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For a homogeneous cosmic fluid, the physical volume bounded by fixed comoving coordinates obeys $V\propto a^3$. The cosmological continuity equation therefore gives $d(\rho c^{2}V)=-PdV$. This is the first law of thermodynamics with conserved entropy and no additional chemical work. The internal energy of that volume need not be constant: a component with negative pressure gains internal energy as the universe expands. Local stress-energy conservation does not imply a global conserved energy for an arbitrary expanding spacetime.

In natural units, conservation of the stress-energy tensor for a homogeneous perfect fluid in an FLRW metric gives the displayed equation for its energy density and pressure. It is equivalent to $d(\rho a^3)=-Pd(a^3)$. For a constant equation-of-state parameter $w=P/\rho$, separation gives $\rho \propto a^{-3(1+w)}$. If $\rho$ instead denotes mass density, the pressure term is $P/c^2$.

This is local stress-energy conservation; it does not in general define a globally conserved total energy for an expanding universe. The adiabatic work in a comoving volume makes this distinction explicit.

For scalar perturbations in cosmic time with $d{s}^2=-d{t}^2+a^2({\delta }_{ij}+h_{ij})d{x}^{i}d{x}^j$, define the covariant fluid velocity potential $u$ by $\delta u_i={\partial }_{i}u$. The linear momentum density is $T^0{}_i=(\bar{\rho }+\bar{P}){\partial }_{i}u$. Normalize scalar anisotropic stress bywhere derivatives inside these parentheses are Cartesian comoving derivatives. Then ${\partial }_j{\pi }^j{}_i=2{\partial }_i{\nabla }^2\Pi /(3a^2)$. Spatial stress-energy conservation gives, for nonzero scalar Fourier modes,Writing ${\pi }^S=2\Pi /(3a^2)$ reproduces the form with ${\nabla }^2{\pi }^S$; the coefficient depends on that stress normalization. The time derivative acts on the entire momentum-density product, not only on $u$.

For pressureless matter with zero scalar stress, background $\dot{\bar{\rho }}+3H\bar{\rho }=0$ reduces the equation to $\dot{u}=0$. The physical peculiar velocity is $\mathbf{v}=\nabla u/a$ and therefore decays as $a^{-1}$. This checks the difference between covariant velocity potential and physical velocity and the expansion factor in the conservative equation.

Simultaneously varying a metric tensor and a scalar field by the Lie derivative along a compactly supported vector field leaves a covariant matter action invariant. Integration by parts gives the displayed Noether identity. It implies stress-energy conservation when the scalar equation holds without needing to impose the metric field equation.

Karim A. Malik developed the treatment of gauge-invariant cosmological curvature with David Wands, David H. Lyth and Andrew R. Liddle. Their 2000 analysis derived large-scale evolution from stress-energy conservation, exposing how a non-adiabatic pressure perturbation can change the uniform-density curvature perturbation.^[ref]

The Linearized Einstein equations retain first-order terms in a metric perturbation $g_{\mu \nu }={\eta }_{\mu \nu }+h_{\mu \nu }$ about the Minkowski metric. Use metric signature $(-,+,+,+)$, coordinates $x^0=ct$, andIn Lorenz gauge in linearized gravity, ${\partial }^{\mu }{\bar{h}}_{\mu \nu }=0$, the first-order Einstein tensor is $G_{\mu \nu }^{(1)}=-\square {\bar{h}}_{\mu \nu }/2$. HenceThe source must satisfy background stress-energy conservation, ${\partial }^{\mu }{T}_{\mu \nu }=0$, since taking the divergence of the left-hand side gives zero. In units $G=c=1$ this becomes $\square {\bar{h}}_{\mu \nu }=-16\pi T_{\mu \nu }$.

The equation is a sourced wave equation, not just a static Poisson equation. A retarded solution additionally needs a radiation boundary condition for linearized gravity; homogeneous gravitational waves can otherwise be added. The expansion is controlled only where the neglected higher-order metric perturbations remain small.

For constant $P=w\rho$, the perfect fluid stress tensor and inverse perturbed spatial metric give $T^{ij}=a^{-2}w\bar{\rho }[(1+\delta ){\delta }^{ij}-h^{ij}]$ at first order. Normalizing the velocity makes $u^0=a^{-1}$ at this order. Stress-energy conservation yields the synchronous perfect-fluid density equation and Cosmological Euler equation in synchronous gauge; the pressure projection includes $v_i{\bar{P}}^{\prime }$, which changes the velocity damping from $\mathcal{H}$ to $(1-3w)\mathcal{H}$.

