Part B · the action · spine

Noether: every conservation law you have met is one theorem

A one-parameter family of transformations that leaves the Lagrangian alone, or changes it only by a total time derivative, hands you a conserved quantity — and the proof is three lines.

needs 12~2 h slides + 4 h problemsproblems: ./mr problems 13

So far the course has proved three conservation laws by three unrelated tricks. Energy (unit 00): multiply $m\ddot x = F(x)$ by $\dot x$ and notice a derivative. Angular momentum (unit 08): $\mathrm d(\vec r\times\vec p)/\mathrm dt = \vec r\times\vec F$, which vanishes when $\vec F$ is central. Momentum: add the equations of motion and let Newton's third law cancel the internal forces. Each proof works, and none says why the quantity exists. This unit replaces all three with one theorem, and then uses it to find conserved quantities nobody would have guessed.

The language is unit 12's: a Lagrangian $L(q,\dot q,t)$, its Euler–Lagrange equations, and the fact that adding a total derivative $\mathrm d\Lambda/\mathrm dt$ to $L$ changes nothing. The only new mathematical object is a one-parameter group of transformations, which is unit 02's flow of a vector field looked at from a different angle.

Why this unit is on the spine: unit 29 proves the same theorem for fields and gets the energy–momentum tensor out of it, which is the right-hand side of Einstein's equations; unit 36 uses it on the Schwarzschild metric to get the conserved energy and angular momentum of an orbit around a black hole; unit 37 feeds those into unit 08's Binet equation to get the 43 arcseconds. Each is this unit's three lines in a new setting.

How to read this deck

One system runs through it: the Sun and Jupiter as two point masses. Section 1 makes "symmetry" precise, section 2 proves the theorem, and sections 3 to 5 run it four times on the Sun–Jupiter Lagrangian — time translation, rotations, the Galilean boost, and a hidden symmetry that is not a transformation of space at all. Section 3 also settles a question usually glossed over: the conserved quantity from time translation is not always the energy $T + V$.

Proofs are on the slides where they are one line, and in these notes under Full proof where they are not. Results not proved here are marked cited, with where the proof is. Terms with a dotted underline are glossary entries; G lists them all, N opens these notes.

13 · overview

Four symmetries, four charges

the running example

The Sun and Jupiter as two point masses, with Lagrangian

$$L = \tfrac12 m_S\lvert\dot{\vec r}_S\rvert^2 + \tfrac12 m_J\lvert\dot{\vec r}_J\rvert^2 + \frac{G m_S m_J}{\lvert\vec r_S - \vec r_J\rvert}.$$

$m_J/m_S = 9.546\times10^{-4}$, semi-major axis $a = 5.2044$ AU, eccentricity $e = 0.0489$.

transformation	what $L$ does	conserved
shift time, $t\mapsto t + s$	unchanged	energy $E$
shift both bodies, $\vec r_a\mapsto\vec r_a + s\,\hat n$	unchanged	momentum $\vec P$
rotate both bodies about $\hat n$	unchanged	angular momentum $\vec L$
boost, $\vec r_a\mapsto\vec r_a + s t\,\hat n$	changes by $\mathrm d(\cdot)/\mathrm dt$	$\vec G = \vec P t - M\vec R$
hidden, velocity-dependent	changes by $\mathrm d(\cdot)/\mathrm dt$	Runge–Lenz $\vec A$

the theorem, and why it is true

Let $\delta q = \xi$ change $L$ by $\mathrm dF/\mathrm dt$ for every curve. With $p_i = \partial L/\partial\dot q_i$, the charge $\;Q = p\cdot\xi - F\;$ is constant on every solution:

1. Product rule: $\dot Q = \dot p\cdot\xi + p\cdot\dot\xi - \dot F$.

2. On a solution, Euler–Lagrange says $\dot p = \partial L/\partial q$, so the first two terms are the change in $L$, $\ \delta L = \tfrac{\partial L}{\partial q}\cdot\xi + \tfrac{\partial L}{\partial\dot q}\cdot\dot\xi$.

3. The hypothesis says $\delta L = \dot F$. So $\dot Q = 0$.

In the table, $M = m_S + m_J$ is the total mass, $\vec R$ the centre of mass, $\vec P$ the total momentum.

Getting around. A dotted underline is a glossary term: click it, or press G. N opens the notes, where the longer proofs live. A solid underline, like the Sun–Jupiter pair, shows the running example on hover.

The sketch, step by step. Each step is a slide later on. Step 1 needs only that $p$, $\xi$ and $F$ are differentiable along the curve. The expression $\delta L$ in step 2 — the rate at which $L$ changes as the curve is pushed along $\xi$ — is defined on Moving curves, and what the Lagrangian does, and step 2 is the one place the equations of motion enter (unit 12's Euler–Lagrange equations). Step 3 is the hypothesis, which is a property of $L$ alone: Four candidate symmetries of the running example checks it for each row of the table. The three steps together are Noether's theorem in section 2, and The same theorem, seen from the action gives the second proof.

The table is the whole unit in miniature, and each row is proved on its own slide. Rows one to three are the three laws you already knew. Row four is the one that looks new, and is not: it says the centre of mass moves in a straight line at constant speed, which you have always assumed. Row five is the one that should surprise you. Unit 08 found $\vec A$ by guessing and differentiating, and noted there that it is extra — most central forces have no such thing. Section 5 explains where it comes from.

Why the Sun and Jupiter rather than unit 08's single body round a fixed centre: a fixed centre breaks translation and boost symmetry by fiat, since the centre is nailed to the origin and not transformed with the rest. Keeping both bodies lets every symmetry of empty space act, and then the momentum and boost charges are exactly what lets you throw the centre of mass away and return to unit 08's relative Lagrangian. The numbers — masses from the measured $GM$ values $1.32712440018\times10^{20}$ and $1.26686534\times10^{17}\ \mathrm{m^3\,s^{-2}}$ with $G = 6.67430\times10^{-11}$ — are used throughout and every one was computed, not quoted.

section 1

Symmetry, made precise

A family of transformations, what it does to a curve, and what it does to $L$.

13 · symmetry

Three laws, three unrelated proofs

energy · unit 00

$m\ddot x = -V'(x)$. Multiply by $\dot x$:

$$\frac{\mathrm d}{\mathrm dt}\Big(\tfrac12 m\dot x^2 + V\Big) = 0.$$

Works because nothing depends on $t$ explicitly.

momentum · Newton III

$\dot{\vec p}_S = \vec F_{JS}$, $\dot{\vec p}_J = \vec F_{SJ}$. Add, with $\vec F_{JS} = -\vec F_{SJ}$:

$$\frac{\mathrm d}{\mathrm dt}\big(\vec p_S + \vec p_J\big) = \vec 0.$$

Works because only $\vec r_S - \vec r_J$ appears.

angular momentum · unit 08

$\dfrac{\mathrm d}{\mathrm dt}(\vec r\times\vec p) = \dot{\vec r}\times\vec p + \vec r\times\vec F$. The first term vanishes always, the second when $\vec F$ is central:

$$\dot{\vec L} = \vec 0.$$

Works because only $\lvert\vec r\rvert$ appears.

the pattern

Each "works because" names a change you could make to the system without it noticing: shift the clock (homogeneity of time), shift everything in space (homogeneity of space), turn everything (isotropy). The theorem says that is not a coincidence: a continuous symmetry is a conservation law.

Read each proof again and ask what it used. The energy proof used that $V$ has no explicit $t$ — shift the origin of time and the equation is unchanged. The momentum proof used the third law, which is assumed in Newton; section 4 will derive it from the fact that the potential depends only on the difference of positions. The angular-momentum proof used that the force is central, which is what a potential depending only on distance produces.

The word continuous matters and will be made precise on the next slide. A mirror reflection is a symmetry of the running example too, but it comes in no family: you cannot reflect "a little". Discrete symmetries give no conserved quantity in classical mechanics (in quantum mechanics they give conserved parity, a different kind of thing).

History. The link between invariance and conservation was used case by case by Lagrange and by Jacobi, who derived momentum and angular-momentum conservation from translation and rotation invariance of the Lagrangian in his lectures on dynamics of 1842–43. Emmy Noether's paper Invariante Variationsprobleme (Göttingen, 1918) proved the general theorem for any variational problem, fields included, and a second theorem about infinite-dimensional symmetry groups; she was answering a question of Hilbert and Klein about what energy conservation means in general relativity, the subject that ends this course. The extension to transformations that change $L$ by a total derivative is due to Bessel-Hagen (1921).

13 · symmetry

One-parameter groups of transformations

definition · one-parameter group

On an open configuration space $Q\subseteq\mathbb R^n$, a family of maps $\Phi_s : Q\times\mathbb R\to Q$, $(q,t)\mapsto\Phi_s(q,t)$, with $(s,q,t)\mapsto\Phi_s(q,t)$ of class $C^2$ and

$$\Phi_0 = \mathrm{id},\qquad \Phi_{s}\big(\Phi_{s'}(q,t),t\big) = \Phi_{s+s'}(q,t).$$

definition · infinitesimal generator $$\xi(q,t) := \frac{\partial}{\partial s}\Phi_s(q,t)\Big|_{s=0}.$$

The velocity with which the family starts moving $q$.

Each point is carried along its own curve, an orbit of the family. Moving by $s'$ and then by $s$ lands where moving by $s+s'$ does — the group law — and $\xi$ is the tangent arrow at the start. Schematic: $\Phi_s(x,y) = (x+s,\ y - \cos(x+s) + \cos x)$.

Reading the figure. The family drawn is $\Phi_s(x,y) = (x + s,\ y - \cos(x+s) + \cos x)$ on the plane, with $t$ playing no part. Check the group law: applying $\Phi_{s'}$ and then $\Phi_s$ gives first coordinate $x + s' + s$ and second $y - \cos(x+s') + \cos x - \cos(x+s'+s) + \cos(x+s')= y - \cos(x+s+s') + \cos x$, which is $\Phi_{s+s'}$. The quantity $y + \cos x$ is unchanged by every $\Phi_s$, so each point stays on its own curve $y = -\cos x + c$: those curves are the orbits. Differentiating at $s = 0$ gives $\xi(x,y) = (1, \sin x)$, the tangent to the orbit, drawn at the three marked points. It is longer where the orbit is steeper, because the family moves every point by exactly $s$ in $x$.

Why a group, and not just a family. A single symmetry is one map. Noether's theorem needs a derivative with respect to the size of the transformation, so it needs a family through the identity, $\Phi_0 = \mathrm{id}$. The group law is what makes the family the same at every point along it: moving by $s+h$ is moving by $s$ and then by the small amount $h$, so the derivative at any $s$ is the derivative at $0$, transported. The next slide turns that remark into a theorem.

13 · symmetry

The generator determines the group

the generator determines the group

For each fixed $(q, t)$, $s\mapsto\Phi_s(q,t)$ solves $\;\dfrac{\mathrm d}{\mathrm ds}\Phi_s = \xi(\Phi_s, t)$, $\Phi_0 = q$.

Proof. $\partial_s\Phi_s = \partial_h\Phi_{h}(\Phi_s)\big|_{h=0} = \xi(\Phi_s)$, by the group law $\Phi_{s+h} = \Phi_h\circ\Phi_s$. ∎

three generators

Translation $q\mapsto q + s\,c$: $\ \xi = c$.
Boost $\vec r\mapsto\vec r + s\,t\,\hat n$: $\ \xi = t\,\hat n$.
Scaling $q\mapsto e^{s}q$: $\ \xi = q$.

The three generators as arrows. A translation pushes every point the same way; a boost pushes harder the later it is; a scaling pushes each point away from the origin in proportion to its distance.

Reading the figure. Each arrow is $\xi$ at its base point: the first-order displacement per unit $s$. In the middle panel time runs to the right and the one space coordinate up, so the boost's generator $\xi = t\hat n$ is zero at $t = 0$ and grows linearly; the dashed lines are worldlines of bodies at rest, and the boost tilts each of them by the same slope $s$ — which is the statement that a boost adds the velocity $s\hat n$ to everything. In the right panel the arrows $\xi = q$ point straight out from the origin, and integrating them gives $\Phi_s(q) = e^sq$ by the theorem on this slide.

This is unit 02's flow, read backwards. There, a vector field was given and its flow $\varphi_s$ was the object built from it, with the group property $\varphi_{s+s'} = \varphi_s\circ\varphi_{s'}$ proved from uniqueness. Here the family is given and the vector field is extracted by differentiating at $s = 0$. The theorem on the slide says the two constructions are inverse: by unit 01's Picard–Lindelöf theorem (which applies because $\xi$ is $C^1$, hence locally Lipschitz), the initial-value problem $\mathrm d\Phi/\mathrm ds = \xi(\Phi)$, $\Phi_0 = q$, has exactly one solution, and $s\mapsto\Phi_s(q)$ is it. So a one-parameter group is known once its generator is known, and everything below is phrased in terms of $\xi$.

Full proof of the displayed step. Fix $q$ and $s$. By the group law, $\Phi_{s+h}(q) = \Phi_h(\Phi_s(q))$ for every $h$. Differentiate in $h$ at $h = 0$: the left side gives $\partial_s\Phi_s(q)$, the right side gives $\partial_h\Phi_h(p)|_{h=0}$ at the point $p = \Phi_s(q)$, which is $\xi(p)$ by definition. ∎

The role of $t$. It is a parameter carried along: the boost moves a point by an amount proportional to the time, so $\Phi_s$ is allowed to depend on $t$, and the group law holds at each fixed $t$. Transformations that move $t$ itself — time translation — are handled in section 3 by a different device.

Only the generator is used. Noether's theorem below needs nothing but $\xi$ and one identity it satisfies. The group structure is what makes "a symmetry" a meaningful object, and what makes the infinitesimal condition equivalent to the finite one; but the proof itself is first order in $s$.

13 · symmetry

Rotations, and why their generator is a cross product

rotation by $s$ about the unit vector $\hat n$ $$\begin{aligned}R_s\vec x = {}&\cos s\;\vec x + \sin s\;\hat n\times\vec x\\ &+ (1-\cos s)(\hat n\cdot\vec x)\,\hat n.\end{aligned}$$

In an orthonormal basis $(\vec e_1,\vec e_2,\hat n)$ it is the plane rotation by $s$ on the first two axes and fixes $\hat n$; the addition formulas for $\cos$ and $\sin$ give $R_sR_{s'} = R_{s+s'}$.

its generator

Differentiate at $s=0$: $\ \xi(\vec x) = \hat n\times\vec x$.

