Work, and the arrival of vector calculus

Idea. The definition is the whole content: substitute the parametrisation and integrate one variable. The interest is in what the four numbers do and do not prove.

1. The four integrals. Throughout, \(\vec F(\vec\gamma)\cdot\dot{\vec\gamma} = (\gamma_2)^2\dot\gamma_1 + \gamma_1\dot\gamma_2\).

• Straight, \(\vec\gamma = (t,t)\), \(\dot{\vec\gamma} = (1,1)\): \(\int_0^1 (t^2 + t)\,\mathrm dt = \tfrac13 + \tfrac12 = \tfrac56\).
• Parabola \(y = x^2\), \(\vec\gamma = (t,t^2)\), \(\dot{\vec\gamma} = (1,2t)\): \(\int_0^1 (t^4 + 2t^2)\,\mathrm dt = \tfrac15 + \tfrac23 = \tfrac{13}{15}\).
• Parabola \(x = y^2\), \(\vec\gamma = (t^2,t)\), \(\dot{\vec\gamma} = (2t,1)\): \(\int_0^1 (2t^3 + t^2)\,\mathrm dt = \tfrac12 + \tfrac13 = \tfrac56\).
• Right then up: on \(y = 0\) the integrand is \(0\cdot 1 + x\cdot 0 = 0\); on \(x = 1\) it is \(y^2\cdot 0 + 1\cdot 1 = 1\), giving \(\int_0^1 1 = 1\). Total \(\mathbf 1\).

Numerically: \(0.8333,\ 0.8667,\ 0.8333,\ 1\).

2. Nothing. Path-independence is a statement about every pair of paths, and two agreeing is a coincidence — here an exact one, produced by the symmetry \((x,y)\mapsto(y,x)\) acting on the pair of parabolas while the field is not symmetric. One disagreement, on the other hand, settles the question for good: since the straight line and \(y = x^2\) disagree, \(\vec F\) is not conservative on any open set containing both.

3. \(\vec\sigma(s) = (s^2,s^2)\), \(\dot{\vec\sigma} = (2s,2s)\): \[ \int_0^1\big(s^4\cdot 2s + s^2\cdot 2s\big)\mathrm ds = \tfrac26 + \tfrac24 = \tfrac13+\tfrac12 = \tfrac56 , \] unchanged, as parametrisation invariance promised: \(\varphi(s) = s^2\) is \(C^1\) and increasing on \([0,1]\) with \(\varphi(0)=0\), \(\varphi(1)=1\).

4. Out along the straight line and back along the two-leg path: \(\tfrac56 - 1 = -\tfrac16\). Explicitly, \((0,0)\to(1,1)\) along \(y=x\), then \((1,1)\to(1,0)\), then \((1,0)\to(0,0)\). That circuit has the triangle on its right, so it is negatively oriented — clockwise.

Check. Green's theorem, on the triangle with vertices \((0,0),(1,0),(1,1)\): \(\partial_xF_2-\partial_yF_1 = 1-2y\), and \[ \int_0^1\!\!\int_0^x (1-2y)\,\mathrm dy\,\mathrm dx = \int_0^1 (x - x^2)\,\mathrm dx = \tfrac12-\tfrac13 = \tfrac16 . \] Anticlockwise the circulation is \(+\tfrac16\); the loop above runs the other way and gives \(-\tfrac16\) ✓. The other loop — out along \(y=x^2\), back along the line — encloses the region between the two curves and gives \(\tfrac{13}{15}-\tfrac56 = \tfrac1{30}\), which the deck checks against Green too.

Idea. Energy is conserved, the potential vanishes at infinity, and the minimum launch is the one that just runs out of speed there.

1. \(E = \tfrac12 mv^2 - GM_\oplus m/r\) is constant. To reach \(r=\infty\) one needs \(\tfrac12 mv^2 = E + GM_\oplus m/r \ge 0\) for all \(r\), and since the left side tends to \(E\), the condition is \(E \ge 0\); \(E = 0\) is the cheapest. At \(r = R_\oplus\) that reads \(\tfrac12 v_{\mathrm{esc}}^2 = GM_\oplus/R_\oplus\), so \[ v_{\mathrm{esc}} = \sqrt{\frac{2GM_\oplus}{R_\oplus}} = \sqrt{\frac{2(3.986004418\times10^{14})}{6.378137\times10^{6}}} = 1.11799\times10^{4}\ \mathrm{m\,s^{-1}} = 11.18\ \mathrm{km\,s^{-1}} . \] It does not depend on \(m\), because \(m\) divides out of the equation of motion.

2. Equatorial surface speed \(\omega_\oplus R_\oplus = (7.292115\times10^{-5})(6.378137\times10^{6}) = 465.10\ \mathrm{m\,s^{-1}}\). The rocket must supply \(11\,179.88 - 465.10 = 10\,714.78\ \mathrm{m\,s^{-1}}\).

• Speed saving: \(465.10/11\,179.88 = 4.160\%\).
• Energy saving: \(1 - (10\,714.78/11\,179.88)^2 = 8.147\%\).

The factor of about two is differentiation: \(E \propto v^2\), so \(\Delta E/E \approx 2\,\Delta v/v\) for a small fractional change, and \(2\times4.160 = 8.32\), close to \(8.15\) with the second-order term accounting for the rest.

3. At latitude \(\lambda\) the surface speed is \(\omega_\oplus R_\oplus\cos\lambda\).

Energy per unit mass: \(\tfrac12 v_{\mathrm{esc}}^2 = 6.2495\times10^{7}\ \mathrm{J\,kg^{-1}}\) with no help; \(5.7424\times10^{7}\) from Kourou and \(5.8009\times10^{7}\) from Canaveral. The difference is \(5.85\times10^{5}\ \mathrm{J\,kg^{-1}}\), that is \(0.94\%\) of the total.

4. Two omissions, both making the real requirement larger: atmospheric drag, which does negative work all the way up and is worst where the air is densest and the vehicle fastest; and gravity losses, because a real vehicle accelerates over minutes rather than instantaneously, and gravity keeps taking energy from it throughout the climb. (A third, smaller: steering losses from not thrusting along the velocity.)

Check. Dimensions: \(\sqrt{\mathrm{m^3s^{-2}}/\mathrm m} = \mathrm{m\,s^{-1}}\) ✓. Limit: at the poles \(\cos\lambda = 0\), the spin is worth nothing, and the saving is zero ✓. Sanity: \(11.18\ \mathrm{km\,s^{-1}}\) is the quoted figure for the Earth, and \(\sqrt2\) times the low-orbit circular speed \(\sqrt{GM/R} = 7.91\ \mathrm{km\,s^{-1}}\), as \(v_{\mathrm{esc}} = \sqrt2\,v_{\mathrm{circ}}\) demands ✓.

Idea. Test closedness first — it is one differentiation and it kills two of the five. Then, for the survivors, either the domain is simply connected and the Poincaré lemma finishes it, or the domain has a hole and a loop must be found.

