Stress-Energy Tensor as the Conserved Source

Derivation

Stress-Energy Tensor as the Conserved Source

D-390 Home PU-403 Threads energy · symmetry · matter Depends on Contracted Bianchi Identity and the Einstein Tensor

Statement

For a field theory whose action is invariant under rigid spacetime translations, Noether's theorem produces a conserved rank-2 current, the canonical stress-energy tensor \( T^{\mu}{}_{\nu} \) satisfying \( \partial_\mu T^{\mu}{}_{\nu} = 0 \). This canonical object is generally neither symmetric nor gauge-invariant, but the Belinfante-Rosenfeld improvement adds an identically conserved superpotential term built from the spin current to yield a symmetric tensor \( T^{\mu\nu} = T^{\nu\mu} \) with the same conserved charges. Coupling the matter to a metric \( g_{\mu\nu} \) and defining \( T^{\mu\nu} \equiv \frac{2}{\sqrt{-g}}\,\frac{\delta S_{\text{m}}}{\delta g_{\mu\nu}} \) reproduces this symmetric tensor, and general covariance of \( S_{\text{m}} \) forces its covariant conservation \( \nabla_\mu T^{\mu\nu} = 0 \). This last identity is exactly what the contracted Bianchi identity \( \nabla_\mu G^{\mu\nu} = 0 \) demands of the Einstein source in \( G^{\mu\nu} = \kappa\, T^{\mu\nu} \).

Why it matters

The stress-energy tensor is the single object that tells spacetime how to curve: it is the entire right-hand side of Einstein's equations. Understanding that it is forced to be symmetric and covariantly conserved — not by fiat, but by the geometry of the left-hand side — is what makes general relativity self-consistent rather than an arbitrary coupling of two unrelated tensors.

The same tensor governs energy density, momentum density, momentum flux (stress), and their local conservation in every field theory from electromagnetism to the Standard Model. The passage from the canonical Noether current to the symmetric Hilbert tensor also resolves a century-old puzzle — why the "obvious" translation current is the wrong one to couple to gravity — and connects rotational (Lorentz) symmetry directly to the symmetry of \( T^{\mu\nu} \).

Assumptions

The action is a local functional of fields and their first derivatives, \( S=\int d^4x\,\mathcal{L}(\phi_a,\partial_\mu\phi_a) \).Without locality Noether's first theorem does not apply and no local conserved current exists; the charge might still be conserved but there is no continuity equation \( \partial_\mu T^\mu{}_\nu=0 \) to localize energy flow.

The Lagrangian has no explicit spacetime dependence, \( \partial_\mu\mathcal{L}|_{\text{explicit}}=0 \), so the four rigid translations \( x^\mu\to x^\mu+a^\mu \) are symmetries.If \( \mathcal{L} \) depends explicitly on \( x \) (e.g. an external time-dependent potential) the translation current acquires a source term \( \partial_\mu T^\mu{}_\nu=-\partial_\nu\mathcal{L}|_{\text{explicit}} \) and energy-momentum is not conserved.

The theory is also Lorentz invariant, giving a conserved total angular-momentum current whose orbital and spin parts combine.Drop Lorentz invariance and there is no reason for \( T^{\mu\nu} \) to be symmetric; the antisymmetric part \( T^{[\mu\nu]} \) equals the divergence of the spin current and need not vanish, so it cannot serve as the source of a symmetric metric.

The fields satisfy their Euler-Lagrange equations (the current is conserved on-shell).Off-shell the Noether identity retains an equation-of-motion term \( \frac{\delta S}{\delta\phi_a}\,\delta\phi_a \); conservation holds only when this vanishes, so \( \partial_\mu T^\mu{}_\nu=0 \) is a statement about physical solutions, not identities.

The matter action can be written covariantly on a curved background and is a scalar under diffeomorphisms.Without general covariance the Hilbert prescription \( T^{\mu\nu}=\frac{2}{\sqrt{-g}}\frac{\delta S_{\text{m}}}{\delta g_{\mu\nu}} \) is not diffeomorphism-covariant and its divergence need not vanish, breaking compatibility with \( \nabla_\mu G^{\mu\nu}=0 \).

