Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 58 / 2

For a uniform-density stellar model, $\rho =3M/(4\pi R^3)$ and $m(r)=4\pi \rho r^3/3$. With zero surface pressure, integrate the stellar hydrostatic equation:Using the ideal gas equation of state, $T(r)=P(r)/(\mathcal{R}\rho )$. ThereforeEquivalently $P_c=3G{M}^2/(8\pi R^4)$ and $T_c=GM/(2\mathcal{R}R)$. The constant mass density is imposed as a structural approximation; a perfect gas can realize it hydrostatically by allowing its temperature and entropy to vary with radius.

Assemble the star from concentric shells. A shell of mass $dm$ at radius $r$ has interaction energy $-Gm(r)dm/r$ with the already assembled interior. This counts each gravitational pair once. Hence the Newtonian gravitational potential energy isUsing $M=4\pi \rho R^3/3$ givesThe negative sign expresses gravitational binding. A factor of one half should not be inserted again: the shell assembly already avoids double counting.

For a spherical hydrostatic star with negligible surface pressure, the stellar virial theorem isFor a monatomic nonrelativistic perfect gas, $U=\frac{3}{2}\int PdV$, so $2U+\Omega =0$ and the total energy is $E=U+\Omega =\Omega /2$. Half the released binding energy raises the internal energy; only the other half can be radiated. With fixed mass, no nuclear supply and no external work, conservation of energy givesFor the uniform-density stellar model, this becomes the contraction-luminosity evolution equationIt relates luminosity to the contraction rate; it does not by itself prescribe $L(t)$. Given $L(t)$,For constant luminosity, $R(t)=R_0/[1+t/t_0]$ with $t_0=3G{M}^2/(10R_{0}L)$. This describes quasi-static Kelvin-Helmholtz contraction, not dynamical free fall. If the gas has a different constant specific-heat ratio ${\gamma }_g$, the appropriate stellar virial theorem is $3({\gamma }_g-1)U+\Omega =0$, so the radiated fraction of binding release is $(3{\gamma }_g-4)/[3({\gamma }_g-1)]$ rather than universally one half.

The conventional Kelvin-Helmholtz cooling time is of order $G{M}^2/(RL)$. For the monatomic uniform-density stellar model, its precise accessible total-energy reservoir is $|E|=3G{M}^2/(10R)$, soUsing $M_{\odot }\simeq 1.9885\times 10^{30}\mathrm{kg}$, $R_{\odot }\simeq 6.957\times 10^8\mathrm{m}$ and $L_{\odot }\simeq 3.828\times 10^{26}\mathrm{W}$, the usual order-of-magnitude normalization without the structural factor is $t_{\mathrm{KH}}\simeq 3.1\times 10^7\mathrm{yr}$. The full gravitational binding reservoir $|\Omega |/L$ is about $1.9\times 10^7\mathrm{yr}$, but that overcounts the radiatable energy in virial equilibrium.

The Sun and the Solar system are approximately $4.6\times 10^9\mathrm{yr}$ old, much older than any of these contraction estimates. Gravitational contraction cannot sustain the Sun's long-lived present power. The dominant long-term source is stellar nuclear fusion. Contraction can still supply transient luminosity, especially in a pre-main-sequence star. The timescale comparison rules out a contraction-only explanation over the observed age; it does not assert that all gravitational energy release is absent.

For a stellar polytrope of index $n>0$, write $P=K{\rho }^{1+1/n}$ and combine hydrostatic equilibrium with mass conservation to obtainIntroduce Lane-Emden variables for a stellar polytrope, $\rho ={\rho }_c{\theta }^n$, $P=P_c{\theta }^{n+1}$ and $r=\alpha \xi$, whereSince ${\rho }^{-1}dP/dr=(n+1)P_c{\theta }^{\prime }/{\rho }_c\alpha$, the mechanical equation reduces to the Lane-Emden equationA constant-density interior corresponds to the formal polytrope of index zero, with $\rho ={\rho }_c{\theta }^0={\rho }_c$ where $\theta >0$. At $n=0$, regularity first gives ${\xi }^2{\theta }^{\prime }=-{\xi }^3/3$, and a second integration givesHere $P=P_c\theta =P_c(1-r^2/R^2)$ and ${\alpha }^2=P_c/(4\pi G{\rho }_c^2)$, reproducing $P_c=2\pi G{\rho }_c^2{R}^2/3$. The mass density jumps from its constant interior value to zero at the surface, while pressure vanishes continuously. The relation $P=K{\rho }^{1+1/n}$ is singular at $n=0$; the regular dimensionless pressure/mass density formulation defines this incompressible structural limit. It does not mean that the gas's perturbative stellar adiabatic exponent is infinite: the hydrostatic ideal-gas toy model and its adiabatic response are distinct choices.