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The event rate counts unordered pairs of reacting nuclei. For unlike species there are $n_i{n}_j$ pairs per unit-volume normalization; for identical species there are approximately $n_i^2/2$ pairs, since counting ordered pairs counts each collision twice. The Kronecker delta therefore supplies exactly the required divisor $1+{\delta }_{ij}$. It is the event rate, not the destruction rate of identical nuclei: one identical-pair reaction destroys two nuclei.

The temperature dependence combines the high-energy tail of the Maxwell-Boltzmann distribution with quantum tunnelling through the Coulomb barrier. The principal exponential factor in the energy integral is $\exp [-E/(k_{B}T)-b/\sqrt{E}]$. Its stationary point, the Gamow peak, has $E_0\propto T^{2/3}$ and exponent $\eta =3E_0/(k_{B}T)\propto T^{-1/3}$. Integrating around this peak gives a prefactor proportional to $T^{-2/3}$, or ${\eta }^2$, when the nuclear factor varies slowly. Thus the supplied thermonuclear reaction rate is locally proportional to ${\eta }^2{e}^{-\eta }$. The slow weak interaction in the proton-proton reaction also makes its overall coefficient very small; its local thermal slope follows the supplied formula.

At fixed number densities, define the local temperature exponent of a thermonuclear reaction by a logarithmic derivative. Since $d\eta /d\log T=-\eta /3$,For proton-proton reactions, $A=1/2$ and $Z_1=1$. At $T_6=15$, ${\eta }_{11}=13.67$ and $\alpha =3.89$. For two helium-3 nuclei, $A=3/2$ and $Z_3^2{Z}_3^2=16$, giving ${\eta }_{33}=49.68$ and $\beta =15.89$. For helium-3 with helium-4, $A=12/7$ and the same charge product gives ${\eta }_{34}=51.95$ and $\gamma =16.65$. HenceThese are local powers near the specified temperature, not constant exponents over an arbitrarily large range. The greater reduced nuclear mass in the $34$ reaction increases its tunnelling exponent and its thermal sensitivity.

For the effective proton-proton reaction network, denote the event rates byThe short-lived beryllium intermediate can be eliminated, so production through $R_{34}$ feeds $n_7$, representing predominantly lithium-7. This rapid conversion is electron capture. Count the nuclei consumed and produced in each reaction: $11$ consumes two protons and makes one deuterium; $21$ consumes a proton and a deuterium and makes one helium-3; $33$ consumes two helium-3 and makes one helium-4 and two protons; $34$ consumes one of each helium isotope; $17$ consumes a proton and lithium-7 and makes two helium-4. It follows thatFor closure, ${\dot{n}}_7={\lambda }_{34}{n}_3{n}_4-{\lambda }_{17}{n}_1{n}_7$. These nuclear terms conserve $n_1+2n_2+3n_3+4n_4+7n_7$, providing an independent stoichiometric check. They describe a fixed local volume; compression or mixing would add transport terms.

The roughly one-second destruction time of deuterium is negligible compared with the proton-burning and helium-3 evolution times. Its abundance rapidly adjusts to the slowly evolving production rate. Set ${\dot{n}}_2\simeq 0$, giving the deuterium quasi-equilibrium in proton-proton burningSubstitution, without setting ${\dot{n}}_1=0$, givesThe helium-3 relaxation time of about $6\times 10^5$ years is likewise short compared with the central proton-burning time and the age of the Sun. On this relaxation time the proton and helium-4 abundances, and the background temperature, can be treated as fixed. Put $A={\lambda }_{33}$, $B={\lambda }_{34}{n}_4$, $C={\lambda }_{11}{n}_1^2/2$. Then ${\dot{n}}_3=C-A{n}_3^2-B{n}_3$, and its nonnegative equilibrium isThe other root is negative and unphysical. Write $n_3=n_{3e}+x$ and subtract the equilibrium equation. Exactly, for a fixed background,Dropping the quadratic displacement givesThus small displacements decay exponentially. A slowly moving equilibrium introduces an additional $-{\dot{n}}_{3e}$ term, negligible to leading order when $\tau$ is short compared with background evolution.

To estimate the temperature for helium-3 freeze-out, use the cooler pp-I-dominated region and neglect $B$ compared with $2A{n}_{3e}$. Then $n_{3e}\simeq n_1\sqrt{{\lambda }_{11}/(2{\lambda }_{33})}$ and ${\tau }^{-1}\simeq n_1\sqrt{2{\lambda }_{11}{\lambda }_{33}}$. With the local powers just obtained and fixed proton number density, $n_{3e}\propto T^{-6}$ and $\tau \propto T^{-10}$. Adopting $t_{\odot }\simeq 4.6\times 10^9\mathrm{yr}$ givesThis is an order-of-magnitude extrapolation of local exponents. Keeping the supplied exponential factors gives the helium-3 freeze-out estimate with Gamow factorswhich yields about $6.8\times 10^6\mathrm{K}$ with the same normalization and fixed-density approximation. Thus the robust estimate is several million kelvin. The supplied data do not fix the local density, composition or pp-II fraction away from the centre, so they cannot determine an exact transition temperature or radius. In particular, using $\tau \propto T^{-16}$ alone would omit the temperature dependence of the equilibrium helium-3 abundance.

The solar hydrogen and helium-3 abundance profiles follow from burning, finite equilibration time and convection. Central hydrogen has been consumed, so $X_1$ is lowest in the centre and increases toward its much less processed outer value. In the hot centre, rapid destruction keeps helium-3 small. Moving outward initially raises its equilibrium abundance because ${\lambda }_{11}/{\lambda }_{33}$ rises strongly as temperature falls. Further out, equilibration is no longer reached within the solar age, and still farther out even production is slow. Hence $X_3$ has an off-centre maximum and then declines toward its small outer abundance. The outer convective envelope mixes each abundance toward a constant value.

The sketch shows these qualitative shapes; the helium-3 ordinate is scaled to its own maximum. The illustrated amplitudes, peak radius and envelope boundary are schematic, not a numerical solar-model solution.