Canonical Stress-Energy Tensor

Derivation

Canonical Stress-Energy Tensor

D-368 Home PU-402 Threads energy · symmetry · fields Depends on Noether's First Theorem

Statement

For a field theory whose Lagrangian density \(\mathcal{L}(\phi_a,\partial_\mu\phi_a)\) carries no explicit dependence on the spacetime coordinates, invariance of the action under the four rigid translations \(x^\mu \to x^\mu + a^\mu\) yields, via Noether's first theorem, a rank-two conserved current — the canonical energy–momentum (stress–energy) tensor \[ T^{\mu\nu} \;=\; \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi_a)}\,\partial^\nu \phi_a \;-\; \eta^{\mu\nu}\,\mathcal{L}, \qquad \partial_\mu T^{\mu\nu} = 0, \] where a sum over the field label \(a\) is understood. Its conserved charges \(P^\nu = \int d^3x\, T^{0\nu}\) are the total energy (\(\nu=0\)) and momentum (\(\nu=i\)) carried by the field.

Why it matters

The stress–energy tensor is where "energy" and "momentum" acquire a local, covariant meaning for a continuous field. Rather than a single number \(E\), one has densities and fluxes packaged into one object: \(T^{00}\) is the energy density, \(T^{0i}\) the energy flux (equivalently \(c^2\) times the momentum density), and \(T^{ij}\) the momentum-flux, i.e. the stress. Conservation \(\partial_\mu T^{\mu\nu}=0\) is the continuity equation that makes total energy and momentum bookkeepable.

It is also the bridge to gravity: the (symmetrised, gauge-corrected) stress–energy tensor is precisely the source on the right-hand side of Einstein's equations. Understanding where the canonical tensor comes from — and why it must sometimes be improved — is the first step toward that coupling.

Assumptions

The action is invariant under spacetime translationsequivalently, \(\mathcal{L}\) has no explicit \(x^\mu\); if an external, position-dependent background is present the divergence acquires a source, \(\partial_\mu T^{\mu\nu} = -\,\partial^\nu \mathcal{L}\big|_{\text{explicit}}\ne 0\), and energy–momentum leaks to whatever fixes that background.

The fields obey the Euler–Lagrange equations (on-shell)the identity used in the derivation trades \(\partial\mathcal{L}/\partial\phi_a\) for a divergence; off-shell field configurations do not give a conserved current.

Fields and their derivatives fall off fast enough at spatial infinityneeded so that \(P^\nu=\int d^3x\,T^{0\nu}\) converges and its time derivative reduces to a vanishing surface term; without it the total charges need not be finite or constant even when the local law \(\partial_\mu T^{\mu\nu}=0\) holds.

The Lagrangian is first order (depends on \(\phi_a\) and \(\partial_\mu\phi_a\) only)higher-derivative theories carry extra Ostrogradsky momenta and the current formula gains additional terms \(\propto \partial\mathcal{L}/\partial(\partial\partial\phi_a)\).

Derivation

\[ \partial_\mu \mathcal{L} \;=\; \frac{\partial \mathcal{L}}{\partial \phi_a}\,\partial_\mu \phi_a \;+\; \frac{\partial \mathcal{L}}{\partial(\partial_\nu \phi_a)}\,\partial_\mu\partial_\nu \phi_a \]

Total spacetime derivative of \(\mathcal{L}\) by the chain rule. Legal because \(\mathcal{L}\) depends on \(x^\mu\) only through \(\phi_a\) and \(\partial_\nu\phi_a\) — the translation-invariance assumption. A

\[ \frac{\partial \mathcal{L}}{\partial \phi_a} \;=\; \partial_\nu\!\left( \frac{\partial \mathcal{L}}{\partial(\partial_\nu \phi_a)} \right) \]

Euler–Lagrange equation of motion; substitute it for the first term of Step 1. Legal on-shell. B

\[ \partial_\mu \mathcal{L} \;=\; \partial_\nu\!\left( \frac{\partial \mathcal{L}}{\partial(\partial_\nu \phi_a)} \right)\partial_\mu \phi_a \;+\; \frac{\partial \mathcal{L}}{\partial(\partial_\nu \phi_a)}\,\partial_\nu\partial_\mu \phi_a \;=\; \partial_\nu\!\left( \frac{\partial \mathcal{L}}{\partial(\partial_\nu \phi_a)}\,\partial_\mu \phi_a \right) \]

