Classical field theory: Noether again, and where \(T^{\mu\nu}\) actually comes from

Idea. Everything is the Euler–Lagrange equation applied once. The only interesting physics is that an undifferentiated \(u\) in \(\mathcal L\) becomes an undifferentiated \(u\) in the equation, and a constant added to \(\omega^2\) is exactly what makes a wave dispersive.

1. The field equation. With the three derivatives from rung 3,

\[ \partial_t(\mu u_t) + \partial_x(-Tu_x) = -\kappa u \quad\Longrightarrow\quad \mu u_{tt} - Tu_{xx} + \kappa u = 0 . \]

Dimensions: \(\mathcal L\) is \(\mathrm{J\,m^{-1}}\), so \(\tfrac12\kappa u^2\) is \(\mathrm{J\,m^{-1}}\) and \([\kappa] = \mathrm{J\,m^{-3}} = \mathrm{N\,m^{-2}}\) — a spring constant per unit length, as it should be. Each term of the equation is then \(\mathrm{N\,m^{-1}}\): a force per unit length.

2. Dispersion. Substituting \(u = e^{i(kx-\omega t)}\) gives \(-\mu\omega^2 + Tk^2 + \kappa = 0\), that is

\[ \omega^2 = c_s^2k^2 + \omega_0^2 ,\qquad c_s^2 = \frac T\mu,\qquad \omega_0^2 = \frac\kappa\mu . \]

Below the cutoff, \(\omega \lt \omega_0\), the relation forces \(k^2 \lt 0\): \(k = i\varsigma\) with \(\varsigma\) real, and the solution is \(e^{-\varsigma x}e^{-i\omega t}\). Nothing propagates — the disturbance is evanescent, decaying with distance while oscillating in place, and the driver does no net work on it in steady state. This is exactly what an ionospheric radio wave does below the plasma frequency, and what a waveguide mode does below its cutoff.

3. Numbers. \(c_s = \sqrt{73/4.0\times10^{-4}} = \sqrt{1.825\times10^{5}} = \mathbf{427.2\ \mathrm{m\,s^{-1}}}\). \(\omega_0 = \sqrt{1.0\times10^{4}/4.0\times10^{-4}} = \sqrt{2.5\times10^{7}} = 5000\ \mathrm{rad\,s^{-1}}\), so \(f_0 = 5000/2\pi = \mathbf{795.8\ Hz}\). For the group velocity, \(v_{\text{gr}} = c_s(1+\omega_0^2/c_s^2k^2)^{-1/2} = c_s/2\) needs \(\omega_0^2/c_s^2k^2 = 3\), so \(k = \omega_0/(c_s\sqrt3) = 5000/(427.2\times1.7321) = 6.757\ \mathrm m^{-1}\) and

\[ \lambda = \frac{2\pi}{k} = \mathbf{0.930\ m} . \]

4. The comparison. Divide the field equation by \(\mu\): \(u_{tt} - c_s^2u_{xx} + \omega_0^2 u = 0\), which is \(\Box u + (\omega_0/c_s)^2u = 0\) with \(c_s\) in the place of \(c\). The slides' \(\kappa_{\text{KG}} = mc/\hbar\) corresponds to \(\omega_0/c_s = \sqrt{\kappa/T}\): the stiffness of the foundation relative to the tension plays the part of the mass. Mass is not a mysterious ingredient in a field theory; it is a term in \(\mathcal L\) proportional to \(\varphi^2\), and its only effect is to add a constant to \(\omega^2\).

Check. Dimensions: \(\omega_0^2 = \kappa/\mu\) is \(\mathrm{N\,m^{-2}}/\mathrm{kg\,m^{-1}} = \mathrm{s^{-2}}\) ✓. Limit: \(\kappa\to0\) gives \(\omega = c_sk\), unit 18's string, non-dispersive ✓, and \(\lambda\to\infty\) for the half-speed wavelength, as it must since there is then no scale to set one. Number: without the foundation, a \(0.65\) m length of this string has fundamental \(c_s/2L = 328.6\) Hz, about five cents flat of E4 at \(329.63\) Hz — a plausible guitar string, so the parameters are not invented nonsense.

Idea. One Lagrangian, run through every machine in the unit. The punchline is part 3: for a scalar field the canonical and metric-variation tensors are equal on the nose, which is why the Maxwell case is the one that teaches the lesson.

1. \(\pi^\mu = \partial\mathcal L/\partial(\partial_\mu\varphi) = -\partial^\mu\varphi\) and \(\partial\mathcal L/\partial\varphi = -\kappa^2\varphi\), so the Euler–Lagrange equation \(\partial_\mu\pi^\mu = \partial\mathcal L/\partial\varphi\) reads \(-\partial_\mu\partial^\mu\varphi = -\kappa^2\varphi\). Since \(\partial_\mu\partial^\mu = -\Box\),

\[ \Box\varphi + \kappa^2\varphi = 0,\qquad \Box = \frac1{c^2}\partial_t^2 - \nabla^2 . \]

2. The canonical tensor. \(T^{\mu\nu}_{\text{can}} = \eta^{\mu\nu}\mathcal L - (-\partial^\mu\varphi)\partial^\nu\varphi = \partial^\mu\varphi\,\partial^\nu\varphi + \eta^{\mu\nu}\mathcal L\), manifestly symmetric. Componentwise, with \(\partial_0 = c^{-1}\partial_t\) and \(\mathcal L = \frac12(\partial_0\varphi)^2 - \frac12|\nabla\varphi|^2 - \frac12\kappa^2\varphi^2\):

\[ T^{00} = (\partial_0\varphi)^2 - \mathcal L = \tfrac{1}{2c^2}\dot\varphi^2 + \tfrac12|\nabla\varphi|^2 + \tfrac12\kappa^2\varphi^2 \ \ge 0\ ✓ ,\qquad T^{0i} = -\tfrac1c\dot\varphi\,\partial_i\varphi . \]

Three non-negative terms: kinetic, gradient and potential energy density. (Sign trail for \(T^{00}\): \(\partial^0\varphi\partial^0\varphi = (-\partial_0\varphi)^2 = (\partial_0\varphi)^2\) and \(\eta^{00}\mathcal L = -\mathcal L\).)

3. By metric variation. Write \(\sqrt{-g}\mathcal L = -\frac12\sqrt{-g}\,(g^{\mu\nu}\partial_\mu\varphi\partial_\nu\varphi + \kappa^2\varphi^2)\). The inverse metric appears once, giving \(-\frac12\partial_\mu\varphi\partial_\nu\varphi\,\delta g^{\mu\nu}\) times \(\sqrt{-g}\); and \(\delta\sqrt{-g} = -\frac12\sqrt{-g}g_{\mu\nu}\delta g^{\mu\nu}\) multiplies \(\mathcal L\). Hence

\[ \delta(\sqrt{-g}\mathcal L) = -\tfrac12\sqrt{-g}\big[\partial_\mu\varphi\,\partial_\nu\varphi + g_{\mu\nu}\mathcal L\big]\delta g^{\mu\nu} , \]

and comparison with the definition gives \(T_{\mu\nu} = \partial_\mu\varphi\,\partial_\nu\varphi + g_{\mu\nu}\mathcal L\), which at \(g = \eta\) is exactly part 2 ✓. No improvement term appeared because the canonical tensor was already symmetric and already gauge invariant — there is nothing for an improvement to repair. A scalar field carries no spacetime index, so it has no spin current, and Belinfante's construction produces \(B = 0\).