Derivation

\[ x^\mu \to x'^\mu = x^\mu + a^\mu,\qquad \phi_a(x)\to\phi_a'(x')=\phi_a(x),\qquad \delta\phi_a=-a^\mu\,\partial_\mu\phi_a \]

Rigid translation acts on coordinates; a scalar field's value is carried along, so its total variation vanishes and its functional variation at fixed \( x \) is minus the transported gradient. A

\[ \delta S=\int d^4x\left[\frac{\partial\mathcal{L}}{\partial\phi_a}\delta\phi_a+\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_a)}\partial_\mu(\delta\phi_a)\right] \]

Vary the action; the chain rule on \( \mathcal{L}(\phi_a,\partial_\mu\phi_a) \) with summation over all fields \( a \). A

\[ \delta S=\int d^4x\left[\underbrace{\left(\frac{\partial\mathcal{L}}{\partial\phi_a}-\partial_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_a)}\right)}_{=\,0\ \text{on-shell}}\delta\phi_a+\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_a)}\delta\phi_a\right)\right] \]

Integrate the second term by parts. The bracket is the Euler-Lagrange expression and vanishes on solutions; only the total-derivative piece survives. B

\[ 0=\delta S=\int d^4x\,\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_a)}\,\delta\phi_a\right) \quad\Longrightarrow\quad \partial_\mu j^\mu=0 \]

For a symmetry \( \delta S=0 \) with the surface term as the only contribution, so the bracketed object is a conserved current. This is Noether's first theorem. B

\[ \delta\mathcal{L}=-a^\nu\partial_\nu\mathcal{L}=-\partial_\mu\!\left(a^\nu\,\delta^\mu{}_\nu\,\mathcal{L}\right) \]

Because \( \mathcal{L} \) has no explicit \( x \)-dependence, a translation changes it only through the fields, giving a total derivative. The right-hand side of the Noether balance is not zero but this surface term. B

\[ \partial_\mu\!\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_a)}\,(-a^\nu\partial_\nu\phi_a)\right)=\partial_\mu\!\left(-a^\nu\,\delta^\mu{}_\nu\,\mathcal{L}\right) \]

Equate the on-shell surface term from step 3 with the intrinsic change of \( \mathcal{L} \) from step 5; substitute \( \delta\phi_a=-a^\nu\partial_\nu\phi_a \). B

\[ a^\nu\,\partial_\mu\!\left[\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_a)}\,\partial_\nu\phi_a-\delta^\mu{}_\nu\,\mathcal{L}\right]=0 \quad\forall\,a^\nu \]

Move both terms to one side and factor the arbitrary constant \( a^\nu \). Since \( a^\nu \) is arbitrary, each of the four brackets is separately conserved. B

\[ \boxed{\,T^\mu{}_\nu\equiv\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_a)}\,\partial_\nu\phi_a-\delta^\mu{}_\nu\,\mathcal{L}\,},\qquad \partial_\mu T^\mu{}_\nu=0 \]

Define the canonical (Noether) stress-energy tensor as the conserved current for translations. Its four conserved charges are \( P_\nu=\int d^3x\,T^0{}_\nu \), the total energy-momentum. A

\[ T^{\mu\nu}_{\text{can}}=\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_a)}\,\partial^\nu\phi_a-\eta^{\mu\nu}\mathcal{L}\quad\text{is generally}\quad T^{\mu\nu}_{\text{can}}\neq T^{\nu\mu}_{\text{can}} \]

Raising the index shows the canonical tensor need not be symmetric whenever the fields carry spin (vector or tensor indices), since the derivative structure singles out \( \mu \) over \( \nu \). C

\[ \partial_\mu M^{\mu\nu\rho}=0,\qquad M^{\mu\nu\rho}=x^\nu T^{\mu\rho}_{\text{can}}-x^\rho T^{\mu\nu}_{\text{can}}+S^{\mu\nu\rho} \]

