Question 2.162: Starting with the virial theorem for an equilibrium configur...

For a finite gaseous configuration, the virial theorem gives

2 \bar{K}+\overline{\sum_{i} r _{i} \cdot F _{i}}=0.

\bar{K} \text { is the average total kinetic energy, } F _{i} is the total force acting on molecule i by all the other molecules of the gas. If the interactions are Newtonian gravitational of potentials V\left(r_{i j}\right) \sim \frac{1}{r_{i j}} we have

\sum_{i} r _{i} \cdot F _{i}=-\sum_{j \neq i} r_{i j} \frac{\partial V\left(r_{i j}\right)}{\partial r_{i j}}=\sum_{j<i} V\left(r_{i j}\right).

r_{i j}=\left| r _{i}- r _{j}\right|.

\text { Hence } 2 \bar{K}+\bar{V}=0 \text {, where } \bar{V} is the average total potential energy.
We can consider the gas in each small region of the configuration as ideal, for which the internal energy density \bar{u}( r ) and the kinetic energy \text { density } \bar{K}( r ) \text { satisfy }

\bar{u}( r )=\frac{2}{3} \frac{1}{\gamma-1} \bar{K}( r ) \quad \text { with } \bar{K}( r )=\frac{3}{2} k T( r ).

\text { Hence the total internal energy is } \bar{U}=2 \bar{K} / 3(\gamma-1) .

\text { When } \gamma=\frac{5}{3}, \bar{U}=\bar{K} \text {. In general the Virial theorem gives } 3(\gamma-1) \bar{U}+ \bar{V}=0 , so the total energy of the system is

E=\bar{U}+\bar{V}=(4-3 \gamma) \bar{U}.

For the system to be in stable equilibrium and not to diverge infinitely, we require E<0 . \text { Since } \bar{U}>0, we must have

\gamma<\frac{4}{3}.