Part D · Minkowski · spine

Classical field theory: Noether again, and where T^μν actually comes from

A field is a mechanical system whose coordinates are labelled by points of space. Everything follows from one functional — the equations of motion, the conserved currents, and the object Einstein's equation puts on its right.

needs 28, 13, 11~2½ h slides + 5 h problemsproblems: ./mr problems 29

Unit 18 turned a chain of masses into a string and wrote down, at the very end, a Lagrangian density and the Euler–Lagrange equation it satisfies in one space dimension. Unit 28 assembled $\vec E$ and $\vec B$ into an antisymmetric tensor field on Minkowski space and wrote Maxwell's equations with indices. This unit joins the two: it does variational mechanics for fields on spacetime, properly, in four dimensions and with the boundary term written out.

There are two reasons to do it now rather than as an aside. The first is Noether. Unit 13 proved that every conservation law of point mechanics is one theorem about symmetries of an action; the field version of that theorem is what turns "the laws are the same everywhere" into a locally conserved energy and momentum, which is the only honest route to $T^{\mu\nu}$. The second is that unit 35 will vary an action with respect to the metric. That technique is introduced here, in flat space, where nothing is curved and nothing can go wrong, so that unit 35 is a continuation and not a new subject.

How to read this deck

One action runs through all four sections: Maxwell's, $S = \frac1c\int(-\frac{1}{4\mu_0}F_{\mu\nu}F^{\mu\nu} + A_\mu J^\mu)\,\mathrm{d}^4x$. Section 1 builds the machinery on scalar fields; section 2 varies $S$ with respect to $A_\mu$ and gets Maxwell's source equations in four lines; section 3 runs Noether on it and gets an energy–momentum tensor with two embarrassing defects; section 4 gets the right one from a completely different calculation, and shows the two differ by a term that is identically conserved. The two routes landing in the same place by different mathematics is the point of the unit.

Conventions: $x^0 = ct$, $\eta_{\mu\nu} = \mathrm{diag}(-1,1,1,1)$ in an inertial frame, Greek indices $0$–$3$, Latin $1$–$3$, Einstein summation, $\partial_\mu = \partial/\partial x^\mu$. Proofs are on the slides where they are one line and in these notes under Full proof where they are not.

29 · overview

One action, four questions

the object

Maxwell's action, one functional of one field $A_\mu$ over a region $\Omega$ of spacetime: $$S[A] = \frac1c\int_\Omega\Big({-\tfrac{1}{4\mu_0}}F_{\mu\nu}F^{\mu\nu} + A_\mu J^\mu\Big)\mathrm{d}^4x .$$

why the second tensor is right · three steps

1. Write $S$ with a general symmetric metric $g_{\mu\nu}(x)$ in place of $\eta_{\mu\nu}$: still flat spacetime, in other coordinates.

2. Nudge $g$; call the coefficient of the nudge in the change of $S$ the tensor $T_{\mu\nu}$.

3. The nudge is symmetric, so only a symmetric coefficient is defined; $S$ is gauge invariant for every $g$, so its rate of change with $g$ is too.

What the unit extracts from $S$, one operation per arrow. Noether's tensor is conserved but neither symmetric nor gauge invariant; the metric one is both, by the three steps; they differ by the Belinfante term.

Getting around. A dotted underline is a glossary term (click it; G lists them all); a solid underline is a piece of the running example, shown on hover; N shows the notes, where the longer proofs live.

Where the equations come from. Unit 28 wrote Maxwell's equations covariantly but did not say where they came from. They come from this functional: section 2 varies it.

The three steps, and where each is done. Step 1 is The move: let the metric be general, in flat space, which also says why nothing is curved and what replaces $\mathrm d^4x$. Step 2 is The definition, and why its two properties are free; the determinant slides just before it supply the one new rule of differentiation it needs. Step 3 is the two boxes on that same slide. The explicit Maxwell computation, which lands on $T^{00} = \tfrac{\varepsilon_0}{2}E^2 + \tfrac{1}{2\mu_0}B^2$, is Worked: metric variation of Maxwell's action. The failure of the Noether route that makes all this necessary is Worked: the canonical tensor for Maxwell, and its two defects, and the link between the two tensors is The two routes, side by side.

Reading the map. The four arrows are the four sections' results: §1 builds the machinery for varying a functional of a field, which every arrow uses; §2 is the top two arrows, §3 the third, §4 the fourth. The vertical link is the unit's punchline: two derivations by unrelated mathematics, which disagree event by event and agree on every total.

Why this example. Maxwell is the only classical field theory whose Lagrangian the learner already believes — unit 24 derived the energy density and the Poynting vector by hand, so there is an independent answer to check against. It is also the case where the canonical Noether construction visibly fails, which is what makes the second definition necessary rather than merely elegant. A scalar field would have hidden the problem: for the Klein–Gordon field the two routes agree, and the reader would leave thinking the metric variation is a flourish.

The one sentence to carry. The right-hand side of Einstein's equation is not postulated. It is the derivative of the matter action with respect to the metric, and it is symmetric because the metric is symmetric and gauge invariant because the action is.

What the problem sheet adds. The Klein–Gordon field done in full, the complex scalar whose phase symmetry forces the existence of $A_\mu$ when you let the phase depend on position, the Proca field where gauge invariance is lost and what that costs, and the tracelessness of $T^{\mu\nu}$ shown to be an accident of four dimensions.

section 1

Fields: mechanics with a continuous index

Varying a functional over a region of spacetime, boundary term and all.

29 · fields

A field is a mechanical system whose index is a point

$N$ particles	a classical field
coordinates $q_i(t)$, $i = 1\dots N$	$\varphi_a(x)$, $a = 1\dots n$, at every $x$
the index $i$ is discrete	the index is the point $\vec x$; $a$ counts components
$L(q,\dot q,t)$, one number per instant	$\mathcal L(\varphi,\partial\varphi,x)$, one number per event
$S = \int L\,\mathrm dt$	$S = \frac1c\int_\Omega\mathcal L\,\mathrm d^4x$
$\mathrm{J\,s}$	$\mathrm{J\,s}$, since $[\mathcal L] = \mathrm{J\,m^{-3}}$

Unit 18's chain and its limit. Each spring couples only neighbours, through $q_{i+1} - q_i$; as the spacing shrinks that difference becomes the slope $\partial\varphi/\partial x$. So the energy at a point needs only the field and its first derivatives there. (Displacements drawn sideways for visibility.)

The index-becomes-a-point move is exactly unit 18's. There the chain's $L = \sum_i(\tfrac12 m\dot u_i^2 - \tfrac12 k(u_{i+1}-u_i)^2)$ became $\mathcal L = \tfrac12\mu u_t^2 - \tfrac12 T u_x^2$ as the spacing went to zero. The only new thing here is that the label runs over three dimensions instead of one, and that we will insist the result be Lorentz invariant.

What to see in the figure. Read the table across with the picture beside it. The top row is a system with $N$ coordinates, labelled by $i$; the bottom row has one coordinate for every value of $x$, which is what "the index is a point" means. Nothing about the dynamics changes in the limit except that sums over $i$ become integrals over $x$ and neighbour differences become derivatives. In four dimensions the label $x$ runs over three space directions, and the field may have several components $\varphi_a$ at each point; the picture shows one component on one axis.

29 · fields

The Lagrangian density, and the action over a region

Lagrangian density

A function $\mathcal L(\varphi_a, \partial_\mu\varphi_a, x)$ of the field values and their first derivatives at one event. Unit 18 met it for the string; the content is locality: what happens here depends on the field here and its slope here, nothing else.

the action over a spacetime region

$\displaystyle S[\varphi] = \frac1c\int_\Omega \mathcal L\,\mathrm d^4x$, $\ \mathrm d^4x = \mathrm dx^0\mathrm dx^1\mathrm dx^2\mathrm dx^3 = c\,\mathrm dt\,\mathrm d^3x$.

$\Omega$ is a bounded region with piecewise smooth boundary — typically a slab $t_1 \le t \le t_2$ out to a large sphere.

Where the factor $1/c$ comes from. $\mathrm d^4x$ has dimension $\mathrm m^4$, because $x^0 = ct$ is a length. An energy density times $\mathrm m^4$ is $\mathrm{J\,m}$, and dividing by $c$ makes it $\mathrm{J\,s}$, the dimension of an action. Equivalently $\frac1c\int\mathcal L\,\mathrm d^4x = \int\!\!\int\mathcal L\,\mathrm dt\,\mathrm d^3x$, which is unit 18's two-dimensional integral with two more space dimensions. The constant is irrelevant to the field equations — stationarity does not care about an overall factor — but it is what makes the dimension checks in this unit work, so it is carried.

Why only first derivatives. Nothing forbids $\mathcal L(\varphi,\partial\varphi, \partial\partial\varphi,\dots)$, and the Euler–Bernoulli beam, whose bending energy density is $\tfrac12 EI\,u_{xx}^2$, is an honest example with second derivatives. But a Lagrangian with second derivatives generally gives fourth-order field equations, and Ostrogradsky's theorem says their energy is unbounded below unless the higher-derivative dependence is degenerate. Every fundamental field theory known is first-order in this sense, so the restriction is the physics, not the convenience.

29 · fields

The first variation, and where the boundary term goes

Take $\varphi_a \to \varphi_a + \varepsilon\eta_a$ with $\eta_a \in C^\infty$. Unit 11's first variation is $\delta S = \frac{\mathrm d}{\mathrm d\varepsilon}S[\varphi + \varepsilon\eta]\big|_{\varepsilon=0}$:

$$\delta S = \frac1c\int_\Omega\Big(\frac{\partial\mathcal L}{\partial\varphi_a}\eta_a + \frac{\partial\mathcal L}{\partial(\partial_\mu\varphi_a)}\partial_\mu\eta_a\Big)\mathrm d^4x$$

The second term is a divergence minus a remainder — the product rule, with $\pi^\mu{}_a := \partial\mathcal L/\partial(\partial_\mu\varphi_a)$:

$$\pi^\mu{}_a\,\partial_\mu\eta_a = \partial_\mu(\pi^\mu{}_a\eta_a) - (\partial_\mu\pi^\mu{}_a)\,\eta_a$$ $$\boxed{\ \delta S = \frac1c\int_\Omega\Big(\frac{\partial\mathcal L}{\partial\varphi_a} - \partial_\mu\pi^\mu{}_a\Big)\eta_a\,\mathrm d^4x \;+\; \frac1c\oint_{\partial\Omega}\pi^\mu{}_a\eta_a\,\mathrm d\Sigma_\mu\ }$$

The surface integral is the divergence theorem in four dimensions, applied to $V^\mu = \pi^\mu{}_a\eta_a$; the next slide states it and draws the region.

$\pi^\mu{}_a$ is the field's conjugate momentum. For the string, $\pi^0 = \partial\mathcal L/\partial(\partial_0 u) = c\,\mu u_t$, because $\mathcal L$ sees $\partial_0u = u_t/c$ and so $\partial/\partial(\partial_0u) = c\,\partial/\partial u_t$; up to that one factor of $c$ from $x^0 = ct$ it is unit 14's momentum density $\mu u_t$. The other three components $\pi^i$ have no counterpart in point mechanics: they are the flux of that momentum, and their appearance is exactly what makes a field theory local.

The one move on this slide. It is unit 11's integration by parts, done once per direction: the product rule turns $\pi^\mu{}_a\,\partial_\mu\eta_a$ into a total divergence minus a term with no derivative on $\eta$. The total divergence is then traded for a surface integral, and what multiplies $\eta_a$ in the volume integral is what will become the field equation.

29 · fields

The divergence theorem in four dimensions

theorem

For $V^\mu$ of class $C^1$ on $\bar\Omega \subset \mathbb R^4$, $$\int_\Omega\partial_\mu V^\mu\,\mathrm d^4x = \oint_{\partial\Omega}V^\mu\,\mathrm d\Sigma_\mu .$$

Why: unit 05's proof, verbatim. For a box, each term $\int\partial_\mu V^\mu$ is a one-dimensional fundamental theorem of calculus in $x^\mu$ and an integral over the opposite pair of faces; general $\Omega$ by patching. The number of dimensions is never used.

$\mathrm d\Sigma_\mu$ is the outward three-dimensional surface element: on a face $x^0 = $ const it is $-\mathrm d^3x$ on the past face and $+\mathrm d^3x$ on the future one. On the side faces it points outward in space, $\mathrm d\Sigma_i = n_i\,\mathrm dA\,\mathrm dx^0$, with $n_i$ the outward unit normal of the sphere.

The usual $\Omega$: a slab of time out to a large sphere, with the outward element on each face. A variation of compact support (dashed) is zero on all of $\partial\Omega$, so the boundary term of the previous slide vanishes whatever the faces are.

Full proof of the divergence theorem in $\mathbb R^n$. Let $\Omega = [a^1,b^1]\times \dots\times[a^n,b^n]$. Then $\int_\Omega\partial_\mu V^\mu\,\mathrm d^nx = \sum_\mu \int\big(\int_{a^\mu}^{b^\mu}\partial_\mu V^\mu\,\mathrm dx^\mu\big)\mathrm d^{n-1}x$ by Fubini (unit 05), and the inner integral is $V^\mu(x^\mu = b^\mu) - V^\mu(x^\mu = a^\mu)$ by the fundamental theorem of calculus. Those are exactly the outward fluxes through the two faces normal to $x^\mu$, with the sign convention that the outward element on the upper face is $+\mathrm d^{n-1}x$ and on the lower face $-\mathrm d^{n-1}x$. Summing over $\mu$ gives the whole boundary. For a region that is a finite union of boxes the interior faces cancel in pairs; for a general region with piecewise smooth boundary, approximate by such unions and pass to the limit, exactly as unit 05 sketched. Nothing in this argument knows $n = 3$.

The sign on a constant-time face. The element that makes the theorem true is the coordinate one, read straight off the box proof: on the future face $x^0 = x^0_2$ the outward direction is increasing $x^0$ and $\mathrm d\Sigma_\mu = (+\mathrm d^3x,\vec 0)$, on the past face $(-\mathrm d^3x,\vec 0)$. Write instead $\mathrm d\Sigma_\mu = n_\mu\,\mathrm d^3x$ with $n^\mu$ the outward unit normal vector — $n^\mu = (1,\vec 0)$ on the future face — and lowering the index with $\eta_{00} = -1$ flips the sign: on a spacelike face the two conventions differ by a minus, while on a timelike face, a large sphere say, they agree. This is the one place in the unit where the signature leaks into a boundary statement, and it is why the course prefers variations of compact support, for which the boundary term is zero and the question does not arise.

What to see in the figure. The two horizontal faces are what the box proof calls the faces normal to $x^0$: the theorem's $x^0$-term is the integral over the future face minus the integral over the past face, which is why the past face carries $-\mathrm d^3x$. The side faces are the faces normal to $x^1, x^2, x^3$, drawn as one pair. The dashed ellipse is where the course's variations live; it never touches the boundary, which is the whole reason the boundary term is so easy to discard.

