 Solution

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

Use fixed masses in an isolated Newtonian

N

-body system, with origin at its centre of mass. Define the

xx

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 gives

\frac{1}{2} \ddot{I}_{xx} = \sum_{i} m_{i} v_{x, i}^{2} + \sum_{i} m_{i} x_{i} \overset{v}{˙}_{x, i} = 2 K_{xx} + W_{xx} .

Insert the gravitational equation of motion and pair each ordered term

(i, k)

with

(k, i)

W_{xx} = - G i \sum k \neq = i \sum \frac{m _{i} m _{k} x _{i} ( x _{i} - x _{k} )}{r _{ik}^{3}} = - \frac{G}{2} i \sum k \neq = i \sum \frac{m _{i} m _{k} ( x _{i} - x _{k} ) ^{2}}{r _{ik}^{3}} = - G i < k \sum \frac{m _{i} m _{k} ( x _{i} - x _{k} ) ^{2}}{r _{ik}^{3}} .

Thus one component of the tensor virial theorem is

\frac{1}{2} \ddot{I}_{xx} = 2 K_{xx} - G i < k \sum \frac{m _{i} m _{k} ( x _{i} - x _{k} ) ^{2}}{r _{ik}^{3}} .

A 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,

2 K_{xx} = 2 T_{xx} + Π_{xx}

, where

T_{xx} = \frac{1}{2} \int ρ u_{x}^{2} d V

and

Π_{xx} = \int ρ σ_{xx}^{2} d V

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