// Target: cosmology.bigb

Assume a metric theory in four dimensions, a Levi-Civita connection, covariant stress-energy conservation and field equations linear in second metric derivatives. The contracted Bianchi identity selects $G_{ab}=R_{ab}-g_{ab}R/2$. Its trace-reversed equation is $R_{ab}=\kappa (T_{ab}-g_{ab}T/2)+\Lambda g_{ab}$. In the convention $R_{00}\simeq -{\nabla }^2\phi$ and signature $(+---)$, nonrelativistic matter gives $R_{00}\simeq \kappa \rho /2$. The Poisson equation ${\nabla }^2\phi =4\pi G\rho$ fixes $\kappa =-8\pi G$. Reversing the curvature convention reverses this displayed coupling sign without changing the physical theory. The strong equivalence principle alone does not establish uniqueness without the stated additional assumptions.

For a matter-dominated fluid with constant $c_s^2=w\ll 1$, combine stress-energy conservation in synchronous gauge with the scalar trace Einstein equation. Discard pressure corrections to the background gravitational and velocity-damping terms while retaining the potentially large pressure-gradient term $w{k}^2\delta$. This produces the displayed leading nonrelativistic equation. It is not an exact equation at finite $w$. At negligible pressure the growing and decaying matter-era solutions are ${\tau }^2$ and ${\tau }^{-3}$.

// Target: cosmology.bigb

The number current is $J^a=n{u}^a$, where $n$ is the rest-frame number density and $u^a$ the unit four-velocity. Its covariant continuity equation is ${\nabla }_a{J}^a=0$, equivalently $u^a{\nabla }_{a}n+n{\nabla }_a{u}^a=0$. Integration over a spacetime volume and the divergence theorem equate the particle numbers crossing its boundary. This conservation law is separate from stress-energy conservation.

Define the momentum density $q_i=(\bar{\rho }+\bar{P})v_{\perp i}$, lowering its spatial label with ${\delta }_{ij}$, so the mixed stress-energy tensor obeys $T^i{}_0=q_i$ and $T^0{}_i=-q_i$. All products of $J_{ij}$ or its derivatives with $q_i$ are second order and must be dropped in linear cosmological perturbation theory.

For $\nu =0$, the three terms in stress-energy conservation areHere the trace $J_{ii}^{\prime }$ vanishes, ${\Gamma }_{00}^i={\Gamma }_{i0}^0=0$, and ${\partial }_i{q}_i=(\bar{\rho }+\bar{P}){\partial }_i{v}_{\perp }^i=0$. ThusFor $\nu =i$, spatial pressure gradients vanish because this is a purely vector cosmological perturbation. The first two terms areIn the remaining contraction, the two background-connection times momentum density terms cancel:The other terms are ${\Gamma }_{0i}^0\bar{\rho }=0$ and $-\bar{P}{\Gamma }_{ji}^j=0$. ConsequentlyExpanding this conservation law and using the background cosmological continuity equation gives the equivalent peculiar velocity equation, when $\bar{\rho }+\bar{P}\ne 0$,

Use metric signature $(-,+,+,+)$ and a bulk four-velocity normalized by $u^{\alpha }{u}_{\alpha }=-c^2$. A perfect fluid in general relativity is isotropic in its local rest frame, with no rest-frame energy flux or anisotropic stress. In an orthonormal frame in spacetime comoving with that fluid,Here $\rho$ is rest-frame energy density divided by $c^2$, including random kinetic energy, and $p$ is isotropic pressure. For a particle gas, $p=\frac{1}{3}\int \mathbf{p}\cdot \mathbf{v}dn$ and $\rho c^2=\int Edn$, with $dn$ the rest-frame particle distribution. Isotropy makes the off-diagonal spatial stresses and net energy flux vanish.

Introduce the spatial projector of a timelike congruenceThe only isotropic decomposition with the preceding components isSince both sides are tensors, equality in the comoving orthonormal frame in spacetime establishesCoordinate-independent definitions of the two scalar fields areThus $\rho$ is not, in general, only the sum of the particle rest masses per volume.

