Part B · unit 13 · problems

Noether: every conservation law you have met is one theorem

the slides · the curriculum

Checkpoint. Given any Lagrangian \(L(q,\dot q,t)\) and any one-parameter symmetry of it, you write down the conserved quantity — generator \(\xi\), \(\delta L\) computed for an arbitrary curve, the \(F\) with \(\delta L = \dot F\), and \(Q = p\cdot\xi - F\) — and verify it is constant on-shell, including at least one quasi-invariant case in which \(L\) itself is not invariant and the charge acquires the extra term \(-F\).

Work each problem on paper first. Under it, Hints opens a ladder one rung at a time — each rung appears only once the one before it is open, and the last gives the first line of the working — and Worked solution has the whole thing: the idea, the working, then a check. Read the solution after your attempt, including when you got it right. F folds everything back; D switches dark and light.

Eleven problems, about four hours. Work them on paper first. Conventions throughout: \(p = \partial L/\partial\dot q\); a family with generator \(\xi\) is a (quasi-)symmetry when \(\delta L = \frac{\partial L}{\partial q}\cdot\xi + \frac{\partial L}{\partial\dot q}\cdot\dot\xi = \dot F\) holds for every curve, and its Noether charge is \(Q = p\cdot\xi - F\). The energy function is \(h = \dot q\cdot\partial L/\partial\dot q - L\).

Constants: \(G = 6.67430\times10^{-11}\ \mathrm{m^3\,kg^{-1}\,s^{-2}}\), \(GM_\odot = 1.32712440018\times10^{20}\ \mathrm{m^3\,s^{-2}}\), \(GM_{\mathrm J} = 1.26686534\times10^{17}\ \mathrm{m^3\,s^{-2}}\), \(GM_\oplus = 3.986004418\times10^{14}\ \mathrm{m^3\,s^{-2}}\), \(1\ \mathrm{AU} = 1.495978707\times10^{11}\) m, \(R_\odot = 6.957\times10^{8}\) m, year \(= 365.25\) d, \(g = 9.81\ \mathrm{m\,s^{-2}}\), electron mass \(m_e = 9.1093837\times10^{-31}\) kg and charge \(-e\) with \(e = 1.602176634\times10^{-19}\) C.

Symmetries and their charges

P1 Reading symmetries off a Lagrangian routine

exercises: finding generators, computing \(\delta L\) off-shell, strict against quasi-symmetry, checking a charge on-shell

For each Lagrangian, find the stated family's generator \(\xi\), compute \(\delta L\) for an arbitrary curve, decide whether it is a symmetry, a quasi-symmetry or neither, write the Noether charge if there is one, and verify directly from the equations of motion that it is constant.

  1. A particle in an axially symmetric potential, \(L = \tfrac12m(\dot x^2 + \dot y^2 + \dot z^2) - U(x^2 + y^2)\), under (a) rotation about the \(z\)-axis and (b) translation along \(z\).
  2. A particle in a helical potential, in cylindrical coordinates \(L = \tfrac12m(\dot\rho^2 + \rho^2\dot\varphi^2 + \dot z^2) - U(\rho,\ \varphi - z/b)\) with \(b\) a constant length, under the screw motion \(\varphi\mapsto\varphi + s\), \(z\mapsto z + bs\).
  3. A charge \(q\) in a uniform electric field \(\mathcal E\hat x\): \(L = \tfrac12m\lvert\dot{\vec r}\rvert^2 + q\mathcal E x\), under translation along \(\hat x\).
Hints · 3 rungs
Rung 1

Each answer is a function of the state \((q,\dot q,t)\) — possibly with explicit \(t\) — and a one-line check. The method is the same four steps every time: \(\xi\), then \(\delta L\) with no equation of motion used, then \(F\), then \(Q = p\cdot\xi - F\).

Rung 2

Write the generators in the coordinates the Lagrangian is given in. For 1(a) in Cartesian coordinates the rotation generator is \(\hat z\times\vec r = (-y, x, 0)\). For 2 the generator in \((\rho,\varphi,z)\) is a constant vector. In 3, \(\delta L\) is a nonzero constant — ask whether a constant is a time derivative.

Rung 3

First lines: in 1(a), \(\delta(x^2 + y^2) = 2x(-y) + 2y(x) = 0\); in 2, \(\xi = (0, 1, b)\) and \(\varphi - z/b\) changes by \(1 - b/b = 0\); in 3, \(\delta L = q\mathcal E = \frac{\mathrm d}{\mathrm dt}(q\mathcal Et)\).

Worked solution

Idea. A symmetry is visible in the Lagrangian as a combination of coordinates that does not change. When the change is a constant, it is still a total derivative — of a linear function of time — and the charge picks up a term linear in \(t\).

1(a) Rotation about \(z\). \(\xi = (-y, x, 0)\), \(\dot\xi = (-\dot y, \dot x, 0)\). Kinetic part: \(m(\dot x(-\dot y) + \dot y\dot x) = 0\). Potential part: \(U\) depends on \(w = x^2+y^2\) and \(\delta w = 2x(-y) + 2y(x) = 0\). So \(\delta L = 0\): a symmetry, \(F = 0\), and \[ Q = m\dot x(-y) + m\dot y(x) = m(x\dot y - y\dot x) = L_z . \] Check: the equations of motion are \(m\ddot x = -2xU'(w)\), \(m\ddot y = -2yU'(w)\), so \(\dot L_z = m(x\ddot y - y\ddot x) = -2U'(xy - yx) = 0\) ✓.

1(b) Translation along \(z\). \(\xi = \hat z\), \(\dot\xi = 0\), and \(U\) does not involve \(z\). \(\delta L = 0\), \(Q = m\dot z = p_z\); \(m\ddot z = 0\) ✓. The rotations about \(\hat x\) and \(\hat y\) are not symmetries unless \(U\) is constant, since they change \(x^2 + y^2\); so only \(L_z\) and \(p_z\) are promised.

2. Screw motion. In coordinates \((\rho,\varphi,z)\), \(\xi = (0, 1, b)\), constant, so \(\dot\xi = 0\) and the kinetic part of \(\delta L\) is \(\partial_\varphi T\cdot 1 + \partial_z T\cdot b = 0\), since \(T\) involves neither \(\varphi\) nor \(z\). The potential's second argument \(u = \varphi - z/b\) changes by \(1 - b/b = 0\). So \(\delta L = 0\) and \[ Q = p_\varphi\cdot 1 + p_z\cdot b = m\rho^2\dot\varphi + b\,m\dot z = L_z + b\,p_z . \] Check: the Euler–Lagrange equations for \(\varphi\) and \(z\) are \(\frac{\mathrm d}{\mathrm dt}(m\rho^2\dot\varphi) = -\partial_uU\) and \(\frac{\mathrm d}{\mathrm dt}(m\dot z) = +\partial_uU/b\), so \(\dot Q = -\partial_uU + b\,\partial_uU/b = 0\) ✓. Neither \(L_z\) nor \(p_z\) is conserved separately — a screw thread converts one into the other.

3. Translation in a uniform field. \(\xi = \hat x\), \(\dot\xi = 0\), and \(\delta L = \partial_xL = q\mathcal E\). A nonzero constant, so not a symmetry; but \(q\mathcal E = \frac{\mathrm d}{\mathrm dt}(q\mathcal E t)\), a quasi-symmetry with \(F = q\mathcal Et\). The charge is \[ Q = m\dot x - q\mathcal E t . \] Check: \(m\ddot x = q\mathcal E\), so \(\dot Q = q\mathcal E - q\mathcal E = 0\) ✓. Without \(F\) one would claim \(m\dot x\) is conserved, which is false; \(Q\) says instead that \(m\dot x\) grows linearly, which is the impulse–momentum theorem for a constant force.

Check. Dimensions: \([L_z] = [bp_z] = \mathsf{ML^2T^{-1}}\) ✓, as it must be since the screw parameter \(s\) is an angle. \([q\mathcal Et] = \mathsf{N\cdot s} = \mathsf{MLT^{-1}}\) ✓. Limits: in 2, \(b\to\infty\) makes \(U\) independent of \(z\), and \(Q/b\to p_z\), the pure translation charge ✓; in 3, \(\mathcal E\to0\) gives the free-particle momentum ✓.

P2 A charge in a magnetic field: translation is only quasi-invariant medium

exercises: a quasi-symmetry with a velocity-dependent potential, the extra term in the charge, gauge (in)dependence, a number

A particle of mass \(m\) and charge \(q\) moves in a uniform magnetic field \(\vec B = B\hat z\). Unit 12's Lagrangian is \(L = \tfrac12m\lvert\dot{\vec r}\rvert^2 + q\,\dot{\vec r}\cdot\vec A(\vec r)\), with the Lorentz force \(m\ddot{\vec r} = q\dot{\vec r}\times\vec B\) as its Euler–Lagrange equation. Take the symmetric vector potential \(\vec A = \tfrac12\vec B\times\vec r\).

  1. Show that the translation \(\vec r\mapsto\vec r + s\hat a\), for any fixed unit vector \(\hat a\), is a quasi-symmetry but not a symmetry, and find \(F\).
  2. Find the charge, and show that as \(\hat a\) varies the charges assemble into the conserved vector \(\vec K = m\dot{\vec r} + q\vec B\times\vec r\). Verify \(\dot{\vec K} = 0\) from the Lorentz force.
  3. Show that on a circular orbit \(\vec K = q\vec B\times\vec r_c\), where \(\vec r_c\) is the centre, so \(\vec K\) locates the centre of the circle.
  4. An electron starts at the origin with velocity \(10^6\ \mathrm{m\,s^{-1}}\,\hat x\) in \(B = 0.010\) T. Where is the centre of its circle, and what is the radius?
  5. Redo part 1 for translation along \(\hat x\) in the gauge \(\vec A = (-By, 0, 0)\). Is it now a symmetry? Is the charge the same function of the state?
Hints · 3 rungs
Rung 1

The answer to 2 is a conserved vector — a function of position and velocity, not of \(t\) — and the answer to 4 is a point and a length. The whole problem is the four-step method; only the \(\dot{\vec r}\cdot\vec A\) term can produce anything in \(\delta L\).

Rung 2

With \(\xi = \hat a\) constant, \(\dot\xi = 0\) and \(\delta L = q\,\dot{\vec r}\cdot(\tfrac12\vec B\times\hat a)\), because \(\vec A\) is linear in \(\vec r\). A velocity dotted into a constant vector is the time derivative of the position dotted into it.

Rung 3

First line: \(\delta L = \frac{\mathrm d}{\mathrm dt}\big(\tfrac q2(\vec B\times\hat a)\cdot\vec r\big)\), so \(F = \tfrac q2(\vec B\times\hat a)\cdot\vec r\). The canonical momentum is \(\vec p = m\dot{\vec r} + \tfrac q2\vec B\times\vec r\). Then use \((\vec B\times\hat a)\cdot\vec r = \hat a\cdot(\vec r\times\vec B)\).

Worked solution

Idea. A uniform magnetic field has no preferred point, but every vector potential for it does: \(\tfrac12\vec B\times\vec r\) vanishes at the origin. So translation changes \(L\) — by a total derivative, because the physics does not change — and the charge carries the extra term. That extra term turns out to be the most useful part: it is what makes \(\vec K\) point at the centre of the orbit.

1. \(\xi = \hat a\), \(\dot\xi = 0\). The kinetic term does not involve \(\vec r\); \(\vec A(\vec r + s\hat a) = \vec A(\vec r) + \tfrac s2\vec B\times\hat a\), so \[ \delta L = q\,\dot{\vec r}\cdot\tfrac12(\vec B\times\hat a) = \frac{\mathrm d}{\mathrm dt}\Big(\tfrac q2(\vec B\times\hat a)\cdot\vec r\Big). \] This is nonzero on a generic curve (any velocity with a component along \(\vec B\times\hat a\), which is nonzero unless \(\hat a\parallel\vec B\)), so it is not a symmetry; it is a quasi-symmetry with \(F = \tfrac q2(\vec B\times\hat a)\cdot\vec r = \tfrac q2\hat a\cdot(\vec r\times\vec B)\). (For \(\hat a\parallel\vec B\) it is a strict symmetry.)

2. \(\vec p = \partial L/\partial\dot{\vec r} = m\dot{\vec r} + q\vec A = m\dot{\vec r} + \tfrac q2\vec B\times\vec r\). The charge: \[ Q = \vec p\cdot\hat a - F = \hat a\cdot\Big(m\dot{\vec r} + \tfrac q2\vec B\times\vec r - \tfrac q2\vec r\times\vec B\Big) = \hat a\cdot\big(m\dot{\vec r} + q\vec B\times\vec r\big) = \hat a\cdot\vec K . \] The two halves of \(q\vec B\times\vec r\) come one from \(\vec p\) and one from \(-F\): without \(F\) the answer would be wrong by a factor of two in the magnetic term. Check: \(\dot{\vec K} = m\ddot{\vec r} + q\vec B\times\dot{\vec r} = q\dot{\vec r}\times\vec B + q\vec B\times\dot{\vec r} = \vec 0\) ✓.

3. For uniform circular motion about \(\vec r_c\) in the plane perpendicular to \(\vec B\), \(m\dot{\vec r} = q(\vec r - \vec r_c)\times\vec B\) (differentiate: \(m\ddot{\vec r} = q\dot{\vec r}\times\vec B\), the Lorentz force, and a solution is fixed by its initial data, so every solution with velocity perpendicular to \(\vec B\) has this form for the appropriate \(\vec r_c\)). Then \(\vec K = q(\vec r - \vec r_c)\times\vec B + q\vec B\times\vec r = -q\vec B\times(\vec r - \vec r_c) + q\vec B\times\vec r = q\vec B\times\vec r_c\).

4. At \(t = 0\): \(\vec K = m_ev_0\hat x\). With \(\vec r_c = (a, b, 0)\), \(q\vec B\times\vec r_c = qB(-b, a, 0)\), so \(a = 0\) and \(b = -m_ev_0/(qB)\). With \(q = -e\): \[ b = \frac{m_ev_0}{eB} = \frac{(9.1093837\times10^{-31})(10^{6})}{(1.602176634\times10^{-19})(0.010)} = 5.686\times10^{-4}\ \mathrm m . \] The centre is at \((0,\ +0.569\ \mathrm{mm},\ 0)\) and the radius is \(0.569\) mm, the distance from the origin. Sense check: the force at \(t=0\) is \(q\dot{\vec r}\times\vec B = (-e)(v_0B)(\hat x\times\hat z) = +ev_0B\,\hat y\), pointing towards \(+\hat y\), towards the centre ✓.

5. In the gauge \(\vec A = (-By, 0, 0)\), \(L = \tfrac12m\lvert\dot{\vec r}\rvert^2 - qBy\dot x\) does not contain \(x\), so translation along \(\hat x\) is a strict symmetry, \(F = 0\), and \(Q = p_x = m\dot x - qBy\). Compare \(K_x = m\dot x + q(\vec B\times\vec r)_x = m\dot x + q(-By) = m\dot x - qBy\). The same function of the state. What changed between gauges is the split between \(p\cdot\xi\) and \(F\), not the charge — which is the gauge invariance proved in P4. (In this gauge, translation along \(\hat y\) is the quasi-symmetry: \(\delta L = -qB\dot x\), \(F = -qBx\), \(Q = m\dot y + qBx = K_y\) ✓.)

Check. Dimensions: one tesla is \(\mathrm{kg\,C^{-1}\,s^{-1}}\), so \([qBr] = \mathrm{C}\cdot\mathrm{kg\,C^{-1}s^{-1}}\cdot\mathrm m = \mathrm{kg\,m\,s^{-1}}\), a momentum ✓. Limit: \(B\to0\) gives \(\vec K = m\dot{\vec r}\), the free-particle momentum ✓, and the radius \(m_ev_0/eB\to\infty\), a straight line ✓.

P3 The boost charge, and how a star gives away its planets medium

exercises: the Galilean boost for \(N\) bodies, quasi-invariance, what \(\vec G\) says, numbers

\(N\) bodies interact through a potential depending only on differences of positions: \(L = \sum_a\tfrac12m_a\lvert\dot{\vec r}_a\rvert^2 - V(\{\vec r_a - \vec r_b\})\).

  1. Show that the boost \(\vec r_a\mapsto\vec r_a + st\hat n\) (the same for every body) is a quasi-symmetry, and derive the charge \(\vec G = \vec Pt - M\vec R\), with \(M = \sum m_a\), \(M\vec R = \sum m_a\vec r_a\).
  2. Show that for unit 08's one-body Lagrangian \(\tfrac12\mu\lvert\dot{\vec r}\rvert^2 + k/\lvert\vec r\rvert\) the same boost is not even a quasi-symmetry, and say physically why.
  3. Treat the Sun and Jupiter as an isolated pair on Jupiter's orbit, semi-major axis \(a = 5.2044\) AU. In the frame where \(\vec G = \vec P = 0\), how far is the Sun's centre from the barycentre when the separation equals \(a\), in solar radii, and how fast does the Sun move there?
  4. Repeat the speed for the Sun and the Earth alone (\(a = 1.00000261\) AU). Which planet would a distant astronomer measuring the Sun's Doppler shift find first?
Hints · 3 rungs
Rung 1

Part 1 is the four-step method with \(\xi_a = t\hat n\), which depends on time; the charge is a vector that depends explicitly on \(t\) and is still conserved. Parts 3 and 4 are two lengths and two speeds, each a mass ratio times a quantity from unit 08.

Rung 2

\(\dot\xi_a = \hat n\), and \(\xi_a - \xi_b = 0\) for every pair, so the potential does not notice. The kinetic term gives \(\hat n\cdot\sum m_a\dot{\vec r}_a\) — a derivative of what? For part 3, \(\vec R = 0\) gives \(\vec r_S = -(m_J/M)\vec r\) with \(\vec r = \vec r_J - \vec r_S\).

Rung 3

First line of 1: \(\delta L = \hat n\cdot M\dot{\vec R} = \frac{\mathrm d}{\mathrm dt}(M\hat n\cdot\vec R)\), so \(F = M\hat n\cdot\vec R\). For 3, get the relative speed at \(r = a\) from vis-viva, \(v^2 = G(m_S + m_J)(2/r - 1/a)\), and multiply by \(m_J/M\).

Worked solution

Idea. A boost changes every velocity by the same amount, so the kinetic energy changes by a term linear in the velocities — the total momentum dotted into \(\hat n\) — which is a total derivative. The potential sees only differences and does not change at all. The charge then says the centre of mass moves in a straight line, and the numbers turn that into an observable wobble.

1. \(\xi_a = t\hat n\), \(\dot\xi_a = \hat n\). \(\delta V = \sum_{a}\nabla_aV\cdot t\hat n = 0\), because \(V\) depends only on differences and each difference changes by \(t\hat n - t\hat n = 0\). \(\delta T = \sum_am_a\dot{\vec r}_a\cdot\hat n = \hat n\cdot\vec P = \frac{\mathrm d}{\mathrm dt}(M\hat n\cdot\vec R)\). So \(F = M\hat n\cdot\vec R\), and \[ Q = \sum_a m_a\dot{\vec r}_a\cdot t\hat n - M\hat n\cdot\vec R = \hat n\cdot(\vec Pt - M\vec R). \] Three directions \(\hat n\) give the vector \(\vec G\). On-shell check: \(\dot{\vec G} = \vec P + \dot{\vec P}t - M\dot{\vec R} = \vec P + 0 - \vec P = 0\), using \(\dot{\vec P} = 0\) (translation symmetry) ✓. Hence \(M\vec R(t) = \vec Pt - \vec G\): uniform straight-line motion.