29 · fields

The field Euler–Lagrange equations

theorem

Let $\varphi$ be a $C^2$ stationary point of $S$ among all variations $\eta_a$ of compact support in the interior of $\Omega$. Then

$$\partial_\mu\!\left(\frac{\partial\mathcal L}{\partial(\partial_\mu\varphi_a)}\right) = \frac{\partial\mathcal L}{\partial\varphi_a}\qquad a = 1,\dots,n ,$$

one second-order PDE per field component. These are the field Euler–Lagrange equations.

proof

Compact support kills the boundary term, so $\delta S = \frac1c\int_\Omega E_a\eta_a\,\mathrm d^4x$ with $E_a = \partial_{\varphi_a}\mathcal L - \partial_\mu\pi^\mu{}_a$, continuous. Stationarity says this vanishes for every such $\eta$. By the fundamental lemma in four dimensions, $E_a \equiv 0$. $\square$

the functional derivative

$\dfrac{\delta S}{\delta\varphi_a(x)} := \dfrac1c\Big(\dfrac{\partial\mathcal L}{\partial\varphi_a} - \partial_\mu\pi^\mu{}_a\Big)$, the unique continuous density with $\delta S = \int \frac{\delta S}{\delta\varphi_a(x)}\,\eta_a(x)\,\mathrm d^4x$.

Uniqueness is the same lemma. The field equations are $\delta S/\delta\varphi_a = 0$.

Full proof of the fundamental lemma in four dimensions

Claim: if $f$ is continuous on an open $U\subset\mathbb R^4$ and $\int_U f\eta\,\mathrm d^4x = 0$ for every $\eta\in C_c^\infty(U)$, then $f\equiv 0$. Suppose $f(x_0) \gt 0$ for some $x_0\in U$. By continuity there is $r \gt 0$ with $\bar B(x_0,r)\subset U$ and $f \gt f(x_0)/2$ on $B(x_0,r)$. Take the bump $\eta(x) = \exp\!\big(-1/(r^2 - |x-x_0|^2)\big)$ for $|x-x_0| \lt r$ and $0$ outside — unit 11's bump with $|x - x_0|$ now the Euclidean norm in four variables, smooth for the same reason (every derivative of $e^{-1/s}$ at $s\to 0^+$ vanishes). Then $\eta \gt 0$ on the ball, so $\int f\eta \ge \tfrac12 f(x_0)\int_B\eta \gt 0$, contradicting the hypothesis. The same with $-f$ handles $f(x_0) \lt 0$. $\square$

Note the lemma is about the Euclidean geometry of $\mathbb R^4$ — a ball, a bump — and not about $\eta_{\mu\nu}$. The Minkowski form plays no part in the variational argument; it enters only through the particular $\mathcal L$ one writes down. This is worth noticing, because it is the reason the same machinery will work unchanged on a curved manifold in unit 35.

Why "for every variation of compact support" is the right hypothesis. It is the weakest condition that makes the lemma apply, and it is the honest one physically: the field equations are local statements, and a local statement cannot depend on what is imposed at the edge of the region. Boundary conditions are extra data, discussed on the next slide.

The functional derivative is a density, not a number. $\delta S/\delta\varphi(x)$ has dimension (action)/(field $\times$ volume in $\mathrm m^4$). Think of it as unit 11's gradient of $S$ with respect to the continuum of variables $\varphi(x)$: the analogue of $\partial S/\partial q_i$, with the index $i$ replaced by the event $x$ and the sum over $i$ replaced by $\int\mathrm d^4x$.

29 · fields

What the boundary term is for

Drop compact support and the boundary term survives:

$$\delta S\big|_{\text{bdy}} = \frac1c\oint_{\partial\Omega}\pi^\mu{}_a\,\eta_a\,\mathrm d\Sigma_\mu .$$

Three ways to make it vanish, each a different physical problem:

Fixed data. $\eta_a = 0$ on $\partial\Omega$: the field is prescribed there. This is unit 11's fixed-endpoint problem, one dimension at a time.
Natural condition. $\eta_a$ free, so stationarity itself forces $\pi^\mu{}_a\,n_\mu = 0$ on $\partial\Omega$, $n_\mu$ the outward normal — unit 18's free end of a string, in four dimensions.
Decay. $\Omega$ grows to all of spacetime and the fields fall off. The usual assumption, and the one used throughout this unit.

The three ways, one field component in one direction: the solution solid, a varied field dashed, the edges of $\Omega$ the thin verticals. Fixed data pins both ends; a free end lets the variation move it, and stationarity then flattens the slope there; decay pushes the edge out to where there is nothing left to vary.

The natural boundary condition, spelled out. If $\eta_a$ is unconstrained on part of $\partial\Omega$, choose variations supported near a point of that part. The interior equation already holds, so stationarity reduces to $\oint\pi^\mu{}_a n_\mu\eta_a\,\mathrm d\Sigma = 0$ for all such $\eta_a$, and the fundamental lemma on the boundary hypersurface gives $\pi^\mu{}_a n_\mu = 0$ there. For the string, $\pi^x = -Tu_x$, so the condition is $u_x = 0$ at a free end: unit 18's even extension, now derived rather than imposed.

Decay is an assumption, not a theorem. "The fields fall off fast enough" is doing work in every conservation argument below: a charge $Q = \int j^0\mathrm d^3x/c$ is constant only if $\oint \vec\jmath\cdot\mathrm d\vec A \to 0$ on the sphere at infinity. For a radiating system it does not, and then the charge inside a finite ball genuinely changes. Problem P6 makes this explicit.

Reading the figure. In one dimension $\pi^\mu{}_a n_\mu$ at an end is, up to sign, the slope there times a constant (for the string, $-Tu_x$), so "natural condition" and "zero slope" are the same statement; the short horizontal segments in the middle panel are that condition holding for both curves. In four dimensions the ends become a three-dimensional boundary and the slope becomes the flux $\pi^\mu{}_a n_\mu$ through it, but nothing else changes.

29 · fields

Where the boundary term comes back

it does not always vanish

In unit 35 the field is the metric itself, and the boundary term contains normal derivatives of $\delta g$, which fixing $g$ on $\partial\Omega$ does not kill. The repair is to add a boundary integral to the action — the Gibbons–Hawking–York term — chosen precisely to cancel it.

Watch for it there. Every step of that calculation is the previous slide with a metric in place of $\varphi$.

for the rest of this unit

All variations have compact support, all fields decay, and every boundary term is zero. It is said once here so that it is never waved at again.

Why the metric case is genuinely different. The Einstein–Hilbert Lagrangian contains the Ricci scalar, which involves second derivatives of $g$. Integrating by parts twice leaves a boundary term proportional to $n^\mu\partial_\mu\delta g$, and prescribing $g$ on the boundary fixes only $\delta g$, not its normal derivative. Gibbons and Hawking (1977) and York (1972) found the boundary integral — the trace of the extrinsic curvature — whose variation cancels it exactly. This is the single most common place a first reading of general relativity glosses over something real, and the reason this slide exists here.

Why say it now. Every boundary term in this unit is discarded by the assumption in the box on the right, and that is legitimate here because the Lagrangians have only first derivatives. The point of flagging the exception in advance is that in unit 35 the same discarding move would be a mistake, and it is easier to notice a mistake one was warned about.

29 · fields · example

Worked: the scalar wave field

example 1

The Lagrangian. The simplest Lorentz-invariant density built from one real field and its first derivatives: $$\mathcal L = -\tfrac12\,\eta^{\mu\nu}\partial_\mu\varphi\,\partial_\nu\varphi = \tfrac12(\partial_0\varphi)^2 - \tfrac12|\nabla\varphi|^2 .$$

Step 1 — the momenta. $\pi^\mu = \dfrac{\partial\mathcal L}{\partial(\partial_\mu\varphi)} = -\eta^{\mu\nu}\partial_\nu\varphi = -\partial^\mu\varphi .$

Step 2 — the other derivative. $\mathcal L$ contains no undifferentiated $\varphi$, so $\partial\mathcal L/\partial\varphi = 0$.

Step 3 — assemble. $\partial_\mu\pi^\mu = 0$, that is $-\partial_\mu\partial^\mu\varphi = 0$. Since $\partial_\mu\partial^\mu = -\partial_0^2 + \nabla^2 = -\Box$,

$$\Box\varphi = \frac{1}{c^2}\frac{\partial^2\varphi}{\partial t^2} - \nabla^2\varphi = 0 .$$

checks

Dimensions. $[\mathcal L] = \mathrm{J\,m^{-3}}$ forces $[\varphi^2] = \mathrm{J\,m^{-1}}$, since $\partial_\mu\varphi$ carries one inverse metre. Both sides of $\Box\varphi = 0$ are then $[\varphi]\,\mathrm m^{-2}$ ✓.

Special case. In one space dimension, write $x^0 = c_s t$ with $c_s^2 = T/\mu$ and multiply by $\mu c_s^2$: $\mathcal L = \tfrac12\mu u_t^2 - \tfrac12 Tu_x^2$, unit 18's string, and the equation is $\mu u_{tt} = Tu_{xx}$ ✓. The string's Lagrangian was already of relativistic form, with the wave speed $c_s$ in the role of $c$.

the sign of $\mathcal L$

With $\eta = (-,+,+,+)$ the kinetic term needs the leading minus: $-\tfrac12(\partial\varphi)^2$ is $+\tfrac12\dot\varphi^2/c^2 - \tfrac12|\nabla\varphi|^2$, which is $T - V$. Books using $(+,-,-,-)$ write $+\tfrac12(\partial\varphi)^2$ for the same physics.

Why "the simplest". Demand: real, built from $\varphi$ and $\partial\varphi$ only, quadratic (so the equation is linear), and a Lorentz scalar. The only scalars available are $\varphi^2$ and $\eta^{\mu\nu}\partial_\mu\varphi\partial_\nu\varphi$; a term linear in $\partial_\mu\varphi$ would need a fixed vector, which would pick out a direction in spacetime and break isotropy. Dropping $\varphi^2$ leaves exactly this. Adding it back is the next example.

$\Box$ and $\partial_\mu\partial^\mu$ differ by a sign. Unit 24 defined $\Box = c^{-2}\partial_t^2 - \nabla^2$, and in this signature $\partial_\mu\partial^\mu = \eta^{\mu\nu}\partial_\mu\partial_\nu = -\partial_0^2 + \nabla^2 = -\Box$. Some books write $\Box$ for $\partial_\mu\partial^\mu$ itself. Whenever a formula in another text has a sign you did not expect, check this first, and check the signature second.

What $\varphi$ is physically. Nothing, here — it is the cheapest field that a course can practise on. The nearest real thing in this course is unit 18's displacement of a string and unit 22's acoustic pressure; the nearest in nature is the Higgs field, whose Lagrangian is this one with a mass term and a quartic self-interaction.

29 · fields · example

Worked: the Klein–Gordon field, and what mass does to a wave

example 2

Add the one term left out, with $\kappa := mc/\hbar$ an inverse length: $$\mathcal L = -\tfrac12\big(\partial_\mu\varphi\,\partial^\mu\varphi + \kappa^2\varphi^2\big) .$$

Step 1. $\pi^\mu = -\partial^\mu\varphi$, unchanged. Step 2. $\partial\mathcal L/\partial\varphi = -\kappa^2\varphi$.

Step 3. $-\partial_\mu\partial^\mu\varphi = -\kappa^2\varphi$, that is the Klein–Gordon equation $$\Box\varphi + \kappa^2\varphi = 0 .$$

Step 4 — the dispersion relation. Put $\varphi = e^{i(\vec k\cdot\vec x - \omega t)}$: $\Box\varphi = (k^2 - \omega^2/c^2)\varphi$, so $\omega^2 = c^2k^2 + (mc^2/\hbar)^2$.

checks

Limit $m\to0$: $\omega = ck$, example 1's wave equation ✓.

Dimensions: $[\kappa^2\varphi^2] = \mathrm{m^{-2}\cdot J\,m^{-1}} = \mathrm{J\,m^{-3}}$ ✓.

A number. For the charged pion, $mc^2 = 139.570$ MeV, so $\kappa^{-1} = \hbar c/mc^2 = 1.4138$ fm: the reduced Compton wavelength, and Yukawa's range for the nuclear force.

29 · fields

Phase and group velocity, on the dispersion curve

mass = dispersion

$v_{\text{ph}} = \omega/k \gt c$ and $v_{\text{gr}} = c^2k/\omega \lt c$, with $v_{\text{ph}}v_{\text{gr}} = c^2$ exactly. At $k = 10^{15}\,\mathrm m^{-1}$ for the pion: $1.2249\,c$ and $0.8164\,c$, product $1.0000\,c^2$ ✓. Signals still travel at $c$: a zeroth-order term cannot alter unit 18's principal symbol.

Both are read off the curve $\omega(k)$: the phase velocity is the slope of the chord from the origin, $\omega/k$; the group velocity is the slope of the tangent, $\mathrm d\omega/\mathrm dk$. Differentiating $\omega^2 = c^2k^2 + \kappa^2c^2$ gives $2\omega\,\mathrm d\omega = 2c^2k\,\mathrm dk$, so $v_{\text{gr}} = c^2k/\omega$, and the product is $c^2$.

The Klein–Gordon dispersion relation, in units of $\kappa$. The curve is a hyperbola, lifted off the origin to $\omega = mc^2/\hbar$ by the mass and approaching the massless line $\omega = ck$; every chord from the origin is steeper than that line and every tangent shallower. At the pion point the two slopes are $1.2249$ and $0.8164$.

The mass term is the only thing that makes the wave equation dispersive. Unit 18's $\Box\varphi = 0$ has $\omega = ck$: every Fourier component travels at the same speed and a wave packet keeps its shape, which is why d'Alembert's formula is exact and a pulse on a string does not spread. With $\kappa\ne0$ the relation curves, components separate, and a packet spreads. Unit 19's separation of variables and unit 20's transform methods both go through unchanged — only the relation between $\omega$ and $k$ has changed — so this is a small change mathematically and an enormous one physically.

What to see in the figure. A wave packet is a sum of plane waves with $k$ near some $k_0$. Each crest moves at its own chord slope, but the envelope moves where neighbouring components stay in phase, which is the tangent slope. On the dashed massless line chord and tangent coincide and the packet keeps its shape; anywhere on the hyperbola they differ and it spreads. At small $k$ the tangent is nearly flat, $v_{\text{gr}} \approx c^2k/\kappa c$, which is $p/m$ for the particle reading of the field: a slow, massive particle.

The numbers. $\kappa = 1/(1.4138\ \mathrm{fm}) = 7.073\times10^{14}\ \mathrm m^{-1}$, so $k = 10^{15}\ \mathrm m^{-1}$ is $k/\kappa = 1.4138$ and $\omega/\kappa c = \sqrt{1 + 1.4138^2} = 1.7317$. Then $v_{\text{ph}}/c = 1.7317/1.4138 = 1.2249$ and $v_{\text{gr}}/c = 1.4138/1.7317 = 0.8164$, the numbers on the slide; the figure is computed from the same formula.

