Part I · The Action Principle — Chapter 1.4

Noether's Theorem

Every continuous symmetry of the action is a conservation law. It is the most important theorem in this book, and the proof is half a page.

Tools you'll need  — Chapter 0.6: the multivariable chain rule, and the equality of mixed partial derivatives (the proof turns on it twice). Chapter 0.7: the continuity equation $\partial_{t}\rho+\nabla\cdot\vv J=0$ and the divergence theorem, since §6 shows that the field version of this theorem produces exactly that equation. Chapter 1.2: the action $S=\int L\,\dd t$, the Euler–Lagrange equation, and above all its Problem 4, which established that adding a total time derivative $\dv{F}{t}$ to $L$ changes nothing physical. That freedom is not a footnote here. It is part of the hypothesis of the theorem, and §1 shows a symmetry that fails without it. Chapter 1.3: the Poisson bracket and the fact that a phase-space function generates an infinitesimal transformation. §7 needs it, and states what it needs.

1 · What a symmetry is, precisely

The word "symmetry" in ordinary speech means a shape looks the same after you flip it. The word in physics means something narrower and more useful, and getting the definition exactly right is most of the work. Once the hypothesis is stated properly, the proof in §2 runs to four lines. So this section is all definition, and everything afterwards is spending it.

1.1 · A continuous family of transformations

Start with a system described by coordinates $q_{1}(t),\dots,q_{n}(t)$. A continuous transformation is a recipe that takes any path and deforms it, smoothly, by an amount controlled by a single real parameter $\epsilon$:

The condition $q_{i}(t,0)=q_{i}(t)$ says that $\epsilon=0$ is the do-nothing transformation. That is what makes the family continuous in the sense we need: we can turn the transformation on gradually, and therefore we can differentiate with respect to how much of it we have turned on.

Only the derivative at $\epsilon=0$ will matter. Define

$K_{i}$ is called the generator of the transformation. It is a vector field on configuration space, telling you which way each coordinate moves when you nudge $\epsilon$. Here are three examples, all of which we will use:

Transformation	Acting on	Generator $K$
Translate by $\epsilon$ along $\vv n$	$\vv x \mapsto \vv x + \epsilon\,\vv n$	$\vv K = \vv n$
Rotate by $\epsilon$ about $\vv n$	$\vv x \mapsto \vv x + \epsilon\,\vv n\times\vv x + O(\epsilon^{2})$	$\vv K = \vv n\times\vv x$
Boost by $\epsilon\vv u$	$\vv x \mapsto \vv x + \epsilon\,\vv u\,t$	$\vv K = \vv u\,t$

The rotation entry is the only one that needs an argument, so let's give it one. A rotation by a small angle $\epsilon$ about the axis $\vv n$ moves the tip of $\vv x$ along a circle of radius $\abs{\vv x}\sin\theta$. That motion is perpendicular to both $\vv n$ and $\vv x$, and it covers an arc length $\epsilon\abs{\vv x}\sin\theta$. The vector with exactly that direction and that magnitude is $\epsilon\,\vv n\times\vv x$. So the infinitesimal version of a rotation is a cross product, which is why angular momentum will turn out to be one.

1.2 · Invariance of the action — with one crucial loosening

Now for the question the rest of the chapter rests on. What should it mean for the transformation to be a symmetry? The naive answer is "$L$ doesn't change". That answer is too strong. Adopt it and you will conclude, wrongly, that Newtonian mechanics is not invariant under a change of reference velocity.

The right condition comes from remembering what $L$ is for. Nothing physical depends on $L$ directly. The physics is in the stationary points of $S=\int L\,\dd t$. And Chapter 1.2's Problem 4 established that two Lagrangians differing by a total time derivative,

have identical Euler–Lagrange equations. The reason is that the extra term integrates to $F(q(t_{2}),t_{2})-F(q(t_{1}),t_{1})$, which is the same number for every path in the competition and therefore invisible to the variation. So a transformation that changes $L$ by a total derivative has not changed the physics at all, and refusing to call it a symmetry would be a mistake of bookkeeping.

1.3 · Why the total derivative is not a technicality: Galilean boosts

Here is the case that forces the loosening. Take $N$ particles interacting through a potential that depends only on their separations. That covers every non-relativistic system of mutually interacting bodies you have ever seen:

Now change to a reference frame moving at a small constant velocity $-\epsilon\vv u$. In the new frame every particle's position is shifted by $\epsilon\vv u\,t$:

Galilean relativity says this had better be a symmetry. That principle is the statement that you cannot detect uniform motion by any mechanical experiment inside a sealed cabin. So let's check whether the boost qualifies. The potential is untouched, since every separation $\vv x_{a}-\vv x_{b}$ is unchanged by shifting all positions equally. But the kinetic energy is not:

So $\delta L\neq0$: the Lagrangian is not invariant under a boost. If "symmetry" meant "$L$ unchanged" we would be forced to say that Newtonian mechanics violates Galilean relativity, which is absurd.

