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For $h_{ab}=\delta g_{ab}$, the Ricci scalar varies asThe last two terms form a covariant divergence and hence contribute only a boundary term to a compactly supported variation of the Einstein-Hilbert action.

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.

For a perfect fluid in general relativity, let $\rho$ be rest-frame total energy density divided by $c^2$, $P$ its isotropic pressure, and $U^{\mu }$ its four-velocity, normalized by $U^{\mu }{U}_{\mu }=-c^2$. ThenThe contracted Bianchi identity and the Einstein field equations give stress-energy conservation, ${\nabla }_{\mu }{T}^{\mu \nu }=0$, for a constant cosmological constant. Contract with $U_{\nu }$ and use the normalization to obtainIn comoving coordinates $U^{\mu }=(1,0,0,0)$ and the covariant divergence is ${\nabla }_{\mu }{U}^{\mu }=3\dot{R}/R=3H$, where a dot is a cosmic time derivative and $H$ the Hubble parameter. ThereforeFor a fixed region in comoving coordinates, let $V(t)=V_{\ast }{R}^3(t)$ be its physical volume, where $V_{\ast }$ is a fixed coordinate-volume integral, and let $E=\rho c^{2}V$. Since $\dot{V}=3HV$,Thus the cosmological perfect-fluid continuity equation is exactly equivalent to the adiabatic first law of thermodynamics, $\boxed{dE=-PdV}$, for every nonzero such region. Conversely, expanding that first law of thermodynamics and dividing by $c^{2}V$ recovers the same equation. No heat or particle flux crosses the comoving boundary; interacting components separately need source terms, although the total still obeys this conservation equation.

There is a sign inconsistency in the supplied formula sheet worth isolating. Its Riemann curvature tensor and Ricci tensor conventions, together with its stated $-+++$ metric signature, give $G_{tt}=-3H^2$ for $R(t)=e^{Ht}$ with constant $H$. Its vacuum Einstein field equations with $+\Lambda g_{\mu \nu }$ then require $H^2=-\Lambda c^2/3$, whereas its printed vacuum Friedmann equation gives $H^2=+\Lambda c^2/3$. For example, $H>0$, $\rho =P=0$ and $\Lambda =3H^2/c^2$ satisfy the latter but leave the former's $tt$ component equal to $-6H^2$. Changing the geometric cosmological constant term to $-\Lambda g_{\mu \nu }$ repairs this mismatch. The conservation derivation is unaffected; below take the geometric cosmological constant to be zero and include any vacuum energy as a fluid, so no ambiguous sign enters.

A semicolon denotes the covariant derivative for the Levi-Civita connection. For example, the four lower indices of the Riemann curvature tensor contribute four connection terms:Here ${\partial }_{\eta }=\partial /\partial x^{\eta }$ is a partial derivative, ${\Gamma }_{b\eta }^a$ are Christoffel symbols, and repeated indices are summed from $0$ to $3$. Write $g_{\mu \nu }$ for the metric tensor, $g^{\mu \nu }$ for its inverse metric, $R_{\mu \nu }=g^{\lambda {\nu }^{\prime }}{R}_{\lambda \mu {\nu }^{\prime }\nu }$ for the Ricci tensor, and $R=g^{\mu \nu }{R}_{\mu \nu }$ for the Ricci scalar. This tensor contraction agrees with the paper's $R_{\mu \nu }=-R^a{}_{\mu \nu a}$.

By metric compatibility, the inverse metric may pass through each covariant derivative. Contracting the second Bianchi identity with $g^{\lambda \nu }$ giveswhere ${\nabla }_{\eta }$ is the same covariant derivative as the semicolon. Next contract with $g^{\mu \kappa }$. Antisymmetry of the first pair gives $g^{\mu \kappa }{R}_{\lambda \mu \kappa \eta }=-R_{\lambda \eta }$, soThus the contracted Bianchi identity is ${\nabla }^{\mu }{R}_{\mu \eta }=\frac{1}{2}{\partial }_{\eta }R$. Raising the remaining index and using metric compatibility provesThe Einstein tensor $G^{\mu \nu }$ therefore has identically zero covariant divergence, without imposing any field equation. This makes the Einstein field equations compatible with stress-energy conservation, ${\nabla }_{\mu }{T}^{\mu \nu }=0$, where $T^{\mu \nu }$ is the stress-energy tensor. The connection terms describe how energy and momentum are compared at neighboring points; ordinary componentwise partial derivatives would not give a coordinate-independent conservation law in curved spacetime.

