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Use units $c=1$, and let $r$ be the areal radial coordinate of the FRW metric, rather than the radial comoving distance when $k\ne 0$. On a constant-cosmic time slice the spatial metric tensor isThe determinant of the spatial metric tensor supplies the factor $(1-k{r}^2{)}^{-1/2}$. Integrating over the solid angle gives $d{V}_{\mathrm{shell}}=4\pi a^3{r}^2|dr|/\sqrt{1-k{r}^2}$. An incoming radial null geodesic obeysConsequently the proper shell volume sampled by an emission-time interval of positive width $dt$ is $4\pi a^2{r}^{2}dt$. Multiplying by the proper number density proves the source counts on an FLRW past light cone:Here $r(t)$ is evaluated on the observer's past light cone, not along a fixed source worldline. The curvature factor cancels between the spatial volume and the radial light speed; it has not been set to one.

Cosmic inflation addresses the horizon problem by extending the causal past of the observed region, the flatness problem by damping the curvature contribution, and the cosmological monopole problem by diluting unwanted relics. A canonical inflaton can drive this expansion when its potential energy density exceeds twice its kinetic energy density.

Use the unreduced Planck mass $m_P=G^{-1/2}$, consistent with the coefficient in the Friedmann equation, and take $V_0>0$, $c\ne 0$. Seek the exponential-potential power-law inflation solution $\varphi =A\log (\beta t)$ and $a=a_{\ast }(t/t_{\ast }{)}^p$. The kinetic energy density is proportional to $t^{-2}$. The inflaton potential must have the same scaling, soSubstitution into the inflaton equation of motion givesThe Friedmann equation now reduces toConsequently the nontrivial positive-potential solution isA positive $V_0$ requires $p>1/3$. Accelerating expansion occurs exactly for $p>1$, givingThis is an exact solution, not a use of the slow-roll approximation. The first Hubble slow-roll parameter and equation-of-state parameter are constant:If the reduced Planck mass $M=m_P/\sqrt{8\pi }$ is used instead, the dimensionless exponential slope is $\lambda =c/\sqrt{8\pi }$ and the exponent becomes $p=2/{\lambda }^2$; these are the same solution.

For $p>1$, the comoving Hubble radius decreases as $(aH{)}^{-1}\propto t^{1-p}$. A region initially in causal contact can grow beyond that radius before the hot radiation-dominated universe begins. In the ideal power law extrapolated all the way to $t=0$, the particle horizon integral ${\int }_0^{t}d{t}^{\prime }/a(t^{\prime })$ diverges, so the solution has no finite past particle horizon. A finite, physically trustworthy inflationary interval instead requires a sufficiently large number of e-folds to enlarge an initially causal patch.

For a small curvature contribution, the Friedmann equation gives $|\Omega -1|=|k|/(a^2{H}^2)\propto t^{2-2p}\to 0$, solving the flatness problem. Conserved nonrelativistic magnetic monopoles have number density proportional to $a^{-3}$, while the inflaton energy density is proportional to $t^{-2}$; their energy-density ratio therefore falls as $t^{2-3p}$. This gives the required dilution, provided later reheating does not recreate them. Inflationary shear damping similarly suppresses anisotropic expansion.

The main failure is no automatic graceful exit or reheating. The first Hubble slow-roll parameter never reaches one, and the exponential inflaton potential has no finite minimum about which the field can oscillate. Additional dynamics are needed for a graceful exit from inflation and transfer to a hot radiation-dominated universe. The ideal solution also retains a curvature singularity at $t=0$; solving the horizon problem does not itself explain the beginning of the universe. For $c=0$, the constant inflaton potential instead permits constant-field de Sitter spacetime expansion; the logarithmic field solution is inapplicable.

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.

The factor called $N$ in the formula must be the stellar number density $n$, not a dimensionless count. Indeed ${\sigma }^3/(G^2{m}^2)$ has dimensions time per volume, so a density in the denominator is essential. If $N_{\ast }$ is the population in a representative volume $V$, then $n=N_{\ast }/V$.

For a weak encounter between equal-mass stars, let $u$ be the asymptotic relative speed and $b$ the impact parameter. In the straight-line impulse approximation, one star's transverse kick isThe other receives the opposite kick, so the relative-velocity change is $\Delta u_{\perp }=4Gm/(bu)$. Encounters in $(b,b+db)$ occur at rate $2\pi bdbnu$. Independent random kicks add in squared magnitude, givingThis is the random-walk derivation of stellar relaxation. The Coulomb logarithm in stellar dynamics is $\ln \Lambda =\ln (b_{\max }/b_{\min })$. The upper cutoff is the size or inhomogeneity scale beyond which independent local encounters are inappropriate. The lower cutoff is of order the strong-deflection scale $G(2m)/u^2$, or a larger stellar-size/softening scale if applicable. A self-gravitating system commonly has $\Lambda$ of order its total number of stars; that population count is distinct from $n$ in the diffusion rate.

To state the numerical convention explicitly, use representative encounter speed $u=\sqrt{2}\sigma$ and define the relaxation estimate from the relative-velocity diffusion rate with reference squared velocity $3{\sigma }^2$. This givesThis recovers the printed coefficient under a specified convention. There is no convention-independent exact numerical prefactor determined by a representative one-dimensional velocity alone. Using the single-star diffusion rate with the same $3{\sigma }^2$ reference instead gives four times this time; averaging over a full velocity distribution also changes the prefactor. The relaxation time conventions in stellar dynamics must therefore distinguish single-star and relative kicks. The physical scaling is $T_R\propto {\sigma }^3/(n{G}^2{m}^2\ln \Lambda )$, and the printed $N$ must be interpreted as number density.

There is a genuine flux typo in the original PDF. A received bolometric radiative flux must have inverse-square luminosity distance, so the physically consistent thermal bremsstrahlung expression is $S_X\propto d_L^{-2}\int n_e^2{T}_e^{1/2}dV$. The printed single power of $d_L$ cannot imply the requested Hubble constant scaling.

For an approximately uniform, spherical galaxy cluster let its depth be $\ell =q\theta d_A$, where $q$ is a fixed geometrical factor. With fixed profile coefficients $A,B$, the Rayleigh-Jeans law limit of the thermal Sunyaev-Zeldovich effect and the bolometric X-ray radiative flux giveEliminate the electron number density and use $d_L=(1+z{)}^2{d}_A$:The cluster distance from Sunyaev-Zeldovich and X-ray signals is consequentlyAt low cosmological redshift, $d_A\simeq cz/H_0$, sowhen the other observables and profile parameters are held fixed. With the PDF's literal $d_L^{-1}$ instead, $S_X/D^2\propto \ell /(T_e^{3/2}{d}_L)=q\theta /[T_e^{3/2}(1+z{)}^2]$, independent of distance: this explicitly shows why the correction is necessary.

Real galaxy clusters are neither uniform nor generally spherical. The inferred depth depends on the ratio of line of sight depth to transverse extent; elongation and orientation selection therefore bias the distance. density clumps enhance X-ray emission, which weights $n_e^2$, more than the thermal Sunyaev-Zeldovich effect, which weights $n_e$. An unmodelled gas clumping factor greater than one makes the inferred distance too small and the inferred Hubble constant too large in this simple model. Temperature gradients and multiple thermal components bias the X-ray spectral temperature relative to the pressure-weighted Compton y parameter. Metals alter the cooling spectrum, and finite X-ray bands require a spectral and redshift correction to the bolometric approximation. Relativistic thermal corrections, the kinetic Sunyaev-Zeldovich effect, radio sources and primary Cosmic microwave background anisotropy contaminate the signal. Finally, calibration, angular-profile fitting and low-redshift peculiar velocities affect the observables and the relation between measured redshift and expansion distance.

Define neutral-hydrogen column density along a ray by $N=\int n_{\mathrm{HI}}d\ell$, in atoms per unit area. The neutral-hydrogen column-density distribution per redshift isso $f_{z}dNdz$ is the expected number of absorbers in those intervals along one sightline. A logarithmic column bin instead has distribution $N\ln (10)f_z$; the bin convention must be specified.

Let $n_c$ be a constant comoving number density and ${\sigma }_p$ a constant physical interception area. The corresponding physical number density is $n_p=n_c(1+z{)}^3$. Proper light-path length satisfies $d\ell =c|dt|=cdz/[(1+z)H(z)]$. The expected intersection count, or first-order probability in infinitesimal $dz$, is thereforeThe Friedmann equation, the constant-equation-of-state density scaling for pressureless matter, and the present cosmological density parameters giveCombining them yields the absorber incidence and comoving number density relationHere the constant cross-section is physical, as required for this result; a fixed comoving cross-section would instead decrease in physical area as $(1+z{)}^{-2}$. It is convenient to define absorption distance by $dX/dz=(H_0/H(z))(1+z{)}^2$ and $f_X=f_z/(dX/dz)$, so a population with these non-evolution assumptions has constant incidence per $X$.

Use absorption distance $X$ and its distribution $f_X$. Summing hydrogen columns per unit $X$ gives $\int N{f}_{X}dN$. The physical light-path length per $X$ isThe mean neutral-hydrogen number density is therefore $(H_0/c)(1+z{)}^3\int N{f}_{X}dN$. Divide by $(1+z{)}^3$ to obtain its comoving density, multiply by hydrogen atomic mass, and divide by today's mass critical density ${\rho }_{\mathrm{crit},0}=3H_0^2/(8\pi G)$. This derives the neutral gas density from an absorption distribution:If “neutral gas” includes helium, replace $m_H$ by ${\mu }_H{m}_H$, with a stated abundance correction approximately ${\mu }_H\simeq 1.3$. Equivalently, for the distribution per redshift the prefactor inside the integral is $m_{H}H(z)/[c{\rho }_{\mathrm{crit},0}(1+z{)}^2]$ multiplying $N{f}_z$. In the PDF's energy-density convention, both the mass density and critical density are multiplied by $c^2$, leaving their ratio unchanged. The usual quoted abundance uses comoving mass density relative to today's critical density; a ratio of physical density to the epoch's critical density instead multiplies this answer by $(1+z{)}^3{H}_0^2/H(z{)}^2$.

To infer total gas, one needs the neutral fraction $x_{\mathrm{HI}}=N_{\mathrm{HI}}/N_H$ for each absorber. The ionized gas density from an absorption distribution isThis is an ionization-corrected integral, not an inference available from the observed columns alone. For optically thin, almost fully ionized hydrogen, photoionization equilibrium gives $\Gamma n_{\mathrm{HI}}=\alpha (T)n_e{n}_{\mathrm{HII}}\simeq \alpha (T)n_H^2$. If the absorbing depth $L$ is specified, $N=\alpha (T)n_H^{2}L/\Gamma$ andThus photoionization rate, temperature and depth are all needed.

One useful physical closure sets $L$ to a local Jeans length, $L\propto (f_{g}T/n_H{)}^{1/2}$, with local gas fraction $f_g$. Solving $N\propto [\alpha (T)/\Gamma ](f_{g}T{)}^{1/2}{n}_H^{3/2}$ then gives the Jeans-length ionization correction for the Lyman-alpha forest:This illustrates why weak, highly ionized forest lines can dominate the total gas inventory even though damped Lyman-alpha systems dominate the neutral-column-weighted inventory. It is a model for locally confined optically thin gas; optically thick Lyman limit systems and damped Lyman-alpha systems require a separate radiative transfer treatment. The physical basis and observational application of this closure are developed in Schaye's absorption-based matter-density calculation.

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.