Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 62 / 3 / a / Solution

For an isolated collisionless stellar system, define the scalar second mass moment and total kinetic energy byTake moments of the Collisionless Boltzmann equation, with vanishing surface fluxes. For any time-independent phase-space function $A$, integration by parts givesApply this first to $A=r^2$ and then to $A=\mathbf{r}\cdot \mathbf{v}$. The result isThe potential is self-consistent Newtonian gravity, soSymmetrize the force integral under $\mathbf{r}\leftrightarrow {\mathbf{r}}^{\prime }$:Thus the dynamical virial theorem is $\ddot{I}/2=2T+W$. For a steady system, $\ddot{I}=0$, and consequentlyFor bounded statistically steady motion, averaging instead gives $2⟨T⟩+⟨W⟩=0$ provided $[\dot{I}(\tau )-\dot{I}(0)]/\tau \to 0$. Collisionlessness alone does not imply instantaneous virial equilibrium: a collapsing system retains the $\ddot{I}$ term. The displayed derivation assumes a finite second mass moment, or a limiting version with vanishing boundary terms, and no confining boundary-pressure contribution.

In virial equilibrium, the total energy is $E_{\mathrm{tot}}=T+W=-T$. Define the kinetic temperature $\Theta$ at fixed particle number $N$ by $T=3N{k}_B\Theta /2$. Along an equilibrium sequence,This is the negative heat capacity of a virialized gravitational system: removing total energy makes the kinetic temperature rise. A collisionless stellar system need not be in thermal equilibrium, so without that additional assumption this is a kinetic-temperature interpretation of heat capacity, rather than an assertion that every stellar distribution has a thermodynamic temperature.