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For relativistic dust, $T^{\alpha \beta }=\rho u^{\alpha }{u}^{\beta }$, with $u^{\alpha }{u}_{\alpha }=-c^2$. Expanding stress-energy conservation givesContract with $u_{\beta }$. Metric compatibility and the fixed normalization implyConsequentlyThis is the covariant mass conservation equation. Substituting it back, wherever $\rho >0$, givesThus the dust's fluid elements follow timelike geodesics. With $u^{\alpha }=d{x}^{\alpha }/d\tau$, the proper time $\tau$ is an affine parameter, and explicitlyThe absence of pressure removes the force produced by pressure gradients; gravity is encoded by the Levi-Civita connection. At a point with $\rho =0$ this calculation makes no claim about a fluid velocity.

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.

Write the equation of state as $P=w\rho c^2$. Integrating the cosmological perfect-fluid continuity equation for constant $w$ gives $\rho \propto R^{-3(1+w)}$. With spatial flatness and the geometric cosmological constant set to zero, the Friedmann equation is $H^2=8\pi G\rho /3$.

In S1, $w=-1$ describes vacuum energy: $T_{\mu \nu }=-\rho c^2{g}_{\mu \nu }$ and $\rho$ is constant. Its negative pressure drives accelerated expansion, $\ddot{R}/R=8\pi G\rho /3=H_1^2$, and the expanding solution is $R\propto e^{H_{1}t}$. This is the ideal de Sitter spacetime approximation to cosmic inflation. An exactly constant vacuum component cannot end its phase unaided; a phase transition or reheating mechanism is required for the assumed change to S2.

In S2, $w=1/3$ describes an isotropic bath of ultrarelativistic particles, such as photons, with negligible rest mass relative to their energy. The number per physical volume falls as $R^{-3}$, while the energy per particle redshifts as $R^{-1}$, so $\rho \propto R^{-4}$. This radiation-dominated universe has $R\propto (t-t_{\ast }{)}^{1/2}$ on its expanding branch.

In S3, $w=0$ describes pressureless matter, or relativistic dust, with negligible random kinetic energy. Conserved particle number and approximately fixed rest energy give $\rho \propto R^{-3}$. A flat universe containing only this component is the Einstein-de Sitter universe, with $R\propto (t-t_{\ast }{)}^{2/3}$. The constants $t_{\ast }$ are integration offsets in the individual phases; they need not represent one common start time when phases are matched.

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 relativistic dust, $p=0$, the relativistic Euler equation gives $\rho a_b=0$. The nowhere-vanishing density therefore makes the four-acceleration zero: $u$ is a timelike geodesic vector field. Let $\alpha =u^{♭}$ be the differential one-form obtained by the musical isomorphism. Its exterior derivative satisfiesHere ${\iota }_u$ is the interior product. By hypersurface orthogonality and the Frobenius theorem, $\alpha \wedge d\alpha =0$. Taking the interior product and using $\alpha (u)=-1$ givesThus $\boxed{d{u}^{♭}=0}$. By the Poincaré lemma, locally $\alpha =-dt$, and then $u(t)=-\alpha (u)=1$. Choose spatial coordinate functions $x^i$ on one level set of $t$ and extend them constantly along the integral curves of a vector field $u$, using its smooth flow. This constructs $u={\partial }_t$. Since $dt({\partial }_i)=0$,The metric on the orthogonal complement of a timelike vector is positive-definite, so the synchronous coordinates for irrotational dust haveThis construction is local; crossing geodesics can prevent these synchronous coordinates from covering a larger region.

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.

For geodesic motion of uncharged dust, use the pressureless stress-energy form, $T^{\mu \nu }=\rho u^{\mu }{u}^{\nu }$ and $u_{\nu }{u}^{\nu }=c^2$. stress-energy conservation expands toContracting with $u_{\nu }$ kills the second term, since $u_{\nu }{\nabla }_{\mu }{u}^{\nu }=\frac{1}{2}{\nabla }_{\mu }(c^2)=0$. Hence ${\nabla }_{\mu }(\rho u^{\mu })=0$, and wherever $\rho >0$,On a flow line $u^{\mu }=d{x}^{\mu }/d\tau$, so this becomes $d^2{x}^{\nu }/d{\tau }^2+{\Gamma }^{\nu }{}_{\alpha \beta }(d{x}^{\alpha }/d\tau )(d{x}^{\beta }/d\tau )=0$. Each relativistic dust element therefore follows a timelike geodesic with proper time as an affine parameter. pressure gradients or external nongravitational forces would add an acceleration term.

