Belinfante-Rosenfeld Improvement

Derivation

Belinfante-Rosenfeld Improvement

D-369 Home PU-402 Threads energy · symmetry · fields Depends on Canonical Stress-Energy Tensor, Noether's First Theorem

Statement

The canonical energy–momentum tensor \(T_c^{\mu\nu}\) obtained from translation invariance is conserved, \(\partial_\mu T_c^{\mu\nu}=0\), but is generally neither symmetric nor gauge invariant. Adding a divergence of a superpotential, \(T_B^{\mu\nu}=T_c^{\mu\nu}+\partial_\lambda B^{\lambda\mu\nu}\) with \(B^{\lambda\mu\nu}=-B^{\mu\lambda\nu}\), leaves the conserved charges and the conservation law untouched. Choosing the superpotential from the spin current \(S^{\lambda\mu\nu}\) fixed by Lorentz (rotation–boost) invariance, \(B^{\lambda\mu\nu}=\tfrac12\!\left(S^{\lambda\mu\nu}+S^{\mu\nu\lambda}+S^{\nu\mu\lambda}\right)\), yields the Belinfante–Rosenfeld tensor \(T_B^{\mu\nu}=T_B^{\nu\mu}\): symmetric, gauge invariant, and identical to the Hilbert tensor \(\tfrac{2}{\sqrt{-g}}\,\delta S/\delta g_{\mu\nu}\) that sources Einstein's equations.

Why it matters

Symmetry of the stress tensor is not optional book-keeping. It is the local statement of angular-momentum conservation, it is what makes \(T^{\mu\nu}\) a legitimate source in \(G^{\mu\nu}=8\pi G\,T^{\mu\nu}\) (the Einstein tensor is symmetric), and it is required for a gauge-invariant, physically meaningful momentum density. The canonical tensor for the electromagnetic field is famously asymmetric and gauge dependent; the Belinfante procedure is the systematic cure.

The construction shows that the "spin" carried by a field is not an independent bolt-on but is repackaged into the orbital moment \(x^{\mu}T_B^{\nu\lambda}-x^{\nu}T_B^{\mu\lambda}\) of a single symmetric tensor. Total angular momentum is unchanged; only its accounting is made covariant and geometric.

Assumptions

The action is a spacetime scalar built from local fields and first derivatives, \(\mathcal{L}(\phi_a,\partial_\mu\phi_a)\).Without Poincaré invariance there is no conserved \(T_c^{\mu\nu}\) and no spin current to improve; the whole Noether machinery collapses.

Global Poincaré invariance: the action is invariant under both translations and Lorentz transformations.Translations alone give a conserved but generally asymmetric \(T_c^{\mu\nu}\); it is Lorentz invariance that supplies \(S^{\lambda\mu\nu}\) and pins its antisymmetric part, so dropping it removes the object that fixes \(B^{\lambda\mu\nu}\).

The fields satisfy their Euler–Lagrange equations (on shell).Off shell the currents are not conserved and \(\partial_\lambda S^{\lambda\mu\nu}\) does not equal the antisymmetric part of \(T_c\); the improvement then fails to symmetrise the tensor.

Fields transform in a finite-dimensional Lorentz representation with generators \((\Sigma^{\mu\nu})_{ab}\).For higher-derivative or non-linearly-realised symmetries the spin current acquires extra terms; the compact three-term superpotential must be rederived and no longer takes this closed form.

Improvement terms are pure divergences of a local \(B^{\lambda\mu\nu}\) antisymmetric in its first two indices.If \(B\) lacks that antisymmetry, \(\partial_\mu\partial_\lambda B^{\lambda\mu\nu}\neq0\) and the "improved" tensor is no longer conserved, so it cannot represent energy–momentum at all.