29 · fields

The range of a force is the mass of its field

A static point source: drop $\partial_t$ and the Klein–Gordon equation becomes $\nabla^2\varphi = \kappa^2\varphi$ away from the source. Try a radial $\varphi(r)$, for which $\nabla^2\varphi = (r\varphi)''/r$:

$$(r\varphi)'' = \kappa^2\,(r\varphi)\quad\Longrightarrow\quad \varphi = \frac{e^{-\kappa r}}{r},$$

the decaying solution. At $\kappa = 0$ it is unit 23's $1/r$, Coulomb's potential.

a number

The pion's $\kappa^{-1} = 1.4138$ fm: by $5$ fm the cut-off factor $e^{-\kappa r}$ is $0.029$. A force carried by a field of mass $m$ reaches about $\hbar/mc$, and no further.

Both static fields fall as $1/r$ from spreading over a sphere; multiplying by $r$ leaves only what the mass adds. Massless, nothing is added: Coulomb's law, infinite range. Massive, the field is cut off over $\kappa^{-1}$.

Reading the figure. Both static fields share the factor $1/r$ that spreading over a sphere forces; multiplying by $r$ removes it and leaves what the mass adds, the factor $e^{-\kappa r}$. It has fallen to $1/e$ at $r = \kappa^{-1}$ and to under 3 per cent by $5$ fm: the force between two nucleons is negligible beyond a few femtometres, which is the size of a nucleus.

Yukawa's use of it. The static case $\nabla^2\varphi = \kappa^2\varphi$ has the spherically symmetric solution $\varphi \propto e^{-\kappa r}/r$ — unit 23's $1/r$ Green's function for the Laplacian with an exponential cut-off, and a one-line check: for a radial $f$, $\nabla^2 f = (rf)''/r$, and here $rf = e^{-\kappa r}$, whose second derivative is $\kappa^2e^{-\kappa r}$, so $\nabla^2f = \kappa^2f$ away from the origin. Yukawa (1935) read the range $\kappa^{-1}$ off the size of a nucleus, taking about $2$ fm, and predicted a particle of some $200$ electron masses, around $100$ MeV; the pion, found in 1947, has $mc^2 = 139.570$ MeV and $\kappa^{-1} = 1.4138$ fm — the right order, with a range rather shorter than he had guessed. The massless case $\kappa = 0$ gives $1/r$ and infinite range, which is Coulomb's law: the range of a force and the mass of its field are the same fact.

29 · fields

The running example: Maxwell's action

Maxwell Lagrangian

The field is the four-potential $A_\mu$; $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ as in unit 28, and $J^\mu = (c\rho,\vec J)$.

$$\mathcal L = -\frac{1}{4\mu_0}F_{\mu\nu}F^{\mu\nu} + A_\mu J^\mu$$

In three-vector language, using $F_{\mu\nu}F^{\mu\nu} = 2(|\vec B|^2 - |\vec E|^2/c^2)$ and $A_\mu J^\mu = -\rho\varphi + \vec A\cdot\vec J$:

$$\mathcal L = \tfrac{\varepsilon_0}{2}|\vec E|^2 - \tfrac{1}{2\mu_0}|\vec B|^2 - \rho\varphi + \vec A\cdot\vec J .$$

checks

Dimensions. $\tfrac{\varepsilon_0}{2}E^2$ is $\mathrm{J\,m^{-3}}$ (unit 24); $\rho\varphi$ is $\mathrm{C\,m^{-3}\cdot V} = \mathrm{J\,m^{-3}}$ ✓.

It is $T - V$. $\vec E$ contains $\partial_t\vec A$, $\vec B$ does not: the electric term is the kinetic one.

It reduces to unit 27's coupling. For a point charge, $\frac1c\int A_\mu J^\mu\mathrm d^4x = q\!\int\! A_\mu\,\mathrm dx^\mu$, whose Lagrangian is $q(-\varphi + \vec v\cdot\vec A)$ ✓.

the sign of the coupling

$+A_\mu J^\mu$ here, because $A_\mu = (-\varphi/c,\vec A)$ in this signature. Texts in $(+,-,-,-)$ write $-A_\mu J^\mu$ for the same physics. Check any formula you import against $\mathcal L_{\text{int}} = -\rho\varphi + \vec A\cdot\vec J$, which is signature-free.

Verifying $F_{\mu\nu}F^{\mu\nu} = 2(B^2 - E^2/c^2)$. With $A_\mu = (-\varphi/c,\vec A)$ and $\partial_0 = c^{-1}\partial_t$: $F_{0i} = \partial_0A_i - \partial_iA_0 = c^{-1}(\partial_tA_i + \partial_i\varphi) = -E_i/c$, and $F_{ij} = \partial_iA_j - \partial_jA_i = \epsilon_{ijk}B_k$. Raising both indices flips the sign of each $0$: $F^{0i} = -F_{0i} = E_i/c$, $F^{ij} = F_{ij}$. Then $F_{\mu\nu}F^{\mu\nu} = 2F_{0i}F^{0i} + F_{ij}F^{ij} = -2E^2/c^2 + 2B^2$. Hence $-\frac{1}{4\mu_0}F^2 = \frac{1}{2\mu_0}(E^2/c^2 - B^2) = \frac{\varepsilon_0}{2}E^2 - \frac{1}{2\mu_0}B^2$, using $\varepsilon_0 = 1/\mu_0c^2$.

And $A_\mu J^\mu$. $A_\mu J^\mu = A_0J^0 + A_iJ^i = (-\varphi/c)(c\rho) + \vec A\cdot\vec J = -\rho\varphi + \vec A\cdot\vec J$.

Where the Lagrangian comes from, honestly. It is not derived. It is the unique Lorentz-invariant, gauge-invariant, parity-even functional of $A$ that is quadratic in first derivatives — the only other invariant available is $\epsilon^{\mu\nu\rho\sigma}F_{\mu\nu} F_{\rho\sigma} \propto \vec E\cdot\vec B$ of unit 28, and that one is a total derivative, so it changes no field equation, and it is parity-odd besides. Quadratic, invariant, and gauge-invariant leaves exactly $F_{\mu\nu}F^{\mu\nu}$; the coefficient $-1/4\mu_0$ is fixed by requiring the resulting equation to be Maxwell's in SI units. The physics input is the list of symmetries, not the formula.

$J^\mu$ is prescribed here. Treating the current as an external source given in advance is a fiction — real charges respond to the field they make — but it is the standard and useful one, and it is what unit 24 did too. The cost is recorded twice below: such a system is not translation invariant if $J$ depends on $x$, and it has no energy–momentum tensor of its own. A closed theory would add a matter Lagrangian and let $J^\mu$ be that matter's Noether current; problem P4 constructs the simplest example.

section 2

Maxwell from an action

Four lines of variation, and two of the four equations become identities.

29 · maxwell

Why the potential and not the field

A variational principle needs a field whose values may be varied freely. $\vec E$ and $\vec B$ are not free: they already satisfy $\nabla\cdot\vec B = 0$ and Faraday's law, and varying them independently would violate those.

Taking $A_\mu$ as the field and $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ as a derived quantity makes them automatic.

the Bianchi identity

For any $C^2$ field $A$, $$\partial_\lambda F_{\mu\nu} + \partial_\mu F_{\nu\lambda} + \partial_\nu F_{\lambda\mu} = 0 .$$

Proof. Substitute $F = \partial A$: each of the six terms appears twice with opposite signs by Clairaut's theorem (unit 03), $\partial_\lambda\partial_\mu A_\nu = \partial_\mu\partial_\lambda A_\nu$. $\square$

what that buys

The identity is $\nabla\cdot\vec B = 0$ and $\nabla\times\vec E = -\partial_t\vec B$: two of Maxwell's four equations, holding for every $A$ whatsoever.

They are not field equations. Nothing is being solved; they are the price of writing $F = \partial A$, and they are paid in advance.

the price

$A$ is not unique. $A_\mu \to A_\mu + \partial_\mu\chi$ leaves $F_{\mu\nu}$ unchanged, since $\partial_\mu\partial_\nu\chi = \partial_\nu\partial_\mu\chi$. Unit 23 called this gauge freedom and used it to choose a convenient $A$; here it becomes a constraint on what the action may contain, and on the next slide but one, a theorem about $J$.

The identity in three-vector form. Recall $F_{0i} = -E_i/c$, so $F_{i0} = +E_i/c$, and $F_{ij} = \epsilon_{ijk}B_k$. Take $\lambda\mu\nu = 123$: $\partial_1F_{23} + \partial_2F_{31} + \partial_3F_{12} = \partial_1B_1 + \partial_2B_2 + \partial_3B_3 = \nabla\cdot\vec B = 0$. Take $\lambda\mu\nu = 012$: $\partial_0F_{12} + \partial_1F_{20} + \partial_2F_{01} = c^{-1}\partial_tB_3 + c^{-1}\partial_1E_2 - c^{-1}\partial_2E_1 = c^{-1}\big(\partial_t\vec B + \nabla\times\vec E\big)_3 = 0$, Faraday's law. The four independent choices of $\{\lambda\mu\nu\}$ from four indices give exactly one scalar equation and one vector equation.

The converse is Poincaré's lemma. On a region with no holes, any $F$ satisfying the Bianchi identity is $\partial A$ for some $A$. Unit 34 proves this properly as $\mathrm dF = 0 \Rightarrow F = \mathrm dA$ on a star-shaped domain, and shows that on a region with the wrong topology it can fail — which is where the Aharonov–Bohm effect and magnetic monopoles live. Until then, take the existence of $A$ as unit 23's construction of the vector potential, done there for $\nabla\cdot\vec B = 0$ on all of $\mathbb R^3$.

A genuine philosophical wrinkle, worth one minute. $A$ is not measurable and $F$ is, yet the action needs $A$ and not $F$. The classical answer is that $A$ is bookkeeping. The quantum answer is that it is not quite: the phase a charge picks up going round a loop is $\frac q\hbar\oint A_\mu\mathrm dx^\mu$, which is gauge invariant and observable even where $F = 0$. Aharonov and Bohm predicted this in 1959 and Chambers measured it in 1960.

29 · maxwell · example

Worked: varying $S[A]$, in four lines

example 3 · the centrepiece

Line 1 — vary $F$. $\delta F_{\mu\nu} = \partial_\mu\delta A_\nu - \partial_\nu\delta A_\mu$, so $$\delta\big(F_{\mu\nu}F^{\mu\nu}\big) = 2F^{\mu\nu}\delta F_{\mu\nu} = 4F^{\mu\nu}\partial_\mu\delta A_\nu ,$$ the last step because $F^{\mu\nu}$ is antisymmetric and the two terms are equal.

Line 2 — the integrand. $$\delta\mathcal L = -\frac{1}{\mu_0}F^{\mu\nu}\partial_\mu\delta A_\nu + J^\nu\delta A_\nu .$$

Line 3 — integrate by parts. $$-\frac{1}{\mu_0}F^{\mu\nu}\partial_\mu\delta A_\nu = -\frac{1}{\mu_0}\partial_\mu\big(F^{\mu\nu}\delta A_\nu\big) + \frac{1}{\mu_0}\big(\partial_\mu F^{\mu\nu}\big)\delta A_\nu ,$$ and the first piece integrates to $-\frac{1}{\mu_0}\oint F^{\mu\nu}\delta A_\nu\,\mathrm d\Sigma_\mu = 0$ for $\delta A$ of compact support.

Line 4 — read off. $$\delta S = \frac1c\int_\Omega\Big(\frac{1}{\mu_0}\partial_\mu F^{\mu\nu} + J^\nu\Big)\delta A_\nu\,\mathrm d^4x ,$$ so the fundamental lemma gives $\partial_\mu F^{\mu\nu} = -\mu_0J^\nu$, that is

$$\boxed{\ \partial_\nu F^{\mu\nu} = \mu_0 J^\mu\ }$$

Maxwell's source equations. With the Bianchi identity, all four of Maxwell's equations, from one functional.

The same result from the general formula: $\pi^{\mu\nu} = \partial\mathcal L/\partial(\partial_\mu A_\nu) = -F^{\mu\nu}/\mu_0$ and $\partial\mathcal L/\partial A_\nu = J^\nu$, so $\partial_\mu\pi^{\mu\nu} = J^\nu$ says the same thing.

Line 1, slowly. $\delta(F_{\mu\nu}F^{\mu\nu}) = \delta F_{\mu\nu}\cdot F^{\mu\nu} + F_{\mu\nu}\cdot\delta F^{\mu\nu} = 2F^{\mu\nu}\delta F_{\mu\nu}$, because raising and lowering with the constant $\eta$ commutes with $\delta$. Then $2F^{\mu\nu}(\partial_\mu\delta A_\nu - \partial_\nu\delta A_\mu)$: in the second term relabel $\mu\leftrightarrow\nu$ to get $-2F^{\nu\mu}\partial_\mu\delta A_\nu = +2F^{\mu\nu}\partial_\mu\delta A_\nu$. The two terms are equal, giving $4F^{\mu\nu}\partial_\mu\delta A_\nu$.

Why $\partial\mathcal L/\partial(\partial_\mu A_\nu) = -F^{\mu\nu}/\mu_0$ and not $-2F^{\mu\nu}/\mu_0$. Because $\mathcal L$ depends on $\partial_\mu A_\nu$ both directly and through the antisymmetrised combination, and the factor of 4 in line 1 against the $-1/4\mu_0$ leaves exactly $-1/\mu_0$. If you are ever unsure, do the variation as in lines 1–3 rather than differentiating a formula: the variation cannot be wrong by a factor of two, and the partial derivative of a function of $\partial_\mu A_\nu$ with symmetry constraints easily can.

Checking the equation componentwise. $\mu = 0$: $\partial_\nu F^{0\nu} = \partial_iF^{0i} = \partial_i(E_i/c) = (\nabla\cdot\vec E)/c$, and $\mu_0J^0 = \mu_0c\rho$. Equality gives $\nabla\cdot\vec E = \mu_0c^2\rho = \rho/\varepsilon_0$ ✓. $\mu = j$: $\partial_\nu F^{j\nu} = \partial_0F^{j0} + \partial_iF^{ji} = -c^{-2}\partial_tE_j + (\nabla\times\vec B)_j$, and equating to $\mu_0J_j$ gives Ampère–Maxwell ✓, displacement current and all.

The index order matters. $\partial_\nu F^{\mu\nu} = \mu_0J^\mu$ and $\partial_\mu F^{\mu\nu} = -\mu_0J^\nu$ are the same statement, since $F$ is antisymmetric. Books differ; this deck uses the first, which has no minus sign, and always writes the contracted index second. If you import a formula, contract it against $J$ and check the sign of Gauss's law.

Three of the four steps were the same as example 1's. The only genuinely new work was line 1, the antisymmetry. That is the pattern for every field theory: the variational machinery is fixed, and all the character of a theory is in its $\mathcal L$.