The two terms are exactly the product-rule expansion of a single divergence (using \(\partial_\nu\partial_\mu\phi_a = \partial_\mu\partial_\nu\phi_a\)). Reverse the Leibniz rule. B

\[ \partial_\mu \mathcal{L} \;=\; \partial_\nu\big(\delta^\nu_{\ \mu}\,\mathcal{L}\big) \quad\Longrightarrow\quad \partial_\nu\!\left( \frac{\partial \mathcal{L}}{\partial(\partial_\nu \phi_a)}\,\partial_\mu \phi_a \;-\; \delta^\nu_{\ \mu}\,\mathcal{L} \right) = 0 \]

Write the left side as a divergence via \(\partial_\mu = \delta^\nu_{\ \mu}\partial_\nu\), then move everything under one \(\partial_\nu\). Pure index bookkeeping. A

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

Name the conserved current. The free lower index \(\mu\) labels which of the four translations \(a^\mu\) we varied — this is Noether's first theorem applied four times over. B

\[ T^{\mu\nu} \;=\; \eta^{\nu\rho}\,T^{\mu}{}_{\rho} \;=\; \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi_a)}\,\partial^\nu \phi_a \;-\; \eta^{\mu\nu}\,\mathcal{L} \]

Raise the second index with the inverse metric \(\eta^{\nu\rho}\) (constant, so it passes through \(\partial_\mu\)). Conservation is preserved: \(\partial_\mu T^{\mu\nu}=0\). A

\[ P^\nu \;=\; \int d^3x\; T^{0\nu}, \qquad \frac{dP^\nu}{dt} \;=\; -\!\int d^3x\; \partial_i T^{i\nu} \;=\; -\oint_{S_\infty} T^{i\nu}\,dS_i \;=\; 0 \]

Integrate the \(\nu\)-current's time component over a spatial slice; the local law plus fall-off at infinity turns \(dP^\nu/dt\) into a vanishing surface flux (Noether charge conservation). C

Result

\[ \boxed{\,T^{\mu\nu} \;=\; \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi_a)}\,\partial^\nu \phi_a \;-\; \eta^{\mu\nu}\,\mathcal{L}, \qquad \partial_\mu T^{\mu\nu}=0, \qquad P^\nu=\int d^3x\,T^{0\nu}\,}\]

Reading. The first term is the "flux-of-field-momentum" piece — the canonical momentum density \(\pi^\mu_a=\partial\mathcal{L}/\partial(\partial_\mu\phi_a)\) contracted with the field's own gradient; the second subtracts the Lagrangian along the diagonal so that the \(\nu=0\) charge is the Hamiltonian. Explicitly \(T^{00}=\pi_a\dot\phi_a-\mathcal{L}=\mathcal{H}\) is the energy density, \(T^{0i}\) the momentum density (\(P^i=\int d^3x\,T^{0i}\)), and \(T^{ij}\) the stress: the \(i\)-momentum flowing in the \(j\)-direction. The single equation \(\partial_\mu T^{\mu\nu}=0\) is four continuity equations, one per conserved component of four-momentum.

Units check. In natural units \([\mathcal{L}]=[\text{energy}]^4\) (energy density in 4D), and \(\partial\mathcal{L}/\partial(\partial_\mu\phi_a)\) times \(\partial^\nu\phi_a\) has dimension \([\mathcal{L}]/[\partial\phi]\times[\partial\phi]=[\mathcal{L}]\); the \(\eta^{\mu\nu}\mathcal{L}\) term matches term-by-term. Restoring SI, every component has the dimension of energy density = \(\mathrm{J\,m^{-3}} = \mathrm{Pa}\): \(T^{00}\) is \(\mathrm{J\,m^{-3}}\), \(c\,T^{0i}\) is an energy flux \(\mathrm{W\,m^{-2}}\), and \(T^{ij}\) is a pressure/stress \(\mathrm{Pa}\). Consistent, since a momentum flux and an energy density carry the same units.

