Angular Momentum Tensor of the Field

Put a charged cylinder inside a solenoid. Nothing moves. The electric field is static, the magnetic field is static, the whole apparatus sits on the bench doing nothing at all. Now switch off the solenoid current — and the cylinders start to spin. Nobody touched them. No external torque acted. Angular momentum, supposedly conserved, has apparently been created out of a switch. The only way out is the one Maxwell and Poynting were already pointing at: the angular momentum was in the field the whole time, stored in the crossed $\mathbf{E}$ and $\mathbf{B}$ , and switching off the solenoid merely handed it back to the matter. A static field, in which not one charge is in motion, can hold angular momentum — and you can weigh it. To say that covariantly you need one object. Relativity fuses the three rotations and the three boosts into a single six-parameter group, so the quantity conserved by that symmetry is not a three-vector $\mathbf{L}$ but an antisymmetric tensor $J^{\nu\rho}$ — whose local density is the rank-three tensor $M^{\mu\nu\rho}$ , built out of the stress–energy tensor and nothing else. Its conservation hinges on a property of $T^{\mu\nu}$ that looks like bookkeeping and is not: symmetry. An asymmetric stress tensor does not conserve angular momentum — which is why the Belinfante–Rosenfeld repair is physics, not tidying.

Key takeaways

Signature $(+,-,-,-)$ , $\eta_{\mu\nu} = \mathrm{diag}(+1,-1,-1,-1)$ , so $ds^2 = c^2dt^2 - dx^2$ . The electromagnetic stress tensor is where getting this wrong is fatal.
A field's angular momentum is carried by $M^{\mu\nu\rho} = x^{\nu}T^{\mu\rho} - x^{\rho}T^{\mu\nu}$ — the tensor form of $\mathbf{r}\times\mathbf{p}$ .
Owned result: $\partial_{\mu}M^{\mu\nu\rho} = T^{\nu\rho} - T^{\rho\nu}$ whenever $\partial_\mu T^{\mu\nu}=0$ . Angular momentum is conserved if and only if the stress tensor is symmetric — and the antisymmetric part of a stress tensor is an internal torque density. Symmetry is a physical hypothesis in the costume of an index convention.
Integrating the $\mu=0$ slice gives six charges: three $J^{ij}$ are ordinary angular momentum; three $J^{0i}$ say the energy centroid moves in a straight line at velocity $c^2\mathbf{P}/U$ — the relativistic centre-of-mass theorem, free of charge.
Owned result: in $(+,-,-,-)$ the symmetric electromagnetic stress tensor is $T^{\mu\nu} = \tfrac{1}{\mu_0}\big[\tfrac14\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta} - F^{\mu\alpha}F^{\nu}{}_{\alpha}\big]$ , giving $T^{00} = +\tfrac12\big(\varepsilon_0 E^2 + B^2/\mu_0\big) > 0$ . The commoner form with the opposite sign belongs to $(-,+,+,+)$ ; transcribed into ours it returns a negative energy density. This article's AllQFT base does exactly that, with $\varepsilon_0$ where $1/\mu_0$ belongs into the bargain. Both repaired here.
In three-vector form, $\mathbf{L}_{\text{field}} = \varepsilon_0\!\int \mathbf{r}\times(\mathbf{E}\times\mathbf{B})\,d^3x$ — mechanical, and measurable. Beth weighed it in 1936.
Key idea: angular-momentum conservation is not a law bolted onto field theory — it is Lorentz invariance, read as a local continuity equation. The price of admission is a symmetric stress tensor, and electromagnetism does not hand you one for free.

The core insight

Everything here follows from one line of calculus. Differentiate $M^{\mu\nu\rho} = x^{\nu}T^{\mu\rho} - x^{\rho}T^{\mu\nu}$ on its first index: the $x$ -weighted pieces vanish by momentum conservation, and what survives is $\delta_\mu^\nu T^{\mu\rho}$ antisymmetrised — which is $T^{\nu\rho} - T^{\rho\nu}$ . Nothing else. So conservation of angular momentum is precisely the vanishing of the antisymmetric part of the stress tensor — and that antisymmetric part is, physically, the torque a piece of the medium exerts on itself. No internal torque, conserved angular momentum. That is the whole theorem.

Where this lives today

The optical tweezer is equation (6) with a price tag. A focused circularly polarised beam transfers $\hbar$ per photon of spin to a birefringent particle; a Laguerre–Gaussian beam with an azimuthal phase winding $e^{i\ell\varphi}$ transfers $\ell\hbar$ of orbital angular momentum on top. The two are separable, micron-scale objects are routinely spun and sorted with them, and the $\ell$ index has become an information channel in its own right — an unbounded orthogonal alphabet. In field theory the Belinfante–Rosenfeld tensor is what stands on the right of Einstein's equations, and that gravity couples to the symmetric tensor is no accident: the source of curvature must be conserved and symmetric, exactly as $\partial_\mu M^{\mu\nu\rho}=0$ demands. The proton's spin decomposition into quark, gluon and orbital pieces runs onto the same rock — the gauge-invariant split of $M^{\mu\nu\rho}$ is not unique, and that ambiguity is an open programme, not a convention.

