Stress-Energy Tensor in Special Relativity

Energy and momentum refuse to be scalars. Boost into a moving frame and a body's energy and momentum both change, and — because $E$ is the time component of the four-momentum $p^\mu=(E/c,\,\mathbf p)$ established in aq-gr-four-vectors — the two mix into each other exactly the way the coordinates $x^0$ and $x^i$ do. So the densities and fluxes of energy and momentum cannot each be a single number either; they must assemble into an object with two spacetime indices. That object is the stress-energy tensor (or energy-momentum tensor) $T^{\mu\nu}$ : the flux of the $\mu$ -component of four-momentum across a surface of constant $x^\nu$ . In one stroke it packs energy density, energy flux, momentum density, pressure, and shear stress into a single Lorentz-covariant array — precisely as the field-strength tensor $F^{\mu\nu}$ unifies $\mathbf E$ and $\mathbf B$ . Two facts organize everything that follows. First, four-momentum is locally conserved, so in the flat Minkowski spacetime of special relativity the tensor is divergence-free, $\partial_\mu T^{\mu\nu}=0$ — four continuity equations, one for energy and three for momentum, that collapse to the continuity and Euler equations of fluid mechanics in the slow limit. Second, the tensor is symmetric, $T^{\mu\nu}=T^{\nu\mu}$ , which forces energy flux to equal momentum density and encodes local conservation of angular momentum. We build the whole structure with the constant metric $\eta_{\mu\nu}$ and ordinary partial derivatives, because this flat-space construction fixes the physical meaning of each of the ten components before gravity is switched on. When it is — via the rule $\partial\to\nabla$ — this same $T^{\mu\nu}$ becomes the entire right-hand side of Einstein's equations, the thing that tells spacetime how to curve.

Key takeaways

The stress-energy tensor $T^{\mu\nu}$ is a symmetric rank-two tensor whose ten components are the densities and fluxes of energy and momentum: $T^{00}$ energy density, $T^{0i}$ energy flux (= momentum density $T^{i0}$ ), $T^{ij}$ the stresses.
In flat spacetime conservation of energy and momentum is the single statement $\partial_\mu T^{\mu\nu}=0$ — four local continuity laws that integrate to a conserved total four-momentum $P^\nu=\int T^{0\nu}\,\mathrm d^3x$ .
Symmetry $T^{\mu\nu}=T^{\nu\mu}$ follows from local angular-momentum conservation and makes $T^{\mu\nu}$ a legitimate symmetric source for gravity.
The canonical matter models are special cases of the perfect fluid $T^{\mu\nu}=(\rho+p)u^\mu u^\nu+p\,\eta^{\mu\nu}$ : dust ( $p=0$ ), radiation ( $p=\rho/3$ , traceless), and vacuum energy ( $p=-\rho$ ).
Projecting $\partial_\mu T^{\mu\nu}=0$ recovers Newtonian fluid dynamics, and applying it to dust forces geodesic motion.
Key idea: $T^{\mu\nu}$ is the flux of the $\mu$ -component of four-momentum across a surface of constant $x^\nu$ — the $00$ slot is energy density, the $0i$ slots are energy flux and momentum density, and the $ij$ slots are stresses; its conservation is the four-fold continuity equation $\partial_\mu T^{\mu\nu}=0$ .

The core insight

$T^{\mu\nu}$ is the flux of the $\mu$ -component of four-momentum across a surface of constant $x^\nu$ : $T^{00}$ is energy density, $T^{0i}$ is energy flux (equal to the momentum density $T^{i0}$ ), and $T^{ij}$ is the spatial stress. Local conservation of four-momentum is the single tensor equation $\partial_\mu T^{\mu\nu}=0$ .

