Hamilton–Jacobi: mechanics was optics all along, and free motion is already geodesic motion

Idea. Everything is explicit: the true path is known in closed form, so \(S\) is one integral of a polynomial, and the theorem of the slides can be checked by differentiation.

1. The Euler–Lagrange equation is \(\ddot q = -g\), whose solutions are \(q(s) = q_0 + v_0 s - \tfrac12 gs^2\). The condition \(q(t) = q\) reads \(v_0t = q - q_0 + \tfrac12gt^2\), which has exactly one solution \(v_0 = (q - q_0)/t + gt/2\) for every \(t > 0\). So there is never a conjugate point, and \(S\) is defined on all of \(\{t > 0\}\).

2. Integrating \(L\) along this path (the terms are polynomials in \(s\); collect after substituting \(v_0\)):

To see it without brute force: \(\int_0^t\tfrac12 m\dot q^2\,\mathrm ds = \tfrac12 m\big(v_0^2t - v_0gt^2 + \tfrac13g^2t^3\big)\) and \(\int_0^t mgq\,\mathrm ds = mg\big(q_0t + \tfrac12v_0t^2 - \tfrac16gt^3\big)\); subtract and substitute \(v_0 = (q-q_0)/t + gt/2\).

3. The arrival velocity is \(\dot q(t) = v_0 - gt = (q - q_0)/t - gt/2\), so \(p(t) = m(q - q_0)/t - mgt/2\). Differentiating,

And \(-H = -\dfrac{p^2}{2m} - mgq\) with \(\dfrac{p^2}{2m} = \dfrac m2\Big[\dfrac{(q-q_0)^2}{t^2} - g(q - q_0) + \dfrac{g^2t^2}{4}\Big]\) gives \(-\dfrac{m(q-q_0)^2}{2t^2} + \dfrac{mg(q - q_0)}{2} - \dfrac{mg^2t^2}{8} - mgq\), which is the same ✓. Since \(\partial_t S = -H(q, \partial_q S)\), \(S\) solves \(\partial_t S + (\partial_q S)^2/2m + mgq = 0\).

4. At \(g = 0\), \(S = m(q - q_0)^2/2t\), the free particle's principal function from the slides ✓.

Check. Dimensions: \(m(q-q_0)^2/t\), \(mgtq\) and \(mg^2t^3\) are all \(\mathrm{kg\,m^2\,s^{-1}} = \mathrm{J\,s}\) ✓. Symmetry: \(S\) is symmetric in \(q \leftrightarrow q_0\), as it must be for a time-reversible system with a time-independent \(H\) (the reversed path is also a true path, with the same action).

Idea. \(H\) has no \(t\), so time separates; the remaining ODE for \(W\) is integrated once, and the trajectory comes out of one derivative in \(E\).

1. \(S = -Et + W\) solves \(\partial_tS + H = 0\) iff \(\tfrac1{2m}W'^2 + mgq = E\). On the branch \(W' = +\sqrt{2m(E - mgq)}\) (valid where \(q \lt E/mg\)),

as differentiating confirms: \(\partial_q\big(2m(E-mgq)\big)^{3/2} = \tfrac32\big(2m(E - mgq)\big)^{1/2}\cdot(-2m^2g)\). It is complete: \(\partial^2 S/\partial q\,\partial E = m/\sqrt{2m(E - mgq)} \ne 0\).

2. \(\partial_E\big(2m(E - mgq)\big)^{3/2} = 3m\big(2m(E-mgq)\big)^{1/2}\), so

Hence \(\sqrt{2m(E - mgq)} = -mg(t + \beta)\), which needs \(t \le -\beta\) on this branch. Squaring,

For \(t > -\beta\) the other branch, \(W' = -\sqrt{\cdots}\), gives the same formula continued — the falling half. And \(p = m\dot q\) since \(\dot q = -g(t+\beta)\) ✓.

3. The apex is at \(t = -\beta\) at height \(E/mg\): \(-\beta\) is the time of the apex and \(E\) the energy. For the stone, \(E/m = \tfrac12(12.0)^2 = 72.0\ \mathrm{J\,kg^{-1}}\), the apex height is \(72.0/9.81 = 7.34\) m, and from \(q(0) = 0\): \((t+\beta)^2 = 2E/mg^2\) at \(t = 0\), so \(\beta = -\sqrt{2\cdot72.0}/9.81 = -1.223\) s — the apex comes \(1.223\) s after the throw.