1. \(\vec F = (2xy+z^2,\ x^2+2yz,\ y^2+2xz)\) on \(\mathbb R^3\). Closed: \(\partial_yF_1 = 2x = \partial_xF_2\); \(\partial_zF_1 = 2z = \partial_xF_3\); \(\partial_zF_2 = 2y = \partial_yF_3\). \(\mathbb R^3\) is a box — a product of open intervals — so by the Poincaré lemma it is conservative. Guess and verify: \[ V = -(x^2y + y^2z + xz^2),\qquad -\nabla V = (2xy + z^2,\ x^2+2yz,\ y^2+2xz) = \vec F . \]

2. \(\vec F = (y,-x)\) on \(\mathbb R^2\). Not closed: \(\partial_xF_2 - \partial_yF_1 = -1-1 = -2\). Loop: the unit circle \(\vec\gamma(t) = (\cos t,\sin t)\), \[ \oint = \int_0^{2\pi}\big[\sin t\,(-\sin t) + (-\cos t)(\cos t)\big]\mathrm dt = -\int_0^{2\pi}\!1\,\mathrm dt = -2\pi \ne 0 . \] (Green agrees: \(-2\times\text{area} = -2\pi\).)

3. \(\vec F = -GMm\,\vec r/r^3\) on \(\mathbb R^3\setminus\{0\}\). Conservative, with \(V = -GMm/r\): since \(\nabla r = \vec r/r\), \(\nabla V = GMm\,r^{-2}\nabla r = GMm\,\vec r/r^{3}\), so \(-\nabla V = \vec F\) ✓. (No appeal to the domain is needed once \(V\) is exhibited.)

4. \(\vec F = (-y,x,0)/(x^2+y^2)\) on \(\mathbb R^3\) minus the \(z\)-axis. Closed: with \(u = x^2+y^2\), \(\partial_xF_2 = (u-2x^2)/u^2 = (y^2-x^2)/u^2 = \partial_yF_1\), and every derivative involving the third component vanishes because \(F_3 \equiv 0\) and \(F_1, F_2\) do not depend on \(z\). Not conservative: the unit circle in the plane \(z = 0\) lies in the domain, and on it \(\vec F = (-\sin t,\cos t,0) = \dot{\vec\gamma}\), so \[ \oint\vec F\cdot\mathrm d\vec r = \int_0^{2\pi}\!1\,\mathrm dt = 2\pi \ne 0 . \] The domain is not simply connected — that loop cannot be contracted without crossing the missing axis.

5. \(\vec F = (e^x\sin y,\ e^x\cos y)\) on \(\mathbb R^2\). Closed: \(\partial_yF_1 = e^x\cos y = \partial_xF_2\). \(\mathbb R^2\) is a box, so conservative, with \(V = -e^x\sin y\) and \(-\nabla V = (e^x\sin y,\ e^x\cos y) = \vec F\) ✓.

The odd one out. Field 4 is closed and not conservative. Restrict it to the half-space \(\{x > 0\} = (0,\infty)\times\mathbb R\times\mathbb R\), a box: there \(V = -\arctan(y/x)\) works, since \(\partial_x\arctan(y/x) = -y/(x^2+y^2)\) and \(\partial_y\arctan(y/x) = x/(x^2+y^2)\), so \(-\nabla V = \vec F\) ✓. The field has not changed; the domain has, and with it the answer. That is the checkpoint.

Check. Dimensions are only meaningful in cases 3 and 4; in 3 both \(V\) and the work are joules ✓. Limits: in case 1, setting \(z = 0\) leaves \((2xy, x^2)\) with \(V = -x^2y\), the two-dimensional sub-case, and it is consistent ✓. In case 4 a loop of radius \(a\) about the axis gives \(2\pi\) for every \(a\), which is the size-independence that signals a topological count ✓.

Idea. Compute both sides. Then notice that Green's theorem read backwards is a way of measuring an area from its boundary, and read carelessly is a way of proving something false.

1. \(\vec F = (xy, x^2)\) on the triangle \((0,0),(1,0),(1,1)\). Anticlockwise, the three edges:

• \((0,0)\to(1,0)\): \(\vec\gamma = (t,0)\), \(\mathrm dy = 0\), and \(P = xy = 0\). Contribution \(0\).
• \((1,0)\to(1,1)\): \(\vec\gamma = (1,t)\), \(\mathrm dx = 0\), \(Q = x^2 = 1\). Contribution \(\int_0^1 1\,\mathrm dt = 1\).
• \((1,1)\to(0,0)\): \(\vec\gamma = (1-t,1-t)\), \(\dot{\vec\gamma} = (-1,-1)\), \(P = Q = (1-t)^2\). Contribution \(\int_0^1 -2(1-t)^2\mathrm dt = -\tfrac23\).

Total \(0 + 1 - \tfrac23 = \tfrac13\). The other side: \(\partial_xQ - \partial_yP = 2x - x = x\), and \[ \int_0^1\!\!\int_0^x x\,\mathrm dy\,\mathrm dx = \int_0^1 x^2\,\mathrm dx = \tfrac13 . \] Both sides give \(\tfrac13\) ✓.

2. The area formula. This uses the general version of Green's theorem — the one the deck cites rather than proves, valid for a region with a piecewise-\(C^1\) boundary. Take \(P = -y/2\), \(Q = x/2\); then \(\partial_xQ - \partial_yP = \tfrac12 + \tfrac12 = 1\), and Green gives \(\tfrac12\oint(x\,\mathrm dy - y\,\mathrm dx) = \int\!\!\int_\Omega 1 = \text{area}(\Omega)\). For the ellipse, \(\vec\gamma(t) = (a\cos t,\ b\sin t)\): \[ \tfrac12\int_0^{2\pi}\big[a\cos t\cdot b\cos t - b\sin t\cdot(-a\sin t)\big]\mathrm dt = \tfrac{ab}{2}\int_0^{2\pi}\!1\,\mathrm dt = \pi ab . \]

3. Which hypothesis fails. Green's theorem requires \(\vec F\) to be \(C^1\) on an open set containing the closed region bounded by the loop. For the vortex and the unit circle, that region is the closed unit disc, and the vortex is not defined at its centre — it is not even bounded there. Nothing can be said about \(\partial_xF_2 - \partial_yF_1\) at the origin, because neither term exists. So the theorem does not apply, and its conclusion is false: the circulation is \(2\pi\).

Check. Dimensions: with \(\vec F\) dimensionless on a dimensionless plane, both sides of part 1 are pure numbers ✓. Limit for part 2: a circle of radius \(a\) gives \(\pi a^2\), and setting \(b = a\) in the ellipse formula agrees ✓. Part 3 sanity: replacing the unit circle by a circle of radius \(1\) about \((3,0)\) does satisfy the hypothesis, and the circulation is indeed \(0\) ✓.

Idea. One differentiation gives the whole family at once, and the exponent \(n\) is visible in the answer as a single factor \((1-n)\).