As $s$ grows, the tip of $\vec x$ runs round a circle about $\hat n$, of radius $\lvert\vec x\rvert\sin\alpha$, where $\alpha$ is the angle between $\hat n$ and $\vec x$. The generator is the tip's starting velocity: tangent to that circle, so perpendicular to both $\hat n$ and $\vec x$, with length $\lvert\vec x\rvert\sin\alpha$ — the cross product $\hat n\times\vec x$.

Reading the figure. Everything the cross product needs is visible. Its direction: the tip moves round a circle centred on the axis, so its velocity is perpendicular to the radius of that circle and to the axis — hence to $\hat n$ and, since $\vec x$ is the radius plus a multiple of $\hat n$, to $\vec x$ as well. Its length: one radian of rotation moves the tip one radius, $\lvert\vec x\rvert\sin\alpha = \lvert\hat n\times\vec x\rvert$. Its sense: counterclockwise seen from the tip of $\hat n$, the right-hand rule. A vector along the axis ($\alpha = 0$) does not move, and $\hat n\times\hat n = \vec 0$ agrees.

Full check of the formula. Take the orthonormal basis $(\vec e_1,\vec e_2,\hat n)$ with $\vec e_1\times\vec e_2 = \hat n$, so $\hat n\times\vec e_1 = \vec e_2$ and $\hat n\times\vec e_2 = -\vec e_1$. For $\vec x = \vec e_1$: $R_s\vec e_1 = \cos s\,\vec e_1 + \sin s\,\vec e_2$. For $\vec x = \vec e_2$: $R_s\vec e_2 = \cos s\,\vec e_2 - \sin s\,\vec e_1$. For $\vec x = \hat n$: $R_s\hat n = \cos s\,\hat n + 0 + (1-\cos s)\hat n = \hat n$. That is the plane rotation by angle $s$ on $\mathrm{span}(\vec e_1,\vec e_2)$, fixing $\hat n$. The composition $R_sR_{s'}$ is the plane rotation by $s$ after the one by $s'$; multiplying the $2\times2$ matrices and using $\cos(s+s') = \cos s\cos s' - \sin s\sin s'$ and $\sin(s+s') = \sin s\cos s' + \cos s\sin s'$ gives the plane rotation by $s+s'$. So this is a one-parameter group, and it is orthogonal with determinant $1$ for every $s$.

13 · symmetry

Every rotation family starts as a cross product

lemma

Any differentiable family of orthogonal matrices with $R_0 = I$ has $\dot R_0$ antisymmetric, and every antisymmetric $3\times3$ matrix acts as $\vec x\mapsto\vec\omega\times\vec x$ for exactly one $\vec\omega$.

So it is not special to the formula on the previous slide: every rotation family starts off as a cross product. That is why angular momentum is a cross product.

proof

Differentiate $R_s^{\top}R_s = I$ at $0$: $\dot R_0^{\top} + \dot R_0 = 0$. An antisymmetric matrix has three free entries; writing $\Omega_{32} = \omega_1$, $\Omega_{13} = \omega_2$, $\Omega_{21} = \omega_3$ gives $\Omega\vec x = \vec\omega\times\vec x$ entry by entry. ∎

The lemma's second half, written out. Antisymmetry forces zero diagonal and $\Omega_{ij} = -\Omega_{ji}$, so $\Omega = \begin{pmatrix}0&-\omega_3&\omega_2\\ \omega_3&0&-\omega_1\\ -\omega_2&\omega_1&0\end{pmatrix}$ and $\Omega\vec x = (\omega_2x_3 - \omega_3x_2,\ \omega_3x_1 - \omega_1x_3,\ \omega_1x_2 - \omega_2x_1) = \vec\omega\times\vec x$ by unit 08's component formula for the cross product. Uniqueness: two such $\vec\omega$ would give the same matrix, whose entries are the $\omega_i$.

Why this matters beyond this slide. Unit 03 met the same fact as the angular-velocity vector of a rotating frame. Unit 16 will say it properly: the antisymmetric matrices are the tangent space to the rotation group at the identity, and the cross product is a coincidence of three dimensions — in $n$ dimensions there are $n(n-1)/2$ independent rotation generators, and "angular momentum" is an antisymmetric matrix, not a vector. In four dimensions there are six, which is why section 5's hidden symmetry has room for $\vec L$ and $\vec A$ together.

13 · symmetry

Moving curves, and what the Lagrangian does

A transformation of $Q$ moves every curve: $q(t)\mapsto q_s(t) := \Phi_s\big(q(t),t\big)$. Its velocity comes from the chain rule (unit 03), the lift to velocities:

$$\dot q_s(t) = D_q\Phi_s\,\dot q(t) + \partial_t\Phi_s .$$

the variation and the induced variation of $L$ $$\delta q := \partial_s q_s\big|_{s=0} = \xi\big(q(t),t\big),$$ $$\delta L := \partial_s L(q_s,\dot q_s,t)\big|_{s=0} = \frac{\partial L}{\partial q}\cdot\xi + \frac{\partial L}{\partial\dot q}\cdot\dot\xi,$$

where $\dot\xi := \dfrac{\mathrm d}{\mathrm dt}\,\xi\big(q(t),t\big)$ along the curve. The second equality is the chain rule plus the lemma on the next slide.

The family pushes the whole curve, so each point moves by $s\,\xi$ to first order — including the two ends, which unit 11's variations kept fixed. Drawn for the scaling $q_s = e^{s}q$, $\ \xi = q$, with $s = 0.32$.

Reading the figure. The solid curve is $q(t)$ on $[t_0, t_1]$; the dashed one is the moved curve $q_s(t) = e^{s}q(t)$, and the arrows join $q(t)$ to $q_s(t)$ at the same time $t$. For this family each arrow has length $(e^{s}-1)q(t)\approx s\,q(t) = s\,\xi$: longer where $q$ is larger. Nothing pins the ends, and that is the difference from unit 11 that the whole theorem turns on.

The lift, in words. A map of positions induces a map of velocities, because a velocity is the derivative of a curve of positions and the chain rule says how the derivative transforms. For a rotation, $D_q R_s = R_s$ and $\partial_tR_s = 0$, so velocities are rotated by the same matrix. For the boost $\vec r\mapsto\vec r + st\hat n$, $D_q\Phi_s = I$ and $\partial_t\Phi_s = s\hat n$, so every velocity gains $s\hat n$ — which is what a boost is. Unit 31 will call the lifted map the tangent map of $\Phi_s$ and the space of $(q,\dot q)$ the tangent bundle; nothing here needs those words. In the figure the lift is the slope of the dashed curve at each $t$, which here is $e^{s}$ times the slope of the solid one.

Notation. $\delta$ is not a new kind of derivative; it is $\partial_s$ at $s = 0$. The symbol is kept because it is universal in physics and because unit 11 used it for exactly this. $D_q\Phi_s$ is the Jacobian matrix of $\Phi_s$ in the $q$ variables.

13 · symmetry

Varying the action along the family

$\delta$ commutes with $\mathrm d/\mathrm dt$

$\partial_s\dot q_s\big|_{0} = \dfrac{\mathrm d}{\mathrm dt}\,\partial_s q_s\big|_{0} = \dot\xi$.

Why: $(s,t)\mapsto q_s(t)$ is $C^2$, so the mixed partials agree — Clairaut, unit 03.

That is the step the formula for $\delta L$ on the previous slide used: the change in the velocity is the velocity of the change.

Integrating over an interval gives the variation of the action that unit 11 differentiated:

$$\frac{\mathrm d}{\mathrm ds}S[q_s]\Big|_{s=0} = \int_{t_0}^{t_1}\delta L\;\mathrm dt .$$

Unit 11 took variations vanishing at the ends. A symmetry moves the ends too, and the boundary term that unit 11 threw away is where the conserved quantity will come from.

Full proof of the formula for $\delta L$. $L$ is $C^2$ and $s\mapsto(q_s(t),\dot q_s(t))$ is $C^1$, so the chain rule gives $\partial_sL(q_s,\dot q_s,t) = \sum_i\big(\partial_{q_i}L\, \partial_sq_{s,i} + \partial_{\dot q_i}L\,\partial_s\dot q_{s,i}\big)$. At $s = 0$, $\partial_sq_s = \xi$ by definition and $\partial_s\dot q_s = \dot\xi$ by the commuting lemma. ∎

The regularity bookkeeping, once. Curves are $C^2$ (so the Euler–Lagrange equations make sense), $\Phi$ is $C^2$ in all its arguments, $L$ is $C^2$. Then $q_s(t)$ is $C^2$ in $(s,t)$ jointly and Clairaut applies. Nothing later needs more.

Differentiating under the integral. $S[q_s] = \int_{t_0}^{t_1}L(q_s,\dot q_s,t)\,\mathrm dt$ over a fixed interval, and the integrand is $C^1$ in $(s,t)$ jointly on a compact set, so the $s$-derivative passes inside the integral (the rule unit 11 used for the same step).

13 · symmetry

Symmetry, and symmetry up to a total derivative

definition · symmetry

$\Phi_s$ is a symmetry of $L$ if, for every $C^2$ curve and every $s$,

$$L(q_s,\dot q_s,t) = L(q,\dot q,t).$$

Infinitesimally: $\delta L = 0$ for every curve.

definition · quasi-symmetry

$\Phi_s$ is a quasi-symmetry if there are $C^2$ functions $\Lambda_s(q,t)$, $\Lambda_0 = 0$, with

$$L(q_s,\dot q_s,t) = L(q,\dot q,t) + \frac{\mathrm d}{\mathrm dt}\Lambda_s\big(q(t),t\big).$$

Infinitesimally: $\delta L = \dfrac{\mathrm dF}{\mathrm dt}$ with $F = \partial_s\Lambda_s|_{s=0}$.

why a total derivative is allowed

Unit 12: $L$ and $L + \mathrm d\Lambda/\mathrm dt$ have the same Euler–Lagrange equations, because $\int_{t_0}^{t_1}\dot\Lambda\,\mathrm dt = \Lambda(t_1) - \Lambda(t_0)$ does not change under variations fixed at the ends. So a quasi-symmetry maps solutions to solutions exactly as a symmetry does.

"for every curve", not "for every solution"

The identity must hold off-shell — on curves that do not solve the equations. On-shell, every $\xi$ satisfies $\delta L = \frac{\mathrm d}{\mathrm dt}(p\cdot\xi)$, and the theorem would say nothing.

Why "for every curve" is the whole content. Along a solution, the Euler–Lagrange equations give $\partial L/\partial q = \dot p$ with $p = \partial L/\partial\dot q$, so $\delta L = \dot p\cdot\xi + p\cdot\dot\xi = \frac{\mathrm d}{\mathrm dt}(p\cdot\xi)$ for any $\xi$ whatsoever. If we only asked for $\delta L = \dot F$ on solutions, $F = p\cdot\xi$ would always work and the "charge" $p\cdot\xi - F$ would be identically $0$. The theorem has content because $F$ is required to be a fixed function of $(q,\dot q,t)$ satisfying the identity on all curves, which is a checkable property of $L$ that knows nothing about solutions.

Why the finite version implies the infinitesimal one. Differentiate the defining identity in $s$ at $s = 0$. For the quasi-symmetry, $\partial_s\frac{\mathrm d}{\mathrm dt}\Lambda_s = \frac{\mathrm d}{\mathrm dt}\partial_s\Lambda_s$ by Clairaut again. The converse — infinitesimal symmetry at every point of the family implies finite symmetry — is true for groups and is proved by integrating in $s$ using the group law; nothing below needs it.

Terminology. Some books call quasi-symmetries divergence symmetries (the name from field theory, where the total derivative becomes a divergence, unit 29) and some say "$L$ is invariant up to a gauge term". The words matter less than the fact: the boost of section 4, the translation of a charge in a magnetic field (problem P2) and section 5's hidden symmetry are all quasi-symmetries and not symmetries, and each gives a conserved quantity that a strict-invariance version of the theorem would miss.

13 · symmetry · example

Four candidate symmetries of the running example

test each on the Sun–Jupiter Lagrangian, with $V = -Gm_Sm_J/\lvert\vec r_S - \vec r_J\rvert$

family	$\xi_a$ (each body $a$)	$\dot\xi_a$	$\delta T$	$\delta V$	verdict
translate, $\vec r_a + s\hat n$	$\hat n$	$\vec 0$	$0$	$\nabla V\cdot(\hat n - \hat n) = 0$	symmetry
rotate about $\hat n$	$\hat n\times\vec r_a$	$\hat n\times\dot{\vec r}_a$	$\sum m_a\dot{\vec r}_a\cdot(\hat n\times\dot{\vec r}_a) = 0$	$0$, depends on $\lvert\vec r_S-\vec r_J\rvert$ only	symmetry
boost, $\vec r_a + st\hat n$	$t\hat n$	$\hat n$	$\sum m_a\dot{\vec r}_a\cdot\hat n$	$0$	quasi
scale, $e^{s}\vec r_a$	$\vec r_a$	$\dot{\vec r}_a$	$2T$	$-V$	neither

Boost: $\delta L = \sum_a m_a\dot{\vec r}_a\cdot\hat n = \dfrac{\mathrm d}{\mathrm dt}\big(M\,\hat n\cdot\vec R\big)$, with $M = m_S + m_J$ and $\vec R$ the centre of mass. A total derivative: $F = M\,\hat n\cdot\vec R$.

Scaling: $\delta L = 2T + V$. Any $\dot F$ with $F = F(q,t)$ is affine in the velocities, and $2T$ is quadratic, so no $F$ works (notes). Neither a symmetry nor a quasi-symmetry.

Every step of the table. Write $T = \frac12\sum_a m_a\lvert\dot{\vec r}_a\rvert^2$, so $\delta T = \sum_a m_a\dot{\vec r}_a\cdot\dot\xi_a$, and $V = V(\vec r_S - \vec r_J)$, so $\delta V = \nabla V\cdot(\xi_S - \xi_J)$.

