Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 321 / 2

For unit mass, the Lagrangian in cylindrical coordinates isIntroduce shearing sheet coordinates by $r=r_0+x$ and $\phi ={\Omega }_{0}t+y/r_0$. Evaluate all derivatives of $\Phi$ at $(r_0,0)$. The reference circular orbit obeys ${\Phi }_r=r_0{\Omega }_0^2$, while midplane symmetry gives ${\Phi }_z={\Phi }_{rz}=0$. Its Taylor expansion isExpanding the kinetic energy to the same order givesThe terms linear in $x$ cancel by circular-orbit balance. Discard the constant and the term $r_0{\Omega }_0\dot{y}$ by total-time-derivative invariance of a Lagrangian; these do not change the Euler-Lagrange equations. The particle Lagrangian in a shearing sheet is thereforewith shearing-sheet tidal potentialThe second form follows from ${\Phi }_r=r{\Omega }^2$, the orbital shear parameter $q_0=-d\log \Omega /d\log r$ and the vertical epicyclic frequency ${\Omega }_z^2={\Phi }_{zz}$.

The Euler-Lagrange equations of the particle Lagrangian in a shearing sheet areThe terms coupling $x$ and $y$ are the Coriolis acceleration. The cyclic coordinate $y$ has conserved canonical momentumFor a Newtonian potential of a point mass, $q_0=3/2$ and ${\Omega }_z={\kappa }_r={\Omega }_0$. Set $p_y={\Omega }_0{x}_0/2$ and substitute $\dot{y}=p_y-2{\Omega }_{0}x$ in the radial equation. It becomes $\ddot{x}+{\Omega }_0^2(x-x_0)=0$, a harmonic oscillator equation about the epicyclic guiding center $x_0$. Integration givesThe four real constants in $A,y_0,x_0$ and the two in $B$ account for the six initial position and velocity data.

Expanding the inertial specific angular momentum gives $h=h_0+r_0{p}_y+O(2)$. Thus $p_y$ measures the angular-momentum offset from the reference circular orbit, and $x_0$ specifies the radius of its associated epicyclic guiding center.

The conserved horizontal energy in the rotating frame isIt is the horizontal part of the local Jacobi energy in a shearing sheet, rather than the inertial specific orbital energy. Its positive term is the epicyclic energy; its negative term is the energy of the background shear at guiding-center position $x_0$. Independently, the vertical harmonic oscillator has conserved energy

Initially every particle has $A=B=0$, so its Jacobi energy in a shearing sheet is $-3{\Omega }_0^2{x}_0^2/8$. During an inelastic collision, positions are fixed at the instant of impact, while momentum conservation preserves the sum of the tangential velocities. Hence the sum of $p_y=\dot{y}+2{\Omega }_{0}x$, and therefore the sum of the epicyclic guiding center positions $x_0=2p_y/{\Omega }_0$, is unchanged. Between collisions, these quantities are individually conserved.

The collision dissipates kinetic energy without changing the instantaneous tidal potential. Consequently the total Jacobi energy in a shearing sheet decreases. At any later time, writing $x_{0j}$ for the particles' current epicyclic guiding center positions givesIf $D\ge 0$ is the accumulated energy dissipated in the inelastic collisions, comparison with the initial circular orbits givesThus the mean guiding-center position stays fixed, while its variance grows. The initial ensemble has no preference for positive or negative $x$, and the local equations and collision law preserve the symmetry $(x,y,z)\mapsto (-x,-y,z)$. The spreading is therefore symmetric in the ensemble average: angular momentum transport moves some particles inward and others outward. A particular finite random realization need not be exactly symmetric. This is the microscopic energy argument for dissipative spreading of a planetary ring.