The Stress-Energy Tensor

In Newtonian gravity the source of the field is a single scalar, the mass density ρ. Special relativity dismantles that simplicity: mass and energy are equivalent, energy density is frame-dependent, and under a boost it mixes with momentum density, energy flux, and the mechanical stresses — pressure, tension, shear. No scalar, and no four-vector, can hold all of these at once. The object that can is a symmetric rank-two tensor, the stress-energy tensor $T^{\mu\nu}$ (also called the energy-momentum tensor), and this article is the synthesis that makes it the seat of gravity. Three threads run through everything below. The first is operational: $T^{\mu\nu}$ is the flux of the μ-component of four-momentum across a surface of constant $x^\nu$ , so a single definition delivers energy density ( $T^{00}$ ), energy flux and momentum density ( $T^{0i}=T^{i0}$ ), and stress ( $T^{ij}$ ) all at once. The second is dynamical: in the Einstein field equations the geometric tensor $G_{\mu\nu}$ is set proportional to $T_{\mu\nu}$ , so the full distribution of energy, momentum, pressure, and stress — not merely rest mass — curves spacetime, and pressure gravitates. The third is conservation: because the Einstein tensor obeys the contracted Bianchi identity identically, the field equations force $\nabla_\mu T^{\mu\nu}=0$ , making local energy-momentum conservation a consequence of geometry rather than a separate postulate. Alongside these sits the deep fact that $T^{\mu\nu}$ can be derived, not just tabulated: varying the matter action with respect to the metric produces exactly the symmetric, conserved object the field equations require.

Key takeaways

The stress-energy tensor $T^{\mu\nu}$ is a symmetric rank-two tensor whose components are the densities and fluxes of energy and momentum: $T^{00}$ energy density, $T^{0i}=T^{i0}$ momentum density and energy flux, $T^{ij}$ stress.
Its one operational definition — the flux of $p^\mu$ across a surface of constant $x^\nu$ — reproduces every component and every matter model, from dust $T^{\mu\nu}=\rho\,u^\mu u^\nu$ to the perfect fluid to the Maxwell field.
It is the source of gravity: $G_{\mu\nu}=8\pi G\,T_{\mu\nu}$ replaces the Newtonian mass density ρ, and in the weak-field limit the active gravitating density is $\rho+3p$ , so pressure gravitates.
Local conservation $\nabla_\mu T^{\mu\nu}=0$ is forced by the contracted Bianchi identity $\nabla_\mu G^{\mu\nu}=0$ — it is a consistency requirement of the theory, not an extra axiom.
The Hilbert (metric) definition $T^{\mu\nu}=-\tfrac{2}{\sqrt{-g}}\,\delta S_\mathrm{m}/\delta g_{\mu\nu}$ makes symmetry and on-shell conservation automatic, because $g_{\mu\nu}$ is symmetric.
Key idea: $T^{\mu\nu}$ is the flux of four-momentum component $p^\mu$ through a constant- $x^\nu$ surface; it packages energy density, momentum density/energy flux, and stress into one symmetric tensor, sources curvature via $G_{\mu\nu}=8\pi G\,T_{\mu\nu}$ , and is conserved by $\nabla_\mu T^{\mu\nu}=0$ .

Where this lives today

The stress-energy tensor is the working interface between matter and geometry across modern gravitational physics. In cosmology it is the entire right-hand side of the Friedmann equations, and pinning down the dark-energy equation of state — is $w$ exactly $-1$ , or does it drift? — is the headline goal of surveys such as DESI, Euclid, and the Vera Rubin Observatory, whose 2024 results are consistent with a cosmological constant to a few percent. In compact-object astrophysics the perfect-fluid tensor feeds the Tolman-Oppenheimer-Volkoff equation, where pressure both supports a neutron star and, through $\rho+3p$ , deepens the well it must climb out of; the neutron-star equation of state is now being read off jointly from NICER's X-ray timing and the tidal signature in the LIGO-Virgo event GW170817. Gravitational waves themselves carry stress-energy — the Isaacson effective tensor of the ripples — which is how a merging binary loses energy and inspirals. And at the foundational frontier, inequalities on $T^{\mu\nu}$ (the weak, null, strong, and dominant energy conditions) gate the Penrose-Hawking singularity theorems and the (im)possibility of wormholes and warp drives; the measured Casimir effect and the accelerating expansion are the two clean demonstrations that the simplest of these conditions can be violated, which is exactly why quantum energy inequalities are an active research topic.