Translation. $\xi_S = \xi_J = \hat n$ is constant, so $\dot\xi = 0$ and $\delta T = 0$; $\xi_S - \xi_J = 0$, so $\delta V = 0$.
Rotation. $\dot\xi_a = \hat n\times\dot{\vec r}_a$ and $\dot{\vec r}_a\cdot(\hat n\times\dot{\vec r}_a) = 0$ (a triple product with a repeated vector). For $V$: $V$ depends on $\rho = \lvert\vec r_S - \vec r_J\rvert$ only, and $\delta\rho^2 = 2(\vec r_S - \vec r_J)\cdot\hat n\times(\vec r_S - \vec r_J) = 0$.
Boost. $\xi_a = t\hat n$, so $\dot\xi_a = \hat n$ and $\xi_S - \xi_J = 0$. $\delta L = \hat n\cdot\sum_a m_a\dot{\vec r}_a = \hat n\cdot M\dot{\vec R}$, where $M\vec R = m_S\vec r_S + m_J\vec r_J$. The finite version is exact too: $L_s - L = s\,\hat n\cdot M\dot{\vec R} + \frac12Ms^2 = \frac{\mathrm d}{\mathrm dt}\big(sM\hat n\cdot\vec R + \tfrac12Ms^2t\big)$.
Scaling. $\xi_a = \vec r_a$, $\dot\xi_a = \dot{\vec r}_a$, so $\delta T = 2T$; $V$ is homogeneous of degree $-1$ in the positions, so $\delta V = -V$ (Euler's theorem, proved in section 3). Hence $\delta L = 2T - (-V) = 2T + V$.

Full proof that scaling is not a quasi-symmetry. Suppose $2T + V = \frac{\mathrm d}{\mathrm dt}F(q,t) = \partial_tF + \nabla_qF\cdot\dot q$ for every curve. The right side is affine in the velocities; the left side has the quadratic part $2T$. Take a curve through a fixed $(q,t)$ with velocity $\lambda v$: the left side grows like $\lambda^2$ and the right side like $\lambda$, a contradiction as $\lambda\to\infty$. ∎ (Section 5 extends exactly this argument to show that the Runge–Lenz vector comes from no transformation of configuration space.)

The scaling is not useless: with time rescaled as well, $t\mapsto\lambda^3 t$, $\vec r\mapsto\lambda^2\vec r$, it maps solutions to solutions, and problem P9 extracts Kepler's third law from it. It is a symmetry of the equations of motion but not of $L$, and that difference is the subject of the warning slide in section 2.

section 2

The theorem

Stated with velocity-dependent generators allowed, because the proof does not care.

13 · theorem

Noether's theorem

theorem · Noether's theorem for point mechanics

Let $L(q,\dot q,t)$ be $C^2$, and let $\xi(q,\dot q,t)$ and $F(q,\dot q,t)$ be $C^1$ functions such that, for every $C^2$ curve $q(t)$,

$$\frac{\partial L}{\partial q}\cdot\xi + \frac{\partial L}{\partial\dot q}\cdot\dot\xi \;=\; \frac{\mathrm dF}{\mathrm dt}.$$

Then along every solution of the Euler–Lagrange equations the Noether charge

$$Q \;:=\; \sum_{i=1}^n\frac{\partial L}{\partial\dot q_i}\,\xi_i \;-\; F \;=\; p\cdot\xi - F$$

is constant.

proof $$\frac{\mathrm dQ}{\mathrm dt} \;=\; \dot p\cdot\xi + p\cdot\dot\xi - \dot F \;=\; \frac{\partial L}{\partial q}\cdot\xi + \frac{\partial L}{\partial\dot q}\cdot\dot\xi - \dot F \;=\; \delta L - \dot F \;=\; 0 .\qquad\blacksquare$$

Product rule; the Euler–Lagrange equation $\dot p = \partial L/\partial q$; the hypothesis.

That is the whole theorem. Three equalities, each with one reason, and the second is the only place the equations of motion enter. The hypothesis is about $L$ alone and can be checked without solving anything; the conclusion is about solutions.

Why $\xi$ may depend on $\dot q$. Nothing in the three lines used that $\xi$ is a function of position only. If $\xi = \xi(q,\dot q,t)$, then $\dot\xi$ along a curve involves $\ddot q$, and the hypothesis is an identity in $(q,\dot q,\ddot q,t)$ that must hold for all curves; the proof goes through word for word. Such a $\xi$ is a generalised symmetry: it does not come from a transformation of configuration space, because it moves a point differently depending on how fast it is moving. Time translation (section 3) and the hidden symmetry (section 5) are of this kind. When $\xi = \xi(q,t)$, the hypothesis has no $\ddot q$ on the left, so $F$ can have no $\dot q$ dependence: $F = F(q,t)$, matching the quasi-symmetry definition.

Regularity. $Q$ is $C^1$ along a $C^2$ solution because $p = \partial L/\partial\dot q$ is $C^1$ in its arguments and $\xi$, $F$ are $C^1$. That is all the product rule needs.

Several coordinates, several symmetries. Each independent generator gives its own charge. Rotations about three independent axes give three charges, which section 4 assembles into a vector. Linearity helps: if $\xi_1$ and $\xi_2$ satisfy the hypothesis with $F_1$ and $F_2$, then $c_1\xi_1 + c_2\xi_2$ does with $c_1F_1 + c_2F_2$, and its charge is $c_1Q_1 + c_2Q_2$.

Uniqueness. $F$ is determined by $\xi$ only up to an additive constant, so $Q$ is too — harmless, since a constant is conserved anyway.

13 · theorem

The same theorem, seen from the action

Let $q$ solve the Euler–Lagrange equations on $[t_0,t_1]$ and vary it by $\delta q = \xi$, not zero at the ends. Unit 11's integration by parts, keeping the boundary term:

$$\delta S = \int_{t_0}^{t_1}\Big(\frac{\partial L}{\partial q} - \frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial\dot q}\Big)\cdot\xi\,\mathrm dt + \Big[p\cdot\xi\Big]_{t_0}^{t_1}.$$

On a solution the integral is $0$: $\ \delta S = [\,p\cdot\xi\,]_{t_0}^{t_1}$.

If $\xi$ is a (quasi-)symmetry, compute $\delta S$ a second way:

$$\delta S = \int_{t_0}^{t_1}\frac{\mathrm dF}{\mathrm dt}\,\mathrm dt = \Big[F\Big]_{t_0}^{t_1}.$$

equate $$\big[\,p\cdot\xi - F\,\big]_{t_0}^{t_1} = 0\quad\text{for all } t_0 \lt t_1 .$$

The charge is the boundary term that unit 11 discarded.

Full proof of the boundary formula

$\delta S = \int_{t_0}^{t_1}\big(\partial_qL\cdot\xi + \partial_{\dot q}L\cdot\dot\xi\big)\,\mathrm dt$ by the previous section. Integrate the second term by parts: $\int p\cdot\dot\xi\,\mathrm dt = [p\cdot\xi]_{t_0}^{t_1} - \int\dot p\cdot\xi\,\mathrm dt$. This is legitimate because $p(t) = \partial_{\dot q}L(q(t),\dot q(t),t)$ is $C^1$ along a $C^2$ curve. Collect terms. ∎ In unit 11 the variations vanished at $t_0$ and $t_1$, the bracket was zero, and the fundamental lemma turned "$\delta S = 0$ for all such $\xi$" into the Euler–Lagrange equation. Here the logic runs the other way: the equation is known, the integral vanishes, and the bracket is all that is left.

Why the two proofs are the same proof. The first slide differentiated $Q$; this one integrated $\dot Q$ over $[t_0,t_1]$. Equal values at every pair of times is constancy.

13 · theorem

Where the charge lives: at the ends

Left: unit 11's variations vanish at $t_0$ and $t_1$, so integration by parts leaves no boundary term. Right: a symmetry moves the whole solution, ends included, and the only thing left of $\delta S$ is $p\cdot\xi$ at each end — minus $F$, for a quasi-symmetry. Equal at both ends, for every choice of ends: conserved.

This is the version that generalises: in unit 29 the interval becomes a region of spacetime, the boundary term becomes a flux through its boundary, and the charge becomes a conserved current.

The picture. A symmetry slides a solution sideways into another solution (a quasi-symmetry does too, by unit 12). The action of the slid curve differs from the original only by what happens at the two ends, and conservation says the end contributions are the same function of time at both ends.

Reading the figure. Both panels show the same solution (solid) and a varied curve (dashed). On the left the variation is a bump that is zero at both ends, the kind unit 11 used to derive the Euler–Lagrange equation; on the right it is the symmetry's $\xi$, which has no reason to vanish anywhere. Since the solid curve is a solution, the integral term of $\delta S$ is zero in both panels; on the left nothing else is left, and on the right the two arrows at the ends carry $[\,p\cdot\xi\,]_{t_0}^{t_1}$. The symmetry gives the second computation $[\,F\,]_{t_0}^{t_1}$, and since $t_0$ and $t_1$ were arbitrary, $p\cdot\xi - F$ has the same value at every time.

13 · theorem

What the hypotheses are for

a symmetry of the equations is not enough

$L = e^{\gamma t}\big(\tfrac12m\dot q^2 - \tfrac12m\omega^2q^2\big)$ gives the damped oscillator $\ddot q + \gamma\dot q + \omega^2q = 0$ (unit 06), whose equation is unchanged by $t\mapsto t+s$. But $L$ becomes $e^{\gamma s}L$, and the change $(e^{\gamma s}-1)L$ is not a total derivative: the energy is not conserved. Problem P8 finds the symmetry this $L$ does have.

discrete symmetries give nothing

A reflection $\vec r\mapsto-\vec r$ leaves the running example's $L$ unchanged, but there is no $s$ to differentiate in, so no $\xi$ and no charge.

the charge is gauge-invariant

Replace $L$ by $L' = L + \dot\Lambda(q,t)$. Then $p' = p + \partial_q\Lambda$ and $F' = F + \partial_q\Lambda\cdot\xi$, so

$$Q' = p'\cdot\xi - F' = p\cdot\xi - F = Q .$$

Momenta are not gauge-invariant (unit 12); Noether charges are.

non-holonomic systems

Everything here assumes the equations of motion are Euler–Lagrange equations. For a rolling constraint they are not (unit 12), and the theorem can fail: a symmetric rolling body need not conserve the naive charge.

Full proof of gauge invariance

$p' = \partial_{\dot q}(L + \partial_t\Lambda + \partial_q\Lambda\cdot\dot q) = p + \partial_q\Lambda$. For $\delta$ of the extra term: $\delta\dot\Lambda = \partial_q(\dot\Lambda)\cdot\xi + \partial_{\dot q}(\dot\Lambda)\cdot\dot\xi = \frac{\mathrm d}{\mathrm dt}(\partial_q\Lambda)\cdot\xi + \partial_q\Lambda\cdot\dot\xi = \frac{\mathrm d}{\mathrm dt}(\partial_q\Lambda\cdot\xi)$, using $\partial_q\dot\Lambda = \frac{\mathrm d}{\mathrm dt}\partial_q\Lambda$ (Clairaut). So $\delta L' = \frac{\mathrm d}{\mathrm dt}(F + \partial_q\Lambda\cdot\xi)$, which is the $F'$ on the slide, and the two extra terms in $Q'$ cancel. ∎ Problem P4 asks for this with the physics attached.

The damped oscillator, a little more. With $p = me^{\gamma t}\dot q$, the Euler–Lagrange equation is $\frac{\mathrm d}{\mathrm dt}(me^{\gamma t}\dot q) = -me^{\gamma t}\omega^2q$, i.e. $\ddot q + \gamma\dot q + \omega^2 q = 0$. The time shift sends $L$ to $e^{\gamma s}L$: a symmetry of the equations (multiplying $L$ by a constant never changes them) but not of the action, which it rescales. Section 3's energy function for this $L$ is $h = e^{\gamma t}(\frac12m\dot q^2 + \frac12m\omega^2q^2)$, and it is not conserved: $\dot h = -\partial_tL = -\gamma L$. Neither is the mechanical energy. The lesson: the hypothesis is invariance of $L$ up to $\dot F$, and invariance of the equations it produces is weaker.

Why the reflection gives nothing. Conserved quantities come from derivatives at the identity, and a discrete group has no elements near the identity other than the identity.

Non-holonomic constraints, for completeness: for velocity constraints that cannot be integrated to position constraints, the correct equations of motion come from d'Alembert's principle and are not the Euler–Lagrange equations of any constrained action (unit 12 says this honestly). Noether's theorem then does not apply, and there are standard examples — the rolling disc, the Chaplygin sleigh — where a symmetry does not give the charge the unconstrained formula predicts. Cited: Bloch, Nonholonomic Mechanics and Control, ch. 5.

13 · theorem

Checking a charge: dimensions

$Q = p\cdot\xi - F$ and $\xi = \partial_s\Phi_s$, so $[Q] = [p][q]/[s]$. Since $[p][q]$ is an action, $\mathsf{ML^2T^{-1}}$:

$$[Q] = \frac{\mathsf{ML^2T^{-1}}}{[s]} .$$

parameter $s$	$[s]$	$[Q]$	charge
time shift	$\mathsf T$	$\mathsf{ML^2T^{-2}}$	energy
space shift	$\mathsf L$	$\mathsf{MLT^{-1}}$	momentum
rotation angle	$1$	$\mathsf{ML^2T^{-1}}$	angular momentum
boost velocity	$\mathsf{LT^{-1}}$	$\mathsf{ML}$	mass × position

a free consistency check

Every charge in this unit is checked against the table. If $[Q]$ is wrong, $\xi$ or $F$ is.

The pairs — time and energy, position and momentum, angle and angular momentum — will reappear in unit 14 as conjugate variables: a charge generates, in phase space, the motion along the parameter it came from.

Why $[p][q]$ is an action. $p = \partial L/\partial\dot q$ has dimensions $[L][t]/[q]$, so $[p][q] = [L][t] = [S]$, the dimension of $\int L\,\mathrm dt$, whatever units the generalised coordinate $q$ has. That is why the table works for coordinates that are angles as well as for lengths.

