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The G-dwarf problem is the observed shortage of metal-poor, long-lived disc stars relative to a simple closed-box model of galactic chemical evolution. In the fully evolved comparison, the half-rate proportional-infall model of galactic chemical evolution lowers the low-$Z$ fraction from $1-e^{-Z/y}$ to $Z/(2y)$, and prevents the closed-box tail from extending to arbitrarily high stellar metallicity. It therefore reduces the relative weight of the earliest metal-poor generation and the very metal-rich tail.

This particular constant-proportional-infall model is an improvement, not a complete fit. It still forms a metal-poor population immediately from its initially pristine reservoir, with nonzero constant $dN/dZ$ all the way down to zero, and predicts a broad linear distribution followed by an abrupt upper cutoff instead of a generally smooth, concentrated observed distribution. Comparison with observed abundance histograms must first use the same coordinate: its logarithmic metallicity distribution rises in proportion to $Z$ up to that cutoff, rather than remaining flat in logarithmic bins. Local G-dwarf measurements exhibit a prominent logarithmic-abundance peak and a paucity of low-abundance stars; translating iron abundances into total $Z$ also requires an abundance-ratio model. More realistic histories can vary the galactic gas inflow and star formation rate, allow recycling and delayed enrichment, and account for migration and selection of different stellar populations; simply tuning a constant stellar yield does not remove all these shape deficiencies.

Putting the whole assumed gas reservoir into stars during the nominal galactic dynamical time givesAn order-unity shorter collapse clock would raise this to a few thousand solar masses per year. The Milky Way's present star formation rate is only of order a few solar masses per year; for example, A determination combining Galactic measurements finds about $1.7M_{\odot }{\mathrm{yr}}^{-1}$ with a specified initial mass function. The extreme rapid-collapse rate is therefore roughly three orders of magnitude larger, resembling an intense short-lived starburst galaxy phase rather than ongoing ordinary disc star formation. This estimate assumes all the quoted mass can form stars; it would be smaller if some mass were nonbaryonic dark matter or if the conversion efficiency were below unity.

Metal absorption lines depend on the temperatures, ionization states and luminosity weights of the contributing stellar populations, as well as their stellar metallicity. A young stellar population dominated by hot luminous stars can have weak metal absorption lines even at substantial abundance, so weak lines alone do not establish low metallicity, an aspect of the age-metallicity degeneracy.

Low-mass main sequence stars live for much of the Milky Way's age, retaining representatives of many generations, while phase mixing disperses their birth associations around their orbits. In an approximately axisymmetric galactic disk, a sufficiently complete local census can therefore sample the mixed annular stellar population at the Sun's radius despite its small volume. This is a representative local-radius sample, not automatically an unbiased whole-disc census: different vertical scale heights, selection effects and radial migration require corrections when inferring integrated population fractions.

Older disc stellar populations generally have larger velocity dispersions, partly through accumulated galactic disk heating, so random-motion stresses supply more of their support against gravity. The radial Jeans equation then requires a smaller mean rotation than the circular speed, producing larger stellar asymmetric drift than in dynamically colder young stars.

Typical present elliptical galaxies are dominated by old stellar populations: their brightest short-lived stars have disappeared, leaving faint long-lived stars and stellar remnants, so luminosity per unit stellar mass is low. Ongoing star formation in many spiral galaxies and irregular galaxies supplies luminous young stars, giving a smaller stellar mass-to-light ratio in a fixed band; dark matter, the initial mass function and aperture must also be specified if the quoted ratio is dynamical.

Many open clusters are weakly bound and are dispersed by two-body relaxation, stellar evaporation and tidal heating, including encounters with molecular clouds in the galactic disk, so their present age histogram preferentially retains young survivors. More massive compact globular clusters can survive much longer, while the released low-mass stars remain as old field stellar populations: disruption removes clusters from the catalogue without removing their stars from the disc.