4. Conservation. \(\partial_\mu T^{\mu\nu} = (\partial_\mu\partial^\mu\varphi)\partial^\nu\varphi + \partial^\mu\varphi\,\partial_\mu\partial^\nu\varphi + \partial^\nu\mathcal L\). The middle term is \(\partial^\nu\big(\frac12\partial^\mu\varphi\partial_\mu\varphi\big)\), and \(\partial^\nu\mathcal L = -\partial^\nu\big(\frac12\partial^\mu\varphi\partial_\mu\varphi\big) - \kappa^2\varphi\,\partial^\nu\varphi\), so those two cancel and

\[ \partial_\mu T^{\mu\nu} = \big(\partial_\mu\partial^\mu\varphi - \kappa^2\varphi\big)\partial^\nu\varphi = -\big(\Box\varphi + \kappa^2\varphi\big)\partial^\nu\varphi = 0 \]

on shell ✓ — and only on shell, as Noether's theorem promised.

5. Dispersion and numbers. \(\omega^2 = c^2k^2 + \omega_0^2\) with \(\omega_0 = c\kappa = mc^2/\hbar\). Then \(v_{\text{ph}} = \omega/k\) and \(v_{\text{gr}} = \mathrm d\omega/\mathrm dk = c^2k/\omega\), whose product is \(c^2\) identically ✓. For the pion,

\[ \kappa^{-1} = \frac{\hbar c}{mc^2} = \frac{197.3269804\ \mathrm{MeV\,fm}}{139.57039\ \mathrm{MeV}} = \mathbf{1.4138\ fm} ,\qquad \omega_0 = \frac{mc^2}{\hbar} = \mathbf{2.1204\times10^{23}\ rad\,s^{-1}} . \]

(The second from \(139.57039\times10^{6}\times1.602176634\times10^{-19}/1.054571817\times10^{-34}\).)

6. The trace. \(T^\mu{}_\mu = \partial_\mu\varphi\,\partial^\mu\varphi + 4\mathcal L = (\partial\varphi)^2 - 2(\partial\varphi)^2 - 2\kappa^2\varphi^2 = -(\partial\varphi)^2 - 2\kappa^2\varphi^2\), writing \((\partial\varphi)^2 = \partial_\mu\varphi\partial^\mu\varphi\). At \(m = 0\) it is \(-(\partial\varphi)^2\), which is not zero: the massless scalar's metric tensor is not traceless, unlike Maxwell's. (There is a modified "conformally improved" tensor whose trace does vanish on shell, differing from this one by an improvement term; building it needs the conformal group, which this course does not reach. P9 pursues the trace further.)

Check. Dimensions: \([\varphi^2] = \mathrm{J\,m^{-1}}\) from \([\mathcal L] = \mathrm{J\,m^{-3}}\), so \(T^{00} = \frac12|\nabla\varphi|^2\) is \(\mathrm{J\,m^{-1}\,m^{-2}} = \mathrm{J\,m^{-3}}\) ✓. Limit: \(\kappa\to0\) recovers the wave equation and \(\omega = ck\) with \(v_{\text{ph}} = v_{\text{gr}} = c\), consistent with \(v_{\text{ph}}v_{\text{gr}} = c^2\) ✓. Number: \(1.4138\) fm is the pion's range. Yukawa in 1935 took about \(2\) fm off the size of a nucleus and predicted a particle of some \(200\) electron masses, around \(100\) MeV; the pion was found in 1947 at \(139.57\) MeV — the right order, with a range rather shorter than he had guessed.

Another way

Part 4 in one line, from Noether: translations are a symmetry off shell because \(\mathcal L\) has no explicit \(x\), so the theorem delivers \(\partial_\mu T^{\mu\nu} = 0\) on shell with no further computation. Doing it by hand, as above, is worth it once — it shows exactly where the field equation enters, and it is the same manipulation you will need in P9.

Idea. The Euler–Lagrange operator annihilates total divergences. That is the whole of part 1, and part 3 says the corresponding freedom in \(T^{\mu\nu}\) is precisely the improvement freedom — so the two ambiguities of the unit are one ambiguity.

1. Direct. Write \(\Lambda^\mu_{,\varphi} = \partial\Lambda^\mu/\partial\varphi\) and \(\Lambda^\mu_{,\mu}\) for the explicit \(x\)-derivative. Then \(\tilde{\mathcal L}_\Lambda = \Lambda^\mu_{,\varphi}\,\partial_\mu\varphi + \Lambda^\mu_{,\mu}\), so

\[ \frac{\partial\tilde{\mathcal L}_\Lambda}{\partial(\partial_\mu\varphi)} = \Lambda^\mu_{,\varphi} , \qquad \frac{\partial\tilde{\mathcal L}_\Lambda}{\partial\varphi} = \Lambda^\mu_{,\varphi\varphi}\,\partial_\mu\varphi + \Lambda^\mu_{,\mu\varphi} . \]

Taking the total divergence of the first,

\[ \partial_\mu\big(\Lambda^\mu_{,\varphi}\big) = \Lambda^\mu_{,\varphi\varphi}\,\partial_\mu\varphi + \Lambda^\mu_{,\varphi\mu} , \]

and \(\Lambda^\mu_{,\mu\varphi} = \Lambda^\mu_{,\varphi\mu}\) by Clairaut's theorem (unit 03), since \(\Lambda\) is smooth. Subtracting, \(E(\tilde{\mathcal L}_\Lambda) = 0\). By linearity of \(E\), \(E(\tilde{\mathcal L}) = E(\mathcal L)\). \(\square\)

2. Via the boundary. \(\tilde S = S + \frac1c\int_\Omega\partial_\mu\Lambda^\mu\,\mathrm d^4x = S + \frac1c\oint_{\partial\Omega}\Lambda^\mu\mathrm d\Sigma_\mu\), and the surface integral depends only on \(\varphi\) and \(x\) on \(\partial\Omega\). A variation of compact support leaves it untouched, so \(\delta\tilde S = \delta S\) and the stationarity conditions coincide. \(\square\)

The hypothesis the second proof needs and the first does not: the variations must vanish near \(\partial\Omega\) (or the boundary data must be fixed). Part 1 is a pointwise algebraic identity between two differential operators and holds with no reference to any region at all. That matters whenever boundary terms are the point — which is exactly the Gibbons–Hawking–York situation flagged on the slides.

3. The tensor of a total divergence. With \(\Lambda^\mu = \Lambda^\mu(\varphi)\), \(\tilde{\mathcal L}_\Lambda = \Lambda^\mu_{,\varphi}\partial_\mu\varphi = \partial_\mu\Lambda^\mu\) and \(\partial\tilde{\mathcal L}_\Lambda/\partial(\partial_\mu\varphi) = \Lambda^\mu_{,\varphi}\), so

\[ T^{\mu\nu}_\Lambda = \eta^{\mu\nu}\partial_\alpha\Lambda^\alpha - \Lambda^\mu_{,\varphi}\,\partial^\nu\varphi = \eta^{\mu\nu}\partial_\alpha\Lambda^\alpha - \partial^\nu\Lambda^\mu , \]

the last step because \(\partial^\nu\Lambda^\mu = \Lambda^\mu_{,\varphi}\partial^\nu\varphi\) when \(\Lambda\) depends on \(x\) only through \(\varphi\). Now set

\[ B^{\lambda\mu\nu} := \eta^{\mu\nu}\Lambda^\lambda - \eta^{\lambda\nu}\Lambda^\mu . \]

Then \(\partial_\lambda B^{\lambda\mu\nu} = \eta^{\mu\nu}\partial_\lambda\Lambda^\lambda - \partial^\nu\Lambda^\mu = T^{\mu\nu}_\Lambda\) ✓, and swapping the first two indices gives \(B^{\mu\lambda\nu} = \eta^{\lambda\nu}\Lambda^\mu - \eta^{\mu\nu}\Lambda^\lambda = -B^{\lambda\mu\nu}\) ✓.

