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Use Gaussian units, uniform number density $n$, and ion mass density $\rho =n{m}_i$. Here $e>0$ is the magnitude of the elementary charge, as fixed by the printed current relation. Neglect Electron inertia and displacement current. ThenFreezing the magnetic flux into this Electron velocity, while retaining the ion momentum equation, gives the closed Hall magnetohydrodynamics systemThe pressure enforces the incompressible flow constraint; taking the divergence of momentum determines it up to the usual boundary-dependent constant. The induction equation preserves the solenoidal magnetic-field constraint. One can absorb $B^2/(8\pi )$ into magnetohydrodynamic total pressure, replacing the Lorentz term by magnetic tension. Keeping ion inertia here is important: electron magnetohydrodynamics with fixed ions will be a later limit, not the complete two-field system.

Let the uniform background number density be $n_{\infty }>0$ and set $\mu =U{n}_{\infty }$. The subtraction of this chemical potential requires a phase rotation as well as a rescaling:Substitution into the physical Gross–Pitaevskii equation givesConsequently the requested coefficients follow fromThe length unit is the specified healing length. Removing tildes produces the paper's dimensionless equation; in particular the coefficient of its time derivative fixes $t_0$, not its reciprocal. Without the phase rotation the linear $+\Psi$ term would not appear.

The dimensionless ground state is $\Psi =e^{i\chi }$ for any constant phase $\chi$, conventionally $\Psi =1$. Its subtracted energy is proportional to $\int [|\nabla \Psi {|}^2+(1-|\Psi {|}^2{)}^2/2]d^{3}x$, which is nonnegative and vanishes for this state. In physical variables it has density $n_{\infty }$ and phase rotating at $\mu /\hslash$.

For an occupied uniform two-component Bose-Einstein condensate, the gradient and trapping terms vanish. Dividing each stationary equation by its nonzero field givesThe same relations follow by differentiating the uniform interaction-energy density $e=U_{11}{n}_1^2/2+U_{22}{n}_2^2/2+U_{12}{n}_1{n}_2$ with respect to each number density. When $D=U_{11}{U}_{22}-U_{12}^2\ne 0$, the inverse relations areThese describe a genuinely overlapping mixture only when both densities are nonnegative. An unoccupied component cannot be divided out of its stationary field equation; it requires the appropriate constrained equilibrium condition instead.

Chemical equilibrium for $p+e^{-}\leftrightarrow H+\gamma$ implies ${\mu }_H={\mu }_p+{\mu }_e$, because equilibrium photons have zero chemical potential. Dividing the three nonrelativistic Maxwell-Boltzmann number densities therefore givesThe Hydrogen binding energy is $I=m_p+m_e-m_H=13.6\mathrm{eV}$ in natural units. For ground-state hydrogen, with hyperfine splittings unresolved, $g_H=4$ and $g_p=g_e=2$, so the degeneracy ratio is one. Equivalently one can omit nuclear-spin degeneracy consistently from both the proton and atom. Excited hydrogen levels are exponentially suppressed at recombination temperatures. Also $m_H/m_p=1+O(m_e/m_p)$, so its effect in the prefactor is negligible. ThusThe sign of the exponential follows from the smaller rest mass of the bound atom: cooling favours neutral hydrogen once translational entropy no longer compensates the binding energy.

A local stellar relaxation time depends on the number density $n$ of scatterers. Calling $N$ a total number, as the PDF does, leaves the displayed expression dimensionally wrong: ${\sigma }^3/(G^2{m}^{2}N)$ has units of time per volume. The intended local expression must use $n$, or $N/V$ for a homogeneous sample of volume $V$. We shall also make the normalization of the stellar relaxation time explicit, since an unspecified representative one-dimensional velocity dispersion does not fix an exact numerical coefficient.

In a small-angle gravitational encounter, approximate the relative trajectory by a straight line of speed $u$ and impact parameter $b$. The perpendicular acceleration of one star due to the other is $Gmb/(b^2+u^2{t}^2{)}^{3/2}$. Its impulse is thereforeA star encounters $2\pi bdbnu$ scatterers per unit time in this interval of impact parameters. Independent deflections have random directions, so their variances add as in a random walk. Hence the accumulated squared velocity change has rateThe Coulomb logarithm in stellar dynamics arises because each logarithmic interval of impact parameter contributes equally. The lower cutoff is approximately the strong-deflection scale, $b_{\min }\simeq G(2m)/u^2$ for equal masses, or the physical collision scale if larger. The upper cutoff is a scale on which the local density and straight-line encounter approximation cease to apply, such as the local disk thickness. Thus $\Lambda$ is dimensionless; it is not a number density or an arbitrary dimensional argument of a logarithm.

To display the normalization issue, use the representative relative speed $u=\sqrt{2}\sigma$ and define $T(\kappa )$ by $D_{v}T=\kappa {\sigma }^2$. This givesChoosing $\kappa =3/4$ recovers precisely the printed coefficient after repairing $N$ to $n$:This is an explicitly defined convention, not a uniquely derivable normalization from the supplied data. If instead relaxation means a mean-square change equal to the full initial random-speed variance, $3{\sigma }^2$, the same encounter approximation gives $T_R=4T_R^{\mathrm{printed}\mathrm{convention}}$. Averaging the encounter speed distribution instead of using one representative $u$ changes the coefficient again. The robust result is the scaling $T_R\propto {\sigma }^3/(n{G}^2{m}^2\ln \Lambda )$.

For a transparent solar-neighbourhood estimate, take $n=0.1{\mathrm{pc}}^{-3}$, $m=M_{\odot }$, $\sigma =30{\mathrm{km}\mathrm{s}}^{-1}$ and $b_{\max }=300\mathrm{pc}$. With $G=4.3009\times 10^{-3}\mathrm{pc}{M}_{\odot }^{-1}({\mathrm{km}\mathrm{s}}^{-1}{)}^2$, one obtains $b_{\min }=4.78\times 10^{-6}\mathrm{pc}$, $\ln \Lambda =17.96$, andThese illustrative parameters give an order $10^{13}$--$10^{14}$ year stellar relaxation time, much longer than a galactic lifetime of order $10^{10}$ years. The solar neighbourhood is consequently a collisionless stellar system with respect to encounters between individual stars. Scattering by massive clouds and collective structure is a different process; the estimate does not assert that every mechanism of disk heating is negligible.

In an axisymmetric cylindrical coordinate system, the steady Collisionless Boltzmann equation isLet $\nu =\int f{d}^{3}v$ and $\overline{A}={\nu }^{-1}\int Af{d}^{3}v$. Multiply by $v_R$ and integrate over velocities. The spatial terms give ${\partial }_R(\nu \overline{v_R^2})+{\partial }_z(\nu \overline{v_R{v}_z})$. Integration by parts in $v_R$ supplies $\nu {\Phi }_R-\nu \overline{v_{\varphi }^2}/R$, and the $v_{\varphi }$ derivative supplies $\nu \overline{v_R^2}/R$. The $v_z$ force term vanishes because $v_R$ does not depend on $v_z$. This assumes sufficient decay of the galactic distribution function to discard all velocity boundary terms. The radial Jeans equation is consequentlyReflection symmetry about $z=0$ makes $\nu$ even and $\overline{v_R{v}_z}$ odd. At the midplane the latter vanishes, and ${\nu }^{-1}{\partial }_z(\nu \overline{v_R{v}_z})={\partial }_z\overline{v_R{v}_z}$. Multiplying by $R/\nu$ provesAll quadratic velocities here are population moments, not individual stellar velocities. In particular, symmetry does not make the midplane tilt contribution to asymmetric drift disappear: an odd mixed moment can have a nonzero derivative at zero.

For application to the Milky Way, assume negligible mean radial and vertical flows, and define the circular speed by $v_c^2=R{\Phi }_R(R,0)$. Write $\overline{v_R^2}={\sigma }_R^2$ and $\overline{v_{\varphi }^2}={\overline{v}}_{\varphi }^2+{\sigma }_{\varphi }^2$. Rearranging gives the exact stress-support relationThus the stellar asymmetric drift $v_a=v_c-{\overline{v}}_{\varphi }$ is the right-hand side divided by $v_c+{\overline{v}}_{\varphi }$. For $v_a\ll v_c$, this denominator is approximately $2v_c$. If $\nu \propto e^{-R/h_{\nu }}$ and ${\sigma }_R^2\propto e^{-R/h_{\sigma }}$ locally, the bracket becomes $R/h_{\nu }+R/h_{\sigma }-1+{\sigma }_{\varphi }^2/{\sigma }_R^2$, making the density and dispersion-gradient contributions explicit.

