i{\partial }_j\Phi \delta x^j$. Spatially constant gravitational acceleration can be removed by an accelerating frame because its Hessian vanishes. The Poisson equation for Newtonian gravity becomes $R_{ab}=4\pi G\rho {\tau }_a{\tau }_b$. Fluid mechanics is compatible with the same connection: the covariant continuity equation is the usual mass-conservation equation, and the spatial Euler equation is $u^b{\nabla }_b{u}^i=-{\rho }^{-1}{\partial }_{i}p$, which includes the gravitational force through the connection.

Now replace this absolute-time geometry by the Lorentzian metric of general relativity, with signature $(-,+,+,+)$. A perfect fluid haswhere $\epsilon$ is rest-frame energy density. In units $c=1$, $\epsilon \simeq \rho$ for a nonrelativistic fluid. The Newtonian equation suggests a covariant curvature equation sourced by $T_{ab}$. However $R_{ab}=\kappa T_{ab}$ alone conflicts with generic stress-energy conservation, because the contracted Bianchi identity gives ${\nabla }^a{R}_{ab}=\frac{1}{2}{\nabla }_{b}R$. Among expressions linear in Ricci curvature, conservation therefore selectsA constant multiple of the metric is also divergence-free, giving the possible cosmological constant term. This is a heuristic selection using locality, universal metric coupling and second-order metric equations; Newtonian matching by itself does not prove uniqueness among every possible gravity theory.

To fix the coefficient, first set the cosmological constant to zero. The trace-reversed equation would read $R_{ab}=\kappa (T_{ab}-\frac{1}{2}T{g}_{ab})$. In the fluid rest frame, $T=-\epsilon +3p$, so $R_{00}=\frac{\kappa }{2}(\epsilon +3p)$. In the weak static Newtonian limit, $g_{00}=-(1+2\Phi )$ and $R_{00}\simeq {\nabla }^2\Phi$. For negligible pressure the Poisson equation for Newtonian gravity then requires $\kappa /2=4\pi G$. Restoring $c$ uses $x^0=ct$, $R_{00}\simeq {\nabla }^2\Phi /c^2$ and $T_{00}\simeq \rho c^2$, giving $\kappa =8\pi G/c^4$. ThusPressure contributes to relativistic active gravity, while the low-pressure weak-field limit recovers Newton's Poisson equation. The comparison fixes the Newtonian normalization but does not fix $\Lambda$, which is an independent parameter.

The useful high-density core bound from TOV compactness follows without choosing the unknown high-density equation of state. Let $a$ be the radius at which the decreasing energy density first reaches ${\rho }_0$, and put $\mu =m(a)$, $p_0=p({\rho }_0)$. The Tolman–Oppenheimer–Volkoff equation gives $p^{\prime }<0$ away from a nonempty centre; since $dp/d\rho >0$, energy density decreases outward. HenceFor the local stellar compactness, if $z\ge 0$, $1-z+\sqrt{1+z}\le 2$, since its derivative is $-1+(2\sqrt{1+z}{)}^{-1}<0$. Applying the supplied compactness inequality at the core boundary therefore gives $\mu /a<4/9$. Combining the two bounds yieldsThe pressure-dependent form further restricts the allowed core-boundary data:In particular, positivity of its right side gives $6\pi a^2{p}_0<3$, so $a^2<1/(2\pi p_0)$ when $p_0>0$. These restrictions depend only on known quantities, not on the equation of state above ${\rho }_0$.

The intended stellar-mass argument now attaches the known low-density envelope to each admissible pair $(a,\mu )$ with pressure $p_0$. Its Tolman–Oppenheimer–Volkoff equation involves only the known equation of state. For a physical low-density envelope with a uniformly bounded mass on this bounded set of core data, taking the supremum of the envelope-plus-core masses gives an upper mass limit independent of any high-density continuation. Stars with ${\rho }_c\le {\rho }_0$ are also determined entirely by the known equation of state. Unknown dense-core physics cannot evade the displayed core bounds.