4. Adding a total divergence to \(\mathcal L\) changes nothing physical — not the field equations, by parts 1 and 2 — and changes \(T^{\mu\nu}_{\text{can}}\) by exactly an improvement term, hence not the total four-momentum either. So the canonical tensor is ambiguous at least to the extent that \(\mathcal L\) is, and the two ambiguities are the same one. No amount of care in choosing \(\mathcal L\) will pin \(T\) down; a different question must be asked, and the metric variation is that question.

Check. Special case: \(\Lambda^\mu = \eta^{\mu 0}\Lambda(\varphi)\), a "total time derivative". Then \(B^{\lambda\mu\nu} = \eta^{\mu\nu}\eta^{\lambda0}\Lambda - \eta^{\lambda\nu}\eta^{\mu0}\Lambda\), whose \(\lambda = \mu\) components vanish ✓, and \(\partial_\alpha\Lambda^\alpha = \partial^0\Lambda = -\partial_0\Lambda\) while \(\partial^0\Lambda^0 = \eta^{00}\partial_0(\eta^{00}\Lambda) = \partial_0\Lambda\), so \(T^{00}_\Lambda = \eta^{00}(-\partial_0\Lambda) - \partial_0\Lambda = 0\): adding a total time derivative to a Lagrangian changes the energy density not at all, which is unit 13's statement that the Noether charge is unchanged by adding a total derivative to \(L\) ✓. Dimensions: \([\Lambda^\mu] = [\mathcal L]\cdot\mathrm m\), so \([B^{\lambda\mu\nu}] = \mathrm{J\,m^{-2}}\) and \([\partial B] = \mathrm{J\,m^{-3}}\) ✓, the same as \(T\).

Idea. A global symmetry gives a conserved charge. Demanding the same symmetry locally — independently at each event — is a far stronger demand, and the only way to meet it is to invent a field that transforms so as to absorb \(\partial_\mu\theta\). That field is \(A_\mu\). Electromagnetism is the price of taking a phase symmetry seriously.

1. The global current. \(\mathcal L\) pairs each \(\phi\) with a \(\phi^*\), so \(e^{i\theta}\) cancels against \(e^{-i\theta}\) term by term: invariant exactly, \(K^\mu = 0\). With \(\delta\phi = i\varepsilon\phi\), \(\delta\phi^* = -i\varepsilon\phi^*\) and the momenta from rung 4,

\[ j^\mu = -\partial^\mu\phi^*\,(i\phi) - \partial^\mu\phi\,(-i\phi^*) = i\big(\phi^*\partial^\mu\phi - \phi\,\partial^\mu\phi^*\big) , \]

which is real, being \(i\) times an imaginary quantity. Directly:

\[ \partial_\mu j^\mu = i\big(\phi^*\partial_\mu\partial^\mu\phi - \phi\,\partial_\mu\partial^\mu\phi^*\big) \]

— the cross terms \(\partial_\mu\phi^*\partial^\mu\phi\) cancel — and on shell \(\partial_\mu\partial^\mu\phi = \kappa^2\phi\), likewise for \(\phi^*\), leaving \(i\kappa^2(\phi^*\phi - \phi\phi^*) = 0\) ✓.

2. Local phase breaks it. \(\partial_\mu(e^{i\theta}\phi) = e^{i\theta} (\partial_\mu\phi + i\phi\,\partial_\mu\theta)\), so

\[ \Delta\mathcal L = -i\,\partial_\mu\theta\,\big(\phi\,\partial^\mu\phi^* - \phi^*\partial^\mu\phi\big) \;-\; \partial_\mu\theta\,\partial^\mu\theta\;\phi^*\phi = j^\mu\partial_\mu\theta - (\partial\theta)^2\phi^*\phi , \]

since \(-i(\phi\,\partial^\mu\phi^* - \phi^*\partial^\mu\phi) = j^\mu\).

The mass term is untouched. \(\Delta\mathcal L\) is not zero, and it is not a total divergence either: for constant \(\theta\) it vanishes, but for general \(\theta\) the first term is \(j^\mu\partial_\mu\theta = \partial_\mu(\theta j^\mu) - \theta\,\partial_\mu j^\mu\), a divergence only on shell — and a symmetry has to hold off shell.

3. The repair. Demand \(D'_\mu\phi' = e^{i\theta}D_\mu\phi\) where \(\phi' = e^{i\theta}\phi\) and \(D'_\mu = \partial_\mu - i\frac q\hbar A'_\mu\). Expanding,

\[ D'_\mu\phi' = e^{i\theta}\Big[\partial_\mu\phi + i(\partial_\mu\theta)\phi - i\tfrac q\hbar A'_\mu\phi\Big] , \]

which equals \(e^{i\theta}[\partial_\mu\phi - i\frac q\hbar A_\mu\phi]\) precisely when

\[ \boxed{\ A'_\mu = A_\mu + \frac\hbar q\,\partial_\mu\theta\ } \]

— the gauge transformation of the four-potential, with \(\chi = (\hbar/q)\theta\). Then \(\mathcal L = -(D_\mu\phi)^*(D^\mu\phi) - \kappa^2\phi^*\phi\) is invariant, because every factor carries \(e^{\pm i\theta}\) and they cancel in pairs.

4. What the covariant derivative contains.

\[ -(D_\mu\phi)^*(D^\mu\phi) = -\partial_\mu\phi^*\partial^\mu\phi \;-\; \frac q\hbar A^\mu\,j_\mu \;-\; \frac{q^2}{\hbar^2}A_\mu A^\mu\,\phi^*\phi , \]

using \(i(\phi\,\partial_\mu\phi^* - \phi^*\partial_\mu\phi) = -j_\mu\). Comparing the linear term with the Maxwell coupling \(A_\mu J^\mu\),

\[ J^\mu = -\frac q\hbar\,j^\mu = -\frac{iq}{\hbar}\big(\phi^*\partial^\mu\phi - \phi\,\partial^\mu\phi^*\big) , \]

which fixes the scale and the sign that Noether's theorem left free, and identifies the conserved charge of the phase symmetry, up to that factor, as electric charge in units of \(q\). The minus is bookkeeping, not physics — generating the symmetry with \(\delta\phi = -i\varepsilon\phi\) would make it a plus — and the check that it is the right way round is a plane wave: for \(\phi\propto e^{-i\omega t}\), \(J^0 = 2q\omega|\phi|^2/\hbar c\), a positive charge density for a field of charge \(q\) ✓. The quadratic term — the "seagull" — is not optional: without it \(\mathcal L\) is not invariant, since \(A_\mu A^\mu\) changes under the transformation in exactly the way needed to cancel the \((\partial\theta)^2\) piece of part 2.

5. Honest accounting. Derived: that a field with the transformation law of \(A_\mu\) must exist; that it couples to the conserved current of the phase symmetry and to nothing else; that the coupling is universal, one constant \(q\) per field. Assumed: that \(\mathcal L\) should contain a kinetic term for \(A\) at all — nothing here produces \(-F^2/4\mu_0\), which must be added by hand as the simplest gauge-invariant option, exactly as the slides said. So gauging produces the coupling and the gauge symmetry; it does not produce Maxwell's Lagrangian.

Check. Special case: a real field has \(\phi = \phi^*\), so \(j^\mu = 0\) — no charge, and nothing to gauge ✓. Consistency: \(A'_\mu - A_\mu = (\hbar/q)\partial_\mu\theta\) is a gradient, so \(F_{\mu\nu}\) is invariant ✓, and the two notions of gauge transformation — the one from unit 23 and the one derived here — coincide. Dimensions: \([\hbar\partial_\mu\theta/q] = \mathrm{J\,s\,m^{-1}\,C^{-1}} = \mathrm{V\,s\,m^{-1}}\), the dimension of \(A_\mu\) ✓.

Another way

A slicker route to part 3: define \(D_\mu\) by demanding that it be a derivative under which the combination \(\phi^*D^\mu\phi\) transforms with no extra terms. That is the geometric statement — \(A_\mu\) is a connection, and \(F_{\mu\nu}\) its curvature — which unit 32 makes precise for the metric connection and unit 33 for its curvature. The commutator \([D_\mu, D_\nu]\phi = -i\frac q\hbar F_{\mu\nu}\phi\) is the same computation as the one defining the Riemann tensor, with a one-dimensional phase in place of the tangent space.

Idea. Both arguments give the same equation and mean different things. One is about whether the theory can be written down; the other is about whether it can be solved.

1. Off shell. \(\Delta S = \frac1c\int J^\mu\partial_\mu\chi\,\mathrm d^4x = \frac1c\oint\chi J^\mu\mathrm d\Sigma_\mu - \frac1c\int(\partial_\mu J^\mu)\chi\,\mathrm d^4x\). The surface term vanishes because \(\chi\) has compact support. If \(\Delta S = 0\) for every such \(\chi\), then \(\int(\partial_\mu J^\mu)\chi\,\mathrm d^4x = 0\) for every \(\chi\), and the fundamental lemma gives \(\partial_\mu J^\mu = 0\). Conversely, if \(\partial_\mu J^\mu = 0\) then \(\Delta S = 0\). \(\square\)

2. On shell. Apply \(\partial_\mu\) to \(\partial_\nu F^{\mu\nu} = \mu_0J^\mu\):

\[ \mu_0\,\partial_\mu J^\mu = \partial_\mu\partial_\nu F^{\mu\nu} = 0 , \]

because \(\partial_\mu\partial_\nu\) is symmetric in \(\mu\nu\) (Clairaut) while \(F^{\mu\nu}\) is antisymmetric, and \(S^{\mu\nu}A_{\mu\nu} = -S^{\nu\mu}A_{\nu\mu} = -S^{\mu\nu}A_{\mu\nu}\) forces the contraction to vanish. \(\square\)

3. The difference. Part 1 never uses a field equation; it is a statement about the functional \(S\) and holds for every configuration \(A\). Part 2 uses the field equation and so is a statement about solutions. Handed a current with \(\partial_\mu J^\mu\ne0\): by part 1 the action is not gauge invariant, so \(S\) assigns different values to configurations that are physically identical and the variational principle is ill posed; by part 2 there is no solution of Maxwell's equations with that source at all, gauge invariance aside. The first is the deeper complaint, and the one that generalises: for a local symmetry, invariance of the action always forces an identity on whatever the symmetry's gauge field couples to. Unit 35 meets the gravitational version, \(\nabla_\mu T^{\mu\nu} = 0\) forced by coordinate invariance.

4. Identity versus equation. Substituting \(F = \partial A\) into \(\partial_\lambda F_{\mu\nu} + \partial_\mu F_{\nu\lambda} + \partial_\nu F_{\lambda\mu}\) gives six terms which cancel in pairs by Clairaut, for every \(A\) whatsoever — solutions and non-solutions alike. An equation that every configuration satisfies restricts nothing, so it cannot be a field equation; it is the price of writing \(F\) as a curl, paid in advance.

Counting. The identity is totally antisymmetric in three indices each running over four values, so it has \(\binom43 = \mathbf 4\) independent components — one scalar equation \(\nabla\cdot\vec B = 0\) and one vector equation, Faraday's law. The source equations have a free index \(\mu\), so four components, but part 2 shows that \(\partial_\mu\) of them is an identity given \(\partial_\mu J^\mu = 0\); so only \(\mathbf 3\) are independent as evolution equations, and the fourth — Gauss's law — is a constraint, propagated by the others. That is exactly unit 24's split of Maxwell's equations into two evolution equations and two constraints, now visible as a counting argument rather than an observation.

Check. Consistency: \(4 + 4 = 8\) equations for the six components of \(F\), which is only possible because two combinations are not independent — which is what the counting just established ✓. Special case: for \(J = 0\) part 1 is vacuous (any \(J\) is conserved) and part 2 gives nothing, and indeed the vacuum equations are gauge invariant for free ✓.

Idea. The antisymmetry does all the work twice: it makes the improvement identically conserved, and it removes the time component that would otherwise contribute to the charge. The numbers in part 4 show that "the same total" and "the same density" are very different claims.

1. \(\partial_\mu\partial_\lambda\) is symmetric under \(\mu\leftrightarrow\lambda\) by Clairaut; \(B^{\lambda\mu\nu}\) is antisymmetric in the same pair. Therefore

\[ \partial_\mu\partial_\lambda B^{\lambda\mu\nu} = -\partial_\lambda\partial_\mu B^{\mu\lambda\nu} = -\partial_\mu\partial_\lambda B^{\lambda\mu\nu} , \]

(relabelling the dummies in the last step), so the quantity equals minus itself and vanishes. Hence \(\partial_\mu\tilde T^{\mu\nu} = \partial_\mu T^{\mu\nu} = 0\). \(\square\)

2. \(\tilde P^\nu - P^\nu = \frac1c\int\partial_\lambda B^{\lambda0\nu}\,\mathrm d^3x\). The \(\lambda = 0\) term is \(\partial_0 B^{00\nu}\), and \(B^{00\nu} = -B^{00\nu} = 0\). So the integrand is \(\partial_i B^{i0\nu}\), a spatial divergence, and by the divergence theorem (unit 05)

\[ \int_{|\vec x|\le R}\partial_iB^{i0\nu}\,\mathrm d^3x = \oint_{|\vec x| = R}B^{i0\nu}n_i\,\mathrm dA . \]

The hypothesis: \(B^{i0\nu} = o(R^{-2})\) as \(R\to\infty\), so that the surface integral, whose area grows as \(R^2\), tends to zero. \(\square\)

3. A radiating system. For an accelerating charge the fields fall off as \(1/r\) rather than \(1/r^2\) (unit 24's radiation fields), so \(B \sim FA \sim r^{-1}\cdot r^{-1} = r^{-2}\) and the surface integral does not vanish. Physically the system is losing energy and momentum through the sphere at infinity, so \(P^\nu\) inside a finite ball is genuinely not constant and the two tensors need not agree on it. The clean statement is always the local one, \(\partial_\mu T^{\mu\nu} = 0\); the global one is a corollary with a hypothesis attached.