Look again at what $\delta L$ actually is. The masses are constants, so

where $M=\sum_{a}m_{a}$ and $\vv R_{\rm cm}=\frac{1}{M}\sum_{a}m_{a}\vv x_{a}$ is the centre of mass. So the change in $L$ is a total time derivative, which is exactly the case the definition was widened to admit. A boost is a symmetry, in the only sense that matters, and it is not a symmetry in the naive sense.

The moral is worth stating flatly, because it will save you confusion in three later chapters: if you ever find that a symmetry you believe in appears to fail, check whether $\delta L$ is a total derivative before concluding anything. Worked example 1 extracts the conserved quantity that this particular $F$ produces, and it is a conservation law that nobody thinks to name.

2 · The theorem, proved

Everything is now in place. The proof is short, and each step uses exactly one hypothesis, so it is worth watching which.

2.1 · Step 1 — the two derivatives commute

We will need to know how the velocities respond, and there is a small fact to establish first. The transformed velocity is $\dot q_{i}(t,\epsilon)=\partial_{t}q_{i}(t,\epsilon)$, so

The middle equality is the equality of mixed partial derivatives (Chapter 0.6). In words: varying and then differentiating in time gives the same thing as differentiating and then varying. That is only true because $\delta$ is taken at fixed $t$. It is precisely why §2.5, where the transformation moves $t$ as well, needs extra care rather than more of the same.

2.2 · Step 2 — expand $\delta L$ by the chain rule

We want to know how $L$ itself responds to the nudge. It is a function of $2n+1$ slots, so nudge the first $2n$ of them and apply the multivariable chain rule of Chapter 0.6:

using Step 1 in the second equality. Note that $t$ itself has not moved, so there is no $\partial L/\partial t$ term. Nothing has been assumed yet except smoothness. Equation (1.4.9) is true for every path.

2.3 · Step 3 — use the equations of motion

Now, and only now, restrict attention to a path that actually solves the Euler–Lagrange equation,

Substitute the left side of (1.4.10) in place of $\partial L/\partial q_{i}$ in (1.4.9). The bracket becomes something you have seen before:

2.4 · Step 4 — recognise a total derivative and compare

The second line of (1.4.11) is the product rule read backwards. That is the whole trick: the equation of motion is exactly what is needed to turn $\delta L$ into a total time derivative of something.

But the symmetry hypothesis says $\delta L$ is also a total time derivative, namely $\epsilon\dv{F}{t}$. Two expressions for the same quantity:

Both sides are the rate of change of something, so the difference of those two somethings has a rate of change of zero. That is what the right-hand half of (1.4.12) says. Give the constant quantity a name and the theorem is proved:

with $p_{i}=\partial L/\partial\dot q_{i}$ the canonical momentum of Chapter 1.2. $Q$ is the Noether charge. $\blacksquare$

2.5 · Transformations that also move time

One case is not yet covered, and it happens to be the most important one. Time translation is the statement that the laws are the same today as they were yesterday, and it moves $t$ itself. The derivation above assumed $t$ stood still. Most treatments wave at this step. It is worth doing properly, because the extra terms are exactly where the Hamiltonian comes from.

Allow the general case: a transformation that shifts the clock and the coordinates,

Read the second equation carefully. It compares the new path at the new time with the old path at the old time, which is what a transformation of spacetime does. But Step 1 above needed a variation at fixed $t$. The grind box reconciles the two and grinds out the general result. Here is the answer.

The object $H$ that appears attached to the time-shift $\tau$ is the combination Chapter 1.2 found by the Beltrami identity and Chapter 1.3 calls the Hamiltonian. It did not have to be put in by hand: it fell out of the bookkeeping of moving the clock. Setting $\tau=0$ recovers (1.4.13), as it must.

Grind box — the general variation, when the clock moves too

The two kinds of variation. Define the total variation, comparing new-at-new with old-at-old, and the form variation, comparing new-at-$t$ with old-at-$t$:

$$\Delta q_{i} \equiv q'_{i}(t') - q_{i}(t) = \epsilon K_{i}, \qquad \delta q_{i} \equiv q'_{i}(t) - q_{i}(t).$$

Only $\delta$ commutes with $\dd/\dd t$, so only $\delta$ can be fed into §2.1–§2.4. Relate them by Taylor expansion (Chapter 0.3), keeping first order in $\epsilon$:

$$q'_{i}(t') = q'_{i}(t+\epsilon\tau) = q'_{i}(t) + \epsilon\tau\,\dot q'_{i}(t) + O(\epsilon^{2}) = q'_{i}(t) + \epsilon\tau\,\dot q_{i}(t) + O(\epsilon^{2}),$$

where in the last step $\dot q'=\dot q+O(\epsilon)$ was used, legitimate because it is already multiplied by $\epsilon$. Rearranging,

$$\delta q_{i} \;=\; \epsilon\big(K_{i} - \dot q_{i}\,\tau\big) \;\equiv\; \epsilon\,\bar K_{i}.$$

The generator to use in the fixed-$t$ derivation is not $K$ but $\bar K = K-\dot q\,\tau$. That single substitution is the entire content of this grind box. The rest is showing what happens to the measure $\dd t$.