Let $u_i=⟨v_i⟩$, ${\sigma }_{ij}=⟨(v_i-u_i)(v_j-u_j)⟩$, and $P_{ij}=\rho {\sigma }_{ij}$. The Jeans equation obtained by taking the first velocity moment of the Collisionless Boltzmann equation isFor an equilibrium population multiply by $x_i$ and integrate. Integration by parts givesAssume the integrals exist and the boundary flux vanishes, as for an isolated finite system with adequate decay. With the stellar kinetic-energy tensor, stellar pressure-energy tensor and stellar potential-energy tensor defined bywe obtain the equilibrium tensor virial theoremThe stellar kinetic-energy tensor describes ordered motion, whereas the stellar pressure-energy tensor describes random motion. In particular, $T_{ij}=0$ does not mean the stars have no kinetic energy. If a surface flux is retained, the right side of the displayed equilibrium identity is that surface tensor; it cannot be discarded for an arbitrary finite aperture. For an isolated self-gravitating population, $W_{ij}$ is symmetric by pairwise interchange in the Newtonian interaction integral, and the time-dependent identity is $\frac{1}{2}{\ddot{I}}_{ij}=2T_{ij}+{\Pi }_{ij}+W_{ij}$ with $I_{ij}=\int \rho x_i{x}_j{d}^{3}x$.

For the spherical Newtonian gravitational potential, ${\partial }_j\Phi ={\Phi }^{\prime }(r)x_j/r$. In the pressure-supported case the tensor virial theorem gives ${\Pi }_{ii}=\int \rho {\Phi }^{\prime }(r)x_i^2/r{d}^{3}x$. The notation ${\Pi }_{RR}$ used in this question is the planar virial trace ${\Pi }_{xx}+{\Pi }_{yy}$, not just the integral of the local cylindrical radial velocity dispersion. In cylindrical coordinates this trace includes both $v_R^2$ and $v_{\varphi }^2$. Summing the two Cartesian components therefore givesThe distinction explains the factor of two in the spherical limit.

Use ordinary spherical coordinates, $R=r\sin \theta$, $z=r\cos \theta$. The mass density separates as ${\rho }_0{r}^{-\gamma }h(\theta )$, whereBoth potential-energy integrals have the same radial factor $2\pi {\rho }_0{\int }_0^{\infty }{r}^{3-\gamma }{\Phi }^{\prime }(r)dr$. When this factor is finite and positive it cancels, leaving the scale-free tracer virial ratioThe angular factor in the printed intermediate formula has its sine and cosine interchanged. Keeping its sine/cosine weights while using that printed factor would instead produce the opposite sign in the final small-flattening correction. Merely measuring $\theta$ from the equatorial plane cannot fix this: the measure and both moment weights would have to change as well.

The angular ratio is finite and positive for every fixed $q>0$ and finite real $\gamma$, because $h$ is positive and bounded above and below. However, the claim that the unrestricted global virial integrals are always well-defined needs a hypothesis. In a point-mass Newtonian gravitational potential, the common radial factor is proportional to ${\int }_0^{\infty }{r}^{1-\gamma }dr$: convergence at zero requires $\gamma <2$, while convergence at infinity requires $\gamma >2$. There is no such $\gamma$. If the global energies diverge, the boxed angular ratio is still the ratio of the two potential-energy integrals with identical spherical inner and outer cutoffs, independent of the cutoffs. It is not a quotient of two finite global stellar pressure-energy tensors; applying a finite-aperture tensor virial theorem additionally requires its surface terms. These convergence and boundary qualifications are separate from the printed angular typo.

Set $ϵ=q^{-2}-1$. For modest flattening, the Taylor expansion is $h=1-(\gamma /2)ϵ{\cos }^2\theta +O({ϵ}^2)$. The required elementary angular integrals areThese follow directly by $t=\cos \theta$, or from the supplied gamma function identity. Hence the numerator is $(4/3)[1-\gamma ϵ/10+O({ϵ}^2)]$ and the denominator is $(2/3)[1-3\gamma ϵ/10+O({ϵ}^2)]$. The flattening–anisotropy virial relation becomesFor a decreasing tracer mass density ($\gamma >0$), oblate flattening ($q<1$) requires more random kinetic support per in-plane direction than vertically. In a spherical Newtonian gravitational field, directional orbital anisotropy supplies the flattening rather than an anisotropic force law. A steeper radial mass density amplifies the required velocity dispersion anisotropy. A prolate tracer reverses the leading inequality. The relation constrains integrated velocity dispersions, not a unique local galactic distribution function.

At each fixed semimajor axis, let $B=b_{3/2}^2(\alpha )/b_{3/2}^1(\alpha )$ and $Z_f=B{e}_{\mathrm{pl}}$. The planet's longitude of pericentre is the zero of longitude. The secular perturbation equation becomes $\dot{z}=iA(z-Z_f)$. Integrating it with the initial condition $z(0)=0$ givesThe forced eccentricity vector is the constant real $Z_f$; the free eccentricity vector is $-Z_f{e}^{iAt}$. Consequently the complex eccentricity follows a circle in the Argand plane centered at $Z_f$, with radius $Z_f$, starting at the origin and moving counterclockwise. Its initial velocity is downward, $\dot{z}(0)=-iA{Z}_f$. The secular period is $2\pi /A$, andThe longitude of pericentre is undefined at the origin, but the complex eccentricity passes smoothly through it. At $At=\pi$, $z=2Z_f$ and the orbit is maximally eccentric and aligned with the planet.

Using the small-$\alpha$ Laplace coefficient expansion givesAlso $n=n_{\mathrm{pl}}{\alpha }^{-3/2}$ to leading order in the planet-to-star mass ratio. The secular forcing by a distant outer planet is thereforeA relative $O({\alpha }^2)$ correction to $A$ can accumulate a phase error at very long times; the leading-frequency formula is not a uniform-in-time expansion.

If $a_2=a_1(1+\delta )$ with $\delta =\delta a/a_1\ll 1$, then $Z_{f,2}/Z_{f,1}=1+\delta +O({\delta }^2)$ and $A_2/A_1=1+3\delta /2+O({\delta }^2)$. The outer planetesimal therefore has a slightly larger forced eccentricity circle and a shorter secular period. Its accumulated secular phase lead is $\Delta (At)\simeq 3A_{1}t\delta /2$. Initially both circles start at the origin, but their eccentricity vectors progressively lose alignment. Linearizing the exponential in $\delta$ additionally requires $|A_{1}t\delta |\ll 1$.

The orbit crossing is caused by this accumulating differential secular phase, even though individual orbital eccentricities remain small. To first order in orbital eccentricity, the radius of an orbit at fixed inertial longitude $\theta$ isFor an infinitesimally neighboring orbit,This derivative is evaluated within the first-order orbital-shape model; derivatives of the neglected quadratic terms give relative $O(Z_f)$ corrections near crossing. The least separation over longitude is consequently $\delta a[1-|z+a{\partial }_{a}z|]$. The differential secular orbit-crossing criterion is $|z+a{\partial }_{a}z|=1$. Since $Z_f\propto a$ and $A\propto a^{3/2}$,The first term stays bounded by $4Z_f\ll 1$, whereas the second grows in proportion to time. To leading order in $Z_f$, crossing begins when $(3/2)Z_{f}At=1$. ThusMore precisely, in this first-order orbital-shape approximation the exact first crossing lies within a fractional $O(Z_f)$ window of this value, since the magnitude differs from $(3/2)Z_{f}At$ by at most $4Z_f$. Higher-order Laplace coefficients give additional corrections. The local derivation takes the neighboring-orbit limit before the long-time limit; for a finite separation it requires $\delta a/a_1\ll Z_f$ near crossing. A broad continuous disk contains such neighboring orbits. The crossings first appear in directions selected by the phase of $z+a{\partial }_{a}z$, not simultaneously at every longitude.