The physical picture

Take a small cell of matter and watch its bookkeeping of four-momentum. Energy can sit inside the cell — that is $T^{00}$ , energy density — and energy can stream through its faces — that is $T^{0i}$ , energy flux. Momentum can likewise sit inside ( $T^{i0}$ , momentum density) and stream through the faces ( $T^{ij}$ , momentum flux, i.e. stress). Relativity fuses two of these into one statement: energy flux is momentum density, because energy carries inertia $E/c^2$ , so a flow of energy is a density of momentum. That is why $T^{0i}=T^{i0}$ and, together with the absence of net internal torques ( $T^{ij}=T^{ji}$ ), why the whole array is symmetric. The first figure makes the sixteen components and their symmetry a single picture.

Figure

The stress-energy tensor is the flux of four-momentum pᵘ through a surface of constant xᵛ. Reading the blocks: T⁰⁰ is energy density, the off-diagonal T⁰ⁱ = Tⁱ⁰ is simultaneously energy flux and momentum density, and the spatial block Tⁱʲ is the stress, with pressure on its diagonal and shear off it — and symmetry Tᵘᵛ = Tᵛᵘ makes the whole array symmetric across its diagonal

Why does pressure gravitate? In $G_{\mu\nu}=8\pi G\,T_{\mu\nu}$ every component of $T$ curves spacetime, not just the rest energy, and working out the Newtonian limit shows the effective gravitating density is $\rho+3p$ rather than ρ. The clearest place this bites is cosmology: model the universe as a perfect fluid $T^\mu{}_\nu=\mathrm{diag}(-\rho,p,p,p)$ , and covariant conservation collapses to the continuity law $\dot\rho+3H(\rho+p)=0$ , whose solution $\rho\propto a^{-3(1+w)}$ (with equation of state $p=w\rho$ ) fixes how each constituent thins as the universe expands. The second figure is the punchline that number gives: three different slopes for three kinds of matter, and hence three eras of cosmic history.

Figure

How the cosmic energy density dilutes with expansion. Solving ρ̇ + 3H(ρ+p) = 0 with p = wρ gives ρ ∝ a^−3(1+w), so radiation with w = 1/3 falls as a⁻⁴, pressureless matter with w = 0 as a⁻³, and the cosmological constant with w = −1 stays exactly constant — the three slopes that set the radiation, matter, and dark-energy eras

The core equations

The operational definition is uniform across every kind of matter and is the seed from which the components sprout. It says the tensor measures the flux of four-momentum; reading off which component and across which surface recovers energy density, energy flux, momentum density, and stress in turn.

T^{\mu\nu}=\text{flux of }p^\mu\text{ across a surface of constant }x^\nu \qquad (1)

where: $p^\mu$ — four-momentum, $p^\mu=(E,\mathbf p)$ ; $x^\nu$ — spacetime coordinate, with $x^0=t$ labelling constant-time surfaces; $T^{00}$ — energy density (flux of energy through a constant- $t$ surface); $T^{0i}$ — flux of energy in the $i$ direction, equal to the density of $i$ -momentum; $T^{ij}$ — flux of $i$ -momentum in the $j$ direction, i.e. stress. The diagonal $T^{ii}$ is pressure; the off-diagonal $T^{ij}$ is shear. Two independent facts force the tensor to be symmetric, and relativity unifies them. Vanishing internal torque on a shrinking cell requires the spatial block to be symmetric, while the equivalence of energy flux and momentum density fixes the mixed components.

T^{\mu\nu}=T^{\nu\mu} \qquad (2)

where: $T^{\mu\nu}$ — the stress-energy tensor. The spatial statement $T^{ij}=T^{ji}$ is local angular-momentum conservation; the mixed statement $T^{0i}=T^{i0}$ is energy-flux = momentum-density; the Hilbert definition below makes both automatic. In flat space the ordinary divergence of $T^{\mu\nu}$ vanishes; in curved space it is replaced by the covariant divergence, and its vanishing is the local conservation of energy and momentum. As Derivation B and the discussion below show, this is not an added postulate but a corollary of the field equations.

\nabla_\mu T^{\mu\nu}=0 \qquad (3)

where: $\nabla_\mu$ — the metric-compatible (Levi-Civita) covariant derivative. Contracting with the field equations $G^{\mu\nu}=8\pi G\,T^{\mu\nu}$ and using the contracted Bianchi identity $\nabla_\mu G^{\mu\nu}=0$ makes equation (3) an identity of the theory. The dynamical face of the tensor is its definition as the response of the matter action to a change in the metric. This is the object that actually appears on the right of Einstein's equations, and it is symmetric and conserved on shell by construction.

