Local Conservation of Energy-Momentum

In flat spacetime the accounting of energy and momentum is simple and complete: the stress-energy tensor $T^{\mu\nu}$ obeys $\partial_\mu T^{\mu\nu}=0$ , which says that whatever flows out through the walls of a box equals the drop in what the box contains. Integrate over a slab of space and Gauss's theorem hands you four numbers — total energy and total momentum — that never change. This is the cleanest conservation law in physics, and it is exactly the thing that curvature seems to break. The moment gravity is present there is no flat box, no globally fixed frame to measure "the momentum inside," and the naive integral loses its meaning. General relativity's answer is to promote the partial derivative to the covariant one, $\partial_\mu\to\nabla_\mu$ , giving the single equation this article is about: $\nabla_\mu T^{\mu\nu}=0$ . What looks like a one-symbol change is a deep restructuring. The covariant divergence carries the connection $\Gamma$ , and those extra terms are the bookkeeping of energy and momentum traded between matter and the gravitational field — redshift, blueshift, tidal work. Even more strikingly, the law is not an extra postulate bolted onto Einstein's equations. The Einstein tensor satisfies the contracted Bianchi identity $\nabla_\mu G^{\mu\nu}\equiv0$ as pure geometry, so the field equations $G^{\mu\nu}=8\pi G\,T^{\mu\nu}$ are consistent only if the source obeys $\nabla_\mu T^{\mu\nu}=0$ . Geometry forces matter to conserve. And because the same equation, applied to dust, collapses into the geodesic equation, conservation and the equations of motion of matter turn out to be one statement.

Key takeaways

The local law $\nabla_\mu T^{\mu\nu}=0$ is the curved-spacetime statement that energy-momentum is conserved at every point; it is the covariant promotion of the flat-space $\partial_\mu T^{\mu\nu}=0$ .
It is not an independent axiom of general relativity. The contracted Bianchi identity forces $\nabla_\mu G^{\mu\nu}\equiv0$ , so the field equations self-consistently require $\nabla_\mu T^{\mu\nu}=0$ .
Equivalently it is the Noether identity of diffeomorphism invariance: any generally covariant matter action has a divergence-free stress tensor on-shell.
Because it contains the connection $\Gamma$ , it is generally NOT an integral conservation law — gravity exchanges energy with matter, and no covariant local energy density of the field itself exists. A true conserved charge needs a Killing vector.
Applied to a perfect fluid it yields relativistic Euler and continuity; applied to dust it becomes the geodesic equation — the motion of matter is contained in the conservation law.
Key idea: $\nabla_\mu T^{\mu\nu}=0$ is not imposed on general relativity — it is forced by it through $\nabla_\mu G^{\mu\nu}\equiv0$ , and the same equation that conserves energy-momentum also supplies the equations of motion for matter.

The core insight

$\nabla_\mu T^{\mu\nu}=0$ is not imposed on general relativity — it is forced by it, through the contracted Bianchi identity $\nabla_\mu G^{\mu\nu}\equiv0$ . The covariant divergence simultaneously expresses local conservation of energy-momentum and supplies the equations of motion for matter.

Where this lives today

The conservation law is a daily working tool rather than a museum piece. In numerical relativity — the codes that produce the gravitational waveforms LIGO and Virgo match against binary black-hole and neutron-star mergers — the matter sector must independently satisfy $\nabla_\mu T^{\mu\nu}=0$ , and the Bianchi identities guarantee that a consistently evolved source keeps the Einstein constraints satisfied; monitoring the residual of the divergence is a standard error diagnostic in codes like the Einstein Toolkit. In cosmology the continuity equation $\dot\rho+3H(\rho+p)=0$ is the backbone of every calculation of the thermal history, from Big Bang nucleosynthesis to the cosmic microwave background, and precision measurements of the equation-of-state parameter $w$ of dark energy (DES, Euclid, DESI) are literally measurements of how $\rho$ dilutes with $a$ . In relativistic astrophysics the TOV equation and its rotating, magnetised generalisations set the maximum mass and radius of neutron stars, now pinned down by NICER's X-ray timing and by the tidal deformability extracted from GW170817. And the deepest open thread is still Noether's: the absence of a local gravitational energy density motivates ongoing work on quasilocal energy definitions (Brown-York, Wang-Yau) and underlies the information-theoretic puzzles of black-hole thermodynamics.