The microscopic averaging notation needs a normalization qualification. The normalization of an averaged four-velocity matters because a literal average of normalized microscopic four-velocities is not normally a normalized bulk four-velocity. For example, six equally weighted streams with the same ordinary speed $v$, directed along the positive and negative coordinate axes, have an isotropic velocity distribution andThe physical rest-frame bulk four-velocity is $(c,0,0,0)$, not this unnormalized average. Even taking the source's schematic kinetic expression literally with prefactor ${\rho }_s$ givesSubstituting $u=⟨v⟩$ and the same ${\rho }_s$ into the claimed fluid expression would instead increase $T^{00}$ by $p({\gamma }^2-1)$. The standard result therefore requires a normalized bulk velocity and a carefully defined rest-frame total energy density; it does not follow with both microscopic identifications interpreted literally. More generally the bulk velocity can be defined as the future-directed timelike energy eigenvector of the stress-energy tensor.

The equation ${\nabla }_{\alpha }{T}^{\alpha \beta }=0$ is local stress-energy conservation. In a weak gravitational field and the nonrelativistic limit its leading energy component gives mass conservation,while its spatial components give momentum conservation, or the inviscid Euler equationsThe next energy order gives conservation of internal plus kinetic energy, including pressure work and gravitational work. These are local conservation laws, not an assertion that arbitrary curved spacetimes possess a global conserved total energy.

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 nonrelativistic particles the energy density is dominated by rest energy, while their isotropic kinetic pressure is approximately $p_m=nm⟨v^2⟩/3$. ThereforeThe matter pressure is negligible in this approximation, including in the gravitational combination ${\rho }_m+3p_m/c^2$. For isotropic radiation in cosmology, $E=pc$ implies one third of the energy density in each diagonal spatial stress:

Because the two fluids do not exchange energy or momentum, they are separately conserved cosmological fluids: ${\nabla }_{\alpha }{T}_i^{\alpha \beta }=0$ for each, rather than merely conservation of their sum. In a comoving volume proportional to $a^3$, the first law of thermodynamics givesEquivalently this follows by projecting the individual stress-energy conservation equation along the common comoving four-velocity. Differentiating and dividing by ${\rho }_i{c}^2{a}^3$ yieldsFor constant equation-of-state parameter $w_i$,The extra power for radiation in cosmology represents the cosmological redshift of photon energies in addition to number dilution. Differentiating the total Friedmann equation gives only the total continuity equation; the separate equations also use the stated absence of interactions between components.

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.

For a contravariant vector field $A^{\lambda }$, the covariant derivative isThe Levi-Civita connection is the unique torsion-free connection satisfying metric compatibility. Its Christoffel symbols in a coordinate basis areThere is generally no symmetry interchanging the upper index with a lower one. The Christoffel symbols themselves are not the components of a tensor.

An ordinary partial derivative of the components of a vector field includes the apparent change caused by the varying coordinate basis. The connection term corrects for that change, so ${\nabla }_{\nu }{A}^{\lambda }$ obeys the tensor component transformation law. It gives coordinate-independent ways to express parallel transport, geodesics and stress-energy conservation in curved spacetime.

Set $H=\dot{a}/a$, the Hubble parameter. The first supplied component of the Einstein field equations gives the Friedmann acceleration equationSubstitution into the second supplied equation givesTherefore the first Friedmann equation isDifferentiate this equation with respect to cosmic time:Insert the Friedmann acceleration equation on the left and cancel $8\pi Ga/3$, which is nonzero for $a>0$. This givesThis is the cosmological continuity equation, expressing stress-energy conservation for the perfect fluid in general relativity. No division by $\dot{a}$ was made, so the derivation remains valid at a cosmological bounce or an instant of maximum expansion.

Write $D=u^a{\nabla }_a$, $\theta ={\nabla }_a{u}^a$ for the expansion scalar, $a_b=u^a{\nabla }_a{u}_b$ for the four-acceleration, and $P_b{}^c={\delta }_b{}^c+u_b{u}^c$ for the spatial projector of a timelike congruence. The normalization of the four-velocity gives $u^b{a}_b=0$ and $u^b{\nabla }_a{u}_b=0$. The product rule and metric compatibility give the covariant derivative of the perfect fluid in general relativity stress-energy tensor:Its tensor contraction with the four-velocity isand its projection with the spatial projector of a timelike congruence isThus stress-energy conservation implies the relativistic perfect-fluid energy equation and the relativistic Euler equation:Conversely, the two displayed equations set both projections of $F_b$ to zero. Since $F_b=P_b{}^c{F}_c-u_b{u}^c{F}_c$, they imply $F_b=0$. This establishes equivalence, including points where $\rho +p=0$, without dividing by that quantity.