Limiting cases

Free real scalar \(\mathcal{L}=\tfrac12\partial_\mu\phi\,\partial^\mu\phi-\tfrac12 m^2\phi^2\): the formula gives \(T^{\mu\nu}=\partial^\mu\phi\,\partial^\nu\phi-\eta^{\mu\nu}\mathcal{L}\), already symmetric, with \(T^{00}=\tfrac12\dot\phi^2+\tfrac12(\nabla\phi)^2+\tfrac12 m^2\phi^2\) — the familiar energy density.
Massless plane wave along \(x\) (\(\omega=k\)): on-shell \(\mathcal{L}=0\), so \(T^{xx}=(\partial_x\phi)^2=T^{00}\) while \(T^{yy}=T^{zz}=0\) — pure radiation pressure directed along propagation.
Static, uniform field (\(\partial_\mu\phi=0\)): \(T^{\mu\nu}=-\eta^{\mu\nu}\mathcal{L}=-\eta^{\mu\nu}(-V)\), i.e. a cosmological-constant-like piece \(T^{\mu\nu}=\eta^{\mu\nu}V\) with \(p=-\rho\).
Non-relativistic point limit: \(\int d^3x\,T^{00}\to E=mc^2+\tfrac{p^2}{2m}+\dots\), recovering ordinary particle energy from the field's charge.

Breaks when

Explicit coordinate dependence. If \(\mathcal{L}=\mathcal{L}(\phi_a,\partial\phi_a,x)\) — an external potential, a time-dependent background, a fixed source — translation invariance is lost and \(\partial_\mu T^{\mu\nu}=-\partial^\nu\mathcal{L}\big|_{\text{explicit}}\). The field's four-momentum is exchanged with whatever holds the background fixed; only the closed total system conserves \(P^\nu\).
Fields with spin / gauge fields. For the electromagnetic field the canonical tensor \(T^{\mu\nu}_{\rm can}=-F^{\mu\alpha}\partial^\nu A_\alpha-\eta^{\mu\nu}\mathcal{L}\) is neither symmetric nor gauge-invariant. One must add an identically-conserved improvement term (Belinfante–Rosenfeld) built from the spin current to obtain the symmetric, gauge-invariant \(\Theta^{\mu\nu}\) that couples to gravity.
Off-shell or higher-derivative theories. If the equations of motion do not hold, Step 2 fails and there is no conservation. If \(\mathcal{L}\) contains \(\partial\partial\phi\), the current picks up extra terms and the naïve formula is wrong.

Failure modes

Dropping the \(-\eta^{\mu\nu}\mathcal{L}\) term. Without it \(T^{00}\) is not the Hamiltonian and \(\partial_\mu T^{\mu\nu}\ne0\); the diagonal subtraction is what makes it conserved.
Index-height slips. Writing \(\partial_\nu\phi\) where \(\partial^\nu\phi\) is required after raising, or forgetting that \(\partial^x=-\partial_x\) in the \((+,-,-,-)\) signature, flips signs in \(T^{ij}\).
Assuming symmetry. Taking \(T^{\mu\nu}=T^{\nu\mu}\) for granted. It holds for a single real scalar but fails for vector/tensor fields until improved; angular-momentum conservation is what forces symmetry.
Forgetting the sum over \(a\). Multi-component fields (complex scalar, gauge potential, spinor) contribute one \(\pi^\mu_a\partial^\nu\phi_a\) term each; using only one component undercounts the energy.
Sign of \(\delta\phi\). Using \(\delta\phi_a=+a^\nu\partial_\nu\phi_a\) instead of \(-a^\nu\partial_\nu\phi_a\) (the field is dragged back under an active translation) flips the whole tensor.
Confusing canonical momentum density \(\pi=\partial\mathcal{L}/\partial\dot\phi\) with \(T^{0i}\). \(T^{0i}\) is a spatial momentum density, not the conjugate momentum \(\pi\) to the field.

Discussion

The deep content of this result is that energy and momentum are the Noether charges of spacetime translation invariance. There is nothing more fundamental to "conservation of energy" than the statement that the laws of physics do not care what time it is; likewise momentum conservation is homogeneity of space. Noether's first theorem turns those symmetries into the four currents \(T^\mu{}_\nu\), and the Hamiltonian \(H=\int d^3x\,T^{00}\) emerges not as an assumption but as the charge generating time translations. This is why energy is conserved exactly when \(\mathcal{L}\) has no explicit \(t\)-dependence.

The tensor is called canonical because it is the one the variational machinery hands you directly — and it is frequently not the one you want. A conserved tensor is defined only up to an improvement term \(\partial_\rho B^{\rho\mu\nu}\) with \(B\) antisymmetric in \((\rho,\mu)\), which changes neither \(\partial_\mu T^{\mu\nu}\) nor the total charges \(P^\nu\) but can shift the local densities. The Belinfante–Rosenfeld procedure uses exactly this freedom, feeding in the spin current from Lorentz invariance, to produce a symmetric \(\Theta^{\mu\nu}\). Symmetry is not optional: it is what makes the angular-momentum current \(M^{\mu\nu\rho}=x^\nu\Theta^{\mu\rho}-x^\rho\Theta^{\mu\nu}\) conserved, tying orbital and spin angular momentum together.