The physical picture

Draw a small coordinate box in spacetime and watch energy flow through its walls. In flat space the ledger balances exactly: the net energy-momentum flux out of the walls equals the rate at which the contents drop, and that is the whole content of $\partial_\mu T^{\mu\nu}=0$ . Switch on gravity and a new column appears in the ledger. The connection terms $\Gamma^\nu_{\mu\lambda}T^{\mu\lambda}$ act as a source density — the gravitational force per unit volume — because the local inertial frames themselves are tilting from one point to the next. A photon climbing out of a potential well loses energy to that column (redshift); an infalling fluid element gains it. Nothing is truly lost, but the missing piece has gone into the geometry, which carries no local energy tag of its own.

Figure

Energy-momentum bookkeeping in a small region. In flat spacetime the net flux of T through the walls equals the rate of decrease inside, so ∂_μT^{μν} = 0. In curved spacetime the connection adds a gravitational source density and the balance reads ∂_μ(√−g T^{μν}) = −√−g Γ^ν_{μλ} T^{μλ}

The reason this is not a defect but the heart of the theory is a chain of three ideas that meet in one equation. First, the identity is automatic: $\nabla_\mu G^{\mu\nu}\equiv0$ holds for every metric whatsoever, so once you write $G^{\mu\nu}=8\pi G\,T^{\mu\nu}$ , the divergence of the left side is zero by geometry and the divergence of the right side has no choice but to follow. Einstein gravity polices its own sources. Second, the same conclusion drops out of Noether's theorem the instant you demand that the matter action not care about coordinate labels — conservation is the price of general covariance. Third, and most usefully, the law has teeth: feed it a specific $T^{\mu\nu}$ and it dictates how that matter moves. The second figure shows the most consequential worked case, cosmology, where $\nabla_\mu T^{\mu\nu}=0$ alone fixes how each species of matter dilutes as the universe expands.

Figure

The conservation law fixes how energy density dilutes with the cosmic scale factor a. Radiation p = ρ/3 falls as ρ ∝ a⁻⁴, matter p = 0 as ρ ∝ a⁻³, a cosmological constant p = −ρ stays flat. Matter overtakes radiation at 1 + z ≈ 3400, the epoch of matter-radiation equality

The core equations

The whole subject is one covariant equation read in several ways. Start with the statement itself: the covariant divergence of the stress-energy tensor vanishes. The free index $\nu$ carries the meaning — $\nu=0$ is conservation of energy, the three spatial values are conservation of momentum.

\nabla_\mu T^{\mu\nu}=0 \qquad (1)

where: $\nabla_\mu$ — metric-compatible (Levi-Civita) covariant derivative; $T^{\mu\nu}$ — symmetric stress-energy tensor of matter and all non-gravitational fields; $\nu$ — free index, $\nu=0$ energy, $\nu=i$ momentum. Read it as: the outflow of energy-momentum from a point, measured covariantly, is zero. Expanding the covariant derivative of a rank-two tensor exposes the connection. Two Christoffel terms appear, and their physical roles are utterly different: one repackages into a volume-weighted ordinary divergence, the other is an irreducible source.

\nabla_\mu T^{\mu\nu}=\partial_\mu T^{\mu\nu}+\Gamma^\mu_{\mu\lambda}T^{\lambda\nu}+\Gamma^\nu_{\mu\lambda}T^{\mu\lambda} \qquad (2)

where: $\Gamma^\mu_{\mu\lambda}$ — contracted Christoffel symbol, equal to $\partial_\lambda\ln\sqrt{-g}$ ; $\Gamma^\nu_{\mu\lambda}$ — full Christoffel symbol; $g$ — determinant of $g_{\mu\nu}$ . The first extra term traces the volume element; the second is genuine curvature. Using $\Gamma^\mu_{\mu\lambda}=\partial_\lambda\ln\sqrt{-g}$ collapses the first two terms into a single ordinary divergence of the tensor density $\sqrt{-g}\,T^{\mu\nu}$ . What is left over is the gravitational force density — the rate at which the field feeds momentum to matter.

\frac{1}{\sqrt{-g}}\,\partial_\mu\!\big(\sqrt{-g}\,T^{\mu\nu}\big)=-\,\Gamma^\nu_{\mu\lambda}T^{\mu\lambda} \qquad (3)

where: left side — apparent (coordinate) divergence of the weighted tensor; right side — gravitational source. Only when $\Gamma^\nu_{\mu\lambda}$ can be set to zero everywhere (flat space, Cartesian coordinates) does this become a true continuity equation. The reason equation (1) must hold is not the equivalence principle alone but the structure of the field equations. The Einstein tensor is divergence-free as an identity — pure differential geometry, true for any metric.