The physical picture

The Newtonian statement is that $\mathbf{L} = \mathbf{r}\times\mathbf{p}$ is conserved when the torque vanishes; the deeper one — Noether's — is that this follows from rotational invariance. Relativity does not touch the logic. It enlarges the symmetry: the Lorentz group has six generators, three rotations and three boosts, mixed into one another by the very transformations they generate. A conserved quantity that is a three-vector under rotations cannot be Lorentz covariant, so the true conserved object is antisymmetric and rank two — $J^{\nu\rho} = -J^{\rho\nu}$ , three $J^{ij}$ and three $J^{0i}$ . For a point particle this is familiar: $J^{\nu\rho} = x^{\nu}p^{\rho} - x^{\rho}p^{\nu}$ , whose spatial part is $\mathbf{r}\times\mathbf{p}$ . For a field, momentum is a density and its flux is the stress tensor, so what we need is a local density whose space integral is $J^{\nu\rho}$ . The only object built from $x$ and $T$ with the right index structure is $M^{\mu\nu\rho} = x^{\nu}T^{\mu\rho} - x^{\rho}T^{\mu\nu}$ , antisymmetric in $\nu\rho$ , with $\mu$ the flux index — exactly as $\mu$ is the flux index of $J^\mu$ for charge. And now the fact that carries the article: this thing is conserved on $\mu$ only if $T^{\mu\nu} = T^{\nu\mu}$ . Why should it be? Because the antisymmetric part is a torque. $T^{ij}$ is the flux of $i$ -momentum in the $j$ -direction — a force per unit area — so $T^{ij} - T^{ji}$ is the net couple a little cube of the medium exerts on itself. Fluids, elastic solids and the free electromagnetic field have no such intrinsic torque, so their stress tensors must be symmetric. Allow one, and angular momentum leaks out of the orbital ledger. Electromagnetism poses the problem sharply: the stress tensor that drops out of Noether's theorem — the canonical one — is neither symmetric nor gauge invariant, so taken at face value it says the angular momentum of a light beam depends on your choice of gauge. The Belinfante–Rosenfeld repair, derived below, fixes both at once.

Figure

PUNCHLINE — static fields, nothing moving, and the field still holds angular momentum. A charged shell at radius a sits inside a solenoid of radius R. E is radial, B is axial, so the momentum density g = ε₀E×B circulates azimuthally and the field carries L_z = −½λB(R²−a²) per metre: here −3.75×10⁻⁹ kg m² s⁻¹. Switch the solenoid off and the induced azimuthal E field spins BOTH cylinders, in OPPOSITE directions — the inner shell picks up +½λa²B = +1.25×10⁻⁹, the outer shell −½λR²B = −5.00×10⁻⁹ — and the sum is exactly the −3.75×10⁻⁹ the field had all along. Nothing was created. The books were always balanced; you just had to count the field

The figure is the whole subject in one apparatus. Before the switch, nothing moves and the mechanical angular momentum is zero; after it, the cylinders spin in opposite senses and it is $-\tfrac12\lambda B(R^2-a^2)$ per metre. The books balance only because the field was carrying exactly that, and handed it over. Feynman used a version of this — charges on a disc rim, a solenoid at the hub — to force the point: you cannot do relativistic mechanics with matter alone. The field is a participant.

The core equations

Take the stress–energy tensor as given — $T^{00}$ the energy density, $T^{0i} = S^i/c$ , $T^{ij}$ the Maxwell stress; its construction belongs to `aq-sr-electromagnetic-stress-energy-tensor`. From it and the position four-vector alone:

\boxed{\;M^{\mu\nu\rho} \;=\; x^{\nu}T^{\mu\rho} \;-\; x^{\rho}T^{\mu\nu}\;} \qquad M^{\mu\nu\rho} = -M^{\mu\rho\nu} \qquad (1)

where: $x^\nu = (ct,\mathbf{x})$ , and $\mu$ is the flux index. Its divergence is two lines of calculus with a one-word answer:

\partial_{\mu}M^{\mu\nu\rho} \;=\; \underbrace{x^{\nu}\,\partial_\mu T^{\mu\rho} - x^{\rho}\,\partial_\mu T^{\mu\nu}}_{=\,0 \text{ by momentum conservation}} \;+\; \big(T^{\nu\rho} - T^{\rho\nu}\big) \qquad (2)

where: the first bracket dies only because $\partial_\mu T^{\mu\nu} = 0$ — an assumption doing real work, which we state wherever we use it. What survives is the internal torque density. Hence

\boxed{\;\partial_{\mu}T^{\mu\nu}=0 \;\;\text{and}\;\; T^{\mu\nu}=T^{\nu\mu} \;\;\Longleftrightarrow\;\; \partial_{\mu}M^{\mu\nu\rho}=0\;} \qquad (3)

The six conserved charges are the space integrals of the $\mu=0$ slice, with the $1/c$ that gives them the units of angular momentum:

J^{\nu\rho} \;=\; \frac{1}{c}\int\!\big(x^{\nu}T^{0\rho} - x^{\rho}T^{0\nu}\big)\,d^{3}x, \qquad \frac{dJ^{\nu\rho}}{dt} = 0 \qquad (4)

where: time-independence needs (3) plus fields falling off fast enough at infinity. The split into $J^{ij}$ and $J^{0i}$ is frame-dependent — a boost rotates angular momentum into centre-of-energy motion just as it rotates $\mathbf{E}$ into $\mathbf{B}$ . For electromagnetism the input is the symmetric stress tensor, and this is where the signature bites. In $(+,-,-,-)$ , SI:

\boxed{\;T^{\mu\nu} \;=\; \frac{1}{\mu_0}\Big[\;\tfrac14\,\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta} \;-\; F^{\mu\alpha}F^{\nu}{}_{\alpha}\;\Big]\;} \quad\Longrightarrow\quad T^{00} = +\tfrac12\Big(\varepsilon_0 E^{2} + \frac{B^{2}}{\mu_0}\Big) \;>\; 0 \qquad (5)

where: $F^{\nu}{}_{\alpha} = \eta_{\alpha\beta}F^{\nu\beta}$ ; equivalently $T^{\mu\nu} = \tfrac{1}{\mu_0}\big[F^{\mu\alpha}F_{\alpha}{}^{\nu} + \tfrac14\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}\big]$ , since $F_{\alpha}{}^{\nu} = -F^{\nu}{}_{\alpha}$ . The overall sign is not free. The version with $+F^{\mu\alpha}F^{\nu}{}_{\alpha} - \tfrac14\eta^{\mu\nu}F^2$ is the $(-,+,+,+)$ form; substituted into our $\eta$ it returns a negative energy density. See the warning. Finally, the three-vector form. With momentum density $\mathbf{g} = \varepsilon_0\,\mathbf{E}\times\mathbf{B} = \mathbf{S}/c^2$ :

L_k = \tfrac12\,\epsilon_{kij}J^{ij} = \int (\mathbf{r}\times\mathbf{g})_k\,d^{3}x \quad\Longrightarrow\quad \boxed{\;\mathbf{L}_{\text{field}} = \varepsilon_0\!\int \mathbf{r}\times(\mathbf{E}\times\mathbf{B})\,d^{3}x\;} \qquad (6)

where: $\mathbf{g}$ is the electromagnetic momentum density. This integrand is nonzero for static fields: no wave, no radiation, no motion — just crossed $\mathbf{E}$ and $\mathbf{B}$ .

Figure

The 4×4 block, and the sign trap that guards it. LEFT — the symmetric stress tensor: T⁰⁰ is the energy density, T⁰ⁱ the Poynting flux over c, Tⁱʲ the Maxwell stress, and the trace vanishes. Wedge it with the position four-vector and you get M^μνρ, whose six charges split into three angular momenta and three boost charges. RIGHT — the trap. The two candidate formulae differ by an overall sign, and only one of them is right in (+,−,−,−). The mostly-plus form, transcribed into our signature, returns T⁰⁰ = −½(ε₀E² + B²/μ₀): a NEGATIVE electromagnetic energy density, and a Poynting vector pointing backwards. This article's base prints exactly that form, with ε₀ in place of 1/μ₀ on top — a further factor of c². Never transcribe a tensor whose sign depends on the metric. Recompute T⁰⁰ from components and check that it is positive

How the conservation law works

Derivation AConservation of the angular momentum tensor forces a symmetric stress tensor

The goal is to derive equation (2) honestly and say what it costs. Two claims get conflated: that $M^{\mu\nu\rho}$ is conserved, and that $T^{\mu\nu}$ is symmetric. They are not independent and neither is free — the first needs momentum conservation as an input before symmetry becomes the remaining condition. This matters, because the AllQFT base states the divergence identity in a way that quietly assumes what it is proving. We keep the assumption visible.

Step 1. Differentiate the definition on its flux index, with no assumptions at all: $\partial_{\mu}M^{\mu\nu\rho} = (\partial_\mu x^{\nu})T^{\mu\rho} + x^{\nu}\partial_\mu T^{\mu\rho} - (\partial_\mu x^{\rho})T^{\mu\nu} - x^{\rho}\partial_\mu T^{\mu\nu}$ . Why: pure calculus, holding for any $T$ whatsoever. Everything that follows is about what we are then allowed to delete.

Step 2. Evaluate the only derivative that is a pure number. In Cartesian coordinates $\partial_\mu x^{\nu} = \delta_\mu^{\nu}$ , so $(\partial_\mu x^{\nu})T^{\mu\rho} = T^{\nu\rho}$ and likewise $(\partial_\mu x^{\rho})T^{\mu\nu} = T^{\rho\nu}$ . Hence, still with no assumptions, $\partial_{\mu}M^{\mu\nu\rho} = T^{\nu\rho} - T^{\rho\nu} + x^{\nu}\partial_\mu T^{\mu\rho} - x^{\rho}\partial_\mu T^{\mu\nu}$ . Why: worth writing in full, because the $x$ -weighted terms are the ones textbooks drop silently. They are not zero as algebra — they are zero as physics.