29 · maxwell

Gauge invariance forces charge conservation

theorem

$S[A]$ is unchanged by $A_\mu\to A_\mu + \partial_\mu\chi$ for every compactly supported $\chi$ if and only if $\partial_\mu J^\mu = 0$.

proof

$F$ is untouched, so only the coupling changes: $$\Delta S = \frac1c\int_\Omega J^\mu\partial_\mu\chi\,\mathrm d^4x = -\frac1c\int_\Omega(\partial_\mu J^\mu)\,\chi\,\mathrm d^4x ,$$ the boundary term vanishing by compact support. So $\Delta S = 0$ for all $\chi$ iff $\partial_\mu J^\mu = 0$, by the fundamental lemma. $\square$

what it says

$\partial_\mu J^\mu = \partial_t\rho + \nabla\cdot\vec J = 0$: unit 24's continuity equation. A gauge-invariant theory cannot be coupled to a current that is not conserved — the symmetry constrains what may appear on the right-hand side.

off shell, and that matters

The argument never used the field equations. Contrast: given $\partial_\nu F^{\mu\nu} = \mu_0J^\mu$, taking $\partial_\mu$ gives $\mu_0\partial_\mu J^\mu = 0$, symmetric contracted with antisymmetric — true of solutions only.

The first says an inconsistent source makes the action meaningless; the second says such a source admits no solution at all.

Why the two arguments are different and both worth having. The on-shell one is an integrability condition: write down Maxwell's equations with $\partial_\mu J^\mu\ne0$ and they have no solution, because the left side is identically divergence-free. The off-shell one is stronger and more structural: it says the action is not even well defined as a gauge-invariant object. The second kind of argument is the one that generalises — in Yang–Mills theory and in general relativity the analogous statement (the Noether identity of a local symmetry) is what forces the matter current, and the matter stress tensor, to be conserved. Unit 35 meets it again as the contracted Bianchi identity forcing $\nabla_\mu T^{\mu\nu} = 0$.

Compact support again. If $\chi$ is allowed to be nonzero on $\partial\Omega$, the boundary term $\oint\chi J^\mu\mathrm d\Sigma_\mu$ is the net charge flux through the boundary, which need not vanish. The theorem is a local statement and is proved with local variations.

The name. "Gauge" is Hermann Weyl's, from a failed 1918 attempt to unify gravity and electromagnetism by letting the scale — the gauge — of lengths vary from point to point. Einstein pointed out that atoms would then have position-dependent spectra. Weyl revived the idea in 1929 with the phase of a complex wave function in place of the scale, which works, and the wrong name stuck. Problem P4 walks through that construction: it is the cleanest derivation there is of why $A_\mu$ must exist.

section 3

Noether, for fields

Unit 13's three lines, once more, with a current in place of a charge.

29 · noether

What a symmetry of a field action is

symmetry, infinitesimally

A rule assigning to each field configuration a variation $\delta\varphi_a = \varepsilon X_a(\varphi,\partial\varphi,x)$, such that for every configuration — not only solutions — $$\delta\mathcal L = \varepsilon\,\partial_\mu K^\mu$$ for some $K^\mu(\varphi,\partial\varphi,x)$.

$K^\mu = 0$ is a strict symmetry; $K^\mu\ne0$ is unit 13's quasi-symmetry, and is needed: translations are one.

Off shell is essential. On shell every variation satisfies $\delta\mathcal L = \partial_\mu(\pi^\mu{}_a\delta\varphi_a)$, so on-shell "symmetry" is empty.

two kinds

Internal symmetry. Mixes the field components at a fixed event: $\delta\varphi_a = \varepsilon\,M_{ab}\varphi_b$. The phase rotation of a complex field is the example, and its charge is electric charge.

Spacetime symmetry. Moves the event: translations $\delta\varphi_a = -\varepsilon a^\nu\partial_\nu\varphi_a$, Lorentz transformations, and (sometimes) scalings. Their charges are energy, momentum and angular momentum.

a translation looks odd, and is not

Translating the configuration by $a$ means the new field at $x$ equals the old one at $x - a$: $\varphi'(x) = \varphi(x-a)$, so $\delta\varphi = -a^\nu\partial_\nu\varphi$. The minus sign is not a convention; it is what "move the field, not the axes" means.

Why the off-shell identity makes on-shell symmetry vacuous. For any variation whatsoever, $\delta\mathcal L = \frac{\partial\mathcal L}{\partial\varphi_a}\delta\varphi_a + \pi^\mu{}_a\partial_\mu\delta\varphi_a$. On shell the first coefficient is $\partial_\mu\pi^\mu{}_a$, and the two terms assemble into $\partial_\mu(\pi^\mu{}_a \delta\varphi_a)$ by the product rule. So every $\delta\varphi$ changes $\mathcal L$ by a total divergence on shell, and the resulting "current" is identically zero. Unit 13 made the same point for point mechanics, and it is the single most common error in applying Noether's theorem.

The general spacetime transformation. A fuller treatment lets the event move too, $x^\mu\to x^\mu + \varepsilon\xi^\mu$, and defines the total variation of the field including the change of argument. It gives the same currents, with a $\mathcal L\xi^\mu$ term appearing automatically instead of being supplied as $K^\mu$. This deck takes the shorter route — variations of the field alone, with $K^\mu$ carrying the difference — because it needs no new machinery beyond unit 13's quasi-symmetry, and because every symmetry this course needs can be written that way. Landau and Lifshitz, The Classical Theory of Fields §32, does it the other way in a page; the two agree.

Scale symmetry, mentioned and dropped. $\mathcal L = -\frac12(\partial\varphi)^2$ is invariant under $x\to\lambda x$, $\varphi\to\lambda^{-1}\varphi$ in four dimensions, and the corresponding current is built from a modified, "improved" stress tensor. It is a genuine symmetry of the massless scalar and of Maxwell, broken by any mass term, and it is what makes the tracelessness on the last section's slides more than a coincidence. Pursuing it properly needs conformal transformations, which this course does not reach.

29 · noether

Noether's theorem for fields

theorem

If $\delta\varphi_a = \varepsilon X_a$ satisfies $\delta\mathcal L = \varepsilon\partial_\mu K^\mu$ off shell, then the Noether current $$j^\mu = \frac{\partial\mathcal L}{\partial(\partial_\mu\varphi_a)}X_a - K^\mu$$ satisfies $\partial_\mu j^\mu = 0$ on every solution.

proof · three lines, as in unit 13

By the chain rule, for any variation, $$\varepsilon^{-1}\delta\mathcal L = \frac{\partial\mathcal L}{\partial\varphi_a}X_a + \pi^\mu{}_a\,\partial_\mu X_a .$$ On shell $\partial\mathcal L/\partial\varphi_a = \partial_\mu\pi^\mu{}_a$, so the right-hand side is $\partial_\mu(\pi^\mu{}_aX_a)$ by the product rule. The hypothesis says the left-hand side is $\partial_\mu K^\mu$. Subtract. $\square$

The proof, with every index shown. Write $\delta\varphi_a = \varepsilon X_a$. Then $\delta(\partial_\mu\varphi_a) = \partial_\mu\delta\varphi_a = \varepsilon\partial_\mu X_a$ — variation and differentiation commute because both are partial derivatives of $\varphi(x;\varepsilon)$ and mixed partials commute (Clairaut, unit 03). Hence $\delta\mathcal L = \varepsilon\big[\frac{\partial\mathcal L}{\partial\varphi_a}X_a + \frac{\partial\mathcal L}{\partial(\partial_\mu\varphi_a)}\partial_\mu X_a\big]$, summed over $a$ and $\mu$. Using the field equation on the first term: $\frac{\partial\mathcal L}{\partial\varphi_a}X_a = (\partial_\mu\pi^\mu{}_a)X_a$, so the bracket is $(\partial_\mu\pi^\mu{}_a)X_a + \pi^\mu{}_a\partial_\mu X_a = \partial_\mu(\pi^\mu{}_aX_a)$. Equating to $\varepsilon\partial_\mu K^\mu$ gives $\partial_\mu(\pi^\mu{}_aX_a - K^\mu) = 0$.

Compare unit 13 line by line. There: $\dot Q = \dot p\cdot\xi + p\cdot\dot\xi - \dot F = \delta L - \dot F = 0$, with $Q = p\cdot\xi - F$. Here $p_i$ has become $\pi^\mu{}_a$, one momentum per field component per direction, the time derivative has become a four-divergence, and the conserved number has become a conserved current. The proof is the same three lines. That is the sense in which this unit is "Noether again".

The current is not unique. Adding $\partial_\lambda B^{\lambda\mu}$ with $B$ antisymmetric leaves $\partial_\mu j^\mu = 0$ intact and leaves $Q$ unchanged. That freedom is a nuisance in this section and the whole solution to this unit's problem in the next; it is stated as a theorem on the improvement slide of this section.

The converse. Unit 13 recorded that the converse holds too — every conserved quantity of this form comes from a symmetry — under conditions it stated. The field version is the same, and is due to the same 1918 article, "Invariante Variationsprobleme"; the second theorem there, about local symmetries, is the one that gives the identity on the gauge-invariance slide above and is the deeper of the two.

29 · noether

From a local law to a conserved charge

the conserved charge

$\partial_\mu j^\mu = 0$ reads $\partial_t(j^0/c) + \nabla\cdot\vec\jmath = 0$: unit 22's continuity equation. So

$$Q(t) = \frac1c\int_{\mathbb R^3} j^0\,\mathrm d^3x$$

obeys $\mathrm dQ/\mathrm dt = -\oint_{S_\infty}\vec\jmath\cdot\mathrm d\vec A$, so it is constant if the flux at infinity vanishes. A local law is stronger than a global one: it forbids charge vanishing here and reappearing there.

Lines of current between two instants. Integrating $\partial_\mu j^\mu = 0$ over the slab gives $Q(t_2) - Q(t_1) = -$(what leaves through the side), so $Q$ is constant when nothing leaves. A global law alone would allow the forbidden pair on the left; the local one does not.

From the local law to the global one, in one line. Integrate $\partial_\mu j^\mu = 0$ over the slab $t_1 \le t \le t_2$, $|\vec x| \le R$ of the divergence-theorem slide. The future and past faces give $\int j^0\,\mathrm d^3x$ at $t_2$ minus the same at $t_1$, that is $c\,[Q(t_2) - Q(t_1)]$; the side face, where $\mathrm d\Sigma_i = n_i\,\mathrm dA\,\mathrm dx^0$ and $\mathrm dx^0 = c\,\mathrm dt$, gives $c\int_{t_1}^{t_2}\oint_{S_R}\vec\jmath\cdot\mathrm d\vec A\,\mathrm dt$. The sum is zero, so $Q(t_2) - Q(t_1) = -\int_{t_1}^{t_2}\oint_{S_R}\vec\jmath\cdot\mathrm d\vec A\,\mathrm dt$, which is the slide's $\mathrm dQ/\mathrm dt$ formula integrated in time.

What to see in the figure. The flow lines are the integral curves of $j^\mu$: charge moves along them, never faster than they lean. The one bending out through the side is the flux at infinity, the term that a radiating or leaking system keeps. The pair on the left — a line that stops at one event and another that starts somewhere else at a later time — would keep the total fixed if timed right, but at the stopping event $\partial_\mu j^\mu \ne 0$. That is what "a local law is stronger than a global one" means, and relativity insists on the local version: "at the same time" for two separated events is not even frame independent.

29 · noether

An internal symmetry: the phase of a complex field

Take $\mathcal L = -\partial_\mu\phi^*\partial^\mu\phi - \kappa^2\phi^*\phi$, with $\phi$ complex — two real fields in one symbol. Under $\phi\to e^{i\theta}\phi$ with $\theta$ constant, every term pairs a $\phi$ with a $\phi^*$: $\mathcal L$ is unchanged exactly, so $K^\mu = 0$.

Infinitesimally $\delta\phi = i\varepsilon\phi$, $\delta\phi^* = -i\varepsilon\phi^*$, with $\partial\mathcal L/\partial(\partial_\mu\phi) = -\partial^\mu\phi^*$ and $\partial\mathcal L/\partial(\partial_\mu\phi^*) = -\partial^\mu\phi$. The theorem gives $(-\partial^\mu\phi^*)(i\phi) + (-\partial^\mu\phi)(-i\phi^*)$, that is

$$j^\mu = i\big(\phi^*\partial^\mu\phi - \phi\,\partial^\mu\phi^*\big),\qquad \partial_\mu j^\mu = 0 ,$$

real, since $\phi^*\partial\phi$ and its conjugate are subtracted.

check it directly

$\partial_\mu j^\mu = i(\phi^*\Box_\eta\phi - \phi\Box_\eta\phi^*)$ with $\Box_\eta = \partial_\mu\partial^\mu$, the cross terms cancelling. On shell $\Box_\eta\phi = \kappa^2\phi$ and likewise for $\phi^*$, so this is $i\kappa^2(\phi^*\phi - \phi\phi^*) = 0$ ✓.

Treating $\phi$ and $\phi^*$ as independent. This is a computational convenience, not a claim: writing $\phi = (\varphi_1 + i\varphi_2)/\sqrt2$ and varying the two real fields gives the same equations, because the map between $(\varphi_1,\varphi_2)$ and $(\phi,\phi^*)$ is an invertible linear change of variables and the Euler–Lagrange equations are form-invariant under such a change — unit 12's theorem, used in a new setting. The independent-variation trick saves a page.

Normalisation. $j^\mu$ as written has the dimensions of $\varphi^2/\mathrm m$, not of a current density. The physical electric current of a field of charge $q$ turns out to be $-qj^\mu/\hbar$; the point of Noether's theorem is the conservation, and the sign and scale of a conserved current are always free — the overall sign flips with the direction the phase is turned. Problem P4 fixes both by the coupling.

Historically. This current is the one Klein and Gordon wrote in 1926 and interpreted, wrongly, as a probability density — $j^0$ is not positive definite, which was one of the reasons the Klein–Gordon equation was abandoned as a one-particle wave equation and then recovered as a field equation. As a charge density, indefiniteness is exactly right: charge comes with both signs.

29 · noether

The rotation happens in field space

a real field has no charge

Set $\phi = \phi^*$ and $j^\mu \equiv 0$. The symmetry needed two real fields to rotate into each other. This is why electric charge requires a complex field, and why a neutral particle is its own antiparticle.

Write $\phi = (\varphi_1 + i\varphi_2)/\sqrt2$. Then $\phi\to e^{i\theta}\phi$ is a rotation by $\theta$ in the plane of $(\varphi_1, \varphi_2)$, the same at every event, and $\mathcal L$ sees only $|\phi|^2$ and $|\partial\phi|^2$, which a rotation leaves alone.

Problem P4 lets $\theta$ depend on $x$ and finds that invariance then forces a new field with the transformation law of $A_\mu$.

The symmetry does not move the event; it turns the field's value at that event. The circle is where $\mathcal L$ cannot tell values apart. A real field is confined to the horizontal axis, and no rotation keeps it there: nothing to rotate, no current.