\nabla_\mu G^{\mu\nu}\equiv 0,\qquad G^{\mu\nu}=R^{\mu\nu}-\tfrac12 g^{\mu\nu}R \qquad (4)

where: $G^{\mu\nu}$ — Einstein tensor; $R^{\mu\nu}$ — Ricci tensor; $R$ — Ricci scalar. The identity (4) follows from contracting the second Bianchi identity twice and is independent of any physics. Couple geometry to matter through the field equations, and the identity (4) becomes a constraint on the source. A cosmological constant does not spoil it because the metric is covariantly constant.

G^{\mu\nu}+\Lambda g^{\mu\nu}=8\pi G\,T^{\mu\nu} \qquad (5)

where: $\Lambda$ — cosmological constant; $G$ — Newton's constant (units $c=1$ ); $\nabla_\mu g^{\mu\nu}=0$ by metric compatibility. Taking $\nabla_\mu$ of both sides and using (4) yields (1) with no further assumption. For the most important source — a perfect fluid — the tensor is built from density, pressure, and four-velocity, and conservation splits cleanly into a momentum (Euler) equation and an energy equation by projecting along and across the flow.

T^{\mu\nu}=(\rho+p)\,u^\mu u^\nu+p\,g^{\mu\nu},\qquad (\rho+p)\,u^\mu\nabla_\mu u^\nu=-\,h^{\mu\nu}\nabla_\mu p \qquad (6)

where: $\rho$ — rest-frame energy density; $p$ — isotropic pressure; $u^\mu$ — four-velocity, $u_\mu u^\mu=-1$ ; $h^{\mu\nu}=g^{\mu\nu}+u^\mu u^\nu$ — projector orthogonal to $u^\mu$ . The inertial coefficient is the enthalpy density $\rho+p$ , not $\rho$ — pressure adds to inertia. Finally, a genuine integral conserved charge does exist whenever the spacetime has a symmetry. Contract $T^{\mu\nu}$ with a Killing vector and the covariant divergence becomes an ordinary one, so Gauss's theorem applies again.

J^\mu=T^{\mu\nu}\xi_\nu,\qquad \nabla_\mu J^\mu=0 \qquad (7)

where: $\xi_\nu$ — Killing vector, $\nabla_{(\mu}\xi_{\nu)}=0$ . Then $Q=\int_\Sigma J^\mu\,\mathrm{d}\Sigma_\mu$ is conserved: a timelike $\xi$ gives total energy, a rotational $\xi$ gives angular momentum.

Derivation AConservation forced by the contracted Bianchi identity

The aim is to show that $\nabla_\mu T^{\mu\nu}=0$ is not an assumption we are free to make or drop, but a theorem that the Einstein equations prove for us. The engine is the contracted Bianchi identity, a statement of pure geometry that holds for every metric before any physics is specified; once the field equations equate the Einstein tensor to the matter tensor, the geometric identity descends onto the source and pins its divergence to zero. We take the divergence of the field equations term by term and let two standard identities do the work.

Step 1. Start from the field equations in the form $G^{\mu\nu}+\Lambda g^{\mu\nu}=8\pi G\,T^{\mu\nu}$ and apply the covariant divergence $\nabla_\mu$ to every term, giving $\nabla_\mu G^{\mu\nu}+\Lambda\,\nabla_\mu g^{\mu\nu}=8\pi G\,\nabla_\mu T^{\mu\nu}$ . Why: the field equations are a tensor identity, so their covariant divergence is also an identity — nothing is assumed by differentiating both sides.

Step 2. Invoke the contracted Bianchi identity $\nabla_\mu G^{\mu\nu}\equiv0$ , which follows from contracting the second Bianchi identity of the Riemann tensor twice and is true for any metric. Why: this is the geometric fact that makes gravity self-consistent — the left-hand geometry has identically zero divergence, so it can only source a divergence-free object.

Step 3. Use metric compatibility, $\nabla_\mu g^{\mu\nu}=0$ , so the cosmological term drops out regardless of the value of $\Lambda$ . Why: the Levi-Civita connection is defined precisely so the metric is covariantly constant, which is why a constant $\Lambda$ never disturbs conservation.

Step 4. The left side of Step 1 has vanished identically, leaving $8\pi G\,\nabla_\mu T^{\mu\nu}=0$ , hence $\nabla_\mu T^{\mu\nu}=0$ . Why: the conclusion is forced, not posited — a matter distribution whose divergence did not vanish could not sit on the right of Einstein's equations at all.

Step 5. Read the logic in reverse to see how strong it is: the requirement $\nabla_\mu T^{\mu\nu}=0$ is what originally told Einstein that the source must couple to $G^{\mu\nu}$ and not to the Ricci tensor $R^{\mu\nu}$ alone, since $\nabla_\mu R^{\mu\nu}=\tfrac12\nabla^\nu R\neq0$ in general. Why: conservation is not a downstream consequence but a design constraint that selects the Einstein tensor uniquely among the natural curvature combinations.