A qualification is necessary for a claim about the entire star: the printed monotonicity assumptions alone do not guarantee a finite maximum total mass for an arbitrary known low-density equation of state. An explicit low-density polytropic mass divergence occurs for $p=K{\rho }^{5/4}$ with fixed $K>0$. In the low-density Newtonian limit, this is a stellar polytrope of index $n=4$, whose Lane-Emden equation has a finite first zero. With $\rho ={\rho }_c{\vartheta }^4$ and $r=\ell \xi$, hydrostatic balance fixeswhere ${\mu }_4>0$ is the dimensionless Lane-Emden surface mass constant. Thus $M\to \infty$ as ${\rho }_c\to 0$, although $M/R\propto {\rho }_c^{1/4}\to 0$. The first Lane-Emden zero is simple. The continuous dependence of an ODE solution on parameters therefore preserves this finite surface for sufficiently small relativistic parameter. The relativistic correction is controlled by $p_c/{\rho }_c=K{\rho }_c^{1/4}\to 0$, so the corresponding regular finite-radius Tolman–Oppenheimer–Volkoff equation solutions have the same limiting scalings. They never sample $\rho >{\rho }_0$, and satisfy the local positivity and monotonicity assumptions in their nonvacuum interiors. This shows exactly why a suitable bounded-envelope hypothesis, or an appropriate restriction to physically stable stellar models, must accompany the intended total-mass conclusion. The compactness inequality supplies a bound on the unknown core, not a bound on the mass of every conceivable dilute envelope.

tidal forces measure differences in gravitational acceleration across an extended region. Neighboring freely falling particles each have zero four-acceleration, but their separation need not accelerate uniformly. Let $u^{\mu }$ be the tangent to a reference timelike geodesic and ${\xi }^{\mu }$ its infinitesimal separation from a neighboring one, chosen from an affinely parametrized geodesic congruence. Using the Riemann curvature tensor convention specified in Solution 3,Changing the overall curvature convention changes this displayed sign. The covariant derivatives compare vectors by parallel transport, avoiding a coordinate-dependent subtraction of accelerations.

In the Newtonian limit, expand the force field at two nearby points:For $\Phi =-GM/r$, a local orthonormal frame aligned with the radius givesThere is radial stretching and transverse compression; the trace vanishes outside the mass. Two freely falling test masses and a measurement of their changing separation implement gravity gradiometry. Their common acceleration drops out, while a relative acceleration proportional to separation remains. A local inertial frame removes Christoffel symbols at a point, but cannot remove nonzero curvature over a finite region.

The Einstein tensor and its contractions areTo match the supplied Ricci tensor components, use the Riemann curvature tensor conventionThis convention also fixes the sign used for geodesic deviation above. In vacuum, tracing the Einstein field equations gives $-R+4\Lambda =0$. Substitution therefore yields $R_{\mu \nu }=\Lambda g_{\mu \nu }$.

Differentiate the angular field equation to obtainConsequently $R_{00}=\Lambda A=\Lambda g_{00}$ and $R_{11}=-\Lambda /A=\Lambda g_{11}$. The other angular component follows from $R_{33}={\sin }^2\theta R_{22}$. Thus the single angular equation implies all the displayed vacuum equations wherever this coordinate chart is regular.

Integrating $(rA{)}^{\prime }=1-\Lambda r^2$ gives $A=1+C/r-\Lambda r^2/3$. Match $g_{tt}\simeq 1+2\Phi$ with the Newtonian gravitational potential $\Phi =-GM/r$ in the weak-field overlap region to identify $C=-2GM$. Hence the Schwarzschild-de Sitter spacetime hasNegative $\Lambda$ instead gives the Schwarzschild--anti-de Sitter black hole case. This matching fixes the normalization locally; a spacetime with nonzero $\Lambda$ is not asymptotically flat.

For timelike geodesics in the orbital plane, take proper time as parameter and use $L=(A{\dot{t}}^2-{\dot{r}}^2/A-r^2{\dot{\varphi }}^2)/2$. The cyclic coordinates give $A\dot{t}=k$ and $r^2\dot{\varphi }=h$; the angular canonical momentum is $-h$ in this signature. The four-velocity normalization $2L=1$ then givesHere $k$ is conserved Killing energy per unit rest mass in the chosen time normalization, and $h$ is conserved angular momentum per unit rest mass. There is no general interpretation of $k$ as energy measured at an asymptotically flat infinity when $\Lambda \ne 0$.

For radial motion $h=0$, differentiation gives $2\dot{r}\ddot{r}=-A^{\prime }\dot{r}$. The radial Euler-Lagrange equation extends the resulting expression through a turning point where division by $\dot{r}$ would fail. ThusThis is the proper-time second derivative of the areal radius, not a nonzero four-acceleration of the freely falling particle. Positive $\Lambda$ contributes an outward term; negative $\Lambda$ reinforces inward attraction. Restoring units in the slow-motion, weak-field Newtonian limit givesThe extra term describes the gravitational effect of a cosmological constant, not drag or a contact force. The approximation requires $GM/(r{c}^2)\ll 1$, $|\Lambda |r^2\ll 1$ and speed much less than $c$.

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.