4. The two uniform fields.

(i) Electrostatic. \(A^0 = \varphi/c = -Ez/c\), \(F^{03} = E_3/c = E/c\), all other \(F^{0\lambda} = 0\). So \(B^{300} = \mu_0^{-1}(E/c)(-Ez/c) = -\varepsilon_0E^2z\) and

\[ \partial_\lambda B^{\lambda00} = \partial_3B^{300} = -\varepsilon_0E^2 = -8.854\ \mathrm{J\,m^{-3}}\quad\text{at }E = 1.00\ \mathrm{MV\,m^{-1}} . \]

The metric tensor gives \(T^{00} = \frac{\varepsilon_0}{2}E^2 = +4.427\ \mathrm{J\,m^{-3}}\); the canonical one gives \(T^{00}_{\text{can}} = -\mathcal L = -\frac{\varepsilon_0}{2}E^2 = -4.427\ \mathrm{J\,m^{-3}}\) (the extra term dies because \(\partial^0\) kills a static field), and \(-4.427 = 4.427 - 8.854\) ✓.

(ii) Magnetostatic. \(\varphi = 0\), so \(A^0 = 0\) and \(B^{\lambda00} = \mu_0^{-1}F^{0\lambda}A^0 = 0\) identically: the improvement term vanishes before anything is differentiated. Both tensors give

\[ T^{00} = T^{00}_{\text{can}} = \frac{B^2}{2\mu_0} = \mathbf{3.979\times10^{5}\ \mathrm{J\,m^{-3}}} \quad\text{at } B = 1.00\ \mathrm T \]

(check: \(T^{00}_{\text{can}} = -\mathcal L = +B^2/2\mu_0\) since \(\mathcal L = -B^2/2\mu_0\) here, and \(F^{0i} = E_i/c = 0\) kills the other term).

The sentence: the improvement term is \(\partial_\lambda(\mu_0^{-1}F^{0\lambda}A^\nu)\), which for \(\nu = 0\) is proportional to the scalar potential — so it bites exactly when the chosen gauge has \(\varphi\ne0\), and the electrostatic field cannot avoid that while the magnetostatic one can.

Check. Dimensions: \([B^{\lambda\mu\nu}] = [F][A]/\mu_0\) is \(\mathrm{T\cdot T\,m/(N\,A^{-2})} = \mathrm{J\,m^{-2}}\) ✓, so its divergence is an energy density. Limit: \(E\to0\) in case (i) gives \(0 = 0 - 0\) ✓. Consistency: both cases obey \(T^{00}_{\text{can}} - T^{00} = \partial_\lambda B^{\lambda00}\), which is the identity the slides proved, checked here on two independent configurations ✓.

Idea. Everything unit 24 obtained by manipulating Maxwell's equations is sitting in sixteen components of one tensor. The algebra is short if the dictionary is written down first.

1. Energy flux and momentum density. Since \(\eta^{0i} = 0\), the trace term is absent, and \(F^{00} = 0\) kills \(\alpha = 0\):

\[ T^{0i} = \frac{1}{\mu_0}F^{0j}F^i{}_j = \frac{1}{\mu_0}\frac{E_j}{c}\,\epsilon_{ijk}B_k = \frac{(\vec E\times\vec B)_i}{\mu_0 c} = \frac{S^i}{c} , \]

using \(F^i{}_j = \eta^{ik}F_{kj} = F_{ij} = \epsilon_{ijk}B_k\). Hence \(cT^{i0} = S^i\) ✓ and \(T^{0i}/c = S^i/c^2 = \varepsilon_0(\vec E\times\vec B)^i = \vec g\) ✓. The symmetry \(T^{0i} = T^{i0}\) asserts that energy flux and momentum density are the same quantity up to a factor \(c^2\): \(\vec g = \vec S/c^2\), which unit 24 stated and could not derive. It is \(E = mc^2\) applied to energy in transit.

2. The spatial block. Split \(F^{i\alpha}F^j{}_\alpha\) into \(\alpha = 0\) and \(\alpha = k\):

\[ F^{i0}F^j{}_0 = \Big(-\frac{E_i}{c}\Big)\Big(\frac{E_j}{c}\Big) = -\frac{E_iE_j}{c^2} ,\qquad F^{ik}F^j{}_k = \epsilon_{ikm}\epsilon_{jkn}B_mB_n = \delta_{ij}B^2 - B_iB_j , \]

and \(-\tfrac14\delta^{ij}F^2 = -\tfrac12\delta^{ij}B^2 + \tfrac12\delta^{ij}E^2/c^2\). Adding and dividing by \(\mu_0\), with \(1/\mu_0c^2 = \varepsilon_0\),

\[ T^{ij} = -\varepsilon_0\Big(E_iE_j - \tfrac12\delta_{ij}E^2\Big) - \frac{1}{\mu_0}\Big(B_iB_j - \tfrac12\delta_{ij}B^2\Big) \ ✓ \]

The minus sign: unit 24's \(T^{ij}_{\text{Maxwell}}\hat n_j\) is the force per unit area the field exerts across a surface, whereas \(T^{ij}\) here is the flux of \(j\)-momentum in the \(i\)-direction. Momentum flowing out is momentum lost, so the two differ by a sign; the same convention clash exists between the stress tensor and the momentum flux in unit 22's continuum mechanics.

3. Conservation. \(\partial_\mu T^{\mu0} = \partial_0T^{00} + \partial_iT^{i0} = c^{-1}\partial_t u + c^{-1}\partial_i S^i = 0\), that is

\[ \frac{\partial u}{\partial t} + \nabla\cdot\vec S = 0 , \]

Poynting's theorem in vacuum. With sources the right-hand side becomes \(-\vec J\cdot\vec E\), because the field's \(T\) alone is then not conserved — it exchanges energy with the charges.

4. The numbers.

• Energy density \(u = S_\odot/c = 1361/2.99792458\times10^{8} = \mathbf{4.540\times10^{-6}\ J\,m^{-3}}\).
• Radiation pressure, absorber: \(\langle T^{33}\rangle = u = \mathbf{4.540\ \mu Pa}\).
• Mirror: the light leaves with reversed momentum, so twice as much, \(\mathbf{9.080\ \mu Pa}\).
• Force on \(1.0\ \mathrm{km^2} = 10^6\ \mathrm{m^2}\) of absorber: \(4.540\times10^{-6}\times10^{6} = \mathbf{4.54\ N}\).
• Amplitudes: \(S_\odot = \frac12\varepsilon_0cE_0^2\) gives \(E_0 = \sqrt{2S_\odot/\varepsilon_0c} = \mathbf{1013\ V\,m^{-1}}\) and \(B_0 = E_0/c = \mathbf{3.378\ \mu T}\).

