Past exam of the short options paper shared by Parts A and B of the physics course of the University of Oxford / 2019 / S7

Define the canonical momentum $p_i=\partial L/\partial {\dot{q}}_i$ and the energy function $E={\sum }_i{p}_i{\dot{q}}_i-L$. Along an Euler-Lagrange equation solution,Thus an autonomous Lagrangian has conserved $E$. For a simple harmonic oscillator, $L=m{\dot{x}}^2/2-k{x}^2/2$ and $\boxed{E=m{\dot{x}}^2/2+k{x}^2/2}$ is constant. For ordinary time-independent mechanical coordinates this is the total mechanical energy.

The appropriate invariance is invariance of the action element $Ldt$, including the change of the time measure. Under $\delta t=\lambda \varphi$ and $\delta q_i=\lambda {\psi }_i$, differentiation of $q_i^{\prime }(t^{\prime })$ gives $\delta {\dot{q}}_i=\lambda (D_t{\psi }_i-{\dot{q}}_i{D}_t\varphi )$. ThereforeUsing the Euler-Lagrange equations and $D_{t}E=-L_t$, this becomesAn action symmetry therefore yields the Noether chargeFor $\varphi =0$ this reduces to the usual spatial form of Noether's theorem. The same on-shell calculation permits velocity-dependent generators when they define a legitimate variational symmetry. Merely preserving the numerical value of $L$ is insufficient for transformations of time: the Jacobian matrix and determinant $d{t}^{\prime }/dt$ matters. If the action changes by a boundary term $\lambda D_{t}B$, the conserved quantity is $Q-B$.

For a central potential use $L=m|\dot{\mathbf{r}}{|}^2/2-V(|\mathbf{r}|)$. An infinitesimal rotation about any constant unit vector $\mathbf{n}$ has $\varphi =0$ and $\mathbf{\psi }=\mathbf{n}\times \mathbf{r}$. Noether's theorem gives $Q=m\dot{\mathbf{r}}\cdot (\mathbf{n}\times \mathbf{r})=\mathbf{n}\cdot \mathbf{L}$, where $\mathbf{L}=\mathbf{r}\times m\dot{\mathbf{r}}$. Since $\mathbf{n}$ is arbitrary, every component of angular momentum is conserved. A time-translation symmetry has $\varphi =1$, $\mathbf{\psi }=0$, giving $Q=-E$ andas conserved quantities.

The orbiting planet produces a potential $V(r,\vartheta ,\phi -\Omega t)$. It has explicit time dependence and exerts an azimuthal torque, so generally $\dot{E}=V_t$ and ${\dot{L}}_z=-V_{\phi }$ are nonzero. Nevertheless a simultaneous time-translation symmetry $\delta t=\lambda$ and rotation $\delta \phi =\Omega \lambda$ leaves the Lagrangian and $dt$ unchanged. Noether's theorem gives $Q=\Omega p_{\phi }-E$ and henceEquivalently $V_t=-\Omega V_{\phi }$ implies $d(E-\Omega L_z)/dt=0$. This Jacobi integral assumes a rigidly rotating potential with constant $\Omega$ and negligible back-reaction from the test particle.

Write the geometric vector as $\mathbf{V}={\sum }_i{V}_i^{\prime }{\mathbf{e}}_i^{\prime }(t)$. A rotating orthonormal basis obeys ${\dot{\mathbf{e}}}_i^{\prime }=\mathbf{\Omega }\times {\mathbf{e}}_i^{\prime }$. The product rule gives the transport identityBoth terms must be expressed in the same basis before adding them. If $R(t)$ maps rotating components into inertial components, the precise component equation is $d(R{\mathbf{V}}^{\prime })/dt=R({\dot{\mathbf{V}}}^{\prime }+{\mathbf{\Omega }}^{\prime }\times {\mathbf{V}}^{\prime })$. The printed expression suppresses this basis conversion.

The inertial Lagrangian is $L=m|\dot{\mathbf{r}}{|}^2/2-U(\mathbf{r},t)$. Put $\mathbf{r}=R(t){\mathbf{r}}^{\prime }$ and define $U^{\prime }({\mathbf{r}}^{\prime },t)=U(R(t){\mathbf{r}}^{\prime },t)$. The transport identity and invariance of the Euclidean norm under a rotation giveIts expansion contains the ordinary kinetic energy, $m{\dot{\mathbf{r}}}^{\prime }\cdot ({\mathbf{\Omega }}^{\prime }\times {\mathbf{r}}^{\prime })$, and $m|{\mathbf{\Omega }}^{\prime }\times {\mathbf{r}}^{\prime }{|}^2/2$. These velocity and position terms generate the apparent forces in the rotating frame.

Suppress primes on the angular velocity while retaining rotating components. The canonical momentum is $\partial L^{\prime }/\partial {\dot{\mathbf{r}}}^{\prime }=m({\dot{\mathbf{r}}}^{\prime }+\mathbf{\Omega }\times {\mathbf{r}}^{\prime })$. Varying $L^{\prime }$ with respect to position and applying the Euler-Lagrange equations givesThe terms are the physical force, the velocity-dependent Coriolis force, the outward centrifugal force, and the Euler force caused by changing angular velocity. The common origin is inertial; an accelerating origin would introduce another translational apparent force.