The action, with limits and measure transformed. The transformed action is an integral over the transformed interval:

$$S' \;=\; \int_{t'_{1}}^{t'_{2}} L\Big(q'(s),\;\tfrac{\dd q'}{\dd s}(s),\;s\Big)\,\dd s.$$

Substitute $s=t+\epsilon\tau(t)$, so that $\dd s=(1+\epsilon\dot\tau)\,\dd t$ and the limits become $t_{1},t_{2}$ again. Now expand the integrand about $s=t$ exactly as above, using $G(t+\epsilon\tau)=G(t)+\epsilon\tau\dot G(t)+O(\epsilon^{2})$ for any function $G$. That gives

$$\begin{aligned} S' &= \int_{t_{1}}^{t_{2}}\Big[L\big(q'(t),\dot q'(t),t\big) + \epsilon\tau\,\dv{L}{t}\Big]\big(1+\epsilon\dot\tau\big)\,\dd t + O(\epsilon^{2})\\[4pt] &= S + \int_{t_{1}}^{t_{2}}\Big[\delta L + \epsilon\,\tau\,\dv{L}{t} + \epsilon\,\dot\tau\,L\Big]\dd t + O(\epsilon^{2})\\[4pt] &= S + \int_{t_{1}}^{t_{2}}\Big[\delta L + \epsilon\,\dv{}{t}\big(\tau L\big)\Big]\dd t + O(\epsilon^{2}), \end{aligned}$$

where $\delta L$ is the fixed-$t$ variation driven by $\bar K$, and the last line collected two terms into one by the product rule. So the natural statement of the symmetry hypothesis, that the action changes at most by a boundary term, reads

$$\delta L + \epsilon\,\dv{}{t}\big(\tau L\big) \;=\; \epsilon\,\dv{F}{t}.$$

Finish. Steps 2 and 3 of §2 applied verbatim with generator $\bar K$ give, on shell, $\delta L = \epsilon\,\dv{}{t}\big(\sum_{i}p_{i}\bar K_{i}\big)$. Substituting,

$$\dv{}{t}\left(\sum_{i}p_{i}\bar K_{i} + \tau L - F\right) = 0.$$

Finally unpack $\bar K$ and recognise the Hamiltonian:

$$\sum_{i}p_{i}\bar K_{i} + \tau L = \sum_{i}p_{i}K_{i} - \tau\left(\sum_{i}p_{i}\dot q_{i} - L\right) = \sum_{i}p_{i}K_{i} - \tau H,$$

which is (1.4.15). $\blacksquare$

Why the sign of $H$ looks backwards. In (1.4.15) the Hamiltonian enters with a minus sign, so pure time translation ($\tau=1$, $K=0$, $F=0$) gives the conserved charge $Q=-H$. Since a constant multiple of a conserved quantity is conserved, that is the same statement as "$H$ is conserved". But the minus is not an accident, and §7 shows it is exactly the sign needed for $H$ to generate forward time evolution.

3 · The three classics, each from one formula

Everything in this section is (1.4.15) with different entries. Nothing new gets proved here. The point is that three conservation laws, each with its own unrelated textbook derivation, are three substitutions into one equation.

3.1 · Time translation ⇒ energy

Take $\tau=1$ (shift every clock reading by the same amount) and $K_{i}=0$ (do not touch the coordinates). By the grind box the form variation is $\bar K_{i}=-\dot q_{i}$, so

We need one more ingredient before we can test the symmetry condition, and it is the rate at which $L$ changes along the path all by itself. By the chain rule that is

Add them as the symmetry condition of the grind box requires, $\delta L + \epsilon\dv{}{t}(\tau L)=\delta L+\epsilon\dv{L}{t}$, and the bracketed sums cancel exactly:

So time translation is a symmetry, with $F=0$, if and only if $L$ has no explicit $t$-dependence. Feeding $\tau=1$, $K=0$, $F=0$ into (1.4.15):

That is the Beltrami identity of Chapter 1.2, Problem 2, which was obtained there by differentiating a lucky combination and watching terms cancel. It is also the Hamiltonian of Chapter 1.3, which was obtained there by a Legendre transform. Neither derivation said why that particular combination of $p$'s and $\dot q$'s should be the conserved one. This one does. It is the charge conjugate to moving the clock, and $H$ appears multiplying $\tau$ in (1.4.15) for the same reason that $p$ appears multiplying $K$.

For $L=\sum_{a}\half m_{a}\abs{\dot{\vv x}_{a}}^{2}-V(\{\vv x\})$ the momentum is $p=m\dot x$ and $E=\sum_{a}m_{a}\abs{\dot{\vv x}_{a}}^{2}-\big(T-V\big)=T+V$, the familiar total energy. But note carefully what the theorem conserves. It conserves $\sum p\dot q-L$. Whether that equals $T+V$ is a separate question about the coordinates, and Chapter 1.2's Problem 3 (the bead on the motor-driven hoop) supplied a system where $E$ is conserved and $T+V$ is not.

