Part E · unit 34 · problems

Differential forms and Cartan: Stokes' theorem was always one theorem

the slides · the curriculum

Checkpoint. You express Maxwell's four equations as \(\mathrm dF = 0\) and \(\mathrm d\star F = \mu_0\star J\), expand them back into components recovering the originals exactly, compute \(H^1(\mathbb R^2\setminus\{0\})\) — closing unit 04's open question — and reproduce one of unit 33's curvature computations by Cartan's structure equations.

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.

Twelve problems, about six hours. Work them on paper first. Conventions are the deck's: signature \((-,+,+,+)\); on Minkowski space the orthonormal coframe \(e^0 = c\,\mathrm dt\), \(e^i = \mathrm dx^i\) and \(\mathrm{vol} = e^0\wedge e^1\wedge e^2\wedge e^3\); \(\varepsilon^1 = \mathrm dy\wedge\mathrm dz\), \(\varepsilon^2 = \mathrm dz\wedge\mathrm dx\), \(\varepsilon^3 = \mathrm dx\wedge\mathrm dy\); \(F = B_i\,\varepsilon^i + E_i\,\mathrm dx^i\wedge\mathrm dt\), \(\star F = \tfrac1c E_i\,\varepsilon^i - c\,B_i\,\mathrm dx^i\wedge\mathrm dt\), \(J = -c^2\rho\,\mathrm dt + J_i\,\mathrm dx^i\); for a frame, \(\nabla e_b = \omega^a{}_b\otimes e_a\), \(\mathrm de^a = -\omega^a{}_b\wedge e^b\), \(\Omega^a{}_b = \mathrm d\omega^a{}_b + \omega^a{}_c\wedge\omega^c{}_b\). The vortex is \(\theta = (x\,\mathrm dy - y\,\mathrm dx)/(x^2+y^2)\).

Constants: \(c = 2.99792458\times10^{8}\ \mathrm{m\,s^{-1}}\) exactly, \(\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}\) (the convention of units 23 and 24), \(h = 6.62607015\times10^{-34}\ \mathrm{J\,s}\), \(e = 1.602176634\times10^{-19}\) C.

The algebra and the derivative

P1 Wedges and derivatives by hand routine

exercises: the wedge product, the exterior derivative, the graded Leibniz rule, pullback to a curve

On \(\mathbb R^3\) let \(\alpha = yz\,\mathrm dx + xz\,\mathrm dy + xy\,\mathrm dz\) and \(\beta = x\,\mathrm dy - y\,\mathrm dx\).

  1. Compute \(\alpha\wedge\beta\), in the basis \(\mathrm dy\wedge\mathrm dz\), \(\mathrm dz\wedge\mathrm dx\), \(\mathrm dx\wedge\mathrm dy\).
  2. Compute \(\mathrm d\alpha\) and \(\mathrm d\beta\).
  3. Compute \(\mathrm d(\alpha\wedge\beta)\) directly, and check the Leibniz rule \(\mathrm d(\alpha\wedge\beta) = \mathrm d\alpha\wedge\beta - \alpha\wedge\mathrm d\beta\). Where does the minus sign come from?
  4. Pull both forms back along the helix \(\gamma(t) = (\cos t, \sin t, t)\), \(0\le t\le 2\pi\), and integrate. Say in one line why the answer for \(\alpha\) could have been written down without integrating.
Hints · 3 rungs
Rung 1

Know the degree of every answer before computing it: \(\alpha\wedge\beta\), \(\mathrm d\alpha\) and \(\mathrm d\beta\) are two-forms; \(\mathrm d(\alpha\wedge\beta)\) is a three-form, so on \(\mathbb R^3\) a single function times \(\mathrm dx\wedge\mathrm dy\wedge\mathrm dz\). Part 4 is two numbers.

Rung 2

Expand \(\alpha\wedge\beta\) term by term: six products, of which the two containing \(\mathrm dx\wedge\mathrm dx\) or \(\mathrm dy\wedge\mathrm dy\) die, and every \(\mathrm dy\wedge\mathrm dx\) becomes \(-\mathrm dx\wedge\mathrm dy\). For \(\mathrm d\), use \(\mathrm d(f\,\mathrm dx^I) = \mathrm df\wedge\mathrm dx^I\). Look hard at \(\alpha\) before differentiating it.

Rung 3

The first line: \(\alpha\wedge\beta = yz\cdot x\,\mathrm dx\wedge\mathrm dy + xz\cdot(-y)\,\mathrm dy\wedge\mathrm dx + xy\cdot(-y)\,\mathrm dz\wedge\mathrm dx + xy\cdot x\,\mathrm dz\wedge\mathrm dy\).

Worked solution

Idea. Nothing but the rules: antisymmetry for \(\wedge\), the chart formula for \(\mathrm d\), and the chain rule for pullback. The one thing worth noticing is that \(\alpha = \mathrm d(xyz)\).

1. From rung 3, using \(\mathrm dy\wedge\mathrm dx = -\mathrm dx\wedge\mathrm dy\) and \(\mathrm dz\wedge\mathrm dy = -\mathrm dy\wedge\mathrm dz\), \[ \alpha\wedge\beta = -x^2y\,\mathrm dy\wedge\mathrm dz - xy^2\,\mathrm dz\wedge\mathrm dx + 2xyz\,\mathrm dx\wedge\mathrm dy . \]

2. \(\alpha = \mathrm d(xyz)\), so \(\mathrm d\alpha = \mathrm d\mathrm d(xyz) = 0\). Directly, the \(\mathrm dx\wedge\mathrm dy\) coefficient is \(\partial_x(xz) - \partial_y(yz) = z - z = 0\), and the other two likewise. And \(\mathrm d\beta = \mathrm dx\wedge\mathrm dy - \mathrm dy\wedge\mathrm dx = 2\,\mathrm dx\wedge\mathrm dy\).

3. Only the derivative in the missing direction survives in each term: \[ \mathrm d(\alpha\wedge\beta) = \partial_x(-x^2y)\,\mathrm dx\wedge\mathrm dy\wedge\mathrm dz + \partial_y(-xy^2)\,\mathrm dy\wedge\mathrm dz\wedge\mathrm dx + \partial_z(2xyz)\,\mathrm dz\wedge\mathrm dx\wedge\mathrm dy . \] All three basis three-forms equal \(\mathrm dx\wedge\mathrm dy\wedge\mathrm dz\) (cyclic permutations are even), so \(\mathrm d(\alpha\wedge\beta) = (-2xy - 2xy + 2xy)\,\mathrm dx\wedge\mathrm dy\wedge\mathrm dz = -2xy\,\mathrm dx\wedge\mathrm dy\wedge\mathrm dz\). The right side: \(\mathrm d\alpha\wedge\beta - \alpha\wedge\mathrm d\beta = 0 - 2\alpha\wedge\mathrm dx\wedge\mathrm dy = -2xy\,\mathrm dz\wedge\mathrm dx\wedge\mathrm dy = -2xy\,\mathrm dx\wedge\mathrm dy\wedge\mathrm dz\) ✓. The minus sign is \((-1)^k\) with \(k = \deg\alpha = 1\): \(\mathrm d\) is a one-form-like operator, and reaching \(\beta\) it has to pass \(\alpha\).

4. \(\gamma^*\mathrm dx = -\sin t\,\mathrm dt\), \(\gamma^*\mathrm dy = \cos t\,\mathrm dt\), \(\gamma^*\mathrm dz = \mathrm dt\). So \(\gamma^*\beta = (\cos^2t + \sin^2t)\,\mathrm dt = \mathrm dt\) and \(\int_\gamma\beta = 2\pi\). For \(\alpha\), naturality gives \(\gamma^*\alpha = \mathrm d(\gamma^*(xyz)) = \mathrm d(t\sin t\cos t)\), so \(\int_\gamma\alpha = [t\sin t\cos t]_0^{2\pi} = 0\). Without integrating: \(\alpha\) is exact, so by the gradient theorem (Stokes with \(n = 1\)) its integral is \(xyz\) at the end, \((1)(0)(2\pi) = 0\), minus \(xyz\) at the start, \(0\).

Check. Special case: projected onto the \(xy\)-plane the helix is the unit circle, and \(\tfrac12\oint(x\,\mathrm dy - y\,\mathrm dx)\) is the enclosed area (Green, with \(\mathrm d\beta = 2\,\mathrm dx\wedge\mathrm dy\)), so \(\int_\gamma\beta = 2\times\pi = 2\pi\) ✓. Dimensions: if \(x,y,z\) are lengths, \(\alpha\) is length\(^3\), \(\beta\) length\(^2\), \(\alpha\wedge\beta\) length\(^5\) and \(\mathrm d\) does not change it: \(-2xy\,\mathrm dx\,\mathrm dy\,\mathrm dz\) is length\(^5\) ✓.

P2 d squared in disguise medium

exercises: the exterior derivative in components, Clairaut's theorem, the homogeneous Maxwell pair, charge conservation — a proof

Let \(A = A_x\,\mathrm dx + A_y\,\mathrm dy + A_z\,\mathrm dz - \varphi\,\mathrm dt\) on \(\mathbb R^4\), with arbitrary smooth components, and \(F = \mathrm dA\).

