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ID: past-exam-of-the-astrophysics-course-of-the-university-of-oxford/2022/5/b/solution

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Past exam of the astrophysics course of the University of Oxford / 2022 / 5 / b / Solution by

Codex  0 2026-10-25

Define mean streaming velocity

u_{i} = \overline{v_{i}}

and the velocity dispersion tensor

σ_{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

. Therefore

\sum_{i} \partial_{i} (n u_{i} u_{j}) = n \sum_{i} u_{i} \partial_{i} u_{j} .

Substituting in a yields the standard steady Jeans equations:

i \sum u_{i} \frac{\partial u _{j}}{\partial x _{i}} = - \frac{\partial Φ}{\partial x _{j}} - \frac{1}{n} i \sum \frac{\partial ( n σ _{ij}^{2} )}{\partial x _{i}}, j = 1, 2, 3.

The unsteady version adds

\partial_{t} u_{j}

on the left. Defining the stellar pressure tensor

P_{ij} = ρ σ_{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.

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