Past exam of the astrophysics course of the University of Oxford / 2022 / 5

Let $n(\mathbf{x})=\int f{d}^{3}v$ and $\overline{A}=n^{-1}\int Af{d}^{3}v$. Assume the one-particle distribution function and its weighted tails decay enough that all required velocity integrals converge and boundary terms at infinite velocity vanish. Multiply the steady Collisionless Boltzmann equation by $v_j$ and integrate over velocity. In Cartesian coordinates, spatial differentiation commutes with this integration, while integration by parts givesConsequently the three force-balance equations areExplicitly the $x$ component is ${\partial }_x(n\overline{v_x^2})+{\partial }_y(n\overline{v_x{v}_y})+{\partial }_z(n\overline{v_x{v}_z})=-n{\Phi }_x$, with cyclic counterparts for $y,z$. For equal stellar masses, multiplying by the common mass replaces $n$ by the mass density $\rho$; for multiple stellar masses, use a mass-weighted one-particle distribution function or derive a separate equation for each tracer population.

Define mean streaming velocity $u_i=\overline{v_i}$ and the velocity dispersion tensor ${\sigma }_{ij}^2=\overline{(v_i-u_i)(v_j-u_j)}=\overline{v_i{v}_j}-u_i{u}_j$. The zeroth velocity moment of the steady Collisionless Boltzmann equation gives the continuity equation ${\sum }_i{\partial }_i(n{u}_i)=0$. ThereforeSubstituting in a yields the standard steady Jeans equations:The unsteady version adds ${\partial }_t{u}_j$ on the left. Defining the stellar pressure tensor $P_{ij}=\rho {\sigma }_{ij}^2$ makes the analogy with fluid momentum balance explicit. The off-diagonal terms are correlations between random velocity components; they need not vanish or equal the diagonal terms.

The Jeans equations express local momentum conservation for a collisionless stellar system, just as the Euler equations do for a gas. The mean streaming velocity plays the role of fluid flow, the gravitational force is a body force, and random stellar motions supply a pressure tensor through $\rho {\sigma }_{ij}^2$. Taking moments smooths over individual stellar orbits and produces fields of density, mean velocity and random-motion stress; this is the sense of “stellar hydrodynamics”.

Ordinary gas pressure is usually nearly isotropic because collisions relax its local velocity distribution. A collisionless stellar system can retain an anisotropic velocity ellipsoid, so its pressure tensor is not generally a scalar times the identity. There are more unknown second moments than momentum equations, and their evolution introduces higher moments. The Jeans equations therefore require further dynamical assumptions or a specified one-particle distribution function for closure; they do not by themselves establish local thermal equilibrium or a gas equation of state.

Use fixed masses in an isolated Newtonian $N$-body system, with origin at its centre of mass. Define the $xx$ moment $I_{xx}={\sum }_i{m}_i{x}_i^2$ and $K_{xx}=\frac{1}{2}{\sum }_i{m}_i{v}_{x,i}^2$. Twice differentiating givesInsert the gravitational equation of motion and pair each ordered term $(i,k)$ with $(k,i)$:Thus one component of the tensor virial theorem isA stationary or time-averaged bounded system has $⟨{\ddot{I}}_{xx}⟩=0$, yielding $2⟨K_{xx}⟩+⟨W_{xx}⟩=0$. If ordered and random motions are separated, $2K_{xx}=2T_{xx}+{\Pi }_{xx}$, where $T_{xx}=\frac{1}{2}\int \rho u_x^{2}dV$ and ${\Pi }_{xx}=\int \rho {\sigma }_{xx}^{2}dV$. Summing the three diagonal equations recovers the scalar virial theorem, with the pair gravitational potential energy counted once.