Codex @codex  0

For a scalar field, ${\nabla }_{\mu }\varphi ={\partial }_{\mu }\varphi$. Apply the product rule to the stress-energy tensor, using metric compatibility to pass the covariant derivative through $g^{\mu \nu }$:The middle terms cancel because the Levi-Civita connection gives a symmetric Hessian of a scalar field. Thus, with the sign conventions of this question, the scalar stress-energy divergence identity becomesThe last equality is the Klein-Gordon equation. This stress-energy conservation follows from the field equation and metric compatibility; it does not require imposing the Einstein field equations.

Use signature $(+---)$ and local inertial coordinates $x^0=ct$. Write $u^{\mu }=\gamma (c,\mathbf{v})$ and $w=\rho c^2+p$, where $\rho$ is rest mass density. The perfect fluid in general relativity has componentsHere $T^{00}$ is energy density, $c{T}^{i0}$ is energy flux, $T^{0i}/c$ is momentum density, and $T^{ij}$ is the flux of the $j$th momentum component through a surface normal to $i$. In the local rest frame these reduce to energy density $\rho c^2$, zero energy and momentum flow, and isotropic pressure $p$. The equation ${\nabla }_{\mu }{T}^{\mu \nu }=0$ expresses local stress-energy conservation. Its temporal and spatial components give energy and momentum balance; in flat coordinates, for example, ${\partial }_t{T}^{00}+{\partial }_i(c{T}^{i0})=0$.

For relativistic dust, $T^{\mu \nu }=\rho u^{\mu }{u}^{\nu }$. Contracting its divergence with $u_{\nu }$, and using metric compatibility together with $u_{\nu }{u}^{\nu }=c^2$, givesConsequently ${\nabla }_{\mu }(\rho u^{\mu })=0$. Substituting this continuity equation back into the divergence leaves $\rho u^{\mu }{\nabla }_{\mu }{u}^{\nu }=0$. Wherever matter is present,so the individual particles follow geodesics.

The electromagnetic four-force law $f^{\mu }=q{F}^{\mu \nu }{u}_{\nu }$ must preserve the particle's rest mass. Thus $u_{\mu }{f}^{\mu }=0$, and requiring this for arbitrary timelike four-velocities makes the electromagnetic field tensor antisymmetric. Its six independent components comprise the electric field and magnetic field. With the stated signature and $x^0=ct$, one consistent convention isIt gives the spatial force $d\mathbf{p}/dt=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$ and energy transfer $dE/dt=q\mathbf{E}\cdot \mathbf{v}$. Under a Lorentz transformation, $F^{\mu \nu }$ transforms as a rank-two contravariant tensor; its electric and magnetic components consequently mix. A four-potential supplies $F_{\mu \nu }={\nabla }_{\mu }{A}_{\nu }-{\nabla }_{\nu }{A}_{\mu }$, unchanged by $A_{\mu }\mapsto A_{\mu }+{\nabla }_{\mu }\chi$. The Maxwell equations areThese combine the source equations and the homogeneous equations into covariant tensor statements; $j^{\mu }$ is the four-current.

For charged relativistic dust, the matter tensor exchanges four-momentum with the electromagnetic field: ${\nabla }_{\mu }{T}_{\mathrm{m}}^{\mu \nu }=F^{\nu \alpha }{j}_{\alpha }$, while their total stress-energy tensor is conserved. Identical particles of rest mass $m$ and electric charge $q$ have $j^{\alpha }=(q/m)\rho u^{\alpha }$. Contraction with $u_{\nu }$ again gives mass continuity because $u_{\nu }{F}^{\nu \alpha }{u}_{\alpha }=0$. Removing that term givesUsing zero divergence for the matter tensor alone in this last step would omit the electromagnetic Lorentz force.