Use signature $(+---)$, coordinates $x^0=ct$ and four-velocity $u^{\mu }=d{x}^{\mu }/d\tau$ normalized by $u_{\mu }{u}^{\mu }=c^2$. The derivative is the metric-compatible Levi-Civita connection. Fix the curvature sign by $[{\nabla }_{\alpha },{\nabla }_{\beta }]V^{\mu }=R^{\mu }{}_{\nu \alpha \beta }{V}^{\nu }$. Then the required contractions areWith $G^{\mu \nu }=R^{\mu \nu }-g^{\mu \nu }R/2$, the contracted Bianchi identity gives ${\nabla }_{\mu }{G}^{\mu \nu }=0$. The Einstein field equations $G^{\mu \nu }+\Lambda g^{\mu \nu }=(8\pi G/c^4)T^{\mu \nu }$, with constant $\Lambda$, therefore require covariant conservation of stress-energy. This is a consistency relation between geometry and the total matter/field source. It does not mean ${\partial }_{\mu }{T}^{\mu \nu }=0$ in arbitrary coordinates, nor guarantee a global conserved energy in an arbitrary curved spacetime. The geometric identity alone, without the field equations or matter dynamics, is not an equation of motion for a freely specified material source.

For relativistic dust, $\rho$ is proper rest-mass density, measured in the fluid rest frame, and $u^{\mu }$ is its timelike four-velocity with the $c^2$ normalization above. In a local Cartesian inertial frame $u^{\mu }=\gamma (c,\mathbf{v})$, soAccordingly $T^{0i}/c$ is momentum density and $c{T}^{0i}$ is energy flux in direction $i$. The equality of momentum-density and energy-flux components after the appropriate factors of $c$ reflects the symmetry of the stress-energy tensor. In the comoving frame the spatial stress is zero, as required for pressureless matter.

Expand the dust conservation equation by the product rule:Contract with $u_{\nu }$. metric compatibility and constant normalization give $u_{\nu }{\nabla }_{\mu }{u}^{\nu }={\nabla }_{\mu }(u_{\nu }{u}^{\nu })/2=0$, leaving $c^2{\nabla }_{\mu }(\rho u^{\mu })=0$. Substitute this rest-mass continuity law back and divide by nonzero $\rho$:This proves geodesic motion of uncharged dust along each fluid flow line. pressure forces or another exchange of momentum would change the argument; at zero density there is no dust worldline to constrain.

The electromagnetic field tensor is antisymmetric and has six independent components. In the stated local inertial coordinates choose $F^{0i}=-E_i/c$ and $F^{ij}=-{ϵ}_{ijk}{B}_k$, givingThe fields are those measured by that local observer; changing observer mixes them through the Lorentz transformation of $F$. With ${ϵ}^{0123}=+1$ and ${\tilde{F}}^{\mu \nu }={ϵ}^{\mu \nu \alpha \beta }{F}_{\alpha \beta }/2$, two independent quadratic contractions areThe second is a pseudoscalar: it is invariant under proper Lorentz transformations and changes sign under an orientation reversal. Opposite choices for the orientation tensor change its written sign. A nonzero null electromagnetic field can have both contractions zero, so the two numbers do not determine whether the tensor itself vanishes.

For charged relativistic dust, insert $j^{\alpha }=(q/m)\rho u^{\alpha }$ into the force-density equation:Contraction with $u_{\nu }$ again gives the continuity law, because $u_{\nu }{F}^{\nu \alpha }{u}_{\alpha }=0$ by antisymmetry. After removing the continuity term,In a local inertial frame the spatial components reduce to $d(\gamma m\mathbf{v})/dt=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$ and the temporal component gives $d(\gamma m{c}^2)/dt=q\mathbf{E}\cdot \mathbf{v}$. This is the relativistic Lorentz force, with gravity represented by the connection. The force is orthogonal to $u$ and preserves the rest-mass normalization. Matter alone now exchanges energy and momentum with the electromagnetic field; adding the field tensor's stress-energy restores zero divergence of the combined source.

Use signature $(+---)$ and local inertial coordinates $x^0=ct$. Write $u^{\mu }=\gamma (c,\mathbf{v})$ and $w=\rho c^2+p$, where $\rho$ is rest mass density. The perfect fluid in general relativity has componentsHere $T^{00}$ is energy density, $c{T}^{i0}$ is energy flux, $T^{0i}/c$ is momentum density, and $T^{ij}$ is the flux of the $j$th momentum component through a surface normal to $i$. In the local rest frame these reduce to energy density $\rho c^2$, zero energy and momentum flow, and isotropic pressure $p$. The equation ${\nabla }_{\mu }{T}^{\mu \nu }=0$ expresses local stress-energy conservation. Its temporal and spatial components give energy and momentum balance; in flat coordinates, for example, ${\partial }_t{T}^{00}+{\partial }_i(c{T}^{i0})=0$.

