Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 317 (Structure and Evolution of Stars)

Assume stellar homology: fixed dimensionless profiles, uniform fixed composition, negligible radiation pressure, ideal gas support, and radiative transport throughout the model. Mass conservation gives ${\rho }_c\propto M/R^3$. The hydrostatic pressure support equation gives $P_c\propto G{M}^2/R^4$; combining with $P\propto \rho T$ gives $T_c\propto M/R$.

Integrating specific proton–proton chain energy generation over mass givesFor radiative diffusion in a star, $T_c/R\propto {\kappa }_c{\rho }_{c}L/(R^2{T}_c^3)$, hence $L_{\mathrm{rad}}\propto R{T}_c^4/({\kappa }_c{\rho }_c)$. With the Kramers' opacity law ${\kappa }_c\propto {\rho }_c{T}_c^{-7/2}$,Equating generated and transported luminosity gives $M^{1/2}\propto R^{13/2}$, so the radiative homology with proton-proton burning and Kramers opacity relation isThe proportionality constants are fixed only within the adopted homologous family, not for arbitrary evolving stars.

Using $R\propto M^{1/13}$ in the radiative diffusion in a star scaling from part (i),Therefore the mass-luminosity relation isThe nuclear scaling independently gives $L\propto M^6{R}^{-7}\propto M^{6-7/13}=M^{71/13}$, consistent with radiative homology with proton-proton burning and Kramers opacity.

The Stefan–Boltzmann law defines the effective temperature through $L=4\pi R^2{\sigma }_{\mathrm{SB}}{T}_{\mathrm{eff}}^4$. The stellar homology relations give $R\propto M^{1/13}\propto L^{1/71}$, soThusIn a logarithmic Hertzsprung-Russell diagram, with $\log L$ plotted vertically against $\log T_{\mathrm{eff}}$,The usual horizontal axis increases in effective temperature toward the left, so the hotter, more luminous end lies toward the upper left. The quoted algebraic slope uses increasing $\log T_{\mathrm{eff}}$ as the coordinate.

Retain the same stellar homology and gas-pressure assumptions, but use the specified CNO cycle law $ϵ\propto \rho T^{16}$ and constant electron-scattering opacity. The nuclear luminosity now scales asThe radiative diffusion in a star scaling becomesEquating the two yields $R^{19}\propto M^{15}$. Therefore radiative homology with CNO burning and electron scattering givesSince $R\propto L^{5/19}$, the Stefan–Boltzmann law givesThis is a model comparison using the exponent 16 prescribed here. Actual massive stars may have appreciable radiation pressure and convective cores, which violate the assumptions of this purely radiative gas-pressure family.

Consider an element displaced upward by $\xi$ in hydrostatic equilibrium. Assume it quickly reaches the ambient pressure, but moves adiabatically and retains its composition. Put $\nabla =d\log T/d\log P$, ${\nabla }_{\mu }=d\log \mu /d\log P$, and let ${\nabla }_{\mathrm{ad}}$ be the adiabatic temperature gradient. The equation of state has differentialwhere $\delta =-(\partial \log \rho /\partial \log T{)}_{P,\mu }$ and $\varphi =(\partial \log \rho /\partial \log \mu {)}_{P,T}$.

At the new pressure, the parcel-to-environment density difference isWith positive pressure scale height $H_P=-(d\log P/dr{)}^{-1}$, $\Delta \log P=-\xi /H_P$. The buoyancy acceleration is therefore $\ddot{\xi }=-N^2\xi$, with stellar buoyancy frequencyA negative $N^2$ amplifies the displacement. For a radiative background, $\nabla ={\nabla }_{\mathrm{rad}}$, so the Ledoux criterion for instability isFor uniform composition, ${\nabla }_{\mu }=0$, giving the Schwarzschild criterionEquality is marginal stability. A mean molecular weight increasing inward has ${\nabla }_{\mu }>0$ and supplies a stabilizing composition term when $\varphi ,\delta >0$.

The Eddington closure approximation with the standard surface boundary condition gives the grey atmosphere profileTogether with the specified pressure law, logarithmic differentiation givesConsequently the radiative temperature gradient isFor a monatomic perfect gas, $\gamma =5/3$ and the adiabatic temperature gradient is $(\gamma -1)/\gamma =2/5$. Uniform composition makes the Schwarzschild criterion applicable. The first marginal layer therefore satisfies $\log (1+3{\tau }_c/2)=4/5$, giving the convection onset in a logarithmic-pressure grey atmosphereThe assumed profile becomes convectively unstable for $\tau >{\tau }_c$; further inward, a self-consistent model must account for convective transport.

Assume zero temperature, complete ionization, constant composition, noninteracting electrons, negligible ion thermal pressure, and Newtonian stellar gravity. Ions supply almost all the mass, while electron degeneracy pressure supplies support. By the Pauli exclusion principle, the two electron spin states fill a momentum sphere up to Fermi momentum $p_F$. Counting states giveswhere ${\mu }_e$ is mean molecular weight per electron. The momentum flux of this isotropic Fermi gas isWriting $x=p_F/(m_{e}c)$ and performing the integral gives the equation of state of a cold electron gasIn the nonrelativistic limit $p_F\ll m_{e}c$, $v(p)\simeq p/m_e$, soIn the ultrarelativistic limit $p_F\gg m_{e}c$, $v(p)\simeq c$, soThese are polytropic equations of state with indices $3/2$ and $3$, respectively. The relativistic softening underlies the Chandrasekhar mass limit; Coulomb corrections, thermal effects, rotation, and general relativity are excluded from this idealized derivation.

For the nonrelativistic white dwarf, combine the hydrostatic pressure support equation and enclosed mass equation:Eliminating $m$ yieldsIntroduce the Lane-Emden variables for a stellar polytrope,The resulting Lane-Emden equation isThe white dwarf surface is at the first zero ${\xi }_1$ of $\theta$, with $R=a{\xi }_1$.

Let $\omega =-{\xi }_1^2{\theta }^{\prime }({\xi }_1)$ be the Lane-Emden surface mass constant. The Lane-Emden mass formula givesThus $R\propto {\rho }_c^{-1/6}$ while $M\propto {\rho }_c^{1/2}$ at fixed composition. Eliminating central mass density gives the polytropic mass-radius relationMore explicitly,Increasing the mass of a nonrelativistic white dwarf decreases its radius. This scaling ceases to apply when relativistic electron degeneracy pressure becomes important.

For $P=K{\rho }^2$, hydrostatic equilibrium gives $2K\nabla \rho =-\nabla \Phi$. The Poisson equation for Newtonian gravity therefore gives the Helmholtz equationFor spherical symmetry, the regular polytrope of index one solution isThe singular $\cos (kr)/r$ solution is excluded by finite central mass density. Taking the first zero as the stellar surface yieldsThe mass integral isDividing by the volume $4\pi R^3/3$ gives

The same polytrope of index one interior equation is ${\nabla }^2\rho +(2\pi G/K)\rho =0$. A positive separated solution with zero density on all six cube faces is the cubic polytropic interiorIts Laplacian is $-3{\pi }^2\rho /L^2$, so the required side length isWith $\Phi =C-2K\rho$ and $P=K{\rho }^2$, it satisfies the interior hydrostatic equilibrium and Poisson equation for Newtonian gravity. Integrating each sine factor yieldsThis formal interior solution is not an isolated physical cubic star. The cubic polytrope fails isolated gravitational matching: at a vertex, the product of sines has $\nabla \rho =0$, so the interior potential predicts zero gravitational acceleration. At the vertex $(0,0,0)$, however, the field generated by its own positive mass isand each component is strictly positive. There can be no continuous matching to the isolated external field without additional forces or mass sources. Thus solving the interior density equation and imposing zero face values is insufficient. Fluid stars also have no rigid structure to maintain sharp cubic faces, and observed stellar shapes are approximately spherical or rotationally flattened rather than cubic.