Step 3. Impose the input. Assume $\partial_\mu T^{\mu\rho} = 0$ — for the free field a theorem, but one that fails when charges are present, where $\partial_\mu T^{\mu\nu} = -f^{\nu}$ with $f^\nu$ the Lorentz four-force density. The $x$ -weighted terms vanish, leaving $\partial_{\mu}M^{\mu\nu\rho} = T^{\nu\rho} - T^{\rho\nu}$ . Why: angular-momentum conservation is strictly stronger than momentum conservation — you must have the second before the first can be asked. And with charges present, the same identity says exactly how much angular momentum the field hands the matter.

Step 4. Read off the condition: $\partial_{\mu}M^{\mu\nu\rho} = 0$ iff $T^{\nu\rho} = T^{\rho\nu}$ . Why: $T^{ij}-T^{ji}$ is the couple a fluid element exerts on itself; its vanishing says the medium carries no intrinsic torque density. Symmetry of the stress tensor is a physical hypothesis wearing the costume of an index convention.

Step 5. Integrate to charges. Integrating $\partial_\mu M^{\mu\nu\rho}=0$ over space and splitting $\mu$ into $0$ and $i$ gives $\frac{1}{c}\frac{d}{dt}\!\int\! M^{0\nu\rho} d^3x = -\frac{1}{c}\!\oint\! M^{i\nu\rho}\,dS_i \to 0$ for fields falling off fast enough, so $dJ^{\nu\rho}/dt = 0$ . Why: the same divergence theorem that turns $\partial_\mu J^\mu = 0$ into charge conservation. The fall-off condition is not pedantry — for a radiating system the flux through the sphere at infinity is the angular momentum the radiation carries away, which is how a rotating dipole spins itself down. *Verified symbolically in verify_aq-sr-field-angular-momentum-tensor.py, which builds a generic symbolic $T^{\mu\nu}$ , confirms the exact identity of Step 2 term by term, shows it collapses to the $x$ -weighted divergences alone when $T$ is symmetric, and exhibits a conserved-but-asymmetric $T$ for which $\partial_\mu M^{\mu\nu\rho} \ne 0$ .*

Go deeper · Derivation B · 2 minThe six charges: three angular momenta and the centre-of-energy theorem

Derivation BThe six charges: three angular momenta and the centre-of-energy theorem

Equation (4) produces six conserved numbers, and only three are the ones the reader expects. The other three are usually skipped and should not be, because they contain a theorem that is genuinely relativistic: the energy centroid of an isolated system moves in a straight line at constant velocity $c^2\mathbf{P}/U$ . The goal is to unpack both halves of $J^{\nu\rho}$ and show that the boost charges are the relativistic replacement for "the centre of mass moves uniformly", not a curiosity.

Step 1. Take the purely spatial components. With $T^{0j} = c\,g^{j}$ , where $\mathbf{g} = \varepsilon_0\mathbf{E}\times\mathbf{B}$ , equation (4) gives $J^{ij} = \int (x^i g^j - x^j g^i)\,d^3x$ , and contracting with the Levi-Civita symbol, $L_k = \tfrac12\epsilon_{kij}J^{ij} = \int(\mathbf{r}\times\mathbf{g})_k\,d^3x$ — equation (6). Why: the factors of $c$ cancel exactly — the check that the $1/c$ in (4) was the right normalisation. What is left is literally the moment of the momentum density, and the tensor says it in a way that survives a boost.

Step 2. Now the mixed components, where the new physics is. Put $\nu=0$ , $\rho=i$ and use $x^0 = ct$ : $J^{0i} = \frac{1}{c}\int\big(ct\,T^{0i} - x^{i}T^{00}\big)d^3x = c\,t\,P^{i} - \frac{U}{c}X^i_E$ , defining the energy centroid $X^i_E = \frac{1}{U}\int x^{i}T^{00}\,d^3x$ with $U$ the total energy and $P^i$ the total field momentum. Why: the centroid is weighted by energy, not mass — so for a light beam, with no rest mass at all, it is still perfectly well defined. That is why the theorem generalises.

Step 3. Impose conservation. $J^{0i}$ , $U$ and $P^i$ are all constant, so differentiating gives $0 = cP^{i} - \frac{U}{c}\dot{X}^i_E$ , hence $\boxed{\dfrac{dX^i_E}{dt} = \dfrac{c^{2}P^{i}}{U}}$ — constant. Why: the relativistic centre-of-mass theorem, free with the boost generators. Note what it says about the figure's static configuration: $\mathbf{L}\ne 0$ but $\mathbf{P}=0$ , so the centroid stays put while the angular momentum does not vanish. Angular momentum without motion is exactly what the apparatus demonstrates.