Derivation

\[ T_c^{\mu\nu}=\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_a)}\,\partial^{\nu}\phi_a-\eta^{\mu\nu}\mathcal{L},\qquad \partial_\mu T_c^{\mu\nu}=0. \]

Import the canonical stress–energy tensor and its conservation from Noether's first theorem applied to the four rigid translations \(\delta\phi_a=-a^{\nu}\partial_\nu\phi_a\). A

\[ T_c^{\mu\nu}-T_c^{\nu\mu}\;\neq\;0 \quad\text{in general}. \]

A term-by-term inspection shows no reason for the two open indices — one from \(\partial\mathcal{L}/\partial(\partial_\mu\phi)\), one from \(\partial^{\nu}\phi\) — to be interchangeable. The asymmetry is the obstruction we must remove. A

\[ M^{\lambda\mu\nu}=x^{\mu}T_c^{\lambda\nu}-x^{\nu}T_c^{\lambda\mu}+S^{\lambda\mu\nu},\qquad S^{\lambda\mu\nu}=-\frac{\partial\mathcal{L}}{\partial(\partial_\lambda\phi_a)}\,(\Sigma^{\mu\nu})_{ab}\,\phi_b. \]

Noether's theorem for the six Lorentz transformations \(\delta\phi_a=-\tfrac12\omega_{\mu\nu}\big(x^{\mu}\partial^{\nu}-x^{\nu}\partial^{\mu}\big)\phi_a-\tfrac12\omega_{\mu\nu}(\Sigma^{\mu\nu})_{ab}\phi_b\) gives a conserved angular-momentum current: an orbital piece plus the intrinsic spin current \(S^{\lambda\mu\nu}=-S^{\lambda\nu\mu}\). B

\[ 0=\partial_\lambda M^{\lambda\mu\nu}=\big(T_c^{\mu\nu}-T_c^{\nu\mu}\big)+\partial_\lambda S^{\lambda\mu\nu}\;\Longrightarrow\; T_c^{\mu\nu}-T_c^{\nu\mu}=-\,\partial_\lambda S^{\lambda\mu\nu}. \]

Differentiate \(M^{\lambda\mu\nu}\): \(\partial_\lambda(x^{\mu}T_c^{\lambda\nu})=\delta^{\mu}_{\lambda}T_c^{\lambda\nu}+x^{\mu}\partial_\lambda T_c^{\lambda\nu}=T_c^{\mu\nu}\) using \(\partial_\lambda T_c^{\lambda\nu}=0\); likewise for the second term. Conservation of \(M\) then equates the antisymmetric part of \(T_c\) to \(-\partial_\lambda S^{\lambda\mu\nu}\). This is the key identity. B

\[ T_B^{\mu\nu}=T_c^{\mu\nu}+\partial_\lambda B^{\lambda\mu\nu},\qquad B^{\lambda\mu\nu}=-B^{\mu\lambda\nu}\;\Rightarrow\;\partial_\mu T_B^{\mu\nu}=\partial_\mu T_c^{\mu\nu}=0. \]

Any superpotential antisymmetric in its first two indices adds a term \(\partial_\mu\partial_\lambda B^{\lambda\mu\nu}\) that vanishes identically (symmetric \(\partial_\mu\partial_\lambda\) contracted with antisymmetric \(B\)). Conservation and the total charge \(P^{\nu}=\int d^3x\,T^{0\nu}\) are therefore preserved for any such \(B\). B

\[ B^{\lambda\mu\nu}=\tfrac12\!\left(S^{\lambda\mu\nu}+S^{\mu\nu\lambda}+S^{\nu\mu\lambda}\right). \]

Demand two things at once. (i) Antisymmetry in \(\lambda\mu\): swapping and adding, \(B^{\lambda\mu\nu}+B^{\mu\lambda\nu}\) pairs each term with its \(\nu\!\leftrightarrow\!\)(last) partner and cancels by \(S^{\,\cdot\,\alpha\beta}=-S^{\,\cdot\,\beta\alpha}\). (ii) The antisymmetric-in-\(\mu\nu\) part reproduces the spin divergence. This unique combination of the three cyclic orderings of \(S\) is the Belinfante superpotential. C

\[ B^{\lambda\mu\nu}-B^{\lambda\nu\mu}=S^{\lambda\mu\nu}\;\Rightarrow\;\partial_\lambda B^{\lambda[\mu\nu]}=\tfrac12\,\partial_\lambda S^{\lambda\mu\nu}=-\,T_c^{[\mu\nu]}. \]