For the collision energy estimate adopt the phase-averaged mean orbital eccentricity $\overline{e}=4Z_f/\pi \simeq 5\alpha e_{\mathrm{pl}}/\pi$. The specified velocity estimate givesUsing the projectile kinetic energy per target mass convention for specific impact energy, equal-size and equal-density bodies have equal masses and $Q=\frac{1}{2}{v}_{\mathrm{col}}^2$. A catastrophic planetesimal collision requires $Q\gtrsim Q_D^{\ast }$, soThe numerical coefficient is only illustrative because the collision speed was specified only to order of magnitude. If the disruption threshold is instead defined using the centre of mass kinetic energy per combined mass, equal masses give $Q_R=v_{\mathrm{col}}^2/8$ and the coefficient becomes $8{\pi }^2/25$ for a threshold expressed in that convention. The robust result is a lower bound of order $Q_D^{\ast }{a}_{\mathrm{pl}}/(G{M}_{\star }{e}_{\mathrm{pl}}^2)$.

Within the inner-disk approximation, catastrophic collisions require sufficiently large $a_1/a_{\mathrm{pl}}$, not arbitrarily small radii. Although the Keplerian speed rises inward, the secularly induced orbital eccentricity falls faster: $v_{\mathrm{col}}^2\propto \alpha$. The lower bound must be much smaller than one to leave a domain compatible with $\alpha \ll 1$. A small planet orbital eccentricity, a large disruption threshold, or a distant planet can eliminate that domain. The planet mass controls the time to crossing but cancels from the eventual forced eccentricity amplitude and this collision-energy criterion. The threshold alone does not ensure that crossing has occurred within the disk's available lifetime.

Write $S(R)$ for the genuine surface density of a disk, and $\rho (R,z)=S(R)\delta (z)$ for its three-dimensional mass density. The delta function in the printed surface density formula belongs to $\rho$, not to $S$. Introduce a reference length $R_{\ast }$ to make logarithms dimensionless. The intended model is the infinite, self-gravitating, scale-free Mestel disk, with no extra source or imposed gravitational field.

The circular speed satisfies $v_c^2=R{\partial }_R\Phi (R,0)$. Thus the flat galaxy rotation curve givesA reflection-symmetric harmonic function with this midplane boundary value isVerify this continuation using the Poisson equation for Newtonian gravity. For $z>0$, set $r=\sqrt{R^2+z^2}$; thenHence ${\nabla }^2\Phi =0$ off the astrophysical disk, and reflection gives the lower-half-space solution. Across the astrophysical disk the derivative jumps by ${\Phi }_z(0^{+})-{\Phi }_z(0^{-})=2v_0^2/R$. The distributional Poisson equation for Newtonian gravity therefore givesAt the origin the enclosed disk mass tends to zero linearly with radius, so there is no additional central point mass. This verifies the Mestel disk potential-density pair. The astrophysical disk has infinite total mass and a logarithmic Newtonian gravitational potential, so it is not an isolated finite-mass model with Newtonian gravitational potential zero at infinity. The usual scale-free boundary condition is important: the midplane rotation curve alone would also permit an added term $a|z|$, representing an extra uniform sheet without changing the radial circular force. That contribution is excluded in the intended Mestel disk model.

Use a mass-weighted planar galactic distribution function, so $F(\mathbf{x},\mathbf{v})d^{2}x{d}^{2}v$ is the stellar mass in a small planar phase space element. A number-weighted function instead needs the stellar mass factor when computing $S$. Assume a steady collisionless stellar system and isotropy in the two in-plane velocity components. Put $u=|\mathbf{v}{|}^2/2$ and write $F=f(\mathbf{x},u)$. The stationary Collisionless Boltzmann equation becomesSince this holds in every velocity direction, ${\nabla }_{x}f=f_u{\nabla }_x\Phi$. In coordinates $(\mathbf{x},E)$ with $E=u+\Phi$, this is exactly ${\nabla }_{x}f{|}_E=0$. Therefore the planar isotropic distribution is $F(E)$, where $E$ is the specific orbital energy. Stationarity is essential; instantaneous isotropy alone would not imply this result.