Lorentz invariance gives a conserved angular-momentum current with an orbital part and a spin part \( S^{\mu\nu\rho}=-S^{\mu\rho\nu} \). Its conservation is the key input. C

\[ 0=\partial_\mu M^{\mu\nu\rho}=T^{\rho\nu}_{\text{can}}-T^{\nu\rho}_{\text{can}}+\partial_\mu S^{\mu\nu\rho}\ \Longrightarrow\ T^{[\nu\rho]}_{\text{can}}=-\tfrac12\partial_\mu S^{\mu\nu\rho} \]

Differentiate the orbital terms using \( \partial_\mu x^\nu=\delta_\mu^\nu \) and \( \partial_\mu T^\mu{}_\nu=0 \). The antisymmetric part of the canonical tensor equals a spin-current divergence — nonzero for spinning fields. C

\[ T^{\mu\nu}\equiv T^{\mu\nu}_{\text{can}}+\partial_\lambda B^{\lambda\mu\nu},\qquad B^{\lambda\mu\nu}=\tfrac12\!\left(S^{\lambda\mu\nu}+S^{\mu\nu\lambda}+S^{\nu\mu\lambda}\right),\ B^{\lambda\mu\nu}=-B^{\mu\lambda\nu} \]

Belinfante-Rosenfeld improvement: add the divergence of a superpotential antisymmetric in its first pair. Antisymmetry guarantees \( \partial_\mu\partial_\lambda B^{\lambda\mu\nu}=0 \) identically, so conservation and the charges \( P^\nu \) are untouched. C

\[ \partial_\mu T^{\mu\nu}=0,\qquad T^{\mu\nu}=T^{\nu\mu} \]

The superpotential is engineered precisely to cancel \( T^{[\mu\nu]}_{\text{can}} \) from step 11, leaving a conserved symmetric tensor — the Belinfante tensor. C

\[ \delta S_{\text{m}}=\frac12\int d^4x\,\sqrt{-g}\;T^{\mu\nu}\,\delta g_{\mu\nu}\quad\Longrightarrow\quad T^{\mu\nu}=\frac{2}{\sqrt{-g}}\,\frac{\delta S_{\text{m}}}{\delta g_{\mu\nu}} \]

Hilbert definition: couple the matter to a metric and vary. This automatically yields a symmetric, gauge-invariant tensor equal to the Belinfante one, because \( g_{\mu\nu} \) is symmetric. C

\[ 0=\delta_{\xi}S_{\text{m}}=\frac12\int d^4x\,\sqrt{-g}\;T^{\mu\nu}\,(\nabla_\mu\xi_\nu+\nabla_\nu\xi_\mu)=\int d^4x\,\sqrt{-g}\;T^{\mu\nu}\nabla_\mu\xi_\nu \]

Under an infinitesimal diffeomorphism \( \delta g_{\mu\nu}=\nabla_\mu\xi_\nu+\nabla_\nu\xi_\mu \) (Lie derivative). A diffeomorphism-invariant scalar action must be stationary; symmetry of \( T^{\mu\nu} \) collapses the symmetrized derivative to one term. C

\[ 0=-\int d^4x\,\sqrt{-g}\;(\nabla_\mu T^{\mu\nu})\,\xi_\nu\quad\forall\,\xi_\nu\ \Longrightarrow\ \boxed{\nabla_\mu T^{\mu\nu}=0} \]

Integrate by parts (covariant Stokes theorem, boundary term dropped for compact-support \( \xi \)). Arbitrariness of \( \xi_\nu \) forces covariant conservation. This is the curved-space upgrade of \( \partial_\mu T^{\mu\nu}=0 \). C

\[ G^{\mu\nu}=\kappa\,T^{\mu\nu},\qquad \nabla_\mu G^{\mu\nu}\equiv 0\ \text{(contracted Bianchi)}\ \Longrightarrow\ \nabla_\mu T^{\mu\nu}=0 \]

The Einstein tensor obeys the contracted Bianchi identity identically (a geometric fact, proved in the assumed prior result). Consistency of the field equation then requires the source to be covariantly conserved — matching step 16 exactly. C

Result

\[ T^{\mu\nu}=\frac{2}{\sqrt{-g}}\,\frac{\delta S_{\text{m}}}{\delta g_{\mu\nu}}=T^{\nu\mu},\qquad \nabla_\mu T^{\mu\nu}=0\ \Longleftarrow\ \nabla_\mu G^{\mu\nu}\equiv 0 \]

Reading. Translation invariance produces a conserved canonical current, the stress-energy tensor. Lorentz invariance and the Belinfante-Rosenfeld improvement symmetrize it without changing its charges; the metric-variation (Hilbert) definition delivers the same symmetric object directly. General covariance of the matter action then forces \( \nabla_\mu T^{\mu\nu}=0 \) — precisely the identity the contracted Bianchi identity demands of the Einstein source. The geometry (left side of Einstein's equations) and the matter (right side) are mutually consistent by construction, not by coincidence.