Antisymmetrising the superpotential over \(\mu\nu\): the \(S^{\mu\nu\lambda}\) and \(S^{\nu\mu\lambda}\) terms cancel and \(S^{\lambda\mu\nu}-S^{\lambda\nu\mu}=2S^{\lambda\mu\nu}\) survives, giving \(B^{\lambda[\mu\nu]}=\tfrac12 S^{\lambda\mu\nu}\). Its divergence is exactly minus the antisymmetric part of \(T_c\) from Step 4. C

\[ T_B^{[\mu\nu]}=T_c^{[\mu\nu]}+\partial_\lambda B^{\lambda[\mu\nu]}=T_c^{[\mu\nu]}-T_c^{[\mu\nu]}=0. \]

Add the improvement of Step 7 to \(T_c\): the antisymmetric parts cancel term for term. The Belinfante tensor is symmetric, and being a sum of \(T_c\) and a divergence, it remains conserved. B

Result

\[ \boxed{\;T_B^{\mu\nu}=T_c^{\mu\nu}+\tfrac12\,\partial_\lambda\!\left(S^{\lambda\mu\nu}+S^{\mu\nu\lambda}+S^{\nu\mu\lambda}\right),\qquad T_B^{\mu\nu}=T_B^{\nu\mu},\quad \partial_\mu T_B^{\mu\nu}=0.\;} \]

Reading. The intrinsic spin content \(S^{\lambda\mu\nu}\) is folded back into the stress tensor as a total divergence. The result is a single symmetric, conserved, gauge-invariant tensor whose orbital moment \(x^{\mu}T_B^{\nu\lambda}-x^{\nu}T_B^{\mu\lambda}\) already carries the full angular momentum — spin and orbital are unified. Because it is symmetric and gauge invariant, \(T_B^{\mu\nu}\) coincides with the Hilbert tensor obtained by varying the action with respect to the metric, and it is this tensor, not \(T_c\), that couples to gravity.

Units check. \(T_c^{\mu\nu}\) has dimension of energy density, \([\mathrm{J\,m^{-3}}]=[\mathrm{Pa}]\). The spin current \(S^{\lambda\mu\nu}\sim x\cdot T\) carries an extra length, \([\mathrm{J\,m^{-2}}]\); the derivative \(\partial_\lambda\) removes it, so \(\partial_\lambda B^{\lambda\mu\nu}\) is again \([\mathrm{J\,m^{-3}}]\), matching \(T_c^{\mu\nu}\). Dimensions are consistent.

Limiting cases

Scalar field. \((\Sigma^{\mu\nu})=0\Rightarrow S^{\lambda\mu\nu}=0\), so \(T_B^{\mu\nu}=T_c^{\mu\nu}\): the canonical tensor is already symmetric and needs no improvement.
Electromagnetism. \(S^{\lambda\mu\nu}=F^{\mu\lambda}A^{\nu}-F^{\nu\lambda}A^{\mu}\) gives \(\partial_\lambda B^{\lambda\mu\nu}=\partial_\lambda(F^{\mu\lambda}A^{\nu})\), converting the asymmetric, gauge-dependent \(T_c^{\mu\nu}\) into \(T_B^{\mu\nu}=-F^{\mu\alpha}F^{\nu}{}_{\alpha}+\tfrac14\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}\), the symmetric gauge-invariant Maxwell tensor.
Dirac field. The spin current \(S^{\lambda\mu\nu}=-\tfrac{i}{4}\bar\psi\{\gamma^{\lambda},\sigma^{\mu\nu}\}\psi\) is totally antisymmetric; Belinfante symmetrisation reproduces the standard \(T_B^{\mu\nu}=\tfrac{i}{4}\bar\psi(\gamma^{\mu}\!\overset{\leftrightarrow}{\partial^{\nu}}+\gamma^{\nu}\!\overset{\leftrightarrow}{\partial^{\mu}})\psi\).
Flat-space / no gravity. Only \(P^{\nu}=\int T^{0\nu}d^3x\) and \(J^{\mu\nu}\) are observable; since improvements are pure divergences, \(T_c\) and \(T_B\) give identical total charges and are physically equivalent for global quantities.