For relativistic dust, $T^{\mu \nu }=\rho u^{\mu }{u}^{\nu }$. Contracting its divergence with $u_{\nu }$, and using metric compatibility together with $u_{\nu }{u}^{\nu }=c^2$, givesConsequently ${\nabla }_{\mu }(\rho u^{\mu })=0$. Substituting this continuity equation back into the divergence leaves $\rho u^{\mu }{\nabla }_{\mu }{u}^{\nu }=0$. Wherever matter is present,so the individual particles follow geodesics.

The electromagnetic four-force law $f^{\mu }=q{F}^{\mu \nu }{u}_{\nu }$ must preserve the particle's rest mass. Thus $u_{\mu }{f}^{\mu }=0$, and requiring this for arbitrary timelike four-velocities makes the electromagnetic field tensor antisymmetric. Its six independent components comprise the electric field and magnetic field. With the stated signature and $x^0=ct$, one consistent convention isIt gives the spatial force $d\mathbf{p}/dt=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$ and energy transfer $dE/dt=q\mathbf{E}\cdot \mathbf{v}$. Under a Lorentz transformation, $F^{\mu \nu }$ transforms as a rank-two contravariant tensor; its electric and magnetic components consequently mix. A four-potential supplies $F_{\mu \nu }={\nabla }_{\mu }{A}_{\nu }-{\nabla }_{\nu }{A}_{\mu }$, unchanged by $A_{\mu }\mapsto A_{\mu }+{\nabla }_{\mu }\chi$. The Maxwell equations areThese combine the source equations and the homogeneous equations into covariant tensor statements; $j^{\mu }$ is the four-current.

For charged relativistic dust, the matter tensor exchanges four-momentum with the electromagnetic field: ${\nabla }_{\mu }{T}_{\mathrm{m}}^{\mu \nu }=F^{\nu \alpha }{j}_{\alpha }$, while their total stress-energy tensor is conserved. Identical particles of rest mass $m$ and electric charge $q$ have $j^{\alpha }=(q/m)\rho u^{\alpha }$. Contraction with $u_{\nu }$ again gives mass continuity because $u_{\nu }{F}^{\nu \alpha }{u}_{\alpha }=0$. Removing that term givesUsing zero divergence for the matter tensor alone in this last step would omit the electromagnetic Lorentz force.

A perfect fluid in general relativity has isotropic rest-frame pressure and no rest-frame heat flux or anisotropic stress. In units $c=1$ and signature $(-,+,+,+)$, let $u^a{u}_a=-1$, let $\rho$ be rest-frame energy density and $p$ rest-frame pressure. Its stress-energy tensor isThe energy measured with the fluid is $T_{ab}{u}^a{u}^b=\rho$. The orthogonal projection $h^a{}_b={\delta }^a{}_b+u^a{u}_b$ isolates spatial directions and gives isotropic projected stress $p{h}_{ab}$. A fluid moving relative to another observer can have nonzero momentum density even though its own rest-frame stress is isotropic.

Projecting stress-energy conservation along and perpendicular to the four-velocity givesThese are the comoving energy balance and relativistic Euler equation. The inertial factor is enthalpy density $\rho +p$. An equation of state is needed to close the evolution; relativistic dust sets $p=0$, while an isotropic radiation fluid has $p=\rho /3$. Perfect-fluid behavior does not require irrotational motion: the potential-flow action below describes a restricted subset.

Relativistic dust is a continuum of noninteracting massive particles with one local four-velocity $u^{\mu }$ and negligible pressure. With signature $(+---)$, $u^{\mu }{u}_{\mu }=c^2$ and rest mass density $\rho$, its stress-energy tensor has the displayed form. Contracting stress-energy conservation with $u_{\nu }$ and using metric compatibility gives $c^2{\nabla }_{\mu }(\rho u^{\mu })=0$. Substituting back yields $u^{\mu }{\nabla }_{\mu }{u}^{\nu }=0$ wherever $\rho >0$: each particle follows a geodesic. The cosmological pressureless matter model is a homogeneous application.

For relativistic dust with nowhere-zero density, stress-energy conservation implies zero four-acceleration. If its normalized four-velocity $u$ has hypersurface orthogonality, then $u^{♭}\wedge d{u}^{♭}=0$ and ${\iota }_{u}d{u}^{♭}=0$. The interior product of the first identity gives $d{u}^{♭}=0$. The Poincaré lemma supplies $u^{♭}=-dt$ locally; spatial coordinate functions constant along the smooth flow of $u$ give $u={\partial }_t$, $g_{tt}=-1$ and $g_{ti}=0$. The resulting synchronous coordinates describe the dust until the local coordinate construction fails.