2. \(\xi = t\hat n\): \(\delta L = \mu\dot{\vec r}\cdot\hat n - \frac{k}{r^3}\vec r\cdot t\hat n\). The first term is \(\frac{\mathrm d}{\mathrm dt}(\mu\hat n\cdot\vec r)\); the second would have to be \(\frac{\mathrm d}{\mathrm dt}\) of some \(F_2(\vec r,t)\), i.e. \(\partial_tF_2 + \nabla F_2\cdot\dot{\vec r} = -kt\,\hat n\cdot\vec r/r^3\) for all velocities, forcing \(\nabla F_2 = 0\), so \(F_2 = F_2(t)\) and \(\partial_tF_2\) cannot depend on \(\vec r\). Contradiction (for \(k\ne0\)). Physically: the force centre is an external body pinned at the origin and not boosted with the particle, so the system is not isolated and has no reason to be boost-invariant.

3. \(m_S = GM_\odot/G = 1.98841\times10^{30}\) kg, \(m_J = 1.89812\times10^{27}\) kg, \(m_J/M = 9.53684\times10^{-4}\). At separation \(a = 7.78567\times10^{11}\) m the Sun is at distance \[ \frac{m_J}{M}\,a = 7.42507\times10^{8}\ \mathrm m = \frac{7.42507}{6.957}\,R_\odot = 1.067\,R_\odot \] from the barycentre — just outside its own surface. At \(r = a\) vis-viva gives relative speed \(\sqrt{G(m_S+m_J)/a} = \sqrt{1.32839\times10^{20}/7.78567\times10^{11}} = 13\,062\ \mathrm{m\,s^{-1}}\), and the Sun's share is \(13\,062\times9.53684\times10^{-4} = 12.46\ \mathrm{m\,s^{-1}}\). (It ranges from \(11.86\) at aphelion to \(13.08\ \mathrm{m\,s^{-1}}\) at perihelion, with \(e = 0.0489\).)

4. \(m_\oplus/(m_S + m_\oplus) = 3.0035\times10^{-6}\); relative speed \(\sqrt{G(m_S + m_\oplus)/a_\oplus} = 29\,785\ \mathrm{m\,s^{-1}}\); Sun's share \(0.0895\ \mathrm{m\,s^{-1}}\), about \(9\ \mathrm{cm\,s^{-1}}\). Jupiter's signal is \(139\) times larger and would be found first, even though its period is twelve times longer.

Check. Dimensions: \([\vec Pt] = [M\vec R] = \mathsf{ML}\) ✓. Limits: \(m_J/m_S\to0\) puts the barycentre at the Sun's centre and the wobble to zero ✓. Consistency: momentum balance \(m_Sv_S = m_Jv_J\) with \(v_J = (m_S/M)\times13\,062 = 13\,050\ \mathrm{m\,s^{-1}}\) gives \(v_S = 13\,050\times m_J/m_S = 12.46\ \mathrm{m\,s^{-1}}\) ✓.

P4 Proof: the charge from the action, and why it is gauge-invariant medium

exercises: the first variation with free ends, Noether's theorem by the action route, gauge invariance of charges

  1. Let \(q\) solve the Euler–Lagrange equations of \(L\) on \([t_0,t_1]\) and let \(\xi(q,\dot q,t)\) be any \(C^1\) function. Prove that \(\displaystyle\int_{t_0}^{t_1}\delta L\,\mathrm dt = \big[p\cdot\xi\big]_{t_0}^{t_1}\), where \(\delta L = \partial_qL\cdot\xi + \partial_{\dot q}L\cdot\dot\xi\) is evaluated along \(q\).
  2. Deduce Noether's theorem from part 1, without differentiating \(Q\).
  3. Let \(L' = L + \frac{\mathrm d}{\mathrm dt}\Lambda(q,t)\). Prove that if \(\xi\) is a quasi-symmetry of \(L\) with function \(F\), it is one of \(L'\) with \(F' = F + \partial_q\Lambda\cdot\xi\), and that the two charges are equal as functions of the state.
  4. Unit 12 showed that the canonical momentum of a charge in a magnetic field depends on the gauge. Is the charge \(\vec K\) of P2 gauge-invariant? Answer using part 3, without recomputing.
Hints · 3 rungs
Rung 1

Three short proofs and a one-line application. Part 1 is unit 11's integration by parts with the boundary term kept; part 3 is a computation of \(p'\) and of \(\delta\) applied to a total derivative.

Rung 2

In part 1, integrate \(p\cdot\dot\xi\) by parts and use the equation of motion on what is left. In part 3, you need \(\delta\big(\dot\Lambda\big) = \frac{\mathrm d}{\mathrm dt}\big(\delta\Lambda\big)\); that is, \(\partial_q\dot\Lambda = \frac{\mathrm d}{\mathrm dt}\partial_q\Lambda\) and \(\partial_{\dot q}\dot\Lambda = \partial_q\Lambda\).

Rung 3

First line of part 1: \(\int p\cdot\dot\xi\,\mathrm dt = [p\cdot\xi] - \int\dot p\cdot\xi\,\mathrm dt\). First line of part 3: \(\dot\Lambda = \partial_t\Lambda + \partial_q\Lambda\cdot\dot q\), so \(p' = p + \partial_q\Lambda\).

Worked solution

Idea. The charge is a boundary term: the part of the first variation that unit 11 threw away because its variations vanished at the ends. Gauge transformations add a boundary term to the action, and the two changes they cause — to \(p\) and to \(F\) — are the same boundary term counted twice with opposite signs.

1. Along \(q\), \(p(t) = \partial_{\dot q}L(q,\dot q,t)\) is \(C^1\) (since \(L\) is \(C^2\) and \(q\) is \(C^2\)), and \(\xi(t)\) is \(C^1\). So integration by parts is legitimate: \[ \int_{t_0}^{t_1}\delta L\,\mathrm dt = \int_{t_0}^{t_1}\partial_qL\cdot\xi\,\mathrm dt + \big[p\cdot\xi\big]_{t_0}^{t_1} - \int_{t_0}^{t_1}\dot p\cdot\xi\,\mathrm dt = \int_{t_0}^{t_1}\big(\partial_qL - \dot p\big)\cdot\xi\,\mathrm dt + \big[p\cdot\xi\big]_{t_0}^{t_1}. \] The Euler–Lagrange equation is \(\dot p = \partial_qL\), so the integral vanishes. ∎

2. If \(\delta L = \dot F\) for every curve, then in particular along \(q\), so \(\int_{t_0}^{t_1}\delta L\,\mathrm dt = [F]_{t_0}^{t_1}\). With part 1, \([p\cdot\xi - F]_{t_0}^{t_1} = 0\) for every \(t_0 \lt t_1\) in the interval: \(Q\) takes the same value at any two times. ∎

3. \(\dot\Lambda = \partial_t\Lambda + \partial_q\Lambda\cdot\dot q\) is affine in \(\dot q\), so \(p' = p + \partial_q\Lambda\). For \(\delta\) of the extra term: \(\partial_q\dot\Lambda = \partial_q\partial_t\Lambda + (\partial_q\partial_q\Lambda)\dot q = \frac{\mathrm d}{\mathrm dt}(\partial_q\Lambda)\) (Clairaut, since \(\Lambda\) is \(C^2\)), and \(\partial_{\dot q}\dot\Lambda = \partial_q\Lambda\). Hence \[ \delta\dot\Lambda = \frac{\mathrm d}{\mathrm dt}(\partial_q\Lambda)\cdot\xi + \partial_q\Lambda\cdot\dot\xi = \frac{\mathrm d}{\mathrm dt}\big(\partial_q\Lambda\cdot\xi\big), \] so \(\delta L' = \frac{\mathrm d}{\mathrm dt}(F + \partial_q\Lambda\cdot\xi)\) for every curve, which is the claim with \(F' = F + \partial_q\Lambda\cdot\xi\). Then \(Q' = p'\cdot\xi - F' = p\cdot\xi + \partial_q\Lambda\cdot\xi - F - \partial_q\Lambda\cdot\xi = Q\). ∎ Note that the proof never assumed \(\xi\) independent of \(\dot q\): it holds for generalised symmetries too.

4. Two vector potentials for the same \(\vec B\) differ by a gradient, \(\vec A' = \vec A + \nabla\chi\) (unit 04: a curl-free field on \(\mathbb R^3\) is a gradient), and then \(L' = L + q\dot{\vec r}\cdot\nabla\chi = L + \frac{\mathrm d}{\mathrm dt}(q\chi)\): a gauge transformation with \(\Lambda = q\chi\). By part 3 every Noether charge is unchanged. So \(\vec K\) is gauge-invariant even though \(\vec p\) is not; P2 part 5 saw this in one instance.

Check. Dimensions in part 3: \([\partial_q\Lambda\cdot\xi] = [\Lambda]/[s]\) and \([\Lambda] = [L][t]\), an action, so \([F'] = [F]\) ✓. Special case: \(\Lambda\) independent of \(q\) (a function of \(t\) alone) changes nothing at all — \(p' = p\), \(F' = F\) — as it must, since then \(L' - L\) is a function of \(t\) only ✓.

Energy is not always \(T + V\)

P5 Proof: when the energy function is the energy routine

exercises: Euler's theorem on homogeneous functions, \(h = L_2 - L_0\), recognising each case

  1. Prove Euler's theorem: if \(f:\mathbb R^n\to\mathbb R\) is \(C^1\) and \(f(\lambda v) = \lambda^kf(v)\) for all \(\lambda\gt0\), then \(v\cdot\nabla f(v) = kf(v)\). Deduce that if \(L = L_2 + L_1 + L_0\) with \(L_j\) homogeneous of degree \(j\) in \(\dot q\), then \(h = L_2 - L_0\).
  2. Compute \(h\) for each of the following, say whether it is conserved, and whether it equals \(T+V\):
    1. the spherical pendulum, \(L = \tfrac12m\ell^2(\dot\theta^2 + \sin^2\theta\,\dot\varphi^2) + mg\ell\cos\theta\);
    2. a charge in static fields, \(L = \tfrac12m\lvert\dot{\vec r}\rvert^2 - q\phi(\vec r) + q\dot{\vec r}\cdot\vec A(\vec r)\);
    3. a pendulum whose string is reeled in at a constant rate \(w\), length \(\ell(t) = \ell_0 - wt\): \(L = \tfrac12m(\dot\ell^2 + \ell^2\dot\theta^2) + mg\ell\cos\theta\) with \(\dot\ell = -w\).
  3. In case (c), show that \(\dot h\) equals the power delivered by whoever pulls the string, \(w\) times the tension \(m(\ell\dot\theta^2 + g\cos\theta)\).
Hints · 3 rungs
Rung 1

Part 1 is a one-variable chain-rule argument; part 2 is three short classifications, each answering two independent questions — is \(h\) conserved, and does it equal \(T+V\)? — which are settled by different features of \(L\); part 3 is an identity between two expressions.

Rung 2

For Euler, fix \(v\) and differentiate \(g(\lambda) = f(\lambda v)\) two ways. For part 2, sort each term of \(L\) by its degree in the velocities. In (c) the time dependence is hidden inside \(\ell(t)\), and \(\dot\ell^2 = w^2\) is degree 0.

Rung 3

First line of Euler: \(g'(\lambda) = \nabla f(\lambda v)\cdot v = k\lambda^{k-1}f(v)\); set \(\lambda = 1\). Then \(\dot q\cdot\partial_{\dot q}L = 2L_2 + L_1\). For (c): \(L_2 = \tfrac12m\ell^2\dot\theta^2\), \(L_1 = 0\), \(L_0 = \tfrac12mw^2 + mg\ell\cos\theta\), and \(\partial_tL \ne 0\) because \(\ell\) depends on \(t\).

Worked solution

Idea. \(h\) sees the velocity-degree of each term: quadratic terms count once with a plus sign, linear terms not at all, velocity-free terms once with a minus sign. Conservation is a separate question, settled by \(\partial_tL\).

1. Fix \(v\), let \(g(\lambda) = f(\lambda v)\). The chain rule gives \(g'(\lambda) = \nabla f(\lambda v)\cdot v\); homogeneity gives \(g(\lambda) = \lambda^kf(v)\), so \(g'(\lambda) = k\lambda^{k-1}f(v)\). At \(\lambda = 1\): \(v\cdot\nabla f(v) = kf(v)\). ∎ Applying it with \((q,t)\) fixed to each piece, \(\dot q\cdot\partial_{\dot q}L = 2L_2 + L_1 + 0\), so \(h = 2L_2 + L_1 - (L_2 + L_1 + L_0) = L_2 - L_0\). ∎

2(a). \(L_2 = T\), \(L_1 = 0\), \(L_0 = mg\ell\cos\theta = -V\). So \(h = T + V\), and \(\partial_tL = 0\): conserved and equal to the energy. The Cartesian position \(\ell(\sin\theta\cos\varphi, \sin\theta\sin\varphi, -\cos\theta)\) has no explicit \(t\) ✓.

2(b). \(L_2 = \tfrac12m\lvert\dot{\vec r}\rvert^2\), \(L_1 = q\dot{\vec r}\cdot\vec A\), \(L_0 = -q\phi\). So \(h = \tfrac12m\lvert\dot{\vec r}\rvert^2 + q\phi\): the magnetic term drops out of \(h\) entirely. Conserved, since static fields make \(\partial_tL = 0\). It equals kinetic plus electrostatic potential energy — the magnetic force \(q\dot{\vec r}\times\vec B\) is perpendicular to the velocity and does no work, and \(h\) knows it.

2(c). \(L_2 = \tfrac12m\ell^2\dot\theta^2\), \(L_1 = 0\), \(L_0 = \tfrac12mw^2 + mg\ell\cos\theta\). So \(h = \tfrac12m\ell^2\dot\theta^2 - \tfrac12mw^2 - mg\ell\cos\theta\), while \(E = T + V = \tfrac12m(w^2 + \ell^2\dot\theta^2) - mg\ell\cos\theta = h + mw^2\). Here \(\partial_tL \ne 0\) (through \(\ell(t)\)), so \(h\) is not conserved; and \(h \ne T + V\), though they differ only by the constant \(mw^2\) because \(T_0 = \tfrac12mw^2\) happens to be constant.

3. \(\dot h = -\partial_tL\). With \(\ell\) the only source of explicit \(t\), \(\partial_tL = \dot\ell\,\partial_\ell L = -w\,(m\ell\dot\theta^2 + mg\cos\theta)\) (the \(\dot\ell^2\) term is the constant \(w^2\)). So \(\dot h = w\,m(\ell\dot\theta^2 + g\cos\theta)\). The string tension \(\mathcal T\) is found from the radial equation for the bob: \(m(\ddot\ell - \ell\dot\theta^2) = mg\cos\theta - \mathcal T\), and \(\ddot\ell = 0\) gives \(\mathcal T = m(\ell\dot\theta^2 + g\cos\theta)\). The hand pulls the string in at speed \(w\) against tension \(\mathcal T\), delivering power \(w\mathcal T\) ✓. Since \(E = h + mw^2\), \(\dot E = \dot h\) too, which is the work–energy theorem.

Check. Dimensions of \(w\mathcal T\): \(\mathsf{m\,s^{-1}\cdot N} = \mathsf W\) ✓. Limit \(w\to0\): the pendulum has fixed length, \(h = E\), and \(\dot h = 0\) ✓. Table of the three cases: (a) conserved, \(= T + V\); (b) conserved, \(= T + V\) though \(L\) has a linear term; (c) neither.

P6 The bead on the rotating hoop, and the motor that drives it medium

exercises: a conserved \(h\) that is not the energy, the work of a driving motor, reconciling the two, numbers

Unit 12's bead of mass \(m = 0.05\) kg slides without friction on a hoop of radius \(R = 0.20\) m that a motor turns at the constant rate \(\Omega = 10\ \mathrm{s^{-1}}\) about its vertical diameter, with \(\theta\) measured from the bottom: \(L = \tfrac12mR^2(\dot\theta^2 + \Omega^2\sin^2\theta) + mgR\cos\theta\).

  1. Write down \(h\) and the inertial-frame energy \(E = T + V\). Which is conserved?
  2. The bead is released from rest (relative to the hoop) at \(\theta_0 = 30^\circ\). Find the other turning point \(\theta_1\).
  3. Compute \(E\) at \(\theta_0\) and at \(\theta_1\), and the work the motor does on the bead between them.
  4. Compute the change of the bead's angular momentum about the axis, \(L_z = mR^2\Omega\sin^2\theta\), and show that \(\Omega\,\Delta L_z\) equals the answer to 3. Explain why.
Hints · 3 rungs
Rung 1

One conserved quantity, one angle, two energies and a work, and a consistency check between two ways of computing that work. The conserved one determines the turning point.

Rung 2

Split \(L\) by degree in \(\dot\theta\): \(h = L_2 - L_0\). At a turning point \(\dot\theta = 0\), so equating \(h\) at \(\theta_0\) and \(\theta_1\) is a quadratic in \(u = \cos\theta\), one of whose roots you already know.

Rung 3

First line: \(-\tfrac12mR^2\Omega^2(1 - u^2) - mgRu = -\tfrac12mR^2\Omega^2(1 - u_0^2) - mgRu_0\), which factorises as \((u - u_0)\big(\tfrac12R\Omega^2(u + u_0) - g\big) = 0\).

Worked solution

Idea. The Lagrangian has no explicit \(t\), so \(h\) is conserved; but the Cartesian position depends on \(t\) (the hoop turns), so \(h\ne T+V\). The difference \(E - h = \Omega L_z\) is exactly the bookkeeping of the motor's work.

1. \(L_2 = \tfrac12mR^2\dot\theta^2\), \(L_0 = \tfrac12mR^2\Omega^2\sin^2\theta + mgR\cos\theta\): \[ h = \tfrac12mR^2\dot\theta^2 - \tfrac12mR^2\Omega^2\sin^2\theta - mgR\cos\theta,\qquad E = \tfrac12mR^2\dot\theta^2 + \tfrac12mR^2\Omega^2\sin^2\theta - mgR\cos\theta . \] \(h\) is conserved (\(\partial_tL = 0\)); \(E\) is not, since \(E - h = mR^2\Omega^2\sin^2\theta\) varies.

2. With \(u = \cos\theta\), the rung-3 factorisation gives \(u_1 = 2g/(R\Omega^2) - u_0 = 0.98100 - 0.86603 = 0.11497\), so \(\theta_1 = 83.40^\circ\). (Here \(\Omega^2 = 100 \gt g/R = 49.05\), so the bead oscillates about the off-axis equilibrium \(\cos\theta_* = g/R\Omega^2 = 0.4905\), and \(u_0\), \(u_1\) sit symmetrically about it.)

3. With \(mR^2\Omega^2 = 0.2\) J and \(mgR = 0.0981\) J:

  • at \(\theta_0 = 30^\circ\): \(E_0 = \tfrac12(0.2)(0.25) - 0.0981(0.86603) = 0.02500 - 0.08496 = -0.05996\) J;
  • at \(\theta_1 = 83.40^\circ\): \(\sin^2\theta_1 = 0.98678\), \(E_1 = \tfrac12(0.2)(0.98678) - 0.0981(0.11497) = 0.09868 - 0.01128 = +0.08740\) J.

The bead's energy rises by \(0.14736\) J. Nothing else acts on it that can do work — gravity is in \(V\), and the hoop is frictionless — except through the hoop's rotation, so the motor does \(+0.1474\) J of work on the bead. (And \(h = -0.10996\) J at both ends ✓.)

4. \(L_z\): \(0.02\times0.25 = 0.005000\) at \(\theta_0\), \(0.02\times0.98678 = 0.019736\) at \(\theta_1\). \(\Omega\,\Delta L_z = 10\times0.014736 = 0.14736\) J ✓. Why: the hoop pushes the bead with a force whose only component able to do work is the horizontal one perpendicular to the hoop's plane, \(N_\varphi\); its power is \(N_\varphi\cdot R\sin\theta\,\Omega = \tau_z\Omega\), and \(\tau_z = \dot L_z\) (gravity exerts no torque about the vertical). Equivalently, \(h = E - \Omega L_z\) is constant, so \(\Delta E = \Omega\,\Delta L_z\) — the conservation of \(h\) is the energy balance with the motor included.

Check. Dimensions: \([\Omega L_z] = \mathsf{s^{-1}\cdot kg\,m^2\,s^{-1}} = \mathsf J\) ✓. Limit \(\Omega\to0\): \(L_z\to0\), \(h\to E\), no work, and the bead is a pendulum ✓. A second number: the slides' release from \(90^\circ\) gives \(\theta_1 = 11.19^\circ\) and \(\Delta E = -0.1925\) J, the motor absorbing energy on the inward swing; here, released inside the equilibrium, the bead swings outwards and the motor supplies it.

P7 A pendulum on a moving train routine

exercises: \(h\) depends on the coordinates, a moving support, \(h = E - uP_x\)

A simple pendulum of length \(\ell\) and mass \(m\) hangs from a support moving horizontally at constant speed \(u\), so the bob is at \((ut + \ell\sin\theta,\ -\ell\cos\theta)\).

  1. Write \(L(\theta,\dot\theta)\) and show it has no explicit \(t\).
  2. Compute \(h\) and the ground-frame energy \(E\). Show that \(h = E - uP_x\), where \(P_x\) is the bob's horizontal momentum, and that \(h = E' - \tfrac12mu^2\) where \(E'\) is the energy measured in the train.
  3. \(E\) is not conserved. Who does work on the bob, and at what rate?
Hints · 3 rungs
Rung 1

Everything is an explicit function of \(\theta\) and \(\dot\theta\). Part 2 is algebra with \(h = L_2 - L_0\); part 3 is one physical sentence and one derivative.

Rung 2

Square the velocity \((u + \ell\dot\theta\cos\theta,\ \ell\dot\theta\sin\theta)\). You get a term of each degree 2, 1, 0 in \(\dot\theta\); the degree-1 term does not appear in \(h\).

Rung 3

First line: \(T = \tfrac12m(u^2 + 2u\ell\dot\theta\cos\theta + \ell^2\dot\theta^2)\), so \(L_2 = \tfrac12m\ell^2\dot\theta^2\), \(L_1 = mu\ell\dot\theta\cos\theta\), \(L_0 = \tfrac12mu^2 + mg\ell\cos\theta\).

Worked solution

Idea. The coordinates move with the train, so \(h\) is conserved but is not the ground-frame energy. It is the train-frame energy, shifted by a constant — the boost analogue of the rotating hoop's \(h = E - \Omega L_z\).

1. Velocity \((u + \ell\dot\theta\cos\theta,\ \ell\dot\theta\sin\theta)\), so \(T = \tfrac12m(u^2 + 2u\ell\dot\theta\cos\theta + \ell^2\dot\theta^2)\), \(V = -mg\ell\cos\theta\), and \(L = T - V\) involves \(t\) nowhere ✓. (Its Euler–Lagrange equation is \(\ell\ddot\theta = -g\sin\theta\): the \(L_1\) term is \(\frac{\mathrm d}{\mathrm dt}(mu\ell\sin\theta)\), a total derivative, as Galilean relativity requires.)

2. \(h = L_2 - L_0 = \tfrac12m\ell^2\dot\theta^2 - \tfrac12mu^2 - mg\ell\cos\theta\). \(E = T + V = \tfrac12m(u^2 + 2u\ell\dot\theta\cos\theta + \ell^2\dot\theta^2) - mg\ell\cos\theta\). Their difference: \(E - h = mu^2 + mu\ell\dot\theta\cos\theta = u\cdot m(u + \ell\dot\theta\cos\theta) = uP_x\) ✓. In the train the bob has velocity \((\ell\dot\theta\cos\theta, \ell\dot\theta\sin\theta)\), so \(E' = \tfrac12m\ell^2\dot\theta^2 - mg\ell\cos\theta = h + \tfrac12mu^2\) ✓.

3. The support: it holds the string's top end moving at \(u\), and the string's tension has a horizontal component. Power delivered \(= u\times\) (horizontal force on the bob) \(= u\dot P_x\). From \(h = E - uP_x\) constant, \(\dot E = u\dot P_x\) ✓. Explicitly \(\dot P_x = m\ell(\ddot\theta\cos\theta - \dot\theta^2\sin\theta)\), which oscillates in sign: the train alternately gives energy to the bob and takes it back.

Check. Dimensions: \([uP_x] = \mathsf{m\,s^{-1}\cdot kg\,m\,s^{-1}} = \mathsf J\) ✓. Limit \(u\to0\): \(h = E = E'\) ✓. Compare the slides' rotating case: there \(h = E - \Omega L_z\), with (angular speed, angular momentum) in place of (speed, momentum) — the conjugate pairs of the dimensions slide.

P8 A careful counterexample: the damped oscillator medium

exercises: a symmetry of the equations that is not a symmetry of \(L\), proving no \(F\) exists, finding the symmetry that does exist

Let \(L = e^{\gamma t}\big(\tfrac12m\dot q^2 - \tfrac12m\omega^2q^2\big)\) with constants \(\gamma, \omega \gt 0\).

  1. Show that its Euler–Lagrange equation is \(\ddot q + \gamma\dot q + \omega^2q = 0\), which is unchanged by \(t\mapsto t + s\).
  2. Prove that time translation, \(\xi = \dot q\), is not a quasi-symmetry of \(L\): no \(F(q,\dot q,t)\) satisfies \(\delta L = \dot F\) for every curve.
  3. Compute \(h\) and \(\dot h\) on-shell. Is it conserved?
  4. Show that \(\xi = \dot q + \tfrac\gamma2q\) — time translation combined with a rescaling of \(q\) — does satisfy the theorem's hypothesis, with \(F = L\), and find its charge \(I\). (Like time translation itself it is a generalised symmetry: both \(\xi\) and \(F\) involve \(\dot q\).)
  5. Show that when \(\gamma \lt 2\omega\) the quadratic form in \(I\) is positive definite, and deduce that the mechanical energy \(\tfrac12m\dot q^2 + \tfrac12m\omega^2q^2\) is bounded above and below by constant multiples of \(e^{-\gamma t}\).
Hints · 3 rungs
Rung 1

Part 2 is an impossibility proof: assume \(F\) exists and derive a contradiction from something you know about the Euler–Lagrange expression of a total derivative. Part 4 gives a quadratic form in \((q,\dot q)\) times \(e^{\gamma t}\).

Rung 2

For \(\xi = \dot q\), \(\delta L = \dot L - \partial_tL = \dot L - \gamma L\). If this were \(\dot F\), then \(\gamma L\) would be a total derivative — and a total derivative has identically zero Euler–Lagrange expression (unit 12). For part 4 use that \(L\) is homogeneous of degree 2 in \((q,\dot q)\) jointly.

Rung 3

First line of 4: \(\delta L = \dot L - \gamma L + \tfrac\gamma2\big(q\,\partial_qL + \dot q\,\partial_{\dot q}L\big) = \dot L - \gamma L + \tfrac\gamma2\cdot2L = \dot L\), since \(\delta\) of the \(\tfrac\gamma2q\) part is \(\partial_qL\cdot\tfrac\gamma2q + \partial_{\dot q}L\cdot\tfrac\gamma2\dot q\).

Worked solution

Idea. Shifting time multiplies \(L\) by \(e^{\gamma s}\). That leaves the equations alone and rescales the action — not a Noether symmetry, and the energy duly decays. But rescaling \(q\) by \(e^{-\gamma s/2}\) undoes the factor, so time shift plus shrink does satisfy the hypothesis, and its charge controls the decay.

1. \(p = me^{\gamma t}\dot q\), \(\partial_qL = -me^{\gamma t}\omega^2q\). The Euler–Lagrange equation \(\frac{\mathrm d}{\mathrm dt}(me^{\gamma t}\dot q) = -me^{\gamma t}\omega^2q\) is \(me^{\gamma t}(\ddot q + \gamma\dot q + \omega^2q) = 0\) ✓. It has constant coefficients, so if \(q(t)\) solves it so does \(q(t+s)\).

2. For \(\xi = \dot q\): \(\delta L = \partial_qL\,\dot q + \partial_{\dot q}L\,\ddot q = \dot L - \partial_tL = \dot L - \gamma L\). Suppose \(\delta L = \dot F\) for every curve, with \(F = F(q,\dot q,t)\). Then \(G := L - F\) satisfies \(\dot G = \gamma L\) for every curve. The right side contains no \(\ddot q\), while \(\dot G = \partial_tG + \partial_qG\,\dot q + \partial_{\dot q}G\,\ddot q\); since \(\ddot q\) can be chosen freely at any point of a curve, \(\partial_{\dot q}G = 0\), so \(G = G(q,t)\) and \(\dot G\) is affine in \(\dot q\). But \(\gamma L\) contains \(\tfrac12\gamma me^{\gamma t}\dot q^2\), quadratic in \(\dot q\) with nonzero coefficient. Contradiction. ∎ (The same degree argument appears in P11 and on the slides' scaling example.)

3. \(h = p\dot q - L = e^{\gamma t}\big(\tfrac12m\dot q^2 + \tfrac12m\omega^2q^2\big)\), and \(\dot h = -\partial_tL = -\gamma L = -\gamma e^{\gamma t}\big(\tfrac12m\dot q^2 - \tfrac12m\omega^2q^2\big)\), which is not identically zero on solutions. Not conserved.

4. For \(\xi = \dot q + \tfrac\gamma2q\): the \(\dot q\) part contributes \(\dot L - \gamma L\), and the \(\tfrac\gamma2q\) part contributes \(\partial_qL\cdot\tfrac\gamma2q + \partial_{\dot q}L\cdot\tfrac\gamma2\dot q = \tfrac\gamma2\cdot2L = \gamma L\) by Euler's theorem (\(L\) is homogeneous of degree 2 in \((q,\dot q)\) at fixed \(t\)). Total \(\delta L = \dot L\): the hypothesis holds with \(F = L\), a generalised symmetry exactly as time translation is. The charge: \[ I = p\,\xi - L = me^{\gamma t}\dot q\big(\dot q + \tfrac\gamma2q\big) - e^{\gamma t}\big(\tfrac12m\dot q^2 - \tfrac12m\omega^2q^2\big) = e^{\gamma t}\big(\tfrac12m\dot q^2 + \tfrac12m\gamma q\dot q + \tfrac12m\omega^2q^2\big). \] Direct check: \(\frac{\mathrm d}{\mathrm dt}\big[e^{\gamma t}(\dot q^2 + \gamma q\dot q + \omega^2q^2)\big]\), with \(\ddot q = -\gamma\dot q - \omega^2q\), is \(e^{\gamma t}\big[\gamma\dot q^2 + \gamma^2q\dot q + \gamma\omega^2q^2 + 2\dot q\ddot q + \gamma\dot q^2 + \gamma q\ddot q + 2\omega^2q\dot q\big]\), and substituting \(\ddot q\) every term cancels ✓.

5. The form \(\tfrac12m(\dot q^2 + \gamma q\dot q + \omega^2q^2)\) has matrix \(\tfrac12m\begin{pmatrix}1 & \gamma/2\\ \gamma/2 & \omega^2\end{pmatrix}\) in \((\dot q, q)\), with determinant \(\tfrac14m^2(\omega^2 - \gamma^2/4) \gt 0\) iff \(\gamma \lt 2\omega\): positive definite exactly in the underdamped case. Two positive definite forms on \(\mathbb R^2\) are comparable: there are constants \(0 \lt c_1 \le c_2\) with \(c_1\mathcal E \le \tfrac12m(\dot q^2 + \gamma q\dot q + \omega^2q^2) \le c_2\mathcal E\), where \(\mathcal E = \tfrac12m\dot q^2 + \tfrac12m\omega^2q^2\) (take the extreme values of their ratio on the unit circle, which is compact). So \(I e^{-\gamma t}/c_2 \le \mathcal E(t) \le Ie^{-\gamma t}/c_1\).

Check. Dimensions: \([\gamma q\dot q]\) is \(\mathsf{s^{-1}\,m\,m\,s^{-1}} = [\dot q^2]\) ✓. Limit \(\gamma\to0\): \(L\) becomes the undamped oscillator, \(\xi\to\dot q\), and \(I\to h = \mathcal E\), the conserved energy ✓. The moral: invariance of the equations is not the hypothesis; invariance of \(L\) up to \(\dot F\) is, and when it fails the right move is to look for a different transformation that restores it.

Hidden symmetries, scaling, and the converse

P9 The scaling symmetry of Kepler medium

exercises: a symmetry of the equations that rescales the action, the balance law it gives, Kepler's third law, the virial theorem, a number

Take unit 08's relative Lagrangian \(L = \tfrac12\mu\lvert\dot{\vec r}\rvert^2 + k/r\).

  1. Show that if \(\vec r(t)\) solves the equations of motion, so does \(\vec r_\lambda(t) = \lambda^2\vec r(t/\lambda^3)\) for every \(\lambda \gt 0\), and that the action over corresponding intervals is multiplied by \(\lambda\).
  2. With \(\lambda = e^s\), the generator is \(\xi = 2\vec r - 3t\dot{\vec r}\). Show that \(\delta L = \frac{\mathrm d}{\mathrm dt}(-3tL) + L\) for every curve — not a total derivative — and deduce the on-shell balance law \(\frac{\mathrm d}{\mathrm dt}\big(2\vec r\cdot\vec p - 3Et\big) = L\).
  3. Integrate the balance law over one period \(\mathcal T\) of a bound orbit. Deduce \(S_{\mathrm{orbit}} = -3E\mathcal T\), where \(S_{\mathrm{orbit}}\) is the action of one orbit, and hence the virial theorem \(\langle T\rangle = -E\), \(\langle V\rangle = 2E\) for time averages over a period. (Throughout, \(\mathcal T\) is the period and \(T\) the kinetic energy.)
  4. Recognise Kepler's third law: using part 1's scalings of \(S\), \(E\) and lengths, show that \(\mathcal T = -S_{\mathrm{orbit}}/3E\) forces \(\mathcal T^2/a^3\) to be the same for every orbit in the family.
  5. Predict Jupiter's period from the Earth's, both as a pure scaling (\(a_J = 5.2044\) AU, \(a_\oplus = 1.00000261\) AU, \(\mathcal T_\oplus = 365.2578\) d) and with the correct masses; the scaling holds at fixed \(k/\mu = G(m_1 + m_2)\).
Hints · 3 rungs
Rung 1

This is a symmetry of the equations, not of \(L\), so the answer to part 2 is not a conserved charge but a balance law whose right-hand side is \(L\). The rest follows by integrating over a period, where anything periodic drops out.

Rung 2

Part 1: each side of \(\mu\ddot{\vec r} = -k\vec r/r^3\) picks up \(\lambda^{-4}\). For \(\xi\): differentiate \(\lambda^2\vec r(t/\lambda^3)\) in \(\lambda\) at \(1\). For \(\delta L\), \(\dot\xi = -\dot{\vec r} - 3t\ddot{\vec r}\); compare with \(\frac{\mathrm d}{\mathrm dt}(tL) = L + t\dot L\).

Rung 3

First line of 2: \(\delta L = -\frac{k}{r^3}\vec r\cdot(2\vec r - 3t\dot{\vec r}) + \mu\dot{\vec r}\cdot(-\dot{\vec r} - 3t\ddot{\vec r}) = -\frac{2k}{r} - \mu\lvert\dot{\vec r}\rvert^2 - 3t\big(\mu\dot{\vec r}\cdot\ddot{\vec r} - k\,\vec r\cdot\dot{\vec r}/r^3\big)\), and the bracket is \(\dot L\).

Worked solution

Idea. Stretching lengths by \(\lambda^2\) and times by \(\lambda^3\) keeps \(k/\mu\), which has dimensions \(\mathsf{L^3T^{-2}}\), fixed — so it maps Kepler orbits to Kepler orbits. It multiplies \(L\) by \(\lambda^{-2}\) and \(\mathrm dt\) by \(\lambda^3\), so the action scales by \(\lambda\): Noether's theorem does not give a conserved quantity, but its proof gives a balance law, and periodicity turns that into the virial theorem and Kepler's third law.

1. With \(\tau = t/\lambda^3\): \(\dot{\vec r}_\lambda = \lambda^{-1}\vec r\,{}'(\tau)\), \(\ddot{\vec r}_\lambda = \lambda^{-4}\vec r\,{}''(\tau)\), and \(-k\vec r_\lambda/r_\lambda^3 = \lambda^{-4}(-k\vec r/r^3)\). Both sides scale by \(\lambda^{-4}\) ✓. Then \(L(\vec r_\lambda,\dot{\vec r}_\lambda) = \lambda^{-2}L(\vec r,\vec r\,{}')\) and \(\mathrm dt = \lambda^3\mathrm d\tau\), so \(S\to\lambda S\).

2. \(\partial_\lambda[\lambda^2\vec r(t/\lambda^3)]_{\lambda=1} = 2\vec r - 3t\dot{\vec r}\), and with \(\lambda = e^s\), \(\partial_s = \lambda\partial_\lambda\), the same at \(s=0\). From rung 3, \(\delta L = -\frac{2k}{r} - \mu\lvert\dot{\vec r}\rvert^2 - 3t\dot L\). Since \(-3\frac{\mathrm d}{\mathrm dt}(tL) = -3L - 3t\dot L\), \[ \delta L = \frac{\mathrm d}{\mathrm dt}(-3tL) + 3L - \frac{2k}{r} - \mu\lvert\dot{\vec r}\rvert^2 = \frac{\mathrm d}{\mathrm dt}(-3tL) + \tfrac12\mu\lvert\dot{\vec r}\rvert^2 + \frac kr = \frac{\mathrm d}{\mathrm dt}(-3tL) + L . \] On-shell, \(\delta L = \frac{\mathrm d}{\mathrm dt}(\vec p\cdot\xi)\) (the Euler–Lagrange equation, as on the off-shell warning slide). So \(\frac{\mathrm d}{\mathrm dt}(\vec p\cdot\xi + 3tL) = L\), and \(\vec p\cdot\xi + 3tL = 2\vec r\cdot\vec p - 3t(\mu\lvert\dot{\vec r}\rvert^2 - L) = 2\vec r\cdot\vec p - 3tE\), since \(\mu\lvert\dot{\vec r}\rvert^2 - L = h = E\). ∎ (So \(2\vec r\cdot\vec p - 3Et - \int_0^tL\) is conserved.)

3. Over one period \(\vec r\cdot\vec p\) returns to its value, so \(-3E\mathcal T = \int_0^{\mathcal T}L\,\mathrm dt = S_{\mathrm{orbit}}\), i.e. \(\langle L\rangle = \langle T\rangle - \langle V\rangle = -3E\) (here \(\mathcal T\) is the period and \(T\) the kinetic energy). With \(\langle T\rangle + \langle V\rangle = E\): \(\langle T\rangle = -E\), \(\langle V\rangle = 2E\). For Jupiter, \(E = -1.61775\times10^{35}\) J and \(\mathcal T = 11.8674\) yr give \(S_{\mathrm{orbit}} = 1.8176\times10^{44}\ \mathrm{J\,s}\).

4. Along the family, \(S_{\mathrm{orbit}}\to\lambda S_{\mathrm{orbit}}\) and \(E\to\lambda^{-2}E\) (energy is \(L\)-like), so \(\mathcal T = -S_{\mathrm{orbit}}/3E\to\lambda^3\mathcal T\), while every length, \(a\) included, goes to \(\lambda^2a\). Hence \(\mathcal T^2/a^3\to\lambda^6\mathcal T^2/\lambda^6a^3\): invariant. The scaling preserves shape, so this says \(\mathcal T^2/a^3\) is the same for all orbits of a given eccentricity. That it is also independent of the eccentricity is not a scaling fact: unit 08's areal-velocity argument supplies it, and the slides' hidden-symmetry section explains it (the energy depends only on \(L^2 + M^2\), hence only on \(a\)). Together: Kepler's third law, with the constant \(4\pi^2\mu/k\) read off from the circular orbit.

5. Pure scaling: \(\mathcal T_J = \mathcal T_\oplus(a_J/a_\oplus)^{3/2} = 365.2578\ \mathrm d\times11.87283 = 11.8731\) yr. With masses: multiply by \(\sqrt{(GM_\odot + GM_\oplus)/(GM_\odot + GM_J)} = 0.999525\), giving \(11.8674\) yr — the direct value \(2\pi\sqrt{a_J^3/G(m_S+m_J)}\) ✓. The \(0.048\%\) difference is Jupiter's mass: Jupiter and the Earth are in different scaling families.

Check. Dimensions: \([\vec r\cdot\vec p] = [Et] = \mathsf{ML^2T^{-1}}\), an action, as the charge of a dimensionless parameter must be ✓. Circular orbit: \(T = k/2r\), \(V = -k/r\), \(E = -k/2r\), so \(\langle T\rangle = -E\) and \(\langle V\rangle = 2E\) hold pointwise ✓.

P10 Finding the Runge–Lenz generator hard

exercises: generalised (velocity-dependent) symmetries, off-shell total derivatives, where the \(1/r\) potential is used

For \(L = \tfrac12\mu\lvert\vec v\rvert^2 + k/r\), with \(\vec v = \dot{\vec r}\), \(\vec a = \ddot{\vec r}\) and a fixed unit vector \(\hat e\), try the most general generator linear separately in \(\hat e\), \(\vec r\) and \(\vec v\):

\[ \xi = \mu\big(\alpha(\hat e\cdot\vec r)\,\vec v + \beta(\hat e\cdot\vec v)\,\vec r + \gamma(\vec r\cdot\vec v)\,\hat e\big). \]
  1. Show that the terms of \(\delta L\) containing \(\vec a\) can equal those of some \(\dot F\), with \(F = F(\vec r,\vec v)\), only if \(\gamma = \beta\).
  2. Show that the remaining velocity-cubic terms vanish only if \(\beta = -\alpha/2\).
  3. With \(\alpha = 2\), show that the potential terms are a total derivative on their own, find \(F\), and show the charge is \(\hat e\cdot\vec A\).
  4. Redo part 3 with the potential \(k/r^n\), \(n\ne1\), and show it fails.
Hints · 3 rungs
Rung 1

The answer is two linear constraints on \((\alpha,\beta,\gamma)\), a function \(F\), and a charge you already know. The organising fact: if \(\dot F\) contains \(\vec a\) only through \(\partial_{\vec v}F\cdot\vec a\), then the coefficient of \(\vec a\) in \(\delta L\) must be a gradient in \(\vec v\).

Rung 2

The \(\vec a\)-terms of \(\mu\vec v\cdot\dot\xi\) are \(\mu^2\big[\alpha(\hat e\cdot\vec r)(\vec v\cdot\vec a) + \beta(\vec r\cdot\vec v)(\hat e\cdot\vec a) + \gamma(\hat e\cdot\vec v)(\vec r\cdot\vec a)\big] = \mu^2\,\vec W\cdot\vec a\). A vector field \(\vec W(\vec v)\) is a gradient only if its Jacobian \(\partial W_j/\partial v_k\) is symmetric.

Rung 3

First line: \(\partial W_j/\partial v_k = \alpha(\hat e\cdot\vec r)\delta_{jk} + \beta\,\hat e_jr_k + \gamma\,r_j\hat e_k\), symmetric iff \((\beta - \gamma)(\hat e_jr_k - \hat e_kr_j) = 0\). Then the gradient is of \(\mu^2\big[\tfrac\alpha2\lvert\vec v\rvert^2(\hat e\cdot\vec r) + \beta(\vec r\cdot\vec v)(\hat e\cdot\vec v)\big]\).

Worked solution

Idea. Demanding "\(\delta L\) is a total derivative for every curve" is a strong condition, because \(\vec a\) is arbitrary. It is linear in the unknown coefficients and splits by degree: the \(\vec a\)-terms, the terms cubic in \(\vec v\), and the potential terms. Each gives one condition, and the last is satisfied only by the inverse-square law.

1. \(\dot\xi/\mu = \alpha(\hat e\cdot\vec v)\vec v + \alpha(\hat e\cdot\vec r)\vec a + \beta(\hat e\cdot\vec a)\vec r + \beta(\hat e\cdot\vec v)\vec v + \gamma(\lvert\vec v\rvert^2 + \vec r\cdot\vec a)\hat e\). Dot with \(\mu\vec v\):

\[ \mu\vec v\cdot\dot\xi = \mu^2\Big[(\alpha + \beta + \gamma)\lvert\vec v\rvert^2(\hat e\cdot\vec v) + \vec W\cdot\vec a\Big],\quad \vec W = \alpha(\hat e\cdot\vec r)\vec v + \beta(\vec r\cdot\vec v)\hat e + \gamma(\hat e\cdot\vec v)\vec r . \]

The potential term \(-\frac{k}{r^3}\vec r\cdot\xi\) has no \(\vec a\). And \(\dot F = \partial_{\vec r}F\cdot\vec v + \partial_{\vec v}F\cdot\vec a\), so matching the \(\vec a\)-terms for all \(\vec a\) requires \(\partial_{\vec v}F = \mu^2\vec W\): \(\vec W\) must be a gradient in \(\vec v\) (at fixed \(\vec r\)), so its Jacobian must be symmetric. \(\partial W_j/\partial v_k = \alpha(\hat e\cdot\vec r)\delta_{jk} + \beta\hat e_jr_k + \gamma r_j\hat e_k\), whose antisymmetric part is \(\tfrac12(\beta - \gamma)(\hat e_jr_k - \hat e_kr_j)\). For \(\vec r\) not parallel to \(\hat e\) this vanishes only if \(\gamma = \beta\). Then \(F = \mu^2\big[\tfrac\alpha2\lvert\vec v\rvert^2(\hat e\cdot\vec r) + \beta(\vec r\cdot\vec v)(\hat e\cdot\vec v)\big] + f(\vec r)\).

2. With \(\gamma = \beta\), the \(\vec a\)-free part of \(\dot F\) from its quadratic piece is \(\mu^2\big[\tfrac\alpha2\lvert\vec v\rvert^2(\hat e\cdot\vec v) + \beta\lvert\vec v\rvert^2(\hat e\cdot\vec v)\big]\) (differentiating \(\hat e\cdot\vec r\) and \(\vec r\cdot\vec v\) through their \(\vec r\)), while \(\delta L\) has \(\mu^2(\alpha + 2\beta)\lvert\vec v\rvert^2(\hat e\cdot\vec v)\). The difference, \(\mu^2(\tfrac\alpha2 + \beta)\lvert\vec v\rvert^2(\hat e\cdot\vec v)\), is cubic in \(\vec v\) and must be matched by \(\nabla f\cdot\vec v\), which is linear. So \(\beta = -\alpha/2\).

3. Take \(\alpha = 2\), \(\beta = \gamma = -1\). Then \(\vec r\cdot\xi = \mu\big[2(\hat e\cdot\vec r)(\vec r\cdot\vec v) - r^2(\hat e\cdot\vec v) - (\vec r\cdot\vec v)(\hat e\cdot\vec r)\big] = \mu\big[(\hat e\cdot\vec r)(\vec r\cdot\vec v) - r^2(\hat e\cdot\vec v)\big]\), so \[ -\frac{k}{r^3}\vec r\cdot\xi = \mu k\Big[\frac{\hat e\cdot\vec v}{r} - \frac{(\hat e\cdot\vec r)(\vec r\cdot\vec v)}{r^3}\Big] = \mu k\,\frac{\mathrm d}{\mathrm dt}\frac{\hat e\cdot\vec r}{r}, \] using \(\dot r = \vec r\cdot\vec v/r\). So \(f = \mu k\,\hat e\cdot\vec r/r\) and \(F = \mu^2\big[\lvert\vec v\rvert^2(\hat e\cdot\vec r) - (\vec r\cdot\vec v)(\hat e\cdot\vec v)\big] + \mu k\,\hat e\cdot\vec r/r\). The charge: \(\mu\vec v\cdot\xi = \mu^2\big[2\lvert\vec v\rvert^2(\hat e\cdot\vec r) - 2(\hat e\cdot\vec v)(\vec r\cdot\vec v)\big]\), so \[ Q = \mu^2\big[\lvert\vec v\rvert^2(\hat e\cdot\vec r) - (\vec r\cdot\vec v)(\hat e\cdot\vec v)\big] - \mu k\frac{\hat e\cdot\vec r}{r} = \hat e\cdot\big(\vec p\times\vec L - \mu k\hat r\big) = \hat e\cdot\vec A, \] by BAC–CAB, \(\vec p\times(\vec r\times\vec p) = \vec r\lvert\vec p\rvert^2 - \vec p(\vec r\cdot\vec p)\).