Breaks when

Higher-derivative or non-minimal Lagrangians. When \(\mathcal{L}\) depends on \(\partial\partial\phi\) or the symmetry is realised non-linearly, the spin current picks up additional terms and the three-term superpotential no longer symmetrises \(T_c\); the improvement must be rederived from scratch.
Local (gauge) or conformal improvements confused with Belinfante. In gravitational or conformal contexts one may also need to add Weyl/curvature-improvement terms (e.g. \(\propto(\partial^{\mu}\partial^{\nu}-\eta^{\mu\nu}\Box)\phi^2\)) to make the trace vanish; the Belinfante procedure fixes only the antisymmetric part, not the trace, so a conformally-improved tensor requires a further, distinct step.
Off shell / with sources. If the equations of motion fail (external currents, constraints not imposed), \(\partial_\lambda S^{\lambda\mu\nu}\neq T_c^{[\nu\mu]}\cdot 2\) and the antisymmetric parts no longer cancel, so \(T_B\) is neither symmetric nor conserved.
Torsionful spacetime. If matter couples to an independent spin connection (Einstein–Cartan gravity), spin sources torsion and cannot be fully absorbed into a symmetric metric tensor; the Belinfante identification with the Hilbert tensor breaks and \(S^{\lambda\mu\nu}\) becomes an independent geometric source.

Failure modes

Symmetrising by hand. Replacing \(T_c^{\mu\nu}\) with its symmetric part \(\tfrac12(T_c^{\mu\nu}+T_c^{\nu\mu})\). This changes the conserved charges and is generally not conserved — the improvement must be a specific divergence, not an ad hoc average.
Wrong index symmetry of \(B\). Using a \(B^{\lambda\mu\nu}\) that is not antisymmetric in \(\lambda\mu\). Then \(\partial_\mu\partial_\lambda B^{\lambda\mu\nu}\neq0\) and conservation is destroyed.
Dropping the on-shell condition. Trying to prove \(T_B^{[\mu\nu]}=0\) off shell; the cancellation in Step 8 relies on \(\partial_\lambda T_c^{\lambda\nu}=0\), which holds only on the equations of motion.
Sign/ordering slips in the three cyclic \(S\) terms. Mis-ordering the indices of \(S^{\mu\nu\lambda}\) versus \(S^{\nu\mu\lambda}\); only the specific combination cancels the antisymmetric part — a swapped index gives a non-symmetric \(T_B\).
Confusing spin current with spin density. Reading \(S^{\lambda\mu\nu}\) as "the spin" rather than a current whose divergence matters; the physical spin density is \(S^{0\mu\nu}\).
Assuming EM \(T_c\) is already the physical tensor. Quoting \(T_c^{\mu\nu}=-F^{\mu\alpha}\partial^{\nu}A_{\alpha}+\ldots\) as the stress tensor — it is gauge dependent and asymmetric and gives a wrong momentum density.

Discussion

The deep content of the Belinfante–Rosenfeld construction is that a field theory does not hand you a unique local energy–momentum tensor. The canonical tensor is defined only up to the addition of an identically-conserved "improvement" divergence, an ambiguity that leaves all Poincaré charges invariant. Physics selects among these representatives by an external criterion: coupling to gravity. General relativity sources spacetime curvature with a symmetric tensor obtained by metric variation, \(T_H^{\mu\nu}=\tfrac{2}{\sqrt{-g}}\,\delta S/\delta g_{\mu\nu}\), and the theorem's punchline is that \(T_B^{\mu\nu}=T_H^{\mu\nu}\). The abstract requirement "be a good gravitational source" and the algebraic requirement "be symmetric" are the same requirement.

Symmetry is equivalent to local angular-momentum conservation. Once \(T_B\) is symmetric, the angular-momentum current is purely orbital, \(M_B^{\lambda\mu\nu}=x^{\mu}T_B^{\lambda\nu}-x^{\nu}T_B^{\lambda\mu}\), and \(\partial_\lambda M_B^{\lambda\mu\nu}=T_B^{\mu\nu}-T_B^{\nu\mu}=0\) follows automatically. The spin has not disappeared — it has been re-expressed as the moment of a redistributed momentum density. This is why the electromagnetic Poynting vector, the "correct" momentum density, emerges only after improvement: the canonical momentum density is gauge dependent and physically wrong at a point, even though its integral is right.

