Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 66 (Structure and Evolution of Stars)

Use radiative homology with proton-proton burning and inverse-fourth-power opacity, retaining the composition factors throughout. To distinguish the stellar radius from the stellar gas constant, write them as $R$ and $\mathcal{R}$. With gas pressure alone, introduce the scale factorsThe ideal gas equation gives $\rho ={\rho }_{\ast }b$, $P=P_{\ast }p$ and $T=T_{\ast }p/b$. Substitution into hydrostatic equilibrium and mass conservation givesso $p^{\prime }=-bq/x^2$ and $q^{\prime }=x^{2}b$.

Set $\zeta =Z(1+X)$. The radiative diffusion in a star equation, with the stipulated opacity, becomesHere ${\rho }^2{T}^{-7}={\rho }_{\ast }^2{T}_{\ast }^{-7}{b}^9{p}^{-7}$. ThusThe luminosity equation follows from the specific heating rate of the Proton–proton chain:ConsequentlyThese give all four dimensionless stellar structure equations with their numerical coefficients.

The stellar homology assumption means the dimensionless profiles are shared across the models, not merely that an individual model can be written in dimensionless coordinates. The displayed equations then require $D$ and $E$ to be fixed. Holding the universal constants and ${\kappa }_0,{ϵ}_0$ fixed givesEquating the two cancels $M^6$ and yields $R^6\propto X^2\zeta {\mu }^{-4}$. Therefore the radius and mass-luminosity relation areThe mass-independent radius is a consequence of the particular heating and opacity exponents in this model; it is not a general radius law for main-sequence stars.

Initially $X=X_0$ is the same for all stars. The adopted mean molecular weight $\mu =4/(5X+3)$ also has the same initial value, neglecting its small direct dependence on $Z$. Thus $R\propto Z^{1/6}$. At fixed luminosity, the Stefan–Boltzmann law $L=4\pi R^2\sigma T_e^4$ givesThe stars compared here can have different masses; the luminosity law adjusts the mass to keep $L$ fixed.

For the evolution, take the stellar mass as constant and maintain homogeneous composition. The available energy in the remaining hydrogen is $M{E}_{H}X$, so conservation of energy in the nuclear-powered model givesThis includes the entire mixed hydrogen reservoir, not just the mass in a central burning region.

To determine the turnoff mass of a homogeneously mixed proton-proton-burning population, write the luminosity law as $L=C{M}^6{Z}^{-7/6}H(X)$, whereSeparating variables, the time to evolve from $X_0$ to $f{X}_0$ isFor a common $X_0$ and $0\le f<1$, this integral is a positive finite constant independent of $M,Z,t$. It follows that the turning-off stars satisfyThe integration retains the evolving luminosity and mean molecular weight; replacing $L$ by its initial value is unnecessary.

For the Schwarzschild criterion, displace a small gas parcel upwards. Assume its composition is unchanged, it remains in pressure balance with its surroundings, and it moves quickly enough that heat exchange is negligible. The parcel obeys the adiabatic equation of state $P{\rho }^{-\gamma }=\mathrm{constant}$. Combining this with the ideal gas equation gives $T\propto P^{(\gamma -1)/\gamma }$, so its adiabatic temperature gradient is ${\nabla }_{\mathrm{ad}}=(\gamma -1)/\gamma =2/5$.

Let the surrounding temperature gradient be $\nabla =d\log T/d\log P$. For an upward displacement $\delta \log P<0$, the parcel's temperature relative to its new surroundings isIf $\nabla <{\nabla }_{\mathrm{ad}}$, this is negative: the parcel is cooler and, at equal pressure and composition, denser than its surroundings. It therefore sinks back; a downward displacement likewise gives a restoring buoyancy force. If the inequality is reversed, the displaced parcel continues to rise. The stability condition is thusEquality is the neutral boundary. There is no composition-gradient term because the ideal-gas parcel and the background have the same mean molecular weight.

In the thin upper stellar atmosphere, take a plane-parallel approximation with constant $g$, optical depth increasing inward, and zero overlying pressure $P(0)=0$. Hydrostatic equilibrium in optical depth gives $dP/d\tau =g/\kappa$. The ideal gas equation converts the stipulated opacity intoDefine $u=1+3\tau /2$, so the given grey atmosphere relation is $T^4=T_e^{4}u/2$. Integrating from the surface givesThis proves the required pressure law. In particular $P^2\propto g{T}_e^{-4\beta }$ times a function only of $u$ and $\beta$.