1. With \(u = x^2+y^2\), \((F_n)_1 = -y\,u^{-n}\) and \((F_n)_2 = x\,u^{-n}\): \[ \partial_x (F_n)_2 = u^{-n} + x(-n)u^{-n-1}(2x) = u^{-n-1}\big(u - 2nx^2\big), \] \[ \partial_y (F_n)_1 = -u^{-n} - y(-n)u^{-n-1}(2y) = u^{-n-1}\big(-u + 2ny^2\big). \] Subtracting, \[ \partial_x(F_n)_2 - \partial_y(F_n)_1 = u^{-n-1}\big(2u - 2n(x^2+y^2)\big) = 2(1-n)\,u^{-n} . \] Since \(u > 0\) on \(U\), this vanishes identically exactly when \(n = 1\).

2. On the circle of radius \(a\), \(\vec\gamma(t) = a(\cos t,\sin t)\), \(\dot{\vec\gamma} = a(-\sin t,\cos t)\), and \(u = a^2\), so \(\vec F_n(\vec\gamma) = a^{1-2n}(-\sin t,\cos t)\). Hence \[ \oint = \int_0^{2\pi} a^{1-2n}\cdot a\,\big(\sin^2 t + \cos^2 t\big)\,\mathrm dt = 2\pi\,a^{2-2n} . \] For \(n = 3\), \(a = 2\): \(2\pi\cdot 2^{-4} = \pi/8 = 0.3927\).

3. Never. For \(n \ne 1\) the field is not closed, and a conservative field with a \(C^2\) potential must be closed, so it fails at the first hurdle. For \(n = 1\) it is closed but the circle of any radius has circulation \(2\pi \ne 0\), so condition (iii) of P6 fails and there is no potential on \(U\).

4. Only for \(n = 1\). The half-plane \(\{x>0\} = (0,\infty)\times\mathbb R\) is a box, so for \(n = 1\) the Poincaré lemma applies directly and gives \(V = -\arctan(y/x)\). For \(n \ne 1\) the field is still not closed there — closedness is a pointwise condition and shrinking the domain cannot create it — so it is still not conservative. No computation of a loop integral is needed in either direction; that is what the theorem is for.

Check. Dimensions: everything is a pure number ✓. Limits: \(n = 1\) in the part-2 formula gives \(2\pi a^0 = 2\pi\), independent of \(a\), which is the deck's vortex ✓. \(n = 0\) gives \(\vec F_0 = (-y,x)\), whose curl is \(2\) and whose circulation is \(2\pi a^2 = 2\times\text{area}\) ✓ — Green's theorem with a constant integrand. And \(n = \tfrac12\) gives circulation \(2\pi a\), matching a field of unit magnitude tangent to a circle of circumference \(2\pi a\) ✓.

Idea. The cheap implications are (i)⇒(ii) and (ii)⇔(iii). The content is (ii)⇒(i), where a potential has to be manufactured — and the only material to hand is the integral itself.

(i) ⇒ (ii). The fundamental theorem of line integrals: for \(V\in C^1\) and \(\vec\gamma\) piecewise \(C^1\) in \(U\), the chain rule makes \(\frac{\mathrm d}{\mathrm dt}V(\vec\gamma) = \nabla V(\vec\gamma)\cdot\dot{\vec\gamma}\) on each smooth piece, so \(\int_{\vec\gamma}\vec F\cdot\mathrm d\vec r = -\int_{\vec\gamma}\nabla V\cdot\mathrm d\vec r = V(\vec\gamma(a)) - V(\vec\gamma(b))\), which sees only the endpoints.

(ii) ⇒ (iii). A closed loop has \(\vec\gamma(a) = \vec\gamma(b)\). By (ii) its integral equals that of the constant path at that point, whose derivative is zero, so the integral is \(0\).

(iii) ⇒ (ii). Let \(\vec\alpha, \vec\beta\) share endpoints. The concatenation of \(\vec\alpha\) with \(\vec\beta\) reversed is a closed piecewise-\(C^1\) loop, so by (iii) \(\int_{\vec\alpha} - \int_{\vec\beta} = 0\), the minus sign coming from parametrisation invariance under reversal.

(ii) ⇒ (i). First, \(U\) is polygonally connected. Fix \(\vec p\in U\) and let \(A\) be the set of points of \(U\) reachable from \(\vec p\) by a polygonal path inside \(U\). \(A\) is open: if \(\vec q\in A\), take a ball \(B(\vec q,\varepsilon) \subseteq U\) (openness of \(U\)); every point of that ball is reached from \(\vec q\) by one extra segment. \(U\setminus A\) is open by the identical argument. \(A\) is nonempty, and \(U\) is connected, so \(A = U\).

Now define, for \(\vec x\in U\), \[ V(\vec x) := -\int_{\vec p}^{\vec x}\vec F\cdot\mathrm d\vec r \] along any polygonal path in \(U\); by (ii) the value is independent of the path, so \(V\) is a well-defined function. Fix \(\vec x\) and \(\varepsilon>0\) with \(B(\vec x,\varepsilon)\subseteq U\). For \(|s|<\varepsilon\), reach \(\vec x + s\vec e_i\) by a path to \(\vec x\) followed by the segment \(u\mapsto\vec x + u\vec e_i\), \(u\) between \(0\) and \(s\). Then \[ V(\vec x + s\vec e_i) - V(\vec x) = -\int_0^s F_i(\vec x + u\vec e_i)\,\mathrm du . \] The integrand is continuous in \(u\) (continuity of \(\vec F\)), so the right-hand side is a \(C^1\) function of \(s\) with derivative \(-F_i(\vec x + s\vec e_i)\); at \(s = 0\), \(\partial_i V(\vec x) = -F_i(\vec x)\). Every partial derivative of \(V\) therefore exists on \(U\) and equals the continuous function \(-F_i\), so \(V\in C^1\) by unit 03's criterion, and \(\nabla V = -\vec F\). \(\square\)

Where each hypothesis is used.

• Openness twice: to get the balls in the polygonal-connectedness argument, and to have room for the segment \(\vec x + u\vec e_i\) when differentiating \(V\).
• Connectedness once, in (ii)⇒(i), and only to let a single base point \(\vec p\) serve the whole of \(U\).
• Continuity of \(\vec F\) once, to apply the fundamental theorem of calculus to \(s\mapsto\int_0^s F_i\). Without it the construction still defines \(V\), but \(V\) need not be differentiable.

If \(U\) is not connected, apply the argument to each connected component — each is open and connected — with its own base point. The resulting \(V\) is still \(C^1\) with \(\nabla V = -\vec F\), but now the additive constant may be chosen independently on each component, so \(V\) is determined by \(\vec F\) up to one constant per component rather than one overall.