Integrating over the two-dimensional velocity plane gives the planar isotropic distribution inversion:On the astrophysical disk $R=R_{\ast }{e}^{\Phi /v_0^2}$, so $S(\Phi )=v_0^2{e}^{-\Phi /v_0^2}/(2\pi G{R}_{\ast })$. Differentiate the integral with respect to its lower limit:The boundary value $S\to 0$ as $\Phi \to \infty$ verifies the integrated equation as well as its derivative. Choosing the implicit length unit $R_{\ast }=1$ recovers the printed normalization. Changing the additive energy zero changes this prefactor accordingly.

At a fixed radius, the normalized velocity density isIt is a product of centered Gaussian distributions. Differentiating the supplied Gaussian integral with respect to its coefficient gives the second moments, and odd moments vanish. Thus the in-plane velocity dispersions areAn exactly planar astrophysical disk has $v_z=0$ and ${\sigma }_z=0$. The nonzero circular speed is a property of the force field, not a statement that this hot stellar distribution has net rotation.

Reversing every retrograde star folds the azimuthal Gaussian to a half-normal distribution. Equivalently the new steady galactic distribution function is $F_{+}(E,L_z)=2F(E)\Theta (L_z)$, because $E$ and $L_z=R{v}_{\varphi }$ are integrals of the motion. Its density and even velocity moments are unchanged. Its streaming velocity isThis maximally prograde stellar distribution still has radial motion and a spread of azimuthal speeds; it does not place every star on a circular orbit. In particular $⟨v_{\varphi }^2{⟩}_{+}=v_0^2$ while ${\sigma }_{\varphi ,+}^2=v_0^2(1-2/\pi )$. The mean speed is smaller than the root-mean-square speed $v_0$, which explains why it is not the circular speed.

For one target galaxy, an encounter with relative speed $u$ sweeps a cylinder of volume $\pi R^{2}udt$ in time $dt$. Multiplication by the number density gives the encounter rate $\Gamma =\pi R^{2}N⟨v_{\mathrm{rel}}⟩$. Thus the geometric galaxy-encounter rate givesThis is the expected number of mergers in the geometric model. For independent encounters, the Poisson process probability of at least one is $P_{\ge 1}=1-e^{-\mu }$, so the displayed linear probability is valid for rare encounters, $\mu \ll 1$. There is no factor of one-half for a single target; that factor would enter a count of distinct pairs across the whole population.

For an illustrative group environment, take $R=20\mathrm{kpc}=0.02\mathrm{Mpc}$, $N\sim 1{\mathrm{Mpc}}^{-3}$, $⟨v_{\mathrm{rel}}⟩\sim 300\mathrm{km}{\mathrm{s}}^{-1}$ and a Hubble time of order $10^{10}$ years, or $3\times 10^{17}\mathrm{s}$. The path length is about $9\times 10^{19}\mathrm{km}\simeq 3\mathrm{Mpc}$, using the stated distance conversion. HenceThe estimate scales linearly with environment density and quadratically with the adopted merger radius. It is an order-of-magnitude model estimate, not a universal observed merger fraction; reducing the illustrative density by a factor of one hundred reduces the estimate by the same factor.

For the tidal calculation, use the relative potential convention in which acceleration is $\nabla \psi$. Expand the point-mass potential about the target centre:The constant does not exert a force. The linear term accelerates the target centre and is subtracted in its freely falling frame. The remaining leading quadrupolar point-mass tidal potential isThe condition $r<r_p$ permits the convergent expansion; using only this term is the leading tidal approximation, accurate when the target radius is small compared with the closest separation.

The given impact and velocity directions imply ${\mathbf{r}}_p(t)=(p,0,vt)$. This trajectory is in the $x,z$ plane, correcting the incompatible printed $x,y$ plane. In the impulse approximation, hold the stellar position fixed while integrating the tidal tensor:For $s=vt/p$, the needed integrals arewith the mixed integral zero by oddness. Thus the integrated tidal tensor of a straight-line flyby is $2G{m}_p\mathrm{diag}(1,-1,0)/(v{p}^2)$ andThere is stretching along the impact direction, compression in the other transverse direction and no net kick along the path.

Write $\mathbf{u}$ for the star's initial velocity, distinguishing it from the perturber speed $v$. Its specific energy change is $\mathbf{u}\cdot \Delta \mathbf{v}+\frac{1}{2}|\Delta \mathbf{v}{|}^2$. The no-correlation assumption removes the first term on averaging; it is not an identity for every individual star. The mean specific heating at a given position is thereforeFor a spherical target, the mass-weighted averages satisfy $⟨x^2⟩=⟨y^2⟩=⟨z^2⟩=⟨r^2⟩/3$. The tidal impulse heating of a spherical galaxy is consequentlyHere $⟨r^2⟩=m_g^{-1}\int r^2\rho d^{3}r$ is assumed finite. Notice the change from specific energy to total energy after multiplying by $m_g$.

For two equal targets, each gains the same internal energy, so $\Delta U_{\mathrm{tot}}=8G^2{m}_g^3⟨r^2⟩/(3v^2{p}^4)$. Their relative-motion reduced mass is $m_g/2$. If $v$ denotes their relative speed at infinity, their initial orbital energy is $E_{\mathrm{orb}}=\frac{1}{2}(m_g/2)v^2=m_g{v}^2/4$. Internal tidal heating comes from this orbital energy. In the tidal capture of galaxies model, capture occurs if $\Delta U_{\mathrm{tot}}>E_{\mathrm{orb}}$, which gives the equal-mass tidal-capture thresholdCapture allows further passages and eventual merger in this simplified picture.

The encounter lasts roughly $p/v$. The impulse approximation needs $\Omega p/v\ll 1$, so that a star barely moves during the tide. When $v/p<\Omega$, stellar orbits respond during the perturbation and the kicks can cancel. Adiabatic invariance of an orbital action produces adiabatic shielding of tidal encounters, rather than the impulsive heating used above. The capture inequality cannot be extrapolated into that slow-encounter regime.

A small-angle gravitational encounter between a test star and a field star of mass $m$ gives a transverse kick $\Delta v_{\perp }\simeq 2Gm/(bu)$, where $b$ is the impact parameter and $u$ the relative speed. For number density $n$, encounters in $b,b+db$ occur at rate $2\pi bdbnu$. Independent kicks add in mean square, giving the random-walk derivation of stellar relaxationThe Coulomb logarithm in stellar dynamics has $b_{\max }$ of order system size $r_s$, and $b_{\min }$ of order the strong-deflection scale $Gm/u^2$. For a virialized $N$-star system, $u^2\sim GNm/r_s$, so $\Lambda$ is of order $N$. The stellar relaxation time is the time for accumulated velocity variance to become comparable with $u^2$:Using $n\sim 3N/(4\pi r_s^3)$ gives $t_{\mathrm{rel}}\sim [N/(6\ln N)]r_s/u$. With a diameter-crossing convention $T_{\mathrm{cross}}\sim 2r_s/u$, this is $[N/(12\ln N)]T_{\mathrm{cross}}$, conventionally rounded toThe coefficient is approximate: spatial profile, velocity averages and crossing-time convention change it by factors of order unity. The robust result is $N/\ln N$ crossing times.

For $N\sim 10^{10}$ and $T_{\mathrm{cross}}\sim 10^8$ years, the estimate is about $4\times 10^{15}$ years, enormously longer than a Hubble time. Most galaxies therefore behave as collisionless stellar systems. Dense nuclei and star clusters can relax more rapidly. Negligible stellar encounters do not mean negligible collective gravitational instabilities.

Define the mass-weighted galactic distribution function by $dM=F(\mathbf{x},\mathbf{v},t)d^{3}x{d}^{3}v$, so $\rho =\int F{d}^{3}v$. Hamiltonian gravitational motion preserves phase space volume by the Liouville theorem in Hamiltonian mechanics. In the collisionless regime no encounter term redistributes stars between neighbouring phase space trajectories, so $dF/dt=0$. With acceleration $\nabla \psi$, the Collisionless Boltzmann equation is

For a steadily rotating galactic bar, write the inertial polar angle as ${\varphi }_{\mathrm{I}}=\varphi +{\Omega }_{b}t$. The inertial radial and angular equations are $\ddot{R}-R{\dot{\varphi }}_{\mathrm{I}}^2={\psi }_R$ and $R{\ddot{\varphi }}_{\mathrm{I}}+2\dot{R}{\dot{\varphi }}_{\mathrm{I}}={\psi }_{\varphi }/R$. Substitution gives the equations of motion in a rotating frame:Here the velocities are measured in the bar frame. Set $v_R=\dot{R}$, $v_{\varphi }=R\dot{\varphi }$ and define the positive-force rotating-frame relative effective potentialThe centrifugal sign is positive in this relative-potential convention. The characteristics areConsequently the rotating-frame collisionless Boltzmann equation isFor these planar equations, integrate over $d{v}_{R}d{v}_{\varphi }$ and interpret $\rho$ as the planar or vertically integrated mass density. Define $\rho ⟨A⟩=\int AFd{v}_{R}d{v}_{\varphi }$. Boundary terms in velocity vanish for a sufficiently decaying distribution.

Let the velocity accelerations above be $A_R,A_{\varphi }$. Their velocity divergence is ${\partial }_{v_R}{A}_R+{\partial }_{v_{\varphi }}{A}_{\varphi }=-v_R/R$. Thus integration by parts in the zeroth moment contributes $\rho ⟨v_R⟩/R$, producing the cylindrical continuity equation for a rotating stellar systemThis is conservation of mass: change of density balances flux through the radial and azimuthal sides of a small cylindrical element. The factor $R$ is geometric; the physical phase space measure is $RdRd\varphi d{v}_{R}d{v}_{\varphi }$.

For the radial first moment, multiply the rotating-frame collisionless Boltzmann equation by $v_R$. Its velocity integrations are $\int v_R{A}_R{F}_{v_R}{d}^{2}v=-\rho ⟨A_R⟩$ and $\int v_R{A}_{\varphi }{F}_{v_{\varphi }}{d}^{2}v=\rho ⟨v_R^2⟩/R$. Hence the radial cylindrical Jeans equations in a rotating frame areFor the azimuthal first moment, the $A_R$ velocity integral vanishes, while ${\partial }_{v_{\varphi }}(v_{\varphi }{A}_{\varphi })={\psi }_{\mathrm{eff},\varphi }/R-2{\Omega }_b{v}_R-2v_R{v}_{\varphi }/R$. This givesThe Jeans equations are local momentum-balance equations. The second moments include streaming momentum flux and random-velocity stress, while gravity, centrifugal acceleration and Coriolis acceleration supply the frame-dependent forces. They do not by themselves close the full distribution: a stress prescription or a galactic distribution function is also needed.

False. Stellar evaporation removes stars above the escape energy after two-body relaxation has redistributed their energies. The remaining self-gravitating cluster need not become less dense; losing mass is not the same as holding its radius fixed.

For a slowly evolving isolated cluster, apply the virial theorem at each approximate equilibrium: $E=T+W=W/2=-T$. Stars evaporating with nearly zero energy at infinity remove little total energy, so the bound cluster has nearly constant negative $E$. With gravitational radius $R_g$ defined by $W=-G{M}^2/R_g$,For homologous structure, its characteristic mean density therefore scales asIt increases as mass is lost. Escaping stars carrying positive energy make the bound energy more negative and strengthen the contraction tendency. This is virial contraction under stellar evaporation, related to the negative heat capacity of bound gravitational systems. A tidally limited globular cluster has additional radius constraints, so this scaling is not universal, but the claimed inevitable density decrease is false even in the simplest isolated evaporation model.

False as a claimed change between isolated steady endpoints. Conservation of energy by itself would allow a more negative $W$ and a larger $T$, but both endpoints must also obey the virial theorem. For a finite isolated equilibrium with no surface term,If total $E$ is conserved between the two equilibria, the fixed-energy virial constraint between steady states givesThis $T$ includes both ordered rotation and random stellar motion. A galactic bar or other instability can redistribute angular momentum and concentrate a central component, but compensating expansion or rearrangement elsewhere must preserve the global energy constraint. Central concentration alone does not prove that the whole system has a more negative $W$.

A remnant could have different energies if stars escape, or if an external component exchanges energy with it; then the conserved total energy would include that other component. Those cases do not justify treating the bound disc simultaneously as a closed fixed-energy system and as a new equilibrium with increased $T$.

Use the separation vector $\mathbf{r}$ from the star to the comet, and let $m_{\mathrm{red}}=m{M}_{\star }/(m+M_{\star })$. The two-body problem separates into uniform centre of mass motion andTaking a cross product with $\mathbf{r}$ proves conservation of angular momentum; taking the scalar product with $\dot{\mathbf{r}}$ proves conservation of energy. The relative constants areThe total relative angular momentum and energy are $m_{\mathrm{red}}\mathbf{h}$ and $m_{\mathrm{red}}\epsilon$, respectively. The conserved eccentricity vector is $\mathbf{e}=\dot{\mathbf{r}}\times \mathbf{h}/\mu -\hat{\mathbf{r}}$. Its scalar product with $\hat{\mathbf{r}}$ gives $r=h^2/[\mu (1+e\cos f)]$, with $f$ the true anomaly. The periapsis and apoapsis distances are therefore $h^2/[\mu (1+e)]$ and $h^2/[\mu (1-e)]$. Their sum is twice the semi-major axis, soAt either apsis $v=h/r$; substituting an apsidal radius gives $\epsilon =-\mu /(2a)$. Thus the specific orbital energy implies the vis-viva equationThese are exact relative-coordinate formulas for an elliptic orbit, $0\le e<1$. In the barycentric frame the comet's speed is $M_{\star }/(M_{\star }+m)$ times the relative speed; the printed identification with the comet's stellar orbital speed uses $M_{\star }\gg m$.

On the corrected steady phase-mixed orbit, each fragment crosses a fixed orbital longitude once per orbital period. Hence the cross-sectional-area current through that point isFrom the fractional luminosity of a phase-mixed eccentric wire, ${\sigma }_{\mathrm{tot}}=4\pi a^2\sqrt{1-e^2}f=2Phf$. Therefore the area rate at the foreground crossing of the line of sight isOnly the foreground segment blocks the star; the far-side intersection does not add another occultation current. The orbital area current is constant because the line density varies as $1/v$, even though the local orbital speed varies.

Radiative drag is only one of several loss mechanisms. Around a star, radiation-pressure blowout can eject small fragments; its $\beta =1/2$ threshold applies specifically to zero-kick release from a circular parent orbit. Stellar-wind drag and gas drag can drive planetary migration, while sublimation destroys grains approaching high-temperature regions. Collisional cascades destroy or fragment grains and can feed the unbound size range. Planetary scattering can cause ejection, collision with a planet, or a stellar impact; resonant trapping of dust can instead delay planetary migration.

For circumplanetary orbits, collisions with the planet or its satellites, disruption in collisions, and escape under stellar tidal forces are additional losses. Orbits near or outside the Hill sphere need not remain planet-bound. Radiation pressure on circumplanetary dust can excite planetocentric orbital eccentricity or unbind very small grains; it need not act only through slow Poynting–Robertson drag. For charged grains, the Lorentz force in stellar or planetary magnetic fields can alter or destabilize an orbit. Shadowing of circumplanetary dust changes the radiation-force average and can reduce the quoted decay rate. Which mechanism dominates depends on grain size and composition, environment, orbit orientation and the available collision or gas density.