At the level of the covariant phase space, the improvement ambiguity is precisely the freedom to add exact forms to the Noether current — a cohomological statement. In the first-order (Palatini/tetrad) formulation the two notions genuinely split: the metric tensor \(\delta S/\delta e^{a}_{\mu}\) is Belinfante's symmetric object, while \(\delta S/\delta\omega^{ab}_{\mu}\) is the spin current, which sources torsion in Einstein–Cartan theory. There the Belinfante identity \(\partial_\lambda S^{\lambda\mu\nu}=T_c^{[\nu\mu]}\times 2\) is nothing but the on-shell content of the torsion equation, and the "improvement" that symmetrises \(T_c\) is exactly the geometric contribution of torsion integrated out. Belinfante–Rosenfeld is thus the flat-space shadow of the tetrad–spin-connection structure of gravity.

Common misconceptions. (1) "The improvement changes the physics." It does not change any measurable charge or the gravitational coupling; it changes only the unobservable local distribution, selecting the representative that is symmetric and gauge invariant. (2) "Belinfante makes the tensor traceless / conformal." No — it fixes the antisymmetric part only. Making the trace vanish (for a conformal theory) is an independent improvement. (3) "Spin is destroyed." Spin angular momentum is conserved throughout; it is merely re-bookkept from an explicit \(S^{\lambda\mu\nu}\) into the orbital moment of \(T_B\).

Worked examples

Example 1 — Electromagnetic plane wave: canonical vs symmetric momentum density.

\[ \vec{E}=E_0\cos(kz-\omega t)\,\hat{x},\qquad \vec{B}=\frac{E_0}{c}\cos(kz-\omega t)\,\hat{y}. \]

A linearly polarised vacuum wave travelling in \(+z\). Choose \(E_0=3.0\times10^{6}\ \mathrm{V\,m^{-1}}\). A

\[ T_B^{00}=u=\tfrac12\varepsilon_0\!\left(E^2+c^2B^2\right)=\varepsilon_0 E^2,\qquad \langle u\rangle=\tfrac12\varepsilon_0 E_0^2. \]

The symmetric Maxwell tensor gives the standard energy density; time-averaging \(\cos^2\) gives the factor \(\tfrac12\). A

\[ T_B^{0z}=c\,g_z=(\varepsilon_0\vec{E}\times\vec{B})_z\,c=\frac{u}{1}\;\Rightarrow\;\langle g_z\rangle=\frac{\langle u\rangle}{c}. \]

For a plane wave the symmetric (Belinfante/Poynting) momentum density equals energy density over \(c\); it is manifestly gauge invariant. B

\[ T_c^{0z}-T_B^{0z}=\partial_\lambda B^{\lambda 0z}=\partial_\lambda\!\left(F^{0\lambda}A^{z}\right),\qquad \int d^3x\,\partial_i(F^{0i}A^{z})=0. \]

The canonical momentum density differs from the physical one by a total spatial divergence (a boundary term for a localised packet), so the total momentum is identical while the pointwise density differs and is gauge dependent. B

\[ \langle u\rangle=\tfrac12(8.854\times10^{-12})(3.0\times10^{6})^2=39.8\ \mathrm{J\,m^{-3}},\quad \langle g_z\rangle=\frac{39.8}{3.0\times10^{8}}. \]

Insert numbers. A

\[ \langle u\rangle\approx 39.8\ \mathrm{J\,m^{-3}},\qquad \langle g_z\rangle\approx 1.33\times10^{-7}\ \mathrm{kg\,m^{-2}\,s^{-1}}. \]

Reading. Belinfante improvement replaces the gauge-dependent canonical momentum density with the Poynting result \(g=u/c\); the total momentum in any packet is unchanged because the correction integrates to zero.

Units check. \(\varepsilon_0 E^2\) has units \(\mathrm{(F\,m^{-1})(V\,m^{-1})^2}=\mathrm{J\,m^{-3}}\); dividing by \(c\ [\mathrm{m\,s^{-1}}]\) gives \(\mathrm{J\,s\,m^{-4}}=\mathrm{kg\,m^{-2}\,s^{-1}}\), a momentum density. Consistent.