3.2 · Space translation ⇒ momentum

Translate every particle by the same displacement, $\vv x_{a}\mapsto\vv x_{a}+\epsilon\vv n$, so $\vv K_{a}=\vv n$ for all $a$ and $\tau=0$. Since the velocities are untouched, only $V$ can respond:

This vanishes for every $\vv n$ precisely when $\sum_{a}\partial V/\partial\vv x_{a}=0$, which is the statement that $V$ is unchanged by moving everything together. In other words, $V$ depends only on differences $\vv x_{a}-\vv x_{b}$. Then $F=0$, and (1.4.15) gives

Since $\vv n$ is arbitrary, all three components of $\vv P$ are conserved. Compare this with the Newtonian derivation, which invokes the third law and cancels internal forces in pairs. The third law is nowhere in sight here. What replaced it is the observation that a two-body potential can only depend on the separation. That is the same physical input, now recognisable as a symmetry statement rather than an axiom about forces.

The one-coordinate version is worth naming, because it is how you will use the theorem in practice. Suppose some coordinate $q_{j}$ is entirely absent from $L$, making it a cyclic or ignorable coordinate. Then $\delta q_{j}=\epsilon$ is a symmetry with $F=0$, and $Q=p_{j}$. That is the remark Chapter 1.2 made in passing after deriving $\dot p_{i}=\partial L/\partial q_{i}$. It is now a special case of a theorem.

3.3 · Rotation ⇒ angular momentum

Rotate everything by $\epsilon$ about the fixed axis $\vv n$, so that from the table in §1.1

Before reading off a charge we have to check that this really is a symmetry, so take the two pieces of $L$ in turn. The kinetic term responds as

because a cross product is perpendicular to both its factors. The potential responds through the separations, and

for the same reason. So if $V$ depends only on the lengths of the separations, which is the case of central forces, then $\delta L=0$ exactly, with $F=0$. Rotational symmetry is strict here, not merely up to a total derivative. Now read off the charge, using the cyclic property of the scalar triple product $\vv a\cdot(\vv b\times\vv c)=\vv b\cdot(\vv c\times\vv a)$:

Arbitrary $\vv n$ again, so the whole vector $\vv L=\sum_{a}\vv x_{a}\times\vv p_{a}$ is conserved. And now the definition of angular momentum stops being an arbitrary-looking cross product that someone decided to write down: $\vv x\times\vv p$ is what you get when you contract the momentum with the generator of rotations. The cross product is there because the infinitesimal rotation was.

Grind box — rotations in index notation, and where $\varepsilon_{ijk}$ comes from

The vector notation above hides the structure that generalises. Write the rotation generator in components using the Levi-Civita symbol $\varepsilon_{ijk}$ of Chapter 0.7 (antisymmetric in every pair, $\varepsilon_{123}=+1$):

$$\delta x_{i} = \epsilon\,\varepsilon_{ijk}\,n_{j}x_{k}, \qquad \delta \dot x_{i} = \epsilon\,\varepsilon_{ijk}\,n_{j}\dot x_{k},$$

with repeated indices summed. The kinetic term:

$$\delta\!\left(\half m\dot x_{i}\dot x_{i}\right) = \epsilon\,m\,\dot x_{i}\varepsilon_{ijk}n_{j}\dot x_{k} = 0,$$

because $\varepsilon_{ijk}$ is antisymmetric under $i\leftrightarrow k$ while $\dot x_{i}\dot x_{k}$ is symmetric, and the full contraction of a symmetric with an antisymmetric pair vanishes identically. That two-line argument is worth internalising. It is the reason rotations preserve lengths, and in Chapter 2.3 the same manoeuvre with a different symbol shows that Lorentz transformations preserve the spacetime interval.

The charge:

$$Q = p_{i}\,\varepsilon_{ijk}n_{j}x_{k} = n_{j}\,\varepsilon_{jki}\,x_{k}p_{i} = n_{j}L_{j}, \qquad L_{j} \equiv \varepsilon_{jki}x_{k}p_{i},$$

where the middle step used $\varepsilon_{ijk}=\varepsilon_{jki}$ (a cyclic permutation of three indices is even). Choosing $\vv n=\hat{\vv z}$ picks out

$$L_{3} = \varepsilon_{312}x_{1}p_{2} + \varepsilon_{321}x_{2}p_{1} = xp_{y} - yp_{x},$$

the familiar $L_{z}$.

The three rotations do not commute, and that matters. Rotating about $x$ then $y$ is not the same as $y$ then $x$. The discrepancy is, at second order, a rotation about $z$. Since the generators are what survive at first order, that failure to commute must show up as an algebraic relation among $L_{x},L_{y},L_{z}$, and §7 computes it to be $\{L_{i},L_{j}\}=\varepsilon_{ijk}L_{k}$. Translations, by contrast, commute with each other in any order, and correspondingly $\{P_{i},P_{j}\}=0$. The conserved charges inherit the group structure of the symmetries that produced them, which is the subject of Chapter 6.1 and the reason quantum angular momentum is quantised in Chapter 4.11.