$F$ carries the same dimensions as $p\cdot\xi$, necessarily, since the two are subtracted. For the boost, $F = M\hat n\cdot\vec R$ has dimensions $\mathsf{ML}$, matching $[p\cdot t\hat n] = \mathsf{MLT^{-1}}\cdot\mathsf T$. ✓

A note ahead. In quantum mechanics these pairs are the ones whose uncertainties multiply to at least $\hbar/2$, and $\hbar$ has the dimensions of action — which is the same dimensional fact as this table. This course does not do quantum mechanics; the remark is only there so that the coincidence, when you meet it, is recognised as not one.

section 3

Energy is time translation

— and the conserved quantity is $T + V$ only when the coordinates do not move.

13 · energy

Time translation as a variation

Shifting the clock sends a curve $q(t)$ to $q_s(t) = q(t+s)$. It moves $t$, not $q$, but its effect on curves is a variation:

$$\delta q = \partial_s q(t+s)\big|_{s=0} = \dot q .$$

A velocity-dependent $\xi$, which the theorem allows. Then

$$\delta L = \frac{\partial L}{\partial q}\cdot\dot q + \frac{\partial L}{\partial\dot q}\cdot\ddot q = \frac{\mathrm dL}{\mathrm dt} - \frac{\partial L}{\partial t}.$$

So when $L$ has no explicit $t$, $\delta L$ is a total derivative with $F = L$.

Sliding the graph left by $s$ changes the value at a fixed time by $s\,\dot q$, to first order — steep where the curve is steep. So the generator of a time shift is the velocity itself.

Reading the figure. The dashed curve is the solid one moved left by $s$: its value at time $t$ is the solid curve's value at $t + s$. At the marked time the vertical gap is $q(t+s) - q(t) = s\,\dot q(t) + O(s^2)$ by Taylor's theorem, and the dotted line is the tangent whose slope $\dot q$ sets it. Where the solid curve is flat — at its peaks — the two curves momentarily agree to first order and $\xi = \dot q = 0$ there.

Why not treat time translation as a transformation of $(q, t)$? One can, and the general theory does (it gives the charge $p\cdot\xi - h\,\tau - F$ for a transformation moving $t$ by $s\tau$ and $q$ by $s\xi$). For this unit the variation $\delta q = \dot q$ is cleaner: it says what time translation does to a curve, which is all the theorem needs, and it shows at once why velocity-dependent generators cannot be excluded — the most important conservation law of all comes from one.

13 · energy

The energy function, and when it is conserved

definition · energy function $$h(q,\dot q,t) := \sum_i\dot q_i\,\frac{\partial L}{\partial\dot q_i} - L .$$

It is the Noether charge $p\cdot\xi - F$ of the previous slide, with $\xi = \dot q$ and $F = L$.

theorem

Along every solution, $\ \dfrac{\mathrm dh}{\mathrm dt} = -\dfrac{\partial L}{\partial t}$. So if $L$ has no explicit $t$, $h$ is conserved.

Proof. If $\partial_tL = 0$, $\delta L = \dot L$: Noether with $\xi = \dot q$, $F = L$ gives $Q = p\cdot\dot q - L = h$. In general, $\dot h = \dot p\cdot\dot q + p\cdot\ddot q - \dot L$, and substituting $\dot L$ and $\dot p = \partial_qL$ leaves $-\partial_tL$. ∎

checks

Dimensions: $[\dot q\,\partial L/\partial\dot q] = [L]$, so $h$ is an energy ✓. One particle, $L = \tfrac12m\dot x^2 - V(x)$: $h = \tfrac12m\dot x^2 + V$, unit 00's energy ✓.

Full proof of the general formula. Along a solution, $\dot L = \partial_qL\cdot\dot q + \partial_{\dot q}L\cdot\ddot q + \partial_tL$ (chain rule) and $\dot p = \partial_qL$ (Euler–Lagrange). So $\dot h = \dot p\cdot\dot q + p\cdot\ddot q - \dot L = \partial_qL\cdot\dot q + p\cdot\ddot q - \partial_qL\cdot\dot q - p\cdot\ddot q - \partial_tL = -\partial_tL$. ∎

You have seen this before. Unit 11's Beltrami identity says that for an integrand $F(y,y')$ with no explicit $x$, the quantity $F - y'F_{y'}$ is constant on extremals. That is $-h$, with $x$ playing time. The brachistochrone was solved with the energy function of a variational problem, before the word "energy" was needed.

The one-particle check. $\partial L/\partial\dot x = m\dot x$, so $h = m\dot x^2 - \tfrac12m\dot x^2 + V = \tfrac12m\dot x^2 + V$. The next slide says exactly when this "$h = T + V$" survives in other coordinates.

13 · energy

When is $h$ the energy?

Euler's theorem on homogeneous functions

If $f(\lambda v) = \lambda^kf(v)$ for all $\lambda \gt 0$ and $f$ is $C^1$, then $v\cdot\nabla f(v) = k\,f(v)$.

Proof. Differentiate in $\lambda$ at $\lambda = 1$, by the chain rule. ∎

Split $L$ into pieces homogeneous of degree 2, 1, 0 in $\dot q$: $\ L = L_2 + L_1 + L_0$. Then $\dot q\cdot\partial_{\dot q}L = 2L_2 + L_1$, and

the energy function, decoded $$h = L_2 - L_0 .$$

For $L = T - V$ with $V = V(q,t)$, write the Cartesian positions as $\vec r_a = \vec r_a(q,t)$. Then

$$T = \underbrace{\tfrac12\dot q^{\top}M(q,t)\,\dot q}_{T_2} + \underbrace{b(q,t)\cdot\dot q}_{T_1} + \underbrace{T_0(q,t)}_{T_0},$$ $$h = T_2 - T_0 + V .$$

so $h = T + V$ only if

$T_1 = T_0 = 0$, which holds when $\partial_t\vec r_a = 0$: the coordinates are fixed in an inertial frame. With a moving constraint or moving coordinates, $h$ is not the energy — and whether it is conserved is the separate question of whether $\partial_tL = 0$.

Full proof of the decomposition

With $\vec r_a = \vec r_a(q,t)$, the chain rule gives $\dot{\vec r}_a = \sum_i\partial_{q_i}\vec r_a\,\dot q_i + \partial_t\vec r_a$. Then $T = \frac12\sum_am_a\lvert\dot{\vec r}_a\rvert^2$ expands into $T_2 = \frac12\sum_{a,i,j}m_a\,\partial_{q_i}\vec r_a\cdot\partial_{q_j}\vec r_a\,\dot q_i\dot q_j$, $T_1 = \sum_{a,i}m_a\,\partial_{q_i}\vec r_a\cdot\partial_t\vec r_a\,\dot q_i$, and $T_0 = \frac12\sum_am_a\lvert\partial_t\vec r_a\rvert^2$. So $L_2 = T_2$, $L_1 = T_1$ (plus any velocity-linear potential), $L_0 = T_0 - V$, and $h = L_2 - L_0 = T_2 - T_0 + V$. If $\partial_t\vec r_a = 0$ for all $a$, then $T_1 = T_0 = 0$, $T = T_2$ and $h = T + V$. ∎

Euler's theorem in full. Fix $v$ and set $g(\lambda) = f(\lambda v) = \lambda^kf(v)$. By the chain rule (unit 03), $g'(\lambda) = \nabla f(\lambda v)\cdot v$; also $g'(\lambda) = k\lambda^{k-1}f(v)$. At $\lambda = 1$: $v\cdot\nabla f(v) = kf(v)$. ∎ Here it is applied with $v = \dot q$ and $(q,t)$ held fixed.

The two conditions are separate. $h$ is conserved when $\partial_tL = 0$. $h = T + V$ when the coordinates are not moving. The bead on the rotating hoop, next, has the first and not the second; a pendulum whose length is being pulled in by a string has neither; a charge in a static magnetic field has both, with $L_1 = q\dot{\vec r}\cdot\vec A$ dropping out of $h$ entirely — the magnetic force does no work, and $h = \frac12m\lvert\dot{\vec r}\rvert^2 + q\phi$ (problem P5).

Why the course needs this later. In unit 36 the Schwarzschild metric has no explicit time, the corresponding charge is conserved, and it is the energy "at infinity" — not the kinetic plus potential energy of anyone's frame. Recognising $h$ as a charge rather than as "the energy" is what makes that result unsurprising.

The running example. For the Sun and Jupiter in Cartesian coordinates, $T$ is purely quadratic, so $h = T + V$. In the frame of the centre of mass it is $E = -k/2a = -1.61775\times10^{35}$ J, with $k = Gm_Sm_J = 2.51905\times10^{47}\ \mathrm{kg\,m^3\,s^{-2}}$ and $a = 5.2044$ AU $= 7.78567\times10^{11}$ m.

13 · energy · example

The bead on the rotating hoop: $h$ conserved, $E$ not

Unit 12's bead, mass $m$, on a hoop of radius $R$ driven at constant $\Omega$ about its vertical diameter; $\theta$ from the bottom. The Cartesian position depends on $t$, so expect $h\ne T+V$.

$$L = \tfrac12mR^2\big(\dot\theta^2 + \Omega^2\sin^2\theta\big) + mgR\cos\theta .$$

step by step

1. $\partial_tL = 0$, so $h$ is conserved.
2. $L_2 = \frac12mR^2\dot\theta^2$, $\ L_1 = 0$, $\ L_0 = \frac12mR^2\Omega^2\sin^2\theta + mgR\cos\theta$.
3. $h = L_2 - L_0 = \frac12mR^2\dot\theta^2 - \frac12mR^2\Omega^2\sin^2\theta - mgR\cos\theta$.
4. $E = T + V = \frac12mR^2\dot\theta^2 + \frac12mR^2\Omega^2\sin^2\theta - mgR\cos\theta$.
5. $E - h = mR^2\Omega^2\sin^2\theta$, which changes as the bead moves.

The hoop turns at the fixed rate $\Omega$; the bead slides freely on it. Released at $\theta_0 = 90^\circ$ with the numbers of the next slide, it swings over the highlighted arc, between $\theta_1$ and $\theta_0$, about the off-axis equilibrium $\theta_*$.

Setting up $L$. With the hoop's axis along $\hat z$ and the hoop at azimuth $\Omega t$, the bead is at $\vec r = R(\sin\theta\cos\Omega t,\ \sin\theta\sin\Omega t,\ -\cos\theta)$. Then $\lvert\dot{\vec r}\rvert^2 = R^2\dot\theta^2 + R^2\Omega^2\sin^2\theta$ (the two velocity components, along the hoop and around the axis, are perpendicular), and $V = -mgR\cos\theta$. The Cartesian position depends explicitly on $t$ — exactly the case of the previous slide — and here $T_2 = \frac12mR^2\dot\theta^2$, $T_1 = 0$, $T_0 = \frac12mR^2\Omega^2\sin^2\theta$.

Reading the figure. The axis is the dashed vertical line, and the bead's distance from it is $R\sin\theta$, the lever arm of everything that follows: the bead's speed round the axis is $R\Omega\sin\theta$, which is where the $\tfrac12mR^2\Omega^2\sin^2\theta$ in $L$ comes from. The three marked angles are computed on the next slide.

13 · energy · example

The hoop in numbers: $h$ stays, $E$ moves

with numbers: $m = 0.05$ kg, $R = 0.20$ m, $\Omega = 10\ \mathrm{s^{-1}}$

Release from rest at $\theta_0 = 90^\circ$. Then $h = -\frac12mR^2\Omega^2 = -0.1000$ J for all time.

Turning points: $\dot\theta = 0$ and $h$ fixed give $\cos\theta\,\big(\tfrac12R\Omega^2\cos\theta - g\big) = 0$, so $\cos\theta_1 = 2g/R\Omega^2 = 0.981$: $\ \theta_1 = 11.19^\circ$.

	$\theta_0 = 90^\circ$	$\theta_1 = 11.19^\circ$
$h$ (J)	$-0.1000$	$-0.1000$
$E$ (J)	$+0.1000$	$-0.0925$

$E$ falls by $0.1925$ J. Where did it go?

$h$ and $E$ along the bead's swing, computed from steps 3 and 4. $h$ is flat; $E$ rises by $0.1925$ J from $\theta_1$ to $\theta_0$. The gap between them is $mR^2\Omega^2\sin^2\theta$. Below $\theta_1$ (shaded) the bead never goes.

The turning-point algebra. At rest on the hoop, $h = -\frac12mR^2\Omega^2\sin^2\theta - mgR\cos\theta$. Setting this equal to its value at $\theta_0 = 90^\circ$, which is $-\frac12mR^2\Omega^2$, and using $\sin^2 = 1 - \cos^2$: $\frac12mR^2\Omega^2\cos^2\theta - mgR\cos\theta = 0$. The root $\cos\theta = 0$ is the starting point; the other is $\cos\theta_1 = 2g/(R\Omega^2) = 2(9.81)/(0.20\cdot100) = 0.981$, so $\theta_1 = 11.187^\circ$.

The energies. $mR^2\Omega^2 = 0.05\cdot0.04\cdot100 = 0.2$ J and $mgR = 0.0981$ J. At $\theta_0$: $E = \frac12(0.2)(1) - 0 = 0.1000$ J. At $\theta_1$: $\sin^2\theta_1 = 1 - 0.981^2 = 0.037639$, so $E = \frac12(0.2)(0.037639) - 0.0981(0.981) = 0.0037639 - 0.0962361 = -0.0924722$ J. The change is $-0.1924722$ J. That $h$ really is constant is one line from the equation of motion $\ddot\theta = \Omega^2\sin\theta\cos\theta - (g/R)\sin\theta$: $\dot h = mR^2\dot\theta\big[\ddot\theta - \Omega^2\sin\theta\cos\theta + (g/R)\sin\theta\big] = 0$. And the bead really does swing between the two roots: with $u = \cos\theta$ and $h = -0.1$ J, $\frac12mR^2\dot\theta^2 = h + \frac12mR^2\Omega^2(1-u^2) + mgRu = 0.1\,u\,(0.981 - u)$, which is positive for $0 \lt u \lt 0.981$ and vanishes at the ends — so the motion is confined between $\theta_1 = 11.187^\circ$ and $\theta_0 = 90^\circ$ and reaches both.