Go deeper · Derivation B · 2 minThe Noether identity from diffeomorphism invariance

Derivation BThe Noether identity from diffeomorphism invariance

The second route to $\nabla_\mu T^{\mu\nu}=0$ never mentions the field equations at all; it follows from the single demand that the matter action be blind to how we label spacetime points. This is Noether's second theorem in action: a continuous local symmetry, here the freedom to relabel coordinates by an arbitrary infinitesimal diffeomorphism, produces an identity satisfied by the matter equations of motion. The stress tensor enters as the response of the action to a change of metric, and covariance turns that response into a conservation law. We define $T^{\mu\nu}$ variationally, generate a diffeomorphism, and integrate by parts.

Step 1. Define the stress tensor as the metric-variation of the matter action, $T_{\mu\nu}=-\dfrac{2}{\sqrt{-g}}\dfrac{\delta S_\mathrm{m}}{\delta g^{\mu\nu}}$ , where $S_\mathrm{m}[g,\psi]$ depends on the metric and the matter fields $\psi$ . Why: this is the definition that couples matter to gravity in the action, and it is symmetric by construction, matching the source in Einstein's equations.

Step 2. Perform an infinitesimal diffeomorphism generated by a vector $\xi^\mu$ ; the metric changes by minus its Lie derivative, $\delta g^{\mu\nu}=-\big(\nabla^\mu\xi^\nu+\nabla^\nu\xi^\mu\big)$ , and the matter fields by $\delta\psi=\mathcal{L}_\xi\psi$ . Why: a coordinate relabelling drags every field along the flow of $\xi^\mu$ , so its effect on each field is a Lie derivative — this is what "generally covariant" means operationally.

Step 3. Because the action is a scalar, its total change under this relabelling is identically zero: $\delta S_\mathrm{m}=\displaystyle\int\!\Big(\frac{\delta S_\mathrm{m}}{\delta g^{\mu\nu}}\delta g^{\mu\nu}+\frac{\delta S_\mathrm{m}}{\delta\psi}\delta\psi\Big)\mathrm{d}^4x=0$ . Why: invariance under coordinate change is not a dynamical statement but a property of writing the action covariantly, so it holds off-shell for the metric piece.

Step 4. Impose the matter equations of motion, $\delta S_\mathrm{m}/\delta\psi=0$ , killing the second term. The first term becomes $\displaystyle\int\sqrt{-g}\,T_{\mu\nu}\,\nabla^\mu\xi^\nu\,\mathrm{d}^4x=0$ after inserting the definition of $T_{\mu\nu}$ and using its symmetry. Why: "on-shell" for matter means the field equations for $\psi$ hold, which is exactly when $T_{\mu\nu}$ is the physical stress tensor.

Step 5. Integrate by parts and discard the boundary term for compactly supported $\xi^\nu$ : $\displaystyle\int\sqrt{-g}\,\big(\nabla^\mu T_{\mu\nu}\big)\xi^\nu\,\mathrm{d}^4x=0$ . Since $\xi^\nu$ is arbitrary, the integrand vanishes, $\nabla^\mu T_{\mu\nu}=0$ . Why: an integral that vanishes for every test vector forces its coefficient to vanish pointwise — the conservation law is the Noether identity of general covariance.

Go deeper · Derivation C · 1 minFrom conservation to the geodesic equation for dust

Derivation CFrom conservation to the geodesic equation for dust

The most physical face of the law is that it moves matter. For pressureless dust the stress tensor is simply $T^{\mu\nu}=\rho\,u^\mu u^\nu$ , and imposing $\nabla_\mu T^{\mu\nu}=0$ does two jobs at once: it conserves rest mass along the flow and it forces each dust worldline to be a geodesic. We compute the divergence, then project it along the four-velocity to extract the continuity equation, and the leftover piece is precisely the geodesic equation — the equation of motion of free matter, obtained without ever postulating it.

Step 1. Expand the divergence with the product rule, $\nabla_\mu(\rho\,u^\mu u^\nu)=u^\nu\,\nabla_\mu(\rho u^\mu)+\rho\,u^\mu\nabla_\mu u^\nu=0$ . Why: separating the two factors isolates a scalar transport piece and a directional-acceleration piece, which will project onto orthogonal parts of the flow.