T^{\mu\nu}=-\frac{2}{\sqrt{-g}}\frac{\delta(\sqrt{-g}\,\mathcal{L}_\mathrm{m})}{\delta g_{\mu\nu}}=-\frac{2}{\sqrt{-g}}\frac{\delta S_\mathrm{m}}{\delta g_{\mu\nu}} \qquad (4)

where: $S_\mathrm{m}=\int\mathrm d^4x\,\sqrt{-g}\,\mathcal L_\mathrm{m}$ — the matter action; $\mathcal L_\mathrm{m}$ — the matter Lagrangian density (a scalar); $g$ — the determinant of $g_{\mu\nu}$ ; $\delta/\delta g_{\mu\nu}$ — the functional (variational) derivative. Symmetry is inherited from the symmetry of $g_{\mu\nu}$ . The workhorse matter model of astrophysics and cosmology is the perfect fluid — isotropic in its rest frame, with no shear or heat flow. Its stress-energy is built from just the energy density, the pressure, and the four-velocity.

T^{\mu\nu}=(\rho+p)\,u^\mu u^\nu+p\,g^{\mu\nu} \qquad (5)

where: ρ — rest-frame energy density; $p$ — isotropic pressure; $u^\mu$ — fluid four-velocity, $u_\mu u^\mu=-1$ ; $g^{\mu\nu}$ — inverse metric (signature $-+++$ ). In the rest frame this is $\mathrm{diag}(\rho,p,p,p)$ ; setting $p=0$ gives dust $T^{\mu\nu}=\rho\,u^\mu u^\nu$ . Finally, the weak-field limit of the field equations shows which combination of the components actually gravitates. Pressure is not a bystander: it adds three times over to the Newtonian source.

\rho_\text{active}=\rho+3p\qquad(c=1) \qquad (6)

where: $\rho_\text{active}$ — the effective gravitating density in the Newtonian-limit Poisson equation $\nabla^2\Phi=4\pi G(\rho+3p)$ ; $p$ — isotropic pressure. Radiation ( $p=\rho/3$ ) gravitates as $2\rho$ ; a cosmological constant ( $p=-\rho$ ) gravitates as $-2\rho$ and pushes outward.

Derivation AThe stress-energy of a point particle, smoothed into dust

The goal is to build $T^{\mu\nu}$ from the ground up for the simplest source — one particle on a worldline — check that its $T^{00}$ is the relativistic energy density, then coarse-grain over many particles to recover the dust tensor $T^{\mu\nu}=\rho\,u^\mu u^\nu$ that seeds every cosmological model. This shows the abstract flux definition and the concrete fluid form are one object at two resolutions, and that conservation of the tensor is nothing more exotic than free-particle motion.

Step 1. A particle of rest mass $m$ tracing the worldline $X^\mu(\tau)$ carries four-momentum $p^\mu=m\,\mathrm dX^\mu/\mathrm d\tau$ , and its stress-energy is a delta-function distribution smeared only along that worldline: $T^{\mu\nu}(x)=\dfrac{m}{\sqrt{-g}}\displaystyle\int\mathrm d\tau\,\dfrac{\mathrm dX^\mu}{\mathrm d\tau}\dfrac{\mathrm dX^\nu}{\mathrm d\tau}\,\delta^{(4)}\!\big(x-X(\tau)\big)$ . Why: the tensor must be concentrated where the particle is, and be built from the only vector available on the worldline, the four-velocity — the product $\dot X^\mu\dot X^\nu$ is the unique symmetric rank-two combination.

Step 2. Check the energy density by setting $\mu=\nu=0$ and doing the τ-integral against $\delta(t-X^0)$ , which contributes a Jacobian $\mathrm d\tau/\mathrm dt=1/\gamma$ . One factor of $\mathrm dt/\mathrm d\tau=\gamma$ survives, leaving $T^{00}=\gamma m\,\delta^{(3)}(\mathbf x-\mathbf X)$ in flat space. Why: $\gamma m$ is exactly the relativistic energy of the particle, so the component the flux definition calls "energy density" really is the energy per unit volume — the construction passes its first physical test.

Step 3. Coarse-grain over a swarm of identical particles sharing a common four-velocity $u^\mu$ with rest-frame number density $n$ . The delta functions add up to a smooth density, and writing $\rho=n\,m$ for the rest energy density gives $T^{\mu\nu}=\rho\,u^\mu u^\nu$ . Why: this is the pressureless dust tensor — cold, non-interacting matter such as galaxies or dark matter on large scales — and it emerges without any new assumption beyond averaging.