Check. Elementary kinematics: apex height \(v^2/2g = 144/19.62 = 7.339\) m and time \(v/g = 1.2232\) s ✓. Dimensions: \(\sqrt{2mE}/(mg)\) is momentum over force, a time ✓, as \(\beta = \partial S/\partial E\) must be (action over energy).

Idea. A solution of the HJ equation is a momentum field \(p = \partial_qS\) that is consistent with the dynamics: flowing along the velocity it prescribes keeps you on it. This is the "method of characteristics" for this PDE, and the reason the level sets of \(W\) are wavefronts.

1. The \(q\)-equation \(\dot q = \partial H/\partial p\) holds by construction, since \(p(t) = \partial_qS\) is exactly the momentum argument used. For \(p\): by the chain rule, \(\dot p_i = \partial_t\partial_{q_i}S + \sum_k\partial_{q_k}\partial_{q_i}S\,\dot q_k\). Differentiating the HJ equation with respect to \(q_i\) — \(H\) depends on \(q_i\) both directly and through \(\partial_qS\) — gives

By Clairaut (unit 03; \(S\) is \(C^2\)) the mixed partials commute, and \(\partial H/\partial p_k = \dot q_k\) along the curve. Subtracting, \(\dot p_i = -\partial H/\partial q_i\). \(\square\)

2. \(\frac{\mathrm d}{\mathrm dt}S(q(t),t) = \partial_tS + \sum_i\partial_{q_i}S\,\dot q_i = -H + p\cdot\dot q = L\), the last step being the Legendre relation between \(H\) and \(L\) (unit 14), valid because \(\dot q = \partial H/\partial p\). \(\square\) So \(S\) increases along these curves by exactly the action: it is the action along them, up to a constant.

3. Here \(\partial H/\partial\vec p = \vec p/m\), so \(\dot{\vec x} = \nabla W/m\): the velocity is parallel to \(\nabla W\). If \(\vec c(\sigma)\) is any curve in a level set of \(W\), then \(0 = \frac{\mathrm d}{\mathrm d\sigma}W(\vec c) = \nabla W\cdot\vec c\,{}'\), so \(\nabla W\) is orthogonal to every tangent of the level set. \(\square\)

4. \(\partial_qS = mq/t\), so \(\dot q = q/t\), whose solutions are \(q = ct\): every straight line through the origin of the \((t,q)\) plane. These are exactly the free trajectories that leave \(q = 0\) at \(t = 0\) — the family whose envelope the slides built this \(S\) from. Along \(q = ct\), \(p = mc\) is constant ✓, and \(\frac{\mathrm d}{\mathrm dt}S = \frac{\mathrm d}{\mathrm dt}\big(mc^2t/2\big) = \tfrac12mc^2 = L\) ✓.

Check. Part 1 gives \(n\) first-order equations and so an \(n\)-parameter family — one trajectory through each point, the ones "belonging to" this \(S\). To get every trajectory you need a family of \(S\)'s: that is what the complete integral in Jacobi's theorem supplies. Dimensions in part 4: \(\partial_qS = mq/t\) is a momentum ✓.

Idea. Spherical symmetry makes the angular part independent of \(V\); the radial integral is the Kepler lemma in the variable \(r^2\). The factor \(2\) in \(H\) is Bertrand's other closed-orbit case.

1. With \(S = -Et + L_z\phi + \Theta(\theta) + R(r)\), multiplying the HJ equation by \(2mr^2\) gives \(r^2R'^2 + m^2\Omega^2r^4 - 2mEr^2 + \big(\Theta'^2 + L_z^2/\sin^2\theta\big) = 0\). The bracket depends on \(\theta\) only and the rest on \(r\) only, so both are constant; call the bracket \(L^2\). The \(\theta\) and \(\phi\) equations are identical to Kepler's, so the same computations (the angular lemma) give \(I_\phi = L_z\), \(I_\theta = L - |L_z|\).