Example 2 — Crossed static fields: field momentum from the symmetric tensor.

\[ \vec{E}=E_x\,\hat{x}=1.0\times10^{5}\ \mathrm{V\,m^{-1}}\,\hat{x},\qquad \vec{B}=B_y\,\hat{y}=0.50\ \mathrm{T}\,\hat{y}. \]

A region of uniform crossed static fields (e.g. overlap of a capacitor and a solenoid). We compute the momentum density stored in the field. A

\[ T_c^{0i}=-F^{0\alpha}\partial^{i}A_{\alpha}\quad\text{(gauge dependent, asymmetric)}. \]

The canonical mixed component depends on the vector potential explicitly and cannot be the physical momentum density; for static fields with a suitable gauge it can even be made to vanish, which is clearly wrong physics. B

\[ g_i=\frac{T_B^{0i}}{c}=\varepsilon_0(\vec{E}\times\vec{B})_i,\qquad \vec{E}\times\vec{B}=E_xB_y\,(\hat{x}\times\hat{y})=E_xB_y\,\hat{z}. \]

The symmetric Belinfante tensor gives the gauge-invariant field momentum density \(\vec{g}=\varepsilon_0\vec{E}\times\vec{B}\), directed along \(+z\). B

\[ g_z=\varepsilon_0 E_x B_y=(8.854\times10^{-12})(1.0\times10^{5})(0.50). \]

Insert numbers. A

\[ g_z\approx 4.4\times10^{-7}\ \mathrm{kg\,m^{-2}\,s^{-1}}\ \hat{z}. \]

Reading. Although nothing moves, the crossed static fields carry a real momentum density along \(\hat{z}\). Only the symmetric, gauge-invariant Belinfante tensor exposes it; the canonical tensor hides it in a gauge artefact. This is the field momentum that balances the "hidden mechanical momentum" of the sources.

Units check. \(\varepsilon_0[\mathrm{F\,m^{-1}}]\cdot E[\mathrm{V\,m^{-1}}]\cdot B[\mathrm{T}]\); using \(\mathrm{F\,V=C}\), \(\mathrm{T=kg\,s^{-1}\,C^{-1}\,m^{0}}\)… combining gives \(\mathrm{kg\,m^{-2}\,s^{-1}}\), a momentum density. Consistent.