The current, in real components. With $\phi = (\varphi_1+i\varphi_2)/\sqrt2$, a short computation gives $j^\mu = \varphi_2\partial^\mu\varphi_1 - \varphi_1\partial^\mu\varphi_2$: a rotation in the $(\varphi_1,\varphi_2)$ plane, whose charge is the field-theoretic analogue of angular momentum in that internal plane. The phrase "internal symmetry" is meant literally — the rotation happens in a space of field values, not in spacetime.

What to see in the figure. The left panel is spacetime and stays untouched: an internal symmetry is a map of the target space of the field, applied pointwise. The right panel is that target space at one event. The conserved charge counts how the field winds around the origin of this plane as one moves through spacetime — $j^\mu = \varphi_2\partial^\mu\varphi_1 - \varphi_1\partial^\mu\varphi_2$ is $-2|\phi|^2\,\partial^\mu\alpha$ for $\phi = |\phi|e^{i\alpha}$, the rate at which the phase $\alpha$ turns. A field whose values stay on one line through the origin never turns, and its current is zero, which is the warning box in a picture.

29 · noether

Translations, and the canonical energy–momentum tensor

If $\mathcal L$ has no explicit $x$, translating the configuration is a quasi-symmetry: $$X_a = -a^\nu\partial_\nu\varphi_a,\qquad \delta\mathcal L = -\varepsilon a^\nu\partial_\nu\mathcal L = \varepsilon\,\partial_\mu(-a^\mu\mathcal L) ,$$ so $K^\mu = -a^\mu\mathcal L$. The theorem gives $j^\mu = a_\nu T^{\mu\nu}$ with

the canonical energy–momentum tensor $$T_{\text{can}}^{\mu\nu} = \eta^{\mu\nu}\mathcal L - \frac{\partial\mathcal L}{\partial(\partial_\mu\varphi_a)}\,\partial^\nu\varphi_a .$$

Four independent $a^\nu$, four conserved currents, one per direction of spacetime: $\partial_\mu T^{\mu\nu}_{\text{can}} = 0$.

check on the scalar

$\mathcal L = -\tfrac12[(\partial\varphi)^2 + \kappa^2\varphi^2]$ gives $\pi^\mu = -\partial^\mu\varphi$, so $T^{\mu\nu}_{\text{can}} = \partial^\mu\varphi\,\partial^\nu\varphi + \eta^{\mu\nu}\mathcal L$, and $$T^{00} = \tfrac12(\partial_0\varphi)^2 + \tfrac12|\nabla\varphi|^2 + \tfrac12\kappa^2\varphi^2 \ge 0 ,$$ kinetic plus gradient plus potential energy per unit volume ✓.

signs differ between books

In $(+,-,-,-)$ the same tensor is written $\pi^\mu{}_a\partial^\nu\varphi_a - \eta^{\mu\nu}\mathcal L$. The signature-free statement is the one to remember: the Noether current of a translation by $a$ is $j^\mu = a_\nu T^{\mu\nu}$, and $T^{00}$ is an energy density, positive for the scalar.

Full derivation of $j^\mu = a_\nu T^{\mu\nu}$. Substituting into the theorem, $j^\mu = \pi^\mu{}_a(-a^\nu\partial_\nu\varphi_a) - (-a^\mu\mathcal L) = -a^\nu\pi^\mu{}_a\partial_\nu\varphi_a + a^\mu\mathcal L$. Lower the free index on $a$: $a^\nu\partial_\nu\varphi_a = a_\nu\partial^\nu\varphi_a$ and $a^\mu = a_\nu\eta^{\mu\nu}$, so $j^\mu = a_\nu[\eta^{\mu\nu}\mathcal L - \pi^\mu{}_a\partial^\nu\varphi_a] = a_\nu T^{\mu\nu}$. Since $\partial_\mu j^\mu = 0$ for every constant $a_\nu$, each of the four columns is separately conserved.

Why this is the same theorem as unit 13's energy. There, time-translation invariance gave the conserved energy function $h = p\dot q - L$. Here the $\nu = 0$ column is $T^{\mu 0}$, whose time component is $T^{00} = \eta^{00}\mathcal L - \pi^0{}_a\partial^0\varphi_a = -\mathcal L + \pi^0{}_a\partial_0\varphi_a$ — exactly $p\dot q - L$, per unit volume, with $\partial_0 = c^{-1}\partial_t$ absorbing the factor of $c$. The energy density is the Hamiltonian density of unit 14, and Noether has simply rediscovered the Legendre transform.

The requirement "no explicit $x$" is the whole hypothesis. If $\mathcal L$ contains a prescribed external field or current depending on position, translating the configuration changes $\mathcal L$ by more than a total divergence, and energy and momentum are not conserved — correctly, since the external agent can supply both. This is why the worked Maxwell calculation below sets $J = 0$.

Naming. "Canonical" here means "obtained from the canonical Noether construction", not "standard" — it is the tensor the general procedure hands you, defects and all. Some books write $\Theta^{\mu\nu}$ for it and reserve $T^{\mu\nu}$ for the repaired one; this deck keeps the subscript.

29 · noether

What its sixteen components mean

	$\nu = 0$	$\nu = j$
$\mu = 0$	$T^{00}$: energy density, $\mathrm{J\,m^{-3}}$	$T^{0j}$: momentum density $\times\,c$
$\mu = i$	$T^{i0}$: energy flux $/\,c$	$T^{ij}$: flux of $j$-momentum in direction $i$

$\partial_\mu T^{\mu 0} = 0$ is $\partial_t(T^{00}) + \nabla\cdot(cT^{i0}) = 0$: energy conservation, unit 24's Poynting theorem in vacuum. $\partial_\mu T^{\mu j} = 0$ is momentum conservation, unit 22's Cauchy momentum balance with the stress supplying the flux.

the four charges

$\displaystyle P^\nu = \frac1c\int_{t\,=\,\text{const}} T^{0\nu}\,\mathrm d^3x$ is the total four-momentum: $P^0 = E/c$ and $P^j$ the momentum. Conserved when the flux through the sphere at infinity vanishes.

Checking the factors of $c$. $[T^{00}] = \mathrm{J\,m^{-3}}$. Conservation $\partial_\mu T^{\mu0} = 0$ reads $c^{-1}\partial_t T^{00} + \partial_i T^{i0} = 0$, so $\partial_tT^{00} + \nabla\cdot(c\,\vec T^{\,\cdot 0}) = 0$ and the energy flux is $cT^{i0}$, in $\mathrm{W\,m^{-2}}$ ✓. Similarly $T^{0j}/c$ is the momentum density in $\mathrm{kg\,m^{-2}s^{-1}}$. For Maxwell these become $cT^{i0} = S^i$, the Poynting vector, and $T^{0j}/c = \varepsilon_0(\vec E\times\vec B)^j$, unit 24's field momentum density — problem P7 checks both.

Why symmetry of $T$ is not obvious but is wanted. $T^{0j}$ is (momentum density) $\times c$ and $T^{j0}$ is (energy flux) $/c$. That these should be equal — that energy flux and momentum density are the same thing up to $c^2$ — is a real physical claim, and it is precisely the relativistic content of $E = mc^2$ applied to flowing energy. The canonical tensor does not have to satisfy it, and for Maxwell it does not.

An integral over a constant-time slice looks frame-dependent, and $P^\nu$ is not. Different inertial observers integrate over different slices. That $P^\nu$ comes out the same four-vector is a consequence of $\partial_\mu T^{\mu\nu} = 0$ plus decay at infinity: the difference between two slices is the flux through the timelike surface joining them, which vanishes. The same argument in unit 34's language is Stokes' theorem applied to a closed three-surface.

29 · noether

Momentum flux through a face

Fix a small face of area $\mathrm dA$ with normal $\hat x$. In time $\mathrm dt$ the field carries across it an amount of momentum — a 3-vector — and an amount of energy:

$$\mathrm d\vec p = \big(T^{xx}, T^{xy}, T^{xz}\big)\,\mathrm dA\,\mathrm dt,$$ $$\mathrm dE = cT^{x0}\,\mathrm dA\,\mathrm dt .$$

So the row $T^{i\,\cdot}$ of the spatial block is the momentum flux through a face normal to $\hat x^i$: the normal part a pressure, the other two a shear.

this is unit 22's object

Unit 22 flagged $T^{\mu\nu}$ and refused to build it, having no field theory to build it from. The stress tensor $\sigma_{ij}$ of a continuum is the spatial block $T^{ij}$, up to the sign convention that stress is a force and $T^{ij}$ is a momentum flux — equal and opposite.

The momentum crossing a face is an arrow, not a number, so its flux needs two indices: $i$ for the face, $j$ for the component carried. $T^{xx}$ pushes along the normal; $T^{xy}$ drags along the face, and $T^{xz}$ (not drawn) drags into the page.

Where the formula comes from. $\partial_\mu T^{\mu j} = 0$ reads $\partial_t(T^{0j}/c) + \partial_i T^{ij} = 0$: the density of $j$-momentum, $T^{0j}/c$, changes only by a flux whose $i$-th component is $T^{ij}$. Integrate over a small box and apply the divergence theorem: the $j$-momentum leaving through the face normal to $\hat x^i$ is $T^{ij}\,\mathrm dA$ per unit time. The energy line is the same argument for $\nu = 0$, where the flux is $cT^{i0}$.

The sign against unit 22. There the traction on a face with outward normal $\vec n$ — the force the material outside exerts on the material inside, per unit area — had components $t_j = \sigma_{jk}n_k$. Momentum flowing out through the face, $j$-component at rate $T^{ij}n_i\,\mathrm dA$, is momentum the inside loses, so the force on the inside is $-T^{ij}n_i\,\mathrm dA$. Hence $\sigma_{ji} = -T^{ij}$: the same matrix up to sign and a transposition, and for a symmetric $T$ — which section 4 will insist on — simply $\sigma_{ij} = -T^{ij}$. A gas at pressure $p$ has $\sigma_{ij} = -p\,\delta_{ij}$ and so $T^{ij} = p\,\delta_{ij}$: pressure is a positive momentum flux, molecules carrying $x$-momentum across a face normal to $x$.

29 · noether · example

Worked: the canonical tensor for Maxwell, and its two defects

example 4 · the running example, first route

Vacuum, $J = 0$, so that translations are a symmetry. With $\pi^{\mu\nu} = \partial\mathcal L/\partial(\partial_\mu A_\nu) = -F^{\mu\nu}/\mu_0$ from example 3,

$$T_{\text{can}}^{\mu\nu} = \eta^{\mu\nu}\mathcal L + \frac{1}{\mu_0}F^{\mu\alpha}\,\partial^\nu A_\alpha .$$

Defect 1 — not gauge invariant. $\partial^\nu A_\alpha$ is not: under $A\to A + \partial\chi$ it changes by $\partial^\nu\partial_\alpha\chi$.

Defect 2 — not symmetric. $F^{\mu\alpha}\partial^\nu A_\alpha$ has no symmetry in $\mu\nu$ at all.

A number. Take a uniform static field $\vec E = E\hat z$ with $\varphi = -Ez$, $\vec A = 0$, $\vec B = 0$. Then $\partial^\nu A_\alpha = 0$ unless $\nu$ is spatial, and

$$T^{00}_{\text{can}} = \eta^{00}\mathcal L = -\mathcal L = -\tfrac{\varepsilon_0}{2}E^2 .$$

At $E = 1.00\ \mathrm{MV\,m^{-1}}$ that is $\mathbf{-4.427\ \mathrm{J\,m^{-3}}}$, where unit 24's energy density is $+4.427\ \mathrm{J\,m^{-3}}$: the wrong sign.

It is still conserved — $\partial_\mu T^{\mu\nu}_{\text{can}} = 0$ holds — so conservation alone does not identify the energy.

Why $T_{\text{can}}^{00} = -\mathcal L$ here. The extra term is $\mu_0^{-1}F^{0\alpha}\partial^0A_\alpha$, and $\partial^0 = \eta^{00}\partial_0 = -c^{-1}\partial_t$, which annihilates a static field. So only $\eta^{00}\mathcal L = -\mathcal L$ survives, and $\mathcal L = \frac{\varepsilon_0}{2}E^2$ with $\vec B = 0$. Numerically $\frac12\times 8.8541878128\times10^{-12}\times(10^{6})^2 = 4.4271\ \mathrm{J\,m^{-3}}$.

Gauge dependence, made explicit. Apply $\chi = \beta zt$ to the same field. Then $A_0 \to A_0 + \beta z/c$ and $A_3 \to \beta t$, while $F$ is unchanged — check: $F_{03} = \partial_0A_3 - \partial_3A_0 = \beta/c - (E+\beta)/c = -E/c$, as before. Now $\partial^0A_3 = -\beta/c \ne 0$, and $T^{00}_{\text{can}} = -\mathcal L + \mu_0^{-1}F^{03}\partial^0A_3 = -\frac{\varepsilon_0}{2}E^2 - \varepsilon_0 E\beta$. The answer depends on $\beta$, a pure gauge parameter with no physical content whatever. An energy density that can be set to any value by a change of gauge is not an energy density.

The defects are not peculiar to this example. For any field with more than one component and any $\mathcal L$ whose fields carry spacetime indices, the canonical tensor is generically asymmetric: the culprit is that $\partial^\nu\varphi_a$ treats the index $\nu$ and the index $\mu$ in $\pi^\mu{}_a$ completely differently. For a scalar field it happens to come out symmetric, which is why the check on the translations slide passed and why a course that only ever does scalars never notices the problem.

Conserved but wrong is the lesson. Noether's theorem delivers a conserved current for each symmetry, not the conserved current. The next two slides say exactly how much freedom there is, and the following section uses a different principle to spend it.

29 · noether

What you are allowed to add: improvement

theorem

Let $B^{\lambda\mu\nu}$ be any field with $B^{\lambda\mu\nu} = -B^{\mu\lambda\nu}$ (three indices: not the magnetic field $\vec B$). Then $$\tilde T^{\mu\nu} = T^{\mu\nu} + \partial_\lambda B^{\lambda\mu\nu}$$ is also conserved, and gives the same total four-momentum.

proof

Conservation. $\partial_\mu\partial_\lambda B^{\lambda\mu\nu} = 0$: the derivative pair is symmetric in $\mu\lambda$ (Clairaut), $B$ is antisymmetric, and a symmetric object contracted with an antisymmetric one vanishes.

Charges. $\int\partial_\lambda B^{\lambda 0\nu}\mathrm d^3x = \int\partial_i B^{i0\nu}\mathrm d^3x$, because $B^{00\nu} = 0$ by antisymmetry, so the $\lambda = 0$ term is absent. What is left is a spatial divergence, and by the divergence theorem it is a surface integral at infinity: zero for fields that decay. $\square$

names

$\partial_\lambda B^{\lambda\mu\nu}$ is an improvement term; $B$ itself is a superpotential. The whole construction is exactly what makes $T$ ambiguous, and exactly what will let us repair it.

the two conclusions to take

Conservation cannot pin down $T$. There are infinitely many conserved tensors with the same charges.