Put $h_{ab}=\delta g_{ab}$, $h=g^{ab}{h}_{ab}$, and $h^{ab}=g^{ac}{g}^{bd}{h}_{cd}$. Throughout this part use the Riemann curvature tensor convention printed in the paper. Varying $g^{ac}{g}_{cb}={\delta }^a{}_b$ gives the inverse-metric variation. The determinant identity $\delta \log |\det g|=g^{ab}{h}_{ab}$ gives the volume form variation. Together these areIn Lorentzian coordinates the latter reads $\delta \sqrt{-g}=\frac{1}{2}\sqrt{-g}h$.

For the connection, vary metric compatibility ${\nabla }_a{g}_{bc}=0$:Add the equations with derivatives $a,b$, subtract that with derivative $c$, and use symmetry of the lower indices of the Levi-Civita connection. This givesThe difference of two connections is a tensor, so this formula is covariant.

Varying the coordinate curvature formula yields the Palatini identityThe contractions of the connection variation areFinally, $\delta R=(\delta g^{ab})R_{ab}+g^{ab}\delta R_{ab}$ gives the metric variation of scalar curvature:The double divergence may equivalently be written ${\nabla }_a{\nabla }_b{h}^{ab}$: relabeling the contracted dummy indices and using symmetry of $h^{ab}$ gives the same expression. The derivative terms are a covariant divergence, useful when varying a curvature-dependent action.

For the torsion-free Levi-Civita connection, cyclically summing the displayed coordinate expression and using ${\Gamma }^a{}_{bc}={\Gamma }^a{}_{cb}$ gives the algebraic first Bianchi identityApply the Jacobi identity to three covariant-derivative commutators, or differentiate the coordinate formula in normal coordinates. The third-derivative terms and the derivatives of quadratic connection terms cancel cyclically, giving the differential second Bianchi identityContracting the first and third curvature indices and then contracting once more yieldsor the contracted Bianchi identity

For the metric variation $h_{ab}=\delta g_{ab}$,while the two double-divergence terms in the supplied metric variation of scalar curvature become a boundary term after integration by parts for tensor fields. The Einstein-Hilbert action therefore contributes $G_{ab}+\Lambda g_{ab}$. Varying the Proca action with respect to the metric gives the Proca stress-energy tensor, and the full Einstein-Proca theory equation is

Since $\delta F_{ab}=2{\nabla }_{[a}\delta A_{b]}$, variation with respect to $A_b$ and one integration by parts give the covariant Proca equationTaking its covariant divergence givesbecause the resulting contractions pair the symmetric Ricci tensor with the antisymmetric electromagnetic field tensor. Since $m\ne 0$, the Lorenz constraint in Proca theory follows:Moreover, $F=dA$ and $d^2=0$, so the Electromagnetic Bianchi identity is

Finally, use that identity to differentiate the matter tensor. The result groups naturally asBoth terms vanish by the Proca equation and its Lorenz constraint in Proca theory. Thus ${\nabla }^a{T}_{ab}=0$, exactly as required by the contracted Bianchi identity applied to the Einstein field equations. The vector equation is therefore consistent with the gravitational equation.

The trace of the electromagnetic stress-energy tensor in four spacetime dimensions isFor its covariant divergence, the source-free Maxwell equations eliminate the derivative of the first factor. Contracting the Bianchi identity ${\nabla }_{[a}{F}_{bc]}=0$ with $F^{ac}$ giveswhich cancels the derivative of the trace term. HenceFor $J^a=T^{ab}{V}_b$, stress-energy conservation and symmetry of $T^{ab}$ now imply

Local energy-momentum conservation is the vanishing covariant divergence of the energy-momentum tensor: