Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 317 (Structure and Evolution of Stars)

For a family in stellar homology, dimensionless radial profiles are fixed. Mass conservation and hydrostatic equilibrium then give the central scalingsThe ideal gas equation of state, with fixed mean molecular weight, consequently giveswhere fixed dimensional constants such as $G$ are suppressed in homology relations.

The proton–proton chain law $ϵ\propto \rho T^{7/2}$ gives the nuclear luminosityOn the other hand, radiative diffusion in a star givesUsing the Kramers opacity law ${\kappa }_c\propto {\rho }_c{T}_c^{-7/2}$,Thermal equilibrium requires $L_{\mathrm{nuc}}=L_{\mathrm{rad}}$. Their common mass factor cancels, leaving $R^{-13/2}\propto R^{-1/2}$, and hence

This model captures the gas-pressure support, pp-chain burning, and strongly mass-dependent luminosity of the lower main sequence, including the Sun approximately. Its fully radiative assumption is an idealization: the Sun has a convective envelope, and sufficiently low-mass red dwarfs become largely or fully convective, so real radii are not exactly constant.

Along the pp-chain homologous sequence, $R$ is constant and therefore $T_c\propto M$. Normalizing to the Sun gives the transition mass

For the CNO cycle law $ϵ\propto \rho T^{23/2}$, the nuclear scaling becomesThe opacity law is unchanged, so $L_{\mathrm{rad}}\propto M^{11/2}{R}^{-1/2}$. Equating them givesand then

The effective temperature follows from the Stefan–Boltzmann law $L=4\pi R^2\sigma T_e^4$. Thus the pp branch has $T_e\propto M^{11/8}$ and $L\propto T_e^4$, whereas the CNO branch hasOn a Hertzsprung-Russell diagram, both branches rise toward larger luminosity and, conventionally, leftward toward larger temperature. They join near $4M_{\odot }/3$; the CNO branch has the steeper $L$--$T_e$ logarithmic slope and lies to the cooler side of the extrapolated constant-radius branch at fixed luminosity.

If $X$ is the hydrogen mass fraction and $X_{\mathrm{CNO}}$ the combined carbon-nitrogen-oxygen mass fraction, the reaction-pair and catalyst abundances give approximatelyThe CNO nuclei are catalysts, so their abundance multiplies the CNO-cycle rate rather than being consumed by the completed cycle.

At fixed hydrogen abundance, the ratio of the two rates near the transition scales asReducing $X_{\mathrm{CNO}}$ by $256=2^8$ therefore requires twice the temperature to recover equality:The metal-poor stars remain on the same pp-chain homology sequence up to this point because the question holds opacity and mean molecular weight fixed. Since that sequence has $T_c\propto M$,

The low-CNO pp branch therefore overlaps the solar-composition pp branch and simply continues farther upward and leftward before turning onto a CNO-dominated branch. The CNO branch retains $R\propto M^{4/7}$ and $L\propto M^{73/14}$, but its normalization changes. Since $R^{14}\propto {ϵ}_0{M}^8$, reducing the CNO coefficient by 256 gives, at fixed mass,It is consequently somewhat smaller, hotter, and more luminous than the solar-CNO homology model of the same mass. The sketch should show its delayed break at $8M_{\odot }/3$ and its CNO branch displaced upward and leftward.

Put $\mu =\cos \theta$ and $S=j/\kappa$. Integrating the radiative transfer equation over solid angle givesThe first integral is the radiative flux $F$. A stationary atmosphere with no local energy source has $dF/d\tau =0$, soBy definition, the mean intensity is $J=(4\pi {)}^{-1}\int Id\Omega$. Thereforewhich is radiative equilibrium: the angle-integrated emission and absorption rates are equal.

For $I=A+C\mu$, the angular moments areandThus this angular form satisfies the Eddington closure approximation, $c{P}_r=4\pi J/3$, and

Part (i) gives $S=J=A$. Substitution in $\mu dI/d\tau =I-S$ yieldsMatching powers of $\mu$ gives $A^{\prime }=C$ and $C^{\prime }=0$, the latter being constant flux. Hence $A=C\tau +A_0$.

At the surface there is no incoming intensity. Applying this condition to the outgoing hemisphere in the moment approximation,so $A_0=F/(2\pi )=2C/3$. ConsequentlyNow $A=J=j/\kappa =\sigma T^4/\pi$ and the effective temperature is defined by $F=\sigma T_e^4$. It follows thatAt the top of the grey atmosphere,

Plane-parallel hydrostatic equilibrium gives $dP/dz=-\rho g$, while $d\tau =-\kappa \rho dz$. HenceFor $\kappa ={\kappa }_0{P}^{\alpha -1}{T}^{4-4\beta }$,The grey-atmosphere relation from part (ii) can be writtenIntegrating from the zero-pressure top, where $T=T_0$, gives

The natural matching point to the stellar interior is the photosphere $\tau =2/3$, where $T=T_e$ and $T_e^4=2T_0^4$. Multiplying the preceding expression by ${\kappa }_0{T}^{4-4\beta }/g$ then yields the stellar surface boundary conditionThe local relation $L_r=4\pi r^2\sigma T^4$ identifies this matching temperature with the local effective temperature.