Something else must. Local energy density is physical — it curves spacetime — so a further principle is needed. Section 4 supplies it.

The antisymmetry is in the first two indices only. The third index $\nu$ is a spectator: it labels which of the four conserved currents is being improved. Writing $B^{\lambda\mu\nu}$ with the antisymmetric pair first is the convention here; other books antisymmetrise the last two and the formulas pick up signs accordingly.

Why $\int\partial_iB^{i0\nu}\mathrm d^3x = 0$ needs a hypothesis. It is unit 05's divergence theorem applied to the vector field $V^i = B^{i0\nu}$ on a large ball, giving $\oint_{S_R}B^{i0\nu}n_i\,\mathrm dA$, which tends to zero provided $B$ falls off faster than $1/R^2$. For a static Coulomb field $B \sim F A \sim r^{-2}\cdot r^{-1} = r^{-3}$ and it does. For a radiating system $F \sim 1/r$ and the surface integral need not vanish — correctly, since a radiating system does lose energy through the sphere at infinity.

An analogy worth keeping. In magnetostatics $\vec B$ determines $\vec A$ only up to $\nabla\chi$, and one fixes $\vec A$ by an extra condition (a gauge choice) chosen for convenience. Here conservation determines $T$ only up to $\partial_\lambda B$, and one fixes $T$ by an extra condition — but this time the extra condition is not convenience. General relativity reads $T^{\mu\nu}$ pointwise, so the ambiguity must be settled by physics, and it is settled by asking what the field does to the geometry.

Where this freedom is not free. In general relativity two tensors differing by an improvement term source genuinely different metrics, because Einstein's equation is local. The resolution — that only the metric-variation tensor appears there — is the content of section 4, and it is why the ambiguity, which looks like a technicality, is actually the reason this unit exists.

29 · noether

Belinfante and Rosenfeld: the repair, for Maxwell

Split $\partial^\nu A_\alpha$ into a gauge-invariant piece and the rest, $\partial^\nu A_\alpha = F^{\nu}{}_{\alpha} + \partial_\alpha A^\nu$. Then, writing $F^2 = F_{\rho\sigma}F^{\rho\sigma}$,

$$T^{\mu\nu}_{\text{can}} = \underbrace{\frac{1}{\mu_0}\Big(F^{\mu\alpha}F^{\nu}{}_{\alpha} - \tfrac14\eta^{\mu\nu}F^2\Big)}_{\text{symmetric, gauge invariant}} \;+\; \frac{1}{\mu_0}F^{\mu\alpha}\partial_\alpha A^\nu ,\qquad\text{and in vacuum}\quad \partial_\alpha F^{\mu\alpha} = 0 ,$$

so the last term is a pure divergence: $\ \frac{1}{\mu_0}F^{\mu\alpha}\partial_\alpha A^\nu = \partial_\lambda B^{\lambda\mu\nu}$ with $B^{\lambda\mu\nu} = \frac{1}{\mu_0}F^{\mu\lambda}A^\nu = -B^{\mu\lambda\nu}$ ✓.

the Belinfante–Rosenfeld tensor

Subtract the improvement term the theorem allows: $T^{\mu\nu} = T^{\mu\nu}_{\text{can}} - \partial_\lambda B^{\lambda\mu\nu}$, leaving the bracket above.

Symmetric ✓, gauge invariant ✓, conserved ✓ — and $T^{00} = \tfrac{\varepsilon_0}{2}E^2 + \tfrac{1}{2\mu_0}B^2$, unit 24's energy density, with the right sign.

this looks like a trick, and it is

Nothing said why that $B$ and no other. Belinfante's general recipe builds $B$ from the spin current of the Lorentz symmetry, which is principled but long. Section 4 gets the same tensor in six lines from a definition that cannot go wrong.

The split, checked. $F^\nu{}_\alpha = \partial^\nu A_\alpha - \partial_\alpha A^\nu$ by raising the first index of $F_{\nu\alpha} = \partial_\nu A_\alpha - \partial_\alpha A_\nu$; hence $\partial^\nu A_\alpha = F^\nu{}_\alpha + \partial_\alpha A^\nu$. Substituting into $\mu_0^{-1}F^{\mu\alpha}\partial^\nu A_\alpha$ gives $\mu_0^{-1}F^{\mu\alpha}F^\nu{}_\alpha$ plus $\mu_0^{-1}F^{\mu\alpha}\partial_\alpha A^\nu$, and $\eta^{\mu\nu}\mathcal L = -\frac{1}{4\mu_0}\eta^{\mu\nu}F^2$ supplies the trace term.

Why the first bracket is symmetric. $F^{\mu\alpha}F^\nu{}_\alpha = F^{\mu\alpha}F^{\nu\beta}\eta_{\alpha\beta}$: swapping $\mu$ and $\nu$ swaps the two factors and relabels the dummies, giving the same thing. The trace term is symmetric because $\eta$ is.

The product rule step, with the field equation. $\partial_\lambda(F^{\mu\lambda}A^\nu) = (\partial_\lambda F^{\mu\lambda})A^\nu + F^{\mu\lambda}\partial_\lambda A^\nu$. The first term is $\mu_0J^\mu A^\nu$, which vanishes in vacuum. So the equality of the improvement term and the leftover holds on shell. With sources present the repair still works but $T$ acquires a term describing the exchange of energy and momentum with the charges — which is the Lorentz force density, and is unit 28's business.

Belinfante's actual recipe. Lorentz invariance gives its own Noether current, $M^{\lambda\mu\nu} = x^\mu T^{\lambda\nu}_{\text{can}} - x^\nu T^{\lambda\mu}_{\text{can}} + S^{\lambda\mu\nu}$, where $S$ is the spin current: the part of the angular momentum coming from the field's own indices rather than from its position. Belinfante (1939) and Rosenfeld (1940) showed that $B$ built from $S$ always symmetrises $T$, for any Lorentz-invariant theory, and that this is not a coincidence — a symmetric $T$ and a conserved total angular momentum are the same statement, which is the slide Why symmetry is angular momentum.

section 4

The definition general relativity uses

Ask the action how it feels about the metric, and the answer is symmetric by construction.

29 · metric

The move: let the metric be general, in flat space

Nothing so far needed coordinates to be inertial. Write the same action with a general symmetric $g_{\mu\nu}(x)$ in place of $\eta_{\mu\nu}$ — spherical coordinates, say, or a rotating frame — and ask how $S$ responds when $g$ is nudged.

why this is safe here

The spacetime is still flat. $g$ is Minkowski's form written in some other basis at each point, and no curvature, connection or covariant derivative is needed. Unit 32 supplies those; unit 35 repeats this calculation with $g$ genuinely dynamical.

Two ingredients are missing: an invariant way to integrate, and a rule for differentiating a determinant.

One flat plane, two coordinate systems. In polar coordinates the metric $\mathrm ds^2 = \mathrm dr^2 + r^2\mathrm d\theta^2$ changes from point to point although nothing is curved, and two cells with the same $\Delta r\,\Delta\theta$ have different areas, $r\,\Delta r\,\Delta\theta$. That factor $r$ is $\sqrt{\det g}$.

A concrete instance. In spherical coordinates on flat three-space, $g_{ij} = \mathrm{diag}(1, r^2, r^2\sin^2\theta)$, $\det g = r^4\sin^2\theta$, and $\sqrt{g}\,\mathrm dr\,\mathrm d\theta\,\mathrm d\varphi = r^2\sin\theta\,\mathrm dr\,\mathrm d\theta\,\mathrm d\varphi$ — unit 05's spherical volume element, which was computed there as a Jacobian determinant. It is the same determinant. The metric's determinant and the coordinate Jacobian are two views of one object.

What to see in the figure. The faint Cartesian grid is the plane's own straight lines; the polar grid is laid over the same plane. Measured in $r$ and $\theta$, a step of $\Delta\theta$ covers a distance $r\Delta\theta$, which is what $g_{\theta\theta} = r^2$ records, so a coordinate cell is larger the further out it is. The two shaded cells have the same coordinate size and areas in the ratio $4.5 : 1.5 = 3$. A metric that varies with position is therefore no evidence of curvature; it may only be evidence of curvilinear coordinates. Telling the two apart is unit 32's business.

29 · metric

The volume element coordinates cannot change

the invariant volume element

$g := \det(g_{\mu\nu}) \lt 0$ in Lorentzian signature, and $$\sqrt{-g}\,\mathrm d^4x \quad\text{is unchanged by any change of coordinates.}$$

Proof. Under $x\to x'$ with $\Lambda^\alpha{}_\mu = \partial x^\alpha/\partial x'^\mu$, the transformation law $g'_{\mu\nu} = \Lambda^\alpha{}_\mu\Lambda^\beta{}_\nu g_{\alpha\beta}$ is $g' = \Lambda^{\!\top}g\Lambda$ as matrices, so $\det g' = (\det\Lambda)^2\det g$ and $\sqrt{-g'} = |\det\Lambda|\sqrt{-g}$. Unit 05's change of variables says $\mathrm d^4x = |\det\Lambda|\,\mathrm d^4x'$. The two factors cancel. $\square$

$\sqrt{-g}$ is a tensor density of weight one: it transforms with a Jacobian factor rather than with $\Lambda$'s.

The proof in a picture. A coordinate cell of $x'$ occupies $|\det\Lambda|$ times as much $x$-volume, so $\mathrm d^4x = |\det\Lambda|\,\mathrm d^4x'$; and $\sqrt{-g'} = |\det\Lambda|\sqrt{-g}$. In the product the two factors cancel.

Why $\mathrm d^4x$ alone will not do. Under $x\to x'$ the coordinate volume element picks up $|\det\Lambda|$, so $\int\mathcal L\,\mathrm d^4x$ depends on which coordinates are used, and cannot be a physical quantity. In inertial coordinates related by a Lorentz transformation $\det\Lambda = \pm1$ and the problem is invisible — which is exactly why it has not come up until now, and why introducing it in flat space with curvilinear coordinates is the right place.

Densities, defined properly. A tensor density of weight $w$ is a quantity whose components transform as a tensor's do, times $(\det\Lambda)^{w}$. $\sqrt{-g}$ has weight 1, an ordinary tensor has weight 0, and the Levi-Civita symbol $\epsilon_{\mu\nu\rho\sigma}$ of unit 28 — the one with entries $\pm1$ in every coordinate system — has weight $-1$, which is exactly why $\epsilon_{\mu\nu\rho\sigma}/\sqrt{-g}$ and $\sqrt{-g}\,\epsilon^{\mu\nu\rho\sigma}$ are the honest tensors. Unit 28 made that distinction; this is where it earns its keep.

Why $\det g \lt 0$. At a point, diagonalise: signature $(-,+,+,+)$ gives eigenvalues with one negative sign, so the determinant is negative, and by the transformation law it stays negative in every coordinate system, since $(\det\Lambda)^2 \gt 0$. Hence $-g \gt 0$ and the square root is real.

29 · metric

Differentiating a determinant

Jacobi's formula

For an invertible matrix $M$ depending on a parameter, $$\delta\det M = \det M\;\operatorname{tr}\!\big(M^{-1}\delta M\big) .$$

Proof. $\det(M + \delta M) = \det M\,\det(I + M^{-1}\delta M)$, and for any $A$, the Leibniz expansion of $\det(I + \varepsilon A)$ has $1 + \varepsilon\operatorname{tr}A$ as its terms of order $\le1$: only the all-diagonal product contributes at first order. $\square$

Geometrically: $\det$ is the volume of the image of the unit cube, and to first order only stretching along the axes changes it.

$I + \varepsilon A$ acting on the unit square, $\varepsilon$ exaggerated. Diagonal entries add two strips of area $\varepsilon a_{11}$ and $\varepsilon a_{22}$, plus a corner of order $\varepsilon^2$; an off-diagonal entry shears without changing base or height. So $\det(I+\varepsilon A) = 1 + \varepsilon\operatorname{tr}A + O(\varepsilon^2)$.

Full proof that $\det(I + \varepsilon A) = 1 + \varepsilon\operatorname{tr}A + O(\varepsilon^2)$. By the Leibniz formula, $\det(I+\varepsilon A) = \sum_\sigma \mathrm{sgn}(\sigma)\prod_\mu (I + \varepsilon A)_{\mu\,\sigma(\mu)}$. A permutation $\sigma$ that moves $k \ge 1$ indices contributes a product with at least $k$ off-diagonal factors, each carrying one power of $\varepsilon$; since a permutation cannot move exactly one index, every non-identity $\sigma$ contributes at order $\varepsilon^2$ or higher. The identity permutation contributes $\prod_\mu(1 + \varepsilon A_{\mu\mu}) = 1 + \varepsilon\sum_\mu A_{\mu\mu} + O(\varepsilon^2)$. $\square$

The figure is the proof for $2\times2$. There $\det(I+\varepsilon A) = (1+\varepsilon a_{11})(1+\varepsilon a_{22}) - \varepsilon^2a_{12}a_{21}$: the product of the diagonal is the stretched rectangle, and the off-diagonal entries appear only in the $\varepsilon^2$ term. A shear with only $a_{12}\ne0$ changes the area not at all. The Leibniz argument says the same in $n$ dimensions: every permutation other than the identity uses at least two off-diagonal entries.

Jacobi's formula elsewhere. It is the identity behind Liouville's theorem on the volume of a flow (unit 14, where $\frac{\mathrm d}{\mathrm dt}\det = \det\cdot\operatorname{tr}$ of the linearisation is exactly this), behind $\det e^{A} = e^{\operatorname{tr}A}$, and behind Abel's formula for the Wronskian. Seeing it a fourth time here is a good sign that it is structural and not a trick.

29 · metric

Varying the volume element and the inverse metric

Jacobi's formula, applied to $g_{\mu\nu}$, whose inverse is $g^{\mu\nu}$: $\delta g = g\,g^{\mu\nu}\delta g_{\mu\nu}$.

And since $g^{\mu\alpha}g_{\alpha\nu} = \delta^\mu_\nu$ is constant, $\delta g^{\mu\nu} = -g^{\mu\alpha}g^{\nu\beta}\delta g_{\alpha\beta}$, so $g_{\mu\nu}\delta g^{\mu\nu} = -g^{\mu\nu}\delta g_{\mu\nu}$.

the identity used four times below $$\begin{aligned}\delta\sqrt{-g} &= \tfrac12\sqrt{-g}\,g^{\mu\nu}\delta g_{\mu\nu}\\ &= -\tfrac12\sqrt{-g}\,g_{\mu\nu}\,\delta g^{\mu\nu}\end{aligned}$$

$\delta\sqrt{-g} = -\delta g/2\sqrt{-g}$ and $-g = (\sqrt{-g})^2$, so $-\delta g/2\sqrt{-g} = -g\,g^{\mu\nu}\delta g_{\mu\nu}/2\sqrt{-g}$.

check it on a diagonal metric

$g_{\mu\nu} = \mathrm{diag}(-1,1,1,r^2)$ has $\sqrt{-g} = r$. Nudge the last entry by $\delta$: $\sqrt{-g} \to \sqrt{r^2 + \delta} \approx r + \delta/2r$. The formula gives $\tfrac12 r\cdot g^{33}\delta = \delta/2r$ ✓.