Step 4. Impose conservation. In flat space $\partial_\nu T^{\mu\nu}=0$ for the single particle is equivalent to $\mathrm dp^\mu/\mathrm d\tau=0$ , and in curved space $\nabla_\nu T^{\mu\nu}=0$ becomes the geodesic equation $u^\nu\nabla_\nu u^\mu=0$ . Why: divergence-freedom of the stress-energy tensor is not an abstract accounting rule but the statement that free matter falls along geodesics — conservation and equation of motion are one fact. *Verified: $T^{00}=\gamma m\,\delta^{(3)}$ reduces to the Newtonian mass density ρ as $\gamma\to1$ ; the dust form matches MTW Ch. 5 and Schutz Ch. 4.*

Go deeper · Derivation B · 1 minLocal conservation gives the relativistic Euler equation

Derivation BLocal conservation gives the relativistic Euler equation

The goal is to show that the single covariant law $\nabla_\mu T^{\mu\nu}=0$ , applied to a perfect fluid, contains all of relativistic hydrodynamics: projecting it along and orthogonal to the four-velocity splits it into a continuity (energy) equation and a force (Euler) equation, and the algebra exposes why the inertia of a relativistic fluid is $\rho+p$ , not ρ. This is what turns "the tensor is conserved" into equations one can integrate for a star or a cosmology.

Step 1. Take the divergence of $T^{\mu\nu}=(\rho+p)u^\mu u^\nu+p\,g^{\mu\nu}$ . Metric compatibility $\nabla_\mu g^{\mu\nu}=0$ lets the pressure term pass through the derivative, giving $\nabla_\mu\!\big[(\rho+p)u^\mu u^\nu\big]+\nabla^\nu p=0$ . Why: separating the pressure gradient from the flow term is the first move that will let the parallel and perpendicular pieces be read off independently.

Step 2. Contract with $u_\nu$ and use $u_\nu u^\nu=-1$ together with $u_\nu\nabla_\mu u^\nu=0$ . The result is the relativistic continuity equation $u^\mu\nabla_\mu\rho+(\rho+p)\nabla_\mu u^\mu=0$ . Why: the component of conservation along the flow tracks how energy density changes as the fluid expands or compresses — in cosmology, with $\nabla_\mu u^\mu=3H$ , this is precisely $\dot\rho+3H(\rho+p)=0$ .

Step 3. Project orthogonal to the flow with $h^\alpha{}_\nu=\delta^\alpha_\nu+u^\alpha u_\nu$ , which annihilates anything parallel to $u$ and isolates the spatial dynamics. Carrying out the projection yields the boxed relativistic Euler equation $(\rho+p)\,u^\mu\nabla_\mu u^\nu=-\big(g^{\mu\nu}+u^\mu u^\nu\big)\nabla_\mu p$ . Why: the operator $g^{\mu\nu}+u^\mu u^\nu$ is the projector onto the space orthogonal to $u$ , so only the part of the pressure gradient that can actually push the fluid sideways drives the acceleration.

Step 4. Read the physics. The coefficient of the acceleration $u^\mu\nabla_\mu u^\nu$ is $\rho+p$ , the inertial (enthalpy) density — pressure adds to inertia because moving a pressurised element does work against its own internal stresses. In the limit $p\ll\rho$ with slow flow, the equation reduces to the ordinary Euler equation $\rho\,\mathrm d\mathbf v/\mathrm dt=-\nabla p$ . Why: recovering Newtonian hydrodynamics in the correct limit is the check that the relativistic generalisation has been assembled correctly. *Verified: reproduces the Newtonian Euler and continuity equations in the slow-motion, low-pressure limit; matches Carroll §1.9 and Wald §4.3.*

Go deeper · Derivation C · 1 minThe Hilbert stress-energy of a scalar field by metric variation

Derivation CThe Hilbert stress-energy of a scalar field by metric variation

The goal is to run the metric-variation definition (4) on a concrete matter action — a single scalar field — and watch a symmetric, conserved stress-energy tensor drop out automatically, with no need to guess the fluxes by hand. This is the template for how every field's stress-energy tensor is obtained, and it shows why the Hilbert prescription, not the older canonical one, is the version that couples to gravity.

Step 1. Start from $S_\mathrm{m}=\displaystyle\int\mathrm d^4x\,\sqrt{-g}\,\big(-\tfrac12 g^{\alpha\beta}\partial_\alpha\phi\,\partial_\beta\phi-V(\phi)\big)$ and vary with respect to the inverse metric, using the determinant identity $\delta\sqrt{-g}=-\tfrac12\sqrt{-g}\,g_{\mu\nu}\,\delta g^{\mu\nu}$ . Why: the determinant identity is the one piece of machinery every such variation needs — it is where the $g_{\mu\nu}\mathcal L_\mathrm{m}$ term in the final tensor comes from.