Check. The construction, applied to the inverse-square field with \(\vec p\) at radius \(r_0\): the integral from \(\vec p\) to \(\vec x\) is \(GMm(1/r - 1/r_0)\), so \(V(\vec x) = GMm(1/r_0 - 1/r) = -GMm/r + GMm/r_0\) — the familiar potential plus a constant ✓. The arbitrary constant appears exactly where the base point was chosen, and letting \(r_0\to\infty\) removes it.

Idea. Parts 1 and 2 are two lines each. Part 3 is the interesting one: an antisymmetry condition on the first derivatives, plus Clairaut, forces all second derivatives to vanish.

1. \((\mathrm{curl}\,\nabla f)_1 = \partial_2(\nabla f)_3 - \partial_3(\nabla f)_2 = \partial_2\partial_3 f - \partial_3\partial_2 f = 0\) by Clairaut's theorem (unit 03), whose hypothesis is exactly \(f \in C^2\); the other two components are the same statement with the indices permuted. Equivalently and more structurally: \(D(\nabla f) = H\) is symmetric, so its antisymmetric part is zero, and the curl is twice that.

2. \(\vec\omega\times\vec r = (\omega_2z - \omega_3y,\ \omega_3x - \omega_1z,\ \omega_1y - \omega_2x)\). Then \[ (\mathrm{curl})_1 = \partial_2F_3 - \partial_3F_2 = \omega_1 - (-\omega_1) = 2\omega_1, \] and by the same computation \((\mathrm{curl})_2 = 2\omega_2\), \((\mathrm{curl})_3 = 2\omega_3\). So \(\mathrm{curl}(\vec\omega\times\vec r) = 2\vec\omega\). Against the definition: the map \(\vec r\mapsto\vec\omega\times\vec r\) is linear and equals \(\hat\omega\), so \(D\vec F = \hat\omega\) everywhere; \(\hat\omega\) is antisymmetric, so it is its own antisymmetric part, whose hat-map vector is \(\vec\omega\); twice that is \(2\vec\omega\) ✓.

3. Write \(R_{ij}\) for the hypothesis \(\partial_j F_i = -\partial_i F_j\), valid at every point of \(U\). Differentiate and apply Clairaut repeatedly: \[ \partial_k\partial_j F_i \overset{R}{=} -\partial_k\partial_i F_j \overset{\text{Cl}}{=} -\partial_i\partial_k F_j \overset{R}{=} +\partial_i\partial_j F_k \overset{\text{Cl}}{=} \partial_j\partial_i F_k \overset{R}{=} -\partial_j\partial_k F_i \overset{\text{Cl}}{=} -\partial_k\partial_j F_i . \] Hence \(\partial_k\partial_j F_i = 0\) for all \(i,j,k\): every second partial of every component vanishes on \(U\).

Therefore each \(\partial_j F_i\) has vanishing gradient on the connected open \(U\), so it is a constant \(A_{ij}\); and then \(F_i(\vec r) - \sum_j A_{ij}r_j\) has vanishing gradient, so it too is a constant \(c_i\). Thus \(\vec F(\vec r) = A\vec r + \vec c\) with \(A = D\vec F\) constant. By hypothesis \(A\) is antisymmetric, and unit 03's hat map is an isomorphism \(\mathbb R^3 \to \mathfrak{so}(3)\), so \(A = \hat\omega\) for a unique \(\vec\omega\), giving \[ \vec F(\vec r) = \vec\omega\times\vec r + \vec c . \qquad\square \]

Check. Dimensions: if \(\vec F\) is a velocity, \(\vec\omega\) is \(\mathrm{s^{-1}}\) and \(\vec c\) is \(\mathrm{m\,s^{-1}}\) — the general rigid motion, a rotation plus a translation ✓. Special case: \(\vec\omega = \vec 0\) gives the constant fields, whose derivative is \(0\), which is antisymmetric ✓. Physical reading: the fields with antisymmetric derivative everywhere are exactly the rigid velocity fields — the ones that strain nothing, since the symmetric part of \(D\vec F\) is the rate of strain. Part 3 is the statement that rigid motions are the only strain-free motions, which is the theorem unit 22 needs.

Idea. The Hessian is the zero matrix, so the second-order test is silent. The homogeneity \(V(t\vec r) = t^3V(\vec r)\) says the whole behaviour is visible on the unit circle, and polar coordinates read it off.

1. \(\nabla V = (3x^2 - 3y^2,\ -6xy)\). Setting both to zero: \(xy = 0\) and \(x^2 = y^2\), whose only common solution is \(x = y = 0\). So the origin is the only critical point. The Hessian is \[ H(x,y) = \begin{pmatrix} 6x & -6y \\ -6y & -6x\end{pmatrix}, \qquad H(0,0) = \begin{pmatrix}0&0\\0&0\end{pmatrix} . \] Degenerate, \(\det H = 0\), and the classification theorem says nothing at all.

2. With \(x = r\cos\theta\), \(y = r\sin\theta\), \[ x^3 - 3xy^2 = r^3\big(\cos^3\theta - 3\cos\theta\sin^2\theta\big) = r^3\big(4\cos^3\theta - 3\cos\theta\big) = r^3\cos3\theta , \] using \(\sin^2 = 1-\cos^2\) and the triple-angle identity.

• \(\{V = 0\}\) is \(\cos3\theta = 0\), that is \(\theta = \pi/6 + k\pi/3\): six rays through the origin, at \(60^\circ\) to one another.
• Between consecutive rays the sign of \(\cos3\theta\) alternates, so there are three open sectors of width \(60^\circ\) on which \(V > 0\) and three on which \(V < 0\).
• Along \(\theta = 0,\ 2\pi/3,\ 4\pi/3\): \(\cos3\theta = 1\) and \(V = +r^3\), rising out of the origin — three ridges. Along \(\theta = \pi/3,\ \pi,\ 5\pi/3\): \(\cos3\theta = -1\) and \(V = -r^3\), falling — three valleys.

The origin is therefore neither a maximum nor a minimum: every neighbourhood contains points with \(V>0\) and points with \(V<0\). It is a saddle with six sectors rather than the usual four, which is why it is called a monkey saddle — a saddle with a place for the tail.

3. The flow of \(\dot{\vec r} = -\nabla V\). The flow decreases \(V\), so it runs down the landscape: trajectories leave the origin along the three valleys \(\theta = \pi/3, \pi, 5\pi/3\) and arrive at the origin along the three ridges \(\theta = 0, 2\pi/3, 4\pi/3\). Three directions of approach, three of departure, and six hyperbolic sectors between them.

Why the rays are the whole story: on each of the six rays \(\theta = k\pi/3\), \(-\nabla V\) is radial (it is \(\mp3r^2\hat r\), from the formula below), so a trajectory that starts on one of these rays stays on it; and every trajectory crosses the level curves of \(V\) at right angles, downhill, so between the rays the flow lines are hyperbola-like curves that come in along a ridge and leave along a neighbouring valley. The linearisation at the origin is the zero matrix and says nothing, which is why the picture has to be read off \(V\) itself.