A radially declining random-motion stress supplies some of the force needed to support a galactic disk, so its stars need less ordered azimuthal motion than a population on circular orbits. Hotter, often older, stellar populations therefore generally rotate more slowly than cold populations in the same gravitational potential; this is the observed stellar asymmetric drift. It is a stress-support effect, not a frictional slowing of each orbit. For a velocity ellipsoid aligned approximately with spherical coordinates, $\overline{v_R{v}_z}\simeq ({\sigma }_R^2-{\sigma }_z^2)z/R$, and the midplane tilt term contributes $-({\sigma }_R^2-{\sigma }_z^2)$ rather than zero.

Local measurements of mean rotation and all components of the velocity dispersion, combined with tracer-density and dispersion gradients, can therefore constrain $v_c$ and the radial Newtonian gravitational field. A hotter sample's mean rotation must not be identified directly with the circular speed. The tracer number density need not be the gravitating mass density. Nonaxisymmetric streaming, a nonstationary disk, poorly measured radial gradients, or a neglected tilt term can bias this inference; these are assumptions of the derivation rather than extra free corrections to the circular force.

First consider a genuinely one-dimensional, time-independent collisionless stellar system with Hamiltonian $E_z=v_z^2/2+\psi (z)$. Choose $\psi (0)=0$. For bound vertical oscillations, a phase-mixed population has equal occupations of the upward and downward branches of each orbit. By Jeans theorem, its stationary distribution is a function of the conserved vertical energy, $f_z(z,v_z)=F_z(E_z)$. Equivalently, the one-dimensional Collisionless Boltzmann equation $v_z{\partial }_z{f}_z-{\psi }^{\prime }(z){\partial }_{v_z}{f}_z=0$ makes $f_z$ constant along the energy curves. Branch-dependent incoming and outgoing populations or time-dependent waves would not satisfy this assumption.

Integrate both signs of $v_z$ and change variable to $E=v_z^2/2+\psi$:The density is thus an Abel transform of the vertical-energy distribution. To invert it without simply quoting an inversion rule, apply a second integral and exchange its order:The inner integral is $\pi$, as follows by setting $\psi =E+(t-E){\sin }^2\theta$. Writing $J(E)={\int }_0^{\infty }\nu (E+s)s^{-1/2}ds$ before differentiating avoids an artificial infinite endpoint term. Under sufficient differentiability and decay,The vertical-energy Abel inversion of a stellar distribution is thereforeThis proves the normalization, including the factor from the two velocity signs. The additive origin of $\psi$ merely changes the origin of energy.

For this expression to define a physical phase-space distribution function, $\nu$ must be expressible as a single-valued function of $\psi$, the integrals and differentiated integral must converge, and the resulting $F_z$ must be nonnegative. Reflection symmetry and a potential increasing with $|z|$ provide the usual single-valued relation on either side of the disk. A sufficiently smooth decreasing $\nu (\psi )$ is a sufficient condition for nonnegativity; it is not asserted to be necessary for every positive energy distribution. If there is a finite escape energy, replace the infinite limits by that energy and impose the appropriate cutoff and boundary conditions. The displayed infinite-limit formula presumes either a confining potential or a distribution extended by zero above its allowed energy, with no omitted boundary population.

A three-dimensional galactic disk does not automatically meet these hypotheses. Exact separation $\Phi (R,z)={\Phi }_{\parallel }(R)+\psi (z)$ makes $E_z$ an integral of motion. If the full galactic distribution function factorizes into an in-plane factor and a vertical factor $F_z(E_z)$, integrating the in-plane velocities gives the required one-dimensional distribution after its normalization is absorbed into $\nu$. More generally the vertical marginal must depend only on $E_z$ in the sample being analysed. Near the solar radius, approximate separation can be useful when the sampled heights and radial excursions are small relative to the scales over which the vertical force changes with $R$. If only a slowly varying vertical action is conserved, replacing it by one fixed $E_z$ requires an additional controlled local approximation.

The population must be sufficiently stationary and phase mixed, with negligible mean vertical flow, and the data must measure a well-defined tracer population with a known selection function. Radial coupling and the tilted velocity ellipsoid must be negligible or modelled. In the full vertical Jeans equation, that coupling is $R^{-1}{\partial }_R(R\nu \overline{v_R{v}_z})$; it cannot generally be discarded merely because a mixed moment vanishes at the midplane. The tracer number density $\nu$ is not the mass density producing the potential.