The sharpest statement of the ambiguity is the comparison with the Hilbert (metric) stress tensor \(\Theta^{\mu\nu}=\tfrac{2}{\sqrt{-g}}\,\delta S/\delta g_{\mu\nu}\), obtained by coupling the theory to a curved background and varying the metric. For a symmetric-improvable theory this coincides with the Belinfante tensor and is automatically symmetric and gauge-invariant — it is the object general relativity sources. The canonical trace \(T^\mu{}_\mu\) is likewise physical: for the free scalar \(T^\mu{}_\mu=\partial_\mu\phi\,\partial^\mu\phi-4\mathcal{L}\), which does not vanish even for \(m=0\). A traceless (conformally invariant) tensor requires an additional curvature-improvement term \(\xi(\partial^\mu\partial^\nu-\eta^{\mu\nu}\Box)\phi^2\), and the failure of tracelessness at the quantum level is the trace anomaly.

Common misconceptions. (i) "The canonical stress tensor is unique." No — it is defined only up to superpotential improvements; the charges are unique, the densities are not. (ii) "\(T^{\mu\nu}\) is always symmetric." Only after improvement, or trivially for a single scalar. (iii) "\(\partial_\mu T^{\mu\nu}=0\) means energy is conserved even in an external field." No — an explicit \(x\)-dependence sources the divergence; conservation is a statement about a translationally-invariant, closed system.

Worked examples

Example 1 — Energy density of a massless-scalar plane wave.

\[ \mathcal{L}=\tfrac12\,\partial_\mu\phi\,\partial^\mu\phi \;=\; \tfrac12\dot\phi^2-\tfrac12(\nabla\phi)^2, \qquad T^{\mu\nu}=\partial^\mu\phi\,\partial^\nu\phi-\eta^{\mu\nu}\mathcal{L} \]

Free massless real scalar; insert into the boxed formula (single field, \(\partial\mathcal{L}/\partial(\partial_\mu\phi)=\partial^\mu\phi\)). A

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

Set \(\mu=\nu=0\), use \(\eta^{00}=+1\); this is the energy density \(\mathcal{H}\). A

\[ \phi=A\cos(\omega t-kx),\ \ \omega=k:\quad \dot\phi^2=A^2\omega^2\sin^2\theta,\ \ (\partial_x\phi)^2=A^2k^2\sin^2\theta \]

Plane-wave solution of \(\Box\phi=0\); differentiate, with \(\theta=\omega t-kx\). Symbols first. A

\[ \langle T^{00}\rangle=\tfrac12 A^2(\omega^2+k^2)\langle\sin^2\theta\rangle=\tfrac14 A^2(\omega^2+k^2)=\tfrac12 A^2\omega^2 \]

Time-average \(\langle\sin^2\rangle=\tfrac12\); massless \(k=\omega\). B

\[ A=2\times10^{-2}\ \text{eV},\quad \omega=5\ \text{eV}:\quad \langle T^{00}\rangle=\tfrac12(2\times10^{-2})^2(5)^2=5\times10^{-3}\ \text{eV}^4 \]

Insert numbers (natural units, \([\phi]=\)eV, \([\omega]=\)eV so \([T^{00}]=\)eV\(^4\)). A

\[ \langle T^{00}\rangle = 5\times10^{-3}\ \text{eV}^4 \approx 0.10\ \mathrm{J\,m^{-3}} \]

Reading. The mean energy density stored by the oscillating wave. Using \(1\ \text{eV}^4\approx 20.9\ \mathrm{J\,m^{-3}}\) (from \((\hbar c)^{-3}\)), this is about a tenth of a joule per cubic metre. The pressure along \(x\) is \(\langle T^{xx}\rangle=\tfrac12 A^2\omega^2=\langle T^{00}\rangle\) (equation of state \(p_x=\rho\)), transverse pressures zero.

Units check. eV\(\times\)eV = eV\(^2\) per factor \((\partial\phi)^2\), giving eV\(^4\); converting with \(1\,\text{eV}^4=20.9\,\mathrm{J\,m^{-3}}\) yields \(0.105\,\mathrm{J\,m^{-3}}\). Correct dimension of energy density.