2. \(R'^2 = 2mE - L^2/r^2 - m^2\Omega^2r^2\). With \(s = r^2\): \(p_r\,\mathrm dr = \sqrt{2mE - L^2/s - m^2\Omega^2 s}\;\mathrm ds/(2\sqrt s) = \tfrac12\sqrt{-m^2\Omega^2 + 2mE/s - L^2/s^2}\;\mathrm ds\). The radial lemma with \(A = m^2\Omega^2\), \(B = mE\), \(C = L^2\) (and \(B^2 > AC\) is \(E > \Omega L\), the condition for a non-circular orbit) gives

3. Solve for \(E\) with \(L = I_\theta + |I_\phi|\): \(E = \Omega(2I_r + I_\theta + |I_\phi|)\). Then \(\omega_r = \partial E/\partial I_r = 2\Omega\) and \(\omega_\theta = |\omega_\phi| = \Omega\). Ratio \(2\), the same for every orbit, so every orbit closes. More: \(r\) completes two oscillations for each revolution in angle, so \(r\) takes its minimum twice per revolution, at opposite points — the symmetry of an ellipse about its centre, which is where the force centre sits. For Kepler the ratio is \(1\): one perihelion per revolution, the force centre at a focus.

4. In Cartesian coordinates \(H = \sum_{x,y,z}\big(p^2/2m + \tfrac12m\Omega^2x^2\big)\) separates into three oscillators, each with \(I = E_x/\Omega\), so \(E = \Omega(I_x + I_y + I_z)\): all frequencies \(\Omega\). The Cartesian and spherical actions are different coordinates on the same tori; the energy agrees because \(2I_r + L = E/\Omega = I_x + I_y + I_z\) ✓. And the radial frequency \(2\Omega\): \(r^2 = x^2 + y^2 + z^2\) with each coordinate a sinusoid of frequency \(\Omega\) is a sinusoid of frequency \(2\Omega\) plus a constant ✓.

Check. Dimensions: \(E/\Omega\) is \(\mathrm{J\,s}\) ✓. Circular orbit: \(I_r = 0\) iff \(E = \Omega L\): for a circle of radius \(r\), \(E = \tfrac12m\Omega^2r^2\cdot2 = m\Omega^2r^2\) and \(L = m\Omega r^2\) ✓.

Idea. The action is an area, and a bounce is a vertical edge of zero width in \(\mathrm dq\). Everything else is one integral and one derivative.

1. The phase curve is the arc \(p = \pm\sqrt{2m(E - mgq)}\), \(0\le q\le E/mg\) — a parabola on its side — closed by the vertical segment from \((0, -\sqrt{2mE})\) to \((0, +\sqrt{2mE})\) at the bounce. So

2. Inverting, \(E = \Big(\dfrac{3\pi mg\,I}{2\sqrt{2m}}\Big)^{2/3} \propto I^{2/3}\), so \(\omega = \dfrac{\mathrm dE}{\mathrm dI} = \dfrac23\dfrac EI = \dfrac23\cdot\dfrac{3\pi mg}{2\sqrt{2m}\,E^{1/2}} = \dfrac{\pi g}{\sqrt{2E/m}}\). Not isochronous: higher bounces take longer, \(T\propto\sqrt E\).

3. \(T = 2\pi/\omega = 2\sqrt{2E/m}/g\). Dropped from \(h\), \(E = mgh\): \(T = 2\sqrt{2h/g} = 2\sqrt{2/9.81} = 0.903\) s. Kinematics: the fall takes \(\sqrt{2h/g} = 0.4515\) s, the rise the same, total \(0.903\) s ✓.

Check. Dimensions: \(\sqrt{2E/m}\) is a speed, \(g\) an acceleration, the ratio a time ✓. The frequency theorem survives the corner because the corner contributes nothing to the area and nothing to the time.

Idea. The frequency is \(\mathrm dE/\mathrm dI\), and \(I(E)\) at first order needs only an average of the perturbation over the unperturbed motion. No nonlinear equation is solved.

