= Solution
The supplied rate law gives a local logarithmic <temperature> sensitivity, evaluated at fixed <number densities>. Since $\eta\propto T^{-1/3}$,
$$
\frac{d\log\lambda_{ij}}{d\log T}=2\frac{d\log\eta}{d\log T}-\frac{d\eta}{d\log T}=\frac{\eta-2}{3}.
$$
Thus a power law here describes the tangent logarithmic slope near the chosen <temperature>, not an exact power law over all temperatures. For the three reactions, the factors $AZ_i^2Z_j^2$ are $1/2$, $24$ and $192/7$, respectively. At $T_6=15$ the <local temperature exponent of a thermonuclear reaction> gives
$$
\begin{array}{c|c|c}
\text{reaction}&\eta&d\log r/d\log T\\\hline
11&13.67&\alpha=3.89\\
33&49.68&\beta=15.89\\
34&51.95&\gamma=16.65
\end{array}
$$
Hence \b[$\alpha\simeq4$, $\beta\simeq16$, $\gamma\simeq16$ and $\gamma>\beta$]. The larger reduced mass for reaction 34 produces the last inequality. Composition changes contribute separately to an evolving reaction rate; they are held fixed for this derivative.
For the <effective proton-proton reaction network>, let the event rates be
$$
R_{11}=\tfrac12\lambda_{11}n_1^2,\quad R_{21}=\lambda_{21}n_2n_1,\quad
R_{33}=\tfrac12\lambda_{33}n_3^2,\quad R_{34}=\lambda_{34}n_3n_4,\quad R_{17}=\lambda_{17}n_1n_7.
$$
The factors of one half count identical pairs once. Event 11 consumes two <protons> and makes one deuteron; event 21 consumes a deuteron and a <proton> and makes <helium-3>; event 33 consumes two <helium-3> nuclei and makes one <helium-4> nucleus and two <protons>. Under the stipulated fast-capture approximation, event 34 consumes <helium-3> and <helium-4> and supplies one mass-seven nucleus, and event 17 consumes that lithium-7 nucleus and a <proton> and makes two <helium-4> nuclei. Therefore
$$
\boxed{\begin{aligned}
\dot n_1&=-\lambda_{11}n_1^2-\lambda_{21}n_2n_1+\lambda_{33}n_3^2-\lambda_{17}n_1n_7,\\
\dot n_2&=\tfrac12\lambda_{11}n_1^2-\lambda_{21}n_2n_1,\\
\dot n_3&=\lambda_{21}n_2n_1-\lambda_{33}n_3^2-\lambda_{34}n_3n_4,\\
\dot n_4&=\tfrac12\lambda_{33}n_3^2-\lambda_{34}n_3n_4+2\lambda_{17}n_1n_7.
\end{aligned}}
$$
For closure, $\dot n_7=\lambda_{34}n_3n_4-\lambda_{17}n_1n_7$. The stoichiometry conserves $n_1+2n_2+3n_3+4n_4+7n_7$, the baryon <number density> at fixed volume. In particular the mass-seven capture produces two <helium-4> nuclei, not one.
The beryllium-to-lithium step is <electron capture>. Eliminating beryllium assumes that its capture flux tracks its production on the slow evolutionary timescale. More generally one retains $\dot n_{\rm Be}=R_{34}-\lambda_e n_{\rm Be}$ and $\dot n_7=\lambda_e n_{\rm Be}-R_{17}$; setting the former to zero gives the effective equation used above. Fast capture relative to slow evolution alone would not prove that lithium exceeds beryllium: if both reach steady state their ratio is $n_{\rm Be}/n_7=\lambda_{17}n_1/\lambda_e$. The mass-seven abundance assertion is thus part of the stipulated schematic limit, not a consequence to impose on every detailed solar model.
The enormous separation between the one-second <deuterium> destruction time and the other stated timescales justifies <deuterium quasi-equilibrium in proton-proton burning>:
$$
\dot n_2\simeq0,\qquad\lambda_{21}n_2n_1\simeq\tfrac12\lambda_{11}n_1^2,\qquad n_2\simeq\frac{\lambda_{11}n_1}{2\lambda_{21}}.
$$
Substitution gives
$$
\boxed{\dot n_1\simeq-\tfrac32\lambda_{11}n_1^2+\lambda_{33}n_3^2-\lambda_{17}n_1n_7,\qquad
\dot n_3\simeq\tfrac12\lambda_{11}n_1^2-\lambda_{33}n_3^2-\lambda_{34}n_3n_4.}
$$
The approximation is to the rapidly adjusting intermediate abundance, not to the slow <proton> abundance.