At latitude $\ell$, the Earth's angular velocity in east–north–up coordinates is $\mathbf{\Omega }=(0,\Omega \cos \ell ,\Omega \sin \ell )$. Assume a small horizontal patch, constant $\Omega$, negligible vertical velocity, and a vertical constraint balancing effective gravity. Absorb the steady centrifugal force into that effective gravity. With $f=2\Omega \sin \ell$ and physical horizontal force per unit mass $F_x^{\prime },F_y^{\prime }$, the Coriolis force givesLarge-scale atmospheric and oceanic flows deflect to their right in the northern hemisphere and left in the southern hemisphere; their balance with a horizontal pressure gradient produces geostrophic flow. A Foucault pendulum has an oscillation plane that precesses relative to the ground at angular rate $-\Omega \sin \ell$. Both effects vanish at the equator within this horizontal approximation.

Let $m$ be the ice mass for which the stated drag coefficients are defined, $K=k_1+k_2$ and $f=2\Omega \sin \ell$. With stationary water, the steady horizontal ice equations areThe linear system givesFor fixed wind speed $V=10\mathrm{m}{\mathrm{s}}^{-1}$, the Cauchy-Schwarz inequality minimizes this component whenThus the wind must blow towards a direction east of south by ${\tan }^{-1}(mf/K)$ in the northern hemisphere, west of south in the southern hemisphere, and due south at the equator. This compensates the Coriolis force; at the optimum $v_x=0$ and the ice moves due south. If the drag coefficients are per unit mass, replace $K/m$ by the corresponding total drag rate. “Blows towards” avoids the opposite convention used when naming winds by their origin.

For a regular Lagrangian, form the canonical momenta $p_i=L_{{\dot{q}}_i}$, invert for ${\dot{q}}_i(\mathbf{q},\mathbf{p},t)$, and take the Legendre transform $H={\sum }_i{p}_i{\dot{q}}_i-L$. On the upper cone let $s>0$ be distance along a generator from the apex and $\varphi$ the azimuth. Then the horizontal radius is $s\sin \theta$ and height is $s\cos \theta$, soConsequently

To match the printed order $df/dt=\{H,f\}$, define the Poisson bracket with the sign conventionHamilton's equations imply $\dot{f}={\sum }_i(f_{q_i}{H}_{p_i}-f_{p_i}{H}_{q_i})=\{H,f{\}}_{\mathrm{paper}}$. With the other common Poisson bracket convention the correct statement is instead $\dot{f}=\{f,H{\}}_{\mathrm{standard}}$. For example $H=p^2/(2m)$, $f=q$ gives $\{H,q{\}}_{\mathrm{standard}}=-p/m$, contradicting the printed order unless its sign convention is adopted. Since $H_t=0$ and $\{H,H\}=0$, the Hamiltonian is conserved. Also ${\dot{p}}_{\varphi }=-H_{\varphi }=0$: this canonical momentum is the axial angular momentum.

For fixed nonzero $l=p_{\varphi }$, the radial effective potential is $V_{\mathrm{eff}}(s)=l^2/(2m{s}^2{\sin }^2\theta )+mgs\cos \theta$. Its stationary point isThe equilibrium is therefore stable. A Taylor expansion with $x=s-s_0$ and $p_x=p_s$ givesQualification: the printed “any” $p_{\varphi }$ excludes a regular interior equilibrium when $p_{\varphi }=0$. In that case $V_{\mathrm{eff}}=mgs\cos \theta$ has no stationary point for $s>0$. Its minimum is the nonsmooth apex, where these coordinates and the smooth-cone dynamics fail.

For $|x|\ll s_0$ the quadratic Hamiltonian describes a simple harmonic oscillator withAn exactly circular orbit closes for every nonzero $l$. A nonzero infinitesimal radial oscillation closes at leading order precisely when $\sqrt{3}\sin \theta$ is rational. This condition concerns the cone angle: changing $s_0$ or $l$ cannot change the leading-order ratio. Thus there are no special nonzero $l$ values selecting closure for a fixed cone at this order.

At finite radial amplitude the system is equivalent, after unrolling the cone with planar angle $\chi =\varphi \sin \theta$, to a planar central force with potential proportional to $s$. Its angle advance per radial period changes continuously from $2\pi /\sqrt{3}$ near a circle towards $\pi$ at very large radial energy. Hence the physical advance changes from $2\pi /(\sqrt{3}\sin \theta )$ towards $\pi /\sin \theta$. This nonzero interval contains rational multiples of $2\pi$, so tuning a larger initial $|p_s|$ can produce closed orbits even when the small-amplitude ratio is irrational. Generic amplitudes still produce a precessing orbit; neither energy conservation nor increasing $|p_s|$ guarantees closure.