Where this lives today

The flat-space stress-energy tensor is a working tool in several active fields. In relativistic heavy-ion physics, the quark-gluon plasma made when gold or lead nuclei collide at RHIC and the LHC is modelled as a nearly ideal relativistic fluid, and fitting the elliptic flow of the debris to relativistic hydrodynamics — the evolution of $T^{\mu\nu}$ under $\partial_\mu T^{\mu\nu}=0$ with a small shear correction to $T^{ij}$ — reveals a shear-viscosity-to-entropy ratio $\eta/s$ astonishingly close to the conjectured bound $\hbar/4\pi k_B$ , the most perfect fluid ever measured. In cosmology, the vacuum-energy case $p=-\rho$ ( $w=-1$ ) is the leading model of dark energy; surveys such as DESI now measure $w$ to a few percent, testing whether it is exactly $-1$ or drifts — a distinction living in the trace of $T^{\mu\nu}$ . And in gravitational-wave astronomy, the neutron-star merger caught in GW170817 is simulated by evolving the perfect-fluid $T^{\mu\nu}$ of nuclear matter, so the equation of state inside a neutron star can be read from the signal.

The physical picture

Picture $T^{\mu\nu}$ as a bookkeeping matrix stretched over a little box of spacetime. The first index names which component of four-momentum you track; the second names through which face it flows. The top-left entry $T^{00}$ is energy flowing through a constant-time face — energy just sitting there, a density. Along the top row, $T^{0i}$ is energy flowing out through the spatial face along $x^i$ . The left column $T^{i0}$ is the density of the $i$ -th momentum component. And the lower-right $3\times3$ block $T^{ij}$ is momentum- $i$ streaming through the face normal to $x^j$ — force per unit area, the ordinary stress tensor of continuum mechanics, with pressure on its diagonal and shear off it. Ten independent numbers, because the matrix is symmetric.

Figure

The stress-energy tensor as a bookkeeping matrix. The first index μ selects which momentum component; the second index ν selects the face it flows through. In a fluid rest frame the tensor collapses to the diagonal T^μν = diag(ρ, p, p, p)

Symmetry says the matrix reads the same across its main diagonal, and that is no accident. Energy flux in direction $i$ is the density of $i$ -momentum, because (in units $c=1$ ) energy is the time-component of momentum. Conservation, $\partial_\mu T^{\mu\nu}=0$ , then reads like inventory control on the box — whatever leaks out through the spatial faces is what the density inside loses per unit time — applied to all four momentum components. One last quantitative fact: boost past a cloud of dust and its energy density rises not by the $\gamma$ of length contraction alone but by $\gamma^2$ , because the energy per particle climbs by one factor of $\gamma$ while the contracted number density climbs by another. That double factor is the fingerprint of a genuine two-index tensor, plotted next.

Figure

Boosted dust. The energy density T⁰⁰ = γ²ρ₀ measured in a frame where the matter streams past at speed β = v/c. At β = 0.6, γ = 1.25 and T⁰⁰ = 1.5625 ρ₀ — one factor of γ from the energy per particle, one from the Lorentz-contracted number density

The core equations

The starting point is a definition, worth stating in words: each component of $T^{\mu\nu}$ is a flux of a momentum component across a face of spacetime. Fixing $\mu$ chooses the momentum tracked; fixing $\nu$ chooses the face.

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

where: $p^\mu$ — four-momentum $(E/c,\,\mathbf p)$ ; $x^\nu$ — spacetime coordinate with $x^0=ct$ ; $T^{\mu\nu}$ — the stress-energy tensor, in units of energy density. Reading: a flux across a constant- $x^0$ surface is a density in space; across a constant- $x^i$ surface it is a genuine current. Applying that definition slot by slot gives the component dictionary — the most useful thing to memorise, turning an abstract tensor into ten laboratory quantities.

\begin{aligned}T^{00}&=\text{energy density},\\ T^{0i}&=\text{energy flux across the } x^i \text{ face},\\ T^{i0}&=\text{density of } i\text{-momentum},\\ T^{ij}&=\text{flux of } i\text{-momentum across the } x^j \text{ face (stress)}\end{aligned} \qquad (2)

where: $i,j$ — spatial indices $1,2,3$ ; $T^{ii}$ (no sum) — pressure along axis $i$ ; $T^{ij}$ , $i\neq j$ — shear stress. Reading: the top row is about energy, the left column about momentum. The tensor is symmetric — a genuine physical constraint, not a convention, cutting the sixteen components of a general $4\times4$ array down to ten.

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

where: the key instance is $T^{0i}=T^{i0}$ — energy flux equals momentum density (in units $c=1$ ; restoring $c$ , momentum density $=$ energy flux $/c^2$ ). Reading: symmetry is enforced by local conservation of angular momentum, derived below. Because four-momentum cannot be created or destroyed locally, the tensor has zero four-divergence — in flat spacetime an ordinary partial derivative, the master conservation law of the subject.

\partial_\mu T^{\mu\nu}=0 \qquad (4)

where: $\partial_\mu=\partial/\partial x^\mu$ ; the $\nu=0$ component is energy conservation, the three $\nu=i$ components are momentum conservation — four continuity equations bundled into one covariant statement. The workhorse matter model is the perfect fluid — matter with a rest-frame energy density and isotropic pressure but no shear or heat flow. Every idealized source in cosmology and stellar structure is a special case.

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

where: $\rho$ — rest-frame energy density; $p$ — isotropic pressure; $u^\mu$ — four-velocity with $u_\mu u^\mu=-1$ ; $\eta^{\mu\nu}=\mathrm{diag}(-1,1,1,1)$ . Reading: in the rest frame $u^\mu=(1,\mathbf 0)$ this collapses to $\mathrm{diag}(\rho,p,p,p)$ — the pressure-in-a-box picture. Fields carry energy and momentum too. The electromagnetic stress-energy tensor is built purely from $F^{\mu\nu}$ , and it unifies field energy, the Poynting flux, and the Maxwell stress.

T^{\mu\nu}_{\text{EM}}=\frac{1}{\mu_0}\left(F^{\mu\alpha}F^{\nu}{}_{\alpha}-\tfrac14\,\eta^{\mu\nu}F^{\alpha\beta}F_{\alpha\beta}\right) \qquad (6)

where: $F^{\mu\nu}$ — field-strength tensor; $\mu_0$ — vacuum permeability; $T^{00}_{\text{EM}}=\tfrac12(\varepsilon_0 E^2+B^2/\mu_0)$ is field energy density and $T^{0i}_{\text{EM}}=S^i/c$ with $\mathbf S=\mathbf E\times\mathbf B/\mu_0$ the Poynting vector. Reading: its trace vanishes, $T^\mu{}_\mu=0$ , the signature of massless matter. The trace $T^\mu{}_\mu$ is a Lorentz scalar — the same in every frame — and for a perfect fluid depends only on the equation of state, the fastest test of whether matter is "radiation-like".

T^\mu{}_\mu=\eta_{\mu\nu}T^{\mu\nu}=-\rho+3p \qquad (7)

where: $T^\mu{}_\mu$ — invariant trace; it vanishes when $p=\rho/3$ (radiation) and is $-4\rho$ for a cosmological constant ( $p=-\rho$ ). Reading: a single scalar fingerprinting the equation of state. Integrating the conserved density $T^{0\nu}$ over a constant-time slice gives the total four-momentum — a conserved four-vector in flat spacetime.

P^\nu=\int_\Sigma T^{0\nu}\,\mathrm d^3x \qquad (8)

where: $P^\nu$ — total four-momentum; $\Sigma$ — a hypersurface $x^0=\text{const}$ ; $\mathrm d^3x$ — spatial volume; units $c=1$ (restoring $c$ , $P^\nu=\tfrac1c\int_\Sigma T^{0\nu}\,\mathrm d^3x$ , so that $P^0=E/c$ ). Reading: $\partial_\mu T^{\mu\nu}=0$ guarantees $\mathrm dP^\nu/\mathrm dt=0$ when nothing flows in from infinity — the theorem that fails once spacetime curves.

Derivation ASymmetry from local conservation of angular momentum

The goal is to prove that $T^{\mu\nu}$ must be symmetric — not an aesthetic preference but a consequence of angular momentum being locally conserved. The strategy is to build the current of orbital angular momentum out of $T^{\mu\nu}$ exactly as $\mathbf r\times\mathbf p$ is built from position and momentum, demand that this current be divergence-free the way every conserved current is, and read off what that demand forces on the tensor — squeezing its antisymmetric part to zero, leaving ten components rather than sixteen.

Step 1. Form the angular-momentum flux $M^{\alpha\mu\nu}=x^\mu T^{\alpha\nu}-x^\nu T^{\alpha\mu}$ , antisymmetric in $\mu\nu$ ; its $\alpha=0$ part integrates to the total angular momentum $J^{\mu\nu}=\int(x^\mu T^{0\nu}-x^\nu T^{0\mu})\,\mathrm d^3x$ . Why: this is "position wedge momentum-density", the relativistic promotion of $\mathbf r\times\mathbf p$ , so if angular momentum is conserved this is the current that is conserved.

Step 2. Impose local conservation, $\partial_\alpha M^{\alpha\mu\nu}=0$ . Why: a conserved charge with density $M^{0\mu\nu}$ requires a divergence-free current, just as electric charge requires $\partial_\alpha J^\alpha=0$ .

Step 3. Expand using $\partial_\alpha x^\mu=\delta_\alpha{}^\mu$ : $\partial_\alpha M^{\alpha\mu\nu}=T^{\mu\nu}-T^{\nu\mu}+x^\mu\partial_\alpha T^{\alpha\nu}-x^\nu\partial_\alpha T^{\alpha\mu}$ . Why: differentiating the explicit $x$ pulls down a Kronecker delta and drops the index onto $T$ , manufacturing the antisymmetric combination $T^{\mu\nu}-T^{\nu\mu}$ for free.

Step 4. Invoke momentum conservation $\partial_\alpha T^{\alpha\nu}=0$ , killing the last two terms and leaving $T^{\mu\nu}-T^{\nu\mu}=0$ . Why: once linear momentum is conserved, angular momentum conservation forces the antisymmetric part to vanish — and a nonsymmetric canonical tensor from a Lagrangian must be repaired (Belinfante–Rosenfeld) before it can couple to gravity.

Go deeper · Derivation B · 1 minConservation law becomes continuity and the Euler equation

Derivation BConservation law becomes continuity and the Euler equation

The goal is to show that the four abstract equations $\partial_\mu T^{\mu\nu}=0$ are, in disguise, the two most familiar laws of fluid dynamics — mass continuity and Euler's equation — confirming that $T^{00}$ really is energy density and $T^{0i}$ really is momentum flux. The strategy is to feed the perfect-fluid tensor into the divergence, then project the four-vector equation twice: along the flow $u^\nu$ (the energy balance) and orthogonal to it (the momentum balance), before taking the slow limit of each.

Step 1. Substitute equation (5) and expand: $\partial_\mu[(\rho+p)u^\mu]\,u^\nu+(\rho+p)u^\mu\partial_\mu u^\nu+\partial^\nu p=0$ , using $\partial_\mu\eta^{\mu\nu}=0$ . Why: the metric is constant, so the only surviving pieces are the flow term and the pressure gradient $\partial^\nu p$ — transport and pushing.

Step 2. Contract with $u_\nu$ . Using $u_\nu u^\nu=-1$ and $u_\nu\partial_\mu u^\nu=0$ , the orthogonal pressure term drops, leaving $\partial_\mu(\rho u^\mu)+p\,\partial_\mu u^\mu=0$ . Why: contracting with $u_\nu$ selects the component along the worldline — the relativistic continuity equation.

Step 3. Slow limit $u^\mu\approx(1,\mathbf v)$ with $p\ll\rho$ : $\partial_t\rho+\nabla\!\cdot(\rho\mathbf v)=0$ . Why: at low speed rest energy dominates pressure, and the balance reduces to mass-continuity — the energy slot of $T$ is ordinary mass density.

Step 4. Project orthogonal with $h^{\nu\mu}=\eta^{\nu\mu}+u^\nu u^\mu$ : $(\rho+p)u^\mu\partial_\mu u^\nu=-h^{\nu\mu}\partial_\mu p$ . Why: the projector strips out the flow direction from Step 2, leaving the spatial force balance — the relativistic Euler equation, with $(\rho+p)$ as inertial mass density.

Step 5. Slow limit again: $\rho(\partial_t\mathbf v+(\mathbf v\cdot\nabla)\mathbf v)=-\nabla p$ . Why: Newton's second law for a fluid element — so the momentum slots $T^{ij}$ are genuine stresses.

Go deeper · Derivation C · 1 minThe dust tensor from a swarm of particles

Derivation CThe dust tensor from a swarm of particles

The goal is to build $T^{\mu\nu}$ from scratch for the simplest matter — a cloud of identical, non-interacting particles ("dust") — directly from the definition "flux of four-momentum", and to verify that the boosted energy density carries the tell-tale two factors of $\gamma$ . The strategy is to write the microscopic tensor as a sum over worldlines weighted by delta functions, coarse-grain into a continuum, identify the rest-frame density, and boost to check the transformation law.

Step 1. Write the microscopic tensor $T^{\mu\nu}(x)=\sum_a\int p_a^\mu\,\dfrac{\mathrm dx_a^\nu}{\mathrm d\tau_a}\,\delta^4(x-x_a(\tau_a))\,\mathrm d\tau_a$ , one term per particle. Why: each particle deposits its four-momentum $p^\mu$ along its worldline, and $\mathrm dx^\nu/\mathrm d\tau=u^\nu$ gives the direction it is carried — the definition "flux of $p^\mu$ " made literal.

Step 2. Coarse-grain over many particles sharing a four-velocity $u^\mu$ ; the delta functions collapse to a proper number density $n$ , giving $T^{\mu\nu}=n\,m\,u^\mu u^\nu$ . Why: because $p^\mu=mu^\mu$ , both free indices come out proportional to $u$ , so the tensor is automatically symmetric.

Step 3. Identify the rest-frame energy density $\rho=nmc^2$ and write $T^{\mu\nu}=\rho\,u^\mu u^\nu$ . Why: in the rest frame $T^{00}$ must equal $nmc^2$ , pinning the coefficient — a pressureless perfect fluid, no pressure because the particles never collide.

Step 4. Boost so the dust streams by, $u^0=\gamma$ , and read $T^{00}=\rho(u^0)^2=\gamma^2\rho$ . Why: one factor of $\gamma$ from each particle's energy, one from the Lorentz contraction packing them denser — the product $\gamma^2$ is what a rank-two tensor's $00$ component must do, and the number on the second figure.

Limiting cases

Every standard source is a specialization of the symmetric tensor — usually of the perfect fluid — obtained by choosing the relation between $\rho$ and $p$ :

Case	Result	Meaning
Pressureless dust	$T^{\mu\nu}=\rho\,u^\mu u^\nu$	$\rho$ is rest-frame mass-energy density; no pressure or shear. Enforcing $\partial_\mu T^{\mu\nu}=0$ forces geodesic (force-free) motion.
Perfect fluid	$T^{\mu\nu}=(\rho+p)u^\mu u^\nu+p\,\eta^{\mu\nu}$	Isotropic pressure added; $u_\mu u^\mu=-1$ in signature $(-,+,+,+)$ . Reduces to dust when $p=0$ .
Radiation / ultrarelativistic gas	$p=\tfrac13\rho,\quad T^\mu{}_\mu=0$	Traceless — the defining property of conformally invariant, massless matter (photon gas, electromagnetic field).
Cosmological constant / vacuum	$T^{\mu\nu}=-\rho_\Lambda\,\eta^{\mu\nu},\quad p=-\rho$	Equation of state $w=-1$ ; pressure and energy density of opposite sign, and the tensor is Lorentz-invariant, which is why it drives accelerated expansion.

Common trapThe curved-space law ∇ₘ Tᵘᵛ = 0 conserves total energy just like the flat-space law

It does not, and this is the most important caveat of the subject. In flat spacetime $\partial_\mu T^{\mu\nu}=0$ integrates over a constant-time slice to a genuinely conserved total four-momentum $P^\nu=\int T^{0\nu}\,\mathrm d^3x$ . In curved spacetime the correct law is $\nabla_\mu T^{\mu\nu}=0$ , and the covariant derivative hides Christoffel terms, $\nabla_\mu T^{\mu\nu}=\partial_\mu T^{\mu\nu}+\Gamma^\mu{}_{\mu\lambda}T^{\lambda\nu}+\Gamma^\nu{}_{\mu\lambda}T^{\mu\lambda}$ , which describe energy and momentum being exchanged with the gravitational field. Those terms spoil the conversion of a local law into a global one: a photon climbing out of a gravitational well redshifts, losing energy that no matter integral accounts for. The matter's $T^{\mu\nu}$ is locally covariantly conserved, but the total energy of matter alone is generally not constant — a lesson that only makes sense once you understand why the flat-space version does conserve it.

Worked examples

1 · Energy density, momentum, and pressure of sunlight (numeric)

At the top of Earth's atmosphere the solar irradiance (magnitude of the Poynting vector) is $S=1361\ \mathrm{W/m^2}$ . Treating sunlight as a plane wave, use the components of $T^{\mu\nu}_{\text{EM}}$ to find its energy density, momentum density, and the radiation pressure on a black (fully absorbing) solar sail.

Step 1. The $00$ component is the field energy density, tied for a plane wave to the Poynting flux by $S=T^{00}c$ . Hence $T^{00}=S/c=1361/(2.998\times10^8)=\mathbf{4.54\times10^{-6}\ J/m^3}$ — a cubic metre of sunlight holds a few microjoules.

Step 2. The $0i$ component is the energy flux, and by symmetry also $c$ times the momentum density: $T^{0i}=S^i/c$ , so the momentum density is $g=T^{0i}/c=S/c^2=\mathbf{1.51\times10^{-14}\ kg\,m^{-2}s^{-1}}$ . The factor $1/c^2$ makes electromagnetic momentum tiny beside its energy — the hallmark of massless matter.

Step 3. Radiation pressure on an absorber is the flux of momentum through the surface, the $T^{zz}$ stress for a wave along $z$ ; for a directed plane wave $T^{zz}=T^{00}=S/c$ , so $P_{\text{abs}}=S/c=\mathbf{4.54\times10^{-6}\ Pa}$ . A perfectly reflecting sail reverses rather than absorbs the momentum, doubling it: $P_{\text{refl}}=2S/c=\mathbf{9.08\times10^{-6}\ Pa}$ . A square-kilometre sail feels only about $9\ \mathrm N$ — feeble, but it never stops, which is why solar sails work. Energy density $4.54\times10^{-6}\ \mathrm{J/m^3}$ , absorber pressure $4.54\ \mu\mathrm{Pa}$ .

2 · The trace test and cosmological dilution (numeric)

Show that the trace of a perfect fluid is $-\rho+3p$ , use it to classify radiation, matter, and vacuum energy, and read off how each dilutes as the universe expands with scale factor $a$ .

Step 1. Contract equation (5) with $\eta_{\mu\nu}$ : $T^\mu{}_\mu=(\rho+p)(u_\mu u^\mu)+p\,\delta^\mu{}_\mu=(\rho+p)(-1)+p(4)=\mathbf{-\rho+3p}$ . The trace is a Lorentz scalar, so the classification is frame-independent.

Step 2. Apply $p=w\rho$ . Radiation $w=\tfrac13$ gives $T^\mu{}_\mu=-\rho+\rho=\mathbf 0$ (traceless, like the electromagnetic tensor); matter $w=0$ gives $-\rho$ ; a cosmological constant $w=-1$ gives $-\rho-3\rho=\mathbf{-4\rho}$ . Three equations of state, three trace signatures.