The turning count. Going once anticlockwise round a circle about the origin, the direction of \(-\nabla V\) turns through \(-4\pi\): winding number \(-2\). Indeed \(x^2-y^2 = r^2\cos2\theta\) and \(2xy = r^2\sin2\theta\), so \[ -\nabla V = 3r^2(-\cos2\theta,\ \sin2\theta) = 3r^2\big(\cos(\pi - 2\theta),\ \sin(\pi-2\theta)\big) , \] whose direction angle \(\pi - 2\theta\) falls by \(4\pi\) as \(\theta\) rises by \(2\pi\). That is \(-2\), against \(-1\) for an ordinary saddle such as \(x^2 - y^2\), whose \(-\nabla V = (-2x, 2y)\) turns through \(-2\pi\) — one more piece of evidence that this is not an ordinary saddle.

4. \(\partial_x^2V + \partial_y^2V = 6x + (-6x) = 0\) at every point. So the trace of the Hessian vanishes identically, and the deck's remark applies everywhere, not only at the critical point: at any critical point of such a \(V\) the eigenvalues of \(H\) sum to zero, so they cannot all be positive, so no critical point of \(V\) is a nondegenerate minimum, and the Hessian test can never hand Lagrange–Dirichlet a minimum to work with. (That such a \(V\) has no local minimum of any kind is the maximum principle for harmonic functions, unit 05.)

Here the unique equilibrium is in fact unstable, and one explicit family of motions shows it. \(V(x,-y) = V(x,y)\), so \(\partial_yV = -6xy\) vanishes on \(y = 0\) and a motion started on the \(x\)-axis with no \(y\)-velocity stays on it, obeying \(m\ddot x = -3x^2\). Start it at rest at \(x = -\varepsilon\), in a valley: energy conservation gives \(\tfrac12 m\dot x^2 = V(-\varepsilon) - V(x) = \lvert x\rvert^3 - \varepsilon^3\) for \(x \lt 0\), and \(\ddot x \lt 0\) throughout, so \(x\) decreases without bound, with a speed that grows as \(\lvert x\rvert^{3/2}\). However small \(\varepsilon\), the motion leaves every fixed neighbourhood of the origin.

Check. Homogeneity: \(V(t\vec r) = t^3V(\vec r)\), consistent with \(r^3\cos3\theta\) ✓. Value on the unit circle: at \(\theta = 0\), \(V(1,0) = 1 = \cos 0\) ✓; at \(\theta = \pi/3\), \((x,y) = (\tfrac12,\tfrac{\sqrt3}2)\) and \(V = \tfrac18 - 3\cdot\tfrac12\cdot\tfrac34 = \tfrac18 - \tfrac98 = -1 = \cos\pi\) ✓.

Another way

Optional, and forward-looking: nothing below is needed for the answer. The complex route is shorter and explains the \(3\). Writing \(z = x+\mathrm iy\), \(z^3 = x^3-3xy^2 + \mathrm i(3x^2y - y^3)\), so \(V = \operatorname{Re}(z^3)\), and \(z^3\) maps each of the six \(60^\circ\) sectors onto a half-plane — which is parts 2 and 3 in one line. The real part of a holomorphic function is also harmonic, which is part 4 for free. Unit 20 is where that machinery arrives properly; the point of doing it by hand here is that the classification theorem's failure is what has to be worked around, and the polar computation shows exactly what replaces it.

Idea. Two attractions along one line balance at one point; the Hessian of each \(1/r\) term at a point on the axis is the same universal \(\mathrm{diag}(2,-1,-1)\) shape, so the total is trace-free and the point cannot be a minimum.

1. With the Earth at the origin, the Moon at \(x = D\) and the test mass at \(0<x<D\), \[ V(x) = -m\Big(\frac{GM_\oplus}{x} + \frac{GM_{\text{M}}}{D-x}\Big),\qquad V'(x) = m\Big(\frac{GM_\oplus}{x^2} - \frac{GM_{\text{M}}}{(D-x)^2}\Big) . \] Setting \(V' = 0\) gives \((D-x)/x = \sqrt{GM_{\text{M}}/GM_\oplus} = 0.1109063\), so \[ x = \frac{D}{1 + 0.1109063} = 0.9001659\,D = 3.46024\times10^{8}\ \mathrm m = \mathbf{346\,024\ km} , \] that is \(90.02\%\) of the way to the Moon. The remaining distance to the Moon is \(38\,376\) km.

2. Write \(\ell_1 = 3.46024\times10^8\) m and \(\ell_2 = 3.83762\times10^7\) m. With \(\vec s\) the displacement from a source, \(\vec s = (\pm\ell,0,0)\) on the axis, the given formula gives \[ \partial_i\partial_j\lVert\vec s\rVert^{-1} = \frac{3s_is_j - \lVert\vec s\rVert^2\delta_{ij}}{\lVert\vec s\rVert^{5}} = \frac{1}{\ell^3}\,\mathrm{diag}(2,-1,-1), \] the same for either sign, since \(s_is_j\) is quadratic. Hence \[ H = -m\Big(\frac{GM_\oplus}{\ell_1^3} + \frac{GM_{\text{M}}}{\ell_2^3}\Big) \mathrm{diag}(2,-1,-1) = m\,K\,\mathrm{diag}(-2,\,1,\,1), \] with \[ K = \frac{3.986004418\times10^{14}}{(3.46024\times10^8)^3}\; +\; \frac{4.9028695\times10^{12}}{(3.83762\times10^7)^3} = 9.6370\times10^{-11}\ \mathrm{s^{-2}} . \] Eigenvalues per unit mass: \(-2K = -1.9274\times10^{-10}\ \mathrm{s^{-2}}\) once, and \(+K = 9.6370\times10^{-11}\ \mathrm{s^{-2}}\) twice. Both signs present: indefinite, so the point is a saddle — unstable along the Earth–Moon line, oscillatory in the two transverse directions.

3. Transverse: \(m\ddot y = -mKy\), so \(\omega = \sqrt K = 9.8168\times10^{-6}\ \mathrm{s^{-1}}\) and \[ T = \frac{2\pi}{\omega} = 6.4004\times10^{5}\ \mathrm s = \mathbf{177.8\ h} = 7.41\ \text{days}. \] Along the line: \(m\ddot x = +2mKx\), whose growing solution is \(\mathrm e^{\sqrt{2K}\,t}\) with \(\sqrt{2K} = 1.3883\times10^{-5}\ \mathrm{s^{-1}}\), so a displacement grows by \(\mathrm e\) in \(1/\sqrt{2K} = 7.2030\times10^{4}\ \mathrm s = \mathbf{20.0\ h}\).

4. \(\operatorname{tr}H = m K(-2+1+1) = 0\). This is not special to the axis: the trace of the supplied formula is \((3\lVert\vec s\rVert^2 - 3\lVert\vec s\rVert^2)/\lVert\vec s\rVert^5 = 0\) at every point, so the Hessian of any sum of such terms is trace-free everywhere. Since the trace is the sum of the eigenvalues, they cannot all be positive, so no critical point of any such potential is a nondegenerate minimum. That is the Hessian half of Earnshaw's theorem; the full statement, that there is no local minimum of any kind and so no stable equilibrium point in the field of fixed masses, needs the maximum principle for harmonic functions (unit 05).