To infer the local mass, measure the tracer density as a function of height and its vertical velocity distribution near the midplane. With $\psi (0)=0$, the latter directly samples $F_z(v_z^2/2)$. Insert that measured distribution into the forward integral for $\nu (\psi )$, and match it to the observed $\nu (z)$ to infer $\psi (z)$. Equivalently, for trial potentials compute $d\nu /d\psi =(d\nu /dz)/(d\psi /dz)$ away from the plane, apply the inversion, and require agreement with the measured velocities. Density data alone do not determine both an unknown potential and an unknown energy distribution.

Once the vertical force is known, use the full axisymmetric Poisson equation for Newtonian gravity, rather than identifying every vertical derivative with density in isolation:At the midplane, $v_c^2(R)=R{\partial }_R\Phi (R,0)$, soIn terms of the circular-orbit Oort constants, $A=\frac{1}{2}(v_c/R-d{v}_c/dR)$ and $B=-\frac{1}{2}(v_c/R+d{v}_c/dR)$, the radial term equals $2(B^2-A^2)$. It vanishes for a locally flat rotation curve, not for an arbitrary rotating disk. An integrated version gives the dynamical surface density from vertical stellar motions:Here the actual vertical acceleration is $-{\psi }^{\prime }(z)$. The sign of the radial correction follows directly from the Poisson equation for Newtonian gravity.

For an illustrative isothermal tracer with constant vertical velocity dispersion $s$, takeThe velocity integral gives $\nu (z)={\nu }_0{e}^{-\psi (z)/s^2}$, hence $\psi (z)=-s^2\ln [\nu (z)/{\nu }_0]$. For a locally flat rotation curve this yieldsThis example shows explicitly how a measured tracer scale and velocity scale constrain the gravitating density; using the tracer's own density as the mass source would instead impose an unjustified self-gravitating model.

The inferred density includes luminous and faint stars, gas and dark matter. Comparing it with an independently estimated baryonic census tests for extra gravitating mass. An excess need not be a dark disk: a dark matter halo also contributes to the local force. Conversely, agreement with the baryonic density locally does not exclude a halo at other radii or heights. Vertical disequilibrium, radial mixing, errors in selection or tracer populations, and neglected velocity tilt can mimic a mass discrepancy, so the one-dimensional equilibrium assumptions must be tested when interpreting such an inference.

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.

At fixed number density $n$, the Electron current givesThe second equality uses Ampère's law without displacement current and stationary ions. A perturbation with characteristic perpendicular wave number $k_{\perp }$ therefore produces $\delta u_e\sim \alpha k_{\perp }\delta B$. Its nonlinear magnetic-advection rate isFor the anisotropic electron magnetohydrodynamics wave, $k\simeq k_{\perp }$ and $|\omega |\sim v_A{d}_i|k_{\parallel }|k_{\perp }$. Equating this to the nonlinear rate by critical balance givesThis is small-amplitude but strong turbulence: a small perturbation can have order-one nonlinear distortion over one wave period because its perpendicular gradients are much larger than its parallel gradients.

At fixed particle number $N_p=\int |\psi {|}^{2}d\mathbf{x}$, an equilibrium wavefunction is a constrained critical point of the energy functional. Introduce the chemical potential $\mu$ as the Lagrange multiplier for $N_p$, and impose $\delta (E-\mu N_p)=0$. The resulting stationary cubic-quintic Gross–Pitaevskii equation isThe time-dependent equilibrium has $\psi (\mathbf{x},t)=e^{-i\mu t/\hslash }{\psi }_0(\mathbf{x})$, so its number density is stationary even though its quantum phase changes uniformly. The specified particle number fixes $\mu$ together with the boundary conditions. In an infinite uniform bulk, particle number is understood through a finite volume or a prescribed bulk density.

Use the central number density $n_c$, radius $R$ and central healing length $\xi$ from the preceding part. The vortex energy means the excess over the no-vortex ground state at fixed particle number. Choose an intermediate matching radius $\xi \ll L\ll R$. Inside $L$, the trapped number density differs little from $n_c$, so the resolved uniform vortex core givesOutside the core, the singly quantized superfluid velocity is $u_{\theta }=\hslash /(mr)$. Its leading flow kinetic energy in the parabolic background is thereforeAdding the regions cancels the arbitrary matching radius. Sending $L/R\to 0$ while $\xi /L\to 0$ gives the matched vortex energy in a parabolic condensateThe $-1/2$ comes from the decrease of background number density away from the centre. It would be missed by treating the whole trap as uniform. To this accuracy, using the unperturbed TF density in the outer region is appropriate at fixed $N$: the first variation of trapping plus interaction energy is $\mu \delta N$ and therefore vanishes after the small normalization adjustment. The resolved core energy supplies the finite core correction $L_{01}$, while remaining cloud-edge and density-relaxation corrections vanish in the small-core TF limit. This is an asymptotic energy estimate, not an exact formula at finite $R/\xi$.

Write $g=V_0$ for the effective coupling and keep $N$ for the number obtained by the stated two-dimensional normalization. A repulsive interaction, $g>0$, permits the Thomas–Fermi approximation for a condensate: density-gradient kinetic energy is small compared with the trapping and interaction terms over most of the cloud. The stationary Gross–Pitaevskii equation then givesand $n_{\mathrm{TF}}=0$ outside the cloud, up to the narrow edge region. HereNormalize the parabolic number density directly:Consequently the two-dimensional Thomas–Fermi condensate parameters areWith $a_{\mathrm{ho}}=\sqrt{\hslash /(m\omega )}$ and central healing length $\xi =\hslash /\sqrt{2m\mu }$, the bulk approximation requiresIndeed $R/\xi =2\mu /(\hslash \omega )\gg 1$. In the coupling convention printed in the paper, formal substitution of $g=4\pi {\hslash }^{2}a/m$ givesThis is the requested relationship in that convention; $\omega$ sets the physical size and chemical potential, while the interaction-strength ratio to $\hslash \omega$ is independent of $\omega$.

There is a dimensional convention to make explicit. A physical three-dimensional scattering length $a$ makes $4\pi {\hslash }^{2}a/m$ a three-dimensional contact coupling, whereas a genuinely two-dimensional particle-number normalization requires coupling units of energy times area. For an axial factor $\chi (z)$ normalized to one, the effective two-dimensional contact coupling is $g_{2D}=(4\pi {\hslash }^{2}a/m)\int |\chi {|}^{4}dz$. For a Gaussian of width $a_z$ this is $4\pi {\hslash }^{2}a/(m\sqrt{2\pi }{a}_z)$, and the physical TF condition becomes $Na/a_z\gg 1$ up to numerical constants. Alternatively the printed coupling can be retained if $N$ is a line density. The paper supplies no axial scale, so its $Na$ expressions require that effective-unit or line-density convention; the formulas in terms of $g$ above are unambiguous. The approximation fails inside vortex cores and at the cloud edge, and it is not a repulsive TF ground-state construction for $g\le 0$.

For a positive uniform number density $n_{\ast }$ and constant superfluid velocity $v\hat{\mathbf{x}}$, current divergence vanishes. The stationary density equation therefore imposes $(\gamma -\Gamma n_{\ast })n_{\ast }=0$, giving $n_{\ast }=\gamma /\Gamma$. The quantum potential vanishes, so the phase equation sets $\mu =m{v}^2/2+V_0\gamma /\Gamma$. Thus the uniform-current pumped polariton condensate isDirect substitution verifies the equation: the kinetic term gives $m{v}^2/2$, the interaction gives $V_0\gamma /\Gamma$, and the gain-loss term is zero. A constant overall phase is arbitrary.

With the stated $\gamma ,\Gamma >0$, this nonzero plane wave exists for any real $v$ and real interaction coefficient $V_0$ when $\mu$ is determined by the solution. If a particular real chemical potential is prescribed instead, the existence condition isEquality gives the uniform condensate at rest. A nonzero density generally requires $\gamma /\Gamma >0$; a zero nonlinear-loss coefficient with positive pumping would not allow this nonzero stationary balance. The vacuum is a separate zero solution, unstable to small perturbations for positive pump.

Existence should not be confused with stability. For completeness, linearize in the comoving frame with $\psi ={\psi }_{\ast }[1+u+iw]$, and let ${ϵ}_k={\hslash }^2{k}^2/(2m)$. The linear real-amplitude and phase equations givesoFor $V_0\ge 0$ both growth rates have nonpositive real parts, including the neutral uniform-phase mode. For $V_0<0$, sufficiently small nonzero $k$ makes the constant term negative, so one growth rate is positive: the plane wave still exists but is unstable in an infinite system. Uniform drift adds only a Doppler phase to these growth rates in this model.

