Stress-energy tensor

The stress-energy tensor describes local energy density, momentum density, and stress. Compatibility with the Einstein field equations requires covariant conservation ${\nabla }^{\mu }{T}_{\mu \nu }=0$.

In four spacetime dimensions, the trace reversal of a stress-energy tensor has the displayed form, where $T=g^{\mu \nu }{T}_{\mu \nu }$. Its trace is $-T$, so applying trace reversal twice returns the original tensor. Taking the trace of the Einstein field equations without a cosmological constant gives $R=-\kappa T$; substitution yields $R_{\mu \nu }=\kappa {\overline{T}}_{\mu \nu }$. In a general dimension $d>2$, solving for the Ricci tensor instead uses $T_{\mu \nu }-g_{\mu \nu }T/(d-2)$.

The metric variational stress tensor is defined by the matter action's response to a metric change. With signature $(+---)$ and the corresponding usual matter-action signs, the displayed convention means $\delta S_m=\frac{1}{2}\int \sqrt{-g}{T}_{ab}\delta g^{ab}=-\frac{1}{2}\int \sqrt{-g}{T}^{ab}\delta g_{ab}$. Fix the independent matter variables while varying the metric. Sign conventions must be matched to the action; copying a mostly-plus stress formula into this convention can reverse its energy density.

For signature $(+---)$ and $S_{\varphi }=\int \sqrt{-g}[(\nabla \varphi {)}^2/2-V(\varphi )]$, the convention $T_{\mu \nu }=2(\delta S_{\varphi }/\delta g^{\mu \nu })/\sqrt{-g}$ gives the displayed tensor. Varying the inverse metric yields the derivative product, while varying the volume element yields the term proportional to $g_{\mu \nu }$. A homogeneous field has energy density ${\dot{\varphi }}^2/2+V$ and pressure ${\dot{\varphi }}^2/2-V$.

Adding ${\partial }_{\rho }{B}^{\rho \mu \nu }$ with $B^{\rho \mu \nu }=-B^{\mu \rho \nu }$ to a stress-energy tensor preserves its divergence identically, since commuting partial derivatives annihilates the antisymmetric pair. The total four-momentum changes only by a spatial boundary term. Improvements can make local densities symmetric or gauge-invariant; equality of integrated charges requires the chosen boundary conditions. Equation-of-motion terms can also relate representatives on shell.

For $d>1$, the displayed stress-energy tensor is symmetric and conserved on the free massless scalar field equation $\square \varphi =0$. Its improvement is identically divergence-free. Since $\square {\varphi }^2=2(\partial \varphi {)}^2$ on shell, its trace vanishes. Contracting it with a Conformal Killing vector field therefore produces a conserved conformal current. In $d=2$ the improvement coefficient is zero: the canonical tensor already has zero trace.

A stress-energy superpotential is an antisymmetric tensor $B^{\rho \mu \nu }=-B^{\mu \rho \nu }$ whose divergence supplies a stress-energy tensor improvement. In the Maxwell field, the potential-dependent expression $B^{\rho \mu }{}_{\nu }=F^{\mu \rho }{A}_{\nu }$ produces the difference between canonical and gauge-invariant stress-energy tensors on shell. Antisymmetry ensures identically conserved improvement terms.

A pressureless fluid with proper number density $n_0$, rest mass $m$ and four-velocity $u$ has tensor $m{n}_0{u}^{\mu }{u}^{\nu }$. In its local rest frame there is energy density but no spatial stress. Particle-number conservation and the Lorentz force make its divergence the electromagnetic force density.

Null dust is matter with a stress-energy tensor $T_{ab}=\epsilon k_a{k}_b$ for a future-directed null vector $k$. For every future-directed timelike vector $t$, the energy current is $-T^a{}_b{t}^b=\epsilon (-k\cdot t)k^a$. Therefore its dominant energy condition holds exactly when $\epsilon \ge 0$, allowing the zero current. Rescaling $k$ rescales $\epsilon$ inversely by the square and leaves the stress-energy tensor unchanged. Ingoing Vaidya metric matter has $k=-{\partial }_r$, $k_a=-(dv{)}_a$, and $\epsilon =\dot{m}/(4\pi r^2)$.

Momentum density is momentum per unit physical volume. In a local orthonormal tetrad of the measuring observer it is $q^{\hat{i}}=T^{\hat{0}\hat{i}}$. A perfect fluid in general relativity moving at a small velocity relative to that observer has $q^{\hat{i}}=(\rho +P)v^{\hat{i}}$ to first order. For isotropic background photons, $P=\rho /3$, so the factor is $4\rho /3$ rather than $\rho$. Integrating the kinetic stress-energy tensor gives $q^{\hat{i}}=\int f{p}^{\hat{i}}{d}^{3}p$.

In a local orthonormal tetrad, anisotropic stress is the trace-free spatial part of the stress-energy tensor, conventionally denoted ${\pi }^{ij}$. Some linear cosmological perturbation theory conventions instead use ${\Pi }^{ij}=-{\pi }^{ij}$ and write $T^{ij}=P{\delta }^{ij}-{\Pi }^{ij}$. The sign must be specified before using a photon Euler equation. The photon quadrupole determines the linear photon anisotropic stress.

For action $S=\int \sqrt{-g}[-\frac{1}{2}(\nabla \Phi {)}^2-\xi R{\Phi }^2]d^{4}x$, metric variation defines $\delta S=\frac{1}{2}\int \sqrt{-g}{T}^{ab}\delta g_{ab}{d}^{4}x$ and gives the displayed tensor. The contracted Bianchi identity and the scalar derivative commutator yield ${\nabla }^a{T}_{ab}=(\square \Phi -2\xi R\Phi ){\nabla }_b\Phi$, so the stress-energy tensor is conserved on the field equation.

Local energy-momentum conservation is the covariant equation ${\nabla }_{\mu }{T}^{\mu \nu }=0$. Its normal and spatial projections give energy and momentum evolution equations in a spacetime foliation.

A Noether identity is an off-shell differential relation among Euler-Lagrange expressions implied by a local gauge symmetry. Diffeomorphism invariance relates the divergence of the stress-energy tensor to the matter field equations.

For a diffeomorphism-invariant pure metric action, write $\delta S=-\int \sqrt{-g}{E}^{ab}\delta g_{ab}$. Substitute $\delta g_{ab}={\nabla }_a{\xi }_b+{\nabla }_b{\xi }_a$ for an arbitrary compactly supported vector field and integrate by parts. Symmetry of $E^{ab}$ gives ${\nabla }_a{E}^{ab}=0$ without using field equations. This Noether identity concerns the gravitational variational tensor alone.

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

A relativistic perfect fluid has energy density $\rho$, isotropic pressure $p$ and unit timelike four-velocity $u$. Projecting ${\nabla }_a{T}^{ab}=0$ along and orthogonally to $u$ gives its energy and Euler equations.

For a perfect fluid, the relativistic stress tensor is $T_{ab}=(\rho +p)u_a{u}_b+p{g}_{ab}$ in signature $(-,+,+,+)$ with $u^a{u}_a=-1$.

In a diagonal static metric, an ordinary perfect fluid in general relativity with a timelike four-velocity has $T^{0i}=(\rho +p/c^2)u^0{u}^i$. Since $u^0\ne 0$, vanishing mixed time-space components imply a static fluid if the enthalpy is nonzero. For vacuum-like matter with $p=-\rho c^2$, the tensor is $p{g}^{ab}$ independently of the velocity; its rest frame cannot be recovered from that tensor.

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

For $T^{\mu \nu }=\rho u^{\mu }{u}^{\nu }$ and constant normalized $u_{\mu }{u}^{\mu }=c^2$, contraction of ${\nabla }_{\mu }{T}^{\mu \nu }=0$ with $u_{\nu }$ gives ${\nabla }_{\mu }(\rho u^{\mu })=0$. At nonzero density the remaining projection gives $u^{\mu }{\nabla }_{\mu }{u}^{\nu }=0$.

For identical particles of rest mass $m$ and electric charge $q$, the four-current is $j^{\mu }=(q/m)\rho u^{\mu }$. The matter stress-energy tensor obeys ${\nabla }_{\mu }{T}_{\mathrm{m}}^{\mu \nu }=F^{\nu \alpha }{j}_{\alpha }$, with the opposite force density in the electromagnetic stress-energy tensor. Antisymmetry of the electromagnetic field tensor makes its contraction with two four-velocities vanish, so mass continuity still holds. Removing that term leaves the displayed covariant Lorentz force law. The matter tensor alone is not divergence-free when an electromagnetic force acts.