Because $H\ll R$ and $M_{\mathrm{env}}\ll M$, the envelope has nearly constant area $4\pi R^2$ and nearly constant surface gravity of a star $g=GM/R^2$. Plane-parallel hydrostatic equilibrium isSince $d\Sigma /dz=-\rho$, one has $dP/d\Sigma =g$. Taking the pressure to vanish with $\Sigma$ at the surface gives

The spherical stellar energy equation is $d{L}_r/dr=4\pi r^2\rho ϵ$. With $r\simeq R$ and $F=L_r/(4\pi R^2)$,Division by $d\Sigma /dr=-\rho$ givesLikewise, radiative diffusion in a star givesand therefore

Write the ideal gas law as $P=\mathcal{R}\rho T/\mu$. Since $P=g\Sigma$,The prescribed opacity and burning laws becomeThe radiative equation is consequentlyWith $y=T^7$ and $x={\Sigma }^2/2$,The energy equation similarly givesDifferentiating the first relation therefore produces the nonlinear ordinary differential equation

At the idealized zero-temperature surface, $x=0$ and $y=0$. The outward flux there is $F_s=L/(4\pi R^2)$, soAt the base, $x_0={\Sigma }_0^2/2$ and $y=T_0^7$. The core supplies no luminosity in this model, so all flux has been generated in the overlying hydrogen envelope and $F(x_0)=0$. Hence

Multiplying $y^{\prime \prime }=-{\omega }^2{y}^2$ by $y^{\prime }$ and integrating giveswhere the base conditions fixed the constant. At the surface, $y=0$ and $y^{\prime }=A{F}_s$, with $F_s=L/(4\pi R^2)$. Thereforeand

Separating variables in the first integral givesSet $y=T_0^7\eta$. ThenSince $x_0={\Sigma }_0^2/2$ and ${\Sigma }_0=M_{\mathrm{env}}/(4\pi R^2)$,Combining this with $F_s\propto T_0^{21/2}$ yields

This inverse relation signals thin-shell instability. Hydrogen burning has extreme temperature sensitivity, $ϵ\propto \rho T^{15}$, while the weight of the thin envelope fixes its pressure and limits thermostatic expansion. As burning consumes envelope mass, the equilibrium luminosity rises sharply, accelerating consumption rather than restoring the original state. A thermal runaway can therefore lead to a shell flash or classical nova rather than steady stable burning.

In the synchronously rotating reference frame, stationary fluid obeyswhere $\Psi$ is the sum of the two gravitational potentials and the centrifugal potential. The circular binary's Kepler third law gives ${\Omega }^2=G(M_1+M_2)/a^3=GM/a^3$, hence

Uniform composition and the assumed central condensation make star 1 approximately barotropic, so one may define the specific enthalpyHydrostatic balance becomes $\nabla (h+\Psi )=0$, or $h+\Psi =\text{constant}$ throughout the star. Since $h$ is a monotone function of $P$ and $\rho$, constant-$P$ and constant-$\rho$ surfaces are equipotential surfaces of $\Psi$.

The Roche lobe of star 1 is the volume around it bounded by the critical closed equipotential passing through the inner Lagrange point $L_1$. Material on a lower potential surface remains confined to star 1; at the critical surface a path opens toward star 2.

Let $\mathbf{g}=-\nabla \Psi$ and take $d\mathbf{S}$ outward. For a nearly spherical interior surface, the divergence theorem givesThe self-gravity term contributes $-4\pi G{M}_1$. The companion lies outside $S$, so its potential is harmonic inside and contributes zero net flux. For the centrifugal term, ${\nabla }^2(-{\Omega }^2{s}^2/2)=-2{\Omega }^2$, so its acceleration has divergence $2{\Omega }^2$ and contributes $2{\Omega }^{2}V$. Using $V=4\pi r^3/3$ and $S=4\pi r^2$,The sign is the outward-normal component; the dominant self-gravity is inward and therefore negative.

Let $g=|\nabla \Psi |$ on the equipotential $S(\Psi )$. The volume between neighboring equipotentials separated by $d\Psi$ isBecause density is constant on each equipotential,The volume-equivalent radius is defined by $V=4\pi r^{\prime 3}/3$, so $dV=4\pi r^{\prime 2}d{r}^{\prime }$. It follows immediately that

Hydrostatic balance gives $dP/d\Psi =-\rho$, whileConsequentlyDefining the geometric correctionputs this in the one-dimensional stellar-structure formFor a sphere, $S=4\pi r^{\prime 2}$ and $g=Gm/r^{\prime 2}$, so $f_P=1$ as required.

Because $T=T(\Psi )$,The outward normal radiative flux is thereforeIntegrating over the equipotential givesPart (iii) gives $dm/d\Psi =\rho S⟨g^{-1}⟩$, and henceSubstitution yieldsThis is the radiative equation in the equipotential stellar-structure approximation.

At the critical Roche surface, $g=0$ at $L_1$, making $⟨g^{-1}⟩$ singular. Once star 1 overfills its Roche lobe, the relevant equipotentials are no longer nested closed surfaces belonging to that star: they open through the $L_1$ saddle and connect to star 2. Matter then undergoes Roche-lobe overflow with finite velocity, so advective momentum and energy transport replace hydrostatic and purely radiative equilibrium near the nozzle. Density, temperature, and composition need not remain uniform on an equipotential. The one-dimensional volume coordinate $r^{\prime }(\Psi )$ and its correction factors therefore cease to describe the three-dimensional mass-transfer flow.