Here equilibrium means a time-independent number density and current, with a common temporal quantum phase. Write $\psi (\mathbf{x},t)=e^{-i\mu t/\hslash }\Psi (\mathbf{x})$ with real $\mu$, since an imaginary temporal frequency would change the density. The stationary pumped Gross–Pitaevskii equation isThe chemical potential is the common phase-rotation frequency times $\hslash$; this open system has no conserved particle-number constraint fixing it as in the conservative problem. Local gain and loss need not cancel separately in an inhomogeneous equilibrium, because a stationary condensate number current can transport particles between gain and loss regions.

Work in natural units $\hslash =c=k_B=1$, at temperatures far below the $W$-boson mass but of order or above the Electron mass. The Fermi interaction then supplies a local four-fermion amplitude. The Fermi constant has dimension energy${}^{-2}$. In scattering of a thermal lepton from a nonrelativistic nucleon, the nucleon current matrix element has size $m_N$, while the lepton current has size $T$. Thus $\mathcal{M}\sim G_F{m}_{N}T$. The two-body relativistic scattering cross-section contains the kinematic factor $s^{-1}\sim m_N^{-2}$; when the final and initial lepton energies are comparable, their momentum ratio is of order one. Consequently $\sigma \sim |\mathcal{M}{|}^2/m_N^2\sim G_F^2{T}^2$. This estimates temperatures above the Electron mass and mass-splitting thresholds; those scales only affect the dimensionless rate function near freeze-out.

The light leptons in the thermal bath have number density $n\sim T^3$ and speed of order unity. Their conversion rate per nucleon is thereforeFor example, $n+{\nu }_e\leftrightarrow p+e^{-}$ and $n+e^{+}\leftrightarrow p+{\bar{\nu }}_e$ change the neutron-to-proton ratio. The bath density here is the lepton density, not the much smaller baryon density. Dimensionless weak couplings, phase-space integrals and the several reaction channels set a numerical coefficient. Electron masses and the neutron-proton splitting also become important for a precision rate near cosmological weak freeze-out.

In the printed convention $\rho$ is energy density, since the gravitational source is $8\pi G\rho /(3c^2)$. For nonrelativistic pressureless matter, ${\rho }_m={\rho }_{m,0}(1+z{)}^3$. Define the cosmological density parameters byThe Friedmann equation then givesLet $n_c$ be the constant comoving number density of absorbers, and let ${\sigma }_p$ be their fixed proper interception area. Their proper number density is $n_p(z)=n_c(1+z{)}^3$, with $a_0=1$. The redshift-time relation gives a local proper photon-path incrementFor an infinitesimal path, the expected number of interceptions, and to first order the probability of an interception, is $d\mathcal{N}=n_p{\sigma }_{p}d\ell$. Thus absorber incidence and comoving number density implyTwo factors survive: proper density brings three powers of $1+z$, whereas proper path length brings the inverse of one power. A fixed comoving area instead of the assumed fixed proper area would change the result. The formula is an incidence rate, not a probability bounded by one over a finite redshift interval; multiple interceptions can occur.

For highly ionized hydrogen in photoionization equilibrium, balance the photoionization rate against radiative recombination:Thus the neutral fraction is $n_{\mathrm{HI}}/n_H\simeq \alpha (T)n_H/{\Gamma }_{\mathrm{HI}}$ and the neutral column density across a proper path $L$ isFor an illustrative diffuse gas with $T\sim 10^4\mathrm{K}$, $\alpha \sim 4\times 10^{-13}\mathrm{c}{\mathrm{m}}^3{\mathrm{s}}^{-1}$, ${\Gamma }_{\mathrm{HI}}\sim 10^{-12}{\mathrm{s}}^{-1}$, and $n_H\sim 10^{-5}\mathrm{c}{\mathrm{m}}^{-3}$, the neutral fraction is only $4\times 10^{-6}$. A proper path of $100$ kiloparsecs then gives $N_{\mathrm{HI}}\sim 10^{13}\mathrm{c}{\mathrm{m}}^{-2}$, enough to produce a weak Lyman-alpha absorption feature even though the gas is almost entirely ionized.