Step 4. Note the frame-dependence, which is why one does this covariantly: under a boost, $J^{ij}$ and $J^{0i}$ mix precisely as $\mathbf{B}$ and $\mathbf{E}/c$ do. Why: "angular momentum" and "centre-of-energy motion" are not two facts about a system. They are one fact, seen from two frames. *Verified in verify_aq-sr-field-angular-momentum-tensor.py: $J^{ij}$ built from $M^{0ij}$ reproduces $\varepsilon_0\int\mathbf{r}\times(\mathbf{E}\times\mathbf{B})$ on the coaxial configuration of Example 1, by brute-force numerical integration of the field, to one part in $10^{6}$ .*

Go deeper · Derivation C · 2 minRepairing the electromagnetic stress tensor: prefactor, sign, and the Belinfante term

Derivation CRepairing the electromagnetic stress tensor: prefactor, sign, and the Belinfante term

This card exists because equation (5) is the most error-prone line in the article, and because this entry's AllQFT base gets it wrong twice — once in the prefactor, once in the overall sign. The goal is to fix both by explicit component computation rather than by quoting a book, since quoting a book is exactly how a mostly-plus formula walks into a mostly-minus article. The Belinfante–Rosenfeld repair comes first, because without it equation (3) has nothing to eat.

Step 1. Start from the canonical tensor, the one Noether's theorem actually hands you. With $\mathcal{L} = -\tfrac{1}{4\mu_0}F_{\alpha\beta}F^{\alpha\beta}$ , translation invariance gives $\Theta^{\mu\nu} = -\frac{1}{\mu_0}F^{\mu\alpha}\,\partial^{\nu}A_{\alpha} + \frac{1}{4\mu_0}\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}$ . Why: conserved, and useless. Not symmetric; and it contains a bare $\partial^\nu A_\alpha$ , so not gauge invariant either. A gauge-dependent energy density is not an observable.

Step 2. Add an identically-conserved improvement: $T^{\mu\nu} = \Theta^{\mu\nu} + \partial_{\lambda}B^{\lambda\mu\nu}$ with $B^{\lambda\mu\nu} = -B^{\mu\lambda\nu}$ , whose antisymmetry makes $\partial_\mu\partial_\lambda B^{\lambda\mu\nu}\equiv 0$ , so conservation survives whatever $B$ is. Take $B^{\lambda\mu\nu} = \frac{1}{\mu_0}F^{\mu\lambda}A^{\nu}$ and use $\partial_\lambda F^{\mu\lambda}=0$ : this converts $-\partial^{\nu}A_\alpha$ into $-\partial_\alpha A^{\nu}$ inside the contraction, assembling the full field strength. Why: not a fudge — the freedom to add a superpotential is real, the charges are unchanged, and the result is unique once you demand symmetry and gauge invariance.

Step 3. The result is equation (5). Note the prefactor: $1/\mu_0$ , not $\varepsilon_0$ . Since $F$ carries components $E/c$ and $B$ , a prefactor of $\varepsilon_0$ would give $\varepsilon_0 E^2/c^2$ — not an energy density, off by $c^2$ . Why: dimensional analysis catches this instantly, the cheapest check in the article. $1/\mu_0 = \varepsilon_0 c^2$ , and that $c^2$ is exactly what turns $(E/c)^2$ back into $E^2$ .

Step 4. Now fix the sign by brute force. With $F^{0i} = -E^i/c$ , $F^{ij} = -\epsilon^{ijk}B^k$ and $F_{\alpha\beta}F^{\alpha\beta} = 2(B^2 - E^2/c^2)$ : the trace term of (5) is $\tfrac14\cdot 1\cdot 2(B^2-E^2/c^2) = \tfrac12 B^2 - \tfrac{E^2}{2c^2}$ , and the field term is $-F^{0\alpha}F^{0}{}_{\alpha} = -\sum_i(-E^i/c)(-1)(-E^i/c) = +E^2/c^2$ . Summing and multiplying by $1/\mu_0$ gives $T^{00} = \tfrac12(\varepsilon_0E^2 + B^2/\mu_0) > 0$ . Why: swap the two signs — the form printed in half the literature, and in this article's base — and every term flips, giving $-\tfrac12(\varepsilon_0E^2+B^2/\mu_0)$ . That form is correct in $(-,+,+,+)$ and catastrophic in ours. <strong>A negative electromagnetic energy density is the loudest alarm in physics, and the base walks straight past it.</strong>

Step 5. Confirm the rest of the block: the same computation gives $T^{0i} = \frac{1}{\mu_0 c}(\mathbf{E}\times\mathbf{B})^i = S^i/c$ — the Poynting flux, right sign — and $T^{\mu}{}_{\mu} = 0$ . Why: tracelessness holds for <strong>either</strong> sign, so it alone would not catch the error; the energy density and the direction of energy flow both would. Check all three. *Verified symbolically in verify_aq-sr-field-angular-momentum-tensor.py: with $\eta = \mathrm{diag}(+1,-1,-1,-1)$ the boxed form of (5) gives $T^{00} = +\tfrac12(\varepsilon_0E^2+B^2/\mu_0)$ , $T^{0i} = (\mathbf{E}\times\mathbf{B})^i/(\mu_0 c)$ and $T^{\mu}{}_{\mu}=0$ ; the base's form is evaluated on a numeric field and asserted to be negative.*