The bead never reaches the bottom. Here $\Omega^2 = 100\ \mathrm{s^{-2}} \gt g/R = 49.05\ \mathrm{s^{-2}}$, above unit 12's bifurcation, so the bottom is unstable and the bead oscillates about the off-axis equilibrium $\cos\theta_* = g/R\Omega^2 = 0.4905$, $\theta_* = 60.63^\circ$. The turning points $\cos\theta = 0$ and $0.981$ sit symmetrically about $\cos\theta_* = 0.4905$ — a coincidence of this $h$ being quadratic in $\cos\theta$, and a check.

13 · energy

Where the motor's work goes

The hoop pushes on the bead sideways to keep it turning at $\Omega$. The bead's angular momentum about the axis is $L_z = mR^2\Omega\sin^2\theta$, so

$$h = E - \Omega L_z .$$

reconciliation $$\frac{\mathrm dh}{\mathrm dt} = 0 \iff \frac{\mathrm dE}{\mathrm dt} = \Omega\,\frac{\mathrm dL_z}{\mathrm dt} = \Omega\,\tau_z ,$$

which is the power the hoop delivers: torque × angular speed.

the numbers close

$L_z$ falls from $mR^2\Omega = 0.02\ \mathrm{kg\,m^2\,s^{-1}}$ at $90^\circ$ to $0.000753$ at $11.19^\circ$. Work by the motor $= \Omega\,\Delta L_z = 10\times(-0.019247) = -0.1925$ J $= \Delta E$ ✓. Limit $\Omega\to0$: $h\to E$, no work, a pendulum ✓.

The hoop seen from above. Of the hoop's force, only the sideways part $N_\phi$ does work, against the bead's sideways velocity $R\Omega\sin\theta$: power $= N_\phi\,R\Omega\sin\theta$. Its torque about the axis is $N_\phi\,R\sin\theta$. The ratio is $\Omega$.

Reading the figure. Looking down the axis, the hoop is a line through it, turning at $\Omega$, and the bead is a point on that line at distance $R\sin\theta$ from the axis. The bead's velocity has a part along the hoop, $R\dot\theta$ (not drawn), and the sideways part $R\Omega\sin\theta$ drawn here. A frictionless hoop exerts no force along itself, and the force's radial part is perpendicular to the velocity, so the sideways part $N_\phi$ is all that works. The proof below is the figure written in components.

Full proof that the hoop's power is $\Omega\tau_z$

The bead's velocity is $R\dot\theta\,\hat e_\theta + R\Omega\sin\theta\,\hat e_\phi$, where $\hat e_\theta$ is along the hoop and $\hat e_\phi$ is horizontal, perpendicular to the plane of the hoop. A frictionless hoop exerts a force $\vec N$ with no $\hat e_\theta$ component, so $\vec N = N_r\hat e_r + N_\phi\hat e_\phi$ with $\hat e_r$ pointing from the centre to the bead. The velocity has no $\hat e_r$ component (the bead stays at distance $R$), so the power is $\vec N\cdot\dot{\vec r} = N_\phi R\Omega\sin\theta$. The torque of $\vec N$ about the vertical axis is $\tau_z = N_\phi\cdot R\sin\theta$ (lever arm $R\sin\theta$); gravity is parallel to the axis and has no $z$-torque. So power $= \Omega\tau_z$, and $\tau_z = \mathrm dL_z/\mathrm dt$ by unit 08's torque law. By the work–energy theorem the power equals $\mathrm dE/\mathrm dt$. ∎

The identity $h = E - \Omega L_z$. $E - \Omega L_z = \frac12mR^2\dot\theta^2 + \frac12mR^2\Omega^2\sin^2\theta - mgR\cos\theta - mR^2\Omega^2\sin^2\theta$, which is the $h$ of the previous slide. It is not special to the hoop: whenever coordinates rotate uniformly at $\Omega$ about a fixed axis, $h = E - \Omega L_z$. That is Jacobi's integral, the conserved quantity of the restricted three-body problem in the frame rotating with the two primaries. The free-particle slide that follows — one particle, two coordinate systems — derives it in rotating coordinates; problem P7 does the translating analogue, where a uniformly moving frame gives $h = E - uP_x$.

Who drives whom. On the swing inwards, from $90^\circ$ to $11.19^\circ$, $L_z$ falls and the work done on the bead is negative: the bead drives the motor. On the swing back out the motor returns the $0.1925$ J. Over a full oscillation the books balance and $E$ returns to its starting value.

Which is "the" energy? Both are honest. $E$ is the bead's energy in the inertial frame and is not conserved because the bead is not isolated: a motor is holding $\Omega$ fixed. $h$ is what Noether's theorem attaches to the time-independence of the Lagrangian in these coordinates. Problem P6 does the same bookkeeping from a different release point, with the motor doing positive work.

13 · energy

$h$ belongs to the coordinates, not to the motion

one free particle, two coordinate systems

Inertial polar coordinates: $L = \frac12m(\dot\rho^2 + \rho^2\dot\varphi^2 + \dot z^2)$, and $h = E$.

Rotating ones, $\varphi = \varphi' + \Omega t$:

$$L' = \tfrac12m\big(\dot\rho^2 + \rho^2(\dot\varphi' + \Omega)^2 + \dot z^2\big),$$

still with no explicit $t$, and

$$h' = E - \Omega L_z .$$

Both conserved, and they differ. "The conserved quantity from time translation" depends on which time translation: at fixed $\varphi$ or at fixed $\varphi'$.

the running example

Cartesian coordinates in the centre-of-mass frame; $T$ is quadratic, so

$$\begin{aligned}h = T + V = E &= -\frac{Gm_Sm_J}{2a}\\ &= -1.618\times10^{35}\ \mathrm J .\end{aligned}$$

In coordinates co-rotating with Jupiter's mean motion, the same Noether argument gives $E - \Omega L_z$ instead — the form the three-body problem uses.

Every step of the rotating case. $\dot\varphi = \dot\varphi' + \Omega$, so $L' = \frac12m(\dot\rho^2 + \rho^2\dot\varphi'^2 + 2\Omega\rho^2\dot\varphi' + \Omega^2\rho^2 + \dot z^2)$. The pieces: $L_2 = \frac12m(\dot\rho^2 + \rho^2\dot\varphi'^2 + \dot z^2)$, $L_1 = m\Omega\rho^2\dot\varphi'$, $L_0 = \frac12m\Omega^2\rho^2$. Then $h' = L_2 - L_0 = \frac12m(\dot\rho^2 + \rho^2\dot\varphi'^2 + \dot z^2 - \Omega^2\rho^2)$. Substituting $\dot\varphi' = \dot\varphi - \Omega$: $\rho^2\dot\varphi'^2 - \Omega^2\rho^2 = \rho^2\dot\varphi^2 - 2\Omega\rho^2\dot\varphi$, so $h' = E - \Omega\,m\rho^2\dot\varphi = E - \Omega L_z$. ∎ Both $E$ and $L_z$ are conserved for a free particle, so $h'$ is too; the point is only that $h'\ne E$.

Why two different "time translations". In inertial coordinates, shifting the clock moves the curve $t\mapsto(\rho,\varphi,z)(t)$ to $(\rho,\varphi,z)(t+s)$. In rotating coordinates, shifting the clock at fixed $\varphi'$ is shifting the clock and rotating by $-\Omega s$, as seen from the inertial frame. So $h' = h + (-\Omega)\times(\text{rotation charge})$ by the linearity remark on the theorem slide: the charges add as the generators do.

The running-example number. $E = -k/2a$ with $k = 2.51905\times10^{47}\ \mathrm{kg\,m^3\,s^{-2}}$, $a = 7.78567\times10^{11}$ m, gives $-1.61775\times10^{35}$ J. The relation between energy and semi-major axis is unit 08's. In the frame where the centre of mass moves, $h$ gains the constant $\frac12M\lvert\dot{\vec R}\rvert^2$, also conserved.

section 4

Space: momentum, angular momentum, and the boost

Three symmetries of the running example, three charges, and the one that is only quasi-invariant.

13 · space

Translations: momentum, and Newton's third law as a theorem

Translate both bodies by $s\hat n$: $\xi_S = \xi_J = \hat n$, $\delta L = 0$ (the example slide), $F = 0$. The charge is

$$Q = \sum_a\frac{\partial L}{\partial\dot{\vec r}_a}\cdot\hat n = \hat n\cdot(m_S\dot{\vec r}_S + m_J\dot{\vec r}_J) = \hat n\cdot\vec P .$$

Three independent $\hat n$: all three components of the total momentum $\vec P$ are conserved.

check

$[Q] = \mathsf{MLT^{-1}}$ ✓. One body in a fixed potential $V(\vec r)$: no translation symmetry unless $V$ is constant, and then $\vec p$ is conserved — Newton I ✓.

Newton III, derived

If $V$ depends only on $\vec r_S - \vec r_J$, then

$$\vec F_{\text{on }S} = -\nabla_{\vec r_S}V = +\nabla_{\vec r_J}V = -\vec F_{\text{on }J}.$$

Chain rule on $V(\vec r_S - \vec r_J)$. Equal and opposite forces are homogeneity of space.

Shift both bodies by the same $s\hat n$ and the separation $\vec r_J - \vec r_S$ — all that $V$ sees — does not change. The forces lie along it, equal and opposite. Schematic.

Reading the figure. Solid discs are the bodies, dashed circles the shifted copies; the two thin arrows from Sun to Jupiter are the separation before and after, the same vector. Because $V$ depends on the positions only through that vector, $\partial V/\partial\vec r_S = -\partial V/\partial\vec r_J$, and the two force arrows are that statement drawn. Positions are schematic, not to scale.

The logic reversed. Newton postulated the third law and derived momentum conservation. Here the postulate is that the Lagrangian of an isolated system does not change when the whole system is moved — space has no preferred point — and both momentum conservation and the third law (for forces derived from a potential) are consequences. That is more economical and it is more general: it survives into field theory, where "action and reaction" between a charge and the field is not even well defined but momentum conservation is (unit 29).

Partial symmetry. If $L$ is invariant under translations in some directions only — a projectile in uniform gravity $V = Mgz$ is invariant under horizontal shifts — then only those components of $\vec P$ are conserved. For the vertical shift, $\delta L = -Mg$, which is $\frac{\mathrm d}{\mathrm dt}(-Mgt)$: a quasi-symmetry after all, with charge $P_z + Mgt$. The checkpoint slide's cold try is this, and problem P1 has the same structure for a uniform electric field.

The cyclic-coordinate view. In coordinates $(\vec R, \vec r)$ — centre of mass and separation — $L = \frac12M\lvert\dot{\vec R}\rvert^2 + \frac12\mu\lvert\dot{\vec r}\rvert^2 - V(\vec r)$, and $\vec R$ does not appear: it is cyclic (unit 12), and its momentum $M\dot{\vec R} = \vec P$ is conserved. Noether's theorem is the coordinate-free version: no need to find the coordinates in which the symmetry is a shift of one of them.

13 · space

Rotations: all three components of angular momentum

Rotate both bodies about $\hat n$: $\xi_a = \hat n\times\vec r_a$, $\delta L = 0$, $F = 0$. The charge, using the triple product $\vec p\cdot(\hat n\times\vec r) = \hat n\cdot(\vec r\times\vec p)$:

$$\begin{aligned}Q &= \sum_a\vec p_a\cdot(\hat n\times\vec r_a)\\ &= \hat n\cdot\sum_a\vec r_a\times\vec p_a = \hat n\cdot\vec L .\end{aligned}$$

Every axis $\hat n$ gives one component: the whole vector $\vec L$ is conserved, and the orbit is planar (unit 08).

the running example's value

In the centre-of-mass frame, $\vec L = \mu\,\vec r\times\dot{\vec r}$ for the relative motion, with

$$\begin{aligned}\lvert\vec L\rvert &= \mu\sqrt{G(m_S+m_J)\,a(1-e^2)}\\ &= 1.926\times10^{43}\ \mathrm{kg\,m^2\,s^{-1}} .\end{aligned}$$

checks

$[Q] = \mathsf{ML^2T^{-1}}$ ✓. A potential depending on $x^2+y^2$ only: rotation about $\hat z$ alone, and only $L_z$ survives ✓ (problem P1).

Why three rotations give a vector. The generator depends linearly on $\hat n$, so the charge $Q(\hat n) = \hat n\cdot\vec L$ is a linear function of $\hat n$. A linear functional on $\mathbb R^3$ is the dot product with one vector, by the Riesz representation in finite dimensions (linear algebra). That vector is $\vec L$. Unit 16 will prefer to say the charge is an antisymmetric bilinear form — the "angular momentum" of a rotation in the plane of two vectors — and the vector is how three dimensions encode it.

The reduction to relative motion. With $\vec r_S = \vec R - (m_J/M)\vec r$ and $\vec r_J = \vec R + (m_S/M)\vec r$, $\sum_a m_a\vec r_a\times\dot{\vec r}_a = M\vec R\times\dot{\vec R} + \mu\vec r\times\dot{\vec r}$; the cross terms cancel because $m_S(m_J/M) - m_J(m_S/M) = 0$. In the centre-of-mass frame $\vec R = 0$.

The number. $\mu = m_Sm_J/M = 1.89631\times10^{27}$ kg, $G(m_S + m_J) = 1.32839\times10^{20}\ \mathrm{m^3\,s^{-2}}$, $a(1-e^2) = 5.1920$ AU (the semi-latus rectum $p$ of unit 08). Then $\lvert\vec L\rvert = 1.92620\times10^{43}\ \mathrm{kg\,m^2\,s^{-1}}$, and $L^2/\mu k = 5.1920$ AU recovers $p$ ✓.

A caution about symbols. $L$ is the Lagrangian and $\vec L$ (with an arrow, or $\lvert\vec L\rvert$) is angular momentum. The clash is universal and the arrow is the only thing separating them on these slides.

13 · space · example

The boost, where $L$ is not invariant

the Galilean boost on the running example

1. Family. $\vec r_a\mapsto\vec r_a + s\,t\,\hat n$, the view from a frame moving at velocity $-s\hat n$.

2. Generator. $\xi_a = t\hat n$, $\ \dot\xi_a = \hat n$.

3. $\delta L$. $V$ sees $\xi_S - \xi_J = 0$; $T$ gives $\sum_am_a\dot{\vec r}_a\cdot\hat n$:

$$\delta L = \hat n\cdot\big(m_S\dot{\vec r}_S + m_J\dot{\vec r}_J\big) = \frac{\mathrm d}{\mathrm dt}\big(M\,\hat n\cdot\vec R\big).$$

4. So $F = M\hat n\cdot\vec R$ with $M\vec R = m_S\vec r_S + m_J\vec r_J$: a quasi-symmetry, not a symmetry.

5. Charge. $Q = \sum_a\vec p_a\cdot t\hat n - M\hat n\cdot\vec R = \hat n\cdot\big(\vec P\,t - M\vec R\big)$.

the boost charge $$\vec G = \vec P\,t - M\vec R\quad\text{is conserved.}$$

Since $\vec P$ is conserved too, $\vec R(t) = \vec R(0) + (\vec P/M)\,t$: the centre of mass moves uniformly.