5. It drops the orbital motion: the Moon is not fixed but goes round in \(27.32\) days, and the honest calculation works in the rotating frame of unit 03 and adds the centrifugal potential \(-\tfrac12 m\Omega^2\rho^2\). Doing so moves the balance point inwards, from \(346\,024\) km to the true \(L_1\) at \(326\,381\) km from the Earth's centre. It does not make it stable: \(L_1\) is still a saddle. The rotating frame does buy stability elsewhere — the Coriolis force, which does no work and so is invisible to an energy landscape, stabilises \(L_4\) and \(L_5\), which are maxima of the effective potential.

Check. Dimensions: \(K\) is \(\mathrm{m^3s^{-2}}/\mathrm{m^3} = \mathrm{s^{-2}}\) ✓, so \(\sqrt K\) is a frequency ✓, and \(mK\) is \(\mathrm{kg\,s^{-2}} = \mathrm{N\,m^{-1}}\), a spring constant ✓. Limit: setting \(GM_{\text{M}} = 0\) sends the balance point to \(x = D\) and \(K\) to \(GM_\oplus/D^3\), the one-body answer ✓. Sanity: the neutral point is much closer to the Moon than to the Earth, in the ratio \(\sqrt{M_{\text{M}}/M_\oplus} = 1/9.02\), and \(38\,376/346\,024 = 1/9.02\) ✓.

Idea. In the rotating frame the statics is a constrained critical-point problem, so Lagrange applies; and the multiplier is not bookkeeping — it is the force the hoop exerts.

1. Unit 03's centrifugal force is \(-m\vec\Omega\times(\vec\Omega\times\vec r)\). Expanding the double cross product, \(\vec\Omega\times(\vec\Omega\times\vec r) = (\vec\Omega\cdot\vec r)\vec\Omega - \Omega^2\vec r\), so the force is \(m\Omega^2\big(\vec r - (\hat\Omega\cdot\vec r)\hat\Omega\big) = m\Omega^2\vec\rho\), where \(\vec\rho\) is the component of \(\vec r\) perpendicular to the axis. Since \(\rho^2 = \lVert\vec r\rVert^2 - (\hat\Omega\cdot\vec r)^2\), \[ \nabla\big(\tfrac12 m\Omega^2\rho^2\big) = m\Omega^2\big(\vec r - (\hat\Omega\cdot\vec r)\hat\Omega\big) = m\Omega^2\vec\rho, \] so the centrifugal force is \(-\nabla\big(-\tfrac12 m\Omega^2\rho^2\big)\). With the axis vertical and the hoop in the \((x,z)\) plane, \(\rho = |x|\) and \(V_{\mathrm{eff}} = mgz - \tfrac12 m\Omega^2x^2\). The Coriolis force \(-2m\vec\Omega\times\vec v\) does not appear because at an equilibrium the bead is at rest in this frame, so \(\vec v = \vec 0\) and the term vanishes identically. (More generally it never enters a potential, being perpendicular to \(\vec v\) and doing no work.)

2. \(\nabla V_{\mathrm{eff}} = (-m\Omega^2x,\ mg)\) and \(\nabla c = (2x, 2z)\), so \(\nabla V_{\mathrm{eff}} = \lambda\nabla c\) reads \[ -m\Omega^2 x = 2\lambda x, \qquad mg = 2\lambda z, \qquad x^2+z^2 = R^2 . \] From the first equation, either \(x = 0\) or \(\lambda = -\tfrac12 m\Omega^2\).

• \(x = 0\): then \(z = \pm R\) — the bottom and the top of the hoop — with \(\lambda = mg/(2z) = \pm mg/(2R)\).
• \(\lambda = -\tfrac12 m\Omega^2\): the second equation gives \(z_* = mg/(2\lambda) = -g/\Omega^2\), and the constraint then gives \(x_* = \pm\sqrt{R^2 - g^2/\Omega^4}\). This pair exists only if \(g/\Omega^2 \le R\), that is \(\Omega^2 \ge g/R\).

3. \(\Omega^2R/g = 36(0.35)/9.81 = 1.2844 > 1\), so the tilted pair exists. \[ z_* = -\frac{9.81}{36} = -0.27250\ \mathrm m, \qquad x_* = \sqrt{0.35^2 - 0.27250^2} = 0.21964\ \mathrm m, \] and the angle from the bottom satisfies \(\cos\theta_* = |z_*|/R = 0.778571\), so \(\theta_* = \mathbf{38.87^\circ}\). The constraint force: \[ \lVert\vec N\rVert = |\lambda|\,\lVert\nabla c\rVert = \tfrac12 m\Omega^2\cdot 2\sqrt{x_*^2+z_*^2} = m\Omega^2R = (0.12)(36)(0.35) = \mathbf{1.512\ N}. \]

Inertial-frame check. In the laboratory the bead travels a horizontal circle of radius \(x_* = 0.21964\) m at \(\Omega = 6.0\ \mathrm{s^{-1}}\), so it needs a net horizontal force \(m\Omega^2x_* = 0.94886\) N towards the axis and zero net vertical force. The only forces are gravity, \(mg = 1.1772\) N down, and the hoop's. So the hoop supplies \(0.94886\) N horizontally and \(1.1772\) N vertically, of magnitude \(\sqrt{0.94886^2 + 1.1772^2} = 1.5120\ \mathrm N\) ✓ — the same number, from a frame in which no fictitious force was used at all.

4. \(\cos\theta = g/(\Omega^2R)\), so \(\theta = 60^\circ\) needs \(\Omega^2 = 2g/R = 56.057\ \mathrm{s^{-2}}\), that is \(\Omega = 7.487\ \mathrm{s^{-1}}\).

5. \(\Omega_c = \sqrt{g/R} = \sqrt{9.81/0.35} = 5.294\ \mathrm{s^{-1}}\). Below it the only equilibria are \(x = 0\): the bottom (stable) and the top (unstable). At \(\Omega_c\) the tilted pair is born at the bottom and moves outwards as \(\Omega\) grows, while the bottom simultaneously turns unstable: one stable equilibrium becomes an unstable one flanked by a stable symmetric pair. That event is called a supercritical pitchfork bifurcation, with \(\Omega\) as the parameter (the name is from unit 02, which classifies such events; nothing here depends on it). Parametrising by the angle \(\theta\) from the bottom confirms it: \(V_{\mathrm{eff}}/m = -gR\cos\theta - \tfrac12\Omega^2R^2\sin^2\theta\), so \(V'_{\mathrm{eff}}/m = R\sin\theta\,(g - \Omega^2R\cos\theta)\) and \(V''_{\mathrm{eff}}(0)/m = R(g - \Omega^2R)\), which changes sign exactly at \(\Omega_c\).