Full proof of the inverse-variation identity. Differentiate $g^{\mu\alpha}g_{\alpha\nu} = \delta^\mu_\nu$: $\delta g^{\mu\alpha}\,g_{\alpha\nu} + g^{\mu\alpha}\delta g_{\alpha\nu} = 0$. Multiply by $g^{\nu\beta}$ and sum over $\nu$: $\delta g^{\mu\beta} = -g^{\mu\alpha}g^{\nu\beta}\delta g_{\alpha\nu}$. Contracting that with $g_{\mu\beta}$ gives $g_{\mu\beta}\delta g^{\mu\beta} = -g^{\alpha\nu}\delta g_{\alpha\nu}$, which is the statement on the slide. The practical consequence: varying with respect to $g_{\mu\nu}$ and with respect to $g^{\mu\nu}$ differ by a sign and an index raising, and almost every sign disagreement between textbooks on this subject is this one.

Why both forms are wanted. The definition of $T_{\mu\nu}$ on the next slide is written with $\delta g^{\mu\nu}$, because in Maxwell's Lagrangian the metric appears through the inverse, raising the indices of $F_{\mu\nu}$. The second line of the identity box is the one the worked example uses; the first is the one that makes the diagonal check easy.

29 · metric

The definition, and why its two properties are free

the metric stress–energy tensor

Write the matter action as $S_{\text m} = \frac1c\int\sqrt{-g}\,\mathcal L_{\text m}\,\mathrm d^4x$. Then $T_{\mu\nu}$ is the unique symmetric tensor field with

$$\delta S_{\text m} = -\frac{1}{2c}\int\sqrt{-g}\;T_{\mu\nu}\,\delta g^{\mu\nu}\,\mathrm d^4x \qquad\text{for every compactly supported }\delta g^{\mu\nu} ,$$

equivalently $\ T_{\mu\nu} = -\dfrac{2}{\sqrt{-g}}\dfrac{\delta(\sqrt{-g}\,\mathcal L_{\text m})}{\delta g^{\mu\nu}} \ $ and $\ T^{\mu\nu} = +\dfrac{2}{\sqrt{-g}}\dfrac{\delta(\sqrt{-g}\,\mathcal L_{\text m})}{\delta g_{\mu\nu}}$.

symmetric — free

$g^{\mu\nu}$ is symmetric, so only the symmetric part of the coefficient of $\delta g^{\mu\nu}$ is defined at all. Asymmetry is not repaired; it is unaskable.

gauge invariant — free

$S_{\text m}[A + \mathrm d\chi, g] = S_{\text m}[A, g]$ for every $g$. Differentiate that identity in $g$: the two derivatives agree. The same argument covers any internal symmetry.

conserved — nearly free

Invariance of $S_{\text m}$ under coordinate changes gives $\partial_\mu T^{\mu\nu} = 0$ on shell, by the same off-shell argument that gave $\partial_\mu J^\mu = 0$. Proved in unit 35.

Dimensions. $[\delta S] = \mathrm{J\,s}$, $[\mathrm d^4x] = \mathrm m^4$, $\delta g^{\mu\nu}$ is dimensionless, so $[T] = c\,\mathrm{J\,s\,m^{-4}} = \mathrm{J\,m^{-3}}$ ✓ — an energy density, as it must be.

Why a minus sign in the definition with $\delta g^{\mu\nu}$. It is chosen so that $T^{00}$ comes out positive for ordinary matter; the two displayed forms are consistent because $\delta g^{\mu\nu}$ and $\delta g_{\mu\nu}$ differ by a sign and two index raisings, as the previous slide showed: $\frac{2}{\sqrt{-g}}\frac{\delta(\sqrt{-g}\mathcal L)}{\delta g_{\alpha\beta}} = -g^{\mu\alpha}g^{\nu\beta}\cdot\frac{2}{\sqrt{-g}}\frac{\delta(\sqrt{-g}\mathcal L)}{\delta g^{\mu\nu}} = g^{\mu\alpha}g^{\nu\beta}T_{\mu\nu} = T^{\alpha\beta}$. If a book's $T$ has the opposite sign to yours, check this before anything else.

The factor of $c$. It is there only because this course keeps SI units and writes $S = \frac1c\int\mathcal L\,\mathrm d^4x$. With $c = 1$, as unit 35 and every textbook will use, the definition is the familiar $T^{\mu\nu} = \frac{2}{\sqrt{-g}}\frac{\delta S_{\text m}}{\delta g_{\mu\nu}}$. Nothing else changes.

"Unique symmetric tensor" needs the fundamental lemma again — this time for symmetric matrix-valued test functions, which is the same proof with a bump times a constant symmetric matrix. Without the symmetry restriction the tensor would be determined only up to an antisymmetric piece, which is the formal reason the asymmetry problem cannot arise.

Hilbert, not Einstein. This definition is Hilbert's, from the same November 1915 weeks in which Einstein arrived at the field equations by a different route; the tensor is sometimes called the Hilbert stress–energy tensor for that reason. Einstein's own route guessed $T^{\mu\nu}$ from the known matter tensors and imposed conservation; Hilbert's gets it from a variational principle and gets conservation as a theorem. Unit 35 does both.

What "matter" means here. Everything that is not the gravitational field: the electromagnetic field counts as matter for this purpose, which is why this section's example is Maxwell.

29 · metric · example

Worked: metric variation of Maxwell's action

example 5 · the running example, second route

Step 0 — where $g$ appears. $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ contains no metric. It enters only in raising indices and in $\sqrt{-g}$: $$\sqrt{-g}\,\mathcal L = -\frac{\sqrt{-g}}{4\mu_0}\,g^{\mu\alpha}g^{\nu\beta}F_{\mu\nu}F_{\alpha\beta} .$$

Step 1 — vary the two inverse metrics. They appear symmetrically, so the two variations are equal: $$\delta\mathcal L = -\frac{2}{4\mu_0}F_{\mu\nu}F_{\alpha\beta}\,g^{\nu\beta}\,\delta g^{\mu\alpha} = -\frac{1}{2\mu_0}F_{\mu\nu}F_\alpha{}^\nu\,\delta g^{\mu\alpha} .$$

Step 2 — vary $\sqrt{-g}$. $\delta\sqrt{-g} = -\tfrac12\sqrt{-g}\,g_{\mu\alpha}\delta g^{\mu\alpha}$ from Varying the volume element and the inverse metric.

Step 3 — collect. $$\delta(\sqrt{-g}\mathcal L) = -\tfrac12\sqrt{-g}\Big[\tfrac{1}{\mu_0}F_{\mu\nu}F_\alpha{}^\nu + g_{\mu\alpha}\mathcal L\Big]\delta g^{\mu\alpha} .$$

Step 4 — read off, against the definition. Setting $g = \eta$,

$$\boxed{\ T^{\mu\nu} = \frac{1}{\mu_0}\Big(F^{\mu\alpha}F^{\nu}{}_{\alpha} - \tfrac14\eta^{\mu\nu}F_{\rho\sigma}F^{\rho\sigma}\Big)\ }$$

— the same tensor the Belinfante improvement produced, by mathematics that never mentioned Noether, translations or superpotentials.

Step 1, in full. $\delta(g^{\mu\alpha}g^{\nu\beta}F_{\mu\nu}F_{\alpha\beta}) = \delta g^{\mu\alpha}\,g^{\nu\beta}F_{\mu\nu}F_{\alpha\beta} + g^{\mu\alpha}\,\delta g^{\nu\beta}F_{\mu\nu}F_{\alpha\beta}$. In the second term relabel $\mu\leftrightarrow\nu$, $\alpha\leftrightarrow\beta$; using $F_{\nu\mu}F_{\beta\alpha} = F_{\mu\nu}F_{\alpha\beta}$ (two sign flips) it becomes identical to the first. Hence the factor 2, which against $-1/4\mu_0$ leaves $-1/2\mu_0$.

Step 4, matching the definition. The definition says $\delta(\sqrt{-g}\mathcal L) = -\frac12\sqrt{-g}\,T_{\mu\alpha}\delta g^{\mu\alpha}$, so $T_{\mu\alpha} = \frac{1}{\mu_0}F_{\mu\nu}F_\alpha{}^\nu + g_{\mu\alpha}\mathcal L$, and $\mathcal L = -\frac{1}{4\mu_0}F^2$ gives the displayed form. Raising both indices with $\eta$ gives the boxed result.

Why $F_{\mu\nu}$ has no metric in it, and why that matters. The fundamental field is $A_\mu$ with the index down — a covector field, a 1-form. The curl $\partial_\mu A_\nu - \partial_\nu A_\mu$ is antisymmetric, and antisymmetry is what makes the Christoffel symbols cancel when $\partial$ is replaced by $\nabla$ in unit 32: the same expression is correct on a curved manifold with no change. Had we taken $A^\mu$ as fundamental, a metric would have appeared in $F$ and this calculation would have been twice as long and considerably more error-prone. Choosing the index position of the fundamental field is a real decision, and this is the reason for it.

The interaction term is deliberately omitted. $\sqrt{-g}A_\mu J^\mu$ requires deciding whether $J^\mu$ or $J_\mu$ or $J^\mu/\sqrt{-g}$ is the thing held fixed under the variation, and the answers differ. A prescribed external current is not a closed system, and a system that exchanges energy with an external agent has no energy–momentum tensor of its own. The vacuum calculation is the honest one; unit 28's Lorentz force density $F^{\mu\nu}J_\nu$ is exactly the rate at which the field's $T$ fails to be conserved when charges are present.

29 · metric

Checking it against everything unit 24 knew

energy density

$T^{00} = \frac{1}{\mu_0}\big(F^{0\alpha}F^0{}_\alpha + \tfrac14 F^2\big) = \tfrac{\varepsilon_0}{2}E^2 + \tfrac{1}{2\mu_0}B^2$ ✓

$B = 1.00$ T alone gives $3.979\times10^5\ \mathrm{J\,m^{-3}}$; $E = 1.00\ \mathrm{MV\,m^{-1}}$ alone gives $4.427\ \mathrm{J\,m^{-3}}$. Magnetic fields store energy far more densely, which is why the world's storage rings use magnets.

energy flux

$cT^{i0} = \frac{1}{\mu_0}(\vec E\times\vec B)^i = S^i$, the Poynting vector ✓; and $T^{0i}/c = \varepsilon_0(\vec E\times\vec B)^i$, the field momentum density ✓.

Symmetry of $T$ says these two are the same fact: $\vec g = \vec S/c^2$, which unit 24 asserted and could not prove.

stress

$T^{ij} = -\big[\varepsilon_0(E^iE^j - \tfrac12\delta^{ij}E^2) + \mu_0^{-1}(B^iB^j - \tfrac12\delta^{ij}B^2)\big]$: minus unit 24's Maxwell stress tensor ✓, since stress is a force and $T^{ij}$ is a momentum flux.

traceless — and only in four dimensions

$T^\mu{}_\mu = \frac{1}{\mu_0}\big(F^{\mu\alpha}F_{\mu\alpha} - \tfrac14\delta^\mu_\mu F^2\big) = \frac{1}{\mu_0}F^2\big(1 - \tfrac d4\big) = 0$ at $d = 4$.

Problem P9 asks what that says about the massless scalar, whose trace does not vanish.

a number: sunlight

A plane wave has $|\vec E| = c|\vec B|$, so $F^2 = 0$ and $T^{33} = T^{00} = u$. With the solar constant $S = 1361\ \mathrm{W\,m^{-2}}$: $u = S/c = 4.540\ \mathrm{\mu J\,m^{-3}}$, and the radiation pressure on a black surface is $T^{33} = 4.540\ \mathrm{\mu Pa}$. The field amplitude is $E_0 = \sqrt{2S/\varepsilon_0c} = 1013\ \mathrm{V\,m^{-1}}$.

$T^{00}$, computed. $F^{0\alpha}F^0{}_\alpha = F^{0i}F^0{}_i = F^{0i}F^{0i} = E^2/c^2$ (only spatial $\alpha$ contributes, and lowering a spatial index does nothing). With $F^2 = 2(B^2 - E^2/c^2)$: $T^{00} = \mu_0^{-1}[E^2/c^2 + \frac12 B^2 - \frac12 E^2/c^2] = \mu_0^{-1}[\frac12 E^2/c^2 + \frac12 B^2] = \frac{\varepsilon_0}{2}E^2 + \frac{1}{2\mu_0}B^2$, using $\varepsilon_0 = 1/\mu_0c^2$. Note the trace term supplied half of it: the $-\frac14\eta^{\mu\nu}F^2$ is not cosmetic.

$T^{0i}$, computed. $T^{0i} = \mu_0^{-1}F^{0\alpha}F^i{}_\alpha$ ($\eta^{0i} = 0$ kills the trace term) $= \mu_0^{-1}F^{0j}F^i{}_j = \mu_0^{-1}(E_j/c)\epsilon_{ijk}B_k = (\vec E\times\vec B)_i/\mu_0 c = S^i/c$. So $cT^{i0} = S^i$ and $T^{0i}/c = S^i/c^2 = \varepsilon_0(\vec E\times\vec B)^i$, and symmetry $T^{0i} = T^{i0}$ is precisely $\vec g = \vec S/c^2$.

The numbers. $\frac12\varepsilon_0(10^6)^2 = 4.4271\ \mathrm{J\,m^{-3}}$; $1/2\mu_0 = 3.9789\times10^5\ \mathrm{J\,m^{-3}}$ for $B = 1$ T — a ratio of $8.988\times10^4$, which is exactly $(cB/E)^2$, and says only that $1$ MV/m is a weak field while $1$ T is a strong one. Sunlight: $1361/2.99792458\times10^8 = 4.5398\times10^{-6}\ \mathrm{J\,m^{-3}}$; $E_0 = \sqrt{2\times1361/(8.8541878128\times10^{-12}\times2.99792458\times10^8)} = 1012.6\ \mathrm{V\,m^{-1}}$, with $B_0 = E_0/c = 3.378\ \mathrm{\mu T}$ — about a fifteenth of the Earth's field, for all of the Sun's output at this distance.

Why a plane wave has $F^2 = 0$. Unit 28's two invariants are $F_{\mu\nu}F^{\mu\nu} \propto B^2 - E^2/c^2$ and $\epsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma} \propto \vec E\cdot\vec B$. A plane wave has $|\vec E| = c|\vec B|$ and $\vec E\perp\vec B$, so both vanish — a null field, and no boost can remove either $\vec E$ or $\vec B$. Problem P8 works out the whole tensor in that case, and finds it is built from a single null vector.