Without $F$ the "charge" would be $\vec Pt$, which is not conserved. The extra term is the whole content.

Why the boost cannot be a strict symmetry. Kinetic energy is not Galilean invariant: a body at rest in one frame is moving in another. So $L$ must change. What Galilean relativity requires is only that the equations of motion be the same in both frames, and unit 12's gauge freedom says that is guaranteed exactly when $L$ changes by a total derivative. The finite version is on the example slide's notes: $L_s - L = \frac{\mathrm d}{\mathrm dt}(sM\hat n\cdot\vec R + \frac12Ms^2t)$.

One body in a fixed potential. For unit 08's $L = \frac12\mu\lvert\dot{\vec r}\rvert^2 + k/r$ the boost $\vec r\mapsto\vec r + st\hat n$ changes $V$: $\delta V = \nabla V\cdot t\hat n \ne 0$, so it is not even a quasi-symmetry. The fixed centre is an external body that was not boosted. That is why the running example keeps both bodies.

13 · space

The centre of mass moves in a straight line

$\vec G = \vec Pt - M\vec R$ conserved, with $\vec P$ conserved, says

$$M\vec R(t) = \vec P\,t - \vec G(0):$$

not a number but a whole trajectory. However the two bodies swing about each other, the point between them weighted by mass moves at constant velocity.

checks

Dimensions: $[\vec Pt] = [M\vec R] = \mathsf{ML}$ ✓, the table's boost row. Verify directly: $\dot{\vec G} = \vec P + \dot{\vec P}t - M\dot{\vec R} = \vec P + 0 - \vec P = \vec 0$ ✓.

One coordinate of each body against time, for a two-body orbit seen from a frame in which the pair drifts. Each body wiggles; the barycentre (dashed) does not. Mass ratio exaggerated to $1:3$ so the Sun's wiggle shows.

Reading the figure. The curves are computed for a circular relative orbit of period 2 (in the plot's units) seen edge-on, so the relative coordinate is $\cos(\pi t)$, with $m_J/M = 0.25$ and the barycentre drifting at constant speed. The Sun's coordinate is $X(t) - \tfrac14\cos\pi t$ and Jupiter's $X(t) + \tfrac34\cos\pi t$: they swing in opposition, with amplitudes in the inverse ratio of the masses, so that $m_Sx_S + m_Jx_J$ is always $MX(t)$, a straight line. For the real pair the Sun's wiggle would be $1/1048$ of Jupiter's and invisible at this scale — the next slide measures it.

Is $\vec G$ really new? It is explicitly time-dependent, and conservation of such a quantity is a statement about the trajectory, not just about a number: $\vec G = \vec G(0)$ says $M\vec R(t) = \vec P t - \vec G(0)$, a straight line at constant speed. Newton would have called it a corollary of the third law; here it is a symmetry of empty space under changes of inertial frame, and that is how relativity will see it. In unit 27 the Lorentz boosts replace the Galilean ones, the boost charge becomes "the centre of energy moves uniformly", and $\vec G$ joins $\vec L$ as the other half of an antisymmetric $4\times4$ array.

13 · space

What the boost charge says about the Sun

In the frame where $\vec P = 0$, $\vec G$ conserved means $\vec R$ is fixed: take it as the origin. Then $m_S\vec r_S + m_J\vec r_J = \vec 0$, so

$$\vec r_S = -\frac{m_J}{M}\,\vec r,\qquad \dot{\vec r}_S = -\frac{m_J}{M}\,\dot{\vec r}.$$

numbers, from the running example

$m_J/M$	$9.537\times10^{-4}$
Sun's distance from the barycentre, at $r = a$	$7.425\times10^{8}$ m $= 1.067\,R_\odot$
Sun's speed about it, at $r = a$	$12.46\ \mathrm{m\,s^{-1}}$
the same, for the Earth alone	$0.0895\ \mathrm{m\,s^{-1}}$

To scale: the barycentre sits $1.07$ solar radii from the Sun's centre, just outside its surface, and the Sun's centre circles it once per Jupiter year. Jupiter itself is about 1100 solar radii off to the right.

A $12\ \mathrm{m\,s^{-1}}$ wobble, repeating every $11.87$ years, is what a distant astronomer would see in the Sun's spectrum: the radial-velocity method by which most early exoplanets were found.

The numbers. $m_S = GM_\odot/G = 1.98841\times10^{30}$ kg and $m_J = 1.89812\times10^{27}$ kg, so $m_J/M = 9.53684\times10^{-4}$. At $r = a = 7.78567\times10^{11}$ m, the Sun's distance from the barycentre is $7.42507\times10^{8}$ m; with $R_\odot = 6.957\times10^{8}$ m that is $1.0673\,R_\odot$ (from $1.015$ at perihelion to $1.120$ at aphelion). The relative speed at $r = a$ is $\sqrt{G(m_S+m_J)/a} = 13.062\ \mathrm{km\,s^{-1}}$ (vis-viva, unit 08), and the Sun's share is $13062\times9.53684\times10^{-4} = 12.457\ \mathrm{m\,s^{-1}}$. For the Earth, $m_\oplus/(m_S + m_\oplus) = 3.0035\times10^{-6}$ times $29.785\ \mathrm{km\,s^{-1}}$ gives $0.08946\ \mathrm{m\,s^{-1}}$. The period from $a$ is $2\pi\sqrt{a^3/G(m_S+m_J)} = 11.867$ yr; the catalogued sidereal period is $11.862$ yr, the difference coming from the other planets and from the mean elements used.

Reading the figure. Drawn to scale at $r = a$: the Sun's disc of radius $R_\odot$ and the barycentre at $1.0673\,R_\odot$ from its centre, so the Sun's centre runs round the dashed circle of that radius. At $r = a$ Jupiter is $a/R_\odot = 1119$ solar radii away, far off the figure in the direction of the arrow; the barycentre divides that distance in the ratio $m_J : m_S$.

Why this is the boost charge and not just "momentum". Momentum conservation alone says $\dot{\vec R}$ is constant. The boost charge says where $\vec R$ is: $M\vec R(t) = \vec Pt - \vec G$. Together they fix the barycentre's whole worldline, and the statement "the Sun circles a fixed point outside itself" uses the second.

History. The radial-velocity method found the first planet round a Sun-like star, 51 Pegasi b, in 1995 (Mayor and Queloz). It is a "hot Jupiter" with a reflex velocity of order $50\ \mathrm{m\,s^{-1}}$, much larger than the Sun's $12$, which is why it was found first. An Earth analogue's $9\ \mathrm{cm\,s^{-1}}$ is still at the edge of what spectrographs can do.

13 · space

Ten symmetries, ten charges

The transformations used so far compose into the Galilean group: every map

$$(t,\vec r)\mapsto(t + t_0,\ R\vec r + \vec v\,t + \vec d),\qquad R\in SO(3),$$

with $1 + 3 + 3 + 3 = 10$ parameters. An isolated system is invariant (or quasi-invariant) under all of it, so it has ten conserved charges:

symmetry	count	charge
time shift $t_0$	1	$E$
space shift $\vec d$	3	$\vec P$
rotation $R$	3	$\vec L$
boost $\vec v$	3	$\vec G$

spending them on the running example

$\vec P$ and $\vec G$ say $\vec R$ moves uniformly. Pass to the frame where it is fixed and change coordinates to $(\vec R,\vec r)$:

$$L = \tfrac12M\lvert\dot{\vec R}\rvert^2 + \underbrace{\tfrac12\mu\lvert\dot{\vec r}\rvert^2 + \frac{k}{r}}_{L_{\mathrm{rel}}} .$$

Six charges spent, and what is left is unit 08's one-body problem, with $E$ and $\vec L$ still available.

That these maps form a group is a direct check: composing $(t,\vec r)\mapsto(t + t_0, R\vec r + \vec vt + \vec d)$ with another such map gives a map of the same form (the composite boost is $R_2\vec v_1 + \vec v_2$, and the composite shift picks up a term $\vec v_2t_{0,1}$). Inverses exist. Each of the four kinds is a one-parameter group along any fixed direction, and the whole group is ten-dimensional in the sense that ten real numbers parametrise it near the identity. Unit 26 derives the Lorentz group by exactly the same kind of counting, and finds ten again.

Why the separation works. With $\vec r_S = \vec R - (m_J/M)\vec r$ and $\vec r_J = \vec R + (m_S/M)\vec r$, the kinetic energy becomes $\frac12M\lvert\dot{\vec R}\rvert^2 + \frac12\mu\lvert\dot{\vec r}\rvert^2$ with $\mu = m_Sm_J/M$: the cross terms are proportional to $m_S(-m_J/M) + m_J(m_S/M) = 0$. This is unit 08's reduction, done once there, and unit 12's form invariance says the Euler–Lagrange equations in the new coordinates are the right ones without further checking.

What remains. The relative Lagrangian has time-translation and rotation symmetry, so $E$ and $\vec L$: four charges. The translation and boost of $\vec r$ alone are no longer symmetries, since $k/r$ has a preferred point. Unit 08 showed four is enough to reduce the orbit to quadrature. It also found a fifth, sixth and seventh — the Runge–Lenz vector — which the Galilean group does not explain. That is section 5.

section 5

The hidden symmetry

A conserved vector that no transformation of space produces, and the transformation that does.

13 · hidden

A conserved vector with no symmetry of space behind it

Unit 08, for the relative Lagrangian: the Runge–Lenz vector

$$\vec A = \vec p\times\vec L - \mu k\,\hat r,\qquad \vec p = \mu\dot{\vec r},$$

is conserved, points at perihelion, and has $\lvert\vec A\rvert = \mu k\,e$. For Jupiter, $e = 0.0489$.

Here $\mu$ is the reduced mass, $k = Gm_Sm_J$, $\vec L = \vec r\times\vec p$ and $e$ the eccentricity. Its two pieces change from point to point of the orbit; their sum does not.

The symmetries of section 4 moved points of space. The next slide shows that no such transformation, even combined with a time shift, can give $\vec A$.

At four points of an orbit, $\vec p\times\vec L$ followed tip to tail by $-\mu k\,\hat r$ lands on the same short vector $\vec A$ (thick arrow), pointing at perihelion. Computed with $\mu = k = a = 1$ and the eccentricity exaggerated to $0.5$; Jupiter's $\vec A$ would be a tenth as long.

Reading the figure. The orbit is $r = a(1-e^2)/(1 + e\cos\theta)$ with $a = 1$, $e = 0.5$, $\mu = k = 1$, traversed counterclockwise, so $\vec L$ points out of the page with $\lvert\vec L\rvert = \sqrt{\mu k a(1-e^2)} = 0.866$. At each marked point, $\vec p\times\vec L$ lies in the plane, perpendicular to the velocity, with length $\lvert\vec p\rvert\lvert\vec L\rvert$ — longest near perihelion, where the body is fastest. The second arrow, $-\mu k\hat r$, always has length $\mu k = 1$ and points back at the Sun. Their sum was computed at each point and is $(0.5, 0)$ to six decimals every time: $\lvert\vec A\rvert = \mu ke$, aimed at perihelion. For Jupiter $e = 0.0489$, about a tenth of the $0.5$ drawn.

Why it points at perihelion, from the picture: at perihelion the velocity is perpendicular to $\vec r$, so $\vec p\times\vec L$ points straight out along $\hat r$, and both arrows lie on the line through the Sun and the perihelion; there $\lvert\vec p\times\vec L\rvert = \mu k(1+e)$ beats $\mu k$ by $\mu ke$.

13 · hidden

No point symmetry produces $\vec A$

no point symmetry gives it

If $\xi = \xi(q,t)$ then $F = F(q,t)$, and $Q = \mu\dot{\vec r}\cdot\xi - F$ is affine in $\dot{\vec r}$. Time translation adds a multiple of $h$, whose velocity-quadratic part is a multiple of $\lvert\dot{\vec r}\rvert^2$. But

$$\begin{aligned}\hat e\cdot\vec A = {}&\mu^2\big(\lvert\dot{\vec r}\rvert^2(\hat e\cdot\vec r) - (\vec r\cdot\dot{\vec r})(\hat e\cdot\dot{\vec r})\big)\\ &- \mu k\,\frac{\hat e\cdot\vec r}{r},\end{aligned}$$

whose quadratic part is not a multiple of $\lvert\dot{\vec r}\rvert^2$.

so the search space was too small

Configuration-space transformations, even with time shifts, cannot produce $\vec A$. The theorem, though, never asked $\xi$ to be a function of position: it allowed $\xi(q,\dot q,t)$.

the plan

Find a velocity-dependent $\xi$ whose $\delta L$ is a total derivative, and check that its charge is $\hat e\cdot\vec A$. It exists; the next two slides give it and verify it.

Full proof that no point symmetry gives $\vec A$

Step 1: $F$ has no velocities. If $\xi = \xi(\vec r,t)$ then $\dot\xi = D\xi\,\dot{\vec r} + \partial_t\xi$ has no $\ddot{\vec r}$, so $\delta L$ is a function of $(\vec r,\dot{\vec r},t)$ alone. If $F$ depended on $\dot{\vec r}$, $\dot F$ would contain $\partial_{\dot{\vec r}}F\cdot\ddot{\vec r}$, and since $\ddot{\vec r}$ can be chosen freely at any point of any curve, the identity $\delta L = \dot F$ forces $\partial_{\dot{\vec r}}F = 0$.

Step 2: the charge is affine. $Q = \mu\dot{\vec r}\cdot\xi(\vec r,t) - F(\vec r,t)$ is affine in $\dot{\vec r}$. Adding a multiple $c\,h$ of the time-translation charge (a combined generator $\xi + c\dot{\vec r}$) adds $c(\frac12\mu\lvert\dot{\vec r}\rvert^2 - k/r)$. So every such charge has velocity-quadratic part $\frac c2\mu\lvert\dot{\vec r}\rvert^2$, i.e. the quadratic form $\frac c2\mu\,I$.

