Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 317 (Structure and Evolution of Stars)

The four stellar structure equations aretogether with the perfect-gas equation of stateHere $a_{\mathrm{rad}}$ is the radiation constant, $ϵ$ is the specific stellar energy-generation rate, and $\kappa$ is the opacity.

Put $x=r/R$. Integrating the prescribed density gives the enclosed massSince $m_R=M$,Integrating hydrostatic equilibrium inward from $P(R)=0$ givesThe ideal-gas law then givesIn particular,

At $x=1/2$, defineDifferentiating the temperature profile givesSince the luminosity is already constant there, the radiative diffusion in a star equation yields

For the Kramers opacity law $\kappa ={\kappa }_0\rho T^{-7/2}$, substitution giveswhereThus $\alpha =11/2$ and $\beta =-1/2$ in the notation of the question.

For constant electron-scattering opacity ${\kappa }_{\mathrm{es}}$,whereHence $\delta =3$ and $\gamma =0$: the radius cancels, and the mass dependence is shallower than under Kramers opacity.

Suppose first that the gas-pressure fraction $\beta =P_g/P$ is spatially constant, with $0<\beta <1$. This is a sufficient condition for a stellar polytrope. Sincethe ratio $P_{\mathrm{rad}}/P_g=(1-\beta )/\beta$ givesThereforewithFor an ordinary positive polytropic index one also assumes $0\le s<3$.

The radiative diffusion in a star equation can be written asBecause $P_{\mathrm{rad}}=(1-\beta )P$ and $\beta$ is constant, hydrostatic equilibrium givesEquating them and using $\eta =(L_r/L)/(m_r/M)$ yields

For $P=K{\rho }^{1+1/n}$, introduce the Lane-Emden equation variablesCombining mass conservation with hydrostatic equilibrium givesRegularity and normalization imposeThe stellar surface is the first positive zero ${\xi }_1$ of $\theta$. Integrating the equation once givesand hence

When $s=0$, part i gives the polytropic index $n=3$. Stellar homology givesbecause gas pressure is the fraction $\beta$ of the pressure required by hydrostatic equilibrium. The Kramers opacity law therefore scales asWith constant $\eta$, the opacity relation from part i gives

For an $n=3$ polytrope, $M\propto (K/G{)}^{3/2}$. The explicit $K(\beta )$ of part i obeysso the Eddington quartic relation isat fixed composition. Substitution gives

A fully convective monatomic perfect-gas star follows an $n=3/2$ adiabatic stellar polytrope. At the photosphere, $\tau =2/3$ and hydrostatic equilibrium in optical depth givesFor an ideal gas on an adiabat,Evaluating this at the photosphere givesThe $n=3/2$ polytropic mass-radius relation is $R\propto K{M}^{-1/3}$, soFinally the Stefan–Boltzmann law givesThis is the specified Hayashi track.

Equate the accretion luminosity to the surface luminosity:Using $R\propto M^3{T}_{\mathrm{eff}}^{-24}$ from part i givesSubstitution into $L\propto M^6{T}_{\mathrm{eff}}^{-44}$ yields the accreting Hayashi track

The specific deuterium-burning energy-generation rate is $ϵ={ϵ}_0\rho T^{15}$. By stellar homology,soSince $L_{\mathrm{acc}}\propto M\dot{M}/R$,Use $R\propto M^3{T}_{\mathrm{eff}}^{-24}$ and $T_{\mathrm{eff}}^{68}\propto M^8/\dot{M}$ from part ii. Then

The two track relations also giveConsequentlyEquality occurs atThe preceding formulae imply $L_{\mathrm{nuc}}>L_{\mathrm{acc}}$ for $L<L_0$, not $L>L_0$. Thus the inequality printed in the question is reversed relative to its requested intermediate scaling; the exponent $964/43$ is consistent.

LetA displaced fluid element remains in pressure balance, changes temperature adiabatically, and retains its composition. Comparing its density with the environment after an upward displacement gives the Ledoux criterion for stability,The reverse inequality causes convection. For uniform composition, or if composition is ignored, ${\nabla }_{\mu }=0$ and this reduces to the Schwarzschild criterionfor stability. A molecular weight increasing inward has ${\nabla }_{\mu }>0$ and stabilizes the stratification.

Write $\mathcal{R}=k_B/(\mu m_u)$. At fixed density,The specific internal energy isApplying the first law of thermodynamics adiabatically givesCombining these logarithmic derivatives yieldsAs gas pressure dominates,the result for a monatomic perfect gas. In the radiation-pressure limit it tends to $1/4$.

Uniform composition and adiabatic convection give $T\propto {\rho }^{2/3}$ for a monatomic perfect gas. Hence near the centreThe luminosity equation isSince ${\rho }^2{T}^{\eta }\propto 1-(2+2\eta /3)\lambda r^2$, integration givesSimilarly,

The radiative temperature gradient isExpanding $\kappa ={\kappa }_0{\rho }^a{T}^{-b}$ and all the preceding factors giveswhereThe Schwarzschild criterion requiresfor the centre to be convectively unstable. If $A>0$, the radiative gradient decreases outward and can cross $2/5$, producing a finite convective core.