Units check. In SI, \( T^{\mu\nu} \) has dimensions of energy density, \( \mathrm{J\,m^{-3}}=\mathrm{Pa}=\mathrm{kg\,m^{-1}\,s^{-2}} \); \( T^{00} \) is energy density, \( T^{0i}c \) is energy flux, \( T^{ij} \) is momentum flux (stress). The conservation law \( \partial_\mu T^{\mu\nu}=0 \) has dimensions of energy density per length, \( \mathrm{J\,m^{-4}} \), i.e. a force density \( \mathrm{N\,m^{-3}} \) once the \( c \)-weighted time index is included, so it reads as "rate of change of momentum density = minus divergence of stress" — Newton's second law for a continuum.

Limiting cases

Scalar field: \( S^{\mu\nu\rho}=0 \), so the canonical tensor is already symmetric and the Belinfante improvement vanishes — canonical, Belinfante, and Hilbert tensors coincide.
Flat spacetime, \( g_{\mu\nu}\to\eta_{\mu\nu} \): covariant derivatives reduce to partials and \( \nabla_\mu T^{\mu\nu}=0 \) becomes the special-relativistic continuity equation \( \partial_\mu T^{\mu\nu}=0 \).
Electromagnetism: the canonical \( T^{\mu\nu}_{\text{can}}=-F^{\mu\lambda}\partial^\nu A_\lambda-\eta^{\mu\nu}\mathcal{L} \) is neither symmetric nor gauge-invariant; the improvement adds \( \partial_\lambda(F^{\mu\lambda}A^\nu) \) to give the symmetric gauge-invariant \( T^{\mu\nu}=-F^{\mu\lambda}F^\nu{}_\lambda+\tfrac14\eta^{\mu\nu}F^2 \).
Perfect fluid: \( T^{\mu\nu}=(\rho+p/c^2)u^\mu u^\nu+p\,g^{\mu\nu} \); \( \nabla_\mu T^{\mu\nu}=0 \) reproduces relativistic Euler plus continuity.
Non-relativistic limit: \( T^{00}\to\rho c^2 \) dominates, \( T^{0i}/c\to \) momentum density, and conservation reduces to the mass-continuity and Cauchy momentum equations of continuum mechanics.

Breaks when

Explicit spacetime dependence in the Lagrangian. An external, time- or position-dependent background (a driven potential, a moving wall) breaks translation invariance; then \( \partial_\mu T^{\mu\nu}=-\partial^\nu\mathcal{L}|_{\text{explicit}}\neq 0 \) and energy-momentum leaks into the background. The tensor still exists but is not conserved.
Loss of Lorentz invariance / spin without improvement. For fields carrying spin, the raw canonical tensor is asymmetric, \( T^{[\mu\nu]}_{\text{can}}=-\tfrac12\partial_\lambda S^{\lambda\mu\nu}\neq 0 \). Using it as the gravitational source is inconsistent — it cannot equal the symmetric \( G^{\mu\nu} \). One must symmetrize first.
Quantum anomalies (broken scale/conformal invariance). The trace \( T^\mu{}_\mu \), classically zero for conformal matter, becomes nonzero at the quantum level (the trace anomaly \( T^\mu{}_\mu\propto \beta(g)\,F^2 \)); classical conservation still holds but the classically-expected tracelessness fails.
Nonlocal or higher-derivative actions. If \( \mathcal{L} \) depends on \( \partial\partial\phi \) or is nonlocal, Noether's first theorem gives extra terms; the simple \( T^\mu{}_\nu=\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\partial_\nu\phi-\delta^\mu_\nu\mathcal{L} \) is incomplete and can even be ill-defined.
Theories with torsion (Einstein-Cartan). The spin current sources torsion and the appropriate conservation law becomes \( \nabla_\mu T^{\mu\nu}=(\text{torsion}\times\text{spin}) \); the symmetric-conserved statement is modified because \( \nabla_\mu G^{\mu\nu}\neq 0 \) in the presence of torsion.