Step 3: $\hat e\cdot\vec A$ is not of that form. By BAC–CAB, $\vec p\times(\vec r\times\vec p) = \vec r\lvert\vec p\rvert^2 - \vec p(\vec r\cdot\vec p)$, which gives the displayed formula. Its quadratic part in $\vec v = \dot{\vec r}$ is the form $\mu^2\big(\lvert\vec v\rvert^2(\hat e\cdot\vec r) - (\vec r\cdot\vec v)(\hat e\cdot\vec v)\big)$, with matrix $\mu^2\big((\hat e\cdot\vec r)I - \frac12(\vec r\hat e^{\top} + \hat e\vec r^{\top})\big)$. This is a multiple of $I$ only if $\vec r\hat e^{\top} + \hat e\vec r^{\top}$ is. But that matrix sends any vector perpendicular to both $\vec r$ and $\hat e$ to $0$, so it would have to be $0$, which forces $\vec r = 0$. At any $\vec r\ne0$ the forms differ. ∎

Even allowing time to be reparametrised by a function $\tau(\vec r,t)$ does not help: that adds $-\tau h$, whose quadratic part is still a multiple of $\lvert\vec v\rvert^2$ at each point. Problem P11 is this proof, set as an exercise with the phase-space rescue attached.

13 · hidden

Letting the transformation see the velocity

a generalised symmetry

A variation $\delta q = \xi(q,\dot q,t)$ with $\delta L = \dot F$ for every curve, $F = F(q,\dot q,t)$. Noether's three lines apply unchanged.

For $L_{\mathrm{rel}}$ and a fixed unit vector $\hat e$, take

$$\xi = \mu\Big(2\,(\hat e\cdot\vec r)\,\dot{\vec r} - (\hat e\cdot\dot{\vec r})\,\vec r - (\vec r\cdot\dot{\vec r})\,\hat e\Big),$$ $$F = \mu^2\Big(\lvert\dot{\vec r}\rvert^2(\hat e\cdot\vec r) - (\vec r\cdot\dot{\vec r})(\hat e\cdot\dot{\vec r})\Big) + \mu k\,\frac{\hat e\cdot\vec r}{r}.$$

claim

$\delta L_{\mathrm{rel}} = \dot F$ for every curve, and the charge is

$$Q = \mu\dot{\vec r}\cdot\xi - F = \hat e\cdot\vec A .$$

Verified on the next slide.

what kind of object this is

$\xi$ moves a point $\vec r$ by an amount depending on $\dot{\vec r}$. So it is not a map of configuration space; it acts on states $(\vec r,\dot{\vec r})$. Its natural home is phase space, unit 14.

dimensions

$[\xi] = \mathsf{M\,L^2T^{-1}}$, so $[s] = \mathsf{L}/[\xi] = \mathsf{M^{-1}L^{-1}T}$ and $[Q] = \mathsf{ML^2T^{-1}}/[s] = \mathsf{M^2L^3T^{-2}} = [\mu k]$ ✓.

Where the formula comes from. Guess $\xi = \mu\big(\alpha(\hat e\cdot\vec r)\dot{\vec r} + \beta(\hat e\cdot\dot{\vec r})\vec r + \gamma(\vec r\cdot\dot{\vec r})\hat e\big)$ — the most general vector built from $\vec r$, $\dot{\vec r}$, $\hat e$ that is linear in $\hat e$, linear in $\dot{\vec r}$ and linear in $\vec r$. Then demand that $\delta L$ be a total derivative. Two conditions come out: the terms containing $\ddot{\vec r}$ must assemble into a derivative, which forces $\gamma = \beta$; and the leftover $\lvert\dot{\vec r}\rvert^2(\hat e\cdot\dot{\vec r})$ term must vanish, which forces $\beta = -\alpha/2$. With $\alpha = 2$ the potential terms then assemble into $\mu k\,\frac{\mathrm d}{\mathrm dt}(\hat e\cdot\vec r/r)$ on their own — and that step works only for the $1/r$ potential. Problem P10 is this derivation.

A tidier form. Using BAC–CAB twice, $\xi = \vec L\times\hat e + \vec r\times(\vec p\times\hat e)$ with $\vec L = \vec r\times\vec p$. The second piece is $\vec r\times(\vec p\times\hat e) = (\hat e\times\vec p)\times\vec r$: a rotation, but about the state-dependent axis $\hat e\times\vec p$. The first, $\vec L\times\hat e$, does not depend on $\vec r$ at all — it is a translation by a state-dependent vector. Neither piece alone is a generalised symmetry.

History. The velocity-dependent generator of the Runge–Lenz vector was written down in this Lagrangian form by Lévy-Leblond, "Conservation laws for gauge-variant Lagrangians in classical mechanics", Am. J. Phys. 39 (1971) 502, whose point was precisely that quasi-invariance is needed. The underlying symmetry was found first in quantum mechanics: Pauli (1926) used the quantum Runge–Lenz vector to get the hydrogen spectrum before Schrödinger's equation was solved for it, and Fock (1935) showed the bound-state symmetry group is $SO(4)$.

13 · hidden · example

The Runge–Lenz generator, checked

every step, off-shell

Compute $\delta L_{\mathrm{rel}} = -\dfrac{k}{r^3}\,\vec r\cdot\xi + \mu\,\vec v\cdot\dot\xi$ and compare it with $\dot F$, where $\vec v = \dot{\vec r}$ and $\vec a = \ddot{\vec r}$ are arbitrary: no equation of motion is used.

1. Differentiate $\xi$. $\dot\xi/\mu = 2(\hat e\cdot\vec v)\vec v + 2(\hat e\cdot\vec r)\vec a - (\hat e\cdot\vec a)\vec r - (\hat e\cdot\vec v)\vec v - (\lvert\vec v\rvert^2 + \vec r\cdot\vec a)\hat e$.

2. Kinetic part. Dot with $\mu\vec v$:

$$\begin{aligned}\mu\vec v\cdot\dot\xi = \mu^2\big[&2(\hat e\cdot\vec r)(\vec v\cdot\vec a) - (\vec r\cdot\vec v)(\hat e\cdot\vec a)\\ &- (\hat e\cdot\vec v)(\vec r\cdot\vec a)\big].\end{aligned}$$

(The $\lvert\vec v\rvert^2(\hat e\cdot\vec v)$ terms cancel: $2 - 1 - 1 = 0$.)

3. That is exactly $\dfrac{\mathrm d}{\mathrm dt}\mu^2\big[\lvert\vec v\rvert^2(\hat e\cdot\vec r) - (\vec r\cdot\vec v)(\hat e\cdot\vec v)\big]$ — expand the derivative and the $\lvert\vec v\rvert^2(\hat e\cdot\vec v)$ terms cancel again.

4. Potential part. $\vec r\cdot\xi = \mu\big[2(\hat e\cdot\vec r)(\vec r\cdot\vec v) - r^2(\hat e\cdot\vec v) - (\vec r\cdot\vec v)(\hat e\cdot\vec r)\big] = \mu\big[(\hat e\cdot\vec r)(\vec r\cdot\vec v) - r^2(\hat e\cdot\vec v)\big]$, so

$$\begin{aligned}-\frac{k}{r^3}\vec r\cdot\xi &= \mu k\Big[\frac{\hat e\cdot\vec v}{r} - \frac{(\hat e\cdot\vec r)(\vec r\cdot\vec v)}{r^3}\Big]\\ &= \mu k\frac{\mathrm d}{\mathrm dt}\frac{\hat e\cdot\vec r}{r}.\end{aligned}$$

5. Steps 3 and 4 add to $\dot F$ ✓. 6. Charge. $\mu\vec v\cdot\xi = \mu^2\big[2\lvert\vec v\rvert^2(\hat e\cdot\vec r) - 2(\hat e\cdot\vec v)(\vec r\cdot\vec v)\big]$, and subtracting $F$:

$$\begin{aligned}Q &= \mu^2\big[\lvert\vec v\rvert^2(\hat e\cdot\vec r) - (\vec r\cdot\vec v)(\hat e\cdot\vec v)\big] - \mu k\frac{\hat e\cdot\vec r}{r}\\ &= \hat e\cdot\vec A\ \checkmark\end{aligned}$$

Step 3 in full. $\frac{\mathrm d}{\mathrm dt}\big[\lvert\vec v\rvert^2(\hat e\cdot\vec r)\big] = 2(\vec v\cdot\vec a)(\hat e\cdot\vec r) + \lvert\vec v\rvert^2(\hat e\cdot\vec v)$ and $\frac{\mathrm d}{\mathrm dt}\big[(\vec r\cdot\vec v)(\hat e\cdot\vec v)\big] = (\lvert\vec v\rvert^2 + \vec r\cdot\vec a)(\hat e\cdot\vec v) + (\vec r\cdot\vec v)(\hat e\cdot\vec a)$. Subtracting, the $\lvert\vec v\rvert^2(\hat e\cdot\vec v)$ terms cancel and what is left is step 2's bracket. ✓

Step 4 in full. $\frac{\mathrm d}{\mathrm dt}\frac{\hat e\cdot\vec r}{r} = \frac{\hat e\cdot\vec v}{r} - \frac{(\hat e\cdot\vec r)\dot r}{r^2}$ and $\dot r = \vec r\cdot\vec v/r$ (differentiate $r^2 = \vec r\cdot\vec r$), giving the bracket on the slide. This is the only step that uses the potential, and it uses $1/r$ exactly: for $V = -k/r^n$ the factor would be $r^{-n-2}$ and the bracket would not be a derivative.

Step 6 and unit 08. $\vec A = \vec p\times\vec L - \mu k\hat r$ with $\vec p\times\vec L = \mu^2\vec v\times(\vec r\times\vec v) = \mu^2\big(\vec r\lvert\vec v\rvert^2 - \vec v(\vec r\cdot\vec v)\big)$ by BAC–CAB, so $\hat e\cdot\vec A$ is the displayed $Q$. As $\hat e$ ranges over three independent directions, all three components of $\vec A$ are Noether charges.

An independent check on $Q$, in one line. Take a circular orbit of radius $r$, where $\vec r\cdot\vec v = 0$ and the force balance gives $\mu\lvert\vec v\rvert^2 = k/r$. Then $Q = \mu^2\lvert\vec v\rvert^2(\hat e\cdot\vec r) - \mu k(\hat e\cdot\vec r)/r = \mu(\hat e\cdot\vec r)\big(\mu\lvert\vec v\rvert^2 - k/r\big) = 0$ for every $\hat e$, so $\vec A = \vec 0$ — which is $\lvert\vec A\rvert = \mu ke$ with $e = 0$ ✓.

Where the hypothesis "every curve" was used. Steps 1 to 5 never used the equation of motion: $\vec a$ was arbitrary. Only when the theorem is applied does $\mu\vec a = -k\vec r/r^3$ enter, through $\dot p = \partial L/\partial q$ — which is unit 08's direct differentiation of $\vec A$, reorganised.

13 · hidden

The hidden SO(4)

For a bound orbit, $E \lt 0$. Rescale $\vec A$ to have the dimensions of angular momentum:

$$\vec M := \frac{\vec A}{\sqrt{-2\mu E}} .$$

the energy knows only $L^2 + M^2$ $$\lvert\vec L\rvert^2 + \lvert\vec M\rvert^2 = -\frac{\mu k^2}{2E}.$$

Proof. Unit 08: $\lvert\vec A\rvert^2 = \mu^2k^2 + 2\mu E\lvert\vec L\rvert^2$. Divide by $-2\mu E$. ∎

checks

$[\vec A/\sqrt{\mu E}] = \mathsf{M^2L^3T^{-2}}/\mathsf{M\,L\,T^{-1}} = \mathsf{ML^2T^{-1}}$ ✓. Circle: $\vec M = 0$, $\lvert\vec L\rvert^2 = \mu ka$ ✓.

Jupiter

$\lvert\vec L\rvert = 1.9262\times10^{43}$, $\ \lvert\vec M\rvert = e\sqrt{\mu ka} = 9.430\times10^{41}$, and $\sqrt{L^2 + M^2} = \sqrt{\mu ka} = 1.9285\times10^{43}\ \mathrm{kg\,m^2\,s^{-1}}$: the angular momentum of the circular orbit with the same energy.

Six conserved components, $\vec L$ and $\vec M$, with equal dimensions; the next slide draws what they say.

Their Poisson brackets (defined in unit 14) close on the Lie algebra of $SO(4)$, rotations in four dimensions; the energy depends only on the combination $L^2 + M^2$ that such rotations preserve.

Cited, proved in unit 14: $\{L_1,L_2\} = L_3$, $\{M_1,L_2\} = M_3$, $\{M_1,M_2\} = L_3$, and the same under cycling $1\to2\to3\to1$.

The numbers. $\mu k a = 1.89631\times10^{27}\cdot2.51905\times10^{47}\cdot7.78567\times10^{11}$ gives $\sqrt{\mu ka} = 1.92851\times10^{43}$; $\lvert\vec M\rvert = \lvert\vec A\rvert/\sqrt{-2\mu E}$ with $\lvert\vec A\rvert = \mu ke = 2.33591\times10^{73}$ and $E = -1.61775\times10^{35}$ J is $9.43040\times10^{41}$, which equals $e\sqrt{\mu ka}$ since $-2\mu E = \mu k/a$. Then $\sqrt{L^2 + M^2} = 1.92851\times10^{43}$ agrees with $\sqrt{\mu k a}$ to all digits, and $-\mu k^2/2(L^2 + M^2)$ returns $E = -1.61775\times10^{35}$ J ✓.

Why this is $SO(4)$ — cited. The Poisson bracket, defined in unit 14, turns conserved quantities into generators of flows on phase space and measures how two such flows fail to commute. The three brackets on the slide are the commutation relations of the six rotation generators of $\mathbb R^4$ (rotations in the planes $x_ix_j$ and $x_ix_4$), which is the claim; unit 14 computes them. Equivalently $\vec J_\pm = \frac12(\vec L\pm\vec M)$ satisfy two independent copies of the angular-momentum relations, and $L^2 + M^2 = 2(J_+^2 + J_-^2)$. For $E \gt 0$ the sign under the root flips and the algebra becomes that of the Lorentz group $SO(3,1)$ — unit 26's group, appearing in Newtonian gravity for scattering orbits.

Why physicists care. In quantum mechanics this symmetry is why the hydrogen energy levels depend on the principal quantum number alone: $n^2$ states share each energy, the dimension of an irreducible representation of $SO(4)$ of the relevant kind. Pauli computed the spectrum this way in 1926. The classical statement on this slide — $E$ depends only on $L^2 + M^2$ — is the same fact before quantisation.

13 · hidden

Orbits of one energy

Left: four orbits with the same semi-major axis $a$, hence the same energy $E = -k/2a$, sharing a focus. Right: each orbit is one point on the quarter circle $L^2 + M^2 = \mu ka$, at $(\sqrt{1-e^2},\ e)$ in units of $\sqrt{\mu ka}$. Jupiter sits next to the circular orbit.

The energy fixes only the radius of that circle, not where on it the orbit sits. Moving along it turns a circle into ever thinner ellipses without changing $E$ — something no rotation of space can do, since rotations keep $\lvert\vec L\rvert$.

Reading the figure. All four orbits have $a = 1$ (in units of Jupiter's $a$, say) and eccentricities $0, 0.5, 0.8, 0.95$; each is $r = a(1-e^2)/(1+e\cos\theta)$ about the common focus. For each, $\lvert\vec L\rvert = \sqrt{\mu ka(1-e^2)}$ (unit 08) and $\lvert\vec M\rvert = \lvert\vec A\rvert/\sqrt{-2\mu E} = \mu ke/\sqrt{\mu k/a} = e\sqrt{\mu ka}$, so the point $(\lvert\vec L\rvert, \lvert\vec M\rvert)/\sqrt{\mu ka} = (\sqrt{1-e^2}, e)$ lies on the unit circle: the previous slide's identity, drawn. Jupiter is the point $(0.99880, 0.0489)$.

What "hidden" means. The group generated by $\vec L$ and $\vec M$ is bigger than the rotation group of space, $SO(3)$, and it does not act on space: its extra generators are the velocity-dependent $\xi$ of the previous slides. It mixes orbits of the same energy but different angular momentum and eccentricity — a circle can be continuously turned into a thin ellipse of the same semi-major axis — which is why $E$ depends on $a$ alone and not on $e$.

13 · hidden

The converse is false, and what rescues it

the converse of Noether's theorem, for point symmetries

"Every conserved quantity is the charge of a transformation of configuration space" is false. $\vec A$ is a counterexample (the slide No point symmetry produces $\vec A$).

why it was never plausible

Conserved quantities are cheap. Any function of conserved quantities is conserved: $h^2$ is, and its generator is $\xi = 2h\,\dot q$, with $F = 2hL + h^2$ — velocity-dependent. And by unit 01 the initial state is a function of the state at time $t$, so every system has $2n$ independent time-dependent ones. Symmetries of configuration space are far scarcer.

the rescue: enlarge the space

Describe a state by $(q,p)$ instead of $(q,\dot q)$, and write the energy $h$ as a function $H(q,p)$ of them — unit 14's phase space and Hamiltonian. Then every function $G(q,p)$ generates a flow, the solutions of $\dot q = \partial G/\partial p$, $\dot p = -\partial G/\partial q$, and that flow leaves $H$ unchanged exactly when $G$ is conserved.

So on phase space the correspondence symmetry ⇔ charge is a bijection. Cited: proved in unit 14 with the Poisson bracket, where it takes one line.

The velocity-dependent $\xi$ for $\vec A$ is the shadow in $(q,\dot q)$ of the phase-space flow generated by $\vec A$.

Check the $h^2$ generator (for an autonomous $L$). With $\xi = 2h\dot q$, $\delta L = 2h(\partial_qL\cdot\dot q + \partial_{\dot q}L\cdot\ddot q) + 2\dot h\,p\cdot\dot q = 2h\dot L + 2\dot h\,p\cdot\dot q$, off-shell. And $\frac{\mathrm d}{\mathrm dt}(2hL + h^2) = 2\dot hL + 2h\dot L + 2h\dot h$. The difference is $2\dot h(p\cdot\dot q - L - h) = 0$ by the definition of $h$. So $F = 2hL + h^2$ and $Q = 2h\,p\cdot\dot q - 2hL - h^2 = 2h^2 - h^2 = h^2$ ✓. The same trick makes any function $f(Q)$ of a charge into a charge, of the generator $f'(Q)\,\xi$ — and that generator depends on the velocities whenever $Q$ does.

The initial-condition functions, and a caution. For the oscillator $\ddot q = -\omega^2q$, $I = q\cos\omega t - (\dot q/\omega)\sin\omega t$ is the initial position as a function of the state at time $t$: $\dot I = \dot q\cos\omega t - \omega q\sin\omega t - (\ddot q/\omega)\sin\omega t - \dot q\cos\omega t = 0$ on-shell. It is not a counterexample to the converse: it is the charge of the time-dependent translation $\xi = -\sin(\omega t)/(m\omega)$, a quasi-symmetry with $F = m\dot\xi\,q$ (check: $\delta L - \dot F = -mq(\ddot\xi + \omega^2\xi) = 0$ for every curve). The Runge–Lenz vector is the honest counterexample, because it is quadratic in the velocities in a way the no-point-symmetry proof rules out.

The rescue, stated precisely (cited, unit 14). Given $H(q,p)$ and any $G(q,p)$, the flow of $\dot q = \partial G/\partial p$, $\dot p = -\partial G/\partial q$ changes $H$ at rate $\{H, G\}$, and $G$ changes along the true motion at rate $\{G,H\} = -\{H,G\}$. So "$G$ is conserved" and "$G$'s flow preserves $H$" are the same equation. This is Noether's theorem and its converse in one line — the reason unit 14 calls it "Noether's converse, now cheap". The price is that the symmetries are transformations of phase space that mix positions and momenta, which have no picture in configuration space.

What the theorem does and does not say, summarised. It says: continuous (quasi-)symmetry of $L$ ⇒ conserved charge, explicitly. It does not say: every conserved quantity comes from a symmetry of space; a symmetry of the equations of motion gives a charge; a discrete symmetry gives one; or anything at all when the equations are not Euler–Lagrange equations.

13 · summary

One theorem, run five times

symmetry of Sun–Jupiter	$\xi$	$F$	charge $p\cdot\xi - F$	value (barycentric frame)
time shift	$\dot q$	$L$	$h = E$	$-1.618\times10^{35}$ J
space shift along $\hat n$	$\hat n$	$0$	$\hat n\cdot\vec P$	$0$
rotation about $\hat n$	$\hat n\times\vec r_a$	$0$	$\hat n\cdot\vec L$	$\lvert\vec L\rvert = 1.926\times10^{43}\ \mathrm{kg\,m^2\,s^{-1}}$
boost along $\hat n$	$t\hat n$	$M\hat n\cdot\vec R$	$\hat n\cdot(\vec Pt - M\vec R)$	$0$
hidden, along $\hat e$	velocity-dependent	quadratic $+\ \mu k\,\hat e\cdot\hat r$	$\hat e\cdot\vec A$	$\lvert\vec A\rvert/\mu k = 0.0489$

what the unit showed

Three laws that needed three tricks are one theorem. Two charges nobody asks for — $\vec G$ and $\vec A$ — need the total derivative $F$, and one of them needs $\xi$ to see the velocity.

and what to carry forward

$h$ is a charge, not "the energy": equal to $T + V$ only in coordinates that do not move. Unit 36's conserved energy of a black-hole orbit is an $h$.

Closing the running example. The Sun–Jupiter Lagrangian has been run through the theorem five times. The barycentric frame is the one in which $\vec P = 0$ and $\vec G = 0$ (the barycentre at rest at the origin), which is why those rows read $0$ — not because the charges are trivial but because they were spent choosing the frame. The remaining seven numbers, $E$, $\vec L$, $\vec A$, satisfy the two relations $\vec A\cdot\vec L = 0$ and $\lvert\vec A\rvert^2 = \mu^2k^2 + 2\mu E\lvert\vec L\rvert^2$, leaving five independent, which is unit 08's count of why Kepler orbits close.

Unit 14 next moves to phase space, where every conserved quantity is a generator and the so(4) brackets can be computed. Unit 16 uses the rotation generators of this unit to define the inertia tensor. Unit 29 proves Noether's theorem for fields — the same two views, the three-line derivative and the boundary term of the action — and gets the energy–momentum tensor.

13 · reading

Reading, and what comes next

Reading

Landau & Lifshitz, Mechanics, §§6–10: energy, momentum, centre of mass, angular momentum, and mechanical similarity (the scaling of problem P9), each from a symmetry, in twelve pages.

Goldstein, Poole & Safko, Classical Mechanics (3rd ed.), §§2.6–2.7: conservation theorems and the energy function, with the $h \ne T + V$ discussion.

Lévy-Leblond, Am. J. Phys. 39 (1971) 502: the quasi-invariant form of the theorem, with the boost and Runge–Lenz examples.

Kosmann-Schwarzbach, The Noether Theorems (Springer, 2011): the 1918 paper in translation and its history. Its later chapters use manifolds and fields; stop at the end of the translation until units 29 and 31.

14 Hamilton: phase space, where the converse holds · 16 rigid bodies and the inertia tensor · 29 Noether for fields, and $T^{\mu\nu}$ · 36 Schwarzschild's conserved energy and angular momentum.

one caution on reading Arnold

Arnold's Mathematical Methods, §20, states Noether's theorem on the tangent bundle of a manifold. The content is this unit's; the language is unit 31's. Read it after unit 31, or read "manifold" as "open subset of $\mathbb R^n$".

Landau and Lifshitz is the most direct: it derives each conservation law from a homogeneity or isotropy assumption in a paragraph, and §10's mechanical similarity is exactly problem P9's scaling argument. Its §6 asserts $h = T + V$ under the assumption of a closed system in Cartesian coordinates, and that assumption is the one this unit's section 3 makes explicit.

Goldstein's §2.7 is the standard place for the energy function and its decomposition $h = T_2 - T_0 + V$; its later chapter on field theory (ch. 13) uses tools of unit 29 and should wait.

Noether's own paper is readable in translation (M. A. Tavel, Transport Theory and Statistical Physics 1 (1971) 186–207, also reprinted in Kosmann-Schwarzbach). It is written for multiple integrals — fields — from the start, so read it after unit 29 for full value; its first section is accessible now.

13 · checkpoint

Done when

checkpoint

Given any Lagrangian $L(q,\dot q,t)$ and any one-parameter symmetry of it, you can, on paper and without the deck,

write down the conserved quantity: find the generator $\xi$, compute $\delta L$ for an arbitrary curve, identify the $F$ with $\delta L = \dot F$, and form $Q = p\cdot\xi - F$;
verify it is constant on-shell, by differentiating $Q$ and using the Euler–Lagrange equations; and
do both for at least one quasi-invariant case, where $L$ itself is not invariant and the charge acquires the extra term $-F$ — the Galilean boost, or a translation in a magnetic field.

Falsifiable: a charge that omits $F$ in a quasi-invariant case, or a $\delta L$ computed only on solutions, fails it. Record it with ./mr done 13.

Problems

./mr problems 13 — 11 problems, about 4 hours. Hints and worked solutions fold under each problem on the same page.

Try it cold

A particle in uniform gravity, $L = \frac12m\lvert\dot{\vec r}\rvert^2 - mgz$, under the vertical shift $z\mapsto z + s$. Answer in the notes.

The cold try, worked. $\xi = \hat z$, $\dot\xi = 0$, so $\delta L = -mg$. That is not zero, but it is $\frac{\mathrm d}{\mathrm dt}(-mgt)$: a quasi-symmetry with $F = -mgt$. The charge is $Q = m\dot z - (-mgt) = m\dot z + mgt$. On-shell, $\ddot z = -g$, so $\dot Q = m\ddot z + mg = 0$ ✓. Dimensions: $[mgt] = \mathsf{M\,LT^{-2}\,T} = \mathsf{MLT^{-1}}$, a momentum ✓. Limit $g\to0$: $Q$ is the vertical momentum of a free particle ✓. If you wrote $Q = m\dot z$ and then found it was not conserved, you have met the checkpoint's failure mode.

If the checkpoint stalls, it usually stalls in one of two places. Computing $\delta L$ only along solutions (using the equation of motion while computing it) makes every $\xi$ look like a symmetry with charge $0$; go back to the off-shell warning. Forgetting the $F$ term gives $\vec Pt$ instead of $\vec G$ for the boost; go back to the boost example. Problems P1, P2 and P3 are the direct practice for the checkpoint; P6 and P8 are the energy-function traps; P10 and P11 are the hidden symmetry.

Noether: every conservation law you have met is one theorem

Four symmetries, four charges

Symmetry, made precise

Three laws, three unrelated proofs

One-parameter groups of transformations

The generator determines the group

Rotations, and why their generator is a cross product

Every rotation family starts as a cross product

Moving curves, and what the Lagrangian does

Varying the action along the family

Symmetry, and symmetry up to a total derivative

Four candidate symmetries of the running example

The theorem

Noether's theorem

The same theorem, seen from the action

Where the charge lives: at the ends

What the hypotheses are for

Checking a charge: dimensions

Energy is time translation

Time translation as a variation

The energy function, and when it is conserved

When is $h$ the energy?

The bead on the rotating hoop: $h$ conserved, $E$ not

The hoop in numbers: $h$ stays, $E$ moves

Where the motor's work goes

$h$ belongs to the coordinates, not to the motion

Space: momentum, angular momentum, and the boost

Translations: momentum, and Newton's third law as a theorem

Rotations: all three components of angular momentum

The boost, where $L$ is not invariant

The centre of mass moves in a straight line

What the boost charge says about the Sun

Ten symmetries, ten charges

The hidden symmetry

A conserved vector with no symmetry of space behind it

No point symmetry produces $\vec A$

Letting the transformation see the velocity

The Runge–Lenz generator, checked

The hidden SO(4)

Orbits of one energy

The converse is false, and what rescues it

One theorem, run five times

Reading, and what comes next

Reading

Next

Done when

Problems

Try it cold