1. At \(\lambda = 0\), \(I = E/\omega_0\). By the lemma, \(\partial_\lambda I|_{E,\lambda=0} = -\langle x^4\rangle/\omega_0\), averaged over \(x = A\cos\omega_0t\), \(A^2 = 2E/m\omega_0^2\). Using \(\cos^4u = \tfrac38 + \tfrac12\cos2u + \tfrac18\cos4u\), \(\langle x^4\rangle = \tfrac38A^4 = \tfrac38\cdot\dfrac{4E^2}{m^2\omega_0^4}\). So

2. Invert: \(E = \omega_0I + \dfrac{3\lambda E^2}{2m^2\omega_0^4} + O(\lambda^2) = \omega_0I + \dfrac{3\lambda I^2}{2m^2\omega_0^2} + O(\lambda^2)\), replacing \(E\) by \(\omega_0 I\) inside the \(O(\lambda)\) term at the cost of \(O(\lambda^2)\). Differentiate:

using \(I = \tfrac12m\omega_0A^2\) at leading order. Hardening (\(\lambda > 0\)) raises the frequency with amplitude.

3. \(\omega_0 = \sqrt{20.0/0.200} = 10.0\ \mathrm{rad\,s^{-1}}\). The shift is \(\dfrac{3\lambda A^2}{2m\omega_0} = \dfrac{3\cdot200\cdot0.0025}{2\cdot0.200\cdot10.0} = 0.375\ \mathrm{rad\,s^{-1}}\), so \(\omega = 10.375\ \mathrm{rad\,s^{-1}}\), a fractional shift \(3\lambda A^2/2k = 3.75\%\). The exact \(10.3672\) differs by \(0.008\ \mathrm{rad\,s^{-1}}\), i.e. by \(0.08\%\) of \(\omega_0\): of order the square of the first-order shift (\(0.0375^2 = 0.14\%\)), as a second-order error should be.

A subtlety. "First order" at fixed \(A\) and at fixed \(E\) differ at second order: with \(E = \tfrac12kA^2 + \lambda A^4 = 0.02625\) J, the formula \(\omega_0 + 3\lambda E/(m^2\omega_0^3)\) gives \(10.394\ \mathrm{rad\,s^{-1}}\). Both are correct to first order; they disagree by \(O(\lambda^2)\) because \(E\) and \(\tfrac12kA^2\) differ at \(O(\lambda)\). The amplitude form is the one to compare with experiment.

Check. Dimensions: \(\lambda A^2/(m\omega_0)\) is \(\mathrm{N\,m^{-3}\,m^2/(kg\,s^{-1})} = \mathrm{s^{-1}}\) ✓. Limit \(\lambda\to0\): \(\omega\to\omega_0\) ✓; \(A\to0\): the shift vanishes, as small oscillations must be harmonic ✓. Unit 21 derives the same \(\omega_0 + 3\lambda A^2/(2m\omega_0)\) by the Poincaré–Lindstedt method, as an independent check (cited forward; for \(\ddot x + \omega_0^2x + \varepsilon x^3 = 0\) it is \(\omega_0 + 3\varepsilon A^2/8\omega_0\), and here \(\varepsilon = 4\lambda/m\)).

Another way

The slides' general formula \(E(I) = \omega_0I + \lambda\langle U\rangle_I\) with \(U = x^4\) and \(A^2 = 2I/m\omega_0\): \(\langle x^4\rangle_I = \tfrac38(2I/m\omega_0)^2 = 3I^2/(2m^2\omega_0^2)\), which is the same \(E(I)\) in one line.

Idea. To first order only the quartic term of \(V\) matters, and P6 has already done the work.

1. \(V = mg\ell\big(\vartheta^2/2 - \vartheta^4/24\big) + O(\vartheta^6)\). In P6's notation: coordinate \(\vartheta\), mass \(M = m\ell^2\), \(k = mg\ell\), \(\lambda = -mg\ell/24\), and \(\omega_0^2 = k/M = g/\ell\). P6's fractional shift is \(3\lambda A^2/(2k) = 3(-mg\ell/24)\vartheta_0^2/(2mg\ell) = -\vartheta_0^2/16\). The \(\vartheta^6\) term contributes at order \(\vartheta_0^4\). So \(\omega = \omega_0(1 - \vartheta_0^2/16)\): softening, lower frequency at larger amplitude.

2. \(\vartheta_0 = \pi/6 = 0.5236\), \(\vartheta_0^2/16 = 0.017135\). So \(T/T_0 = 1/(1 - 0.017135) \approx 1 + 0.017135 = 1.01713\) to first order. The exact ratio is \(1.01741\): error \(0.00027\), i.e. \(0.027\%\). That is of order \(\vartheta_0^4 = 0.075\) times a small coefficient; the next term of the known series is \(\tfrac{11}{3072}\vartheta_0^4 = 0.00027\) (cited: it follows by carrying P6 to second order, or from the series of \(K\)), which accounts for the difference.

Check. Limit \(\vartheta_0\to0\): Galileo's isochronism ✓. Sign: a pendulum's restoring torque \(mg\ell\sin\vartheta\) is weaker than linear, so it should slow down at large amplitude ✓. At \(30^\circ\) a clock regulated for small swings loses \(0.0171\times86400 = 1480\) s, about 25 minutes, a day — why pendulum clocks keep the amplitude small and constant.

Idea. The period is the slope of \(I(E)\) (times \(2\pi\)); constant slope is an affine function.

1. By the frequency theorem, \(I\) is differentiable on the interval with \(I'(E) = T(E)/2\pi\). If \(T(E) = T\) for all \(E\), then \(I - ET/2\pi\) has zero derivative on an interval, so it is constant (mean value theorem). Conversely if \(I = ET/2\pi + \text{const}\), then \(T(E) = 2\pi I'(E) = T\). \(\square\)

2. The turning points are the zeros of \(2mE - m^2\omega^2x^2 - 2mc/x^2\). With \(s = x^2\), \(\mathrm dx = \mathrm ds/(2\sqrt s)\) and \(\sqrt{2mE - m^2\omega^2s - 2mc/s}/(2\sqrt s) = \tfrac12\sqrt{-m^2\omega^2 + 2mE/s - 2mc/s^2}\). The radial lemma with \(A = m^2\omega^2\), \(B = mE\), \(C = 2mc\) gives

affine in \(E\) with slope \(1/2\omega\). By part 1 the system is isochronous with \(T/2\pi = 1/2\omega\), i.e. \(T = \pi/\omega\) for every energy. \(\square\)

3. \(V\) is not quadratic, yet every orbit has the same period: isochronism does not single out the harmonic oscillator. At the bottom, \(V' = m\omega^2x - 2c/x^3 = 0\) at \(x_*^4 = 2c/m\omega^2\), and \(V''(x_*) = m\omega^2 + 6c/x_*^4 = m\omega^2 + 3m\omega^2 = 4m\omega^2\). Small oscillations have frequency \(\sqrt{V''/m} = 2\omega\), period \(\pi/\omega\) ✓ — the same, as it must be.

Check. Dimensions: \(c\) has dimension \(\mathrm{J\,m^2}\), so \(\sqrt{2mc}\) is \(\mathrm{J\,s}\), an action ✓. Limit \(c\to0\): the wall at \(x = 0\) becomes a hard wall and the motion is half a harmonic oscillation reflected — period half of \(2\pi/\omega\), i.e. \(\pi/\omega\) ✓. This potential is exactly the radial problem of P4 with \(2mc = L^2\), which is why its radial frequency is \(2\omega\). A numerical check of the period integral at three energies gives \(T = \pi/\omega\) to ten digits.

Idea. The action is the rectangle's area; slowness makes it invariant; a direct collision count confirms it.

1. The phase curve is \(p = +mv\) going out and \(p = -mv\) coming back, over \(0\le q\le\ell\), with vertical edges at the walls: area \(2mv\ell\), so \(I = mv\ell/\pi\). Invariance: \(v\ell\) constant. Halving \(\ell\) doubles the speed to \(1000\ \mathrm{m\,s^{-1}}\) and quadruples the energy: with \(m = 28\times1.6605\times10^{-27} = 4.649\times10^{-26}\) kg, \(E\) goes from \(5.81\times10^{-21}\) J to \(2.32\times10^{-20}\) J.

2. Between collisions with the piston the molecule travels \(2\ell\), taking \(2\ell/v\); each collision with a piston moving inward at \(u = -\dot\ell\) adds \(2u\) (in the piston's frame the bounce is elastic). If \(u\ll v\) and \(\ell\) changes little per round trip, the mean rate is \(\dot v = 2u\cdot v/2\ell = -v\dot\ell/\ell\), so \(\mathrm d\ln v = -\mathrm d\ln\ell\) and \(v\ell\) is constant ✓ — the same answer, from mechanics that knows nothing about actions.

3. Momentum \(2mv\) delivered every \(2\ell/v\): \(F = mv^2/\ell = 2E/\ell\). At fixed \(I\), \(v\propto1/\ell\) so \(F\propto\ell^{-3}\): \(\gamma = 3\). This is the adiabatic law \(PV^\gamma = \text{const}\) for a gas with one degree of freedom, where thermodynamics gives \(\gamma = 1 + 2/f = 3\) for \(f = 1\).

Check. Energy balance: the piston does work \(-F\,\mathrm d\ell = -(2E/\ell)\,\mathrm d\ell\), and \(E\propto\ell^{-2}\) gives \(\mathrm dE = -2E\,\mathrm d\ell/\ell\) ✓. Dimensions of \(I\): \(\mathrm{kg\,m\,s^{-1}\,m} = \mathrm{J\,s}\) ✓. Slowness: "slow" means \(u\ll v\); a piston moving at \(1\ \mathrm{m\,s^{-1}}\) is slow for this molecule.

Idea. The adiabatic invariant predicts how the swing grows. Independently, the extra work done against the swinging (the tension exceeds \(mg\) on average) must equal the growth of the swing's energy. Two routes, one answer.

1. \(I = \tfrac12mg\ell\vartheta_0^2/\sqrt{g/\ell} = \tfrac12m\sqrt g\,\ell^{3/2}\vartheta_0^2\), so \(\vartheta_0\propto\ell^{-3/4}\): \(\vartheta_0 = 4.00^\circ\times4^{3/4} = 11.31^\circ\) at the end.

2. The bob is at \(\ell(\sin\vartheta, -\cos\vartheta)\). Its radial equation, with \(\ddot\ell = 0\) for a steady pull, gives tension \(= mg\cos\vartheta + m\ell\dot\vartheta^2\). Over one swing, \(\vartheta = \vartheta_0\cos\omega t\) (the swing is fast compared with the change of \(\ell\)): \(\langle\cos\vartheta\rangle = 1 - \langle\vartheta^2\rangle/2 = 1 - \vartheta_0^2/4\), and \(\langle m\ell\dot\vartheta^2\rangle = m\ell\omega^2\vartheta_0^2/2 = mg\vartheta_0^2/2\). So

The string is massless and the hole frictionless, so the puller's work goes into the bob: \(\mathrm d\big(-mg\ell + \tfrac12mg\ell\vartheta_0^2\big) = -mg\big(1 + \vartheta_0^2/4\big)\,\mathrm d\ell\). The \(-mg\,\mathrm d\ell\) terms cancel, leaving \(\tfrac12\vartheta_0^2\,\mathrm d\ell + \ell\vartheta_0\,\mathrm d\vartheta_0 = -\tfrac14\vartheta_0^2\,\mathrm d\ell\), i.e.

exactly part 1's law ✓. The adiabatic invariant is not an extra assumption: it is what energy conservation says, once the fast motion is averaged.

3. Lifting the bob's equilibrium by \(0.75\) m: \(mg\Delta\ell = 0.50\times9.81\times0.75 = 3.679\) J. The swing energy goes from \(\tfrac12mg\ell\vartheta_0^2 = \tfrac12\times0.50\times9.81\times1.00\times(0.069813)^2 = 0.011953\) J to twice that, \(0.023906\) J (since \(E_{\text{swing}}\propto\ell^{-1/2}\)), an extra \(0.01195\) J. Total work \(3.691\) J, of which \(0.32\%\) goes into the swing. Integrating \(\tfrac14mg\vartheta_0^2\) against \(-\mathrm d\ell\) from \(1.00\) to \(0.25\) m with \(\vartheta_0^2 = \vartheta_i^2(\ell_i/\ell)^{3/2}\) gives the same \(0.011953\) J ✓.