For a canonical scalar with timelike gradient, its stress-energy tensor has perfect-fluid form. In a homogeneous background, $\rho ={\dot{\varphi }}^2/2+V$ and $p={\dot{\varphi }}^2/2-V$. Potential domination gives negative pressure, while the scalar's physical rest-frame sound speed is one. A constant negative background $w$ therefore does not force its perturbations to obey the adiabatic barotropic closure $\delta p=w\delta \rho$.

// Target: cosmology.bigb

For $d{s}^2=a^2(d{\tau }^2-d{\mathbf{x}}^2)$ and a homogeneous canonical scalar, $\rho ={\varphi }^{\prime 2}/(2a^2)+V$ is the physical density, whereas $T_{00}=a^2\rho$. At $w=-1/3$, the kinetic and potential energies are $K=\rho /3$ and $V=2\rho /3$, so ${\varphi }^{\prime 2}=2a^2\rho /3=2T_{00}/3$. A single power of ${\varphi }^{\prime }$ cannot replace the square in this energy identity.

// Target: cosmology.bigb

The canonical vorticity two-form is ${\omega }_{ab}={\nabla }_a(ϵu_b)-{\nabla }_b(ϵu_a)$, with specific enthalpy per particle $ϵ$. Potential flow $ϵu_a={\nabla }_a\chi$ has $\omega =0$ for a torsion-free connection. Contracting this identity with $u^a$ yields $ϵu^a{\nabla }_a{u}_b={\nabla }_bϵ-u_b{u}^a{\nabla }_aϵ$, the isentropic relativistic Euler equation in signature $(+---)$. This canonical condition does not mean the unweighted four-velocity one-form is closed.

In signature $(+---)$, the matter action $S=\int \sqrt{-g}[p(ϵ)+n((\nabla \chi {)}^2-{ϵ}^2)/(2ϵ)]$ uses independent $n,ϵ,\chi$. Variations of $n,ϵ,\chi$ give $(\nabla \chi {)}^2={ϵ}^2$, $p^{\prime }(ϵ)=n$, and particle number conservation in a relativistic fluid for $u_a={\nabla }_a\chi /ϵ$. Its Hilbert stress-energy tensor is $T_{ab}=(\rho +p)u_a{u}_b-p{g}_{ab}$. The action describes isentropic potential flow, with vanishing canonical vorticity of a relativistic fluid, rather than every vortical perfect-fluid solution.

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

For number density $n$, rest-frame energy density $\rho$ and pressure $p$, the enthalpy per particle is $ϵ=(\rho +p)/n$. In an isentropic barotropic fluid, $d\rho =ϵdn$ and $dp=ndϵ$. It differs from enthalpy per unit mass by the particle mass when that mass is fixed. Using $ϵ$ as thermodynamic variable gives $n=p^{\prime }(ϵ)$ and $\rho =ϵp^{\prime }(ϵ)-p(ϵ)$.

In units $c=1$, an isentropic perfect fluid in general relativity has sound speed $c_s^2=dp/d\rho$. Writing $n=p^{\prime }(ϵ)$ gives $d\rho =ϵdn$ and $c_s^2=n/(ϵp^{\prime \prime }(ϵ))$. Linearized particle number conservation in a relativistic fluid and the relativistic Euler equation then yield acoustic waves with this speed. Positive $c_s^2$ is needed for ordinary acoustic stability; a causal fluid model additionally requires $c_s^2\le 1$.

For metric signature $(+---)$, a perfect fluid has $T_{ab}=(\rho +p)u_a{u}_b-p{g}_{ab}$. The Einstein field equations and static Killing field is a Ricci eigenvector imply $(\rho +p)(k\cdot u)u_{[a}{k}_{b]}=0$. Two nonzero timelike vectors cannot be orthogonal, so $k\cdot u\ne 0$. If $\rho +p\ne 0$, the four-velocity is parallel to the static Killing vector. Choosing both future-directed gives the displayed normalization. For $\rho +p=0$, the stress-energy tensor is proportional to the metric tensor and does not determine a preferred fluid velocity; the strict enthalpy hypothesis is therefore material.