3.4 · The table, and what it means

Symmetry of the action	Generator	Conserved charge
Time translation, $t\to t+\epsilon$	$\tau=1$	Energy $H=\sum p_{i}\dot q_{i}-L$
Space translation along $\vv n$	$\vv K=\vv n$	Momentum $\vv n\cdot\vv P$
Rotation about $\vv n$	$\vv K=\vv n\times\vv x$	Angular momentum $\vv n\cdot\vv L$
Galilean boost by $\vv u$	$\vv K=\vv u\,t$, $F=M\vv u\cdot\vv R_{\rm cm}$	$\vv P t-M\vv R_{\rm cm}$ (Worked Ex. 1)
Phase rotation $\phi\to\ee^{\ii\alpha}\phi$	$K=\ii\phi$	Electric charge (§6)

Read the first three rows as physics rather than as bookkeeping. The hypothesis in row one is that the laws are the same at every moment. The hypothesis in row two is that they are the same at every place. The hypothesis in row three is that they are the same in every direction. So:

4 · When symmetry fails, conservation fails

A theorem earns trust by being falsifiable. If symmetry really is the reason for conservation, then removing the symmetry must remove the conservation law. It must also remove that one only, leaving the others intact. This section takes the theorem at its word.

4.1 · The rate at which conservation fails

Start with the quantitative version, which is stronger than "the theorem no longer applies". Differentiate $E=\sum_{i}p_{i}\dot q_{i}-L$ along a solution, with $L$ now allowed an explicit $t$-dependence:

Look at the two braced brackets before reading on. The first one vanishes because the path solves the Euler–Lagrange equation, and the second vanishes because that is how $p_{i}$ was defined in the first place. Only the last term survives:

Read that as an accounting identity. The left side is the rate at which the conservation law is violated. The right side is the rate at which the symmetry is violated. They are equal. Noether's theorem is the special case in which both sides are zero.

Example. A mass on a spring being driven by an external agent, $L=\half m\dot x^{2}-\half kx^{2}+x\,f(t)$, has $\partial L/\partial t = x\dot f$, so $\dv{E}{t}=-x\dot f(t)$. The energy of the oscillator is not conserved, and it should not be, because something outside the system is doing work on it. The Lagrangian's explicit $t$-dependence is the mathematical signature of "there is an outside". Enlarge the system to include whatever is driving $f$, and the enlarged $L$ has no explicit $t$, and total energy is conserved again. Every apparent violation of energy conservation in ordinary physics is of that kind. It is an incomplete system.

4.2 · A crystal: continuous symmetry broken to discrete

Now a case where the symmetry is not destroyed but reduced. An electron moving through a crystal lattice sees a potential that repeats:

but $V(\vv x+\vv\delta)\neq V(\vv x)$ for a general small displacement $\vv\delta$. Translation invariance survives only for a discrete set of displacements.

Noether's theorem has nothing to say here. Its hypothesis was a family $q(t,\epsilon)$ continuous in $\epsilon$, so that $\partial/\partial\epsilon$ exists at $\epsilon=0$. A discrete symmetry has no $\epsilon$ to differentiate with respect to, so there is no generator $K$ to put into (1.4.15). And indeed momentum is not conserved. The lattice exerts forces on the electron, and $\vv p$ changes.

⚑ What survives is quoted rather than derived here, and it belongs to solid-state physics. Because the symmetry group is still nontrivial, the eigenstates can be labelled by a quantity $\hbar\vv k$ called the crystal momentum, and it is conserved modulo a reciprocal lattice vector $\vv G$. Collisions can change $\vv k$ by any $\vv G$ at no cost.

The pattern is general and worth carrying. A continuous symmetry gives an additively conserved quantity taking values in $\R$. Break it down to a discrete subgroup and you are left with a quantity conserved only up to the discrete "dual" of that subgroup. The residual freedom is exactly the amount of symmetry you threw away.

4.3 · An expanding universe: no conserved total energy

The striking case, and the one where the honest answer is more interesting than the tidy one.

⚑ Here is one fact quoted forward from Chapter 3.9. On the largest scales the universe is well described by a spacetime whose spatial distances all grow by a common factor $a(t)$, the scale factor. Matter fields living in that spacetime have an action whose Lagrangian contains $a(t)$ explicitly. So $\partial L/\partial t\neq0$, and not because of some external driving agent that could be absorbed into a larger system. It is because the geometry itself is time-dependent.

By (1.4.27) there is then no conserved energy, and this is not a technicality that a cleverer definition repairs. The most visible consequence is one you have heard described in a misleading way. Light from a distant galaxy arrives redshifted. Its wavelength has been stretched by the factor $a$, so each photon's energy has fallen by $1/a$. The energy is not transferred to the expansion, and it is not stored anywhere. There is no conserved global energy in this spacetime for it to be a part of, so there is nothing that needs to balance. Asking where it went presupposes the very conservation law that the absence of time-translation symmetry has removed.

4.4 · Breaking one symmetry at a time

The figure below is the theorem's own falsification test. A particle orbits in a central potential, and two dials sit next to it, each of which breaks exactly one symmetry.

The system is a particle of mass $m$ in the plane, with

At $\lambda=\mu=0$ this is the isotropic harmonic oscillator. It is rotationally symmetric, so $L_{z}$ is conserved, and it is time-independent, so $E$ is conserved. Each dial spoils one of those two properties and leaves the other alone.