  1. Compute \(\mathrm d(\mathrm dA)\) in components without quoting \(\mathrm d^2 = 0\), and show that each of its four coefficients vanishes by Clairaut's theorem. Name the vector identity each one is.
  2. Deduce that \(\nabla\cdot\vec B = 0\) and \(\nabla\times\vec E + \partial_t\vec B = 0\) hold for any potentials. What, then, is the physical content of the homogeneous pair?
  3. From the preamble's \(\star F\), compute \(\mathrm d(\mathrm d\star F)\) in components and show it vanishes identically. Then apply \(\mathrm d\) to \(\mathrm d\star F = \mu_0\star J\) and read off charge conservation.
  4. Replace Ampère–Maxwell by Ampère's original \(\nabla\times\vec B = \mu_0\vec J\), keeping Gauss's law. Show that this forces \(\nabla\cdot\vec J = 0\), and say why that is incompatible with a charging capacitor.
Hints · 3 rungs
Rung 1

The answer is a proof. \(\mathrm d(\mathrm dA)\) is a three-form on \(\mathbb R^4\): four coefficients, along \(\mathrm dx\wedge\mathrm dy\wedge\mathrm dz\) and the three \(\mathrm dt\wedge\varepsilon^i\). Each must turn out to be a sum of mixed partial derivatives that cancel in pairs.

Rung 2

Example 1 of the deck computed \(\mathrm dA = B_i\varepsilon^i + E_i\,\mathrm dx^i\wedge\mathrm dt\) with \(\vec B = \nabla\times\vec A\) and \(\vec E = -\nabla\varphi - \partial_t\vec A\); example 2 computed \(\mathrm d\) of any two-form of that shape. Use both, with these particular \(\vec B\) and \(\vec E\), and then open the brackets.

Rung 3

The first line: \(\nabla\cdot(\nabla\times\vec A) = \partial_x(\partial_yA_z - \partial_zA_y) + \partial_y(\partial_zA_x - \partial_xA_z) + \partial_z(\partial_xA_y - \partial_yA_x)\).

Worked solution

Idea. \(\mathrm d^2 = 0\) is Clairaut's theorem, and the classical identities "div curl \(= 0\)" and "curl grad \(= 0\)" are its components. Two of Maxwell's equations and charge conservation are those identities, applied to \(A\) and to \(\star F\).

1. Example 2's expansion used nothing about \(F\) but its shape, so \[ \mathrm d(\mathrm dA) = \big(\nabla\cdot(\nabla\times\vec A)\big)\,\mathrm dx\wedge\mathrm dy\wedge\mathrm dz + \big[\partial_t(\nabla\times\vec A) + \nabla\times(-\nabla\varphi - \partial_t\vec A)\big]_i\,\mathrm dt\wedge\varepsilon^i . \] The first coefficient, from rung 3, is \((\partial_x\partial_y - \partial_y\partial_x)A_z + (\partial_y\partial_z - \partial_z\partial_y)A_x + (\partial_z\partial_x - \partial_x\partial_z)A_y = 0\): div curl. The bracket is \(-\nabla\times\nabla\varphi + \big(\partial_t\nabla\times\vec A - \nabla\times\partial_t\vec A\big)\). Its first term has \(x\)-component \((\partial_y\partial_z - \partial_z\partial_y)\varphi = 0\): curl grad. The second vanishes because \(\partial_t\) commutes with \(\partial_y\) and \(\partial_z\) — Clairaut again, now in a time–space pair. Four coefficients, all zero, by Clairaut and nothing else.

2. Since \(F = \mathrm dA\) gives \(\vec B = \nabla\times\vec A\) and \(\vec E = -\nabla\varphi - \partial_t\vec A\), the expansion of \(\mathrm dF\) in example 2 reads \(\nabla\cdot\vec B = 0\) and \(\nabla\times\vec E + \partial_t\vec B = 0\), now as identities. So the homogeneous pair has no content once potentials exist; its content is that they exist. On \(\mathbb R^4\), star-shaped, the Poincaré lemma makes "\(\mathrm dF = 0\)" and "\(F = \mathrm dA\) for some \(A\)" the same statement — no magnetic charge, and Faraday's induction law, are the claim that \(\vec E\) and \(\vec B\) come from one potential.

3. Write \(\mathrm d\star F = \tfrac1c(\nabla\cdot\vec E)\,\mathrm dx\wedge\mathrm dy\wedge\mathrm dz + V_i\,\mathrm dt\wedge\varepsilon^i\) with \(\vec V = \tfrac1c\partial_t\vec E - c\,\nabla\times\vec B\) (example 4). Then \[ \mathrm d(\mathrm d\star F) = \tfrac1c\partial_t(\nabla\cdot\vec E)\,\mathrm dt\wedge\mathrm dx\wedge\mathrm dy\wedge\mathrm dz + \partial_jV_i\,\mathrm dx^j\wedge\mathrm dt\wedge\varepsilon^i . \] Only \(j = i\) survives, and \(\mathrm dx^i\wedge\mathrm dt\wedge\varepsilon^i = -\mathrm dt\wedge\mathrm dx\wedge\mathrm dy\wedge\mathrm dz\), so the coefficient of \(\mathrm dt\wedge\mathrm dx\wedge\mathrm dy\wedge\mathrm dz\) is \[ \tfrac1c\partial_t(\nabla\cdot\vec E) - \nabla\cdot\vec V = \tfrac1c\partial_t\nabla\cdot\vec E - \tfrac1c\nabla\cdot\partial_t\vec E + c\,\nabla\cdot(\nabla\times\vec B) = 0 , \] by Clairaut twice. Applying \(\mathrm d\) to the equation: \(0 = \mu_0\,\mathrm d\star J\), and the deck's computation gives \(\mathrm d\star J = c(\partial_t\rho + \nabla\cdot\vec J)\,\mathrm dt\wedge\mathrm dx\wedge\mathrm dy\wedge\mathrm dz\). So \(\partial_t\rho + \nabla\cdot\vec J = 0\).

4. Take the divergence of \(\nabla\times\vec B = \mu_0\vec J\): the left side is zero identically, so \(\nabla\cdot\vec J = 0\) always. But continuity says \(\nabla\cdot\vec J = -\partial_t\rho\), and on a charging capacitor plate \(\partial_t\rho\ne0\). No field \(\vec B\) satisfies the equation there. With the displacement term, the divergence gives \(\mu_0\nabla\cdot\vec J + \mu_0\varepsilon_0\partial_t\nabla\cdot\vec E = \mu_0(\nabla\cdot\vec J + \partial_t\rho) = 0\), consistent. In forms: the right side of \(\mathrm d\star F = \mu_0\star J\) must be closed, because the left is exact.

Check. Dimensions of part 3: \([\tfrac1c\partial_t\nabla\cdot\vec E] = \mathrm{s\,m^{-1}}\cdot\mathrm{s^{-1}}\cdot\mathrm{V\,m^{-2}} = \mathrm{V\,m^{-3}}\) and \([c\,\nabla\cdot\nabla\times\vec B] = \mathrm{m\,s^{-1}\,T\,m^{-2}} = \mathrm{V\,m^{-3}}\) (as \(\mathrm{T} = \mathrm{V\,s\,m^{-2}}\)) ✓. Special case: statics, \(\partial_t\rho = 0\), requires \(\nabla\cdot\vec J = 0\): steady currents, which is exactly where Ampère's original law is right ✓.

P3 The star, and the signature it does not care about medium

exercises: the Hodge star from its definition, the double-star sign, where the metric enters, sign conventions

Work on Minkowski space with the preamble's coframe and \(\eta = \operatorname{diag}(-1,1,1,1)\).

  1. From \(\alpha\wedge\star\beta = \langle\alpha,\beta\rangle\,\mathrm{vol}\) alone, compute \(\star e^0\), \(\star e^1\), \(\star(e^1\wedge e^2\wedge e^3)\) and \(\star(e^0\wedge e^2\wedge e^3)\). Check \(\star\star = (-1)^{k(n-k)}s\) on one-forms and three-forms.
  2. Now use the opposite signature \(\eta' = -\eta\), the other common convention, with the same orientation. Show \(\mathrm{vol}' = \mathrm{vol}\) and \(\langle\cdot,\cdot\rangle' = (-1)^k\langle\cdot,\cdot\rangle\) on \(k\)-forms, and hence \(\star' = (-1)^k\star\).
  3. The physical current is the vector \(\rho\,\partial_t + J^i\partial_i\); its one-form is obtained by lowering the index, which uses the metric. Show that the one-form changes sign, \(J' = -J\), and conclude that \(\mathrm dF = 0\) and \(\mathrm d\star F = \mu_0\star J\) are word for word the same equations in either signature.
Hints · 3 rungs
Rung 1

One-forms go to three-forms and three-forms to one-forms, and each answer in part 1 is \(\pm\) a single basis element. The whole job is the sign.

Rung 2

For a basis element \(\beta = e^I\), try \(\star\beta = \lambda\,e^{I^c}\) with \(I^c\) the missing indices. \(\langle\beta,\beta\rangle\) is the product of \(\eta^{aa}\) over \(a\in I\), and \(e^I\wedge e^{I^c}\) is \(\pm\mathrm{vol}\) according to the parity of the reordering. For part 2, count how many inverse-metric entries go into a \(k\times k\) determinant.

Rung 3

The first line: \(\langle e^0,e^0\rangle = \eta^{00} = -1\), so we need \(e^0\wedge\star e^0 = -\mathrm{vol}\); with \(\star e^0 = \lambda\,e^1\wedge e^2\wedge e^3\) this is \(\lambda\,\mathrm{vol} = -\mathrm{vol}\).

Worked solution

Idea. The star is fixed by the inner product and the orientation. Flipping the signature changes the inner product by a sign per slot, leaves the volume form alone, and flips the lowering map — and in Maxwell's equations the flips cancel.

1. From rung 3, \(\star e^0 = -e^1\wedge e^2\wedge e^3\). \(\langle e^1,e^1\rangle = +1\); with \(\star e^1 = \lambda\,e^0\wedge e^2\wedge e^3\), \(e^1\wedge e^0\wedge e^2\wedge e^3 = -\mathrm{vol}\) (one transposition), so \(\lambda = -1\): \(\star e^1 = -e^0\wedge e^2\wedge e^3\). \(\langle e^{123},e^{123}\rangle = +1\); with \(\star e^{123} = \lambda e^0\), \(e^1\wedge e^2\wedge e^3\wedge e^0 = -\mathrm{vol}\) (three transpositions), so \(\star(e^1\wedge e^2\wedge e^3) = -e^0\). \(\langle e^{023},e^{023}\rangle = \eta^{00}\eta^{22}\eta^{33} = -1\); with \(\star e^{023} = \lambda e^1\), \(e^0\wedge e^2\wedge e^3\wedge e^1 = +\mathrm{vol}\) (two transpositions), so \(\lambda = -1\): \(\star(e^0\wedge e^2\wedge e^3) = -e^1\).

Double star: \(\star\star e^0 = \star(-e^{123}) = e^0\) and \(\star\star e^1 = \star(-e^{023}) = e^1\); so \(\star\star = +1\) on one-forms, and then on three-forms too (\(\star\star e^{123} = \star(-e^0) = e^{123}\)). The formula: \((-1)^{k(4-k)}s\) with \(s = -1\) is \((-1)^{3}(-1) = +1\) for \(k = 1\) and \(k = 3\) ✓.

2. \(\det\eta' = (-1)^4\det\eta = \det\eta\), so \(\sqrt{|\det\eta'|}\) is unchanged; the orientation is the same, so \(\mathrm{vol}' = \mathrm{vol}\). On one-forms \(\langle\alpha,\beta\rangle' = \eta'^{ab}\alpha_a\beta_b = -\langle\alpha,\beta\rangle\). On decomposable \(k\)-forms the inner product is a \(k\times k\) determinant of one-form inner products, each negated, so it picks up \((-1)^k\); by linearity this holds on all of \(\Lambda^k\). Then for all \(\alpha\), \(\alpha\wedge\star'\beta = (-1)^k\langle\alpha,\beta\rangle\mathrm{vol} = \alpha\wedge\big((-1)^k\star\beta\big)\), and since the pairing \(\Lambda^k\times\Lambda^{n-k}\to\Lambda^n\) is nondegenerate, \(\star'\beta = (-1)^k\star\beta\).

3. Lowering with \(\eta'\) is lowering with \(\eta\) followed by a sign: for the current vector \(j\), \(J' = \eta'(j,\cdot) = -\eta(j,\cdot) = -J\). Check with \(\eta\): lowering \(\rho\,\partial_t\) with \(g_{tt} = -c^2\) gives \(-c^2\rho\,\mathrm dt\), the preamble's \(J\). Now \(\star'J' = (-1)^1\star(-J) = \star J\). And \(A\) is a one-form from the start — the potentials are covector components, not raised from anything — so \(F = \mathrm dA\) and \(\mathrm dF = 0\) involve no metric, while \(\star'F = (-1)^2\star F = \star F\). Hence \(\mathrm d\star'F = \mu_0\star'J'\) is literally \(\mathrm d\star F = \mu_0\star J\).

Check. Special case: in the other signature, a book that forgot to flip \(J\) would find \(\nabla\cdot\vec E = -\rho/\varepsilon_0\), attractive like charges — which is why texts in \((+,-,-,-)\) write \(J_\mu = (c\rho, -\vec J)\) with the minus on the spatial part. The two sign flips in part 3 are exactly that bookkeeping. Dimensions are untouched by a sign ✓.

P4 The two invariants, with numbers routine

exercises: \(F\wedge F\) and \(F\wedge\star F\), frame independence, arithmetic with \(c\)

In a laboratory \(\vec E\) has magnitude \(3.0\times10^6\ \mathrm{V\,m^{-1}}\) — about the breakdown field of air — along \(x\), and \(|\vec B| = 0.020\) T.

  1. Take \(\vec B\) along \(y\). Compute the coefficients of \(\mathrm{vol}\) in \(F\wedge F\) and \(F\wedge\star F\). If some observer measures a purely magnetic field, how large is it? Can any observer measure a purely electric one?
  2. Take \(\vec B\) in the \(xy\)-plane at \(60^\circ\) to \(\vec E\). Compute both coefficients again, and show that no observer measures a purely electric or a purely magnetic field.
  3. Check the units of both coefficients, and say what both are for a plane wave.
Hints · 3 rungs
Rung 1

Each part is a pair of numbers in \(\mathrm T^2\), plus yes-or-no answers argued from their signs.

Rung 2

\(F\wedge F = -\tfrac2c(\vec E\cdot\vec B)\,\mathrm{vol}\) and \(F\wedge\star F = (|\vec B|^2 - |\vec E|^2/c^2)\,\mathrm{vol}\), and both coefficients are the same for every observer. An observer who sees \(\vec E' = 0\) has coefficients \(|\vec B'|^2\) and \(0\).

Rung 3

The first line: \(|\vec E|/c = 3.0\times10^6/2.99792458\times10^8 = 1.0007\times10^{-2}\) T.

Worked solution

Idea. Frame independence turns "what would another observer see" into arithmetic: whatever they see must have the same two coefficients.

1. \((|\vec E|/c)^2 = 1.0014\times10^{-4}\ \mathrm T^2\) and \(|\vec B|^2 = 4.000\times10^{-4}\ \mathrm T^2\), so \[ |\vec B|^2 - |\vec E|^2/c^2 = 2.9986\times10^{-4}\ \mathrm T^2 ,\qquad \vec E\cdot\vec B = 0 . \] The \(F\wedge F\) coefficient is \(0\). An observer seeing only \(\vec B'\) has \(|\vec B'|^2 = 2.9986\times10^{-4}\ \mathrm T^2\), so \(|\vec B'| = 1.7317\times10^{-2}\) T \(= 0.01732\) T. A purely electric field would give \(-|\vec E'|^2/c^2\le0\) for the second invariant, but it is positive in every frame: no.

2. \(\vec E\cdot\vec B = 3.0\times10^6\times0.020\times\cos60^\circ = 3.0\times10^4\ \mathrm{V\,T\,m^{-1}}\), and \[ -\tfrac2c\,\vec E\cdot\vec B = -2.0014\times10^{-4}\ \mathrm T^2 . \] The other coefficient depends only on the magnitudes and is again \(2.9986\times10^{-4}\ \mathrm T^2\). A purely electric or purely magnetic field has \(\vec E'\cdot\vec B' = 0\), so its \(F\wedge F\) coefficient is zero; ours is not, in any frame. No such observer exists.

3. \([\vec E\cdot\vec B/c] = \mathrm{V\,m^{-1}\cdot T\cdot s\,m^{-1}} = (\mathrm{V\,s\,m^{-2}})\,\mathrm T = \mathrm T^2\), and \([|\vec E|^2/c^2] = \mathrm T^2\) likewise ✓; multiplied by \(\mathrm{vol}\), in \(\mathrm m^4\), both give \(\mathrm{T^2\,m^4} = \mathrm{Wb}^2\), the units of \(F\wedge F\) since \([F] = \mathrm{Wb}\) ✓. A plane wave has \(|\vec E| = c|\vec B|\) and \(\vec E\perp\vec B\): both zero.

Check. Limit \(|\vec E|\to0\): the invariants become \(|\vec B|^2\) and \(0\), and the purely magnetic observer of part 1 sees \(|\vec B'| = |\vec B| = 0.020\) T, as she must when she is the laboratory ✓. And the electric contribution here is about a quarter of the magnetic one, \(1.0014/4.000 = 0.2504\), so \(|\vec B'|\) is \(\sqrt{0.7496} = 0.8658\) of \(|\vec B|\): \(0.8658\times0.020 = 0.01732\) ✓.

One theorem

P5 Stokes, checked twice routine

exercises: integrating forms, pullback to a sphere, the induced orientation, the divergence and curl theorems as one

  1. Let \(B\) be the ball of radius \(R\) in \(\mathbb R^3\) and \(\omega = x\,\mathrm dy\wedge\mathrm dz\). Compute \(\int_B\mathrm d\omega\) and \(\int_{\partial B}\omega\) separately, the second by pulling back to spherical angles \((\theta,\phi)\) with the orientation induced by the outward normal. Evaluate for \(R = 0.10\) m.
  2. Let \(S\) be the upper hemisphere \(r = R\), \(z\ge0\), with the outward normal, and \(\beta = x\,\mathrm dy - y\,\mathrm dx\) (from P1). Compute \(\int_S\mathrm d\beta\) by pulling back, and \(\oint_{\partial S}\beta\) with the induced orientation.
  3. Replace \(S\) by the flat disc \(z = 0\), \(x^2+y^2\le R^2\), oriented by \(+z\). Why must the answer be the same? One line of forms.
Hints · 3 rungs
Rung 1

Each part is a number — a volume or an area up to a factor — computed twice, once on each side of \(\int_M\mathrm d\omega = \int_{\partial M}\omega\).

Rung 2

On \(r = R\): \(x = R\sin\theta\cos\phi\), \(y = R\sin\theta\sin\phi\), \(z = R\cos\theta\). The pair \((\theta,\phi)\) is positively oriented for the outward normal, because \((\partial_r,\partial_\theta,\partial_\phi)\) is positive in \(\mathbb R^3\). For the hemisphere's rim, the outward direction within \(S\) is \(+\partial_\theta\).

Rung 3

The first line: \(\mathrm d\omega = \mathrm dx\wedge\mathrm dy\wedge\mathrm dz\), so \(\int_B\mathrm d\omega = \tfrac43\pi R^3\).

Worked solution

Idea. Stokes is checkable, and checking it is how the orientation convention becomes second nature. Part 1 is unit 05's divergence theorem and part 2 the curl theorem; the computation is the same.

1. \(\int_B\mathrm d\omega = \tfrac43\pi R^3\). On the sphere, \(\mathrm dy = R(\cos\theta\sin\phi\,\mathrm d\theta + \sin\theta\cos\phi\,\mathrm d\phi)\) and \(\mathrm dz = -R\sin\theta\,\mathrm d\theta\), so \(\mathrm dy\wedge\mathrm dz = R\sin\theta\cos\phi\,\mathrm d\phi\wedge(-R\sin\theta\,\mathrm d\theta) = R^2\sin^2\theta\cos\phi\,\mathrm d\theta\wedge\mathrm d\phi\). Then \[ \int_{\partial B}\omega = \int_0^{2\pi}\!\!\int_0^\pi R^3\sin^3\theta\cos^2\phi\,\mathrm d\theta\,\mathrm d\phi = R^3\cdot\tfrac43\cdot\pi = \tfrac43\pi R^3\ ✓ \] using \(\int_0^\pi\sin^3\theta\,\mathrm d\theta = \tfrac43\) and \(\int_0^{2\pi}\cos^2\phi\,\mathrm d\phi = \pi\). For \(R = 0.10\) m: \(4.1888\times10^{-3}\ \mathrm m^3\).

2. \(\mathrm d\beta = 2\,\mathrm dx\wedge\mathrm dy\). On the sphere \(\mathrm dx\wedge\mathrm dy = R^2(\cos\theta\cos\phi\sin\theta\cos\phi + \sin\theta\sin\phi\cos\theta\sin\phi)\,\mathrm d\theta\wedge\mathrm d\phi = R^2\sin\theta\cos\theta\,\mathrm d\theta\wedge\mathrm d\phi\), so \(\int_S\mathrm d\beta = 2R^2\cdot\tfrac12\cdot2\pi = 2\pi R^2\). The rim is the equator; "outward normal first" with outward \(= +\partial_\theta\) makes \(+\partial_\phi\) positive, i.e. anticlockwise seen from above. There \(x\,\mathrm dy - y\,\mathrm dx = R^2\,\mathrm d\phi\), so \(\oint\beta = 2\pi R^2\) ✓. For \(R = 0.10\) m: \(0.062832\ \mathrm m^2\).

3. Both are \(\oint_{\partial}\beta\) over the same oriented circle, by Stokes. Equivalently, \(S\) and the disc \(D\) together bound the half-ball \(H\), and \(\int_S\mathrm d\beta - \int_D\mathrm d\beta = \int_H\mathrm d(\mathrm d\beta) = 0\). Directly: \(\int_D 2\,\mathrm dx\wedge\mathrm dy = 2\pi R^2\) ✓.

Check. Dimensions: \(\omega\) is length\(^3\) and part 1 a volume; \(\beta\) is length\(^2\) and part 2 an area times 2 ✓. Limit \(R\to0\): both sides vanish. Sign: had the equator been run clockwise, part 2 would give \(-2\pi R^2\) against \(+2\pi R^2\) — the orientation convention is the only thing preventing that.

P6 Cauchy's theorem is Stokes, and dz/z is the vortex medium

exercises: complex-valued forms, Stokes, Cauchy–Riemann as closedness, the vortex as a residue

Allow complex-valued forms \(\omega_{\mathrm R} + \mathrm i\,\omega_{\mathrm I}\) on \(\mathbb R^2 = \mathbb C\), with \(\wedge\) and \(\mathrm d\) extended complex-linearly, and put \(z = x + \mathrm iy\), \(\mathrm dz = \mathrm dx + \mathrm i\,\mathrm dy\), \(\mathrm d\bar z = \mathrm dx - \mathrm i\,\mathrm dy\).

  1. Show \(\mathrm d\bar z\wedge\mathrm dz = 2\mathrm i\,\mathrm dx\wedge\mathrm dy\), and that for smooth \(f\), \(\mathrm d(f\,\mathrm dz) = \dfrac{\partial f}{\partial\bar z}\,\mathrm d\bar z\wedge\mathrm dz\) with \(\dfrac{\partial}{\partial\bar z} = \tfrac12(\partial_x + \mathrm i\,\partial_y)\).
  2. Show \(f\,\mathrm dz\) is closed if and only if \(f\) satisfies the Cauchy–Riemann equations of unit 20, and deduce Cauchy's theorem \(\oint_{\partial D}f\,\mathrm dz = 0\) for \(f\) holomorphic near a compact region \(D\) with smooth boundary.
  3. Show \(\mathrm dz/z = \mathrm d(\log r) + \mathrm i\,\theta\) on the punctured plane. Compute \(\oint_{|z|=1}\mathrm dz/z\), and say which part is exact and which is not.
  4. Apply Stokes to \(\bar z\,\mathrm dz\) and show \(\oint_{\partial D}\bar z\,\mathrm dz = 2\mathrm i\cdot\mathrm{Area}(D)\). Check it on the unit circle.
Hints · 3 rungs
Rung 1

Parts 1 and 2 are identities, part 3 is a number you have met in unit 20, and part 4 an area formula.

Rung 2

Rewrite \(\mathrm df = f_x\,\mathrm dx + f_y\,\mathrm dy\) in the basis \(\mathrm dz\), \(\mathrm d\bar z\). Then \(\mathrm d(f\,\mathrm dz) = \mathrm df\wedge\mathrm dz\), and \(\mathrm dz\wedge\mathrm dz = 0\) kills one term.

Rung 3

The first line: \(\mathrm dx = \tfrac12(\mathrm dz + \mathrm d\bar z)\) and \(\mathrm dy = -\tfrac{\mathrm i}2(\mathrm dz - \mathrm d\bar z)\), so \(\mathrm df = \tfrac12(f_x - \mathrm i f_y)\,\mathrm dz + \tfrac12(f_x + \mathrm i f_y)\,\mathrm d\bar z\).

Worked solution

Idea. A complex one-form is a pair of real ones, and Stokes applies to each. Holomorphy is closedness of \(f\,\mathrm dz\); the \(2\pi\mathrm i\) of the residue theorem is the vortex's \(2\pi\).

1. \(\mathrm d\bar z\wedge\mathrm dz = (\mathrm dx - \mathrm i\,\mathrm dy)\wedge(\mathrm dx + \mathrm i\,\mathrm dy) = \mathrm i\,\mathrm dx\wedge\mathrm dy - \mathrm i\,\mathrm dy\wedge\mathrm dx = 2\mathrm i\,\mathrm dx\wedge\mathrm dy\). By rung 3, \(\mathrm df = f_z\,\mathrm dz + f_{\bar z}\,\mathrm d\bar z\), and so \(\mathrm d(f\,\mathrm dz) = \mathrm df\wedge\mathrm dz = f_{\bar z}\,\mathrm d\bar z\wedge\mathrm dz\).

2. With \(f = u + \mathrm iv\), \(f_{\bar z} = \tfrac12(u_x - v_y) + \tfrac{\mathrm i}2(v_x + u_y)\), which vanishes if and only if \(u_x = v_y\) and \(u_y = -v_x\) — Cauchy–Riemann. Then Stokes, applied to real and imaginary parts: \(\oint_{\partial D}f\,\mathrm dz = \int_D f_{\bar z}\,\mathrm d\bar z\wedge\mathrm dz = 0\).

3. \(\dfrac{\mathrm dz}{z} = \dfrac{(x - \mathrm iy)(\mathrm dx + \mathrm i\,\mathrm dy)}{x^2+y^2} = \dfrac{x\,\mathrm dx + y\,\mathrm dy}{x^2+y^2} + \mathrm i\,\dfrac{x\,\mathrm dy - y\,\mathrm dx}{x^2+y^2} = \mathrm d(\log r) + \mathrm i\,\theta\). So \(\oint_{|z|=1}\mathrm dz/z = 0 + \mathrm i\cdot2\pi = 2\pi\mathrm i\). The real part is exact — \(\log r\) is a function on the punctured plane — and the imaginary part is the vortex, closed and not exact. A primitive of \(\mathrm dz/z\) would be \(\log r + \mathrm i\phi\), and \(\phi\) does not exist on the punctured plane: that is the branch cut of \(\log z\), seen as the non-exactness of \(\theta\).

4. \(\mathrm d(\bar z\,\mathrm dz) = \mathrm d\bar z\wedge\mathrm dz = 2\mathrm i\,\mathrm dx\wedge\mathrm dy\), so \(\oint_{\partial D}\bar z\,\mathrm dz = 2\mathrm i\int_D\mathrm dx\wedge\mathrm dy = 2\mathrm i\cdot\mathrm{Area}(D)\). On the unit circle \(\bar z = 1/z\), so the left side is \(\oint\mathrm dz/z = 2\pi\mathrm i\), and the right side is \(2\mathrm i\cdot\pi\) ✓.

Check. Dimensions: with \(z\) a length, \(\oint\bar z\,\mathrm dz\) is an area ✓ and \(\mathrm dz/z\) is dimensionless, as an angle must be ✓. Special case \(f = 1\): \(\mathrm dz = \mathrm d(z)\) is exact and \(\oint\mathrm dz = 0\) ✓.

Cohomology

P7 Compute a cohomology: the punctured plane medium

exercises: closed versus exact, periods, \(H^1(\mathbb R^2\setminus\{0\})\) — the checkpoint's third clause

On \(U = \mathbb R^2\setminus\{0\}\) consider

\[ \alpha_1 = \frac{x\,\mathrm dx + y\,\mathrm dy}{x^2+y^2},\quad \alpha_2 = \frac{x\,\mathrm dy - y\,\mathrm dx}{x^2+4y^2},\quad \alpha_3 = \frac{x\,\mathrm dy - y\,\mathrm dx}{(x^2+y^2)^2},\quad \alpha_4 = 3\,\theta + y\,\mathrm dx + x\,\mathrm dy . \]
  1. Decide which are closed.
  2. For each closed one compute its period round the unit circle, write its class as \(c\,[\theta]\), and produce a function \(f\) on \(U\) with \(\alpha - c\,\theta = \mathrm df\).
  3. Prove, in your own words and citing only the Poincaré lemma and Stokes, that \(H^1(U)\cong\mathbb R\) with \([\theta]\) a generator. Say which physical statement the single dimension encodes.
Hints · 3 rungs
Rung 1

Part 1 is four yes-or-no answers; part 2 a period, a number \(c\) and a function per closed form; part 3 is the deck's worked example 3, rebuilt.

Rung 2

Closedness of \(P\,\mathrm dx + Q\,\mathrm dy\) is \(\partial_xQ = \partial_yP\). For \(\alpha_2\), change variables to \(u = x\), \(v = 2y\): it becomes a multiple of the vortex in \((u,v)\). For its \(f\), compare the angle of \((x, 2y)\) with the angle of \((x, y)\) — their difference is a genuine function.

Rung 3

The first line, for \(\alpha_3\): \(\partial_x\big(x/r^4\big) + \partial_y\big(y/r^4\big) = 2/r^4 - 4(x^2+y^2)/r^6\).

Worked solution

Idea. Closedness is a local derivative test; the class is read off one period, because the punctured plane has one hole.

1. \(\alpha_1\): \(\partial_x(y/r^2) - \partial_y(x/r^2) = -2xy/r^4 + 2xy/r^4 = 0\), closed. \(\alpha_2\): with \(q = x^2+4y^2\), \(\partial_x(x/q) + \partial_y(y/q) = (q - 2x^2)/q^2 + (q - 8y^2)/q^2 = (2q - 2x^2 - 8y^2)/q^2 = 0\), closed. \(\alpha_3\): by rung 3 the sum is \(-2/r^4\), so \(\mathrm d\alpha_3 = -2r^{-4}\,\mathrm dx\wedge\mathrm dy\ne0\): not closed, and it has no cohomology class at all — a trap worth falling into once. \(\alpha_4\): \(3\theta\) is closed and \(y\,\mathrm dx + x\,\mathrm dy = \mathrm d(xy)\), closed.

2. \(\alpha_1 = \mathrm d\big(\tfrac12\log(x^2+y^2)\big)\): period \(0\), \(c = 0\), \(f = \log r\). \(\alpha_2\): on the unit circle it is \(\mathrm dt/(\cos^2t + 4\sin^2t)\), and \(\int_0^{2\pi}\mathrm dt/(a^2\cos^2t + b^2\sin^2t) = 2\pi/(ab)\) with \(a = 1\), \(b = 2\) gives period \(\pi = 3.1416\), so \(c = \tfrac12\). In \(u = x\), \(v = 2y\): \(x\,\mathrm dy - y\,\mathrm dx = \tfrac12(u\,\mathrm dv - v\,\mathrm du)\) and \(x^2+4y^2 = u^2+v^2\), so \(\alpha_2 = \tfrac12\theta_{uv}\). The angle of \((x,2y)\) minus the angle of \((x,y)\) is the argument of \((x + 2\mathrm iy)(x - \mathrm iy) = x^2 + 2y^2 + \mathrm i\,xy\), whose real part is positive on \(U\); so \[ f = \tfrac12\arctan\frac{xy}{x^2+2y^2} \] is smooth on \(U\), and \(\alpha_2 - \tfrac12\theta = \mathrm df\) (differentiate to confirm). \(\alpha_4\): period \(3\cdot2\pi = 6\pi\), \(c = 3\), \(f = xy\).

3. Nonzero. If \(\theta = \mathrm df\), Stokes on the circle (no boundary) gives \(\oint\theta = 0\), but it is \(2\pi\). Radius-independent. For closed \(\alpha\) and radii \(r_1\lt r_2\), Stokes on the annulus gives \(\oint_{r_2}\alpha - \oint_{r_1}\alpha = \int\mathrm d\alpha = 0\). Spanning. Let \(c = \tfrac1{2\pi}\oint\alpha\), and pull back by \(\Phi(r,\phi) = (r\cos\phi, r\sin\phi)\) to the convex half-plane \(r \gt 0\). There \(\Phi^*\alpha = \mathrm dg\) (Poincaré) and \(\Phi^*\theta = \mathrm d\phi\). The increment \(g(r,\phi+2\pi) - g(r,\phi)\) is a period, \(2\pi c\), for every \(r\); so \(g - c\phi\) is \(2\pi\)-periodic, descends to \(f\) on \(U\), and \(\alpha - c\theta = \mathrm df\). The map \([\alpha]\mapsto c\) is therefore a linear isomorphism \(H^1(U)\to\mathbb R\). Physically: a curl-free field on the plane with one point removed can circulate, by an amount no local measurement predicts, round exactly one independent loop — unit 04's \(2\pi\).

Check. Two routes to \(\alpha_2\)'s period agree: the integral gives \(\pi\), and \(\alpha_2 = \tfrac12\theta_{uv}\) with \((x,y)\mapsto(u,v)\) orientation-preserving and winding once gives \(\tfrac12\cdot2\pi = \pi\) ✓. The classes add: \(\alpha_2 + \alpha_4\) has period \(7\pi\), class \(\tfrac72[\theta]\) ✓. Special case: \(\alpha_2\) with \(4\) replaced by \(1\) is \(\theta\) itself, and \(2\pi/(1\cdot1) = 2\pi\) ✓.

P8 Compute a cohomology: the torus, and a torque with no potential hard

exercises: forms on a quotient, periods, homotopy invariance, \(H^1(T^2)\) — a proof, and a number

Let \(T^2 = S^1\times S^1\) with angles \((\phi_1,\phi_2)\) mod \(2\pi\) — unit 31's configuration space of the double pendulum — and \(\pi:\mathbb R^2\to T^2\), \(\pi(x,y) = (x \bmod 2\pi,\ y\bmod 2\pi)\). Let \(C_1\) be the loop \(\phi_2 = 0\) and \(C_2\) the loop \(\phi_1 = 0\), each run with its angle increasing.

  1. Show there are closed one-forms \(\mathrm d\phi_1\), \(\mathrm d\phi_2\) on \(T^2\) with \(\pi^*\mathrm d\phi_i = \mathrm dx_i\), though \(\phi_i\) are not functions on \(T^2\). Show \(c_1\,\mathrm d\phi_1 + c_2\,\mathrm d\phi_2\) is exact only if \(c_1 = c_2 = 0\).
  2. Prove every closed \(\alpha\) on \(T^2\) is \(c_1\,\mathrm d\phi_1 + c_2\,\mathrm d\phi_2 + \mathrm df\) for a unique pair \((c_1,c_2)\), so \(H^1(T^2)\cong\mathbb R^2\).
  3. A motor applies a constant torque \(\tau\) to the upper arm, so the generalised force is the one-form \(Q = \tau\,\mathrm d\phi_1\). Is \(Q\) closed? Is there a potential energy \(V\) on the configuration space with \(Q = -\mathrm dV\)? Compute the work per revolution of the upper arm for \(\tau = 0.50\) N m.
  4. Say what physical statement \(b_1 = 1\) for the punctured plane and \(b_1 = 2\) for the torus each encode.
Hints · 3 rungs
Rung 1

Part 1 is about forms and their periods; part 2 is a proof shaped like the punctured-plane argument, with two periods instead of one; part 3 is yes, no and a number in joules.

Rung 2

Lift to \(\mathbb R^2\): there \(\pi^*\alpha = \mathrm dg\) by the Poincaré lemma. Compare \(g(x+2\pi,y) - g(x,y)\) with a period of \(\alpha\), using the homotopy-invariance slide.

Rung 3

The first line of part 2: the curve \(s\mapsto\pi(x+s, y)\), \(s\in[0,2\pi]\), is a loop in \(T^2\) homotopic to \(C_1\) through loops, so \(g(x+2\pi,y) - g(x,y) = \oint_{C_1}\alpha\).

Worked solution

Idea. The torus is the plane with two independent translations identified. A closed form lifts to an exact one upstairs; it comes back down exactly when its two periods vanish.

1. \(\pi\) is a local diffeomorphism, and \(\mathrm dx_i\) is invariant under translation by \(2\pi\mathbb Z^2\), so \((\mathrm d\phi_i)_{\pi(p)}(\mathrm d\pi_p v) := \mathrm dx_i(v)\) is well defined and smooth. It is closed: \(\pi^*\mathrm d(\mathrm d\phi_i) = \mathrm d\mathrm dx_i = 0\) and \(\pi^*\) is injective on forms because \(\pi\) is a local diffeomorphism. Periods: \(\oint_{C_1}\mathrm d\phi_1 = 2\pi\), \(\oint_{C_2}\mathrm d\phi_1 = 0\), and symmetrically. An exact form has all periods zero (Stokes; loops have no boundary), so \(c_1\,\mathrm d\phi_1 + c_2\,\mathrm d\phi_2\) exact forces \(2\pi c_1 = 2\pi c_2 = 0\).

2. Let \(\alpha\) be closed, \(c_i = \tfrac1{2\pi}\oint_{C_i}\alpha\), and \(\beta = \alpha - c_1\,\mathrm d\phi_1 - c_2\,\mathrm d\phi_2\), closed with both periods zero. On \(\mathbb R^2\), \(\pi^*\beta = \mathrm dg\). The segment from \((x,y)\) to \((x+2\pi,y)\) maps to a loop, and \(H(s,u) = \pi\big((1-u)x + s,\ (1-u)y\big)\), \(s\in[0,2\pi]\), \(u\in[0,1]\), deforms it through loops (each \(H(\cdot,u)\) closes up, since its ends differ by \(2\pi\) in the first coordinate) into \(C_1\) at \(u = 1\). By homotopy invariance, \(g(x+2\pi,y) - g(x,y) = \oint_{C_1}\beta = 0\); likewise in \(y\). So \(g\) is \(2\pi\)-periodic in both variables, hence \(g = f\circ\pi\) for a smooth \(f\) on \(T^2\), and \(\pi^*(\beta - \mathrm df) = 0\) gives \(\beta = \mathrm df\). Uniqueness of \((c_1,c_2)\) is part 1. So \([\alpha]\mapsto(c_1,c_2)\) is an isomorphism \(H^1(T^2)\to\mathbb R^2\).

3. \(\mathrm dQ = \tau\,\mathrm d\mathrm d\phi_1 = 0\): closed, so every small loop does zero net work and locally \(V = -\tau\phi_1\). But its period on \(C_1\) is \(2\pi\tau\ne0\), so it is not exact: no potential energy exists on the configuration space. Work per revolution: \(2\pi\tau = 2\pi\times0.50 = 3.1416\) J.

4. \(b_1\) counts the independent loops round which a curl-free force can do net work. Punctured plane: one, round the hole — unit 04's vortex. Torus: two, turning either arm — which is why a torque on either arm, or any combination, is locally conservative and globally not.

Check. Units: \(\tau\) in N m \(=\) J per radian, times \(2\pi\) rad, gives J ✓. Limit \(\tau\to0\): the work vanishes and \(Q = 0 = -\mathrm d(0)\) ✓. Compare the punctured plane: its \(H^1\) is \(\mathbb R\) by the same lift-and-descend argument with one translation instead of two.

P9 The monopole needs two potentials hard

exercises: pullback to spherical coordinates, closed but not exact in degree two, gluing potentials, a number

Write \(g = \mu_0q_m\) and let \(F = \dfrac{g}{4\pi}\,\dfrac{x\,\mathrm dy\wedge\mathrm dz + y\,\mathrm dz\wedge\mathrm dx + z\,\mathrm dx\wedge\mathrm dy}{r^3}\) on \(\mathbb R^3\setminus\{0\}\), the deck's monopole.

  1. Show \(F = \dfrac{g}{4\pi}\sin\theta\,\mathrm d\theta\wedge\mathrm d\phi\) and \(\oint_{S^2}F = g\).
  2. Show \(A_N = \dfrac{g}{4\pi}(1-\cos\theta)\,\mathrm d\phi\) is smooth off the half-line \(\{x = y = 0,\ z\le0\}\) with \(\mathrm dA_N = F\), and \(A_S = -\dfrac{g}{4\pi}(1+\cos\theta)\,\mathrm d\phi\) is smooth off \(\{x = y = 0,\ z\ge0\}\) with \(\mathrm dA_S = F\).
  3. On the overlap, compute \(A_N - A_S\) and identify its cohomology class.
  4. Recover \(\oint_{S^2}F = g\) by Stokes on the two hemispheres, using only \(A_N\) and \(A_S\).
  5. Dirac's quantum argument (outside this course, cited on the monopole slide) requires \(g\) to be a multiple of \(h/e\). For \(g = h/e\), what is the coefficient of the vortex class in part 3, in webers?
Hints · 3 rungs
Rung 1

Parts 1–4 are each an identity or one explicit form; part 5 is a number in Wb.

Rung 2

For part 1, the numerator is \(i_{r\partial_r}(\mathrm dx\wedge\mathrm dy\wedge\mathrm dz)\), and \(\mathrm dx\wedge\mathrm dy\wedge\mathrm dz = r^2\sin\theta\,\mathrm dr\wedge\mathrm d\theta\wedge\mathrm d\phi\). For part 2 use \(1 - \cos\theta = (r - z)/r\), \(x^2+y^2 = (r - z)(r + z)\) and \(\mathrm d\phi = (x\,\mathrm dy - y\,\mathrm dx)/(x^2+y^2)\).

Rung 3

The first line of part 4: \(\int_{\text{north}}F = \int_{\text{north}}\mathrm dA_N = \oint_{\text{equator}}A_N\), the equator run anticlockwise seen from above.

Worked solution

Idea. \(F\) is closed and not exact on \(\mathbb R^3\setminus\{0\}\), so no single potential exists; two potentials on two contractible-enough pieces do, and their difference on the overlap is the vortex. The monopole's \(H^2\) class is the vortex's \(H^1\) class one step down.

1. On the Cartesian side, \(i_{x\partial_x + y\partial_y + z\partial_z}(\mathrm dx\wedge\mathrm dy\wedge\mathrm dz) = x\,\mathrm dy\wedge\mathrm dz - y\,\mathrm dx\wedge\mathrm dz + z\,\mathrm dx\wedge\mathrm dy\), the numerator. On the spherical side \(r\partial_r\) fed into \(r^2\sin\theta\,\mathrm dr\wedge\mathrm d\theta\wedge\mathrm d\phi\) gives \(r^3\sin\theta\,\mathrm d\theta\wedge\mathrm d\phi\). Divide by \(r^3\): \(F = \tfrac{g}{4\pi}\sin\theta\,\mathrm d\theta\wedge\mathrm d\phi\), and \(\oint_{S^2}F = \tfrac{g}{4\pi}\cdot4\pi = g\).

2. \(\mathrm dA_N = \tfrac{g}{4\pi}\sin\theta\,\mathrm d\theta\wedge\mathrm d\phi = F\), and \(\mathrm dA_S = -\tfrac{g}{4\pi}(-\sin\theta)\,\mathrm d\theta\wedge\mathrm d\phi = F\). Smoothness is the real content, because \(\mathrm d\phi\) is singular on the whole \(z\)-axis. In Cartesian form, \[ (1-\cos\theta)\,\mathrm d\phi = \frac{r - z}{r}\cdot\frac{x\,\mathrm dy - y\,\mathrm dx}{(r-z)(r+z)} = \frac{x\,\mathrm dy - y\,\mathrm dx}{r(r+z)} , \] smooth wherever \(r + z\gt0\), which fails only on \(x = y = 0,\ z\le0\). Likewise \((1+\cos\theta)\,\mathrm d\phi = (x\,\mathrm dy - y\,\mathrm dx)/\big(r(r - z)\big)\), smooth off \(x = y = 0,\ z\ge0\).

3. \(A_N - A_S = \tfrac{g}{4\pi}\big[(1-\cos\theta) + (1+\cos\theta)\big]\mathrm d\phi = \tfrac{g}{2\pi}\,\mathrm d\phi = \tfrac{g}{2\pi}\,\theta\), with \(\theta\) the vortex in the \((x,y)\) variables. The overlap is \(\mathbb R^3\) minus the \(z\)-axis, which is the punctured plane times a line, so its \(H^1\) is \(\mathbb R\) and the class is \(\tfrac{g}{2\pi}[\theta]\ne0\). The gauge function relating the two potentials would be \(\lambda = \tfrac{g}{2\pi}\phi\), which is not a function on the overlap.

4. The northern hemisphere's boundary, with the induced orientation, is the equator run anticlockwise from above; the southern one's is the same circle run the other way. So \(\int_{\text{north}}F = \oint A_N = \tfrac{g}{4\pi}\cdot2\pi = \tfrac g2\) and \(\int_{\text{south}}F = -\oint A_S = -\big(-\tfrac{g}{4\pi}\cdot2\pi\big) = \tfrac g2\), using \(\cos\theta = 0\) on the equator. Sum \(g\) ✓ — equivalently \(\oint(A_N - A_S) = \tfrac{g}{2\pi}\cdot2\pi = g\): the flux through the sphere is the period of the transition form round the overlap.

5. \(h/e = 6.62607015\times10^{-34}/1.602176634\times10^{-19} = 4.1357\times10^{-15}\) Wb, and the coefficient is \(g/2\pi = 6.5821\times10^{-16}\) Wb.

Check. Units: \(F\) integrates to webers, so \(g\) and \(A\) are in Wb, and \(\theta\) is dimensionless ✓. Limit \(g\to0\): both potentials vanish and so does the obstruction ✓. The analogy is exact: for the vortex a function (an angle) fails to exist globally; for the monopole a one-form does, and the failure is measured by that same angle.

Cartan's structure equations

P10 Cartan versus Christoffel: the hyperbolic plane medium

exercises: the first and second structure equations, reading curvature off a two-form, counting the arithmetic — the checkpoint's fourth clause

The hyperbolic plane of unit 33's problem sheet: \(y\gt0\) with \(g = a^2(\mathrm dx^2 + \mathrm dy^2)/y^2\).

  1. With \(e^1 = a\,\mathrm dx/y\), \(e^2 = a\,\mathrm dy/y\), compute \(\mathrm de^1\) and \(\mathrm de^2\), find \(\omega^1{}_2\) by inspection, and verify both first structure equations.
  2. Compute \(\Omega^1{}_2\) and read off \(K\).
  3. Redo it by unit 33's route: every Christoffel symbol, then \(R^x{}_{yxy}\), then \(K\).
  4. Count the steps each way, with the deck's rule (one step is one derivative or one product written down), and give the ratio.
Hints · 3 rungs
Rung 1

\(K\) is a single constant, the same everywhere, and its sign is the opposite of the sphere's.

Rung 2

\(\mathrm de^1 = \mathrm d(a/y)\wedge\mathrm dx\) is a multiple of \(e^1\wedge e^2\), and \(\mathrm de^2 = 0\). The antisymmetry \(\omega^2{}_1 = -\omega^1{}_2\) means there is one connection form to find; try a multiple of \(e^1\). In two dimensions \(\Omega^1{}_2 = K\,e^1\wedge e^2\).

Rung 3

The first line: we need \(\mathrm de^1 = -\omega^1{}_2\wedge e^2\); with \(\omega^1{}_2 = \lambda\,e^1\) this reads \(\tfrac1a\,e^1\wedge e^2 = -\lambda\,e^1\wedge e^2\).

Worked solution

Idea. Two exterior derivatives, one guess forced by uniqueness, one more derivative. The coordinate route computes the same number from six symbols that mostly carry no information.

1. \(\mathrm de^1 = -\tfrac{a}{y^2}\,\mathrm dy\wedge\mathrm dx = \tfrac{a}{y^2}\,\mathrm dx\wedge\mathrm dy = \tfrac1a\,e^1\wedge e^2\), since \(e^1\wedge e^2 = \tfrac{a^2}{y^2}\mathrm dx\wedge\mathrm dy\). \(\mathrm de^2 = \mathrm d(a/y)\wedge\mathrm dy = 0\). By rung 3, \(\lambda = -1/a\): \(\omega^1{}_2 = -\tfrac1a e^1 = -\mathrm dx/y\). Verify the second equation: \(-\omega^2{}_1\wedge e^1 = -\tfrac{\mathrm dx}{y}\wedge\tfrac{a\,\mathrm dx}{y} = 0 = \mathrm de^2\) ✓. Uniqueness (the deck's index-cycling proof) makes this the connection.

2. \(\Omega^1{}_2 = \mathrm d\omega^1{}_2 = \mathrm d(-y^{-1})\wedge\mathrm dx = y^{-2}\,\mathrm dy\wedge\mathrm dx = -\tfrac1{y^2}\,\mathrm dx\wedge\mathrm dy = -\tfrac1{a^2}\,e^1\wedge e^2\). So \(K = -1/a^2\).

3. \(g^{xx} = g^{yy} = y^2/a^2\), and only \(\partial_yg_{xx} = \partial_yg_{yy} = -2a^2/y^3\) are nonzero. From \(\Gamma^a{}_{bc} = \tfrac12g^{ad}(\partial_bg_{cd} + \partial_cg_{bd} - \partial_dg_{bc})\): \(\Gamma^x{}_{xy} = -1/y\), \(\Gamma^y{}_{xx} = 1/y\), \(\Gamma^y{}_{yy} = -1/y\), and \(\Gamma^x{}_{xx} = \Gamma^x{}_{yy} = \Gamma^y{}_{xy} = 0\). Then \[ R^x{}_{yxy} = \partial_x\Gamma^x{}_{yy} - \partial_y\Gamma^x{}_{xy} + \Gamma^x{}_{x\lambda}\Gamma^\lambda{}_{yy} - \Gamma^x{}_{y\lambda}\Gamma^\lambda{}_{xy} = 0 - \tfrac1{y^2} + \tfrac1{y^2} - \tfrac1{y^2} = -\tfrac1{y^2} , \] the two products being \(\Gamma^x{}_{xy}\Gamma^y{}_{yy}\) and \(\Gamma^x{}_{yx}\Gamma^x{}_{xy}\). Lower with \(g_{xx}\) and divide by \(g_{xx}g_{yy}\): \(K = R^x{}_{yxy}/g_{yy} = -\tfrac1{y^2}\cdot\tfrac{y^2}{a^2} = -\tfrac1{a^2}\) ✓.

4. Cartan: \(\mathrm de^1\), \(\mathrm de^2\), the inspection, \(\mathrm d\omega^1{}_2\) — 4. Coordinates: 2 inverse-metric entries, 6 symbols of 3 derivative terms each (18), the Riemann formula's 2 derivatives and 4 products (6), then lower and divide (2) — 28. The same count as the sphere in the deck, ratio 7. The gap is not special to these metrics: in \(n\) dimensions there are \(n^2(n+1)/2\) Christoffel symbols against \(n(n-1)/2\) connection forms.

Check. \([K] = \mathrm m^{-2}\) when \(a\) is a length ✓. Sign: negative, the opposite of the sphere, as unit 33's hyperbolic plane requires. Limit \(a\to\infty\): \(K\to0\), and on any fixed region the geometry looks flat. Gauss–Bonnet check: a disc with \(K\lt0\) needs \(\oint\kappa_g\,\mathrm ds \gt 2\pi\) — its rim must curve more than a flat circle's.

P11 The three-sphere by moving frames hard

exercises: structure equations in three dimensions, the \(\omega\wedge\omega\) term, Ricci from curvature forms, counting

The round three-sphere of radius \(a\): \(g = a^2\big[\mathrm d\chi^2 + \sin^2\!\chi\,(\mathrm d\theta^2 + \sin^2\!\theta\,\mathrm d\phi^2)\big]\), \(0\lt\chi\lt\pi\) — the spatial slice of a closed universe in unit 39.

  1. Take \(e^1 = a\,\mathrm d\chi\), \(e^2 = a\sin\chi\,\mathrm d\theta\), \(e^3 = a\sin\chi\sin\theta\,\mathrm d\phi\) and compute \(\mathrm de^a\).
  2. Find the three connection forms by inspection and verify all three first structure equations.
  3. Compute the three curvature two-forms, including the \(\omega\wedge\omega\) terms, and show \(\Omega^a{}_b = \tfrac1{a^2}\,e^a\wedge e^b\).
  4. Read off \(R^a{}_{bcd}\), the Ricci tensor and the scalar curvature, and show every plane has sectional curvature \(1/a^2\).
  5. How many Christoffel symbols would the coordinate route need, and how many are nonzero?
Hints · 3 rungs
Rung 1

The answer is constant curvature: \(R_{abcd} = (\delta_{ac}\delta_{bd} - \delta_{ad}\delta_{bc})/a^2\), Ricci \(\propto\delta_{ab}\).

Rung 2

\(\mathrm de^1 = 0\), \(\mathrm de^2 = \tfrac{\cot\chi}{a}\,e^1\wedge e^2\), \(\mathrm de^3 = \tfrac{\cot\chi}{a}\,e^1\wedge e^3 + \tfrac{\cot\theta}{a\sin\chi}\,e^2\wedge e^3\). Each coefficient suggests one connection form, proportional to the second coframe element in its term.

Rung 3

The first line: try \(\omega^2{}_1 = \cos\chi\,\mathrm d\theta\), \(\omega^3{}_1 = \cos\chi\sin\theta\,\mathrm d\phi\), \(\omega^3{}_2 = \cos\theta\,\mathrm d\phi\).

Worked solution

Idea. The same three moves as in two dimensions, with the one new feature that \(\omega^a{}_c\wedge\omega^c{}_b\) no longer vanishes — and here it is exactly what makes the answer come out constant.

1. \(\mathrm de^1 = 0\); \(\mathrm de^2 = a\cos\chi\,\mathrm d\chi\wedge\mathrm d\theta\); \(\mathrm de^3 = a\cos\chi\sin\theta\,\mathrm d\chi\wedge\mathrm d\phi + a\sin\chi\cos\theta\,\mathrm d\theta\wedge\mathrm d\phi\), which are rung 2's expressions.

2. With rung 3's guesses and \(\omega^a{}_b = -\omega^b{}_a\): \(-\omega^2{}_1\wedge e^1 - \omega^2{}_3\wedge e^3 = -\cos\chi\,\mathrm d\theta\wedge a\,\mathrm d\chi + \cos\theta\,\mathrm d\phi\wedge a\sin\chi\sin\theta\,\mathrm d\phi = a\cos\chi\,\mathrm d\chi\wedge\mathrm d\theta = \mathrm de^2\) ✓. \(-\omega^3{}_1\wedge e^1 - \omega^3{}_2\wedge e^2 = a\cos\chi\sin\theta\,\mathrm d\chi\wedge\mathrm d\phi + a\sin\chi\cos\theta\,\mathrm d\theta\wedge\mathrm d\phi = \mathrm de^3\) ✓. \(-\omega^1{}_2\wedge e^2 - \omega^1{}_3\wedge e^3 = a\sin\chi\cos\chi\,\mathrm d\theta\wedge\mathrm d\theta + a\sin\chi\cos\chi\sin^2\theta\,\mathrm d\phi\wedge\mathrm d\phi = 0 = \mathrm de^1\) ✓.