Limiting cases

Case	Result	Meaning
Static crossed fields	$\mathbf{L}_{\text{field}} \ne 0$ with $\mathbf{P} = 0$	angular momentum with zero motion — the figure, and Feynman's disc
Charges present	$\partial_\mu M^{\mu\nu\rho} = x^{\nu}f^{\rho} - x^{\rho}f^{\nu}$	not conserved alone — the deficit is the torque on the matter
Canonical tensor used raw	$T^{\nu\rho}-T^{\rho\nu}\ne 0$	$M^{\mu\nu\rho}$ not conserved; answer is gauge-dependent

Common trapThe sign of the electromagnetic stress tensor is a signature artefact, and it will give you negative energy

The far commoner form, $\tfrac{1}{\mu_0}\big[F^{\mu\alpha}F^{\nu}{}_{\alpha} - \tfrac14\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}\big]$ , is the mostly-plus version. It is not wrong in its own house. Dragged into ours, it returns

T^{00} \;=\; -\tfrac12\Big(\varepsilon_0E^{2} + \frac{B^{2}}{\mu_0}\Big) \;<\; 0,

a negative electromagnetic energy density — and, since $T^{0i}$ flips too, a Poynting vector pointing the wrong way. This article's AllQFT base prints exactly that form, and writes the prefactor as $\varepsilon_0$ instead of $1/\mu_0$ , which is separately wrong by $c^2$ . Both are asserted, symbolically, in the verify script. Never transcribe a tensor whose sign depends on the metric. Recompute $T^{00}$ from components and check that it is positive. Four lines, and it is the difference between physics and typography.

Worked examples

1 · The apparatus that spins itself

A long solenoid of radius $R = 10.0\ \mathrm{cm}$ carries a uniform axial field $B = 1.00\ \mathrm{T}$ . Coaxial with it, at $a = 5.00\ \mathrm{cm}$ , is a thin shell with linear charge density $\lambda = 1.00\ \mu\mathrm{C/m}$ ; a second shell outside, at $b = 15\ \mathrm{cm}$ , carries $-\lambda$ , so the whole thing is neutral. Everything is static. How much angular momentum does the field hold — and what happens when you switch the solenoid off?

Step 1 — identify. Nothing moves, so the mechanical angular momentum is zero; the field's is equation (6). Between the shells Gauss gives $\mathbf{E} = \dfrac{\lambda}{2\pi\varepsilon_0 s}\hat{\mathbf{s}}$ ; inside the solenoid $\mathbf{B} = B\hat{\mathbf{z}}$ . Both are nonzero only in the overlap $a < s < R$ .

Step 2 — the density. $\mathbf{g} = \varepsilon_0\mathbf{E}\times\mathbf{B} = -\dfrac{\lambda B}{2\pi s}\hat{\boldsymbol{\varphi}}$ , so $\ell_z = (\mathbf{r}\times\mathbf{g})_z = -\dfrac{\lambda B}{2\pi}$ — independent of $s$ , uniform across the annulus.

Step 3 — integrate. Per unit length the annulus has area $\pi(R^2-a^2)$ , so

L_z = -\frac{\lambda B}{2\pi}\cdot\pi\big(R^{2}-a^{2}\big) = -\tfrac12\lambda B\big(R^{2}-a^{2}\big) = -\tfrac12(1.00\times10^{-6})(1.00)(0.0100-0.00250) = \mathbf{-3.75\times10^{-9}\ kg\,m^{2}\,s^{-1}}

per metre of length.

Step 4 — the boundary number. Switch off the solenoid. The changing flux drives an azimuthal field, $E_\varphi = -\tfrac{s}{2}\dot{B}$ inside and $-\tfrac{R^2}{2s}\dot{B}$ outside, which torques both shells. Integrating the angular impulse from $B$ to $0$ : the inner shell gains $+\tfrac12\lambda a^2 B = +1.25\times10^{-9}$ , the outer $-\tfrac12\lambda R^2 B = -5.00\times10^{-9}$ , per metre. They spin in opposite directions, and the sum is $-3.75\times10^{-9}$ — exactly what the field had. If the inner shell weighs $50\ \mathrm{g}$ per metre it ends up turning at $\omega = \Delta L/(ma^2) = 1.0\times10^{-5}\ \mathrm{rad/s}$ : one revolution every $7.3$ days. Slow, absurd, and real. The angular momentum was never created. It was transferred from the field, which had been holding it, motionlessly, all along.

2 · Weighing the spin of light

A $1.00\ \mathrm{W}$ beam of circularly polarised light at $500\ \mathrm{nm}$ passes through a quarter-wave plate, emerging linearly polarised. What torque does it exert on the plate?

Step 1 — identify. The plate absorbs the beam's spin angular momentum flux while passing its energy. From (6) on a plane wave, the ratio of angular-momentum flux to energy flux for circular polarisation is $1/\omega$ — purely classical. So $N = P/\omega$ .

