= Solution
For an <ideal gas> with constant <adiabatic index> $\gamma>1$, let $c_v$ and $c_p$ be specific heat capacities and $\mathcal R=c_p-c_v$. The <ideal gas law> and the <internal energy of an ideal gas> give $P=\rho\mathcal RT$ and specific internal energy $e=c_vT$. Since $\gamma=c_p/c_v$, the <internal energy> per unit volume is $\rho e=P/(\gamma-1)$. Consequently a star with <spherical symmetry> has
$$
\boxed{U=\frac1{\gamma-1}\int P\,dV=\frac{4\pi}{\gamma-1}\int_0^R P(r)r^2\,dr.}
$$
The radial integral requires the volume measure $dV=4\pi r^2dr$: the first printed expression omits this factor if its differential is read literally as $dr$. The mass-coordinate expression confirms the required volume interpretation.
Define the enclosed mass $M(r)$, total mass $M_{\rm tot}$ and dimensionless shell label $m=M(r)/M_{\rm tot}$. <Mass conservation> gives $M_{\rm tot}\,dm=dM=\rho\,dV$, so the <change of variables> gives
$$
\boxed{U=\frac{M_{\rm tot}}{\gamma-1}\int_0^1\frac{P(m)}{\rho(m)}\,dm.}
$$
Here $m$ labels a fixed material shell rather than its radial position. In a mass-preserving <stellar homology>, $r(m)=xr_0(m)$ with $x=R/R_0$. Both a shell's radius and thickness gain a factor $x$, so its volume gains $x^3$ and its <mass density> becomes $\rho(m)=x^{-3}\rho_0(m)$. An <adiabatic process> keeps $P/\rho^\gamma$ fixed separately in each shell, even if that constant differs between shells. Thus $P=x^{-3\gamma}P_0$ and $P/\rho=x^{-3(\gamma-1)}P_0/\rho_0$. Integration at fixed mass labels proves
$$
\boxed{U(x)=U_0x^{-3(\gamma-1)}.}
$$
The <spherical shell theorem> gives the <Newtonian gravitational potential energy> by assembling shells against the mass already inside them:
$$
W=-\int_0^{M_{\rm tot}}\frac{GM}{r(M)}\,dM.
$$
Every radius gains the same factor $x$, while $M$ and $dM$ remain fixed. Hence
$$
\boxed{W(x)=W_0x^{-1},\qquad E(x)=U_0x^{-a}+W_0x^{-1},\quad a=3(\gamma-1).}
$$
This is the <homologous adiabatic stellar stability> test. The derivative with respect to the physical stellar radius is $\partial_R=R_0^{-1}\partial_x$. Stationarity at $x=1$ therefore requires $-aU_0-W_0=0$, or
$$
\boxed{|W_0|=3(\gamma-1)U_0.}
$$
After imposing this equilibrium relation, $E/U_0=x^{-a}-a/x$ and
$$
\frac{dE}{dx}=aU_0\left(x^{-2}-x^{-a-1}\right),\qquad
\left.\frac{\partial^2E}{\partial R^2}\right|_{R_0}=\frac{a(a-1)U_0}{R_0^2}.
$$
Since $a>0$, a strict energy minimum under this homologous displacement requires
$$
\boxed{\gamma>\frac43.}
$$
For $a>1$, contraction sends $E$ to $+\infty$, expansion sends it to $0$ from below, and $x=1$ is the unique minimum, with $E(1)=(1-a)U_0<0$. For $0<a<1$, contraction sends $E$ to $-\infty$, expansion sends it to $0$ from above, and $x=1$ is the unique maximum. When $a=1$, the equilibrium relation makes $E(x)=0$ for every $x$: the scale direction is marginal, not strictly stable. The plotted curves display these three possibilities. This checks homologous radial stability; stability against every stellar displacement needs a fuller perturbation analysis.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/iii/paper-62-energy.png]
{title=Stellar energy under homologous adiabatic expansion: stable minimum, unstable maximum and marginal scaling}
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