For the grey-atmosphere convection onset with density-linear opacity, differentiate the two profiles to obtain the radiative temperature gradient:It increases monotonically from zero, since $\beta >1$, and first reaches the adiabatic temperature gradient at $u_b^{\beta -1}=(4\beta +1)/5$. ThereforeBelow this point the radiative stratification is unstable and convection must transport some of the outward energy flux.

At the outer convection boundary, $u_b$ depends only on $\beta$, so $T_b=T_e(u_b/2{)}^{1/4}\propto T_e$ and $P_b\propto g^{1/2}{T}_e^{-2\beta }$ for fixed $\mu ,{\kappa }_0,\beta$. The matching constant in $P=K{T}^{5/2}$ is $K=P_b/T_b^{5/2}$. ConsequentlyThis temperature form of the polytropic relation follows from an ideal monatomic adiabat; its $K$ is not the density-polytrope coefficient in Question 3.

For the deep radiative boundary beneath a power-law-opacity convective envelope, negligible envelope mass and absent local energy generation give $m\simeq M$ and $L_r\simeq L$. Divide the radiative temperature equation by the pressure equation to get the stellar radiative temperature gradientThis is the gradient that would be needed if radiation alone carried $L$, and it is the quantity to compare with $2/5$ even while the actual region is convective. The deeper Kramers' opacity law gives $\kappa ={\kappa }_1\mu P{T}^{-9/2}/\mathcal{R}$. Substituting $P=K{T}^{5/2}$ yieldsAs $T$ rises inward this decreases, so the transition back to a stable radiative interior is atThus $T_{\mathrm{in}}^{7/2}\propto K^{2}L/M$. Using the outer matching and the Stefan–Boltzmann law,The mass and radius cancel. Hence, at fixed opacity coefficients and composition,This uses the same adiabat through the convection zone and distinguishes its inner transition temperature from its outer onset temperature.

For a regular, spherical star, hydrostatic equilibrium gives $dP/dr=-Gm(r)\rho (r)/r^2$. Multiply by $4\pi r^3$ and integrate to the surface. Integration by parts givesThe central boundary term vanishes for finite central pressure. Since $dm=4\pi r^2\rho dr$, the right-hand side of the integrated hydrostatic equation isThe last expression is the gravitational energy: adding a thin mass shell $dm$ to the interior mass $m$ contributes $-Gmdm/r$, counting each interacting pair once. HenceWith the specified zero surface pressure this proves the stellar virial theorem in the required form,

To derive the nonrelativistic electron-degeneracy pressure in a hydrogen-helium mixture, take fully ionized hydrogen and helium with hydrogen mass fraction $X$, the electron number density isHere helium nuclei have mass $4m_p$ in the approximation used in the question. Filling the two spin states of each momentum mode up to the Fermi momentum $p_0$ givesThe isotropic momentum flux gives pressure $\frac{1}{3}\int pvdn$. For nonrelativistic electrons $v=p/m_e$, so the electron degeneracy pressure isSubstitute the composition-dependent $n_e$ and simplify powers of two:Comparing $5/3=1+1/n$ gives the polytropic index $n=3/2$. The result assumes complete electron degeneracy, $p_0\ll m_{e}c$, and negligible nuclear mass corrections.

Now adopt the stated approximate partially degenerate equation of state, keeping composition and hence $K,\mu$ fixed during contraction. Set $q=m/M$ and define the mass-averaged temperature ${\overline{T}}_M={\int }_0^{1}Tdq$. The stellar virial theorem from part (a) givesFor an index-$3/2$ stellar polytrope, the permitted gravitational-energy formula is $\Omega =-6G{M}^2/(7R)$. Also, using $\bar{\rho }=3M/(4\pi R^3)$,Thus, with $A=3K{\int }_0^1[3\rho /(4\pi \bar{\rho }){]}^{2/3}dq$,The normalized density $\rho /\bar{\rho }$ is a fixed function of $q$ for this homologous polytropic structure. Therefore its integral and $K$ fix $A$, which is independent of $M$ and $R$ for a common composition. The mean here is mass-weighted, rather than a volume average.

For the mass-averaged temperature maximum in a partially degenerate polytrope, put $\alpha =2\mu GM/(7\mathcal{R})$ and $\delta =\mu A{M}^{2/3}/(3\mathcal{R})$, both positive. As a function of $s=1/R$ the temperature is the concave quadraticIt reaches its maximum when $R=2\delta /\alpha =7A{M}^{-1/3}/(3G)$, withAt large radius, contraction increases the temperature as in a thermally supported star. Once electron degeneracy provides a sufficiently large fraction of the support, further contraction can lower the thermal temperature. The formal zero-temperature radius is $R_{\mathrm{deg}}=\delta /\alpha =R_{\max }/2$, with $R_{\mathrm{deg}}\propto M^{-1/3}$. The continuation to negative temperatures would be unphysical; the adopted sequence only describes its positive-temperature part.

The contraction-limited central temperature has the same mass scaling. Indeed, if the total polytropic pressure is $P=K_{\mathrm{tot}}{\rho }^{5/3}$, the approximate equation of state gives $T=(\mu /\mathcal{R})(K_{\mathrm{tot}}-K){\rho }^{2/3}$. The fixed density profile therefore fixes $T_c/{\overline{T}}_M$. A sufficiently small mass never reaches the central temperature required for sustained hydrogen burning, producing a hydrogen-burning minimum mass. Such an object becomes a cooling, degeneracy-supported brown dwarf instead of settling onto the main sequence. More massive contracting objects can ignite hydrogen before reaching this degeneracy-limited maximum. This conclusion uses the nonrelativistic, fixed-composition model; it does not require extrapolation into the relativistic regime.

Treat the stars as point masses for the orbital motion, neglecting their spin angular momenta. Let $M=M_1+M_2$. Their distances from the centre of mass are $a_1=a{M}_2/M$ and $a_2=a{M}_1/M$. The sum of their orbital angular momenta isThus the circular-binary orbital angular momentum is

For conservative binary mass transfer, both $M$ and $J$ are fixed. In terms of the binary mass ratio $q=M_1/M_2$, $M_1=Mq/(1+q)$ and $M_2=M/(1+q)$, henceMass conservation also gives $d{M}_2=-d{M}_1$, soDifferentiating $J\propto M_1{M}_2{a}^{1/2}$ at fixed $M,J$ gives $d\log a/d\log M_1=2(q-1)$. Therefore the stipulated Roche lobe law has the Roche-lobe radius response exponent

Define the overfill by $\Delta =\log (R_1/R_L)$. On a time scale short compared with the slow change of $K$, the donor's response is $d\log R_1=\beta d\log M_1$. ThusDuring mass loss $d\log M_1<0$. If ${\zeta }_L>\beta$, mass loss increases the overfill and therefore drives more mass loss: this is positive feedback, giving more rapid Roche-lobe overflow. If ${\zeta }_L<\beta$, mass loss decreases the overfill and can balance the slow expansion due to stellar evolution. More explicitly, with $k_{\mathrm{ev}}=d\log K/dt>0$, contact requiresA slow negative transfer rate can satisfy this only on the stable side $\beta >{\zeta }_L$. The rapid-transfer condition is thereforeIts time scale depends on the physical donor response represented by $\beta$; it need not be a dynamical time scale if $\beta$ is a thermal-equilibrium mass-radius exponent.

For Roche-lobe overflow feedback with a constant wind torque factor, retain mass conservation to the stated approximation but allow the prescribed angular-momentum loss. Since $d\log J=fd\log M_1$, differentiation givesThe new Roche-lobe radius response exponent isThe same overfill calculation givesA loss of angular momentum has $f>0$ because ${\dot{M}}_1<0$, and it lowers the critical ratio by making the orbit contract more strongly.

For the specified solar-mass donor exponent $\beta =1/2$, the answer isWithout the wind torque this becomes $q>22/23\simeq 0.957$, so donors slightly less massive than their companions can already lie on the rapid-transfer side of this model. For $0<f<11/10$ the lower threshold is $(22-20f)/23$, and for $f\ge 11/10$ every positive $q$ satisfies the strict inequality. There is no finite upper bound on $q$ from this criterion. Since the paper does not specify $f$, the wind case has a conditional range rather than a unique numerical cutoff.