Step 2. The kinetic term contributes $\partial_\mu\phi\,\partial_\nu\phi$ from varying $g^{\alpha\beta}$ , and combining it with the determinant piece and the $-2/\sqrt{-g}$ prefactor gives $T_{\mu\nu}=\partial_\mu\phi\,\partial_\nu\phi-g_{\mu\nu}\big(\tfrac12\partial_\alpha\phi\,\partial^\alpha\phi+V(\phi)\big)$ . Why: the first term is manifestly symmetric and the second is proportional to the symmetric metric, so the whole tensor is symmetric with no improvement terms needed — exactly the advantage of the Hilbert route over the canonical Noether tensor.

Step 3. Check the trace in four dimensions: $T^\mu{}_\mu=g^{\mu\nu}T_{\mu\nu}=\partial_\alpha\phi\,\partial^\alpha\phi-4\big(\tfrac12\partial_\alpha\phi\,\partial^\alpha\phi+V\big)=-\partial_\alpha\phi\,\partial^\alpha\phi-4V$ . Why: the trace encodes the equation of state — it does not vanish in general, and for a slowly rolling field ( $V$ dominant) it approaches $-4V$ , the same signature as a cosmological constant, which is why scalar fields power inflation and quintessence.

Step 4. Take the covariant divergence: $\nabla^\mu T_{\mu\nu}=(\Box\phi-V'(\phi))\,\partial_\nu\phi$ , which vanishes precisely when the Klein-Gordon equation $\Box\phi=V'(\phi)$ holds. Why: conservation is automatic on shell — it follows from the field's own equation of motion, which is itself a consequence of translation invariance, closing the loop between symmetry, dynamics, and conservation. *Verified: symmetric and conserved on shell; the trace $-\partial_\alpha\phi\,\partial^\alpha\phi-4V$ and the massless case match Carroll §4.2 and Wald App. E.*

Limiting cases

The one definition specialises to every standard matter model by choosing the Lagrangian or the symmetry of the source. The table collects the forms that recur throughout general relativity.

Case	Result	Meaning
Pressureless dust	$T^{\mu\nu}=\rho\,u^\mu u^\nu$	Cold, non-interacting matter (galaxies, dark matter); only $T^{00}=\rho$ in the rest frame. Equation of state $w=0$ , so $\rho\propto a^{-3}$ .
Perfect fluid	$T^{\mu\nu}=(\rho+p)u^\mu u^\nu+p\,g^{\mu\nu}$	Isotropic pressure $p$ ; rest frame $\mathrm{diag}(\rho,p,p,p)$ . The basis of stellar structure and cosmology.
Electromagnetic field	$T^{\mu\nu}=F^{\mu\alpha}F^\nu{}_\alpha-\tfrac14 g^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}$	Traceless in 4D ( $T^\mu{}_\mu=0$ ), reflecting conformal invariance; $T^{00}=\tfrac12(E^2+B^2)$ ; equation of state $w=1/3$ .
Vacuum energy / Λ	$T^{\mu\nu}=-\rho_\Lambda\,g^{\mu\nu},\ p=-\rho_\Lambda$	Lorentz-invariant, so it looks the same to every observer; $w=-1$ , drives accelerated expansion (dark energy).
Scalar field	$T_{\mu\nu}=\partial_\mu\phi\,\partial_\nu\phi-g_{\mu\nu}\big(\tfrac12\partial_\alpha\phi\,\partial^\alpha\phi+V\big)$	The canonical example of the Hilbert definition; potential-dominated it mimics Λ, powering inflation and quintessence.

Common trap“∇_μ T^{μν} = 0 means the total energy of matter is conserved in curved spacetime.”

It does not. Covariant conservation is a local law, and it generally cannot be integrated into a globally conserved matter energy. The reason is structural: $\nabla_\mu T^{\mu\nu}=0$ expands to $\partial_\mu T^{\mu\nu}+\Gamma^\mu_{\mu\lambda}T^{\lambda\nu}+\Gamma^\nu_{\mu\lambda}T^{\mu\lambda}=0$ , and the connection (Christoffel) terms cannot be absorbed into an ordinary divergence of a genuine tensor density, so Gauss's theorem does not hand you a conserved integral. Physically, those connection terms describe energy and momentum flowing between matter and the gravitational field — a cosmological photon redshifting as it climbs through expanding space is losing energy to exactly this channel. A global conservation law returns only in special cases: in flat spacetime, where $\Gamma=0$ ; or whenever the spacetime has a Killing vector $\xi^\nu$ , which builds a genuinely conserved current $J^\mu=T^\mu{}_\nu\,\xi^\nu$ with $\nabla_\mu J^\mu=0$ . Absent such symmetry there is no observer-independent "energy of the matter" to conserve, and attempts to define one (gravitational pseudotensors) are inherently coordinate-dependent.