Problems

Conservation of the improvement (grade: A). Show explicitly that for any \(B^{\lambda\mu\nu}=-B^{\mu\lambda\nu}\), the tensor \(T_B^{\mu\nu}=T_c^{\mu\nu}+\partial_\lambda B^{\lambda\mu\nu}\) satisfies \(\partial_\mu T_B^{\mu\nu}=\partial_\mu T_c^{\mu\nu}\).
Solution
\(\partial_\mu T_B^{\mu\nu}=\partial_\mu T_c^{\mu\nu}+\partial_\mu\partial_\lambda B^{\lambda\mu\nu}\). The operator \(\partial_\mu\partial_\lambda\) is symmetric under \(\mu\!\leftrightarrow\!\lambda\), while \(B^{\lambda\mu\nu}\) is antisymmetric under \(\lambda\!\leftrightarrow\!\mu\); the contraction of a symmetric with an antisymmetric pair vanishes. Hence \(\partial_\mu\partial_\lambda B^{\lambda\mu\nu}=0\) and \(\partial_\mu T_B^{\mu\nu}=\partial_\mu T_c^{\mu\nu}=0\).
Equal total charges (grade: A). Prove that \(P_B^{\nu}=\int d^3x\,T_B^{0\nu}\) equals \(P_c^{\nu}=\int d^3x\,T_c^{0\nu}\) for fields vanishing at spatial infinity.
Solution
\(P_B^{\nu}-P_c^{\nu}=\int d^3x\,\partial_\lambda B^{\lambda 0\nu}\). Split \(\lambda=0,i\). The \(\lambda=0\) term is \(\partial_0 B^{00\nu}=0\) because \(B^{00\nu}=-B^{00\nu}=0\) by antisymmetry in the first two indices. The remaining \(\int d^3x\,\partial_i B^{i0\nu}\) is a total spatial divergence, converting by Gauss's theorem to a surface integral at infinity, which vanishes for localised fields. Hence \(P_B^{\nu}=P_c^{\nu}\).
Scalar field (grade: B). For a real scalar with \(\mathcal{L}=\tfrac12(\partial\phi)^2-\tfrac12 m^2\phi^2\), compute \(T_c^{\mu\nu}\) and its spin current, and confirm no improvement is needed.
Solution
\(\partial\mathcal{L}/\partial(\partial_\mu\phi)=\partial^{\mu}\phi\), so \(T_c^{\mu\nu}=\partial^{\mu}\phi\,\partial^{\nu}\phi-\eta^{\mu\nu}\mathcal{L}\), which is manifestly symmetric in \(\mu\nu\). A scalar carries the trivial Lorentz representation, \((\Sigma^{\mu\nu})=0\), so \(S^{\lambda\mu\nu}=0\) and \(B^{\lambda\mu\nu}=0\). Therefore \(T_B^{\mu\nu}=T_c^{\mu\nu}\); the canonical tensor is already the Belinfante (and Hilbert) tensor.
Electromagnetic superpotential (grade: B). Using \(S^{\lambda\mu\nu}=F^{\mu\lambda}A^{\nu}-F^{\nu\lambda}A^{\mu}\), evaluate \(B^{\lambda\mu\nu}=\tfrac12(S^{\lambda\mu\nu}+S^{\mu\nu\lambda}+S^{\nu\mu\lambda})\) and show \(\partial_\lambda B^{\lambda\mu\nu}\) turns \(T_c^{\mu\nu}\) into the symmetric Maxwell tensor.
Solution
Substituting the three cyclic orderings, the terms proportional to \(A^{\lambda}\) cancel (\(F^{\mu\nu}+F^{\nu\mu}=0\)) and the survivors combine to \(B^{\lambda\mu\nu}=F^{\mu\lambda}A^{\nu}\), which is antisymmetric in \(\lambda\mu\) as required. Then \(\partial_\lambda B^{\lambda\mu\nu}=\partial_\lambda(F^{\mu\lambda}A^{\nu})=(\partial_\lambda F^{\mu\lambda})A^{\nu}+F^{\mu\lambda}\partial_\lambda A^{\nu}=F^{\mu\lambda}\partial_\lambda A^{\nu}\) on shell (\(\partial_\lambda F^{\mu\lambda}=0\)). Adding to \(T_c^{\mu\nu}=-F^{\mu\alpha}\partial^{\nu}A_{\alpha}+\tfrac14\eta^{\mu\nu}F^2\) and using \(F^{\mu\lambda}\partial_\lambda A^{\nu}-F^{\mu\alpha}\partial^{\nu}A_{\alpha}=-F^{\mu\alpha}(\partial^{\nu}A_{\alpha}-\partial_{\alpha}A^{\nu})=-F^{\mu\alpha}F^{\nu}{}_{\alpha}\) gives \(T_B^{\mu\nu}=-F^{\mu\alpha}F^{\nu}{}_{\alpha}+\tfrac14\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}\), symmetric and gauge invariant.
Radiation-pressure numeric (grade: C). Sunlight at Earth delivers intensity \(I=1.36\times10^{3}\ \mathrm{W\,m^{-2}}\). Using the symmetric tensor, find the momentum-density and the radiation pressure on a perfectly absorbing surface, and state why the canonical tensor would give an ambiguous answer.
Solution
Energy density \(u=I/c=1.36\times10^{3}/3.0\times10^{8}=4.53\times10^{-6}\ \mathrm{J\,m^{-3}}\). Momentum density \(g=u/c=I/c^{2}=1.36\times10^{3}/(9.0\times10^{16})=1.51\times10^{-14}\ \mathrm{kg\,m^{-2}\,s^{-1}}\). Radiation pressure on an absorber equals the momentum flux \(T_B^{zz}=u\), so \(P_{\mathrm{rad}}=I/c=4.53\times10^{-6}\ \mathrm{Pa}\) (double for a perfect reflector, \(9.1\times10^{-6}\ \mathrm{Pa}\)). The canonical tensor \(T_c^{zz}\) is gauge dependent and asymmetric, so its momentum-flux component is not gauge invariant and cannot be equated with a measurable pressure; only \(T_B^{\mu\nu}\), which is symmetric and gauge invariant, yields the observed value.