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The central equation must be read as $\ddot{\mathbf{r}}=-\mu \mathbf{r}/r^3$; its printed left-hand expression lacks $=0$. Initially $\mu$ is constant. Dotting this equation with the velocity givesTaking its cross product with $\mathbf{r}$ gives $d(\mathbf{r}\times \mathbf{v})/dt=0$. In the fixed orbital plane the magnitude is $h=r^2\dot{\theta }$. These are conservation of specific orbital energy and specific angular momentum.

Let $f=\theta -\varpi$. Differentiating the polar equation of the Kepler orbit, with its elements initially constant, givesSubstitution into the specific orbital energy yields

For isotropic stellar mass loss with no recoil or drag, the force is still central, so $\mathbf{h}$ remains exactly constant, although $C$ does not. NowThe same algebraic relation between $C,h,e$ holds for the instantaneous osculating orbital elements. Differentiating it and using $h^2/(\mu r)=1+e\cos f$ givesand, for $e>0$,A particularly useful nonsingular description uses the eccentricity vectorThe vector equation remains meaningful at a circular orbit, where a longitude of pericentre is undefined. Its components along and perpendicular to $\mathbf{e}$ give the scalar formula above and

The osculating pericentre is $q=h^2/[\mu (1+e)]$. ThereforeFor mass loss, $\dot{\mu }<0$, this is nonnegative: pericentre does not decrease. It is positive except at pericentre itself, where $\cos f=1$ and $\dot{q}=0$. Thus a strictly opposite sign at every instant is not literally true; a pericentre passage with nonzero mass-loss rate is a counterexample to the strict wording. The intended monotonicity follows exactly, without requiring slow mass loss.

For adiabatic orbital expansion under isotropic mass loss, average over the unperturbed Kepler orbit and hold $\dot{\mu }/\mu$ constant to leading order during that orbit. If $E$ is the eccentric anomaly and $n=\sqrt{\mu /a^3}$, thenIt follows immediately thatAveraging with uniform true anomaly instead of uniform time would give the wrong result. Since $\mu a(1-e^2)=h^2$, conserved $h$ and constant secular orbital eccentricity giveThe pericentre and apocentre expand in the same secular proportion. These are leading adiabatic invariants, rather than exact invariants for arbitrary time-dependent $\mu$.

For the instantaneous precession fraction write $\theta =f+\varpi$ and $\dot{\theta }=h/r^2$. The adiabatic apsidal condition is controlled byIt is a signed fraction: apsidal advance contributes positively on the outward leg during mass loss, and negatively on the inward leg. Let ${\tau }_{\mu }=|\mu /\dot{\mu }|$. Ordinary orbit averaging requires ${\tau }_{\mu }\gg P$, together with negligible variation of the mass-loss law over one orbit. If the pericentre direction is also to vary negligibly relative to the instantaneous azimuthal motion, the maximum absolute fraction must be small. A precise condition for $0<e<1$ isThe maximizing cosine is $c=(1-\sqrt{1+8e^2})/(2e)$: differentiating the last factor gives $c+2e-e{c}^2=0$. Thus one can evaluate $F(e)$ explicitly by putting $\sqrt{1-c^2}/(1+ec{)}^2$ into the formula. Its useful limits areFor orbital eccentricity of order unity, mass loss much slower than an orbital period suffices; near a circular orbit, keeping the pericentre direction slowly varying additionally requires $n{\tau }_{\mu }e\gg 1$. The latter divergence is a singularity of the pericentre coordinate, not a divergence of the eccentricity vector dynamics. Indeed, at frozen $\dot{\mu }/\mu$, integrating the scalar evolution to first order gives $\delta e=-(\dot{\mu }/\mu )(1-e^2)\sin E/n$ up to an integration constant. Its absolute amplitude stays small for slow mass loss even when the scalar longitude of pericentre becomes unsuitable.

Use rotating barycentric coordinates $\mathbf{R}=(x,y,z)$, rotating velocity $\mathbf{w}$, and $\mathbf{\Omega }=\hat{\mathbf{z}}$. Multiplying the three equations by the corresponding velocity components and adding cancels the Coriolis force terms:Thus the Jacobi constant is $C=2U-|\mathbf{w}{|}^2$. The inertial barycentric velocity is $\mathbf{V}=\mathbf{w}+\mathbf{\Omega }\times \mathbf{R}$. If $H_z=(\mathbf{R}\times \mathbf{V}{)}_z$ and $E_b=|\mathbf{V}{|}^2/2-{\mu }_1/r_1-{\mu }_2/r_2$, thenThe inertial energy $E_b$ and inertial angular momentum $H_z$ need not separately be constant. Their combination is constant because the two gravitational sources rotate steadily at unit angular speed.

For the Tisserand parameter take ${\mu }_2\ll {\mu }_1\simeq 1$, and compare the osculating orbital elements well outside a close encounter, where ${\mu }_2/r_2$ and barycentre-to-primary corrections are negligible. ThenThis requires a circular secondary orbit, the restricted three-body problem approximation, and no dissipative force or other perturber. $T$ is conserved to leading order between encounter episodes; the osculating $T$ during a close encounter need not be constant. The exact invariant is $C$.

For clarity the exact two-body energy relation is $a={\mu }_1/(2{\mu }_1/r_1-v_1^2)$, rather than the printed formula without the numerator ${\mu }_1$. At the leading order ${\mu }_1=1$ used for $T$, the two agree; retaining finite ${\mu }_1$ in just part of that formula is inconsistent.

At a close encounter put the secondary radius and speed equal to one at leading order, and denote the particle's incoming relative speed outside the secondary's strong-deflection region by $u$. Conservation of relative speed in the short gravitational encounter givesAn incoming orbit with $a=1$ has $|{\mathbf{v}}_1|=1$ at $r_1=1$ by the vis-viva equation; escape there requires an outgoing speed exceeding $\sqrt{2}$. The triangle inequality therefore gives the necessary single-encounter escape velocity boundEquality only gives a marginal parabolic limit. This argument deliberately gives a weak necessary bound, not a sufficient scattering criterion.

The planet-encounter relative velocity is related to the Tisserand parameter byHere $v_{1,\mathrm{tangential}}=h_z$ at the encounter radius, and the formula also applies to inclined passages at the planet's orbital plane. For the intended prograde coplanar incoming orbit, $I=0$ and $a=1$, so $u^2=2-2\sqrt{1-e^2}$. Applying the preceding necessary bound givesCoplanar alone also allows $I=\pi$. For that retrograde case $u^2=2+2\sqrt{1-e^2}$, and even a circular incoming orbit can be scattered onto an escaping orbit; the orbital eccentricity restriction therefore presumes prograde motion. Also, retaining the outgoing relative-velocity geometry gives the sharper necessary condition $|{\mathbf{v}}_{1,\mathrm{out}}|\le 1+u$, hence $u>\sqrt{2}-1$ and $e>\sqrt{4\sqrt{2}-5}/2$ for the same prograde incoming orbit. This stronger bound is consistent with, and implies, the weaker bound above.

For the Tisserand periapsis bound, first writeA bound orbit capable of another close encounter must intersect the secondary's circular radius, so $q\le 1\le Q$. For fixed $q<1$, the largest possible $T$ occurs at $I=0$. The remaining expression decreases with $Q\ge 1$, sinceThus every such orbit obeysOn $0<q<1$, $g(q)$ increases monotonically from $2$ to $3$. For $2<T<3$ the least accessible pericentre is the unique root of $T=g(q)$: the limiting orbit is prograde and coplanar, with apocentre just at the planet. Introduce $s=\sqrt{3-T}$. Its tangential speed at this apocentre is $1-s$, and its vis-viva equation gives the particularly well-behaved answerEquivalently, setting $w=\sqrt{2q/(1+q)}$ gives $T=2-w^2+2w$ and $w=1-s$, with $q=w^2/(2-w^2)$. Rationalizing the previous result yieldsThe apparent singularity at $T=\sqrt{8}$ is removable; the first form gives $q_{\min }=(\sqrt{2}-1)/2$ there. The limits are $q_{\min }\to 0$ as $T\downarrow 2$ and $q_{\min }\to 1$ as $T\uparrow 3$. The same lower envelope also excludes a smaller pericentre on an escaping trajectory when $2<T<3$. To see this without introducing an apocentre, use the planet-frame velocity sphere: $u=\sqrt{3-T}<1$ forces the tangential component at an encounter to be at least $w=1-u>0$, so total specific angular momentum obeys $h\ge w$. At pericentre the stellar energy obeys $h^2=2q+2E{q}^2$, while $E=h_z-T/2\le h-T/2$. ThusIf $q<q_{\min }=w^2/(2-w^2)$, then $q^2<w\le h$, and the left side is increasing with $h$. Hence $w^2\le 2q+(2w-T)q^2$. But $2w-T=w^2-2$, and the right side is smaller than $w^2$ for $0\le q<q_{\min }$, a contradiction. This establishes the barrier for all Kepler trajectories that reach the encounter radius, including unbound ones.