Check. Dimensions: \(m\Omega^2R\) is \(\mathrm{kg\,s^{-2}\,m} = \mathrm N\) ✓; \(\lambda\) is \(\mathrm{J\,m^{-2}}\) and \(\lVert\nabla c\rVert\) is \(\mathrm m\), so their product is \(\mathrm{J\,m^{-1}} = \mathrm N\) ✓. Limit: at \(\Omega = \Omega_c\), \(z_* = -R\) and \(\lVert\vec N\rVert = m\Omega_c^2R = mg\) — the hoop merely carries the weight ✓. Limit: \(\Omega\to\infty\) sends \(z_*\to0\) and \(\theta_*\to90^\circ\), the bead to the equator of the hoop, with \(\lVert\vec N\rVert\to\infty\) ✓.

Idea. The implicit function theorem is a statement about one partial derivative at a time. Its failure is not a failure of the surface but of the coordinate chosen to solve for.

1. \(\nabla G = (2x,\ 2y,\ -2z)\), which vanishes only at the origin. But \(G(0,0,0) = -1 \ne 0\), so the origin is not on \(S\), and \(\nabla G \ne \vec 0\) at every point of \(S\). Hence \(0\) is a regular value, and near each of its points \(S\) is the graph of a \(C^1\) function of two of the coordinates: it is a \(C^1\) surface.

2. The theorem gives \(z = \zeta(x,y)\) wherever \(\partial_zG = -2z \ne 0\), that is wherever \(z \ne 0\). The exceptional set is \(S\cap\{z=0\} = \{x^2+y^2 = 1,\ z = 0\}\), the waist circle of the hyperboloid. Geometrically the tangent plane there contains the \(z\)-direction — it is "vertical" — so \(S\) cannot be a graph over \((x,y)\) near such a point. Concretely, near \((1,0,0)\) the surface has \(z = \pm\sqrt{x^2+y^2-1}\): two values of \(z\) over each \((x,y)\) with \(x^2+y^2 > 1\), and none at all with \(x^2+y^2 < 1\). Neither existence nor uniqueness survives.

3. At a waist point \((x_0,y_0,0)\) we have \(x_0^2+y_0^2 = 1\), so at least one of \(x_0, y_0\) is nonzero. If \(x_0 \ne 0\) then \(\partial_xG = 2x_0 \ne 0\) and the theorem gives \(x = \xi(y,z)\); explicitly \(x = \operatorname{sgn}(x_0)\sqrt{1 + z^2 - y^2}\), which is \(C^1\) near \((y_0,0)\) since \(1+z^2-y^2 = x_0^2 > 0\) there. If instead \(y_0 \ne 0\), the same with \(y = \eta(x,z) = \operatorname{sgn}(y_0)\sqrt{1+z^2-x^2}\). So \(S\) is a graph near every point — just not always over the same pair of coordinates. That is exactly the reason a manifold is defined with an atlas of charts rather than one global graph (unit 31).

4. \(\nabla G_0 = (2x,2y,-2z)\) vanishes at the origin, and \(G_0(0,0,0) = 0\), so the origin is on \(\{G_0 = 0\}\) and \(0\) is not a regular value. The set is the double cone \(z^2 = x^2+y^2\), and the offending point is its apex. Near the apex it is not a graph over any pair of coordinates: over \((x,y)\) there are two branches \(z = \pm\sqrt{x^2+y^2}\) meeting at the apex; over \((x,z)\) one needs \(y = \pm\sqrt{z^2-x^2}\), which has two branches where it is defined and is undefined where \(|x| > |z|\); and the same for \((y,z)\). More strongly, it is not a surface at all there: deleting the apex disconnects every small neighbourhood of it in the cone into two pieces, which deleting a point from a disc never does.

Check. Part 1 against part 4: the two functions differ only by the constant \(1\), and that constant is the whole difference between a smooth surface and a cone with a singular point ✓. Part 2 against part 3: the exceptional set of the first is a circle, of dimension one, inside a surface of dimension two — a set of lower dimension, as it must be, since a surface can fail to be a graph over a fixed pair of coordinates only where its tangent plane is parallel to the remaining axis ✓.

Idea. Solving \(\vec f(\vec x) = \vec y\) is finding a fixed point of a map that differs from the identity by something small. The Banach fixed-point theorem (proved in the deck's notes on the inverse function theorem) then does all the work, and every estimate below exists to satisfy its two hypotheses: a contraction, and a complete space it maps into itself.

1. The mean value inequality. Let \(\varphi\) be \(C^1\) on a convex set with \(\lVert D\varphi\rVert \le k\), and let \(\vec x, \vec x'\) lie in it. Put \(h(t) = \varphi\big(\vec x' + t(\vec x-\vec x')\big)\) for \(t\in[0,1]\); the segment stays in the set by convexity. By unit 03's chain rule \(h\) is \(C^1\) with \(h'(t) = D\varphi(\cdot)(\vec x-\vec x')\), so \(\lVert h'(t)\rVert \le k\lVert\vec x-\vec x'\rVert\) for every \(t\). Integrating componentwise and using \(\lVert\int_0^1 h'\rVert \le \int_0^1\lVert h'\rVert\), \[ \lVert\varphi(\vec x)-\varphi(\vec x')\rVert = \Big\lVert\int_0^1 h'(t)\,\mathrm dt\Big\rVert \le k\,\lVert\vec x-\vec x'\rVert . \] (There is no mean value equality for vector-valued maps — the two components would need the same intermediate point — which is why the inequality is the right statement.)

2. The contraction. \(D\varphi_{\vec y}(\vec x) = I - D\vec f(\vec x)\), whose norm is \(\le\tfrac12\) on \(\bar B_r\) by the choice of \(r\). The ball is convex, so part 1 gives \(\lVert\varphi_{\vec y}(\vec x)-\varphi_{\vec y}(\vec x')\rVert \le \tfrac12\lVert\vec x-\vec x'\rVert\): a \(\tfrac12\)-contraction. It maps \(\bar B_r\) into itself because \(\varphi_{\vec y}(\vec 0) = \vec y\) (using \(\vec f(\vec 0)=\vec 0\)), so for \(\vec x\in\bar B_r\), \[ \lVert\varphi_{\vec y}(\vec x)\rVert \le \lVert\varphi_{\vec y}(\vec x)-\varphi_{\vec y}(\vec 0)\rVert + \lVert\vec y\rVert \le \tfrac12 r + \tfrac12 r = r . \] Completeness is used for the space \(\bar B_r\) with the Euclidean metric: it is a closed subset of the complete space \(\mathbb R^n\), hence complete, and Banach's theorem needs exactly that. (Unit 01 uses the same theorem on a space of continuous functions instead; the theorem does not care which.) The unique fixed point \(\vec x\) satisfies \(\vec x = \vec x + \vec y - \vec f(\vec x)\), that is \(\vec f(\vec x) = \vec y\).