Failure modes

Confusing canonical with symmetric. Quoting \( T^{\mu\nu}_{\text{can}} \) for the electromagnetic field and then asserting it is symmetric or gauge-invariant — it is neither until improved.
Index placement in the canonical tensor. Writing \( \partial^\mu\phi\,\partial^\nu\phi \) but forgetting the \( -\eta^{\mu\nu}\mathcal{L} \) term, or lowering/raising the wrong index so the conserved index and the derivative index get swapped.
Sign of \( \delta\phi \). Taking \( \delta\phi=+a^\mu\partial_\mu\phi \) instead of \( -a^\mu\partial_\mu\phi \), which flips the sign of \( T^\mu{}_\nu \) and mislabels energy as negative.
Assuming the superpotential changes the charges. Believing the Belinfante term alters total energy-momentum; because it is a total divergence of an object antisymmetric in \( (\lambda\mu) \), \( P^\nu \) is unchanged.
Treating \( \nabla_\mu T^{\mu\nu}=0 \) as global energy conservation. In curved spacetime the covariant divergence does not integrate to a conserved total energy without a Killing vector; the connection terms represent exchange with the gravitational field.
Varying \( \sqrt{-g} \) incorrectly. Forgetting \( \delta\sqrt{-g}=\tfrac12\sqrt{-g}\,g^{\mu\nu}\delta g_{\mu\nu}=-\tfrac12\sqrt{-g}\,g_{\mu\nu}\delta g^{\mu\nu} \), producing a spurious \( -g^{\mu\nu}\mathcal{L} \) sign error in the Hilbert tensor.

Discussion

The deep lesson is that symmetry dictates source. Each continuous symmetry of the action yields a conserved current (Noether), and the four spacetime translations yield the four components of energy-momentum bundled into a single rank-2 tensor. That the correct gravitational source is symmetric is not an aesthetic choice: the metric \( g_{\mu\nu} \) is symmetric, so only the symmetric combination \( \frac{\delta S_{\text{m}}}{\delta g_{\mu\nu}} \) can couple to it. Lorentz invariance is what makes this symmetrization possible without spoiling conservation, tying the symmetry of \( T^{\mu\nu} \) to the conservation of angular momentum.

The interplay with the Bianchi identity is the keystone of general relativity's internal consistency. The Einstein tensor is built so that \( \nabla_\mu G^{\mu\nu}\equiv 0 \) is a geometric identity — it holds for any metric, on-shell or off. Setting \( G^{\mu\nu}=\kappa T^{\mu\nu} \) therefore does more than define a coupling: it predicts \( \nabla_\mu T^{\mu\nu}=0 \). Remarkably, the same conservation law follows independently from general covariance of the matter action alone. The two derivations agreeing is the statement that gravity and matter are compatible — you cannot write down a diffeomorphism-invariant matter action whose stress-energy is not conserved, and you cannot build a curvature scalar whose field equation demands anything other than a conserved source.

This also explains why the "wrong" (canonical) tensor was historically confusing. In pre-GR field theory one is free to use \( T^{\mu\nu}_{\text{can}} \) because only its conserved charges \( P^\nu \) are physical, and the improvement terms are invisible to them. But gravity couples to the local tensor, not just its integrated charges, and gravity sees the full symmetric, gauge-invariant Belinfante-Hilbert object. Localization of energy-momentum — which \( T^{0\nu} \) you call the energy density at a point — becomes physically meaningful precisely because gravity resolves the improvement ambiguity.

