Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 317 (Structure and Evolution of Stars)

At the stellar surface boundary condition, match the interior to a stellar atmosphere, rather than setting the temperature and pressure to zero at an arbitrary radius. At a photosphere of radius $R$, the enclosed mass is $M$, the luminosity is $L$, and the outward radiative flux defines the effective temperature through $F=L/(4\pi R^2)={\sigma }_{\mathrm{SB}}{T}_{\mathrm{eff}}^4$. With no external illumination the incoming specific intensity satisfies $I_{\nu }(0,\mu )=0$ for $\mu <0$. A deeper boundary is matched to the nearly isotropic radiative diffusion field. A grey atmosphere commonly matches near $\tau =2/3$; under the Eddington surface boundary condition this gives $T=T_{\mathrm{eff}}$. For roughly constant gravity and opacity, the total pressure rises by $g\tau /\kappa$ relative to its outer boundary value; if radiation supplies appreciable support, the gas-pressure gradient uses the effective gravity $g-\kappa F/c$ instead.

Take $z$ increasing outward and $\mu$ the outward direction cosine. If $j_{\nu }$ is emission per unit path length and solid angle, the radiative transfer equation isDefine inward-increasing optical depth by $d{\tau }_{\nu }=-\rho {\kappa }_{\nu }dz$ and the source function by $S_{\nu }=j_{\nu }/(\rho {\kappa }_{\nu })$. Then $\mu {\partial }_{{\tau }_{\nu }}{I}_{\nu }=I_{\nu }-S_{\nu }$. This states the emission-coefficient convention; if $j_{\nu }$ is defined per unit mass instead, the emission term is $\rho j_{\nu }$.

For frequency-integrated intensity, introduce the radiation-field momentsThe first angular approximation writes $I\simeq J+3\mu H$. Its odd part does not contribute to $K$, soThis is the Eddington closure approximation, not an exact description of the escaping angular distribution. An approximate surface condition takes the outgoing hemisphere to have direction-independent intensity $I_0$ and the incoming hemisphere to be dark. Its moments give $F=\pi I_0$ and $J(0)=I_0/2=F/(2\pi )$. Integrating the moment equations with the closure then gives $J=3F(\tau +2/3)/(4\pi )$; in thermal equilibrium, $J=a_{\mathrm{rad}}c{T}^4/(4\pi )$ gives $T^4=(3/4)T_{\mathrm{eff}}^4(\tau +2/3)$. This derives the usual approximate photospheric matching condition rather than imposing a zero surface temperature. For conservative coherent isotropic scattering, $S=J$. Integrating the transfer equation gives $dH/d\tau =0$ and $dK/d\tau =H$, the moment conditions for radiative equilibrium.

The exponential ansatz in the PDF has a genuine consistency problem. Compute its moments before making any approximation. For a nonsingular real intensity on all $-1\le \mu \le 1$, take $|ϵ|<1$ and defineThe continuous zero limits are $A_0=1$, $A_1=0$, $A_2=1/3$. Direct integration yieldsFor a decaying nontrivial correction, $\delta >0$ and $\gamma \ne 0$. Exact constant flux then requires $A_1=0$. But $\mathrm{atanh}ϵ/ϵ>1$ for every nonzero real $|ϵ|<1$, so $A_1=0$ forces $ϵ=0$. Thus exact radiative equilibrium cannot imply a small nonzero $ϵ$ for this ansatz.

Substitution into the full transfer equation makes the obstruction sharper. The exponential coefficient must satisfy, for every $\mu$,Comparing the constant and linear coefficients gives $A_0=1$ and $\delta =A_0ϵ=ϵ$. Hence regular real coefficients require $\delta =ϵ=0$, eliminating any decaying correction. This is the obstruction to a single smooth exponential mode in conservative grey transfer. There is no exact nontrivial real solution of the stipulated form with a small nonzero decay parameter. Choosing, for example, $ϵ=\delta =0.1$ and $\gamma \ne 0$ supplies an explicit counterexample to the claimed exact flux constancy, because $A_1\ne 0$.

Several requested conclusions remain useful as formal approximations, provided they are identified as such. If $\delta >0$, the large-depth exponential vanishes, giving $F\to 4\pi \alpha /3$, hence $\alpha =3F/(4\pi )$ for the asymptotic physical flux. It also gives $c{P}_{\mathrm{rad}}\to (4\pi /3)J$. Since $A_0=1+{ϵ}^2/3+O({ϵ}^4)$, neglecting second-order angular errors yields the formal relation $\delta \simeq ϵ$. However $A_1=-ϵ/3+O({ϵ}^3)$, so the finite-depth flux still varies at first order. The formal relation is not an exact transfer or constant-flux solution.

There is also a flux-normalization mismatch in the PDF's last displayed relation. With the same physical $F$ used above, the Hopf function of a grey atmosphere is defined byThe PDF's coefficient $3F/4$ would require redefining its $F$ as physical flux divided by $\pi$; it cannot simultaneously use the earlier $\alpha =3F/(4\pi )$ for the same physical flux. With the consistent physical-flux normalization, formal moment matching givesIn the intended small-$ϵ$ approximation this isThe approximate outgoing surface ansatz then gives the limb darkening ratioFor finite $ϵ$, $\gamma =-0.14/A_0$ gives a parameter-dependent ratio, not a unique exact value. The two Hopf endpoint values alone do not determine the exact emergent intensity: the formal solution of the radiative transfer equation uses the entire source profile, $I(0,\mu )={\int }_0^{\infty }J(t)e^{-t/\mu }dt/\mu$ for $\mu >0$. Nor does the simple linear-plus-exponential ansatz satisfy the exact no-incoming-radiation condition for every negative $\mu$. Thus $2.75$ is a clearly labelled approximate disk-centre-to-limb result, not an exact consequence of inconsistent premises.

For a quasistatic spherical star with fixed total mass, the stellar structure equations in radius areHere $s$ is the specific entropy of a moving mass shell, and the entropy term accounts for release or storage of thermal and gravitational energy. In thermal equilibrium it vanishes. An equation of state, opacity, energy-generation law and composition evolution close the equations. The absence of mass loss fixes the outer mass coordinate; it does not require the star to be in thermal equilibrium.

The radiative temperature gradient needed to carry the full luminosity isHere the actual temperature gradient is $\nabla =d\log T/d\log P$. For uniform composition the Schwarzschild criterion says that a radiative layer is stable if ${\nabla }_{\mathrm{rad}}<{\nabla }_{\mathrm{ad}}$, with marginal stability at equality, where the adiabatic temperature gradient is $(\partial \log T/\partial \log P{)}_s$. Use $\nabla ={\nabla }_{\mathrm{rad}}$ there. If ${\nabla }_{\mathrm{rad}}>{\nabla }_{\mathrm{ad}}$, convection carries some of the flux. Efficient convection gives $\nabla \simeq {\nabla }_{\mathrm{ad}}$; inefficient surface convection needs a transport prescription and can be superadiabatic. For composition gradients, use the Ledoux criterion, with threshold ${\nabla }_{\mathrm{ad}}+\varphi {\nabla }_{{\mu }_{\mathrm{mol}}}/{\delta }_{\mathrm{th}}$ instead. Here ${\nabla }_{{\mu }_{\mathrm{mol}}}=d\log {\mu }_{\mathrm{mol}}/d\log P$, $\varphi =(\partial \log \rho /\partial \log {\mu }_{\mathrm{mol}}{)}_{P,T}$ and ${\delta }_{\mathrm{th}}=-(\partial \log \rho /\partial \log T{)}_{P,{\mu }_{\mathrm{mol}}}$, using the mean molecular weight ${\mu }_{\mathrm{mol}}$.

Above the thin burning shell, take constant luminosity $L$, constant mean molecular weight ${\mu }_{\mathrm{mol}}$, ideal-gas pressure $P=\mathcal{R}T\rho$ with $\mathcal{R}=k_B/({\mu }_{\mathrm{mol}}{m}_u)$, and the Kramers' opacity law $\kappa ={\kappa }_0\rho T^{-7/2}$. Then the temperature equation isTo implement the printed assumption that $P,\rho$ and $M_r$ are all power laws while retaining mass conservation, write $M_r\propto r^m$, $\rho \propto r^b$, $P\propto r^p$ and $T\propto r^d$. The mass equation gives $m=b+3$; the ideal gas law gives $d=p-b$; hydrostatic equilibrium gives $d=m-1$. The diffusion equation gives $d-1=2b-2-13d/2$. Solving this linear system gives the self-gravitating Kramers power-law envelope:The coefficients can also be matched consistently. If $r_c$ is the core boundary and $M_c$ its enclosed mass,Radiative transfer fixes the constant luminosity throughwhose right-hand side is independent of radius for these exponents. Thus this is a solution of all four envelope equations, not just a dimensional estimate.

Matching the boundary temperature to the isothermal core and extrapolating the idealized power law to the specified photosphere givesFor reference $M_R/M_c=(R/r_c{)}^{1/11}=2500^{1/10}\simeq 2.19$, so the envelope mass is not negligible in this particular solution. The given $M_c\simeq M_{\odot }$ fixes the core-radius normalization $r_c=(11/42)G{M}_c/(\mathcal{R}10^7\mathrm{K})$; a numerical absolute radius needs the unspecified molecular weight. The exponent ratio also gives $\nabla =d/p=5/21<2/5$, so the fully ionized monatomic version of this idealization is convectively stable.

A different common approximation neglects the envelope's self-gravity and sets $M_r\simeq M_c$ in the force equation. It cannot obey the exact mass equation with a nonzero density and exactly constant $M_r$. Under that additional approximation, the same calculation instead givesThis is the constant-core-mass power-law opacity radiative envelope limit, not the full power-law solution with changing enclosed mass. Stating which approximation is used resolves the otherwise different numerical radius ratios.

A compact nearly isothermal core, a thin luminosity-producing shell and a much larger cool envelope describe a shell-burning red giant. The full power-law profile is a deliberately simple model. The Kramers' opacity law and an ionized ideal gas are not reliable at a $4000\mathrm{K}$ photosphere; partial ionization, other opacity sources and a convective envelope usually modify real cool giants. The quoted radius ratios are extrapolations within the stipulated model.

Assume a stationary spherical outflow with speed $v>0$, mass-loss rate $\dot{M}>0$, an optically thick thermal radiation field and negligible radiation inertia compared with matter inertia. The stellar wind continuity equation givesUse the printed gas fraction $\beta =P_{\mathrm{gas}}/P$, with $P=P_{\mathrm{gas}}+P_{\mathrm{rad}}$, and assume constant mean molecular weight ${\mu }_{\mathrm{mol}}$. Put $\mathcal{R}=k_B/({\mu }_{\mathrm{mol}}{m}_u)$ and $a_g^2=\mathcal{R}T$ for the isothermal sound speed squared. Then $P_{\mathrm{gas}}=\rho a_g^2$ and $P_{\mathrm{rad}}=a_{\mathrm{rad}}{T}^4/3$. The momentum equation including the radiation-pressure gradient isDo not also add a separate radiative force here: the diffusion radiation-pressure gradient already represents that force.

Using radiative diffusion in a star,Write $g=G{M}_r/r^2$ and $\Gamma =L_r/L_{\mathrm{crit}}=\kappa L_r/(4\pi cG{M}_r)$, the local ratio to the Eddington luminosity. Substituting continuity into the gas-pressure gradient gives the gas and radiation pressure stellar wind equationThe denominator changes sign at the sonic point $v=a_g$. A smooth transonic branch must have its right-hand numerator vanish there too. In a subsonic flow the denominator is negative, so acceleration requires a negative numerator; for a supersonic flow it requires a positive numerator. A regular subsonic-to-supersonic branch must therefore pass through a compatible common zero, with a finite slope obtained by differentiating the equations. Boundary conditions and that regularity requirement normally select the mass flux. Other branches can remain subsonic, remain supersonic or encounter a singular gradient. In the isothermal nonradiating limit the numerator is $2a_g^2/r-GM/r^2$, recovering the Parker wind equation and critical radius $GM/(2a_g^2)$.

To expose the role of $\beta$, rewrite the diffusion temperature gradient asHenceAt the stipulated local value $\Gamma =1$, with $0<\beta <1$, its numerator isThe gas-pressure contribution is not removed by setting the luminosity equal to the Eddington value. Instead the negative temperature gradient makes the thermal term in the numerator positive as well. A subsonic outflow then has $dv/dr<0$, and no common numerator/denominator zero is available for smooth sonic passage. This is the sonic-point obstruction at the Eddington luminosity.

Thus a subsonically launched wind cannot accelerate smoothly to supersonic speed under these diffusion assumptions at $L_r/L_{\mathrm{crit}}=1$. It is not a claim that every initially supersonic solution decelerates; those have positive denominator and accelerate in this equation. Even in the radiation-dominated limit $\beta \to 0$, the spherical thermal term remains positive for $a_g^2>0$. The singular pressureless limit is not a finite-temperature sonic crossing. Finally $a_g$ is the thermal sound scale of this diffusion-controlled stationary equation, not the full adiabatic sound speed of a tightly trapped gas-plus-radiation perturbation. Optically thin winds or different heating/driving mechanisms require another transport model.

Let $f(\mathbf{p})d^{3}p$ be particle number per unit volume in a momentum element, including the spin-state count in $f$. This is the local momentum part of a phase-space distribution function, normalized by the number density $n=\int f(\mathbf{p})d^{3}p$. For an isotropic distribution with particle energy $E(p)=\sqrt{m^2{c}^4+p^2{c}^2}$, the kinetic pressure of an isotropic gas is the average momentum flux:The kinetic energy density, explicitly excluding rest mass, isFor $p\ll mc$, $E-m{c}^2=p^2/(2m)$ and $\mathbf{v}=\mathbf{p}/m$, giving $P=2u_{\mathrm{kin}}/3$. For $p\gg mc$, $E\simeq pc$ and $v\simeq c$, giving $P\simeq u_{\mathrm{kin}}/3$. ThusNeither relation assumes a Maxwellian distribution. At intermediate momenta neither constant ratio is exact, and anisotropic distributions require a pressure tensor rather than this scalar pressure.

For a classical Maxwell-Boltzmann distribution, $f=n(2\pi m{k}_{B}T{)}^{-3/2}\exp [-p^2/(2m{k}_{B}T)]$. Three Gaussian component integrals give $⟨p^2⟩=3m{k}_{B}T$, so the ideal gas hasIn a fully ionized mixture, sum over independent species to obtain $P_{\mathrm{gas}}=\rho k_{B}T/({\mu }_{\mathrm{mol}}{m}_u)$, where the mean molecular weight counts ions and free electrons. This excludes partial-ionization and interaction corrections. A dilute classical ultrarelativistic gas still has $P=n{k}_{B}T$ but $u_{\mathrm{kin}}\simeq 3n{k}_{B}T$; its momentum distribution is proportional to $e^{-pc/(k_{B}T)}$ instead of the nonrelativistic Gaussian.

For a thermal photon gas, the two polarization states and zero photon chemical potential give the Planck photon distributionUse $x=pc/(k_{B}T)$ and ${\int }_0^{\infty }{x}^3/(e^x-1)dx={\pi }^4/15$. The radiation constant and photon equation of state are thereforeLikewise $n_{\gamma }=2\zeta (3)(k_{B}T/\hslash c{)}^3/{\pi }^2$. Unlike a gas with a fixed particle number, photons do not have pressure proportional to baryonic mass density.

For fully degenerate electrons, the Fermi-Dirac distribution becomes a filled momentum sphere with two spin states. Counting them gives the Fermi momentumHere ${\mu }_e$ is the mean molecular weight per electron. The equation of state of a cold electron gas follows by integrating momentum flux up to $p_F$:Its two limits areThe corresponding kinetic energy densities are $3n_e[p_F^2/(2m_e)]/5$ and $3n_e{p}_{F}c/4$, respectively. Thus in the high-density relativistic limit the electron pressure scales as ${\rho }^{4/3}$ and is nearly independent of temperature. This is the ideal noninteracting, fixed-composition electron result, not a universal equation of state at nuclear densities where captures, interactions and the composition change.