3. The lower bound. Apply part 1 to \(\varphi_{\vec 0} = \mathrm{id} - \vec f\), which also has \(\lVert D\varphi_{\vec 0}\rVert\le\tfrac12\): \[ \lVert(\vec x-\vec x') - (\vec f(\vec x)-\vec f(\vec x'))\rVert \le \tfrac12\lVert\vec x-\vec x'\rVert . \] By the reverse triangle inequality the left side is at least \(\lVert\vec x-\vec x'\rVert - \lVert\vec f(\vec x)-\vec f(\vec x')\rVert\), so \[ \lVert\vec f(\vec x)-\vec f(\vec x')\rVert \ge \tfrac12\lVert\vec x-\vec x'\rVert . \] Hence \(\vec f(\vec x) = \vec f(\vec x')\) forces \(\vec x = \vec x'\): \(\vec f\) is injective on \(\bar B_r\). Writing \(\vec g\) for its inverse on the image, the same inequality reads \(\lVert\vec g(\vec y)-\vec g(\vec y')\rVert \le 2\lVert\vec y-\vec y'\rVert\), so \(\vec g\) is Lipschitz with constant \(2\), in particular continuous.

4. \(\vec f(x,y) = (\mathrm e^x\cos y,\ \mathrm e^x\sin y)\) has \[ D\vec f = \mathrm e^x\begin{pmatrix}\cos y & -\sin y\\ \sin y & \cos y\end{pmatrix}, \qquad \det D\vec f = \mathrm e^{2x} \ne 0 \ \text{everywhere}, \] yet \(\vec f(x, y+2\pi) = \vec f(x,y)\), so it is not injective on \(\mathbb R^2\). The step that is local is the choice of \(r\): it comes from continuity of \(D\vec f\) at one point, and it guarantees \(\lVert D\vec f - I\rVert\le\tfrac12\) only on a ball around that point. A global version would need that bound everywhere at once, which says the map is a small perturbation of the identity globally — a far stronger hypothesis than "\(D\vec f\) is invertible at each point", and one this map fails. Nothing weaker will do: invertibility of each \(D\vec f(\vec x)\) is a statement about infinitesimal behaviour, and injectivity is a statement about pairs of distant points.

Check. Dimensions are vacuous here, the setting being dimensionless. Sanity on the constant \(\tfrac12\): any \(k<1\) works, with the ball condition becoming \(\lVert\vec y\rVert \le (1-k)r\) and the inverse Lipschitz constant \(1/(1-k)\); \(k = \tfrac12\) is chosen only because it makes both \(\tfrac12 r\) and \(2\) come out tidily ✓. Limit: \(\vec f = \mathrm{id}\) gives \(\varphi_{\vec y}(\vec x) = \vec y\), a \(0\)-contraction whose fixed point is \(\vec y\) itself, and the inverse is the identity ✓.

Idea. Being outside the conservative frame has two quite different causes. One force takes energy out; the other never touches it.

1. The work–energy theorem gives \(\frac{\mathrm d}{\mathrm dt}\big(\tfrac12 m\lVert\vec v\rVert^2\big) = \vec F\cdot\vec v\). For the magnetic force, \(\vec F\cdot\vec v = q(\vec v\times\vec B)\cdot\vec v = 0\), since \(\vec v\times\vec B\) is orthogonal to \(\vec v\) by construction (equivalently, the triple product \(\vec v\cdot(\vec v\times\vec B)\) is the determinant of a matrix with two equal columns). So \(\lVert\vec v\rVert^2\) has zero derivative and the speed is constant, whatever \(\vec B(\vec r)\) is and however complicated the trajectory. Only the direction of \(\vec v\) changes.

2. Along the path, \(\hat v = \dot{\vec\gamma}/\lVert\dot{\vec\gamma}\rVert\), so \[ \oint\vec F\cdot\mathrm d\vec r = \int_0^{T}\big(-\mu N\hat v\big)\cdot\dot{\vec\gamma}\,\mathrm dt = -\mu N\int_0^{T}\lVert\dot{\vec\gamma}\rVert\,\mathrm dt = -\mu N L , \] the arc length \(L\) of the path (unit 03). This is strictly negative for every loop of positive length: the integrand is \(\le 0\) pointwise, with equality only where the block is momentarily at rest. No loop can make it zero, which is a much stronger failure of condition (iii) than the vortex's — there the circulation could be made zero by choosing a loop that misses the hole.

3. Coriolis: \(\vec F_{\mathrm C}\cdot\vec v = -2m(\vec\omega\times\vec v)\cdot\vec v = 0\), for the same reason as part 1. Centrifugal: from \(\vec\omega\times(\vec\omega\times\vec r) = (\vec\omega\cdot\vec r)\vec\omega - \omega^2\vec r\), \[ -m\vec\omega\times(\vec\omega\times\vec r) = m\omega^2\big(\vec r - (\hat\omega\cdot\vec r)\hat\omega\big) = m\omega^2\vec\rho , \] and with \(\rho^2 = \lVert\vec r\rVert^2 - (\hat\omega\cdot\vec r)^2\), \(\nabla\big(\tfrac12 m\omega^2\rho^2\big) = m\omega^2\vec\rho\). Hence the centrifugal force equals \(-\nabla\big(-\tfrac12 m\omega^2\rho^2\big)\): it is conservative, with potential energy \(-\tfrac12 m\omega^2\rho^2\).

4. The magnetic force and the Coriolis force. Both are (charge-like constant) \(\times\ \vec v\times\)(axial vector), both are perpendicular to \(\vec v\), and both therefore do no work. Explicitly \[ \vec F_{\mathrm C} = -2m\,\vec\omega\times\vec v = \vec v\times(2m\vec\omega), \qquad \vec F_{\text{mag}} = q\,\vec v\times\vec B , \] so the substitution is \(q\vec B \longleftrightarrow 2m\vec\omega\). A charged particle in a uniform \(\vec B\) circles at \(\omega_{\mathrm c} = qB/m\); a free particle in a rotating frame, with the centrifugal term balanced (as gravity balances it on the Earth's surface), circles at \(2\omega\) — unit 03's inertial circle, whose period is half the frame's rotation period, and the factor two is exactly the one in the substitution.

Check. Dimensions of the substitution: \(qB\) is \(\mathrm{C\cdot T} = \mathrm{C\,kg\,C^{-1}s^{-1}} = \mathrm{kg\,s^{-1}}\), and \(2m\omega\) is \(\mathrm{kg\,s^{-1}}\) ✓. Limit in part 2: \(\mu\to0\) gives \(\oint\to0\) and friction rejoins the conservative world, trivially, as the zero field ✓. Part 3 special case: on the equator \(\vec r\perp\vec\omega\), so \(\rho = r\) and the centrifugal potential energy is \(-\tfrac12 m\omega^2r^2\), whose gradient has magnitude \(m\omega^2r\) — unit 03's \(0.346\%\) of \(g\) ✓.