A perfect fluid with equation of state $P=\rho$ is called a stiff fluid. For positive energy density, its constant equation-of-state parameter is $w=1$. The cosmological continuity equation gives $\rho \propto a^{-6}$. In a flat universe dominated by this fluid, the Friedmann equation then gives $a(t)\propto (t-t_B{)}^{1/3}$. This statement assumes a nonzero fluid density and negligible other components.

For a spatially homogeneous minimally coupled scalar field with zero scalar potential, the stress-energy tensor has $\rho =P={\dot{\varphi }}^2/2$. Its field equation is $\ddot{\varphi }+3H\dot{\varphi }=0$, so $a^3\dot{\varphi }$ is constant and its energy density decreases as $a^{-6}$. Thus its homogeneous stress is precisely that of a stiff fluid. Spatial gradients would introduce directional stresses, so this equivalence requires homogeneity.

In units $8\pi G=1$, a free homogeneous scalar has $\rho =P={\dot{\varphi }}_0^2/2$ and therefore a stiff equation of state. Its equation gives $a^2{\varphi }_0^{\prime }=\text{constant}$, while the Friedmann equation gives $3{\mathcal{H}}^2={\varphi }_0^{\prime 2}/2$. The displayed contracting branch satisfies both equations for negative conformal time. Constant normalizations and an additive scalar shift can be restored.

In conformal Newtonian gauge, a free scalar has no linear anisotropic stress. Eliminating its perturbation with the momentum constraint gives ${\Phi }_k^{\prime \prime }+3{\Phi }_k^{\prime }/\tau +k^2{\Phi }_k=0$. Substitution of ${\Phi }_k=(-\tau {)}^{-1}y(-k\tau )$ gives the order-one Bessel equation. The displayed Hankel function has positive incoming frequency $e^{-ik\tau }$ at early times.

The momentum constraint relates the incoming scalar mode $a^{-1}\sqrt{\hslash /(2k)}{e}^{-ik\tau }$ to the Newtonian-potential Hankel modes in stiff contraction. Large-argument Hankel asymptotics give $|N_k{|}^2=3\pi \hslash /(32k^2)$. The small-argument behavior $H_1^{(1)}(z)\sim -2i/(\pi z)$ then gives the displayed spectrum. Its dimensional power is $k^{-4}$ and its power per logarithmic interval is proportional to $k^{-1}$. These are Newtonian-potential spectra; a curvature-perturbation spectrum need not have the same exponent.

The timelike projection of stress-energy conservation is $u^a{\nabla }_a\rho +(\rho +p){\nabla }_a{u}^a=0$. It is the local adiabatic first law for a perfect fluid.

The spatial projection of stress-energy conservation is $(\rho +p)u^b{\nabla }_b{u}^a=-(g^{ab}+u^a{u}^b){\nabla }_{b}p$.

For $d{s}^2=(1+2\varphi )d{t}^2-d{\mathbf{x}}^2$ with static weak $\varphi$, normalization gives $u^0=1-\varphi +O({\varphi }^2,v^2)$ and ${\Gamma }^i{}_{00}={\partial }_i\varphi$. Around a homogeneous state, the first-order equations are ${\partial }_t\delta n+n_0\nabla \cdot \mathbf{v}=0$, ${\partial }_t\delta \rho +({\rho }_0+p_0)\nabla \cdot \mathbf{v}=0$, and $({\rho }_0+p_0){\partial }_t\mathbf{v}=-\nabla \delta p-({\rho }_0+p_0)\nabla \varphi$. The inertial density is enthalpy density even though the flow is slow. Only a further nonrelativistic equation of state, $p\ll \rho \simeq mn$, replaces it by mass density.

For metric signature $(+---)$, a stationary perfect fluid follows a timelike Killing vector field $K$ with four-velocity $U=K/N$. Its Killing-flow four-acceleration is $U^b{\nabla }_b{U}_a=-{\nabla }_a\log N$. The relativistic Euler equation and stationarity of the pressure give the displayed hydrostatic equilibrium equation. This uses stationarity along $K$ and does not require hypersurface orthogonality.