Check. Dimensions: \(u = S/c\) is \(\mathrm{W\,m^{-2}}/\mathrm{m\,s^{-1}} = \mathrm{J\,m^{-3}}\) ✓, and a pressure is \(\mathrm{N\,m^{-2}} = \mathrm{J\,m^{-3}}\) ✓ — energy density and pressure share a dimension, which is why \(T\) can hold both. Sanity: \(B_0 = 3.378\ \mathrm{\mu T}\) is about a fifteenth of the Earth's field, and \(4.54\) N over a square kilometre is the weight of a \(463\) g mass at the Earth's surface. Solar sailing is real and very slow: a square-kilometre mirror sail pushes with under \(10\) N. Limit: with \(\vec B = 0\) part 2 gives \(T^{ij} = -\varepsilon_0(E_iE_j - \frac12\delta_{ij}E^2)\), whose \(zz\) component for \(\vec E = E\hat z\) is \(-\frac12\varepsilon_0E^2\) — a negative momentum flux along the field, which is the familiar statement that field lines pull along their length ✓.

Idea. A plane wave is built from one null vector \(k\) and one polarisation \(a\) orthogonal to it. Every statement about it — that the invariants vanish, that \(T\) is traceless, that the energy moves at \(c\) — is a statement about \(k\cdot k = 0\), and the calculation is three lines if you never introduce components.

1. The field. \(\partial_\mu A_\nu = k_\mu a_\nu f'\), so

\[ F_{\mu\nu} = (k_\mu a_\nu - k_\nu a_\mu)\,f'(k\cdot x) . \]

Then \(\partial_\nu F^{\mu\nu} = (k^\mu a^\nu - k^\nu a^\mu)k_\nu f'' = \big[k^\mu(a\cdot k) - a^\mu(k\cdot k)\big]f''\), which vanishes for every \(f\) exactly when \(k\cdot a = 0\) and \(k\cdot k = 0\) ✓. (The Bianchi identity holds automatically, since \(F\) came from an \(A\).)

2. The invariants. \(F_{\mu\nu}F^{\mu\nu} = (k_\mu a_\nu - k_\nu a_\mu)(k^\mu a^\nu - k^\nu a^\mu)f'^2 = 2\big[(k\cdot k)(a\cdot a) - (k\cdot a)^2\big]f'^2 = 0\) ✓. For the second, \(\epsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma}\) contains \(\epsilon^{\mu\nu\rho\sigma}k_\mu a_\nu k_\rho a_\sigma\) up to factors, and \(\epsilon\) is totally antisymmetric while \(k_\mu k_\rho\) is symmetric in \(\mu\rho\): the contraction vanishes ✓. In three-vector language these are \(B^2 = E^2/c^2\) and \(\vec E\cdot\vec B = 0\), the two defining properties of a null field — no boost can remove either \(\vec E\) or \(\vec B\).

3. The tensor. Expanding the product,

\[ F^{\mu\alpha}F^\nu{}_\alpha = \Big[k^\mu k^\nu (a\cdot a) - k^\mu a^\nu (k\cdot a) - a^\mu k^\nu(k\cdot a) + a^\mu a^\nu (k\cdot k)\Big]f'^2 = (a\cdot a)\,k^\mu k^\nu f'^2 , \]

and the trace term is zero by part 2, so \(T^{\mu\nu} = \mu_0^{-1}(a\cdot a)f'^2\,k^\mu k^\nu\) ✓. Consequences: \(T^\mu{}_\mu \propto k\cdot k = 0\), traceless ✓; with \(k^\mu = (\omega/c)(1,0,0,1)\) for propagation along \(z\), every component is \(u\) times the corresponding product of \((1,0,0,1)\), so \(T^{00} = T^{03} = T^{33} = u\) — energy density, energy flux over \(c\), and momentum flux all equal, which is exactly what "the energy moves along \(\hat z\) at speed \(c\)" means. Only the one null direction appears.

4. Three-vector form. \(\vec E = E_0\cos(kz-\omega t)\hat x\) forces \(\vec B = (E_0/c)\cos(kz-\omega t)\hat y\) (from Faraday, or from \(\vec B = \hat k\times\vec E/c\)), so

\[ u = \tfrac{\varepsilon_0}{2}E^2 + \tfrac{1}{2\mu_0}B^2 = \varepsilon_0E_0^2\cos^2(kz-\omega t) , \qquad \langle u\rangle = \tfrac12\varepsilon_0E_0^2 , \]

the two halves being equal because \(B = E/c\). Hence \(I = c\langle u\rangle = \tfrac12\varepsilon_0cE_0^2\).

5. The laser. \(I = P/\pi w^2 = 10.0/(\pi(5.0\times10^{-5})^2) = \mathbf{1.273\times10^{9}\ W\,m^{-2}}\). Then

\[ E_0 = \sqrt{\frac{2I}{\varepsilon_0c}} = \sqrt{\frac{2\times1.2732\times10^{9}}{8.8541878128\times10^{-12}\times2.99792458\times10^{8}}} = \mathbf{9.795\times10^{5}\ V\,m^{-1}} , \]

\(B_0 = E_0/c = \mathbf{3.267\ mT}\), and the radiation pressure on an absorber is \(\langle T^{33}\rangle = I/c = \mathbf{4.247\ Pa}\).

Check. Dimensions: \([T^{\mu\nu}] = [a]^2[k]^2/\mu_0\); with \([a] = \mathrm{V\,s\,m^{-1}}\) and \([k] = \mathrm m^{-1}\) this is \(\mathrm{V^2s^2m^{-4}}/(\mathrm{N\,A^{-2}}) = \mathrm{J\,m^{-3}}\) ✓. Limit: \(a\cdot a \to 0\) (a null polarisation, which for \(k\) null and \(k\cdot a = 0\) means \(a \propto k\), i.e. pure gauge) gives \(T = 0\) ✓ — a pure gauge carries no energy. Number: \(E_0 \approx 0.98\ \mathrm{MV\,m^{-1}}\) is a third of the \(3\ \mathrm{MV\,m^{-1}}\) static breakdown field of dry air — and nothing happens, because a static criterion does not transfer to a field that reverses every couple of femtoseconds: optical breakdown in air needs an intensity some six orders of magnitude above this one. And \(4.2\) Pa is the pressure of about \(0.4\) mm of water: optical tweezers work, and only on very small things.

Another way

Part 3 in one line, with no expansion: \(T^{\mu\nu}\) is built quadratically from \(F\), and \(F\) is built from the antisymmetric product \(k\wedge a\). Any quadratic invariant of \(k\wedge a\) must be a combination of \(k\cdot k\), \(k\cdot a\) and \(a\cdot a\) times \(k^\mu k^\nu\), \(k^{(\mu}a^{\nu)}\) or \(a^\mu a^\nu\); the two conditions kill every term but one. Unit 34's exterior algebra makes this the natural way to say it.

Idea. The trace of \(T\) is the response of the action to a change of scale, and Maxwell's action is scale-blind only in four dimensions, because the number of inverse metrics it contains happens to match the number of coordinates. That coincidence is why light has no length scale of its own.

1. Any \(d\). Nothing in the variation used \(d\): \(F_{\mu\nu}\) is still metric-free, \(\delta\sqrt{-g} = -\frac12\sqrt{-g}g_{\mu\nu}\delta g^{\mu\nu}\) still holds (Jacobi's formula is dimension-blind), and the two inverse metrics still give a factor 2. So

\[ T^{\mu\nu} = \frac{1}{\mu_0}\Big(F^{\mu\alpha}F^\nu{}_\alpha - \tfrac14 g^{\mu\nu}F^2\Big) \quad\text{in any }d ,\qquad T^\mu{}_\mu = \frac{1}{\mu_0}\Big(F^2 - \tfrac d4 F^2\Big) = \frac{F^2}{\mu_0}\Big(1 - \frac d4\Big) , \]

using \(\delta^\mu{}_\mu = d\). Zero for all fields exactly at \(d = 4\); at other \(d\) it vanishes only on null fields, which is why P8's plane wave has a traceless tensor in every dimension.