The boundaries on a stellar equation-of-state regime diagram concern the electron component. Define the electron relativistic density threshold and thermal electron relativistic threshold byA degenerate electron gas is nonrelativistic well below ${\rho }_{\ast }$ and ultrarelativistic well above it: this is an approximately vertical division on a log-density plot. A nondegenerate electron gas instead becomes thermally relativistic near $T_{\ast }$: this is an approximately horizontal division. The regimes have broad crossovers rather than a discontinuity at either line.

The exact electron Fermi temperature, subtracting electron rest energy, isFor electrons with negligible thermal pairs, strong degeneracy requires $T\ll T_F$, whereas $T\gg T_F$ gives a nondegenerate gas. In the pair-rich regime the actual electron and Positron distributions must instead be treated with their chemical potentials, as discussed below. The two limiting degeneracy boundaries areThese slopes $2/3$ and $1/3$ explain the bent degeneracy boundary on the logarithmic graph. A thermal-wavelength test $n_e[2\pi {\hslash }^2/(m_e{k}_{B}T){]}^{3/2}\sim 2$ gives the same nonrelativistic density/temperature scaling, with an order-one definition of the crossover.

The radiation-to-gas pressure boundary in the nondegenerate fully ionized regime isRadiation dominates above this line. Once electrons are strongly degenerate, compare radiation pressure with $P_e(\rho )$ instead of continuing the ideal-gas comparison into that regime. The radiation-to-degeneracy pressure boundary is $T=[3P_e(\rho )/a_{\mathrm{rad}}{]}^{1/4}$, with logarithmic slopes $5/12$ and $1/3$ in the nonrelativistic and ultrarelativistic limits. Degeneracy of the electrons and dominance of their pressure are distinct criteria.

For the pair curve, distinguish baryonic net electrons from thermally created electrons and Positrons. Chemical equilibrium with photons requires opposite electron/Positron chemical potentials when the one-particle energy includes rest energy, as in the distributions below. At low temperature and low degeneracy the zero-chemical-potential density per charge species isCharge neutrality gives $n_{-}-n_{+}=n_b=\rho /({\mu }_e{m}_u)$, while the nondegenerate equilibrium product is $n_{-}{n}_{+}=n_0^2$. HencePairs become important when $n_0$ is comparable with or exceeds $n_b$, approximately the electron-positron thermal pair abundance curve ${\rho }_{\mathrm{pair}}(T)={\mu }_e{m}_u{n}_0(T)$. Below its density at a fixed temperature, pairs dominate over the net charge electrons. The exponential makes the low-temperature portion steep in a log-log plot. For ${\mu }_e=2$, the full zero-potential integral gives approximate pair-marker densities $1.80\times 10^3\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ at $10^9\mathrm{K}$ and $7.52\times 10^5\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ at $3\times 10^9\mathrm{K}$. At relativistic temperature use the full zero-chemical-potential Fermi-Dirac distribution instead; it gives $n_0\to [3\zeta (3)/(2{\pi }^2)](k_{B}T/\hslash c{)}^3$, so the high-temperature pair curve approaches slope three in log-density versus log-temperature. Strong net-electron degeneracy suppresses Positrons and requires the full chemical-potential-dependent distribution.

The original diagram uses fully ionized helium, ${\mu }_e=2$ and ${\mu }_{\mathrm{mol}}=4/3$, only to set numerical locations. It plots the full $T_F$ curve and the zero-chemical-potential pair integral, rather than extending their asymptotes into the crossover. The pair boundary is deliberately an approximate abundance marker, not a phase transition; at high temperature the nonrelativistic ion approximation and fixed-composition model also have limits. For $T\gg T_{\ast }$ with abundant nondegenerate pairs, the combined electron/Positron energy density is $7a_{\mathrm{rad}}{T}^4/4$, so photons plus pairs have $u=11a_{\mathrm{rad}}{T}^4/4$ and $P=u/3$. Real stellar matter adds Coulomb effects, partial ionization, nuclear reactions and, at sufficiently high density, nuclear-matter physics beyond the ideal regime map.