Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 320 / 4

False. Stellar evaporation removes stars above the escape energy after two-body relaxation has redistributed their energies. The remaining self-gravitating cluster need not become less dense; losing mass is not the same as holding its radius fixed.

For a slowly evolving isolated cluster, apply the virial theorem at each approximate equilibrium: $E=T+W=W/2=-T$. Stars evaporating with nearly zero energy at infinity remove little total energy, so the bound cluster has nearly constant negative $E$. With gravitational radius $R_g$ defined by $W=-G{M}^2/R_g$,For homologous structure, its characteristic mean density therefore scales asIt increases as mass is lost. Escaping stars carrying positive energy make the bound energy more negative and strengthen the contraction tendency. This is virial contraction under stellar evaporation, related to the negative heat capacity of bound gravitational systems. A tidally limited globular cluster has additional radius constraints, so this scaling is not universal, but the claimed inevitable density decrease is false even in the simplest isolated evaporation model.

False as a claimed change between isolated steady endpoints. Conservation of energy by itself would allow a more negative $W$ and a larger $T$, but both endpoints must also obey the virial theorem. For a finite isolated equilibrium with no surface term,If total $E$ is conserved between the two equilibria, the fixed-energy virial constraint between steady states givesThis $T$ includes both ordered rotation and random stellar motion. A galactic bar or other instability can redistribute angular momentum and concentrate a central component, but compensating expansion or rearrangement elsewhere must preserve the global energy constraint. Central concentration alone does not prove that the whole system has a more negative $W$.

A remnant could have different energies if stars escape, or if an external component exchanges energy with it; then the conserved total energy would include that other component. Those cases do not justify treating the bound disc simultaneously as a closed fixed-energy system and as a new equilibrium with increased $T$.

False. The singular isothermal sphere is infinite in extent and mass. The isolated finite-system virial theorem cannot be applied to its global divergent energies without boundary terms.

A direct Spherical Jeans equation counterexample is enough. For isotropic velocities with constant one-dimensional dispersion ${\sigma }^2$, its density slope givesThus ${\sigma }^2=v_0^2/2$ and the circular and isotropic speed comparison in a singular isothermal sphere isThis is realized by a positive isotropic distribution $F\propto \exp (E/{\sigma }^2)$, which gives a Maxwell-Boltzmann distribution of velocities and $\rho \propto \exp (\psi /{\sigma }^2)\propto r^{-2}$. Circular orbits have instead $⟨v_r^2⟩=0$ and $⟨v_{\theta }^2⟩=⟨v_{\varphi }^2⟩=v_0^2/2$, so their total mean square speed is $v_0^2$. The isotropic model contains eccentric orbits except for a measure-zero subset.

The boundary virial term of a singular isothermal sphere explains the different energy budgets of a finite spherical region. At radius $R$, its radial-pressure surface term is $4\pi R^3\rho (R){\sigma }_r^2$. It is nonzero in the isotropic model and zero in the circular-orbit model. Equality of the density therefore does not enforce equality of the velocity energy through an incorrectly omitted surface term.

False. Adiabatic invariance of an orbital action preserves actions under sufficiently slow regular evolution, not the ratios of their frequencies. With action-angle variables, ${\Omega }_i=\partial H/\partial J_i$ can change even when $J_i$ does not. A closed bound orbit requires commensurate radial and angular frequencies; rational commensurability is not an adiabatic invariant.

For an explicit counterexample, slowly change the physical potential from Kepler form towith a small, slowly varying positive $\alpha$. The radial epicyclic frequency and circular angular frequency at a nearly circular orbit areInitially their ratio is one. For generic nonzero $\alpha$, $\Omega /\kappa$ is irrational, producing the Kepler-harmonic epicyclic precession. An initially slightly eccentric ellipse with nonzero radial action therefore becomes a precessing nonclosed orbit, while its radial action remains approximately fixed and its angular momentum remains exactly fixed by spherical symmetry.

Adiabatic actions do not preserve orbital closure. Exceptional circular or resonantly commensurate final orbits do not rescue a claim about every initially closed orbit.

False. The enclosed-mass formula is a consequence of the spherical shell theorem. In a spherical galaxy, exterior shells exert no radial force and interior mass acts as though concentrated at the centre, giving $v_c^2=GM(<r)/r$.

A razor-thin axisymmetric disc instead hasBoth interior and exterior annuli contribute to that derivative. Its geometry is not determined by $M(<R)$.

A concrete counterexample compares a central point mass $M_0$ plus an exterior spherical shell with the same point mass plus a thin circular ring of mass $m$ and radius $a$. Their enclosed-mass profiles are identical: $M_0$ inside $a$, and $M_0+m$ outside. The interior gravitational force of an exterior thin ring follows by expanding its angularly averaged relative potential:For $R\ll a$, the ring's acceleration is outward, ${\partial }_R{\psi }_{\mathrm{ring}}=GmR/(2a^3)+\cdots$, whereas the spherical shell's acceleration is zero. With $M_0$ large enough to retain circular orbits, the disc-side speed isA narrow smooth annulus gives the same distinction. This proves that enclosed mass does not determine a disc rotation curve, even when the mass profiles of the spherical and disc models agree.