2. The trace is the response to a rescaling. With \(\delta g^{\mu\nu} = -2\omega g^{\mu\nu}\),

\[ \delta S_{\text m} = -\frac{1}{2c}\int\sqrt{-g}\;T_{\mu\nu}\,(-2\omega g^{\mu\nu})\,\mathrm d^dx = \frac1c\int\sqrt{-g}\;\omega\,T^\mu{}_\mu\,\mathrm d^dx . \]

If \(T^\mu{}_\mu = 0\) this vanishes for every \(\omega\); conversely if it vanishes for every compactly supported \(\omega\), the fundamental lemma gives \(T^\mu{}_\mu = 0\). \(\square\) So tracelessness is not a curiosity of Maxwell's tensor: it is the statement that the theory does not know what a length is.

3. Counting powers. Under \(g_{\mu\nu}\to\Omega^2g_{\mu\nu}\) with \(\Omega\) constant: \(\det g\to\Omega^{2d}\det g\), so \(\sqrt{-g}\to\Omega^{d}\sqrt{-g}\); each inverse metric gives \(\Omega^{-2}\), and there are two; \(F_{\mu\nu}\) and \(\mathrm d^dx\) are untouched. Hence

\[ S \to \Omega^{d-4}S , \]

invariant precisely when \(d = 4\) ✓, in agreement with part 1 — and now visibly for the reason that the two inverse metrics "cost" \(\Omega^{-4}\) while the volume element "pays" \(\Omega^{d}\).

4. The scalar. From P2, \(T_{\mu\nu} = \partial_\mu\varphi\partial_\nu\varphi + g_{\mu\nu}\mathcal L\), so in \(d = 4\)

\[ T^\mu{}_\mu = (\partial\varphi)^2 + 4\mathcal L = -(\partial\varphi)^2 - 2\kappa^2\varphi^2 , \]

which is not zero even at \(m = 0\). No contradiction with part 2: the massless scalar action \(-\frac12\int\sqrt{-g}\,g^{\mu\nu}\partial_\mu\varphi\partial_\nu\varphi\,\mathrm d^4x\) scales as \(\Omega^{4-2} = \Omega^2\) with \(\varphi\) held fixed, and part 2's criterion holds the matter fields fixed. The massless scalar is invariant if \(\varphi\) is also rescaled, as \(\varphi\to\Omega^{-1}\varphi\) in four dimensions; the tensor whose trace vanishes on shell is then not this one but an improved version, differing from it by \(\frac16(\eta^{\mu\nu}\partial^2 - \partial^\mu\partial^\nu)\varphi^2\) — an improvement term of the kind P3 and P6 studied. Building it properly needs the conformal group, which this course does not reach.

Check. Special case \(d = 2\): \(T^\mu{}_\mu = F^2/2\mu_0\), and in two dimensions \(F_{\mu\nu}\) has one component, so the electromagnetic field in a two-dimensional spacetime carries a trace — it behaves like a constant energy density, which is the well-known fact that the Coulomb "force" in \(1+1\) dimensions is constant with distance ✓. Consistency: at \(d = 4\), \(T^\mu{}_\mu = 0\) implies \(T^{00} = \sum_i T^{ii}\), and for the plane wave of P8 that reads \(u = T^{33}\) ✓, the fact that made the radiation pressure equal to the energy density in P7. Dimensions: \(\omega\) is dimensionless, so part 2's integrand has the dimensions of \(\mathcal L\) ✓.

Idea. Angular momentum is built from \(T\) by taking a moment, exactly as in mechanics. The moment is conserved only if the tensor is symmetric, and when it is not, the missing piece is a genuine physical object — the spin carried by the field's own indices.

1. \(\partial_\lambda(x^\mu T^{\lambda\nu}) = (\partial_\lambda x^\mu)T^{\lambda\nu} + x^\mu\partial_\lambda T^{\lambda\nu} = \delta^\mu_\lambda T^{\lambda\nu} + 0 = T^{\mu\nu}\). Antisymmetrising in \(\mu\nu\) gives \(\partial_\lambda M^{\lambda\mu\nu} = T^{\mu\nu} - T^{\nu\mu}\). \(\square\) Hence \(M\) is conserved if and only if \(T\) is symmetric.

2. \(J^{12} = \frac1c\int(x^1T^{02} - x^2T^{01})\mathrm d^3x\). Since the momentum density is \(g^i = T^{0i}/c\),

\[ J^{12} = \int\big(x^1g^2 - x^2g^1\big)\mathrm d^3x = \int(\vec x\times\vec g)_3\,\mathrm d^3x , \]

the \(z\)-component of the total angular momentum: unit 08's \(\vec r\times\vec p\), summed over a continuum. The three \(J^{ij}\) are the three components of angular momentum.

3. \(J^{0i} = \frac1c\int(x^0T^{0i} - x^iT^{00})\mathrm d^3x = ct\,P^i - \frac1c\int x^iT^{00}\mathrm d^3x\) (using \(P^i = \frac1c\int T^{0i}\mathrm d^3x\), and that \(x^0 = ct\) is constant on the slice). Define the centre of energy \(R^i = \big(\int x^iT^{00}\mathrm d^3x\big)/E\) with \(E = \int T^{00}\mathrm d^3x\). Then \(cJ^{0i} = c^2tP^i - ER^i\), and \(\mathrm dJ^{0i}/\mathrm dt = 0\) with \(E\) and \(P\) constant gives

\[ \frac{\mathrm dR^i}{\mathrm dt} = \frac{c^2P^i}{E} , \]

a constant: the centre of energy moves uniformly. This is the relativistic version of unit 13's Galilean boost charge \(\vec G = \vec Pt - M\vec R\), whose conservation said the centre of mass moves at constant velocity. The six charges \(J^{\mu\nu}\) are thus three angular momenta and three "centre of energy" statements, one for each of the six parameters of the Lorentz group.

4. The spin current. The metric tensor is symmetric, so the antisymmetric part of \(T_{\text{can}}\) comes entirely from the improvement term:

\[ T^{\mu\nu}_{\text{can}} - T^{\nu\mu}_{\text{can}} = \partial_\lambda\big(B^{\lambda\mu\nu} - B^{\lambda\nu\mu}\big) = \frac{1}{\mu_0}\partial_\lambda\big(F^{\mu\lambda}A^\nu - F^{\nu\lambda}A^\mu\big) = -\partial_\lambda S^{\lambda\mu\nu} , \]

where, using \(F^{\mu\lambda} = -F^{\lambda\mu}\),

\[ \boxed{\ S^{\lambda\mu\nu} = \frac{1}{\mu_0}\big(F^{\lambda\mu}A^\nu - F^{\lambda\nu}A^\mu\big)\ } \]

which is antisymmetric in \(\mu\nu\) by construction ✓. Then

\[ \partial_\lambda\Big(x^\mu T^{\lambda\nu}_{\text{can}} - x^\nu T^{\lambda\mu}_{\text{can}} + S^{\lambda\mu\nu}\Big) = \big(T^{\mu\nu}_{\text{can}} - T^{\nu\mu}_{\text{can}}\big) + \partial_\lambda S^{\lambda\mu\nu} = 0 ✓ \]

So the canonical construction does deliver a conserved angular-momentum current — it just has two pieces, an orbital one and \(S\), the spin current. Belinfante's repair moves \(S\) into \(T\), after which all the angular momentum looks orbital. Nothing is lost or gained; the same total is bookkept differently.