Worked examples

1 · Radiation pressure and the traceless tensor (fully worked)

A gas of photons — for concreteness, the cosmic microwave background at $T=2.725\ \mathrm K$ — has energy density $\rho$ . Find its pressure, its equation-of-state parameter $w$ , the trace of its stress-energy tensor, and its active gravitating density, then put in a number for ρ.

Step 1. For an isotropic radiation gas the stress-energy tensor is that of a perfect fluid with the extra property that it is traceless (the Maxwell tensor has $T^\mu{}_\mu=0$ in four dimensions). Setting the perfect-fluid trace to zero, $T^\mu{}_\mu=-\rho+3p=0$ , gives immediately $p=\rho/3$ and hence the equation of state $w=p/\rho=\mathbf{1/3}$ .

Step 2. The trace is $T^\mu{}_\mu=-\rho+3p=-\rho+3(\rho/3)=\mathbf 0$ , consistent with the conformal (scale) invariance of free electromagnetism — a photon gas has no rest-mass scale to break it.

Step 3. The active gravitating density from equation (6) is $\rho_\text{active}=\rho+3p=\rho+3(\rho/3)=\mathbf{2\rho}$ . Radiation gravitates twice as strongly as its energy density alone would suggest — pressure doubles its pull. This is one reason the early, radiation-dominated universe decelerated so hard.

Step 4. Put in a number. The CMB energy density is $\rho=aT^4$ with the radiation constant $a=7.566\times10^{-16}\ \mathrm{J\,m^{-3}\,K^{-4}}$ , giving $\rho=aT^4\approx\mathbf{4.17\times10^{-14}\ J\,m^{-3}}$ . The pressure is $p=\rho/3\approx\mathbf{1.39\times10^{-14}\ Pa}$ , and the active density is $2\rho\approx8.3\times10^{-14}\ \mathrm{J\,m^{-3}}$ . Result: $p=\rho/3$ , $w=1/3$ , $T^\mu{}_\mu=0$ , $\rho_\text{active}=2\rho$ ; for the CMB, $p\approx1.4\times10^{-14}\ \mathrm{Pa}$ .

2 · When did radiation stop running the universe?

Today the density parameters are roughly $\Omega_{m,0}\approx0.31$ (matter, $w=0$ ) and $\Omega_{r,0}\approx9.2\times10^{-5}$ (radiation, $w=1/3$ ). Using the dilution law $\rho\propto a^{-3(1+w)}$ from the second figure, find the scale factor $a_\mathrm{eq}$ and redshift $z_\mathrm{eq}$ at which the two densities were equal.

Step 1. Solve conservation for each component. With $p=w\rho$ , the continuity equation $\dot\rho+3H(\rho+p)=0$ integrates to $\rho\propto a^{-3(1+w)}$ , so matter dilutes as $\rho_m\propto a^{-3}$ and radiation as $\rho_r\propto a^{-4}$ — radiation thins one power of $a$ faster because expansion also redshifts each photon.

Step 2. Write the ratio at scale factor $a$ (today $a=1$ ): $\dfrac{\rho_r}{\rho_m}=\dfrac{\Omega_{r,0}\,a^{-4}}{\Omega_{m,0}\,a^{-3}}=\dfrac{\Omega_{r,0}}{\Omega_{m,0}}\,a^{-1}$ . Equality means this ratio equals 1, so $a_\mathrm{eq}=\dfrac{\Omega_{r,0}}{\Omega_{m,0}}$ .

Step 3. Insert numbers: $a_\mathrm{eq}=\dfrac{9.2\times10^{-5}}{0.31}\approx\mathbf{3.0\times10^{-4}}$ , hence $z_\mathrm{eq}=a_\mathrm{eq}^{-1}-1\approx\mathbf{3.4\times10^{3}}$ . Before this redshift the universe was radiation-dominated (and decelerating hard, per Example 1); after it, matter took over, and structure could begin to grow. Result: $a_\mathrm{eq}\approx3.0\times10^{-4}$ , $z_\mathrm{eq}\approx3400$ — the single equation of state carried by $T^{\mu\nu}$ dates the handover between cosmic eras.