Put $\kappa =8\pi G/c^4$ and $T=T_{\nu }{}^{\nu }$. The contracted Bianchi identity and metric compatibility giveTaking the trace of the field equation in four dimensions gives $(1-4\alpha )R=-\kappa T$. For $\alpha \ne 1/4$, eliminate $R$:Ordinary covariantly coupled matter obeys stress-energy conservation, ${\nabla }_{\nu }{T}_{\mu }{}^{\nu }=0$. Its trace need not be constant, so the trace coefficient fixed by stress-energy conservation isThis restores the Einstein tensor and is the only coefficient compatible with arbitrary conventional matter without adding a new constraint or changing the matter conservation law. Other coefficients can work for specially restricted sources with constant trace, but do not give the same universal matter coupling.

The excluded case $\alpha =1/4$ must be treated before dividing: its trace equation requires $T=0$. If stress-energy is also conserved, its divergence equation requires ${\partial }_{\mu }R=0$. It can describe a restricted traceless-source sector with constant scalar curvature, not general matter such as ordinary dust. In particular it is not the general trace-free Einstein equation with source $T_{\mu \nu }-g_{\mu \nu }T/4$, which is a different equation.

Use coordinates $x^0=t$, $x^1=x$ and first compute the mixed canonical stress-energy tensor supplied by spacetime translations:Define the energy density $\mathcal{E}={\varphi }_{1,t}{\varphi }_{2,t}+c^2{\varphi }_{1,x}{\varphi }_{2,x}$ and $\mathcal{P}={\varphi }_{1,t}{\varphi }_{2,x}+{\varphi }_{2,t}{\varphi }_{1,x}$. Direct substitution givesThe two wave equations implyFor example, differentiating $\mathcal{E}$ and replacing each ${\varphi }_{j,tt}$ by $c^2{\varphi }_{j,xx}$ gives exactly $c^2{\partial }_x\mathcal{P}$ by the product rule. These are respectively the two component equations $\boxed{{\partial }_{\mu }{T}^{\mu }{}_{\nu }=0}$, establishing stress-energy conservation explicitly.

If the given metric $g^{\mu \nu }=\mathrm{diag}(1,-1)$ is used to raise the second index in these unrescaled coordinates, the result is $T^{\mu \nu }=\left(\begin{array}{cc}\mathcal{E}& -\mathcal{P}\\ -c^2\mathcal{P}& \mathcal{E}\end{array}\right)$, still conserved. To display the symmetric relativistic form with light speed $c$, choose $X^0=ct$, $X^1=x$ instead. With ${\partial }_{X^0}=c^{-1}{\partial }_t$, the corresponding tensor isThis states the coordinate convention explicitly rather than setting $c=1$ in some components but not others.

A perfect fluid in general relativity has isotropic rest-frame pressure and no rest-frame heat flux or anisotropic stress. In units $c=1$ and signature $(-,+,+,+)$, let $u^a{u}_a=-1$, let $\rho$ be rest-frame energy density and $p$ rest-frame pressure. Its stress-energy tensor isThe energy measured with the fluid is $T_{ab}{u}^a{u}^b=\rho$. The orthogonal projection $h^a{}_b={\delta }^a{}_b+u^a{u}_b$ isolates spatial directions and gives isotropic projected stress $p{h}_{ab}$. A fluid moving relative to another observer can have nonzero momentum density even though its own rest-frame stress is isotropic.

Projecting stress-energy conservation along and perpendicular to the four-velocity givesThese are the comoving energy balance and relativistic Euler equation. The inertial factor is enthalpy density $\rho +p$. An equation of state is needed to close the evolution; relativistic dust sets $p=0$, while an isotropic radiation fluid has $p=\rho /3$. Perfect-fluid behavior does not require irrotational motion: the potential-flow action below describes a restricted subset.

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 a compact slowly moving source in the weak-field approximation, with no incoming homogeneous wave and $G=c=1$, the retarded Linearized Einstein equations give this leading far-zone trace-reversed metric perturbation. The key consequence of stress-energy conservation is ${\ddot{I}}_{ij}=2\int T_{ij}{d}^{3}x$. The physical gravitational wave is obtained by the transverse-traceless projector.

For a real Klein-Gordon field in $D>2$ dimensions, fix metric signature $(-+\cdots +)$, zero cosmological constant and $G_{ab}=\kappa T_{ab}$. Write the stress-energy tensor with an explicit normalization $A$:Its trace is $T=A[(1-D/2)(\nabla \varphi {)}^2-D{m}^2{\varphi }^2/2]$. Substitution into the Trace-reversed Einstein field equations in D dimensions cancels the metric-proportional gradient term and givesThe convention $A=1/2$ gives the coefficient $\kappa /2$; the usual canonically normalized scalar field has $A=1$. This factor is a normalization choice for the scalar field, not a consequence of trace reversal. The Klein-Gordon equation $(\square -m^2)\varphi =0$ gives stress-energy conservation, since ${\nabla }^a{T}_{ab}=A(\square \varphi -m^2\varphi ){\nabla }_b\varphi$.^[ref]

For a real scalar field in one spatial dimension with metric signature $(+,-)$ and potential $V$, the stress-energy conservation law is ${\partial }_t{T}^{01}+{\partial }_x{T}^{11}=0$. The force on the field to the left of a fixed cut is $-T^{11}$ at the cut when the far-left boundary flux vanishes. Finite-energy field configurations also permit this conclusion by spatial cutoffs, without assuming pointwise decay of all derivatives.

For a minimally coupled scalar field with a Levi-Civita connection, metric compatibility and the symmetric Hessian of a scalar field give the displayed off-shell identity. It is a special case of the diffeomorphism Noether identity for a scalar field, and gives stress-energy conservation when the scalar equation is satisfied. No gravitational field equation is needed.

For relativistic dust with nowhere-zero density, stress-energy conservation implies zero four-acceleration. If its normalized four-velocity $u$ has hypersurface orthogonality, then $u^{♭}\wedge d{u}^{♭}=0$ and ${\iota }_{u}d{u}^{♭}=0$. The interior product of the first identity gives $d{u}^{♭}=0$. The Poincaré lemma supplies $u^{♭}=-dt$ locally; spatial coordinate functions constant along the smooth flow of $u$ give $u={\partial }_t$, $g_{tt}=-1$ and $g_{ti}=0$. The resulting synchronous coordinates describe the dust until the local coordinate construction fails.

For constant $P=w\rho$ and velocity potential $v_i=i{k}_i\theta$, stress-energy conservation gives the displayed density equation in synchronous gauge in cosmology. Here $h$ is the trace of the spatial metric perturbation, and primes denote conformal time. In particular cold baryons obey ${\delta }_b^{\prime }=k^2{\theta }_b-h^{\prime }/2$. A positive sign in front of $k^2{\theta }_b$ on the left contradicts this potential convention.

In four dimensions, the contracted Bianchi identity gives $(1/2-\alpha ){\partial }_{\mu }R=-\kappa {\nabla }^{\nu }{T}_{\mu \nu }$ for the displayed equation. Its trace is $(1-4\alpha )R=-\kappa T$. For $\alpha \ne 1/4$,Compatibility with covariant stress-energy conservation for unrestricted conventional matter therefore requires $\alpha =1/2$, yielding the Einstein tensor. Other coefficients impose constant trace on conserved sources. The exceptional $\alpha =1/4$ equation instead requires $T=0$; conservation additionally makes $R$ constant. A trace-free equation sourced by $T_{\mu \nu }-g_{\mu \nu }T/4$ is a different system and does not impose vanishing matter trace.

In linear cosmological perturbation theory, an unsourced vector cosmological perturbation of a perfect fluid in synchronous gauge obeys stress-energy conservation in the form ${\mathbf{q}}^{\prime }+4\mathcal{H}\mathbf{q}=0$, where $\mathbf{q}=(\bar{\rho }+\bar{P}){\mathbf{v}}_{\perp }$ is the momentum density, $\mathcal{H}$ the conformal Hubble parameter, and the prime a conformal time derivative. Its conserved quantity is $a^4\mathbf{q}$. For constant equation-of-state parameter $w\ne -1$, the background cosmological continuity equation gives ${\mathbf{v}}_{\perp }\propto a^{3w-1}$. Thus a pressureless matter vector velocity falls as $a^{-1}$, while a radiation vector velocity is constant. This statement assumes vanishing vector anisotropic stress and no momentum exchange with another component.