Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 70 / 3

For an isotropic degenerate electron gas, pressure is one third of the momentum flux. In the ultrarelativistic limit each occupied mode has speed $c$, so the supplied momentum-space state count givesA fully ionized helium nucleus provides two electrons for approximately $4m_p$ of mass, so $n_e=\rho /(2m_p)$. Eliminating the Fermi momentum $p_0=(3h^3\rho /(16\pi m_p){)}^{1/3}$ therefore givesThe two-spin-state degeneracy is already included in the supplied state count; no further factor 2 is needed.

If electron degeneracy pressure supplies the star's pressure, this is an index-three stellar polytrope. Its central pressure and density obeyThe radius cancels, leaving $M^{2/3}=K(3a_3/(4\pi ){)}^{4/3}/(c_{3}G)$. Thus the Chandrasekhar mass from central polytropic constants isThe dependence on microscopic constants reflects the balance of fermionic pressure and gravity. The absence of radius dependence is special to the limiting $4/3$ pressure exponent. Ordinary partially relativistic white dwarfs can have a range of masses below this Chandrasekhar limit; the formula does not force every white dwarf to have this mass. As a degenerate stellar core grows toward the limiting mass, compression cannot provide an arbitrarily larger supporting pressure relative to gravity. Above it the ideal cold, nonrotating electron-supported configuration cannot remain in equilibrium; nuclear ignition, collapse or other changes in the support mechanism must intervene. Which outcome occurs depends on the core's composition and evolutionary state.

Outside the thin burning shell, the enclosed mass and luminosity are approximately $M_c$ and $L$. Divide the hydrostatic pressure support equation by radiative diffusion in a star with constant opacity:In the constant-opacity radiative-zero giant envelope approximation, the outer pressure and temperature constants are negligible. Integration givesThe radiation pressure is $a{T}^4/3$, so the gas pressure fraction $\beta$ is constant and satisfies $(1-\beta )\Lambda =a/3$. Thus alsoThe physical branch has positive gas pressure, hence $L<4\pi cG{M}_c/{\kappa }_0$, the constant-opacity Eddington luminosity for the core mass.

Using the gas part of the equation of state, define $H=\mu \beta \Lambda /\mathcal{R}$, so $\rho =H{T}^3$. Substitution into hydrostatic balance givesConsequently, in the same deep-envelope radiative-zero approximation,Precisely, a finite outer matching point $(R_s,T_s)$ gives $T=T_s+\mu \beta G{M}_c(1/r-1/R_s)/(4\mathcal{R})$. The displayed $1/r$ profile neglects that additive outer-boundary constant; it is appropriate near a compact core when the envelope extends to much smaller temperatures.

To estimate the thin-shell luminosity, write $K_T=\mu \beta G{M}_c/(4\mathcal{R})$, so $T=K_T/r$. The hydrogen-layer heating law givesThis steep radial dependence strongly concentrates the emission near $R_c$. Continuing the outer profile across the thin emitting layer and taking a distant upper limit givesThe same radiative coefficient implies $H=16\pi ac{K}_T/(3{\kappa }_{0}L)$. Hence the high-power shell burning in a constant-opacity giant envelope satisfiesTaking the cube root gives the requested form,Thus $C$ contains only the specified opacity and heating normalizations and physical constants. The explicit coefficient uses the outer radiative profile as a thin-shell estimate; resolving the variation of $L_r$ within the emitting layer can change that coefficient. Also $\beta$ depends on $L$, so the radiation-pressure-inclusive relation is implicit, not a pure explicit eighth-power core-mass law with independently fixed $\beta$.