At the quantum level this classical tidiness is qualified. The trace \( T^\mu{}_\mu \), which vanishes classically for conformally invariant matter (massless, dimensionless couplings), acquires the trace anomaly \( \langle T^\mu{}_\mu\rangle=\frac{\beta(g)}{2g}F^{a}_{\mu\nu}F^{a\,\mu\nu}+\cdots \), signalling that scale invariance is broken by renormalization. Conservation \( \nabla_\mu T^{\mu\nu}=0 \) survives (diffeomorphism invariance is anomaly-free in four dimensions for sensible theories), but tracelessness does not. In curved backgrounds the expectation value \( \langle T^{\mu\nu}\rangle \) further requires careful regularization, and its finite part sources semiclassical gravity — the arena of Hawking radiation and cosmological particle creation. The classical identity we derived is thus the backbone onto which the subtler quantum structure is grafted.

Common misconceptions. (i) "The stress-energy tensor is unique." It is not: any \( T^{\mu\nu}+\partial_\lambda B^{\lambda\mu\nu} \) with \( B \) antisymmetric in \( (\lambda\mu) \) has the same charges; gravity's coupling to the metric selects the symmetric representative. (ii) "\( \nabla_\mu T^{\mu\nu}=0 \) means total energy is conserved in curved spacetime." No — a globally conserved energy requires a timelike Killing vector \( \xi^\mu \), giving \( \nabla_\mu(T^{\mu\nu}\xi_\nu)=0 \); generic dynamical spacetimes have none. (iii) "Energy is conserved because of the Bianchi identity." More precisely, local covariant conservation is required by the Bianchi identity for consistency, but the physical origin of the current is translation invariance of the matter action.

Worked examples

Real scalar field: build \( T^{\mu\nu} \) and evaluate the energy density for a plane wave. \[ \mathcal{L}=\tfrac12\,\partial_\mu\phi\,\partial^\mu\phi-\tfrac12 m^2\phi^2 \]

Start from the Klein-Gordon Lagrangian (metric signature \( +--- \)). A

\[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}=\partial^\mu\phi\quad\Rightarrow\quad T^\mu{}_\nu=\partial^\mu\phi\,\partial_\nu\phi-\delta^\mu{}_\nu\,\mathcal{L} \]

Insert into the canonical formula; a single scalar has no spin current, so this is already symmetric. A

\[ T^{00}=\tfrac12\dot\phi^2+\tfrac12(\nabla\phi)^2+\tfrac12 m^2\phi^2 \]

Take \( \mu=\nu=0 \); the energy density is manifestly positive — kinetic plus gradient plus mass terms. B

Numbers: massless wave \( \phi=A\cos(kx-\omega t) \), \( A=2.0\times10^{-3}\ \mathrm{units} \), \( \omega=1.0\times10^{15}\ \mathrm{s^{-1}} \), \( k=\omega/c \). \[ \langle\dot\phi^2\rangle=\tfrac12 A^2\omega^2,\quad \langle(\partial_x\phi)^2\rangle=\tfrac12 A^2 k^2=\tfrac12 A^2\omega^2/c^2 \]

Time-average of \( \sin^2 \) is \( \tfrac12 \); for a massless wave \( k=\omega/c \). B

\[ \langle T^{00}\rangle=\tfrac12\!\left(\tfrac12 A^2\omega^2\right)+\tfrac12\!\left(\tfrac12 A^2\omega^2/c^2\right)c^2=\tfrac12 A^2\omega^2 \]

In natural field units where kinetic and gradient terms both carry the same \( c \)-weighting, they add equally for a null wave, doubling the kinetic average. B

\[ \langle T^{00}\rangle=\tfrac12 A^2\omega^2=\tfrac12\,(2.0\times10^{-3})^2(1.0\times10^{15})^2=2.0\times10^{24}\ \text{(field units)} \]

Reading. The averaged energy density of a massless scalar wave scales as amplitude-squared times frequency-squared, exactly as for any harmonic field. The equal kinetic and gradient contributions are the field-theory signature of a null (light-like) excitation.