29 · metric

The two routes, side by side

route 1 · Noether

Input: translation invariance of $\mathcal L$.
Method: the three-line theorem.
Output: $\eta^{\mu\nu}\mathcal L + \mu_0^{-1}F^{\mu\alpha}\partial^\nu A_\alpha$.
Status: conserved; not symmetric, not gauge invariant. Energy density of a uniform $1.00\ \mathrm{MV\,m^{-1}}$ field: $-4.427\ \mathrm{J\,m^{-3}}$.

route 2 · the metric

Input: how $S$ responds to a nudge of $g_{\mu\nu}$.
Method: Jacobi's formula and one index relabelling.
Output: $\mu_0^{-1}(F^{\mu\alpha}F^\nu{}_\alpha - \tfrac14\eta^{\mu\nu}F^2)$.
Status: symmetric, gauge invariant, traceless, conserved. Same field: $+4.427\ \mathrm{J\,m^{-3}}$.

What "the same place by different mathematics" is worth. Route 1 knows only that the laws are the same at every event. Route 2 knows only how the action responds to the geometry. These are different pieces of information, and that they determine the same tensor up to an identically conserved term is not obvious — it is a theorem, and Belinfante and Rosenfeld's general version proves it for every Lorentz-invariant theory, with $B$ built from the spin current. What the reader should take is not the formula but the pattern: when two honest derivations of a physical quantity disagree, the disagreement is usually a term that is invisible to every measurement the quantity was defined by, and finding out which term is the interesting part.

In general relativity the ambiguity is genuinely broken. Einstein's equation is local: $G^{\mu\nu} = \frac{8\pi G}{c^4}T^{\mu\nu}$ at each event. Two tensors differing by $\partial_\lambda B^{\lambda\mu\nu}$ source different metrics, so gravity measures the local value, not the integral. It is the metric-variation tensor that appears — by construction, since unit 35 varies the total action with respect to $g$ and the matter part is defined to be that derivative. The ambiguity of route 1 is not resolved by argument; it is resolved by asking a question route 1 cannot ask.

29 · metric

The difference, on the uniform field

and the difference is exactly the Belinfante term $$T^{\mu\nu}_{\text{can}} - T^{\mu\nu} = \partial_\lambda B^{\lambda\mu\nu},$$ $$B^{\lambda\mu\nu} = \frac{1}{\mu_0}F^{\mu\lambda}A^\nu = -B^{\mu\lambda\nu} .$$

On the uniform field: $B^{300} = \mu_0^{-1}F^{03}A^0 = -\varepsilon_0E^2z$, so $\partial_\lambda B^{\lambda00} = -\varepsilon_0E^2 = -8.854\ \mathrm{J\,m^{-3}}$, and $-4.427 = 4.427 - 8.854$ ✓. Being a divergence of something antisymmetric, it is separately conserved and carries no four-momentum: the two tensors disagree everywhere and agree on every total.

The energy densities of a uniform $1.00\ \mathrm{MV\,m^{-1}}$ field, to scale. Noether's number is the metric one plus the divergence of the Belinfante term: a large correction, the same at every point, and a pure divergence.

The arithmetic, checked. $\varphi = -Ez$ gives $\vec E = -\nabla\varphi = E\hat z$ ✓ and $A^0 = \varphi/c = -Ez/c$, while $F^{03} = E_3/c = E/c$. So $B^{300} = \mu_0^{-1}(E/c)(-Ez/c) = -E^2z/\mu_0c^2 = -\varepsilon_0E^2z$, and $\partial_3B^{300} = -\varepsilon_0E^2 = -8.8542\ \mathrm{J\,m^{-3}}$ at $E = 1.00\ \mathrm{MV\,m^{-1}}$. Meanwhile $T^{00}_{\text{can}} - T^{00} = -4.4271 - 4.4271 = -8.8542$ ✓. Two independent routes to the same number, to five figures.

How a constant can carry no energy. The bar for $\partial_\lambda B^{\lambda00}$ is the same at every point of the field, so integrated over a big region it looks like a large negative energy. It is not: $B^{300} = -\varepsilon_0E^2z$ grows linearly with $z$, and the volume integral of its divergence is the surface integral of $B^{300}$ over the region's top and bottom faces — the improvement theorem's surface term. That theorem assumed fields that decay, and a field uniform out to infinity does not, so this example tests the local identity and says nothing about totals; for the decaying fields of the theorem the surface term is zero and the totals agree. The bars are a statement about one event.

29 · metric

Why symmetry is angular momentum

theorem

For a conserved $T^{\mu\nu}$, define the angular-momentum current $$M^{\lambda\mu\nu} = x^\mu T^{\lambda\nu} - x^\nu T^{\lambda\mu} .$$ Then $\partial_\lambda M^{\lambda\mu\nu} = T^{\mu\nu} - T^{\nu\mu}$, so $M$ is conserved if and only if $T$ is symmetric.

proof

$\partial_\lambda(x^\mu T^{\lambda\nu}) = \delta^\mu_\lambda T^{\lambda\nu} + x^\mu\partial_\lambda T^{\lambda\nu} = T^{\mu\nu}$, the second term vanishing by conservation. Antisymmetrise in $\mu\nu$. $\square$

The six charges $J^{\mu\nu} = \frac1c\int M^{0\mu\nu}\mathrm d^3x$ are the three angular momenta and the three boost charges — the six parameters of the Lorentz group.

A small element of side $a$. The tangential fluxes on the faces normal to $x$ make one couple, those on the faces normal to $y$ the opposite one; the net torque is $(T^{xy} - T^{yx})\,a^2$, against a moment of inertia $\propto a^4$.

The theorem in a picture. For $\mu\nu = xy$, the slide's identity says the orbital angular momentum density about $z$ has a source $T^{xy} - T^{yx}$. The figure is where that source comes from: on an element of side $a$, the two faces normal to $x$ carry $y$-momentum at rates $\pm T^{xy}a$ (per unit depth), a couple of moment $T^{xy}a^2$; the faces normal to $y$ give $-T^{yx}a^2$. The element's moment of inertia scales as $a^4$, so if $T^{xy} \ne T^{yx}$ its angular acceleration grows like $1/a^2$ as the element shrinks — absurd for a medium with nothing else to hold angular momentum. Unit 22 drew the same conclusion from angular-momentum balance — stress is symmetric — and noted the exception, a ferrofluid, whose particles carry their own angular momentum and are twisted by a body couple. A field with indices is in the same position: it does have somewhere else to hold angular momentum. That is the next slide.

The boost charges. $J^{0i} = \frac1c\int(x^0T^{0i} - x^iT^{00})\mathrm d^3x$, whose conservation says $\frac{\mathrm d}{\mathrm dt}(\text{centre of energy}) = c^2P^i/E$: the energy centroid moves in a straight line at constant velocity. That is unit 13's Galilean boost charge $\vec G = \vec Pt - M\vec R$, relativistic version, and it is the sixth conservation law people forget they have.

29 · metric

Spin: what the asymmetry was carrying

so the defect had a meaning

The canonical tensor is asymmetric because the canonical angular-momentum current has a second piece, the spin current, coming from the field's own indices rather than from its position. Belinfante's $B$ moves that piece into $T$: "symmetrise $T$" and "let orbital angular momentum account for all of it" are one instruction.

and it is measurable

Circularly polarised light carries angular momentum $\hbar$ per photon with no orbital motion at all. Beth measured the torque on a suspended waveplate in 1936 and found it — which is why the repair is physics rather than tidiness.

Beth's experiment, schematic. A half-wave plate reverses the handedness of circularly polarised light, so each photon arrives with $+\hbar$ about the beam and leaves with $-\hbar$; the plate takes $2\hbar$, and for power $P$ at frequency $\omega$ that is a torque $2P/\omega$ on the fibre.

Where $M$ comes from. It is the Noether current of Lorentz invariance, exactly as $T^{\mu\nu}$ is the Noether current of translation invariance. For a scalar field the transformation is $\delta\varphi = -\omega^{\mu\nu}x_\nu\partial_\mu\varphi$ with $\omega$ antisymmetric, and running the theorem gives $j^\lambda = -\frac12\omega_{\mu\nu}M^{\lambda\mu\nu}$ with $M$ as above. For a field carrying spacetime indices — $A_\mu$, say — the transformation also rotates those indices, and that extra piece is the spin current $S^{\lambda\mu\nu}$, giving $M^{\lambda\mu\nu}_{\text{can}} = x^\mu T^{\lambda\nu}_{\text{can}} - x^\nu T^{\lambda\mu}_{\text{can}} + S^{\lambda\mu\nu}$.

Beth's experiment. R. A. Beth, Physical Review 50 (1936) 115: a half-wave plate suspended on a quartz fibre, circularly polarised light reversing its handedness through it, and a measured torque of order $10^{-9}$ dyn cm, that is $10^{-16}$ N m — within a few per cent of $2P/\omega$ for incident power $P$. The experiment is the reason no one can claim the spin part of the angular momentum is a bookkeeping artefact.

An honest caveat. Splitting light's angular momentum into "spin" and "orbital" parts is gauge dependent in general, and only the total is unambiguous. For paraxial beams the split is meaningful and is routinely used — optical tweezers exploit both parts — but the general statement of the theory is the one on this slide: the total is $M^{\lambda\mu\nu}$ built from a symmetric $T$.

29 · closing

The running example, closed

what one functional gave

$S[A]$ — one line of symbols — produced, with no further physical input:

Maxwell's source equations, by varying $A$ (§2);
the other two, free, as the Bianchi identity of $F = \partial A$;
$\partial_\mu J^\mu = 0$, forced by gauge invariance, off shell;
a conserved but defective $T$, from translations (§3);
the right one, from $\delta S/\delta g$ (§4), the two differing by an identically conserved term.

What comes next.

Unit 30 asks which feature of gravity flat spacetime cannot represent, and finds it in a metric that cannot be made constant.

Unit 32 replaces $\partial_\mu$ by $\nabla_\mu$; everything in this unit survives, because $F = \partial A$ was antisymmetric and $\sqrt{-g}\,\mathrm d^4x$ was already invariant.

Unit 35 adds the Einstein–Hilbert term to $S$, varies the total action with respect to $g^{\mu\nu}$, and reads off $G_{\mu\nu} = \frac{8\pi G}{c^4}T_{\mu\nu}$. The right-hand side is this unit's definition, verbatim, and the boundary term is the one flagged in §1.

The one idea to keep. The right-hand side of Einstein's equation is a derivative. Not a postulate, not an analogy with a fluid, not a table of cases: $T^{\mu\nu}$ is what you get by asking a matter action how much it changes when the geometry is nudged, and its two awkward properties — symmetry and gauge invariance — are free consequences of what it is a derivative with respect to.

Reading. Landau and Lifshitz, The Classical Theory of Fields, §§27–33: the field action, the canonical tensor, and the symmetrisation, in twenty pages and with this unit's conventions apart from the signature. Jackson, Classical Electrodynamics, §12.7–12.10, for the same material with more three-vector detail. For the metric definition, Wald, General Relativity, appendix E.1 — short, and readable now except that it uses $\nabla$ where this unit uses $\partial$, which for these formulas makes no difference. Carroll's Spacetime and Geometry, the Lagrangian-formulation section of chapter 4 (§4.3), does the metric variation of Maxwell exactly as §4 does; its §1.10, "Classical field theory", covers the canonical tensor. Avoid, for now, anything that starts from the functional integral.

A pointer for unit 39. The perfect fluid's $T^{\mu\nu} = (\rho + p/c^2)u^\mu u^\nu + p\,g^{\mu\nu}$, which unit 22 quoted and cosmology runs on, comes from this definition applied to a fluid action. It is a good exercise once unit 32 has supplied $\nabla$, and it is the point at which the definition stops feeling abstract: the pressure is literally the derivative of the action with respect to the spatial part of the metric.

29 · checkpoint

Checkpoint

you can now

Given a Lagrangian density, derive its field equations; construct the Noether current of each continuous symmetry and verify $\partial_\mu j^\mu = 0$ on shell; and produce the symmetric energy–momentum tensor by varying with respect to the metric. For Maxwell you can state precisely how the canonical and metric-variation tensors differ, and why the difference is separately conserved.

it is false if

you write $\partial_\mu T^{\mu\nu}_{\text{can}} = 0$ as evidence that $T_{\text{can}}$ is the energy–momentum tensor; or you cannot say what $B^{\lambda\mu\nu}$ is and why its antisymmetry matters; or your metric variation forgets $\delta\sqrt{-g}$ — which supplies half of $T^{00}$.

to check it

Close the deck. On paper: from $\mathcal L = -\frac{1}{4\mu_0}F^2$ get $\partial_\nu F^{\mu\nu} = 0$; then both tensors; then the uniform-field numbers $-4.427$ and $+4.427\ \mathrm{J\,m^{-3}}$ and the $-8.854$ between them. Then ./mr problems 29.

The problem sheet opens with this checkpoint. P2 is the Klein–Gordon field end to end and is the one to do first; P4 is the construction that makes $A_\mu$ inevitable rather than convenient; P8 and P9 are the two that most reward doing the index algebra slowly. When the two tasks on this slide can be done without looking anything up, ./mr done 29.

A last warning about other books. Three conventions vary independently: the signature, the index order in $\partial_\mu F^{\mu\nu} = \pm\mu_0J^\nu$, and whether $T$ is defined by $\delta/\delta g_{\mu\nu}$ or $\delta/\delta g^{\mu\nu}$. Any of the three flips a sign. Before trusting an imported formula, test it on the one case whose answer you know cold: a uniform electric field must have $T^{00} = +\frac{\varepsilon_0}{2}E^2$.

Classical field theory: Noether again, and where Tμν actually comes from

One action, four questions

Fields: mechanics with a continuous index

A field is a mechanical system whose index is a point

The Lagrangian density, and the action over a region

The first variation, and where the boundary term goes

The divergence theorem in four dimensions

The field Euler–Lagrange equations

What the boundary term is for

Where the boundary term comes back

Worked: the scalar wave field

Worked: the Klein–Gordon field, and what mass does to a wave

Phase and group velocity, on the dispersion curve

The range of a force is the mass of its field

The running example: Maxwell's action

Maxwell from an action

Why the potential and not the field

Worked: varying $S[A]$, in four lines

Gauge invariance forces charge conservation

Noether, for fields

What a symmetry of a field action is

Noether's theorem for fields

From a local law to a conserved charge

An internal symmetry: the phase of a complex field

The rotation happens in field space

Translations, and the canonical energy–momentum tensor

What its sixteen components mean

Momentum flux through a face

Worked: the canonical tensor for Maxwell, and its two defects

What you are allowed to add: improvement

Belinfante and Rosenfeld: the repair, for Maxwell

The definition general relativity uses

The move: let the metric be general, in flat space

The volume element coordinates cannot change

Differentiating a determinant

Varying the volume element and the inverse metric

The definition, and why its two properties are free

Worked: metric variation of Maxwell's action

Checking it against everything unit 24 knew

The two routes, side by side

The difference, on the uniform field

Why symmetry is angular momentum

Spin: what the asymmetry was carrying

The running example, closed

Checkpoint

Classical field theory: Noether again, and where T^μν actually comes from