Step 2. Contract with $u_\nu$ and use the normalisation $u_\nu u^\nu=-1$ , together with $u_\nu\nabla_\mu u^\nu=\tfrac12\nabla_\mu(u_\nu u^\nu)=0$ . This leaves $-\nabla_\mu(\rho u^\mu)=0$ . Why: the four-velocity has fixed length, so its covariant derivative is orthogonal to it, and the projection along $u$ picks out only the scalar transport term — the continuity of rest mass.

Step 3. Read off the continuity equation $\nabla_\mu(\rho u^\mu)=0$ : the rest-mass current $\rho u^\mu$ is conserved along the flow. Why: this is the covariant version of "matter is neither created nor destroyed," and it is the first half of the conservation law's content.

Step 4. Substitute Step 3 back into Step 1. The first term vanishes, leaving $\rho\,u^\mu\nabla_\mu u^\nu=0$ , and since $\rho\neq0$ we get the geodesic equation $u^\mu\nabla_\mu u^\nu=0$ . Why: with the transport term gone, the only way the full divergence can vanish is for the fluid's four-acceleration to be zero — free dust falls on geodesics.

Step 5. State the conclusion: the equation of motion of free matter is not an extra law of general relativity but a corollary of $\nabla_\mu T^{\mu\nu}=0$ . For a perfect fluid with $p\neq0$ the same projection instead yields the relativistic Euler equation (6), with pressure gradients providing the only non-gravitational force. Why: this is why the theory needs no separate "force law" — telling matter to conserve energy-momentum already tells it how to move.

Limiting cases

The one covariant law specialises, in each regime, to a familiar equation. The table reads off what survives when curvature, pressure, or symmetry is switched on or off.

Case	Result	Meaning
Flat spacetime	$\partial_\mu T^{\mu\nu}=0$	Christoffels vanish in Cartesian coordinates; recovers special-relativistic conservation with globally conserved charges $P^\nu=\int T^{0\nu}\,\mathrm{d}^3x$
Spacetime with a Killing vector $\xi$	$\nabla_\mu(T^{\mu\nu}\xi_\nu)=0$	a genuine conserved current exists; $\int_\Sigma T^{\mu\nu}\xi_\nu\,\mathrm{d}\Sigma_\mu$ is constant — total energy in a static spacetime, angular momentum in an axisymmetric one
Pressureless dust	$u^\mu\nabla_\mu u^\nu=0,\ \ \nabla_\mu(\rho u^\mu)=0$	conservation becomes the geodesic equation plus rest-mass continuity — motion follows from conservation
Perfect fluid, static star	$\dfrac{\mathrm{d}p}{\mathrm{d}r}=-\dfrac{(\rho+p)(m+4\pi r^3 p)}{r(r-2Gm)}$	the $\nu=r$ component gives the Tolman-Oppenheimer-Volkoff equation of hydrostatic equilibrium
FRW cosmology	$\dot\rho+3\dfrac{\dot a}{a}(\rho+p)=0$	fixes the dilution law: $\rho\propto a^{-3}$ (dust), $a^{-4}$ (radiation), const ( $\Lambda$ )
Newtonian limit	$\partial_t\rho+\nabla\!\cdot(\rho\mathbf v)=0,\ \ \rho\,\dot{\mathbf v}=-\nabla p-\rho\nabla\Phi$	continuity plus Euler's equation with the gravitational potential $\Phi$

Common trapTreating ∇_μT^{μν}=0 as a global conservation law

The covariant divergence law does not, in a general spacetime, give you a conserved total energy. The trap is to reach for Gauss's theorem as in flat space, but the connection terms $\Gamma^\nu_{\mu\lambda}T^{\mu\lambda}$ block the conversion of $\nabla_\mu T^{\mu\nu}=0$ into an ordinary divergence $\partial_\mu(\cdots)=0$ , so there is no surface integral to hold fixed. Physically, energy and momentum are continually traded between matter and the gravitational field — a photon redshifts, an infalling cloud speeds up — and gravity keeps no local ledger of its own. There is no covariant local stress-energy tensor for the gravitational field, only non-tensorial pseudotensors (Landau-Lifshitz, Einstein) or asymptotic charges (ADM mass at spatial infinity, Bondi mass at null infinity). A true conserved quantity reappears only with extra structure: a Killing vector $\xi$ gives the conserved current $T^{\mu\nu}\xi_\nu$ , and asymptotic flatness gives a total mass. "Energy of matter plus gravity" is simply not a local concept in general relativity.

Worked examples

1 · How each species dilutes, and when matter overtook radiation (fully worked)

Using only $\nabla_\mu T^{\mu\nu}=0$ in a spatially flat FRW universe, find how the energy density of matter, radiation, and a cosmological constant scale with the scale factor $a(t)$ , and compute the redshift of matter-radiation equality given today's densities $\Omega_{m,0}=0.31$ and $\Omega_{r,0}=9.1\times10^{-5}$ .