Physics of the portrait

The stress-energy tensor is where nineteenth-century field theory, special relativity, and geometry converge: a bookkeeping device for momentum flux that turned out to be the very thing that curves spacetime.

Origins · 1 minstress-energy tensor and how this was discovered

stress-energy tensor

the symmetric source Tᵘᵛ of spacetime curvature · energy, momentum and stress in general relativity

The stress-energy tensor grew out of the Maxwell stress tensor of classical electromagnetism — James Clerk Maxwell's 1873 accounting of the momentum flux carried by electric and magnetic fields, the object that lets a field push on a charge. Hermann Minkowski's 1908 four-dimensional reformulation of electrodynamics lifted that three-by-three stress tensor into a four-by-four spacetime tensor, uniting energy density, momentum density, energy flux, and stress as components of one geometric object. Max von Laue took the decisive conceptual step in 1911, arguing that any closed physical system — not just the electromagnetic field — must be described by such a symmetric "world tensor," and that only its full set of components, pressures included, transforms correctly under Lorentz boosts. When Albert Einstein and David Hilbert arrived at the field equations in November 1915, this was the natural source: Einstein set $G_{\mu\nu}$ proportional to $T_{\mu\nu}$ , and Hilbert, working from a variational principle, produced $T_{\mu\nu}$ by differentiating the matter action with respect to the metric — the definition that guarantees symmetry and conservation. Emmy Noether's 1918 theorems then explained the canonical tensor as the conserved current of spacetime-translation symmetry, and by 1940 Frederik Belinfante and Léon Rosenfeld had shown how to symmetrise that canonical tensor into precisely Hilbert's form. What began as a way to tally the push of light became, in three decades, the seat of gravity itself.

History

1873

— Maxwell formulates the stress tensor for the electromagnetic field

1908

— Minkowski recasts electrodynamics in four-dimensional form, exposing the field stress-energy tensor

1911

— von Laue extends the symmetric stress-energy tensor to any closed system

1915

— Einstein and Hilbert make Tμν the source of curvature in the field equations

1940

— Belinfante and Rosenfeld symmetrise the canonical tensor into the Hilbert form

Going further

References. Misner, Thorne & Wheeler, Gravitation (1973), Ch. 5 · Wald, General Relativity (1984), §4.3, §E.1 · Carroll, Spacetime and Geometry (2004), §1.9, §4.2 · Schutz, A First Course in General Relativity (2009), Ch. 4 · Hartle, Gravity (2003), Ch. 22 · Landau & Lifshitz, The Classical Theory of Fields (1975), §32–33, §94.

Provenance. Symbol conventions and signature $(-+++)$ follow Misner-Thorne-Wheeler and Carroll. A verifier should: (i) confirm the perfect-fluid trace $-\rho+3p$ and electromagnetic tracelessness from the tensors given; (ii) reproduce the point-particle $T^{00}=\gamma m\,\delta^{(3)}$ and the dust form $\rho\,u^\mu u^\nu$ ; (iii) re-derive the relativistic Euler equation by projecting $\nabla_\mu T^{\mu\nu}=0$ (Carroll §1.9, Wald §4.3); (iv) re-derive the scalar-field $T_{\mu\nu}$ by metric variation using $\delta\sqrt{-g}=-\tfrac12\sqrt{-g}\,g_{\mu\nu}\,\delta g^{\mu\nu}$ (Carroll §4.2); (v) verify $\nabla_\mu T^{\mu\nu}=0$ follows from the contracted Bianchi identity (MTW Ch. 5, 13). Worked numbers: radiation $w=1/3$ , $\rho_\text{active}=2\rho$ , CMB $\rho=aT^4\approx4.17\times10^{-14}\ \mathrm{J\,m^{-3}}$ at $2.725\ \mathrm K$ ; radiation-matter equality $a_\mathrm{eq}=\Omega_{r,0}/\Omega_{m,0}\approx3.0\times10^{-4}$ , $z_\mathrm{eq}\approx3400$ , from standard density parameters. The active mass $\rho+3p$ is the Tolman-Whittaker result; equation-of-state values ( $w=0,\,1/3,\,-1$ ) are standard.

Prerequisites. Tensors on a manifold · The Einstein Tensor · Covariant derivative and the Bianchi identity. Leads to. The Einstein Field Equations · Perfect Fluid Stress-Energy · Local Conservation of Energy-Momentum · Variational Definition of Stress-Energy · The Energy Conditions · The Cosmological Constant.

Ask the tutor: Ask me to walk through the metric variation that turns any matter Lagrangian into its stress-energy tensor, or to show how $\nabla_\mu T^{\mu\nu}=0$ becomes the Friedmann continuity equation.

What is the difference between the canonical and the Hilbert stress-energy tensors?

Both come from symmetry, but by different routes. The canonical (Noether) tensor is the conserved current associated with invariance under spacetime translations; it is what a first course in field theory derives, but it is in general neither symmetric nor gauge-invariant — for electromagnetism it is not even the Maxwell tensor you want. The Hilbert (metric) tensor is defined by $T^{\mu\nu}=-\tfrac{2}{\sqrt{-g}}\,\delta S_\mathrm{m}/\delta g_{\mu\nu}$ , the response of the matter action to a change in the metric; it is automatically symmetric and gauge-invariant, and it is the object that actually couples to gravity. The two differ by an identically conserved total-derivative term (the Belinfante-Rosenfeld improvement), so they carry the same total four-momentum but distribute it differently.

Is there any conserved total energy in general relativity, then?

Not in general. Because $\nabla_\mu T^{\mu\nu}=0$ carries connection terms that cannot be turned into an ordinary divergence, there is no coordinate-independent "energy of the matter" that integrates to a constant in an arbitrary spacetime. A conserved quantity does exist whenever the spacetime has a symmetry: a Killing vector $\xi^\nu$ builds the current $J^\mu=T^\mu{}_\nu\,\xi^\nu$ with $\nabla_\mu J^\mu=0$ , which does integrate to a conserved charge (energy for a timelike Killing vector, momentum or angular momentum for others). Asymptotically flat spacetimes also admit well-defined total energies — the ADM and Bondi masses — measured at infinity. But a generic dynamical, expanding, or radiating spacetime simply has no global energy conservation law for matter alone.

Why must the stress-energy tensor be symmetric?

Two reasons that relativity fuses into one. First, the spatial block: an antisymmetric part of $T^{ij}$ would exert a net torque on a fluid element whose magnitude, relative to its moment of inertia, would diverge as the element shrank — so finite, well-defined internal angular momentum demands $T^{ij}=T^{ji}$ . Second, the mixed components: $T^{0i}$ is energy flux and $T^{i0}$ is momentum density, and these are equal because energy carries inertia ( $E/c^2$ ), so any flow of energy is automatically a density of momentum. The Hilbert definition wraps both facts up trivially: since it differentiates with respect to the symmetric metric $g_{\mu\nu}$ , the resulting tensor is symmetric for any matter Lagrangian whatsoever.

Does the gravitational field carry its own stress-energy?

Locally, no — and that is a feature, not a bug. By the equivalence principle you can always transform to a freely falling frame where the gravitational field (the first derivatives of the metric) vanishes at a point, so any purely local "gravitational energy density" would be coordinate-dependent and could be set to zero. That is why $T^{\mu\nu}$ contains only matter and non-gravitational fields, and why attempts to add a gravitational piece produce pseudotensors rather than true tensors. Nonlocally, gravitational energy is real and measurable: gravitational waves carry it (captured by the Isaacson effective stress-energy tensor, averaged over several wavelengths), and asymptotically flat systems have a well-defined total mass at infinity. The energy is there — it just cannot be localised to a point.

Why is the inertial density ρ + p but the gravitating density ρ + 3p?

They answer different questions and appear in different equations. The combination $\rho+p$ is the inertia of a fluid element — the enthalpy density — and it shows up as the coefficient of the acceleration in the relativistic Euler equation, because accelerating a pressurised element requires doing work against its internal pressure as well as moving its energy. The combination $\rho+3p$ is the active gravitational density and appears in the weak-field Poisson equation, where all three spatial pressures add to the source alongside the energy density. One p enters inertia (from the flow term of $T^{\mu\nu}$ ); three enter gravity (one per spatial diagonal component). Keeping the two roles distinct is essential in stellar-structure calculations, where pressure simultaneously supports the star and strengthens the gravity trying to crush it.

What are the energy conditions, and why do they matter?

They are inequalities that restrict $T^{\mu\nu}$ to what physicists regard as reasonable matter, since the field equations by themselves place no bound on the source. The weak energy condition ( $T_{\mu\nu}u^\mu u^\nu\ge0$ ) says every observer measures non-negative energy density; the null, strong, and dominant conditions add related constraints on pressures and energy flux. They matter because the powerful theorems of general relativity — the Penrose-Hawking singularity theorems, the area theorem, the topological censorship results — assume them as hypotheses, and because exotic objects such as traversable wormholes and warp drives require them to be violated. Crucially, they are not laws: the Casimir vacuum has negative energy density, and the cosmological constant violates the strong condition, which is precisely how the universe manages to accelerate.