Step 3. Conservation $\nabla_\mu T^{\mu 0}=0$ in an expanding homogeneous universe becomes $\dot\rho+3\dfrac{\dot a}{a}(\rho+p)=0$ , with solution $\rho\propto a^{-3(1+w)}$ . Matter ( $w=0$ ) gives $\rho\propto a^{-3}$ — pure volume dilution; radiation ( $w=\tfrac13$ ) gives $\rho\propto a^{-4}$ — one extra power from the redshift of each photon; vacuum ( $w=-1$ ) gives $\rho\propto a^{0}$ — constant, no dilution at all. Radiation $a^{-4}$ , matter $a^{-3}$ , vacuum $a^{0}$ — why the early universe was radiation-dominated and the late universe is vacuum-dominated.

Physics of the portrait

The stress-energy tensor has two parents in special relativity: Minkowski, who first wrote a four-dimensional energy-momentum tensor for the electromagnetic field, and Max von Laue, who extended it to matter of every kind and made it a general principle.

Origins · 1 minMax von Laue and how this was discovered

Max von Laue

1879–1960 · the physicist who first cast the energy and momentum of matter and fields into one symmetric spacetime tensor

Max von Laue is remembered by most physicists for his 1914 Nobel Prize — the diffraction of X-rays by crystals. But among relativists his lasting monument is the stress-energy tensor. Building on Hermann Minkowski's 1908 reformulation of Maxwell's theory, in which the field energy, the Poynting flux, and the Maxwell stresses were first packed into a four-dimensional array, von Laue wrote in 1911 the first textbook of relativity, Das Relativitätsprinzip, and there generalized the construction to matter of arbitrary composition. He recognized that any closed, stationary system must obey what is now called von Laue's theorem — that the space integral of the stress components vanishes, $\int T^{ij}\,\mathrm d^3x=0$ — which resolved the notorious " $4/3$ problem" of the electron's electromagnetic mass by insisting that a stable body needs non-electromagnetic (Poincaré) stresses to hold it together. This insistence that energy, momentum, and stress form one indivisible tensorial object is precisely the step that let Einstein, four years later, place $T^{\mu\nu}$ on the right-hand side of the field equations as the source of gravity.

History

1908

— Minkowski recasts Maxwell's theory in four-dimensional form and writes the electromagnetic energy-momentum tensor

1911

— von Laue extends the tensor to matter in the first relativity monograph and proves that a static bounded body satisfies the integral relation for its stresses

1915

— Einstein places the symmetric stress-energy tensor on the right-hand side of the gravitational field equations

1918

— Noether's theorem ties the conservation law to invariance under spacetime translations

1940

— Belinfante and Rosenfeld show the canonical Noether tensor can be symmetrized without changing the total energy and momentum

Going further

Texts. Misner, Thorne & Wheeler, Gravitation, ch. 5 (the definitive component-by-component construction) · Wald, General Relativity, §4.3 · Carroll, Spacetime and Geometry, §1.9 and §4.2 · Hartle, Gravity, ch. 22 · Schutz, A First Course in General Relativity, ch. 4 · Landau & Lifshitz, The Classical Theory of Fields, §32–33 (electromagnetic tensor and its tracelessness). Key self-consistency checks: the trace of the perfect-fluid tensor is $-\rho+3p$ ; in the rest frame equation (5) collapses to $\mathrm{diag}(\rho,p,p,p)$ ; projecting $\partial_\mu T^{\mu\nu}=0$ along and orthogonal to $u^\mu$ recovers the continuity and Euler equations; and the sunlight energy density evaluates to $S/c=4.54\times10^{-6}\,\mathrm{J/m^3}$ while the boosted dust gives $\gamma^2=1.5625$ at $\beta=0.6$ .

Builds on. Four-Vectors · Minkowski Metric and Signature · Lorentz Transformations · Proper Time. Leads to. Einstein Field Equations · Perfect Fluid in General Relativity · Minkowski Spacetime.

Ask the tutor: Ask me to walk through any slot of $T^{\mu\nu}$ — from why sunlight's momentum is so tiny beside its energy to how $\partial_\mu T^{\mu\nu}=0$ becomes the Euler equation.

Why is energy density the 00 component and not just a scalar?

