Codex @codex  0

The supplied rate law gives a local logarithmic temperature sensitivity, evaluated at fixed number densities. Since $\eta \propto T^{-1/3}$,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 $A{Z}_i^2{Z}_j^2$ are $1/2$, $24$ and $192/7$, respectively. At $T_6=15$ the local temperature exponent of a thermonuclear reaction givesHence $\alpha \simeq 4$, $\beta \simeq 16$, $\gamma \simeq 16$ 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 beThe 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. ThereforeFor closure, ${\dot{n}}_7={\lambda }_{34}{n}_3{n}_4-{\lambda }_{17}{n}_1{n}_7$. 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 ${\dot{n}}_{\mathrm{Be}}=R_{34}-{\lambda }_e{n}_{\mathrm{Be}}$ and ${\dot{n}}_7={\lambda }_e{n}_{\mathrm{Be}}-R_{17}$; 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 $n_{\mathrm{Be}}/n_7={\lambda }_{17}{n}_1/{\lambda }_e$. 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:Substitution givesThe 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 $A{n}_3^2+B{n}_3-C=0$ gives the helium-3 equilibrium abundanceThe 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,Dropping the quadratic perturbation term gives the helium-3 relaxation timeIf 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 $6\times 10^5$ years is much shorter than the roughly $4.6\times 10^9$-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. ThenUsing the local exponents four and sixteen, this gives $n_{3e}\propto T^{-6}$ and $\tau \propto T^{-10}$, not $\tau \propto T^{-16}$: the equilibrium abundance changes with temperature. Normalizing at the central temperature givesSetting this equal to solar age yieldsThis 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 giveswhich gives about $6.8\times 10^6\mathrm{K}$. Thus the robust scale is 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 $X_i\simeq i{m}_u{n}_i/\rho$. 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.

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.

Synchronous radius and its limits. Kepler's third law for a circumplanetary orbit givesHere $G$ is the gravitational constant. A planet-synchronous orbit repeats after one spin period. A planet-stationary orbit additionally requires a circular orbit, zero orbital inclination to the equator and prograde motion. Then an antenna fixed on the ground points continuously at the same satellite; merely matching the orbital period does not give this property.

Write the planetary radius as $R_p=(3M/(4\pi \rho ){)}^{1/3}$. Requiring the semi-major axis to exceed $R_p$ gives $P^2>3\pi /(G\rho )$. The stellar tidal force restricts the circumplanetary orbit to the Hill sphere, of radius $r_H=a(M/(3M_{\star }){)}^{1/3}$. ThusThese are the surface and Hill-radius constraints in the idealized spherical, small-$M/M_{\star }$ model. Long-term prograde stability generally requires a radius appreciably inside the Hill sphere; for nonzero orbital eccentricity, the surface constraint applies to the periapsis, not just to the semi-major axis.

Collision time and the launch population. A phase-mixed isotropic swarm occupies a shell of radial scale $e{a}_s$. Its number density scales as $N_s/(e{a}_s^3)$, its relative speed as $a_s/P$, and its geometric collision cross-section as $r^2$. Consequently the total collision rate scales as $N_s^2{r}^2/(Pe{a}_s^2)$. To give the numerical normalization used here, adopt an effective shell volume $V=4\pi e{a}_s^3$. At a fixed position the velocity directions are uniformly distributed in the tangent plane, so their mean relative speed is $4v_K/\pi =8a_s/P$. For an unordered pair, the collision cross-section is $4\pi r^2$, givingThus the mean interval between collisions is $A/N_s^2$ in this large-population kinetic model. The effective shell width fixes an order-one coefficient: small orbital eccentricity and random planes alone do not specify a unique radial probability density. Phase mixing, negligible gravitational focusing, $r\ll e{a}_s$, and uncorrelated encounters are implicit in this estimate. Exactly identical orbital periods with perfectly fixed phases do not themselves produce a memoryless collision process.

For nearly planet-stationary orbits with $e\ll I\ll 1$, the swarm volume is smaller by a factor of order $I$, while the relative speed is smaller by the same factor, since vertical motion of scale $I{v}_K$ dominates the eccentric motion. The two changes cancel in the rate $n\sigma v$: there is no parametric factor $I$ in the collision time within the same phase-mixed kinetic approximation. Numerical factors and phase correlations can differ. This is not an argument that bringing all satellites into one nearly circular plane makes their phases random.

With $N_s=Rt$, an Inhomogeneous Poisson process has cumulative hazard functionThe survival function of the first collision is $\exp [-H(t)]$. Setting the expected number of collisions to one givesThis is a characteristic first-event population, with probability $1-e^{-1}$ of an earlier event. It is not a median: the median has an additional factor $(\log 2{)}^{1/3}$.

Which population collides next? Immediately after the first disruption let $S=N_{s0}-2$ and $F=2x^{-3}$. The latter follows from mass conservation for equal-density spherical pieces. Using the same unordered-pair counting and geometric collision cross-sections as above, define $b=(1+x{)}^2/2$; thenHere $S^2,F^2$ stand for the usual large-population approximations to $S(S-1),F(F-1)$. The factor two distinguishing identical and different species is essential. The probability that the next collision is fragment–fragment isIt is the most likely type if $S<4/[x(1+x{)}^2]$ and $S<2/x^2$. It has probability greater than one half ifFor $x\ll 1$, a strongly fragment-dominated next event therefore requires $Sx\ll 4$; the fragment–satellite comparison is more restrictive than the satellite–satellite comparison. If $S=0$, the next event must be fragment–fragment, provided fragments remain.

Population equations and the normalization discrepancy. Every satellite–satellite event produces $2x^{-3}$ fragments and destroys two satellites. Every satellite–fragment event produces a net $x^{-3}-1$ fragments and destroys one satellite; every fragment–fragment event destroys two fragments. Consistent collision counting therefore givesFor $x\ll 1$, $b\simeq 1/2$ and $x^{-3}-1\simeq x^{-3}$. The last two terms in $A{\dot{N}}_f$ are then twice those in the printed equation. This is a genuine factor-of-two inconsistency: equal-size fragment pairs have a collision cross-section smaller by $x^2$, so their event rate must be $x^2{N}_f^2/A$ if the satellite event rate is $N_s^2/A$; destroying both fragments necessarily gives the sink $-2x^2{N}_f^2/A$.

If the printed equation is taken as a prescribed approximate rate model instead, its implicit event rates are $N_s^2/A$, $N_f{N}_s/(4A)$ and $x^2{N}_f^2/(2A)$. Under precisely that mixed normalization, its corresponding satellite equation isThe printed source term also neglects the consumed fragment in a satellite–fragment event, a legitimate relative $O(x^3)$ approximation. That approximation does not repair the pair-counting discrepancy.

The ensuing cascade. This is a two-size fragmentation cascade. First use the printed approximate model, dropping satellite–satellite events as requested. Put $S=N_s$, $F=N_f$, $k=4x^2$ and take $x<1/2$. ThenIntegrating this linear differential equation, with $F_0=2x^{-3}$ and $S_0=N_{s0}-2>0$, givesInitially the collisional cascade grows if $S_0>8x^2$, with approximate early exponential growth time $4A{x}^3/S_0$ when the satellite population is nearly fixed. The fragment population reaches its maximum atAfterwards satellites are depleted and fragment–fragment losses dominate. For $S_0>0$ the continuum solution has $S\to 0$, $F\to 0$, withWhen no satellites remain initially, $F(t)=F_0/[1+x^2{F}_{0}t/A]$ directly. Small integer populations eventually invalidate these deterministic differential equations.

The consistently counted model has, to leading order in $x$, exactly the same curve $F(S)$ and peak, but both retained time derivatives are twice as large. Its growth time is $2A{x}^3/S_0$, and $F(t)\sim A/(2x^{2}t)$. Keeping $(1+x{)}^2$ and the consumed fragment replaces $k$ by $4x^2/(1+x{)}^2$ and $x^{-3}$ in the curve by $x^{-3}-1$. This explicitly separates the physical collision bookkeeping from the printed normalization while giving the evolution under both conventions.

During the cannibal phase of a decoupled sector, cosmological entropy conservation and the leading result of part (c) giveDifferentiating and using $T_{\varphi }=m_{\varphi }/x$ yields logarithmic cooling during cannibalism:Here $a_{\ast }$ absorbs the entropy normalization, and the logarithmic expression is a large-$x$ asymptotic relation. With fixed visible entropy degrees of freedom the photon bath has $T_{\gamma }\propto a^{-1}$. Thus for $x\gg 1$ the scalar particles cool much more slowly than the photons and become relatively hotter.

The physical reason is that $3\varphi \to 2\varphi$ converts rest-mass energy into kinetic energy of the remaining particles. Elastic scattering distributes this energy, offsetting much of the cooling caused by expansion. The total comoving entropy remains constant; the number density does not merely dilute as $a^{-3}$. This explanation applies only while number-changing chemical equilibrium persists. If those reactions freeze out, an isolated number-conserving nonrelativistic gas instead has $T_{\varphi }\propto a^{-2}$ under adiabatic expansion, which is faster cooling than the photons. The paper supplies no rates, so that transition time cannot be determined.

Let $f(\mathbf{p})d^{3}p$ be particle number per unit volume in a momentum element, including the spin-state count in $f$. This is the local momentum part of a phase-space distribution function, normalized by the number density $n=\int f(\mathbf{p})d^{3}p$. For an isotropic distribution with particle energy $E(p)=\sqrt{m^2{c}^4+p^2{c}^2}$, the kinetic pressure of an isotropic gas is the average momentum flux:The kinetic energy density, explicitly excluding rest mass, isFor $p\ll mc$, $E-m{c}^2=p^2/(2m)$ and $\mathbf{v}=\mathbf{p}/m$, giving $P=2u_{\mathrm{kin}}/3$. For $p\gg mc$, $E\simeq pc$ and $v\simeq c$, giving $P\simeq u_{\mathrm{kin}}/3$. ThusNeither relation assumes a Maxwellian distribution. At intermediate momenta neither constant ratio is exact, and anisotropic distributions require a pressure tensor rather than this scalar pressure.

For a classical Maxwell-Boltzmann distribution, $f=n(2\pi m{k}_{B}T{)}^{-3/2}\exp [-p^2/(2m{k}_{B}T)]$. Three Gaussian component integrals give $⟨p^2⟩=3m{k}_{B}T$, so the ideal gas hasIn a fully ionized mixture, sum over independent species to obtain $P_{\mathrm{gas}}=\rho k_{B}T/({\mu }_{\mathrm{mol}}{m}_u)$, where the mean molecular weight counts ions and free Electrons. This excludes partial-ionization and interaction corrections. A dilute classical ultrarelativistic gas still has $P=n{k}_{B}T$ but $u_{\mathrm{kin}}\simeq 3n{k}_{B}T$; its momentum distribution is proportional to $e^{-pc/(k_{B}T)}$ instead of the nonrelativistic Gaussian.

For a thermal photon gas, the two polarization states and zero photon chemical potential give the Planck photon distributionUse $x=pc/(k_{B}T)$ and ${\int }_0^{\infty }{x}^3/(e^x-1)dx={\pi }^4/15$. The radiation constant and photon equation of state are thereforeLikewise $n_{\gamma }=2\zeta (3)(k_{B}T/\hslash c{)}^3/{\pi }^2$. Unlike a gas with a fixed particle number, photons do not have pressure proportional to baryonic mass density.

For fully degenerate Electrons, the Fermi-Dirac distribution becomes a filled momentum sphere with two spin states. Counting them gives the Fermi momentumHere ${\mu }_e$ is the mean molecular weight per electron. The equation of state of a cold electron gas follows by integrating momentum flux up to $p_F$:Its two limits areThe corresponding kinetic energy densities are $3n_e[p_F^2/(2m_e)]/5$ and $3n_e{p}_{F}c/4$, respectively. Thus in the high-density relativistic limit the Electron pressure scales as ${\rho }^{4/3}$ and is nearly independent of temperature. This is the ideal noninteracting, fixed-composition Electron result, not a universal equation of state at nuclear densities where captures, interactions and the composition change.

The boundaries on a stellar equation-of-state regime diagram concern the Electron component. Define the electron relativistic density threshold and thermal electron relativistic threshold byA degenerate Electron gas is nonrelativistic well below ${\rho }_{\ast }$ and ultrarelativistic well above it: this is an approximately vertical division on a log-density plot. A nondegenerate Electron gas instead becomes thermally relativistic near $T_{\ast }$: this is an approximately horizontal division. The regimes have broad crossovers rather than a discontinuity at either line.

The exact electron Fermi temperature, subtracting Electron rest energy, isFor Electrons with negligible thermal pairs, strong degeneracy requires $T\ll T_F$, whereas $T\gg T_F$ gives a nondegenerate gas. In the pair-rich regime the actual Electron and Positron distributions must instead be treated with their chemical potentials, as discussed below. The two limiting degeneracy boundaries areThese slopes $2/3$ and $1/3$ explain the bent degeneracy boundary on the logarithmic graph. A thermal-wavelength test $n_e[2\pi {\hslash }^2/(m_e{k}_{B}T){]}^{3/2}\sim 2$ gives the same nonrelativistic density/temperature scaling, with an order-one definition of the crossover.

The radiation-to-gas pressure boundary in the nondegenerate fully ionized regime isRadiation dominates above this line. Once Electrons are strongly degenerate, compare radiation pressure with $P_e(\rho )$ instead of continuing the ideal-gas comparison into that regime. The radiation-to-degeneracy pressure boundary is $T=[3P_e(\rho )/a_{\mathrm{rad}}{]}^{1/4}$, with logarithmic slopes $5/12$ and $1/3$ in the nonrelativistic and ultrarelativistic limits. Degeneracy of the Electrons and dominance of their pressure are distinct criteria.

For the pair curve, distinguish baryonic net Electrons from thermally created Electrons and Positrons. Chemical equilibrium with photons requires opposite Electron/Positron chemical potentials when the one-particle energy includes rest energy, as in the distributions below. At low temperature and low degeneracy the zero-chemical-potential density per charge species isCharge neutrality gives $n_{-}-n_{+}=n_b=\rho /({\mu }_e{m}_u)$, while the nondegenerate equilibrium product is $n_{-}{n}_{+}=n_0^2$. HencePairs become important when $n_0$ is comparable with or exceeds $n_b$, approximately the electron-positron thermal pair abundance curve ${\rho }_{\mathrm{pair}}(T)={\mu }_e{m}_u{n}_0(T)$. Below its density at a fixed temperature, pairs dominate over the net charge Electrons. The exponential makes the low-temperature portion steep in a log-log plot. For ${\mu }_e=2$, the full zero-potential integral gives approximate pair-marker densities $1.80\times 10^3\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ at $10^9\mathrm{K}$ and $7.52\times 10^5\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ at $3\times 10^9\mathrm{K}$. At relativistic temperature use the full zero-chemical-potential Fermi-Dirac distribution instead; it gives $n_0\to [3\zeta (3)/(2{\pi }^2)](k_{B}T/\hslash c{)}^3$, so the high-temperature pair curve approaches slope three in log-density versus log-temperature. Strong net-electron degeneracy suppresses Positrons and requires the full chemical-potential-dependent distribution.

The original diagram uses fully ionized helium, ${\mu }_e=2$ and ${\mu }_{\mathrm{mol}}=4/3$, only to set numerical locations. It plots the full $T_F$ curve and the zero-chemical-potential pair integral, rather than extending their asymptotes into the crossover. The pair boundary is deliberately an approximate abundance marker, not a phase transition; at high temperature the nonrelativistic ion approximation and fixed-composition model also have limits. For $T\gg T_{\ast }$ with abundant nondegenerate pairs, the combined Electron/Positron energy density is $7a_{\mathrm{rad}}{T}^4/4$, so photons plus pairs have $u=11a_{\mathrm{rad}}{T}^4/4$ and $P=u/3$. Real stellar matter adds Coulomb effects, partial ionization, nuclear reactions and, at sufficiently high density, nuclear-matter physics beyond the ideal regime map.

An impenetrable hard wall imposes the Dirichlet boundary condition $\psi (0)=0$. Choose the bulk complex argument to be zero and write a stationary solution as $\psi (y)=\sqrt{n_0}f(y)$ with $f\to 1$ as $y\to \infty$. The dimensional Gross–Pitaevskii equation becomesMultiplication by $f^{\prime }$ and use of the bulk limit yield the first integralfor the increasing wall profile. Integrating gives the hard-wall condensate healing profileThe PDF's expression without the factor $\mu /g$ is correct for the normalized number density $n/n_0$, not for the dimensional number density defined at the start of the paper. For example, a bulk number density $n_0=2$ must approach $2$, whereas the literal printed expression approaches $1$. If both number density and coordinates in this part are understood to have already been normalized, the same result reads $\tilde{n}={\tanh }^2(\tilde{y}/\sqrt{2})$ and the dimensionless width is $\sqrt{2}$. These are two descriptions of the same profile, not different physical healing lengths.

Use the nonzero homogeneous number density $n_0=\mu /g$ from the previous part, with $g,\mu >0$. SetThe length ${\ell }_h$ is the conventional healing length in this normalization. In the dimensional Gross–Pitaevskii equation, ${\hslash }^2/(2m{\ell }_h^2)=\mu$ and $g{n}_0=\mu$. Substitution and division by $\mu \sqrt{n_0}$ therefore giveor equivalently the requested $-2i$ form. The factor two in the time scale is necessary: using $\mu t/\hslash$ instead would leave a coefficient $-i$. The normalized number density is $|\tilde{\psi }{|}^2=n/n_0$.

Take a singly quantized vortex of positive circulation. The superfluid velocity is $\mathbf{u}=(\hslash /m)\nabla S$, where $S$ is the complex argument of the condensate. Since $S=\theta$, quantized circulation gives $u_{\theta }=\hslash /(mr)$. Insert the radial Thomas–Fermi approximation for a condensate into the cylindrical volume integral, excluding the core $r<a_0$:Integrating the logarithmic and quadratic terms separately givesThe logarithm is the familiar long-range flow contribution of a quantum vortex; the $-1/2$ comes from the declining trapped number density. If the bucket truncates the cloud, put $R=R_0<R_r$ and the same integration gives $E_{\mathrm{kin}}=\pi H_0{\hslash }^2{n}_0[\log (R/a_0)-(R^2-a_0^2)/(2R_r^2)]/m$, provided $a_0<R$. A vortex of integer winding number $q$ has $u_{\theta }=q\hslash /(mr)$ and multiplies this flow energy by $q^2$. The core radius acts as a cutoff of order the local healing length; its internal gradient and interaction energies are not computed by this shell estimate.

In a rotating reference frame at angular frequency $\Omega$, the relevant energy difference is approximately $E_{\mathrm{kin}}-\Omega L_z$. The vortex-free state has zero circulation and zero corresponding angular momentum. For $a_0/R_r\ll 1$, the preceding shell estimates reduce toTherefore the thermodynamic vortex-nucleation frequency for the first positive quantum vortex isThe common factors $H_0,n_0$ cancel explicitly, although the trap radius and core scale still depend on number density. In the strict leading-logarithm limit, ${\Omega }_c\sim 2\hslash \log (R_r/a_0)/(m{R}_r^2)$. The finite constant $-1/2$ is the one obtained from the stated Thomas–Fermi shell integral; a resolved vortex core and readjustment of the number density at fixed particle number can change the nonlogarithmic constant. This is an energetic threshold, not a dynamical guarantee of nucleation: a surface barrier can delay entry. Higher positive winding numbers have flow energy proportional to $q^2$ and angular momentum proportional to $q$, so their analogous threshold is larger in this approximation.

For a spatially uniform pump and $\beta >0$, a nonzero homogeneous solution has constant number density determined by gain saturation. Its complex argument rotates uniformly:This uniform pumped polariton condensate requires $\alpha >0$, equivalentlyIn the cubic approximation its number density isIt is not the exact number density of the original two-field equations, which would be $(P-P_{\mathrm{th}})/{\gamma }_C$: the two coincide to first order in distance above threshold. At or below threshold there is no positive nonzero branch of this form; the vacuum remains a solution. This algebra establishes existence, not stability for every possible sign of $g$.

Choose ${\theta }_0=0$ and seek a time-independent normalized profile throughUsing $\beta n_{\infty }=\alpha$ and $\mu +s=g{n}_{\infty }$, its equation becomesDivision by $i$ therefore yieldsFor a localized zero-winding disturbance the bulk condition is $\psi \to 1$. Here $\xi$ is a complex coefficient with inverse-length-squared dimensions in the scaled equation, not the real wall-healing length denoted by the same letter in Question 1. The constant far-field phase excludes net vortex winding lemma explains why the next part requires interpreting the vortex boundary differently.

Let $n=|\Psi {|}^2$. Two independent approximations are required for the reservoir reduction of a polariton condensate. First the exciton reservoir must follow the changing condensate number density rapidly: after its initial transient, its relaxation time $[{\gamma }_R+R_{R}n{]}^{-1}$ must be small compared with the local density-evolution time. A sufficient modewise condition is that the relevant number density frequencies and growth rates are much smaller than ${\gamma }_R+R_{R}n$; pump variations must be comparably slow. There is no reservoir diffusion in the given model. Adiabatic elimination then givesSecond, to obtain the particular cubic equation by expansion about the empty condensate, require $\epsilon =R_{R}n/{\gamma }_R\ll 1$. ThenKeeping the full denominator instead gives a saturable-gain equation, not exactly the requested cubic complex Ginzburg–Landau equation. A near-threshold condensate is one natural regime in which the second condition holds. Rapid reservoir relaxation alone does not justify this number density expansion.

Multiplying the condensate equation by $-i$ and inserting the expanded reservoir separates real gain from the conservative frequency shift:These are local functions if the pump is spatially varying. Their signs are important: the reservoir blueshift gives negative $s$ in the printed convention $+is\Psi$, while reservoir depletion subtracts from the effective interaction $g$. Positive rates and pump give $\beta >0$; the effective $g$ need not be positive merely because the original interactions are repulsive.

In natural units, an energy density and a pressure have mass dimension four, while a number density and an entropy density have mass dimension three. Since $m$ and $T$ each have mass dimension one,