4. For \(V = -k/r^n\), \(\partial_{\vec r}L = -nk\vec r/r^{n+2}\), and the potential part of \(\delta L\) becomes \(n\mu k\big[r^{-n}(\hat e\cdot\vec v) - r^{-n-2}(\hat e\cdot\vec r)(\vec r\cdot\vec v)\big]\). It must equal \(\nabla f\cdot\vec v\) for all \(\vec v\), i.e. \(\nabla f = n\mu k\big[r^{-n}\hat e - r^{-n-2}(\hat e\cdot\vec r)\vec r\big]\). A gradient field has symmetric Jacobian; the Jacobian of \(r^{-n}\hat e\) contains \(-nr^{-n-2}\hat e_jr_k\), and that of \(-r^{-n-2}(\hat e\cdot\vec r)\vec r\) contains \(-r^{-n-2}r_j\hat e_k\) and symmetric terms. The antisymmetric part is \(\propto(1 - n)r^{-n-2}(\hat e_jr_k - r_j\hat e_k)\cdot\tfrac12\), zero only for \(n = 1\). So the ansatz fails for every other power, matching unit 08's finding that \(\vec A\) is special to the inverse-square law.

Check. Dimensions: \([Q] = [\mu^2v^2r] = \mathsf{M^2L^3T^{-2}} = [\mu k]\) ✓. A number: for Jupiter, \(\lvert\vec A\rvert/\mu k = e = 0.0489\), and \(\lvert\vec A\rvert = 2.3359\times10^{73}\ \mathrm{kg^2\,m^3\,s^{-2}}\), agreeing with \(\sqrt{\mu^2k^2 + 2\mu E\lvert\vec L\rvert^2}\) from unit 08 to five figures ✓.

P11 Proof: the converse fails hard

exercises: what charges of point symmetries look like, a conserved quantity from no point symmetry, the phase-space rescue

Work with the relative Kepler Lagrangian \(L = \tfrac12\mu\lvert\dot{\vec r}\rvert^2 + k/r\) on \(\mathbb R^3\setminus\{0\}\).

  1. Prove: if \(\xi = \xi(\vec r,t)\) generates a quasi-symmetry, then \(F\) cannot depend on \(\dot{\vec r}\), and the charge is an affine function of \(\dot{\vec r}\) at each \((\vec r,t)\).
  2. Adding time translation with a constant coefficient, \(\xi + c\,\dot{\vec r}\), adds \(c\,h\) to the charge. Show that every charge obtained this way has velocity-quadratic part \(\tfrac c2\mu\lvert\dot{\vec r}\rvert^2\).
  3. Prove that no component \(\hat e\cdot\vec A\) of the Runge–Lenz vector is such a charge.
  4. Explain in a paragraph what enlarging the arena to phase space \((\vec r,\vec p)\) (unit 14) changes, so that \(\hat e\cdot\vec A\) is the charge of a one-parameter group.
Hints · 3 rungs
Rung 1

A proof by comparing polynomial degrees in the velocity, at a fixed point \((\vec r,t)\). The objects to compare are two quadratic forms in \(\vec v = \dot{\vec r}\), i.e. two symmetric \(3\times3\) matrices.

Rung 2

For part 1, \(\ddot{\vec r}\) appears in \(\dot F\) through \(\partial_{\dot{\vec r}}F\cdot\ddot{\vec r}\) but nowhere in \(\delta L\) when \(\xi\) has no \(\dot{\vec r}\). For part 3, write out the velocity-quadratic part of \(\hat e\cdot\vec A\) by BAC–CAB and ask whether its matrix can be a multiple of the identity.

Rung 3

First line of 3: the quadratic part is \(\mu^2\big[\lvert\vec v\rvert^2(\hat e\cdot\vec r) - (\vec r\cdot\vec v)(\hat e\cdot\vec v)\big]\), matrix \(\mu^2\big[(\hat e\cdot\vec r)I - \tfrac12(\vec r\hat e^{\top} + \hat e\vec r^{\top})\big]\); apply \(\vec r\hat e^{\top} + \hat e\vec r^{\top}\) to a vector perpendicular to both \(\vec r\) and \(\hat e\).

Worked solution

Idea. A transformation of positions produces charges of the form "velocity dotted into a function of position, plus a function of position": linear in the velocity. Time translation brings in the energy, whose quadratic part is isotropic in the velocity. The Runge–Lenz vector's quadratic part is not isotropic, so it is out of reach — until positions and momenta are allowed to mix.

1. With \(\xi = \xi(\vec r,t)\), \(\dot\xi = D\xi\,\dot{\vec r} + \partial_t\xi\) contains no \(\ddot{\vec r}\), so \(\delta L\) is a function of \((\vec r,\dot{\vec r},t)\). If \(\delta L = \dot F\) for every curve, fix any point \((\vec r,\dot{\vec r},t)\): curves through it can have any acceleration \(\vec a\), and \(\dot F = \partial_t F + \partial_{\vec r}F\cdot\dot{\vec r} + \partial_{\dot{\vec r}}F\cdot\vec a\) must be independent of \(\vec a\), so \(\partial_{\dot{\vec r}}F = 0\). Then \(Q = \mu\dot{\vec r}\cdot\xi(\vec r,t) - F(\vec r,t)\) is affine in \(\dot{\vec r}\). ∎

2. By linearity of the theorem in the generator, \(\xi + c\dot{\vec r}\) has charge \(Q + ch\) with \(h = \tfrac12\mu\lvert\dot{\vec r}\rvert^2 - k/r\). The velocity-quadratic part of \(Q\) is zero and of \(ch\) is \(\tfrac c2\mu\lvert\dot{\vec r}\rvert^2\), matrix \(\tfrac c2\mu I\). ∎

3. \(\vec p\times\vec L = \mu^2\,\vec v\times(\vec r\times\vec v) = \mu^2\big(\vec r\lvert\vec v\rvert^2 - \vec v(\vec r\cdot\vec v)\big)\), so the velocity-quadratic part of \(\hat e\cdot\vec A\) has matrix \(M = \mu^2\big[(\hat e\cdot\vec r)I - \tfrac12(\vec r\hat e^{\top} + \hat e\vec r^{\top})\big]\). If \(M = \tfrac c2\mu I\), then \(N := \vec r\hat e^{\top} + \hat e\vec r^{\top}\) is a multiple \(\kappa I\). Choose \(\vec w\ne0\) perpendicular to both \(\vec r\) and \(\hat e\) (possible in \(\mathbb R^3\)): \(N\vec w = \vec r(\hat e\cdot\vec w) + \hat e(\vec r\cdot\vec w) = 0\), so \(\kappa = 0\) and \(N = 0\). But \(N\vec r = (\hat e\cdot\vec r)\vec r + r^2\hat e\). If this vanished, then \(\hat e = -(\hat e\cdot\vec r)\vec r/r^2\) would be parallel to \(\vec r\), say \(\hat e = \sigma\hat r\) with \(\sigma = \pm1\); substituting, \(N\vec r = \sigma r^2\hat r + \sigma r^2\hat r = 2\sigma r^2\hat r\ne0\). So \(N\ne0\) at every \(\vec r\ne0\), a contradiction: no choice of \(\xi(\vec r,t)\) and \(c\) produces \(\hat e\cdot\vec A\), even allowing an added constant. ∎ (Allowing the time shift to vary, \(t\mapsto t + s\tau(\vec r,t)\), adds \(-\tau h\), whose quadratic part is \(-\tfrac\tau2\mu I\) at each point — still a multiple of \(I\), so the same proof applies.)

4. On phase space, any smooth function \(G(\vec r,\vec p)\) defines a vector field by Hamilton's equations with \(G\) in place of the Hamiltonian, \(\dot{\vec r} = \partial G/\partial\vec p\), \(\dot{\vec p} = -\partial G/\partial\vec r\), and so a one-parameter group of transformations of phase space (unit 02's flow). Unit 14 proves that this flow preserves \(H\) exactly when \(G\) is conserved by \(H\)'s flow — the same bracket read in both orders. Take \(G = \hat e\cdot\vec A\): its flow moves \(\vec r\) by \(\partial(\hat e\cdot\vec A)/\partial\vec p\), which depends on \(\vec p\) — this is P10's velocity-dependent \(\xi\), up to the factor relating \(\vec p\) and \(\mu\vec v\) — and it changes \(\vec p\) too, in a way no transformation of positions alone can induce. So the rescue is: allow transformations that mix positions and momenta. On that larger space the correspondence between one-parameter symmetries and conserved quantities is a bijection (up to constants).

Check. Special case of part 3's criterion: the energy \(h\) has quadratic matrix \(\tfrac12\mu I\) and is indeed a (time-translation) charge ✓; \(\hat e\cdot\vec L = \mu\,\hat e\cdot(\vec r\times\vec v)\) is linear in \(\vec v\) and is indeed a point-symmetry charge ✓. Dimensions: \([M] = \mu^2[r] = \mathsf{M^2L}\), and \([M\,v^2] = \mathsf{M^2L^3T^{-2}} = [\mu k]\) ✓.