Check. Dimensions: \([M^{\lambda\mu\nu}] = \mathrm{m\cdot J\,m^{-3}} = \mathrm{J\,m^{-2}}\), so \([J^{\mu\nu}] = \mathrm{J\,m^{-2}\,m^3/(m\,s^{-1})} = \mathrm{J\,s}\) ✓ — angular momentum, as it must be. \([S^{\lambda\mu\nu}] = [F][A]/\mu_0 = \mathrm{J\,m^{-2}}\) ✓, the same as \(M\). Special case: for a scalar field \(T_{\text{can}}\) is already symmetric (P2), so \(S \equiv 0\) — a scalar field carries no spin ✓, which is the field-theoretic version of "spin zero". Limit: for a single particle of mass \(m\) at rest, \(E = mc^2\) and \(\vec P = 0\), so part 3 gives \(\mathrm d\vec R/\mathrm dt = 0\) ✓.

Idea. A mass term for \(A\) is the one addition that destroys gauge invariance, and everything that gauge invariance was quietly providing then has to be bought back — or lost. The Lorenz condition stops being a choice and becomes a law; the current's conservation stops being forced and becomes an assumption; and a third polarisation appears.

1. Under \(A_\mu\to A_\mu + \partial_\mu\chi\), \(F\) is unchanged but \(A_\mu A^\mu \to A_\mu A^\mu + 2A^\mu\partial_\mu\chi + \partial_\mu\chi\partial^\mu\chi\), which is not a total divergence for general \(A\) and \(\chi\). So \(\mathcal L\) is not invariant, and P5 part 1 — gauge invariance forcing \(\partial_\mu J^\mu = 0\) — has no hypothesis to run on. Charge conservation is no longer a consequence of the theory's structure; it must be imposed, or derived from the matter sector separately.

2. The Proca equation. The \(F^2\) term varies exactly as in example 3, giving \(\mu_0^{-1}(\partial_\mu F^{\mu\nu})\delta A_\nu\); the mass term gives \(-\mu_0^{-1}\kappa^2A^\nu\delta A_\nu\), since \(\delta(A_\alpha A^\alpha) = 2A^\nu\delta A_\nu\); the coupling gives \(J^\nu\delta A_\nu\). Setting the bracket to zero, \(\partial_\mu F^{\mu\nu} = \kappa^2A^\nu - \mu_0J^\nu\), and relabelling with \(\partial_\nu F^{\mu\nu} = -\partial_\nu F^{\nu\mu}\),

\[ \boxed{\ \partial_\nu F^{\mu\nu} + \kappa^2 A^\mu = \mu_0 J^\mu\ } \]

which reduces to \(\partial_\nu F^{\mu\nu} = \mu_0J^\mu\) at \(\kappa = 0\) ✓.

3. The constraint. Apply \(\partial_\mu\). The first term dies identically (\(\partial_\mu\partial_\nu\) symmetric against \(F^{\mu\nu}\) antisymmetric), leaving

\[ \kappa^2\,\partial_\mu A^\mu = \mu_0\,\partial_\mu J^\mu . \]

In vacuum the right side is zero, so \(\partial_\mu A^\mu = 0\) for \(\kappa\ne0\): what was a gauge choice — the Lorenz condition of unit 23 — is now a consequence of the field equation, and there is no gauge freedom left with which to choose anything. With a source, the constraint holds if and only if \(\partial_\mu J^\mu = 0\); a non-conserved source is now mathematically permitted and simply sets \(\partial_\mu A^\mu\ne0\), letting a scalar mode propagate.

4. Klein–Gordon for each component. In vacuum, \(\partial_\nu F^{\mu\nu} = \partial_\nu(\partial^\mu A^\nu - \partial^\nu A^\mu) = \partial^\mu(\partial_\nu A^\nu) - \partial_\nu\partial^\nu A^\mu = 0 + \Box A^\mu\), using part 3 and \(\partial_\nu\partial^\nu = -\Box\). The field equation then reads

\[ \Box A^\mu + \kappa^2A^\mu = 0 , \]

a Klein–Gordon equation for each of the four components ✓. Counting: four components, one algebraic constraint \(\partial_\mu A^\mu = 0\), and no gauge freedom to remove any more, giving \(4 - 1 = \mathbf 3\) propagating degrees of freedom — the two transverse polarisations of light plus a longitudinal one. The massless theory has \(4 - 1 - 1 = 2\), the extra subtraction being the gauge freedom. Gauge invariance is what removes the third polarisation.

5. Numbers. \(mc^2 = \hbar c\kappa = \hbar c/(1.0\ \mathrm m)\), so

\[ mc^2 = 1.9732698\times10^{-7}\ \mathrm{eV} = \mathbf{0.1973\ \mu eV} ,\qquad m = \frac{\hbar}{c\times1.0\ \mathrm m} = \mathbf{3.518\times10^{-43}\ kg} . \]

What to look for: the static solution of \(\nabla^2\varphi = \kappa^2\varphi\) outside a charge is \(\varphi\propto e^{-\kappa r}/r\), a Yukawa potential, so Coulomb's law would fail at distances of order \(\kappa^{-1}\). Equivalently the magnetic field of a dipole picks up a correction at that range, and light of wavelength comparable to \(\kappa^{-1}\) would become dispersive (\(\omega^2 = c^2k^2 + c^2\kappa^2\)), so pulses of different frequency from a distant source would arrive at different times. Laboratory Coulomb-law tests, planetary magnetic-field measurements and pulse dispersion from astrophysical sources are the three standard routes, and they push \(\kappa^{-1}\) far beyond a metre.

Check. Dimensions: \([\kappa^2A_\mu A^\mu/\mu_0] = \mathrm{m^{-2}(T\,m)^2/(N\,A^{-2})} = \mathrm{J\,m^{-3}}\) ✓, the same as the \(F^2\) term. Limit: \(\kappa\to0\) restores gauge invariance, the Lorenz condition becomes a choice again, the third polarisation decouples, and the Yukawa potential becomes Coulomb's \(1/r\) ✓. Consistency: part 4's dispersion \(\omega^2 = c^2k^2 + (mc^2/\hbar)^2\) is P2's, as it must be — Proca is four Klein–Gordon fields with one constraint ✓.

Another way

Part 4 without solving anything: count in momentum space. A plane-wave solution \(A^\mu = a^\mu e^{ik\cdot x}\) needs \(k\cdot k = -\kappa^2\) and, from part 3, \(k\cdot a = 0\) — one linear condition on the four-component polarisation \(a\), leaving three independent choices. In the massless case \(k\) is null, \(k\cdot a = 0\) admits \(a\propto k\) as a solution, that solution is pure gauge and carries no field (P8's check), so one further polarisation drops out and two remain.