• The dial $\lambda$ squashes the potential into an ellipse. Rotational symmetry gone, time-translation symmetry untouched.
• The dial $\mu$ makes the whole well breathe in time. Time-translation symmetry gone, rotational symmetry untouched.

Before looking, extract a prediction. The angular equation of motion is $\dv{}{t}\big(mr^{2}\dot\theta\big)=-\partial V/\partial\theta$, and $mr^{2}\dot\theta = L_{z}$, so

That is the rate at which the rotational conservation law fails. Now do the same for the energy, where (1.4.27) has already done the work for us and gives

Let's look at what those two lines are saying. Each rate is exactly proportional to its own dial and completely independent of the other. The theorem does not merely predict that conservation fails. It predicts which one fails, and how fast.

λ  breaks rotation

λ = 0.000

μ  breaks time-shift

μ = 0.000

both to zero

max |ΔLz| = 2.3e-13

max |ΔE| = 2.9e-13

One detail in the figure repays attention. At $\lambda=0$ the orbit is a closed ellipse, and the particle retraces its own path exactly. Turn $\lambda$ up and the ellipse begins to precess, so the path never closes. Closure of orbits is not a generic property. It is a symptom of extra conserved quantities, and Worked example 2 is about the most famous case of that.

5 · Discrete symmetries — an honest exception

Nature has symmetries that are not continuous, and the theorem simply does not reach them. The three that matter are parity $P$ (reflect all spatial coordinates, $\vv x\to-\vv x$), charge conjugation $C$ (exchange every particle for its antiparticle), and time reversal $T$ ($t\to-t$). None of them can be applied a little bit.

Parity is the cleanest case to see why. The matrix $-I$ has determinant $-1$ in three dimensions, and the identity has determinant $+1$. The determinant of a rotation-or-reflection is a continuous function taking only those two values, so no continuous path of transformations connects $P$ to the identity without leaving the group. There is therefore no one-parameter family to differentiate. No $\epsilon$, no generator $K$, no (1.4.15), and no conserved current.

What a discrete symmetry gives instead is a multiplicative quantum number, typically $\pm1$, whose product over the particles present is the same before and after a reaction. It gives that only in quantum mechanics, where states can be eigenvectors of the operation. Additive charges come from continuous symmetries and multiplicative ones from discrete symmetries. They are different objects with different arithmetic.

⚑ Here is a fact quoted now and developed in Chapter 6.6. Both $P$ and $CP$ are violated in nature. Parity was found to fail in the weak interaction in 1957, and the combination $CP$ was found to fail in 1964. This was a genuine surprise, and it is instructive about how much of physics is convention and how much is fact. Nobody had ever suspected that the laws could distinguish left from right, because there is no reason within mechanics or electromagnetism why they should.

Notice what does not happen when they fail. There is no Noether charge to lose, and correspondingly no continuity equation quietly breaks. The symmetry just is not there. The one combination that appears to survive is $CPT$, and ⚑ that one is not an accident. It is a theorem, forced by Lorentz invariance and locality, which is why it has a different status from the others.

6 · Fields: symmetry gives a conserved current

Chapter 1.2 §8 showed that the Euler–Lagrange derivation goes through verbatim for fields, at the cost of one substitution. Replace $L(q,\dot q,t)$ by a Lagrangian density $\mathcal{L}(\phi,\partial_{\mu}\phi)$, replace $\int\dd t$ by $\int\dd^{4}x$, and replace $\dv{}{t}$ by $\partial_{\mu}$. The same substitution turns §2 into the field version of Noether's theorem, and the result is more useful than the mechanics version.

Take an internal symmetry, meaning one that transforms the field at each point without moving the point:

The middle equation is Step 1 again, the statement that variations commute with derivatives. The right-hand one is the symmetry hypothesis, with the total time derivative promoted to a four-divergence, since that is now what integrates to a boundary term. Steps 2 and 3 are unchanged. Expand by the chain rule, then use the field Euler–Lagrange equation $\partial_{\mu}\big(\partial\mathcal{L}/\partial(\partial_{\mu}\phi)\big)=\partial\mathcal{L}/\partial\phi$:

Now compare that with the symmetry hypothesis. Once again we have two expressions for the same quantity, so the four-divergence of their difference vanishes, and that difference is the object we were after:

6.1 · This is the continuity equation, and that is a promotion

Split $j^{\mu}=(j^{0},\vv j)$ into its time and space parts. Then $\partial_{\mu}j^{\mu}=\partial_{t}j^{0}+\nabla\cdot\vv j$, and (1.4.34) is

which is precisely the continuity equation of Chapter 0.7, with $j^{0}$ the density of the conserved stuff and $\vv j$ its current. Chapter 0.7 promised that Noether's theorem would deliver one of these, and here it is. Integrating over a region $V$ and applying the divergence theorem, exactly as in that chapter,

Now notice how much stronger this is than the mechanics version. In §2 we proved that a number $Q(t)$ does not change. Here we have proved something local. The charge in any region changes only by flux through that region's boundary. Charge cannot vanish in London and reappear in Sydney with the books still balancing, because to get from one to the other it must cross every surface in between. Field conservation laws forbid teleportation, and mechanical ones do not. That is not a stylistic upgrade. It is the difference between a conservation law compatible with special relativity and one that is not, since a global bookkeeping rule requires a notion of "at the same instant everywhere" that Chapter 2.2 will destroy.

