Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 317 (Structure and Evolution of Stars)

For a homologous star, hydrostatic equilibrium, mass conservation, and the ideal gas equation give the central scalingswhere $\mu$ is the mean molecular weight. Integrating the nuclear energy-generation law over a fixed homologous profile givesRadiative stellar structure gives independentlyEquating the two luminosities yieldsAt fixed zero-age composition, the stars are therefore homologous with

The effective temperature satisfies $L=4\pi R^2\sigma T_e^4$, so $T_e\propto M^{3/8}$. The zero-age main sequence consequently hasIt is a steep line rising toward high luminosity and high temperature on a Hertzsprung-Russell diagram.

For fully ionized hydrogen and helium with $Y=1-X$,while $\kappa \propto 1+X$. At fixed mass,and the corresponding radius relation is

At the pure-hydrogen zero-age point $X=1$,whereasHenceandAs hydrogen is consumed, $X$ falls, so both $L$ and $T_e$ rise. In the usual diagram with temperature increasing leftward, the evolutionary track initially moves upward and leftward from the zero-age main sequence.

Letbe the scalar moment of inertia and total kinetic energy. Sincesubstitution of the Euler momentum equation reduces the stress contribution, by the divergence theorem and isotropic pressure $\mathbf{P}=P1$, toFor self-gravity, $\int \rho \mathbf{r}\cdot \mathbf{F}dV=\Omega$, where $\Omega <0$ is the gravitational potential energy. Thus the stellar virial theorem is

Apply hydrostatic equilibrium to the isothermal core, taking its boundary pressure to be $P_c$. Its ideal-gas pressure integral scales as $\int P,dV\propto M_c{T}_c$, its volume as $R_c^3$, and its gravitational energy as $-G{M}_c^2/R_c$. The virial theorem therefore givesor, after absorbing fixed dimensional and structural factors into positive constants,Hydrogen-burning reactions are extremely temperature-sensitive, so expansion cools and suppresses burning while contraction heats and enhances it. This stellar thermostat keeps the shell and adjoining isothermal core near an approximately fixed $T_c$.

At fixed $M_c,T_c$, differentiating $P_c(R_c)$ gives

For a homologous envelope, hydrostatic balance gives $P_c\propto G{M}^2/R^4$ and the ideal-gas temperature scale gives $T_c\propto GM/R$. Eliminating $R$ yieldsup to composition and gravitational constants common to the sequence. A matching core exists only if this required pressure does not exceed $P_{c,\max }\propto T_c^4/M_c^2$. ThereforeThis maximum fractional isothermal-core mass is the mechanism behind the Schönberg-Chandrasekhar limit.

For $P=K{\rho }^2$, the specific enthalpy isHydrostatic equilibrium says $\nabla (H+\Phi )=0$. Taking a Laplacian and using the gravitational Poisson equation gives the Helmholtz equation

For a spherical star, regularity at the centre selects the $n=1$ stellar polytropeIts first zero is $kR=\pi$, henceDirect integration gives $M=4{\rho }_c{R}^3/\pi$, and therefore

On the cube, the separated positive solutionvanishes on all six faces. It solves the same Helmholtz equation whenThe mean of each sine over $[0,L]$ is $2/\pi$, so

The interior fields formally solve the local structure equations, but an isolated fluid surface must be an equipotential and its interior gravitational field must match a decaying exterior solution with continuous normal derivative. A cube does not satisfy the global free-boundary conditions for a nonrotating self-gravitating barotrope. Such sharp planar faces and edges are also not observed in stars; ordinary pressure and gravity smooth the body toward a sphere.

For a circular binary with fixed total mass $M=M_1+M_2$, the orbital angular momentum isConservative binary mass transfer keeps $J$ and $M$ fixed, so $a\propto (M_1{M}_2{)}^{-2}$. Since $M_1=Mq/(1+q)$ and $M_2=M/(1+q)$,

Conservative transfer gives $d\log M_2/d\log M_1=-q$ and hence $d\log q/d\log M_1=1+q$. It follows thatand, using $R_L\simeq 0.4a{q}^{2/9}$,

Let $f=\log (R_1/R_L)$. While the system is detached, $M_1$ is constant, soDuring Roche-lobe overflow, $\dot{\log M_1}=-f/{\tau }_{\mathrm{dyn}}$, whileThereforeIf $\beta >\alpha$, the stable fixed point isThe corresponding mass-transfer rate isso the donor overfills its Roche lobe by only a tiny amount while losing mass on the slow nuclear timescale.

If $\beta <\alpha$, overflow is unstable. Starting at contact time $t_0$ with $f(t_0)=0$,The overflow and mass-loss rate grow exponentially on a dynamical timescale, leading toward unstable mass transfer or a common-envelope phase.