Because energy is not a scalar. Energy is the time component of the four-momentum $p^\mu$ , so it transforms under boosts — a moving object has more energy than a resting one. A density additionally involves a volume, and volumes Lorentz-contract. So energy density carries two "time-like" tags: one from the energy being $p^0$ , one from measuring it per unit volume on a constant-time slice, itself a flux through a constant- $x^0$ surface. Two tags means two indices, both $0$ — hence $T^{00}$ . The same logic makes momentum density $T^{i0}$ and stress $T^{ij}$ .

What is the difference between the flat-space and curved-space conservation laws?

In flat spacetime $\partial_\mu T^{\mu\nu}=0$ uses ordinary derivatives, and integrating over all space converts it into $\mathrm dP^\nu/\mathrm dt=0$ : the total four-momentum of the matter is strictly conserved. In curved spacetime the correct law replaces the partial with a covariant derivative, $\nabla_\mu T^{\mu\nu}=0$ , carrying extra Christoffel terms that represent energy and momentum flowing between matter and the gravitational field. Those terms obstruct the integration step, so there is generally no conserved total matter energy. The flat law is a true conservation statement; the curved law is a local balance that includes gravity as an unlisted party.

Why must energy flux equal momentum density?

This is the content of the symmetry $T^{0i}=T^{i0}$ , and physically it is almost a tautology once you accept mass-energy equivalence. An energy flux is energy moving with some velocity; but energy has an equivalent mass $E/c^2$ , and mass in motion is a momentum. So a flow of energy is automatically a density of momentum, equal after the $c^2$ conversion. The deeper guarantee is the derivation above: only a symmetric tensor conserves angular momentum locally.

Why does the boosted energy density scale as γ² and not γ?

Boost past a cloud of dust and two independent things happen. First, each particle's energy rises from $mc^2$ to $\gamma mc^2$ — one factor of $\gamma$ . Second, the cloud is Lorentz-contracted along the motion, so the same particles are squeezed into a smaller volume and the number density rises by another factor of $\gamma$ . Energy density is energy per volume, so the two multiply: $T^{00}=\gamma^2\rho_0$ . A Newtonian bookkeeper tracking only the rising energy, or only the contracting volume, would predict a single $\gamma$ and be wrong. The extra factor is the signature that $T^{\mu\nu}$ is a genuine rank-two tensor, with $T^{00}=\rho(u^0)^2$ and $u^0=\gamma$ .

How does conservation of the dust tensor force geodesic motion?

Take $T^{\mu\nu}=\rho\,u^\mu u^\nu$ and impose $\partial_\mu T^{\mu\nu}=0$ . The product rule gives $\partial_\mu(\rho u^\mu)\,u^\nu+\rho\,u^\mu\partial_\mu u^\nu=0$ . Contract with $u_\nu$ : since $u_\nu\partial_\mu u^\nu=0$ , the second term dies and the first gives number conservation $\partial_\mu(\rho u^\mu)=0$ . Substitute back and the first term dies too, leaving $\rho\,u^\mu\partial_\mu u^\nu=0$ , so for $\rho\neq0$ the four-acceleration vanishes — the dust moves in straight lines. Promote $\partial\to\nabla$ and this becomes $u^\mu\nabla_\mu u^\nu=0$ , the geodesic equation. Free matter follows geodesics not as an extra postulate but as a corollary of conservation.

Why is the cosmological constant's stress-energy proportional to the metric?

Vacuum energy must look the same to every observer — there is no preferred rest frame for "empty space", so no boost may change it. The only symmetric two-index tensor invariant under all Lorentz boosts is a multiple of the metric, $T^{\mu\nu}=-\rho_\Lambda\,\eta^{\mu\nu}$ , whose components give energy density $\rho_\Lambda$ and pressure $-\rho_\Lambda$ : the equation of state $p=-\rho$ , $w=-1$ . The negative pressure makes vacuum energy gravitationally repulsive and able to accelerate expansion, and the Lorentz-invariance is why its density does not dilute — the $a^0$ scaling of Worked Example 2.