Grind box — the current is not unique, and the ambiguity is useful

Suppose $j^{\mu}$ is conserved and let $\Sigma^{\nu\mu}$ be any field antisymmetric in its two indices, $\Sigma^{\nu\mu}=-\Sigma^{\mu\nu}$. Define

$$\tilde j^{\mu} \;=\; j^{\mu} + \partial_{\nu}\Sigma^{\nu\mu}.$$

Then $\partial_{\mu}\tilde j^{\mu} = \partial_{\mu}j^{\mu} + \partial_{\mu}\partial_{\nu}\Sigma^{\nu\mu} = 0$, because $\partial_{\mu}\partial_{\nu}$ is symmetric in $\mu\nu$ (equality of mixed partials, Chapter 0.6) while $\Sigma^{\nu\mu}$ is antisymmetric, and a symmetric object fully contracted with an antisymmetric one vanishes. That is the identical argument used in §3.3's grind box. The new current is conserved too.

Does it carry the same charge? The added piece contributes $\int\partial_{\nu}\Sigma^{\nu0}\dd^{3}x$. The $\nu=0$ term vanishes by antisymmetry ($\Sigma^{00}=0$), leaving $\int\partial_{i}\Sigma^{i0}\dd^{3}x$, a total spatial divergence, which by the divergence theorem is a surface integral at infinity. If the fields fall off, it is zero and $\tilde Q = Q$. So the current is ambiguous but the charge is not.

This is not idle. The Noether procedure applied to spacetime translations produces an energy–momentum tensor $T^{\mu\nu}$ that is in general not symmetric in its indices, which is a disaster for general relativity, where $T^{\mu\nu}$ sits on the right-hand side of a manifestly symmetric field equation (Chapter 3.6). The repair is exactly this freedom. One adds a suitable $\partial_{\nu}\Sigma^{\nu\mu\rho}$, called an "improvement term", to symmetrise $T^{\mu\nu}$ without changing a single conserved charge. Chapter 5.2 does it.

6.2 · The example that sets up Part VI: a global phase

Take a complex scalar field $\phi$, with $\phi$ and its conjugate $\phi^{*}$ treated as independent. That is legitimate for the same reason $x$ and $\dot x$ were in Chapter 1.2, namely that they are independent slots of the function, and any complex $\phi$ can be traded for the real pair $(\operatorname{Re}\phi,\operatorname{Im}\phi)$ and back:

Now rotate the phase of the field by the same angle everywhere:

Every term in (1.4.37) pairs a $\phi$ with a $\phi^{*}$, so the phases cancel exactly. That makes $\mathcal{L}$ strictly invariant, with $F^{\mu}=0$. This is a global $U(1)$ symmetry. It is $U(1)$ because the transformations are the unit complex numbers $\ee^{\ii\alpha}$, which form a circle group under multiplication, and global because $\alpha$ is the same at every point of spacetime.

Infinitesimally, $\epsilon=\alpha$ and $K_{\phi}=\ii\phi$, $K_{\phi^{*}}=-\ii\phi^{*}$. The two derivatives needed are $\partial\mathcal{L}/\partial(\partial_{\mu}\phi)=\partial^{\mu}\phi^{*}$ and $\partial\mathcal{L}/\partial(\partial_{\mu}\phi^{*})=\partial^{\mu}\phi$, so (1.4.34) summed over both fields gives

Two checks are worth making. First, the current is real, because complex-conjugating swaps the two terms and the $\ii$ out front, giving back the same expression. Second, it is conserved directly, without invoking the theorem at all. Take the divergence, note that the terms with two derivatives on different factors cancel in pairs, and use the field equation $\Box\phi=-m^{2}\phi$ that follows from (1.4.37):

Multiplying a conserved current by any constant leaves it conserved, so the normalisation and overall sign are conventions. Fix them by demanding that a positive-energy mode carry positive charge, and the physical electromagnetic current of a field of charge $q$ is $-q\,j^{\mu}$. The conserved quantity $Q=\int j^{0}\dd^{3}x$ is, up to that constant, electric charge.

7 · The other direction: charges generate their own symmetries

So far the arrow has run one way, from symmetry to conserved charge. This section runs it backwards, and the result is the real destination of Part I.