The equation of state of isotropic blackbody radiation is $p=\rho /3$. The barotropic hydrostatic first integral gives $\rho N^4=C$. The Stefan–Boltzmann law $\rho \propto T^4$ then yields the Tolman–Ehrenfest relation $TN=\mathrm{constant}$. Here $N=\sqrt{|K^a{K}_a|}$ for the stationary timelike Killing vector field; its normalization only changes the constants.

A connected nonvacuum perfect fluid with equation of state $p=\rho /3$ obeys the stationary radiation density law $\rho =C/N^4$ with $C>0$. On a compact region of a regular spacetime, a smooth Killing vector field has bounded norm. Consequently $\rho$ cannot tend to zero at a finite regular free surface. A vanishing timelike norm instead makes the density diverge. A radiation-filled cavity can have a material wall with nonzero boundary pressure; this is a different boundary condition.

For a barotropic fluid in hydrostatic equilibrium along a Killing vector, assume $\rho +p>0$. Integrating $dp/(\rho +p)=-d\log N$ gives the displayed first integral, constant on each connected fluid region. For the equation of state $p=w\rho$ with constant $w>0$, it gives $\rho N^{(1+w)/w}=C_1$.

The Tolman–Oppenheimer–Volkoff equation governs a static, spherically symmetric perfect fluid in general relativity. In geometrized units, if $m(r)$ is the mass inside areal radius $r$, then

The ratio of enclosed gravitational mass to areal radius in units $G=c=1$. In a spherical star the local value is $m(r)/r$, and the surface value is $M/R$. Bounds on this ratio constrain relativistic stellar models; it is unrelated to topological compactness.

For a smooth centre with density ${\rho }_c$ and pressure $p_c$, the Tolman–Oppenheimer–Volkoff equation gives $m(r)=4\pi {\rho }_c{r}^3/3+O(r^5)$ and $p(r)=p_c-2\pi ({\rho }_c+p_c)({\rho }_c/3+p_c)r^2+O(r^4)$. A fixed barotropic stellar equation of state leaves one physical central parameter; the additive lapse constant fixes time normalization.

The interior Schwarzschild metric is the exact static solution for a spherical fluid of constant density. If the star has mass $M$ and radius $R$, its pressure is

The Interior Schwarzschild metric has central pressureIt diverges when $M/R=4/9$, the limiting compactness in Buchdahl's theorem.

The dominant energy condition for the Interior Schwarzschild metric reduces to its largest pressure satisfying $p_c\le \rho$. Substitution of the exact central pressure gives $\sqrt{1-2M/R}\ge 1/2$, hence $M/R\le 3/8$. Equality has finite central pressure and is allowed by the non-strict condition.

Buchdahl's theorem gives $2M/R\le 8/9$ for a static spherical perfect-fluid star whose density is nonincreasing outwards, subject to its regularity assumptions. Equality would require divergent central pressure, so every regular such star satisfies $M/R<4/9$.

Let ${\rho }_0>0$ be a density threshold and let $a,\mu$ be the radius and enclosed mass where a monotonically decreasing stellar density reaches that threshold. Then $\mu \ge 4\pi {\rho }_0{a}^3/3$. A local compactness bound $\mu /a<4/9$ implies $a<(3\pi {\rho }_0{)}^{-1/2}$ and $\mu <4/(9\sqrt{3\pi {\rho }_0})$, independently of the equation of state above the threshold. A total-mass bound also requires suitable control of the known low-density envelope.

The null energy condition requires $T_{\mu \nu }{k}^{\mu }{k}^{\nu }\ge 0$ for every null vector $k$. Through the Einstein field equations this is equivalent to $R_{\mu \nu }{k}^{\mu }{k}^{\nu }\ge 0$ when the cosmological term is absent.

For $T_{ab}=(\rho +p)u_a{u}_b+p{g}_{ab}$ and nonzero null $k$, contraction gives $T_{ab}{k}^a{k}^b=(\rho +p)(u\cdot k{)}^2$. A timelike fluid velocity has nonzero scalar product with every nonzero null vector, so the null energy condition is exactly the displayed inequality. Individual nonnegativity of pressure or density is unnecessary for null focusing. A cosmological constant term has zero null contraction.

