In Problem 1.55 you used the virial theorem to estimate the heat capacity of a star. Starting with that result, calculate the entropy of a star, first in terms of its average temperature and then in terms of its total energy. Sketch the entropy as a function of energy, and comment on the shape of the graph.

The heat capacity of a star that was estimated using the virial theorem is given as:

$C_V=-\frac{3}{2}NK$

Where,

$N$is the number of particles (typically dissociated protons and electrons).

The negative sign symbolizes that it is a gravitational bound system.

The change in entropy is given as:

$S=\int \frac{C_V\left(T\right)}{T}dT$

Where,

$C_V$= specific heat

$T$= Temperature in Kelvin

By substituting the value of $C_V$in the above equation, we get,

$$\begin{gathered} S=\int -\frac{3}{2}\left(\frac{NK}{T}\right)dT \\ S=-\frac{3}{2}NK\int \left(\frac{1}{T}\right)dT \\ S=-\frac{3}{2}NKT\ln \left(T\right)+f(N,V)..........\left(1\right) \end{gathered}$$

In this equation, $f$is the function of $N$and volume $V$.

Total energy of gravitationally bound system is negative and from the virial theorem, it is found that:

$U=-K=-\frac{3}{2}NKT$

By rearranging the terms, we get,

$T=-\frac{2U}{3NK}$

By substituting this value in equation (1), we get,

$S=-\frac{3}{2}NK\ln \left(\frac{2U}{3NK}\right)+f(N,V)$

For plotting the graph, let us further simplify the above equation,

$$\begin{gathered} S=-\frac{3}{2}NK\ln \left(U\right)-\frac{3}{2}NK\ln \left(\frac{3NK}{2}\right)+f(N,V) \\ S=-\frac{3}{2}NK\ln \left(U\right)+g(N,V) \end{gathered}$$

From the above equation, the graph can be plotted as below:

Hence, the required expression is: $S=-\frac{3}{2}NK\ln \left(\frac{2U}{3NK}\right)+f(N,V)$

The graph of entropy as a function can be sketched as follow: