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Let the closed box have ceiling height $H$ above the source. The printed width $2R$ alone does not specify this height, so it must appear as an additional geometrical parameter. Take $t=0$ here when the plume has first reached the ceiling and its thin outflow has spread across the box. The leading filling box model assumes a negligible source volume, insulated boundaries, fast horizontal spreading, slow evolution compared with the plume transit time, negligible diffusion, and $b\ll R$. Below the interface the ambient still has its original temperature, so the local plume flux remains $q(z)=2\alpha w_{\ast }z$ from part (c).

The ceiling outflow creates a warm upper region. Fluid entrainment draws the original lower fluid into the plume, making that upper region deepen. The filling-box first front is the lower material boundary of fluid already modified by plume discharge, separating it from the initially unmodified ambient; it need not bound a uniformly mixed upper layer. With negligible occupied plume area, volume conservation at its height $h(t)$ givesThereforeIf source switch-on is the time origin instead, replace $t$ by elapsed time after the initial transit and spreading stage; the quasi-steady model does not resolve that stage.

The buoyancy jump requires distinguishing ambient fluid at the front from the plume crossing it. Just after the first ceiling outflow, the ambient buoyancy immediately above the nascent front isOutside the narrow plume, ambient temperature is transported without diffusion. If $G_a$ is its buoyancy relative to the original fluid, its equation isThe first front follows precisely these material trajectories, so the temperature of the fluid immediately above it is conserved. The initially unmodified fluid below has $G_a=0$. Thus the ambient jump isBy contrast, the contemporaneous plume immediately below the front has $G_p(h,t)=w_{\ast }^2/h(t)$, which grows as the front descends. That plume value is not the ambient jump across the material first front. Equating the two would inadvertently impose a uniformly mixed upper layer and discard the stable density stratification of the standard filling box model. Diffusion or extensive interfacial mixing can smooth and alter the ideal jump.

Write $H=L/c_p$, $k=\rho c_p\kappa$ and $ϵ=\sqrt{D/\kappa }$. The requested model neglects density differences and uses the same thermal transport coefficients in the interfacial balances. Let $C_b=C(b(t),t)$ and $T_b=T_m-m{C}_b$. With no imposed heat gradient in the fresh liquid, its temperature remains $T_m$. The prescribed linear temperature in the ice isThe salt and temperature fields in the lower liquid satisfy diffusion equations, $C_t=D{C}_{zz}$ and $T_t=\kappa T_{zz}$. For constant interfacial values and the proposed similarity solution, their decaying solutions areThese follow by substituting a variable $z/(2\sqrt{\chi t})$ into the diffusion equation, which gives $U^{\prime \prime }+2\eta U^{\prime }=0$; integrating gives the complementary error function. Their values at $b$ and at infinity give the required boundary data.

For the upward-moving fresh-water/ice boundary, the Stefan condition is $\rho L\dot{a}=k{T}_{i,z}$: negative $\dot{a}$ means fresh water freezes. At the lower boundary the signed heat balance isSalt cannot enter the pure ice. With no bulk liquid motion its interfacial conservation law isThe derivatives and interface velocities areDefine $F(x)=\sqrt{\pi }x{e}^{x^2}\mathrm{erfc}x$. Substitution into all three balances givesTogether with $T_b=T_m-m{C}_b$, these determine the interfacial constants. In particular $C_b=C_0/[1-F({\lambda }_b)]$.

The physical branch has ${\lambda }_a<{\lambda }_b<0$. Since $F$ is negative at negative arguments, the salt relation gives $0<C_b<C_0$: lower-surface melting dilutes the brine. Combining the two heat balances eliminates the ice gradient and givesThus $T_m>T_b>T_{\infty }>T_L(C_0)$. The interface is at local liquidus equilibrium with diluted brine, while the undiluted far-field brine remains above its own liquidus. The similarity model describes a layer starting with zero thickness, or the long-time limit when an initial finite thickness no longer sets the dominant scale.

For the small-diffusivity-ratio limit, put $A=ϵ{\lambda }_a$ and $\Delta =T_m-T_{\infty }$, keeping $S=H/\Delta$ and the imposed concentrations fixed. Seek $A=O(1)$ and ${\lambda }_b=O(1)$. Then $F(ϵ{\lambda }_b)\sim \sqrt{\pi }ϵ{\lambda }_b$ and ${\lambda }_b-{\lambda }_a\sim -{\lambda }_a$. The upper heat balance givesIn the lower balance the two terms of order ${ϵ}^{-1}$ must cancel, givingConsequentlyand $T_b-T_{\infty }\sim -\sqrt{\pi }HA$. Adding the two temperature differences yields $2A^2-\sqrt{\pi }A=1/S$. The negative root isThe resulting $C_b=(T_m-T_b)/m$ lies strictly between zero and $C_0$. The salt relation therefore determines a finite negative ${\lambda }_b$, confirming its $O(1)$ scaling. The top interface travels on the thermal scale $\sqrt{\kappa t}$, whereas the lower melting interface travels on the shorter solutal scale $\sqrt{Dt}$. This is the brine-ice sandwich with disparate diffusion scales; $S$ is the latent-to-sensible heat ratio, the reciprocal of the usual sensible-to-latent Stefan number convention.

The sketches show constant fresh-water temperature, a declining linear ice temperature, and a much broader thermal than solutal boundary layer in the brine. Concentration is zero in freshwater and pure ice, jumps to $C_b$ across the lower interface, and rises to $C_0$. Accordingly the local liquidus is $T_m$ on the salt-free side, jumps to $T_b$ and falls rapidly to $T_L(C_0)$. In the phase diagram, the brine starts on the liquidus at $(C_b,T_b)$ and ends above it at $(C_0,T_{\infty })$; the pure ice has concentration zero, and the interface tie-line joins it to the diluted brine.

Physically, the cold saline liquid permits freshwater to freeze above, but salt lowers the liquidus and allows the existing ice to melt into brine below even though that brine is colder than $T_m$. Dilution raises the lower interfacial freezing temperature. Heat released by upper freezing supplies the lower melting and the conductive loss to the colder deep brine. With $|{\lambda }_a|\gg |{\lambda }_b|$, upper freezing exceeds lower melting and the ice layer thickens as $b-a\propto \sqrt{t}$.

Let the surface advance downwards at speed $V>0$. Ice is stationary in the laboratory, so in the surface frame its velocity is $-V$. The steady heat equation with volumetric heating is thereforeTake the atmospheric heat flux to be negligible in the stated model. The upward conductive flux at the ice surface supplies the latent heat of sublimation, giving the Stefan condition $k{T}^{\prime }(0)=\rho LV$.

Write $p=V/\kappa$ and $\Delta =T_s-T_{\infty }$. For $p\ne \lambda$, solve the constant-coefficient equation to obtainFor the resonant case $p=\lambda$, its continuous limit isAn especially transparent way to find $V$ is to integrate the heat equation over the half-space. Its far-field gradient vanishes, soUsing the surface heat balance givesAbsorbed radiation supplies both the latent heat and the sensible heat needed to bring incoming cold ice to the surface temperature. This is the radiatively heated sublimation profile.

The surface gradient isThus temperature first rises into the solid, but ultimately falls to the colder far field: it has a subsurface maximum. For a useful dimensionless form, set $x=\lambda z$, $r=p/\lambda$ and $S=L/(c_p\Delta )$. Thenwith limit $[1+(S+1)x]e^{-x}$ when $r=1$. The maximum is at $x_{\ast }=[\log ((rS+1)/(S+1))]/(r-1)$, or $S/(S+1)$ at $r=1$.

Melting begins below the surface if the interior maximum reaches $T_m$. The surface remains constrained to the lower value $T_s<T_m$, and the remote ice is cold. Increasing absorbed heating can raise the interior maximum to the melting point, after which an all-solid steady solution is no longer valid and an internal melt region must be modeled. The maximum, rather than the far-field temperature or surface temperature alone, decides onset.

For the linear stability analysis, perturb the mean surface by $\eta =\hat{\eta }{e}^{\sigma t+i\alpha x}$, with ice on $z>\eta$. Approximate the base temperature locally by $T_s+Gz$. The prescribed radiative source has no perturbation, so the perturbation heat equation is homogeneous. Choose outward curvature $\mathcal{K}\simeq {\eta }_{xx}$, so the Gibbs-Thomson effect gives local equilibrium temperature $T_s-\Gamma \mathcal{K}$. If the perturbation temperature is $\hat{T}(z)e^{\sigma t+i\alpha x}$, its surface condition isIn the quasi-static thermal limit, ${\hat{T}}^{\prime \prime }-{\alpha }^2\hat{T}=0$, henceLinearizing the Stefan condition, with the base gradient taken constant, gives $\rho L\sigma \hat{\eta }=k{\hat{T}}^{\prime }(0)$. ThereforePositive $G$ amplifies long-wave shape perturbations, while capillarity suppresses sufficiently short waves. The unstable band is $0<\alpha <\sqrt{G/\Gamma }$, and the approximate fastest mode is ${\alpha }_{\ast }=\sqrt{G/(3\Gamma )}$. This is the sublimation-front capillary instability.

The qualifications on wavelength matter. Retaining the perturbation's time dependence and mean recession giveswhose decay constant is $p_{\alpha }=[V+\sqrt{V^2+4\kappa (\kappa {\alpha }^2+\sigma )}]/(2\kappa )$. Replacing it by $\alpha$ requiresWriting $H=L/c_p$, the second condition becomes $|G-\Gamma {\alpha }^2|\ll H\alpha$ under the approximate dispersion relation. A simple sufficient range is $V/\kappa \ll \alpha \ll H/\Gamma$; for unstable modes the lower inequality already controls thermal time dependence. Large $S$ alone does not justify arbitrarily small $\alpha$.

Finally the local linear background is appropriate only if the perturbation penetration depth is shorter than its curvature length: $\alpha \gg |T_0^{\prime \prime }(0)|/G$. From the steady equation this ratio is $V/\kappa +\lambda (1+1/S)$. Thus for the actual exponentially heated profile, a sufficient local range at large $S$ is also $\alpha \gg V/\kappa +\lambda$. Otherwise evaluation of the base gradient at the displaced surface contributes an omitted term $k{T}_0^{\prime \prime }(0)/(\rho L)$, and the simple quoted dispersion relation need not describe those long waves.

Measure $z$ upwards from the inlet and let $L$ now denote the specific latent heat of melting of the crust. Use the standard approximation of equal solid and magma densities, negligible axial conduction, negligible heat loss beyond the melting walls, and a horizontally well-mixed turbulent flow. The wall is at $T_m$, so newly added wall melt enters with zero excess temperature $\theta =T-T_m$.

For a slice of dike, total magma mass per unit height and out-of-plane length is $\rho b$. Both melting walls add mass at rate $\rho b_t$. Consequently mass conservation givesThe growth of channel storage and mass supplied by melting cancel; omitting the right-hand source would incorrectly force the inlet flux to fill this volume without accounting for the newly melted rock. Unequal densities would instead give $(bw{)}_z=({\rho }_s/\rho -1)b_t$ and change the similarity model.

Let $q_w={\alpha }_T\rho c_{p}w\theta$ be heat transfer to each wall. As the original crust is already at $T_m$, no sensible heating of unmelted crust is needed. Each wall moves outwards at speed $b_t/2$, so its Stefan condition isExcess sensible-heat conservation, including both walls, isCombining these equations, with $H=L/c_p$, givesThe term $(T-T_m)b_t$ is the heat used to warm freshly melted wall material. It cannot simply be dropped from a conservative balance. This is melt-dilution heat balance in a dike.

At the inlet $bw=Q_0$ and $\theta ={\Delta }_0=T_0-T_m$. The wall balance there gives $b(0,t{)}^2=b_0^2+4{\alpha }_T{Q}_{0}t/S$, where $S=L/(c_p{\Delta }_0)$. At late times neglect $b_0$ in the widened basal region and setThe basal width and the vertical scale both grow as $\sqrt{t}$. Substituting into the wall condition givesConservative heat balance first gives $f\vartheta -\eta (f\vartheta {)}^{\prime }+S{\vartheta }^{\prime }=-S\vartheta /f$. Eliminating $f-\eta f^{\prime }$ with the wall equation yieldsThe boundary conditions are $f(0)=1$, $\vartheta (0)=1$, with $f\to 0$ in the outer idealized limit; the physical cooled branch also has $\vartheta \to 0$. These equations and time-independent boundary data establish the thermal dike-widening similarity. The condition at infinity is an approximation for matching away from the widened basal region, not a claim that the original finite-width dike literally has zero width.

The inlet is a regular singular point of the first equation. Differentiating it and evaluating the second givesA regular expansion can still contain an undetermined higher-order coefficient, for example in $f^{\prime \prime }(0)$; the outer matching condition selects it. Thus specifying the inlet values does not eliminate the boundary-value aspect of the problem. If a computed branch reaches $f=0$ at finite similarity coordinate, that endpoint is treated as the edge of the widened inner region and matched to the unneglected original-width channel. One must not continue $w=Q_0/(Bf)$ through zero width as a physical flow.

For numerical solution, formulate this as a nonlinear two-point boundary-value problem: start from a regular inlet expansion and adjust its free coefficient to meet the outer decay/matching condition, or solve for the profiles and boundary data together on a truncated or compactified similarity interval. Enforce nonnegative width and temperature excess, check any apparent singularity at $S=\eta f$, and refine the outer matching location until the basal profiles are insensitive to it. No specific algorithm is required.

A useful conservation check follows by adding the latent storage to the sensible storage:For a vanishing outer excess-heat flux, integration of the similarity profile giveswhere the integral runs over its active widened region. This checks the heat admitted at the base against latent and sensible storage. All factors of two refer to the two channel walls; $b$ is the full width.

In the specified dust model $H_0=2/(3t_0)$, so ${\rho }_{\mathrm{crit}}=m_{\mathrm{pl}}^2/(6\pi t_0^2)$ and $n_B={\Omega }_{B0}{\rho }_{\mathrm{crit}}/m_B$. Photons and the six neutrino/antineutrino helicity states giveUsing the supplied GeV estimates literally givesThe rounded time/temperature conversions are unusually crude: converting $10$ Gyr to $4.8\times 10^{41}{\mathrm{GeV}}^{-1}$ and $3$ K to $2.6\times 10^{-13}$ GeV instead gives $Y_B\sim 10^{-10}$. Either set gives the same order-of-magnitude mass conclusion below, but the numerical abundance should identify which supplied units are used.

For a stable relic, the relevant comparison is energy density, not merely particle number. The matter density left over after the baryons is at most nine times the baryon density here. ConsequentlyPut $x=(m_{Y\mathrm{nr}}/m_Y{)}^{1/3}$ and use $Y_Y\simeq C_Y{x}^{3/2}{e}^{-x}$ with $C_Y\simeq 10^{-3}$. The condition becomesFor the literal supplied numbers the right side is about $30.5$, giving $x\gtrsim 25.6$ and $m_Y\lesssim 40$–$50$ GeV. With the physically consistent conversions the estimate is about $60$ GeV. The intended rough conclusion is therefore a sufficiently suppressed branch with $m_Y$ no larger than order $10^2$ GeV, within the toy cosmological particle freeze-out law. Comparing the two number yields alone gives a similar rough scale but misses the mass weighting. This is a thermal relic density bound for a tunable fermion.

For $m_Y\gtrsim m_{Y\mathrm{nr}}$, relativistic decoupling would instead leave $Y_Y\simeq 135\zeta (3)g_Y/(8{\pi }^4{g}_{\ast s})\simeq 4\times 10^{-3}$, incompatible with the matter-density bound for such heavy stable particles. These are conditional bounds on a populated, stable thermal relic. A short lifetime or an initially suppressed nonthermal population can change the conclusion; the supplied information alone does not constrain every possible production and decay history.

Above the transition,for $g_{\ast }\sim 100$. The ratio increases during cooling, rather than at higher temperature. A GUT transition at $T_c\sim 10^{15}$–$10^{16}$ GeV occurs well above this scale, so the interaction is too slow to establish or maintain thermal equilibrium throughout $T>T_c$. In fact its integrated interaction probability from an arbitrarily high temperature down to $T_c$ is at most of order $T_{\mathrm{eq}}/T_c\ll 1$, since $dt=-dT/(HT)$.

An initial Fermi-Dirac distribution with zero chemical potential therefore cannot be assumed from this high-temperature rate alone. If imposed initially and collisionlessly redshifted while the particles are massless, its thermal-looking form can persist, but this is not dynamically maintained thermal equilibrium. Other initial states can remain nonthermal. This is the thermalization temperature for a massless gauge mediator.