Example 2 — Energy density of a homogeneous oscillating massive field (axion-like).

\[ \mathcal{L}=\tfrac12\dot\phi^2-\tfrac12 m^2\phi^2\quad(\nabla\phi=0),\qquad T^{00}=\tfrac12\dot\phi^2+\tfrac12 m^2\phi^2 \]

Spatially uniform massive scalar; \(T^{00}=\dot\phi^2-\mathcal{L}\) with no gradient term. A

\[ \phi(t)=\phi_0\cos(mt):\quad \dot\phi=-m\phi_0\sin(mt) \]

Solution of \(\ddot\phi+m^2\phi=0\) for the homogeneous mode. A

\[ T^{00}=\tfrac12 m^2\phi_0^2\sin^2(mt)+\tfrac12 m^2\phi_0^2\cos^2(mt)=\tfrac12 m^2\phi_0^2 \]

Add the kinetic and potential pieces; \(\sin^2+\cos^2=1\) — the density is exactly constant in time. B

\[ \langle T^{ii}\rangle=\langle\tfrac12\dot\phi^2-\tfrac12 m^2\phi^2\rangle=\langle\mathcal{L}\rangle=0 \]

The pressure is \(T^{ii}=-\eta^{ii}\mathcal{L}=\mathcal{L}\) here; \(\langle\sin^2\rangle=\langle\cos^2\rangle\) makes it average to zero — pressureless, matter-like (\(w=0\)). B

\[ m=1\times10^{-5}\ \text{eV},\quad \phi_0=1\times10^{9}\ \text{eV}:\quad \rho=\tfrac12(10^{-5})^2(10^{9})^2=5\times10^{7}\ \text{eV}^4 \]

Insert numbers; \([\rho]=[m]^2[\phi]^2=\)eV\(^4\). A

\[ \rho = T^{00} = \tfrac12 m^2\phi_0^2 = 5\times10^{7}\ \text{eV}^4, \qquad \langle p\rangle = 0 \]

Reading. A coherently oscillating massive scalar has a time-independent energy density set by amplitude and mass, and zero average pressure — it redshifts like non-relativistic matter, the mechanism behind cold scalar (axion) dark matter. The stress–energy tensor delivers both the density and the equation of state \(w=p/\rho=0\).

Units check. \([m^2\phi_0^2]=\text{eV}^2\cdot\text{eV}^2=\text{eV}^4\), an energy density; \(5\times10^{7}\,\text{eV}^4\approx1.0\times10^{9}\,\mathrm{J\,m^{-3}}\) using \(1\,\text{eV}^4=20.9\,\mathrm{J\,m^{-3}}\). Dimensionally an energy density, as required.