The classical stress-energy tensor of a canonical scalar field has a gradient-product term and a metric-proportional term. Contracting with a real null vector removes the latter, leaving the displayed square for a real field. This proves the null energy condition for a minimally coupled canonical scalar, including the massless case. It is not a statement about arbitrary curvature couplings, wrong-sign kinetic terms or a renormalized quantum stress-energy tensor.

In a planar warped spacetime, contraction of the Ricci tensor with a null vector in an orthonormal frame gives $R_{ab}{k}^a{k}^b=-2(\log A{)}^{\prime \prime }(k^3{)}^2$. The Einstein field equations identify this with $8\pi T_{ab}{k}^a{k}^b$; metric terms, including a cosmological constant, vanish in a null contraction. Thus the null energy condition is equivalent to concavity of $\log A$ in the proper transverse coordinate $z$. Null directions confined to a planar slice saturate the condition.

The null convergence condition requires $R_{ab}{k}^a{k}^b\ge 0$ for every null vector. Through the Einstein field equations, it follows from the null energy condition, since all terms proportional to the metric tensor vanish on a null vector.

Contract the Einstein field equations with a null vector. Both the trace term and the cosmological constant term multiply $g_{ab}{\ell }^a{\ell }^b=0$ and disappear. The null energy condition therefore still implies the null convergence condition, independently of the sign of the cosmological constant. A black-hole area theorem still needs the appropriate global horizon assumptions as well as this local focusing condition.

The averaged null energy condition requiresalong every complete affinely parametrized null geodesic. It is weaker than the pointwise null energy condition.

For the curvature-coupled scalar equation considered here, contraction with an affine null tangent and integration by parts gives, for $\xi <0$,when $\Phi {\Phi }^{\prime }/(1-\xi {\Phi }^2)$ vanishes at the endpoints.

For ${\nabla }_{\alpha }{\nabla }^{\alpha }\varphi =m^2\varphi$, considerCovariant conservation for all solutions holds exactly when

For a real scalar satisfying ${\nabla }_a{\nabla }^a\psi -{\mu }^2\psi =0$,is symmetric and covariantly conserved.

The weak energy condition requiresfor every timelike vector $X$, so every observer measures nonnegative energy density.

For a perfect fluid in general relativity in metric signature $(-,+,+,+)$, the weak energy condition is equivalent to $\rho \ge 0$ and $\rho +p\ge 0$. A normalized observer's four-velocity $W$ has Lorentz factor $\gamma =-u\cdot W\ge 1$ relative to the fluid, and $T_{ab}{W}^a{W}^b=\rho +(\rho +p)({\gamma }^2-1)$. The case $\gamma =1$ forces the first inequality and arbitrarily large $\gamma$ forces the second; conversely the same expression proves sufficiency.

For the general scalar ansatz in Conserved Klein-Gordon scalar stress-energy ansatz, work in a unit timelike frame. ThenThe weak energy condition for arbitrary local field data is therefore equivalent to

The strong energy condition requires $(T_{ab}-T{g}_{ab}/2)u^a{u}^b\ge 0$ for every timelike vector $u$ in four spacetime dimensions. Through the Einstein field equations, it implies the timelike convergence condition $R_{ab}{u}^a{u}^b\ge 0$ when the cosmological constant vanishes.

The dominant energy condition requires the energy flux measured by every future-directed timelike observer to be future-directed and causal. Equivalently, $T_{ab}{u}^a{v}^b\ge 0$ for every pair of future-directed causal vectors $u$ and $v$ under the $(-,+,+,+)$ convention.

For a scalar field with $X=-q^2/2$, Lagrangian $F(X)$ and stress-energy tensor $T_{ab}=F\prime q_a{q}_b+F{g}_{ab}$, the inequalities $A=F\prime \ge 0$ and $Y=XF\prime -F\ge 0$ imply the dominant energy condition for any gradient type. In a local Lorentz frame, $\rho =A(q_0^2+{\sum }_i{q}_i^2)/2+Y\ge 0$ and ${\rho }^2-{\sum }_i{T}_{0i}^2=A^2{X}^2+A(q_0^2+{\sum }_i{q}_i^2)Y+Y^2\ge 0$.