3. \(\Omega^2{}_1 = \mathrm d\omega^2{}_1 + \omega^2{}_3\wedge\omega^3{}_1 = -\sin\chi\,\mathrm d\chi\wedge\mathrm d\theta + (-\cos\theta\,\mathrm d\phi)\wedge(\cos\chi\sin\theta\,\mathrm d\phi) = -\sin\chi\,\mathrm d\chi\wedge\mathrm d\theta = \tfrac1{a^2}\,e^2\wedge e^1\). \(\Omega^3{}_1 = -\sin\chi\sin\theta\,\mathrm d\chi\wedge\mathrm d\phi + \cos\chi\cos\theta\,\mathrm d\theta\wedge\mathrm d\phi + \cos\theta\,\mathrm d\phi\wedge\cos\chi\,\mathrm d\theta = -\sin\chi\sin\theta\,\mathrm d\chi\wedge\mathrm d\phi = \tfrac1{a^2}\,e^3\wedge e^1\) — here the wedge term cancels a derivative term. \(\Omega^3{}_2 = -\sin\theta\,\mathrm d\theta\wedge\mathrm d\phi + \cos\chi\sin\theta\,\mathrm d\phi\wedge(-\cos\chi\,\mathrm d\theta) = -\sin^2\!\chi\sin\theta\,\mathrm d\theta\wedge\mathrm d\phi = \tfrac1{a^2}\,e^3\wedge e^2\) — here it turns \(1\) into \(\sin^2\chi\).

4. \(\Omega^a{}_b = \tfrac12R^a{}_{bcd}e^c\wedge e^d = \tfrac1{a^2}e^a\wedge e^b\) gives \(R^a{}_{bcd} = (\delta^a_c\delta_{bd} - \delta^a_d\delta_{bc})/a^2\). Sectional curvature of the plane of orthonormal \(e_a, e_b\) is \(R_{abab} = 1/a^2\); and for any orthonormal pair \(u, v\), \(R(u,v,u,v) = \big(|u|^2|v|^2 - (u\cdot v)^2\big)/a^2 = 1/a^2\), so every plane has \(K = 1/a^2\). Ricci: \(R_{bd} = R^a{}_{bad} = (3-1)\delta_{bd}/a^2 = 2\delta_{bd}/a^2\); scalar \(R = 6/a^2\).

5. \(n^2(n+1)/2 = 18\) symbols; nonzero, up to the lower-index symmetry, six: \(\Gamma^\chi{}_{\theta\theta} = -\sin\chi\cos\chi\), \(\Gamma^\chi{}_{\phi\phi} = -\sin\chi\cos\chi\sin^2\theta\), \(\Gamma^\theta{}_{\chi\theta} = \cot\chi\), \(\Gamma^\theta{}_{\phi\phi} = -\sin\theta\cos\theta\), \(\Gamma^\phi{}_{\chi\phi} = \cot\chi\), \(\Gamma^\phi{}_{\theta\phi} = \cot\theta\). Then six independent Riemann components (\(n^2(n^2-1)/12 = 6\)), each a four-term formula with sums. Cartan used three \(\mathrm de\)'s, three connection forms and three curvature forms.

Check. Special case: freeze \(\chi = \pi/2\); the \((\theta,\phi)\) part is the two-sphere of radius \(a\), and \(\Omega^3{}_2\) there is \(-\sin\theta\,\mathrm d\theta\wedge\mathrm d\phi\), the deck's example 5 ✓. Dimensions: \(R\) in \(\mathrm m^{-2}\) ✓. Limit \(a\to\infty\): flat. And in three dimensions Ricci determines Riemann (unit 33): \(R_{ab} = 2\delta_{ab}/a^2\) is consistent with that.

P12 The symplectic form, used medium

exercises: wedge powers, the interior product, Cartan's magic formula, generating functions as \(\mathrm d^2 = 0\)

  1. On \(\mathbb R^4\) with \(\omega = \mathrm dp_1\wedge\mathrm dq_1 + \mathrm dp_2\wedge\mathrm dq_2\), compute \(\omega\wedge\omega\); show \(\sigma = \mathrm dp_1\wedge\mathrm dq_1\) has \(\sigma\wedge\sigma = 0\). Why is \(\tfrac12\omega\wedge\omega\) a volume form and \(\sigma\) not symplectic?
  2. For the pendulum \(H = p^2/(2ml^2) - mgl\cos q\) on \(\mathbb R^2\) with \(\omega = \mathrm dp\wedge\mathrm dq\), find \(X_H = A\,\partial_q + B\,\partial_p\) from \(i_{X_H}\omega = -\mathrm dH\) and compare with Hamilton's equations. Show \(\mathcal L_{X_H}\omega = 0\) by the magic formula.
  3. The swap \(Q = p\), \(P = -q\): verify \(p\,\mathrm dq - P\,\mathrm dQ = \mathrm dF_1\) with unit 14's \(F_1 = qQ\), and deduce it preserves \(\omega\).
  4. The kick \(Q = q\), \(P = p + k\sin q\): show \(p\,\mathrm dq - P\,\mathrm dQ\) is exact, and hence that the kick is symplectic.
  5. The stretch \(Q = 2q\), \(P = p\): compute \(\mathrm dP\wedge\mathrm dQ\); a blob of phase-space area \(1.0\times10^{-2}\) J s is mapped to what area? Why can no generating function exist?
Hints · 3 rungs
Rung 1

Part 1 is a four-form; part 2 a vector field; parts 3–5 are each a yes-or-no with a one-line reason, and part 5 a number.

Rung 2

\(i_X(\mathrm dp\wedge\mathrm dq) = \mathrm dp(X)\,\mathrm dq - \mathrm dq(X)\,\mathrm dp\). For 3–5, a map preserves \(\omega\) exactly when \(\mathrm dP\wedge\mathrm dQ = \mathrm dp\wedge\mathrm dq\), and applying \(\mathrm d\) to \(p\,\mathrm dq - P\,\mathrm dQ\) gives the difference.

Rung 3

The first line of part 2: \(B\,\mathrm dq - A\,\mathrm dp = -H_q\,\mathrm dq - H_p\,\mathrm dp\).

Worked solution

Idea. Everything unit 14 proved about phase space used only \(\omega\), closedness and \(\mathrm d^2 = 0\). Written in forms, each proof is a line.

1. Two-forms commute, and \(\sigma_i\wedge\sigma_i = 0\) (repeated factors), so \(\omega\wedge\omega = 2\,\mathrm dp_1\wedge\mathrm dq_1\wedge\mathrm dp_2\wedge\mathrm dq_2\ne0\), and \(\tfrac12\omega\wedge\omega\) is a nowhere-zero top form — a volume form. \(\sigma\wedge\sigma = 0\): \(\sigma\) is degenerate, killing \(\partial_{q_2}\) and \(\partial_{p_2}\), so it cannot turn \(\mathrm dH\) into a unique vector field.

2. By rung 3, \(A = H_p = p/(ml^2)\) and \(B = -H_q = -mgl\sin q\): \(\dot q = p/(ml^2)\), \(\dot p = -mgl\sin q\), Hamilton's equations ✓. Then \(\mathcal L_{X_H}\omega = \mathrm d(i_{X_H}\omega) + i_{X_H}\mathrm d\omega = -\mathrm d\mathrm dH + 0 = 0\). Unit 14's divergence check agrees: \(\partial_qA + \partial_pB = 0 + 0 = 0\).

3. \(p\,\mathrm dq - P\,\mathrm dQ = p\,\mathrm dq + q\,\mathrm dp = \mathrm d(qp) = \mathrm d(qQ)\) ✓. Apply \(\mathrm d\): \(\mathrm dp\wedge\mathrm dq - \mathrm dP\wedge\mathrm dQ = \mathrm d\mathrm dF_1 = 0\). Directly: \(\mathrm dP\wedge\mathrm dQ = -\mathrm dq\wedge\mathrm dp = \mathrm dp\wedge\mathrm dq\) ✓.

4. \(p\,\mathrm dq - (p + k\sin q)\,\mathrm dq = -k\sin q\,\mathrm dq = \mathrm d(k\cos q)\), exact, so \(\mathrm dp\wedge\mathrm dq = \mathrm dP\wedge\mathrm dQ\). Directly: \((\mathrm dp + k\cos q\,\mathrm dq)\wedge\mathrm dq = \mathrm dp\wedge\mathrm dq\) ✓. (It has no generating function of type \(F_1(q,Q)\), since \(Q = q\) means \(q, Q\) are not independent coordinates; the one-form statement does not care.) This kick is the building block of unit 17's standard map, for those who take that branch.

5. \(\mathrm dP\wedge\mathrm dQ = 2\,\mathrm dp\wedge\mathrm dq = 2\omega\): areas double, and the blob becomes \(2.0\times10^{-2}\) J s. \(p\,\mathrm dq - P\,\mathrm dQ = -p\,\mathrm dq\), whose \(\mathrm d\) is \(-\mathrm dp\wedge\mathrm dq\ne0\): not closed, so certainly not exact, and no \(F\) exists.

Check. Units: \(\omega\) is momentum times length, \(\mathrm{kg\,m\,s^{-1}\cdot m} = \mathrm{J\,s}\) ✓, the units of the blob's area. Special case \(k = 0\) in part 4: the identity map, \(F = \) const ✓. And part 5 fails Liouville exactly as it must: a map that doubles volume cannot be a Hamiltonian flow.