Question 35.3: Use eqns 35.28 and 35.30 to relate U, Ω and E for a gas of (......
Use eqns 35.28 and 35.30 to relate U, Ω and E for a gas of (i) nonrelativistic particles (γ = \frac{5}{3} ) and (ii) relativistic particles (γ = \frac{4}{3} ).
3(γ −1)U + Ω = 0 , (35.28)
E = (4−3γ)U = \frac{3γ −4 }{3(γ −1)}Ω. (35.30)
(i) For a gas of non-relativistic particles, we have (using γ = \frac{5}{3} in eqns 35.28 and 35.30) that
2U + Ω = 0 , (35.31)
and hence
E = −U = \frac{Ω}{2}. (35.32)
Since the kinetic energy U is positive, the total energy E is negative thus the system is bound. Moreover, this shows that if the total energy E of a star decreases, this corresponds to a decrease in the gravitational potential energy Ω, but an increase in the kinetic energy U. Since U is directly related to the temperature T, we conclude that a star has a ‘negative heat capacity’: as a star radiates energy (E decreases), it contracts and heats up! This allows the nuclear heating process to be, to some extent, self-regulating. If a star loses energy from its surface, it contracts and heats up; therefore nuclear burning can increase, leading to an expansion, which cools the stellar core.
(ii) For a gas of relativistic particles, we have that (using γ = \frac{4}{3} in eqns 35.28 and 35.30)
U + Ω = 0 , (35.33)
and hence
E = 0 . (35.34)
Because the total energy is zero, a gravitationally bound state is not stable.