Step 2 — apply. $\omega = \dfrac{2\pi c}{\lambda} = \dfrac{2\pi(2.998\times10^{8})}{5.00\times10^{-7}} = 3.767\times10^{15}\ \mathrm{rad/s}$ , so

N = \frac{P}{\omega} = \frac{1.00}{3.767\times10^{15}} = \mathbf{2.65\times10^{-16}\ N\,m}.

Step 3 — the quantum cross-check. The photon flux is $P/(\hbar\omega) = 2.52\times10^{18}$ per second, and if each carries $\hbar$ of spin the torque is $2.52\times10^{18}\times 1.055\times10^{-34} = 2.65\times10^{-16}\ \mathrm{N\,m}$ — the same number. The classical field's angular momentum and " $\hbar$ per photon" are not competing theories. They are one flux, counted twice.

Step 4 — the boundary number. A quarter of a femtonewton-metre. Beth measured it in 1936 by hanging a wave plate from a quartz fibre in vacuum, driving the beam at the torsion pendulum's resonance, and reversing the handedness to flip the sign. He got the magnitude, the sign, and the dependence on every variable he could change. Field angular momentum is not a formal device. It goes on a balance.

Physics of the portrait

Angular momentum changes shape more violently than any other conserved quantity when it crosses into relativity: in as a three-vector, out as a six-component antisymmetric tensor whose extra slots encode the motion of the centre of energy. Nothing about $\mathbf{r}\times\mathbf{p}$ prepares you for that.

Origins · 2 minAngular momentum and how this was discovered

Angular momentum

1909–1940 · Poynting predicts the torque of circular light · Noether ties conservation to symmetry · Beth weighs it · Belinfante and Rosenfeld make the stress tensor deserve it

That a field — not a spinning body, not a moving particle, but an arrangement of $\mathbf{E}$ and $\mathbf{B}$ — could carry angular momentum was proposed by John Henry Poynting in 1909 (Proc. Roy. Soc. A 82, 560), reasoning from a mechanical analogy with a rotating screw and concluding that circularly polarised light must torque a quarter-wave plate. The prediction sat unmeasured for a generation because the numbers looked hopeless. Richard A. Beth reached them in 1936 (Phys. Rev. 50, 115), hanging a birefringent plate from a quartz fibre in vacuum and driving the beam at the torsion pendulum's resonance; the torque came out at exactly $\hbar$ per photon. Meanwhile the theory had been rebuilt from underneath. Emmy Noether's 1918 theorem made conservation of angular momentum the shadow of rotational invariance, and the covariant extension to boosts — which makes $J^{\nu\rho}$ a tensor — was already implicit in Minkowski's 1908 geometry. The last piece was the least glamorous and the most necessary: the stress tensor Noether's theorem produces for electromagnetism is asymmetric and gauge-dependent, which would render field angular momentum meaningless. Frederik Belinfante (Physica 6, 887, 1939) and Léon Rosenfeld (Mém. Acad. Roy. Belg. 18, 1536, 1940), independently, showed how to add an identically-conserved improvement term that symmetrises it without changing a single physical charge — and that the repaired tensor is precisely the one sourcing gravity in general relativity. A rare career for a quantity: predicted from a loose analogy, measured on a torsion fibre, and given a formalism that deserved it only afterwards.

History

1909

— Poynting argues that circularly polarised light must carry angular momentum and predicts a torque on a quarter-wave plate (Proc. Roy. Soc. A 82, 560).

1918

— Emmy Noether's "Invariante Variationsprobleme" makes conservation of angular momentum the exact shadow of rotational invariance, and the boost charges the shadow of Lorentz invariance.

1936

— Beth measures the torque on a suspended wave plate and confirms an angular momentum of one hbar per photon (Phys. Rev. 50, 115).

1939

— Belinfante constructs the improvement term that symmetrises the canonical stress tensor without altering the conserved charges (Physica 6, 887).

1940

— Rosenfeld independently obtains the same symmetrised tensor and identifies it with the source of gravity (Mem. Acad. Roy. Belg. 18, 1536).

1964

— Feynman's disc paradox, published in the Lectures, forces a generation of students to admit that a static field can hold angular momentum.

Going further

Texts. Jackson, Classical Electrodynamics, 3rd ed., §12.10 (canonical versus symmetric stress tensor) · Landau & Lifshitz, The Classical Theory of Fields, §§32–33 · Griffiths, Introduction to Electrodynamics, 4th ed., §8.2 and ch. 12 · Belinfante, F. J., Physica 6, 887 (1939); Rosenfeld, L., Mém. Acad. Roy. Belg. 18, 1536 (1940) · Beth, R. A., Phys. Rev. 50, 115 (1936) · Poynting, J. H., Proc. Roy. Soc. A 82, 560 (1909).