Chapter 1.3 supplies the tool. On phase space, coordinates $q_{i}$ and momenta $p_{i}$ are independent, and any two functions of them have a Poisson bracket

with the two facts we need: time evolution is $\dv{f}{t}=\{f,H\}+\partial f/\partial t$, and a function $G$ generates an infinitesimal transformation of phase space through

7.1 · The charge is the generator — in one line

Take a Noether charge of the standard form (1.4.15) with $\tau=0$, namely $Q=\sum_{i}p_{i}K_{i}(q,t)-F(q,t)$, and ask what transformation it generates. Only the first term contains any $p$, and it is linear in them. That is exactly where the restriction to $K_{i}(q,t)$ is spent. §1.1 allowed the wider form $K_{i}(q,\dot q,t)$, and a velocity-dependent generator makes $Q$ quadratic in the momenta, so the bracket below would return something else. For generators of the narrower kind,

That is not an analogy or a coincidence. It is an identity, and it says that the Noether charge of a symmetry, used as a generator, reproduces exactly the symmetry it came from. The theorem and its converse are the same equation read in opposite directions. Here are three instances, worth doing explicitly because the brackets are how you will compute in Part IV.

Momentum generates translations. With $G=p_{x}$, equation (1.4.41) gives $\{x,p_{x}\}=1$ and $\{y,p_{x}\}=\{p_{i},p_{x}\}=0$, so

A rigid translation along $x$, exactly as advertised.

Angular momentum generates rotations. With $G=L_{z}=xp_{y}-yp_{x}$, the derivatives are $\partial L_{z}/\partial p_{x}=-y$, $\partial L_{z}/\partial p_{y}=x$, $\partial L_{z}/\partial x=p_{y}$, $\partial L_{z}/\partial y=-p_{x}$, so

The position rotates by $\epsilon$ about the $z$-axis, and so does the momentum. That second part is a bonus the Lagrangian derivation never showed us, and it is exactly right, since momentum is a vector and vectors rotate.

The Hamiltonian generates time evolution. With $G=H$, equation (1.4.42) reads $\delta f=\epsilon\{f,H\}=\epsilon\,\dv{f}{t}$ for any $f$ without explicit time dependence, which is the statement that the transformation generated by $H$ is advance everything by $\epsilon$. Time evolution is not a different kind of process from a symmetry transformation. It is the symmetry transformation generated by the energy.

This is also where the minus sign in (1.4.15) pays. The Noether charge of time translation is $-H$, which generates $\delta f=-\epsilon\dv{f}{t}$. That is the shift of the graph of $q(t)$ forward by $\epsilon$, which at fixed $t$ shows you the value the old path had at $t-\epsilon$. The two signs describe the same motion from opposite sides.

7.2 · The converse, and why it is not decorative

Now the reverse implication. Suppose $Q$ is conserved and has no explicit time dependence. Then $0=\dv{Q}{t}=\{Q,H\}$, and therefore the transformation $Q$ generates does this to the Hamiltonian:

The Hamiltonian is invariant under the transformation generated by $Q$. So $Q$ is not merely produced by a symmetry. Viewed as a generator, $Q$ is a symmetry. There is nothing left to prove in either direction.

7.3 · The algebra of the rotation charges

Grind box — computing $\{L_{i},L_{j}\}$ by hand

Take $L_{x}=yp_{z}-zp_{y}$ and $L_{y}=zp_{x}-xp_{z}$ and grind (1.4.41) over the three coordinate pairs. Only terms in which the same variable is differentiated in both factors survive. The nonzero partials are

$$\begin{aligned} &\pdv{L_{x}}{y}=p_{z}, && \pdv{L_{x}}{z}=-p_{y}, && \pdv{L_{x}}{p_{y}}=-z, && \pdv{L_{x}}{p_{z}}=y,\\[3pt] &\pdv{L_{y}}{z}=p_{x}, && \pdv{L_{y}}{x}=-p_{z}, && \pdv{L_{y}}{p_{z}}=-x, && \pdv{L_{y}}{p_{x}}=z. \end{aligned}$$

Assemble the bracket term by term. From the $x$-pair: $\partial_{x}L_{x}\,\partial_{p_{x}}L_{y}-\partial_{p_{x}}L_{x}\,\partial_{x}L_{y} = 0\cdot z - 0\cdot(-p_{z}) = 0$. From the $y$-pair: $p_{z}\cdot 0 - (-z)(0) = 0$. From the $z$-pair: $(-p_{y})(-x) - (y)(p_{x}) = xp_{y}-yp_{x}$. Hence

$$\{L_{x},L_{y}\} \;=\; xp_{y}-yp_{x} \;=\; L_{z},$$

and by relabelling $x\to y\to z\to x$, which permutes the three definitions cyclically,

$$\{L_{i},L_{j}\} \;=\; \varepsilon_{ijk}L_{k}.$$

(Verified symbolically for all nine pairs.) Two remarks. First, the right-hand side is not zero, and that non-vanishing is the algebraic residue of the geometric fact that rotations about different axes do not commute. The bracket of generators measures the failure of the transformations to commute. Second, the same computation for translations gives $\{P_{i},P_{j}\}=0$, because translations do commute. The bracket is a faithful record of the group.

Here is one consequence you can already read off, and Chapter 4.12 turns it into the whole theory of spin. Since $\{L_{x},L_{y}\}\neq0$, the three components of angular momentum are not simultaneously "diagonalisable" in the sense that will matter after quantisation. You will be able to know $\abs{\vv L}$ and one component, and no more.

8 · Worked examples

9 · Your turn