Electromagnetic plane wave: confirm the improved tensor is traceless and compute the radiation pressure. \[ T^{\mu\nu}=\frac{1}{\mu_0}\!\left(F^{\mu\lambda}F^\nu{}_\lambda-\tfrac14\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}\right) \]

The Belinfante-improved, gauge-invariant symmetric electromagnetic stress-energy tensor. B

\[ T^{00}=\tfrac12\!\left(\varepsilon_0 E^2+\tfrac{1}{\mu_0}B^2\right)=u,\qquad T^\mu{}_\mu=0 \]

The \( 00 \) component is the familiar EM energy density; the trace vanishes identically because \( \eta^\mu{}_\mu=4 \) cancels the \( -\tfrac14\cdot4 F^2 \) against \( F^{\mu\lambda}F_{\mu\lambda} \). Tracelessness reflects conformal invariance of free EM. C

\[ \text{For a wave along } x:\ T^{xx}=u,\quad T^{0x}=\frac{S_x}{c}=\frac{u\,c}{c}=u \]

For a plane wave \( E=cB \) and the Poynting flux \( S=uc \); the momentum flux (radiation pressure on a perfect absorber) equals the energy density. C

Numbers: solar constant \( I=1.36\times10^{3}\ \mathrm{W\,m^{-2}} \). \[ u=\frac{I}{c}=\frac{1.36\times10^{3}}{3.00\times10^{8}}=4.53\times10^{-6}\ \mathrm{J\,m^{-3}} \]

Energy density from intensity divided by \( c \); numerically evaluate. B

\[ P_{\text{abs}}=T^{xx}=u=4.53\times10^{-6}\ \mathrm{Pa},\qquad P_{\text{refl}}=2u=9.06\times10^{-6}\ \mathrm{Pa} \]

Radiation pressure on an absorber equals \( u \); on a mirror it doubles because momentum reverses. C

\[ P_{\text{abs}}=\frac{I}{c}=4.5\times10^{-6}\ \mathrm{Pa},\qquad P_{\text{refl}}=\frac{2I}{c}=9.1\times10^{-6}\ \mathrm{Pa} \]

Reading. The off-diagonal and spatial components of \( T^{\mu\nu} \) are not abstractions — \( T^{xx} \) is literally the pressure sunlight exerts, about a micropascal, the quantity that drives solar sails. Tracelessness \( T^\mu{}_\mu=0 \) is the covariant statement that free light has no rest frame.

Units check. \( I/c \) has units \( \mathrm{(W\,m^{-2})/(m\,s^{-1})}=\mathrm{J\,m^{-3}}=\mathrm{Pa} \), confirming energy density and pressure share dimensions as \( T^{\mu\nu} \) requires.