Provenance. Signature $(+,-,-,-)$ throughout. Every equation and number above is recomputed in verify_aq-sr-field-angular-momentum-tensor.py, which also greps this article for each printed value. Base defects found and repaired. The AllQFT base's eq[4] prints $T^{\mu\nu} = \varepsilon_0\big(F^{\mu\alpha}F^{\nu}{}_{\alpha} - \tfrac14\eta^{\mu\nu}F^{\alpha\beta}F_{\alpha\beta}\big)$ — two independent errors: the prefactor must be $1/\mu_0$ (with $\varepsilon_0$ it is off by $c^2$ and is not an energy density), and the overall sign is the $(-,+,+,+)$ one, so in our signature it yields $T^{00}<0$ . Both are asserted symbolically, broken form and repaired. The base's eq[1] states $\partial_\mu M^{\mu\nu\rho} = T^{\nu\rho}-T^{\rho\nu}$ as if unconditional; it needs $\partial_\mu T^{\mu\nu}=0$ as an input, made visible wherever used above. The base's eq[3] omits the $1/c$ that $J^{\nu\rho}$ needs to carry the units of angular momentum.

Scope. The symmetric stress tensor is used, not derived: `aq-sr-electromagnetic-stress-energy-tensor`. The field tensor is `aq-sr-electromagnetic-field-tensor`. The spin-versus-orbital decomposition of optical angular momentum is named, not developed — it is not a settled question and this article does not pretend otherwise.

Builds on. The Electromagnetic Field Tensor · Covariant Maxwell Equations · Charge Invariance and Continuity · The Covariant Lorentz Force Law. Leads to. Electromagnetic Field Invariants · The Dual Field Tensor · Transformation of Electric and Magnetic Fields.

If the fields are completely static, how can there be angular momentum? Nothing is going round.

Nothing material is going round, but the momentum density $\mathbf{g} = \varepsilon_0\mathbf{E}\times\mathbf{B}$ is nonzero and azimuthal, so there is a genuine circulating flow of electromagnetic momentum. It has no carrier you can point at, it does not dissipate, and it does not radiate — it simply circulates, like a persistent current. The test that it is real is mechanical: cut off the magnetic field and the stored angular momentum turns up in the cylinders, to the last decimal place. Refuse to count the field and you are left with a violated conservation law — a far worse price than the mild strangeness of a static circulation.

Is the split of light's angular momentum into spin and orbital parts real?

Operationally, yes: circular polarisation spins a birefringent particle about its own axis, an azimuthal phase winding drags it around the beam axis, and these are separately measurable. Formally it is delicate — the decomposition of $M^{\mu\nu\rho}$ into "spin" and "orbital" pieces is not gauge invariant term by term, even though the total is. In the paraxial regime the difficulty is harmless; beyond it, and especially in QCD, the ambiguity is a live research problem. This article claims only what the total conserved tensor gives.

Why does the centre-of-energy theorem come out of the angular momentum tensor rather than the four-momentum?

Because the boost generators sit in the same tensor as the rotation generators. The Lorentz group has six parameters and Noether gives one charge each: three from rotations (angular momentum) and three from boosts, $J^{0i} = ctP^i - (U/c)X_E^i$ , whose conservation is the statement that the energy centroid moves uniformly. The four-momentum $P^\mu$ , by contrast, comes from translation invariance — a different symmetry. Boosts and rotations are the pair that mix under a boost, which is why they must share one antisymmetric tensor.

Does this all break down when there are charges around?

It becomes more informative. With sources, $\partial_\mu T^{\mu\nu} = -f^{\nu}$ , so Step 3 of the first derivation no longer kills the $x$ -weighted terms; instead $\partial_\mu M^{\mu\nu\rho} = x^{\nu}f^{\rho} - x^{\rho}f^{\nu}$ , whose right-hand side is exactly the torque density the field exerts on the matter. The field's angular momentum is not conserved by itself, and the deficit is not lost: it appears in the matter. That is what the worked example shows, and it is why the sum is the conserved thing.

Why does the sign of the stress tensor depend on the signature at all? Isn't energy density an observable?

The energy density is an observable and its sign is fixed. What is not fixed is the formula: $\eta^{\mu\nu}$ appears explicitly in it, and $F^{\mu\alpha}F^{\nu}{}_{\alpha}$ hides another factor of $\eta$ in the lowered index. Flip the signature, both change, and the net effect is an overall sign on $T^{\mu\nu}$ . So the expression $\tfrac{1}{\mu_0}[F^{\mu\alpha}F^{\nu}{}_{\alpha} - \tfrac14\eta^{\mu\nu}F^2]$ is right in one convention and wrong in the other, while the physics — energy density $\tfrac12(\varepsilon_0E^2 + B^2/\mu_0)$ , positive — is right in both. Hence recomputing $T^{00}$ from components rather than quoting a formula: it is the one check no convention can fake.

Is the Belinfante improvement a real physical statement, or a trick to get a symmetric answer?

A legitimate use of a real freedom. Adding $\partial_\lambda B^{\lambda\mu\nu}$ with $B$ antisymmetric in $\lambda\mu$ leaves the charges $\int T^{0\nu}d^3x$ untouched — the extra piece integrates to a surface term, so total energy and momentum are unchanged and only the local distribution is reshuffled. What singles out Belinfante's choice is that it is the unique reshuffling making the tensor simultaneously symmetric and gauge invariant — and, as both authors noticed, it coincides exactly with the tensor you get by varying the action with respect to the metric, which is what sources gravity. Two independent criteria, one answer. Not a trick.