Check. Dimensions: \(mg\vartheta_0^2\,\mathrm d\ell\) is an energy ✓. Limit \(\vartheta_0 = 0\): the puller only lifts the bob ✓. Slowness: pulling over a minute, \(T|\dot\ell|/\ell\le0.05\) (the slides' worked example); a quick jerk would violate it and the swing energy would not follow \(\ell^{-1/2}\). We neglected \(\tfrac12m\dot\ell^2 = 4\times10^{-5}\) J, which the puller supplies at the start and takes back at the end.

Idea. A uniform field is a medium stratified in \(y\); the Beltrami identity is Snell's law across the strata, and Snell's invariant here is the horizontal momentum.

1. \(g_J = 2(E - V)\,m\delta = 2m(E - mgy)(\mathrm dx^2 + \mathrm dy^2)\) on \(y \lt E/mg\), and for a graph \(y(x)\), \(\mathcal L[y] = \sqrt{2m}\int\sqrt{E - mgy}\,\sqrt{1 + y'^2}\,\mathrm dx\).

2. Drop the constant \(\sqrt{2m}\); \(F = \sqrt{E - mgy}\sqrt{1 + y'^2}\), \(F_{y'} = \sqrt{E - mgy}\,y'/\sqrt{1 + y'^2}\), and \(F - y'F_{y'} = \sqrt{E - mgy}/\sqrt{1 + y'^2} = c\). Mechanically, \(\sqrt{2m(E - mgy)} = \lVert\vec p\rVert\) and \(1/\sqrt{1 + y'^2} = \cos\alpha\) (angle to the horizontal), so \(\sqrt{2m}\,c = \lVert\vec p\rVert\cos\alpha = p_x\): conservation of horizontal momentum, which is Snell's \(n\sin(\text{angle to the normal})\) = const. Now solve: \(y'^2 = (E - c^2 - mgy)/c^2\). Put \(Y = E - c^2 - mgy\), so \(Y' = -mgy'\) and \(Y'^2 = m^2g^2Y/c^2\); hence \(\big(\sqrt Y\big)' = \pm mg/2c\), \(\sqrt Y = mg|x - x_0|/2c\), and

a downward parabola with apex at \(x_0\). With \(c^2 = p_x^2/2m = \tfrac12mv_x^2\): apex height \(y_a = (E - \tfrac12mv_x^2)/mg\) and curvature coefficient \(g/2v_x^2\) — the textbook trajectory.

3. \(v_x = 20.0\cos30^\circ = 17.32\ \mathrm{m\,s^{-1}}\), \(E = \tfrac12m(20.0)^2\) at \(y = 0\). Apex: \(y_a = (400 - 300)/(2\times9.81) = 5.097\) m. Coefficient \(g/2v_x^2 = 9.81/600 = 0.01635\ \mathrm m^{-1}\). Range: the parabola meets \(y = 0\) at \(x_0\pm\sqrt{y_a/0.01635}\), and passes through the origin, so the range is \(2\sqrt{5.097/0.01635} = 35.31\) m. Check: \(v^2\sin60^\circ/g = 400\times0.8660/9.81 = 35.31\) m ✓, and \(v_y^2/2g = 100/19.62 = 5.097\) m ✓.

4. The brachistochrone minimises time, \(\int\mathrm ds/v\propto\int\sqrt{1 + y'^2}/\sqrt{E - mgy}\,\mathrm dx\): speed in the denominator. Maupertuis extremises \(\int p\,\mathrm ds\): speed in the numerator. The bead's "index" is \(1/v\), like wave optics; the free particle's is \(v\), like Newton's corpuscles — the corpuscle puzzle of the slides. The bead is constrained to a wire and the question is which wire; the projectile is free and the question is which path nature takes. Different questions, opposite powers.

Check. Dimensions: \(c^2\) and \(E - mgy\) are energies, so \(mg(x - x_0)^2/c^2\) is a length ✓. Limit \(g\to0\): \(g_J\) becomes \(2mE\) times flat, and the parabola flattens into a straight line ✓.

Idea. The conformal factor \(1/r\) of the \(E = 0\) Jacobi metric is exactly undone by the squaring map \(r = \rho^2\), \(\phi = 2\psi\) — the map \(w\mapsto w^2\) written in polar coordinates. A metric that is Euclidean in some coordinates has straight lines as geodesics.