Problems

(Grade A) For the free complex scalar \(\mathcal{L}=\partial_\mu\phi^\ast\partial^\mu\phi-m^2\phi^\ast\phi\), write down the canonical stress–energy tensor and its energy density \(T^{00}\), treating \(\phi\) and \(\phi^\ast\) as independent fields.
Solution
Two field labels \(a\in\{\phi,\phi^\ast\}\). \(\partial\mathcal{L}/\partial(\partial_\mu\phi)=\partial^\mu\phi^\ast\) and \(\partial\mathcal{L}/\partial(\partial_\mu\phi^\ast)=\partial^\mu\phi\). Summing both contributions, \[ T^{\mu\nu}=\partial^\mu\phi^\ast\partial^\nu\phi+\partial^\mu\phi\,\partial^\nu\phi^\ast-\eta^{\mu\nu}\mathcal{L}. \] Then \(T^{00}=\dot\phi^\ast\dot\phi+\dot\phi\,\dot\phi^\ast-\mathcal{L}=2|\dot\phi|^2-(|\dot\phi|^2-|\nabla\phi|^2-m^2|\phi|^2)=|\dot\phi|^2+|\nabla\phi|^2+m^2|\phi|^2\ge0\), manifestly positive. The tensor is symmetric.
(Grade B) Show that the canonical stress tensor of the free real scalar is not traceless for \(m=0\) in general, by computing \(T^\mu{}_\mu\). Reconcile with Example 1, where a massless plane wave gave \(\mathcal{L}=0\).
Solution
\(T^\mu{}_\mu=\eta_{\mu\nu}T^{\mu\nu}=\partial_\mu\phi\,\partial^\mu\phi-\eta_{\mu\nu}\eta^{\mu\nu}\mathcal{L}=\partial_\mu\phi\,\partial^\mu\phi-4\mathcal{L}\) in 4D (\(\eta_{\mu\nu}\eta^{\mu\nu}=4\)). For massless \(\mathcal{L}=\tfrac12\partial_\mu\phi\,\partial^\mu\phi\), so \(T^\mu{}_\mu=2\mathcal{L}-4\mathcal{L}=-2\mathcal{L}\ne0\) in general. It vanishes only where \(\mathcal{L}=0\), which is exactly the on-shell null plane wave of Example 1 (\(\omega=k\Rightarrow\partial_\mu\phi\,\partial^\mu\phi=0\)). A genuinely traceless (conformal) tensor needs the improvement \(\tfrac16(\partial^\mu\partial^\nu-\eta^{\mu\nu}\Box)\phi^2\).
(Grade B) A real scalar sits in an external, static potential: \(\mathcal{L}=\tfrac12\partial_\mu\phi\,\partial^\mu\phi-U(x)\phi\), with \(U\) a fixed function of position. Compute \(\partial_\mu T^{\mu\nu}\) and interpret.
Solution
The Lagrangian now carries explicit \(x\)-dependence through \(U(x)\). Repeating the derivation, the on-shell divergence is the explicit derivative: \(\partial_\mu T^{\mu\nu}=-\partial^\nu\mathcal{L}\big|_{\text{explicit}}=+\,(\partial^\nu U)\,\phi\). Energy (\(\nu=0\), \(\partial^0 U=0\) since \(U\) is static) is still conserved, but spatial momentum is not: \(\partial_\mu T^{\mu i}=(\partial^i U)\phi\ne0\). The external potential exerts a force density on the field; only the field-plus-source system conserves total momentum.
(Grade C) For electromagnetism \(\mathcal{L}=-\tfrac14 F_{\alpha\beta}F^{\alpha\beta}\), obtain the canonical tensor and show it is neither symmetric nor gauge-invariant. State the improvement that fixes both.
Solution
With \(A_\alpha\) the fields, \(\partial\mathcal{L}/\partial(\partial_\mu A_\alpha)=-F^{\mu\alpha}\), so \[ T^{\mu\nu}_{\rm can}=-F^{\mu\alpha}\partial^\nu A_\alpha+\tfrac14\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}. \] It contains \(\partial^\nu A_\alpha\) rather than the gauge-invariant \(F^{\nu}{}_\alpha\), so it changes under \(A_\alpha\to A_\alpha+\partial_\alpha\Lambda\); and the first term is not \(\mu\leftrightarrow\nu\) symmetric. Add the improvement \(\partial_\alpha(F^{\mu\alpha}A^\nu)=(\partial_\alpha F^{\mu\alpha})A^\nu+F^{\mu\alpha}\partial_\alpha A^\nu\); on-shell \(\partial_\alpha F^{\mu\alpha}=0\), so this converts \(-F^{\mu\alpha}\partial^\nu A_\alpha\to -F^{\mu\alpha}F^\nu{}_\alpha\), giving \[ \Theta^{\mu\nu}=-F^{\mu\alpha}F^{\nu}{}_{\alpha}+\tfrac14\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}, \] symmetric, gauge-invariant, traceless, with \(\Theta^{00}=\tfrac12(E^2+B^2)\) and \(\Theta^{0i}=(\mathbf{E}\times\mathbf{B})^i\) the Poynting vector. This is the Belinfante–Rosenfeld improvement.
(Grade A, numeric) A homogeneous massive scalar oscillates with \(m=2\times10^{-6}\ \text{eV}\) and amplitude \(\phi_0=5\times10^{11}\ \text{eV}\). Find its energy density in eV\(^4\) and in \(\mathrm{J\,m^{-3}}\), and state its equation of state.
Solution
Symbolically \(\rho=\tfrac12 m^2\phi_0^2\) (constant in time), \(\langle p\rangle=0\Rightarrow w=0\). Numbers: \(m^2=(2\times10^{-6})^2=4\times10^{-12}\ \text{eV}^2\), \(\phi_0^2=(5\times10^{11})^2=2.5\times10^{23}\ \text{eV}^2\). So \(\rho=\tfrac12(4\times10^{-12})(2.5\times10^{23})=\tfrac12(10^{12})=5\times10^{11}\ \text{eV}^4\). Converting, \(5\times10^{11}\,\text{eV}^4\times20.9\ \mathrm{J\,m^{-3}\,eV^{-4}}\approx1.0\times10^{13}\ \mathrm{J\,m^{-3}}\). Equation of state \(w=p/\rho=0\): pressureless, matter-like.