Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2017 / S7

In cylindrical coordinates the orthogonal velocity components are $\dot{r}$, $r\dot{\theta }$, $\dot{z}$. Their squared magnitudes add, and $\dot{\mathbf{r}}\cdot \mathbf{A}=\dot{z}A(r)$. Thus $\boxed{\mathcal{L}=m({\dot{r}}^2+r^2{\dot{\theta }}^2+{\dot{z}}^2)/2+q\dot{z}A(r)}$. Taking the curl gives $B_{\theta }=-A^{\prime }(r)$, consistent with the prescribed field.

Velocity differentiation gives $\boxed{p_r=m\dot{r},p_{\theta }=m{r}^2\dot{\theta },p_z=m\dot{z}+qA(r)}$. The coordinates $\theta$ and $z$ are cyclic, so $p_{\theta },p_z$ are conserved. The radial momentum generally changes. In particular the conserved axial canonical momentum differs from the axial mechanical momentum by $qA$.

The Legendre transform cancels the velocity-linear magnetic term. Expressing velocities in terms of momenta givesThe Hamiltonian is time independent and hence conserved. It equals kinetic energy because a static magnetic field does no work; its dependence on $A$ is through the mechanical momentum combination.

The initial condition makes $p_{\theta }=0$ at all times. Accessible radii obey $|p_z-qA(r)|\le \sqrt{2mH}$. For $q>0$ and $p_z>\sqrt{2mH}$, monotonicity of $A$ gives exactly the two stated turning equations and confinement to the interval between their unique roots.

The squared condition alone omits a sign qualification. More generally both positive roots require $p_z/q>0$, and are the sorted roots $A(r)=(p_z\pm \sqrt{2mH})/q$. For example $q=-1$, $A=r$, $p_z=-3$, $2mH=1$ confines the orbit to $2\le r\le 4$, but the source's named $r_1,r_2$ equations reverse those labels. This is a convention issue, not a failure of confinement.

Expand $A(r)=A(r_0)+A^{\prime }(r_0)s+O(s^2)$ about $qA(r_0)=p_z$. Then $p_z-qA\simeq -q{A}^{\prime }(r_0)s$ and conservation of energy givesHere $\Omega =q{A}^{\prime }(r_0)/m$ can be signed, but the physical oscillation frequency and amplitude are nonnegative. The small-energy condition ensures that this local expansion samples only a narrow range about $r_0$, provided the field varies smoothly there.

Use rotating basis vectors with ${\dot{\mathbf{e}}}_1=\Omega {\mathbf{e}}_2$, ${\dot{\mathbf{e}}}_2=-\Omega {\mathbf{e}}_1$. The inertial velocity is $({\dot{x}}_1-\Omega x_2){\mathbf{e}}_1+({\dot{x}}_2+\Omega x_1){\mathbf{e}}_2$. The two fixed rotating anchors are $(0,\pm a)$, so the spring lengths are ${\ell }_{\pm }=\sqrt{x_1^2+(x_2\mp a{)}^2}$. Combining kinetic and elastic energies givesA constant elastic offset has no dynamical effect.

Differentiate the quadratic Lagrangian with respect to coordinates and velocities. The equations areThe velocity-cross terms are Coriolis accelerations; moving $-{\Omega }^2{x}_i$ to the force side identifies the outward centrifugal acceleration. The remaining terms are elastic restoring forces. The sign pair is fixed by the rotating-basis convention in part a.

Substitution of the exponential ansatz gives the homogeneous amplitude systemA nonzero mode requires zero determinant. Thus $\boxed{({\omega }_1^2-{\omega }^2)({\omega }_2^2-{\omega }^2)=4{\Omega }^2{\omega }^2}$ with the stated two effective stiffnesses. The off-diagonal product is positive, which determines the sign on the right.

Put $X={\omega }^2$ and $S={\omega }_1^2+{\omega }_2^2+4{\Omega }^2$. Its equation is $X^2-SX+{\omega }_1^2{\omega }_2^2=0$. If both effective stiffnesses are positive, the sum and product of roots are positive and $S^2-4{\omega }_1^2{\omega }_2^2\ge 0$, so both roots are positive. They can coincide in a degenerate zero-rotation isotropic case; “two” need not mean distinct.

If ${\omega }_1^2<0<{\omega }_2^2$, the product is negative, so one root is negative. Its imaginary frequency gives a growing exponential displacement: centrifugal softening has overwhelmed one elastic restoring direction, making the central configuration unstable. Coriolis coupling mixes the two coordinates but does not remove this negative-product instability.

Hamilton's equations define phase-space velocity $({\dot{q}}_i,{\dot{p}}_i)=(H_{p_i},-H_{q_i})$. Its divergence isThus infinitesimal phase-space volumes are preserved, assuming sufficient differentiability. This is Liouville's theorem; spatial density alone need not be incompressible because the statement concerns the combined coordinate-momentum space.

To match the source's order define $[a,b]={\sum }_i(a_{p_i}{b}_{q_i}-a_{q_i}{b}_{p_i})$. The chain rule and Hamilton's equations then giveAntisymmetry implies $[H,H]=0$, hence $dH/dt=H_t$, which vanishes for a time-independent Hamiltonian. With the alternative common convention $\{a,b\}={\sum }_i(a_{q_i}{b}_{p_i}-a_{p_i}{b}_{q_i})$, the same law is $dg/dt=g_t+\{g,H\}$; mixing the two orders would reverse the physical time evolution.

Conservation of probability obeys ${\partial }_{t}f+{\nabla }_{q,p}\cdot (f\mathbf{v})=0$. Expand the divergence and use incompressibility to obtain ${\partial }_{t}f+\mathbf{v}\cdot {\nabla }_{q,p}f=0$, or $\boxed{df/dt=0}$. This is fine-grained transport for Hamiltonian trajectories without sources or collisions. Coarse graining can obscure this constancy without contradicting it.

For $f=F(H)$ with no explicit time dependence, ${\partial }_{t}f=0$ and $[H,F(H)]=F^{\prime }(H)[H,H]=0$. It therefore solves the stationary transport equation. Normalizability and the accessible energy region still have to be checked for a proposed probability distribution.

Axisymmetry supplies conserved $p_{\theta }$ in addition to conserved energy. Consequently $\boxed{f=F(H,p_{\theta })}$ is a more general stationary solution, since both arguments are constant on every trajectory. Additional independent integrals may permit still more general distributions. This is a sufficient construction, not a claim that every stationary distribution must have exactly these two arguments.