Problems

(A) Show that adding a total divergence \( \partial_\lambda B^{\lambda\mu\nu} \) with \( B^{\lambda\mu\nu}=-B^{\mu\lambda\nu} \) to \( T^{\mu\nu} \) leaves the conserved charges \( P^\nu=\int d^3x\,T^{0\nu} \) unchanged.
Solution
The change in charge is \( \Delta P^\nu=\int d^3x\,\partial_\lambda B^{\lambda 0\nu}=\int d^3x\,(\partial_0 B^{00\nu}+\partial_i B^{i0\nu}) \). By antisymmetry in the first pair, \( B^{00\nu}=0 \), killing the time term. The spatial term \( \int d^3x\,\partial_i B^{i0\nu} \) is a total spatial divergence, which by Gauss's theorem becomes a surface integral at spatial infinity and vanishes for fields decaying there. Hence \( \Delta P^\nu=0 \). The improvement is invisible to the charges.
(A) For \( \mathcal{L}=\tfrac12\partial_\mu\phi\,\partial^\mu\phi-V(\phi) \), compute \( T^\mu{}_\mu \) and state the condition on \( V \) for the trace to vanish for static configurations in 4D.
Solution
\( T^\mu{}_\nu=\partial^\mu\phi\,\partial_\nu\phi-\delta^\mu{}_\nu\mathcal{L} \). Trace: \( T^\mu{}_\mu=\partial^\mu\phi\,\partial_\mu\phi-4\mathcal{L}=\partial_\mu\phi\,\partial^\mu\phi-4(\tfrac12\partial_\mu\phi\,\partial^\mu\phi-V)=-\partial_\mu\phi\,\partial^\mu\phi+4V \). For a static field \( \partial_0\phi=0 \), \( \partial_\mu\phi\,\partial^\mu\phi=-(\nabla\phi)^2 \), so \( T^\mu{}_\mu=(\nabla\phi)^2+4V \). This vanishes only if \( (\nabla\phi)^2=-4V \), impossible for \( V\ge0 \) with nonconstant \( \phi \) — a massive/interacting scalar is not conformally invariant, consistent with the general rule that only special improved scalar theories are traceless.
(B) Starting from \( T^\mu{}_\nu=\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_a)}\partial_\nu\phi_a-\delta^\mu{}_\nu\mathcal{L} \), prove \( \partial_\mu T^\mu{}_\nu=0 \) on-shell for translation-invariant \( \mathcal{L} \).
Solution
\( \partial_\mu T^\mu{}_\nu=\partial_\mu\!\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_a)}\right)\partial_\nu\phi_a+\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_a)}\partial_\mu\partial_\nu\phi_a-\partial_\nu\mathcal{L} \). Use the Euler-Lagrange equation \( \partial_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_a)}=\frac{\partial\mathcal{L}}{\partial\phi_a} \) in the first term, giving \( \frac{\partial\mathcal{L}}{\partial\phi_a}\partial_\nu\phi_a+\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_a)}\partial_\nu\partial_\mu\phi_a \). This is exactly the chain-rule expansion of \( \partial_\nu\mathcal{L} \) (since \( \mathcal{L} \) has no explicit \( x \)). Therefore \( \partial_\mu T^\mu{}_\nu=\partial_\nu\mathcal{L}-\partial_\nu\mathcal{L}=0 \).
(B) A laser delivers \( P=5.0\ \mathrm{kW} \) focused to a spot of radius \( r=0.50\ \mathrm{mm} \) onto a perfect mirror. Using \( T^{xx} \), find the force on the mirror.
Solution
Intensity \( I=P/(\pi r^2)=5.0\times10^3/(\pi(5.0\times10^{-4})^2)=5.0\times10^3/(7.85\times10^{-7})=6.37\times10^{9}\ \mathrm{W\,m^{-2}} \). Energy density \( u=I/c=6.37\times10^9/3.00\times10^8=21.2\ \mathrm{J\,m^{-3}} \). For a mirror the momentum-flux pressure is \( P_{\text{rad}}=T^{xx}_{\text{in}}+T^{xx}_{\text{refl}}=2u=42.5\ \mathrm{Pa} \). Force \( F=P_{\text{rad}}\cdot\pi r^2=2I A/c=2P/c=2(5.0\times10^3)/(3.00\times10^8)=3.3\times10^{-5}\ \mathrm{N} \). The neat result \( F=2P/c \) shows the spot size cancels — total force depends only on total power.
(C) On a curved background with a Killing vector \( \xi^\nu \) (so \( \nabla_\mu\xi_\nu+\nabla_\nu\xi_\mu=0 \)), show that \( J^\mu=T^{\mu\nu}\xi_\nu \) is covariantly conserved and hence yields a genuinely conserved charge.
Solution
Compute \( \nabla_\mu J^\mu=\nabla_\mu(T^{\mu\nu}\xi_\nu)=(\nabla_\mu T^{\mu\nu})\xi_\nu+T^{\mu\nu}\nabla_\mu\xi_\nu \). The first term vanishes by \( \nabla_\mu T^{\mu\nu}=0 \). For the second, since \( T^{\mu\nu} \) is symmetric, \( T^{\mu\nu}\nabla_\mu\xi_\nu=T^{\mu\nu}\nabla_{(\mu}\xi_{\nu)}=\tfrac12 T^{\mu\nu}(\nabla_\mu\xi_\nu+\nabla_\nu\xi_\mu)=0 \) by the Killing equation. Hence \( \nabla_\mu J^\mu=0 \). Because \( J^\mu \) is a true vector current, \( \nabla_\mu J^\mu=\frac{1}{\sqrt{-g}}\partial_\mu(\sqrt{-g}\,J^\mu)=0 \), so \( Q=\int_\Sigma d^3x\,\sqrt{-g}\,J^0 \) is time-independent. This is why symmetry of \( T^{\mu\nu} \) plus a Killing vector — not conservation alone — is required for a globally conserved energy in curved spacetime.