1. \(g_J = 2m(E + k/r)(\mathrm dr^2 + r^2\mathrm d\phi^2)\), which at \(E = 0\) is \(\dfrac{2mk}{r}(\mathrm dr^2 + r^2\mathrm d\phi^2)\).

2. With \(r = \rho^2\), \(\phi = 2\psi\): \(\mathrm dr = 2\rho\,\mathrm d\rho\), \(\mathrm d\phi = 2\,\mathrm d\psi\), so \(\mathrm dr^2 + r^2\mathrm d\phi^2 = 4\rho^2\mathrm d\rho^2 + 4\rho^4\mathrm d\psi^2 = 4\rho^2(\mathrm d\rho^2 + \rho^2\mathrm d\psi^2)\). Multiply by \(2mk/\rho^2\): \(g_J = 8mk(\mathrm d\rho^2 + \rho^2\mathrm d\psi^2)\). (Precisely: by the components theorem of the slides, the metric's components in the \((\rho,\psi)\) coordinate basis are \(A^{\mathsf T}GA\) with \(A = \mathrm{diag}(2\rho, 2)\) the derivative of the substitution, and \(G = (2mk/\rho^2)\,\mathrm{diag}(1, \rho^4)\); the product is \(8mk\,\mathrm{diag}(1,\rho^2)\).)

3. The length functional of \(8mk\) times the Euclidean metric is \(\sqrt{8mk}\) times Euclidean length, whose critical points are straight lines. A straight line not through the origin is \(\rho\cos(\psi - \psi_0) = \rho_0\). Undo the substitution: \(\sqrt r\cos\big((\phi - \phi_0)/2\big) = \sqrt{r_0}\) with \(r_0 = \rho_0^2\), \(\phi_0 = 2\psi_0\); square and use \(\cos^2(\chi/2) = (1 + \cos\chi)/2\):

a conic of eccentricity \(1\) with focus at the origin and perihelion distance \(r_0\): a parabola ✓. As the line runs from one end to the other, \(\psi - \psi_0\) sweeps an interval of length \(\pi\), so \(\phi - \phi_0\) sweeps \((-\pi, \pi)\): the two arms of the parabola going off to infinity. (The substitution is two-to-one: \((\rho, \psi)\) and \((\rho, \psi + \pi)\) give the same point, which is why a whole straight line maps to a single parabola traversed once.)

4. For general \(E\): \(g_J = 2m(E + k/\rho^2)\cdot4\rho^2(\mathrm d\rho^2 + \rho^2\mathrm d\psi^2) = 8m(k + E\rho^2)(\mathrm d\rho^2 + \rho^2\mathrm d\psi^2)\). A 2D isotropic oscillator of mass \(M\), frequency \(\Omega\) and energy \(E'\) has Jacobi metric \(2M(E' - \tfrac12M\Omega^2\rho^2)(\mathrm d\rho^2 + \rho^2\mathrm d\psi^2)\). These are proportional when \(E'/k = -\tfrac12M\Omega^2/E\), possible exactly when \(E \lt 0\) (then \(\Omega^2 = -2E'E/(Mk) > 0\) for \(E' > 0\)). A constant factor does not change geodesics, so the squaring map \((\rho,\psi)\mapsto(r,\phi) = (\rho^2, 2\psi)\) carries the oscillator's orbits onto Kepler orbits of energy \(E \lt 0\): the oscillator's centred ellipses become Kepler's focal ellipses. This is the Kepler–oscillator correspondence (Bohlin, 1911; Levi-Civita used it in 1920 to regularise collisions), and it explains P4 and the slides together: the squaring map halves angles, so the oscillator's radial-to-angular frequency ratio \(2\) becomes Kepler's \(1\).

Check. Dimensions: \(mk\) is \(\mathrm{kg\,J\,m}\) and \(\rho^2\) is a length, so \(8mk(\mathrm d\rho^2 + \rho^2\mathrm d\psi^2)\) has the same dimension as \((2mk/r)\,\mathrm dr^2\) ✓. The perihelion: the line's closest point to the origin, \(\rho_0\), maps to the parabola's closest point \(r_0 = \rho_0^2\) ✓. Energy: at \(E = 0\) the speed at perihelion is \(\sqrt{2k/mr_0}\), the escape speed, as a parabola must have.