The below-transition rate stipulated in the model has $\Gamma /H=(T/T_D{)}^3$. If $T_c\gg T_D$ and it includes efficient number-changing reactions, it can restore chemical equilibrium after the transition and erase the initial-state memory, making the estimates in a and b applicable. If it only equilibrates momenta, or if reheating never populates the relevant range, the thermal relic calculation must be reconsidered.

There is a sign error in the printed synchronous line-of-sight integrand. Direct contraction of the stated decomposition gives $h_{ij}^{\prime }{n}^i{n}^j=(h^{\prime }-h_S^{\prime })/3+(\hat{\mathbf{k}}\cdot \mathbf{n}{)}^2{h}_S^{\prime }$, with a plus sign in its last term. In the positive spatial-perturbation convention of part a, the redshift term is half this contraction. The printed minus sign cannot yield the requested Newtonian expression with the supplied transformations.

The consistent identity can be derived explicitly. Work mode by mode with $q=\mathbf{k}\cdot \mathbf{n}$, $E(\tau )=e^{iq(\tau -{\tau }_0)}$ along the unperturbed ray; the shift of the phase origin is immaterial. Put $A=(h-h_S)/6$. Part a gives $A=\Psi -\mathcal{H}T$, while $\Phi =-\mathcal{H}T-T^{\prime }$. Therefore $A^{\prime }={\Phi }^{\prime }+{\Psi }^{\prime }+T^{\prime \prime }$. The corrected synchronous integral issince $[E(T^{\prime }-iqT){]}^{\prime }=E(T^{\prime \prime }+q^{2}T)$. This is the photon redshift identity with positive spatial perturbations.

The supplied fluid shifts are ${\delta }_{\gamma }^N={\delta }_{\gamma }+4\mathcal{H}T$ and ${\mathbf{v}}_{\gamma }^N={\mathbf{v}}_{\gamma }+i\mathbf{k}T$. At emission, the synchronous intrinsic and Doppler terms are consequently $E_d[{\delta }_{\gamma }^N/4+{\mathbf{v}}_{\gamma }^N\cdot \mathbf{n}-\mathcal{H}T-iqT{]}_d$. Adding the lower integral endpoint cancels the velocity shift and leaves $-\mathcal{H}T-T^{\prime }=\Phi$. The upper endpoint is the local observer monopole/dipole $[T^{\prime }-iqT{]}_0$, removed by temperature calibration and the observer-rest-frame convention. With zero anisotropic stress, $\Phi =\Psi$, the observable higher-multipole result isAll fields are evaluated along the ray; the integral contains the time derivative at fixed spatial position. Retaining the printed negative angular term would leave the additional $-2q^2\int ETd\tau$, generally a gauge-dependent, nonzero term. It is therefore not repaired merely by discarding observer monopole and dipole.

The first term is the intrinsic photon temperature perturbation, important in acoustic compressions and rarefactions. The velocity term is the Doppler shift and is prominent near acoustic, roughly degree and subdegree, scales; it is suppressed on very large scales. The potential at emission is the ordinary Sachs-Wolfe effect; combined with the intrinsic term it gives the familiar large-angle adiabatic matter-era result $\Phi /3$. The final term is the Integrated Sachs-Wolfe effect, caused by changing potentials. It vanishes for constant matter-era potentials, receives an early contribution near radiation-matter transition, and has an important late large-angle contribution when curvature or dark energy causes potentials to evolve. The sign of the Doppler term follows the photon-propagation direction convention used in the ray; reversing that direction reverses it consistently.

Use units $k_B=\hslash =c=1$. In critical light-cone gauge in string theory, each chiral sector of the closed bosonic string has 24 transverse string oscillator species. Their generating function isEach factor sums the independent occupation numbers of the string oscillators at level $m$. For $q=e^{-t}$ with $t\downarrow 0$, the modular transformation of the Dedekind eta function givesExtract the coefficient by a contour integral and apply the saddle-point approximation to the exponent $(N-1)t+4{\pi }^2/t$. The saddle is $t_{\ast }=2\pi /\sqrt{N-1}$, with exponent $4\pi \sqrt{N-1}$ and second derivative $(N-1{)}^{3/2}/\pi$. The factor $t^{12}$ gives $(N-1{)}^{-6}$, and the Gaussian width gives $(N-1{)}^{-3/4}/\sqrt{2}$. HenceFor a noncompact closed string, closed-string level matching imposes equal left and right levels, but does not correlate their transverse polarizations. Therefore the large-level degeneracy of a closed bosonic string isThe mass relation is ${\alpha }^{\prime }{M}^2=4(N-1)$. Thus $8\pi \sqrt{N-1}={\beta }_{H}M$ withDistinguish degeneracy per discrete level from a smoothed density per unit rest mass. Since $dN/dM={\alpha }^{\prime }M/2$, the latter isThe exponential growth comes from the many ways of distributing string oscillator energy along a string; it is much faster than a fixed finite collection of particle species. The corresponding high-energy entropy is $S(E)={\beta }_{H}E+O(\log E)$, so its microcanonical temperature tends to $T_H$. Additional energy can go into very long strings rather than raising the temperature substantially.

The Boltzmann suppression competes with this exponential density. For the string oscillator rest-mass sum alone the large-$M$ tail is $\int dM{M}^{-26}{e}^{-(\beta -{\beta }_H)M}$. It converges for $\beta >{\beta }_H$ and fails for $\beta <{\beta }_H$. To include translational states in 25 noncompact spatial dimensions, take the one-string partition function per unit spatial volume. At large $M$,This follows by expanding the energy as $M+p^2/(2M)+\cdots$ and doing a Gaussian momentum integral. Its high-mass contribution is consequently proportional toThis integral is finite at $\beta ={\beta }_H$, although nonanalytic there: its first twelve derivatives in $\beta$ are finite, while the thirteenth diverges. The endpoint expansion contains a term proportional to $(\beta -{\beta }_H{)}^{25/2}$. Thus one must not assert that the canonical partition function necessarily diverges exactly at $T_H$; the polynomial prefactor and the spatial spectrum matter. This is the finite canonical partition function at the bosonic Hagedorn threshold distinction. The multi-string ideal gas has the same leading Hagedorn threshold, since the higher occupation-cycle terms probe $\ell \beta$, $\ell \ge 2$, and are nonsingular at ${\beta }_H$.

There is also a spacetime diagnosis. Euclidean temperature makes time a circle of radius $R_{\beta }=\beta /(2\pi )$. A string oscillator ground state winding once around that circle, with zero momentum number, hasIt becomes massless at ${\beta }_H$ and tachyonic for $T>T_H$. This thermal winding instability of a closed bosonic string signals failure of the perturbative low-temperature gas and motivates a transition to a different high-energy phase, often described in terms of long-string or winding-mode condensation. Free state counting fixes the threshold and its singular behavior, not the interacting phase's endpoint or the order of a transition.

Finally, the ordinary bosonic tachyon is already an infrared instability at zero temperature. The above canonical discussion is a formal analysis of the high-level spectrum with that separate infrared problem regulated. It does not establish a stable thermal vacuum of the unmodified bosonic theory; an actual equilibrium transition requires an appropriate stable setting and control of interactions.

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.

Set $v=\dot{x}$, $\zeta =6\pi \mu a$, and $\tau =m/\zeta$. To keep dimensions and the thermal amplitude explicit, write the force covariance as $⟨F(t)F(s)⟩=A\delta (t-s)$. The printed unit coefficient is a choice of noise normalization, not a general dimensional thermal force. Assume the initial velocity is independent of the future white noise. The Underdamped Langevin dynamics has integrating-factor solutionTaking its expectation and using the delta covariance in the double integral givesHere the noise contribution is $A{m}^{-2}{\int }_0^t{e}^{-2(t-s)/\tau }ds$. The equipartition theorem requires the long-time value $⟨m{v}^2/2⟩=k_{B}T/2$. Consequently the fluctuation-dissipation relation for a Langevin particle isWith an initially thermal velocity, $⟨v_0⟩=0$ and $⟨v_0^2⟩=k_{B}T/m$, so the velocity variance is stationary. If one insists on the printed coefficient $A=1$ in dimensional units, the model has temperature fixed by $k_{B}T=1/(2\zeta )$ in those units; it cannot describe arbitrary temperature while retaining that coefficient.