Step 1. For a comoving perfect fluid, $u^\mu=(1,0,0,0)$ , the $\nu=0$ component of the conservation law in the FRW metric is the continuity equation $\dot\rho+3H(\rho+p)=0$ , where $H=\dot a/a$ . This is the cosmological content of $\nabla_\mu T^{\mu\nu}=0$ — no field equation was used.

Step 2. Insert a linear equation of state $p=w\rho$ . Then $\dot\rho/\rho=-3(1+w)\dot a/a$ , which integrates to $\rho\propto a^{-3(1+w)}$ . Read off the three cases: dust $w=0$ gives $\rho\propto a^{-3}$ (volume dilution); radiation $w=\tfrac13$ gives $\rho\propto a^{-4}$ (volume dilution plus a redshift factor $1/a$ per quantum); vacuum $w=-1$ gives $\rho=\text{const}$ .

Step 3. Equality of matter and radiation means $\rho_m=\rho_r$ . Writing each in terms of today's values, $\Omega_{m,0}\,a^{-3}=\Omega_{r,0}\,a^{-4}$ , so $a_\mathrm{eq}=\Omega_{r,0}/\Omega_{m,0}$ and $1+z_\mathrm{eq}=1/a_\mathrm{eq}=\Omega_{m,0}/\Omega_{r,0}$ .

Step 4. Put in the numbers: $1+z_\mathrm{eq}=0.31/(9.1\times10^{-5})=\mathbf{3.4\times10^{3}}$ , i.e. $z_\mathrm{eq}\approx3400$ . Before this epoch the universe was radiation-dominated and expanded as $a\propto t^{1/2}$ ; after it, matter-dominated with $a\propto t^{2/3}$ . The entire thermal history hinges on one line of the conservation law — the different exponents $a^{-4}$ vs $a^{-3}$ guarantee radiation always wins at early times.

2 · Pressure has weight and inertia (worked, with numbers)

A relativistic fluid element is a photon gas with pressure $p=\rho/3$ . Show quantitatively that pressure adds both to its inertia and to its gravitating mass, and give the two enhancement factors.

Step 1. Inertia. The relativistic Euler equation (6) reads $(\rho+p)\,a^\nu=-h^{\mu\nu}\nabla_\mu p$ , so the coefficient resisting acceleration is the enthalpy density $\rho+p$ , not $\rho$ . For the photon gas $\rho+p=\rho+\rho/3=\tfrac43\rho$ . A relativistic gas is 33% harder to push than its energy density alone would suggest.

Step 2. Gravity. Trace the field equations to find the source of the Newtonian-limit potential: the active gravitational mass density is $\rho+3p$ (the Tolman-Whittaker result, from $R_{00}\propto\rho+3p$ ). For the photon gas $\rho+3p=\rho+\rho=2\rho$ . Radiation gravitates twice as strongly as an equal energy density of dust.

Step 3. Sanity check the signs and Newtonian recovery. Both enhancements vanish when $p\ll\rho$ : a cold, slow gas has $\rho+p\to\rho$ and $\rho+3p\to\rho$ , so the Newtonian mass density is recovered and pressure drops out, exactly as it must. The two coefficients are genuinely different — $\rho+p$ for inertia, $\rho+3p$ for gravity — because one comes from the spatial projection of $\nabla_\mu T^{\mu\nu}=0$ and the other from the time-time part of the field equations.

Step 4. Consequence. In a static star the same two effects reinforce each other in the TOV equation: pressure appears in the numerator as extra gravitating mass ( $m+4\pi r^3 p$ ) and again through the enthalpy $\rho+p$ , so raising the pressure to support a star makes it harder to support. This runaway is why neutron stars have a maximum mass (the Tolman-Oppenheimer-Volkoff limit, near $2\text{–}3\,M_\odot$ ) instead of supporting arbitrary loads like a Newtonian gas.

Physics of the portrait

Conservation laws had been the bedrock of physics for a century when general relativity threw them into crisis: in Einstein's theory the total energy of a closed universe would not sit still, and no one could say whether energy was conserved at all. The person who dissolved the paradox — by proving exactly what kind of "conservation" a generally covariant theory can and cannot have — was a mathematician the Göttingen faculty had to fight to employ.