Near the centre the <helium-3> relaxation time is also short compared with solar age and with the timescale of significant <hydrogen> evolution. Put $A=\lambda_{33}$, $B=\lambda_{34}n_4$ and $C=\lambda_{11}n_1^2/2$. For slowly varying background quantities, the stable positive root of $An_3^2+Bn_3-C=0$ gives the <helium-3 equilibrium abundance>
$$
\boxed{n_{3e}=-\frac{\lambda_{34}n_4}{2\lambda_{33}}+\sqrt{\left(\frac{\lambda_{34}n_4}{2\lambda_{33}}\right)^2+\frac{\lambda_{11}n_1^2}{2\lambda_{33}}}.}
$$
The other root is negative and unphysical. The production-minus-destruction function decreases strictly with positive $n_3$, so this is the unique attracting equilibrium. For $n_3=n_{3e}+x$, with the background held fixed during relaxation,
$$
\dot x=-(2\lambda_{33}n_{3e}+\lambda_{34}n_4)x-\lambda_{33}x^2.
$$
Dropping the quadratic perturbation term gives the <helium-3 relaxation time>
$$
\boxed{\dot x=-\frac x\tau,\qquad\tau=(2\lambda_{33}n_{3e}+\lambda_{34}n_4)^{-1},\qquad x(t)=x(0)e^{-t/\tau}.}
$$
If the equilibrium itself evolves slowly, an additional forcing term $-\dot n_{3e}$ appears; tracking is accurate when this drift is small over one relaxation time. The given central value $6\times10^5$ years is much shorter than the roughly $4.6\times10^9$-year age of the <Sun>.
To estimate the <temperature for helium-3 freeze-out>, take the cooler pp-I-dominated regime and keep the background <proton> <mass density> approximately fixed for this order-of-magnitude scaling. Then
$$
n_{3e}\simeq n_1\sqrt{\frac{\lambda_{11}}{2\lambda_{33}}},\qquad
\tau^{-1}\simeq n_1\sqrt{2\lambda_{11}\lambda_{33}}.
$$
Using the local exponents four and sixteen, this gives \b[$n_{3e}\propto T^{-6}$ and $\tau\propto T^{-10}$], not $\tau\propto T^{-16}$: the equilibrium abundance changes with <temperature>. Normalizing at the central <temperature> gives
$$
\tau(T)\simeq6\times10^5\left(\frac{T}{1.5\times10^7\,\mathrm K}\right)^{-10}\mathrm{yr}.
$$
Setting this equal to solar age yields
$$
\boxed{T\simeq1.5\times10^7\left(\frac{6\times10^5}{4.6\times10^9}\right)^{1/10}\mathrm K\simeq6.1\times10^6\,\mathrm K.}
$$
This is the requested power-law estimate. The exponents were evaluated locally at the central <temperature>, so extrapolation over this large range is approximate. Keeping the supplied full exponential rate factors in the same fixed-density pp-I estimate gives
$$
\frac{\tau(T)}{\tau(T_c)}=\left(\frac{T}{T_c}\right)^{2/3}\exp\left[\frac{\eta_{11,c}+\eta_{33,c}}2\left(\left(\frac{T_c}{T}\right)^{1/3}-1\right)\right],
$$
which gives about $6.8\times10^6\,\mathrm K$. Thus the robust scale is \b[a few million kelvin, roughly six to seven million in these estimates]. A unique precise solar transition <temperature> cannot be obtained from the given numbers without the <mass density>, composition and pp-II contribution as functions of radius.
Finally, use mass fractions $X_i\simeq i m_un_i/\rho$. The <hydrogen mass fraction> is depleted most strongly in the central burning region, so $X_1$ increases outward and approaches the nearly unprocessed envelope value. Convective mixing makes the outer envelope composition approximately uniform.
In the hot core <helium-3> is quickly destroyed and remains close to its small equilibrium abundance. Moving outward, its destruction rates fall much faster than its production rate, so the equilibrium ratio $n_{3e}/n_1\propto T^{-6}$ rises. Where the relaxation time becomes comparable with solar age the abundance ceases to follow that rising equilibrium. Farther out, production itself becomes too slow to accumulate much <helium-3>, so $X_3$ falls again toward the envelope value. The result is a broad off-centre maximum, rather than a central maximum. These are the <solar hydrogen and helium-3 abundance profiles> requested by the sketch.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-53-solar-abundances.png]
{title=Schematic present solar hydrogen and helium-3 radial abundance profiles, showing central hydrogen depletion and the off-centre helium-3 maximum with separate normalizations}
{height=600}
\b[<hydrogen> rises from a depleted centre to an almost uniform envelope; <helium-3> has a small central abundance and an off-centre peak before declining outward.] The figure shows qualitative shapes with independently normalized vertical axes, not a fitted or computed solar model.
Back to article page