Origins · 1 minEmmy Noether and how this was discovered

Emmy Noether

1882–1935 · symmetry and conservation laws in field theory

Emmy Noether was the mathematician who explained why energy conservation looks so strange in general relativity. In 1915 Hilbert and Klein, working through Einstein's new theory at Göttingen, were disturbed to find that its energy conservation law seemed to hold "improperly" — as a mathematical identity rather than a genuine physical balance, unlike every other theory they knew. They asked Noether, then working at Göttingen despite being denied a paid position because she was a woman (Hilbert protested to the faculty senate, "this is a university, not a bathing establishment"), to sort it out. Her 1918 paper Invariante Variationsprobleme contained two theorems that remain foundational. The first ties every continuous global symmetry of an action to a conserved current — the result that gives energy conservation from time-translation symmetry, momentum from space-translation, and so on throughout physics. The second, less famous but decisive here, showed that a theory invariant under a continuous group of local transformations (like general relativity's diffeomorphisms) does not get ordinary conserved charges; instead it gets identities among its field equations. This is precisely why $\nabla_\mu T^{\mu\nu}=0$ is a covariant identity forced by the Bianchi identities rather than a flat-space conservation law, and why the "energy of the gravitational field" refuses to localise. Noether was later driven from Germany by the Nazis in 1933 and taught at Bryn Mawr until her sudden death in 1935; Einstein, in a letter to the New York Times, called her the most significant creative mathematical genius produced since the higher education of women began.

History

1915

— Hilbert and Klein notice that energy conservation in general relativity is "improper"

1917

— Einstein introduces his gravitational stress-energy pseudotensor to track field energy

1918

— Noether's two theorems tie symmetries to conservation laws and identities

1922

— Friedmann derives the expanding-universe models whose continuity equation is the conservation law

1939

— Tolman, Oppenheimer and Volkoff obtain hydrostatic equilibrium from the conservation law

1962

— Arnowitt, Deser and Misner define the ADM mass, recovering total energy asymptotically

Going further

Texts. Misner, Thorne & Wheeler, Gravitation (1973), §5.10, §16.2, Ch. 22 · Wald, General Relativity (1984), §4.3, §E.1–E.2 · Carroll, Spacetime and Geometry (2004), §1.9, §4.1–4.3 · Hartle, Gravity (2003), §22.3 · Schutz, A First Course in General Relativity, 2nd ed. (2009), §4.5–4.7, §7.3. The key results collected in this article are: (1) $\Gamma^\mu_{\mu\lambda}=\partial_\lambda\ln\sqrt{-g}$ and the tensor-density form $\tfrac{1}{\sqrt{-g}}\partial_\mu(\sqrt{-g}\,T^{\mu\nu})=-\Gamma^\nu_{\mu\lambda}T^{\mu\lambda}$ ; (2) the Bianchi-to-conservation argument $\nabla_\mu G^{\mu\nu}\equiv0\Rightarrow\nabla_\mu T^{\mu\nu}=0$ ; (3) the Noether-identity derivation from diffeomorphism invariance; (4) the dust-to-geodesic reduction and relativistic Euler projection; (5) the FRW scalings $\rho\propto a^{-3(1+w)}$ and $1+z_\mathrm{eq}=\Omega_{m,0}/\Omega_{r,0}\approx3400$ ; (6) the two coefficients $\rho+p$ (inertia) and $\rho+3p$ (active mass).

Prerequisites. The Stress-Energy Tensor · The Einstein Tensor · The Einstein Field Equations. Leads to. Perfect Fluid Stress-Energy · The Newtonian Limit · The Energy Conditions · Variational Definition of the Stress-Energy Tensor.

Ask the tutor: Ask me to walk through how projecting $\nabla_\mu T^{\mu\nu}=0$ along and across the four-velocity turns one conservation law into both the continuity equation and the geodesic (or Euler) equation of motion.

Is ∇_μT^{μν}=0 an extra assumption added to Einstein's equations?

No — it is a consequence of them. The Einstein tensor obeys the contracted Bianchi identity $\nabla_\mu G^{\mu\nu}\equiv0$ as a matter of pure geometry, true for every metric, and the metric is covariantly constant. Taking the divergence of $G^{\mu\nu}+\Lambda g^{\mu\nu}=8\pi G\,T^{\mu\nu}$ therefore forces $\nabla_\mu T^{\mu\nu}=0$ automatically. Historically the logic even ran the other way: the requirement that the source be conserved is what told Einstein to couple matter to $G^{\mu\nu}$ rather than to the Ricci tensor, whose divergence does not vanish. So conservation is woven into the very choice of the field equations.

Does general relativity conserve energy or not?

Locally, yes; globally, usually not in the familiar sense. The equation $\nabla_\mu T^{\mu\nu}=0$ holds at every point and expresses that matter neither creates nor destroys energy-momentum on its own — but it includes connection terms that describe energy traded with the gravitational field. Because those terms are not a total derivative, you cannot integrate to a fixed total energy unless the spacetime has a time-translation symmetry (a timelike Killing vector) or is asymptotically flat (so the ADM or Bondi mass is defined). In an expanding universe there is no such symmetry, and the energy of, say, the cosmic photon gas genuinely decreases as it redshifts — with no local pot for the lost energy to flow into. This is not a failure of the theory but a precise feature of it.

Why can't the gravitational field carry its own local energy density?

Because gravity is encoded in the connection, and the connection is not a tensor — you can always make it vanish at any single point by choosing a freely falling frame (the equivalence principle). Any quantity built from first derivatives of the metric, like a candidate gravitational energy density, can therefore be set to zero at a point by a change of coordinates, so it cannot be a genuine local tensor. The best one can do is a pseudotensor (Landau-Lifshitz, Einstein), which is coordinate-dependent, or an integral quantity defined only asymptotically (ADM, Bondi) or quasilocally (Brown-York). This is the same fact, seen from the field's side, that stops $\nabla_\mu T^{\mu\nu}=0$ from being a global conservation law.

How does one equation contain both conservation and the equations of motion?

Feed $\nabla_\mu T^{\mu\nu}=0$ a specific stress tensor and project it. For dust, $T^{\mu\nu}=\rho u^\mu u^\nu$ : projecting the divergence along the four-velocity gives rest-mass continuity $\nabla_\mu(\rho u^\mu)=0$ , and the part orthogonal to it forces the four-acceleration to vanish, $u^\mu\nabla_\mu u^\nu=0$ — the geodesic equation. For a perfect fluid with pressure, the same two projections give the energy equation and the relativistic Euler equation. So the single conservation law simultaneously says "nothing is lost" and "here is how the matter moves." Free matter falls on geodesics not because we assumed it, but because that is the only motion consistent with conserving its energy-momentum.

What is the difference between the coefficients ρ+p and ρ+3p?

They answer two different questions and come from two different parts of the theory. The enthalpy density $\rho+p$ is the fluid's inertia — it is the coefficient in the relativistic Euler equation, the resistance to acceleration, and it comes from the spatial projection of $\nabla_\mu T^{\mu\nu}=0$ . The combination $\rho+3p$ is the active gravitational mass density — the source of the Newtonian potential in the weak-field limit — and it comes from the time-time part of the Einstein equations (the trace that makes $R_{00}\propto\rho+3p$ ). Pressure thus enhances both inertia and gravity, but by different factors: for a photon gas ( $p=\rho/3$ ) the inertia is boosted to $\tfrac43\rho$ and the gravitating mass to $2\rho$ . Both reduce to $\rho$ when $p\ll\rho$ , recovering Newtonian dust.

How does the conservation law give the cosmological continuity equation?

Take the FRW metric and a comoving perfect fluid, $u^\mu=(1,0,0,0)$ . The $\nu=0$ component of $\nabla_\mu T^{\mu\nu}=0$ works out to $\dot\rho+3H(\rho+p)=0$ , where $H=\dot a/a$ is the Hubble rate; the $3H\rho$ term is volume dilution and the $3Hp$ term is the work pressure does as the universe expands. With an equation of state $p=w\rho$ this integrates immediately to $\rho\propto a^{-3(1+w)}$ : matter ( $w=0$ ) thins as $a^{-3}$ , radiation ( $w=\tfrac13$ ) as $a^{-4}$ because each photon also redshifts, and vacuum energy ( $w=-1$ ) stays constant, which is what lets a cosmological constant drive accelerated expansion. None of this needs the Friedmann equations themselves — it is pure conservation.

Is the Newtonian limit of this law just the ordinary Euler equation?

Yes, and recovering it is a good consistency check. Take slow velocities ( $u^\mu\approx(1,\mathbf v)$ ), weak fields, and small pressure ( $p\ll\rho$ ). The time component of $\nabla_\mu T^{\mu\nu}=0$ becomes the mass-continuity equation $\partial_t\rho+\nabla\!\cdot(\rho\mathbf v)=0$ , and the spatial components become $\rho\,\dot{\mathbf v}=-\nabla p-\rho\nabla\Phi$ — Euler's equation of fluid dynamics, with the gravitational force $-\rho\nabla\Phi$ supplied by the connection terms (the Christoffels reduce to $\nabla\Phi$ in the weak field). So classical hydrodynamics in a gravitational field is exactly the low-energy shadow of the covariant conservation law, and the relativistic enthalpy factor $\rho+p$ simply drops back to $\rho$ .