For $0<T\le 2$, bound encounter-crossing orbits can approach $q=0$ (take $Q\to 2/T$), so the positive root extrapolated from the displayed expression is not a universal barrier. For $T>3$ there is no trajectory reaching a close encounter in this approximation, since $u^2=3-T$ would be negative. Multiple encounters can approach the derived lower envelope by changing energy, angular momentum and inclination while preserving the Tisserand parameter; the invariant alone does not guarantee a particular encounter history reaches it.

Assume an axisymmetric thin disc rotating in the fixed potential of a dominant central mass, with $\Omega =(G{M}_{\ast }{)}^{1/2}{r}^{-3/2}$ independent of time and height. Neglect vertical mass loss and vertical angular-momentum flux at the two faces, as well as self-gravity and radial pressure corrections to the rotation law. Define the surface density and density-weighted kinematic viscosity byand let $v_r={\Sigma }^{-1}\int \rho u_{r}dz$. These assumptions give the vertically averaged viscous disk equationswhere $l=r^2\Omega$ is specific angular momentum. Subtract $l$ times conservation of mass from conservation of angular momentum. Since $l$ is fixed in time,Substitution into conservation of mass proves the Keplerian viscous diffusion equationNo assumption of height-independent kinematic viscosity is needed; its density-weighted average is the one appearing in the integrated stress. A wind or surface magnetic stress would add terms and must not be silently discarded.

A strip $dx$ at $x>0$ passes the satellite at relative speed $Sx$, so its encounter mass flux is $Sx\Sigma (x)dx$. Each unit mass gains specific angular momentum $r_0\Delta v_y$. Therefore the one-sided impulse torque on a disk isPutting $\chi =1/2$ gives exactly the requested expression. Inner strips have the opposite signed impulse but the same positive encounter flux. The net disc torque is consequentlyAn even surface density makes the integrand vanish pointwise, so the total torque is zero although each one-sided torque can be nonzero. The cutoff must keep the encounters in the weak-deflection regime; an appropriate physical value is at least of order the thickness or the relevant strong-scattering radius. The infinite-sheet integral also assumes convergence or a specified outer truncation.

Use the supplied cosmological cosmic baryon fraction $f_b={\Omega }_{\mathrm{bar}}/{\Omega }_{\mathrm{mat}}=0.2$, and assume the halo initially has that fraction. Its total baryon mass is then $M_{b,\mathrm{tot}}=6\times 10^{11}{M}_{\odot }$. The selected baryons are the lowest-angular-momentum fractionHere the given cumulative specific angular momentum distribution must be normalized separately for the baryon component: equal baryon and dark-matter distributions mean equal normalized fractions, not equal absolute masses.

The virial velocity and the largest specific angular momentum in the selected inner baryon population areEstimate the outer radius of the settled low-angular-momentum component using circular rotational support and an enclosed-mass approximation to its self-gravity. With negligible dark matter inside the component, $v_d^2\simeq G{M}_b/r_d$ and $j_d=r_d{v}_d$, soThe given rounded gravitational constant produces the same estimate. The very small radius follows from selecting a small low-$j$ fraction, rather than assigning all central baryons the halo-edge angular momentum.

These estimates assume that gas radiates energy, preserves each parcel's specific angular momentum, and settles with negligible pressure support and no strong redistribution or cancellation of its angular-momentum vectors. They also assume that the phrase “innermost baryons” selects the lowest-$j$ material, the original cosmic baryon supply is retained, and the central gas supplies the dominant gravity. A possible central black hole or an exact flattened disk potential changes the numerical coefficient; the stated baryonic mass is used for this estimate.

The low-angular-momentum baryonic disk estimate uses a cumulative distribution uniform in $j$ from zero to $j_d$, so its mean is $j_d/2$. The quoted $r_d$ is the outer radius based on the cutoff angular momentum, not a one-zone radius based on the mean. Within the same enclosed-mass approximation, $M(<j)=M_{b}j/j_d$ and circular balance give $r(j)=j{j}_d/(G{M}_b)$: the rotation curve is approximately flat and the half-mass radius is $r_d/2$. Treating every baryon as one shell with the mean $j$ would instead give $r_d/4$ and twice the velocity, a different radius convention rather than the outer edge of the supplied distribution.

Let $f=0.46$. At Roche-lobe overflow contact, the supplied equilibrium radius relation givesKepler's third law then impliesThus the Roche-lobe period-mass relation for a linear donor radius law isHere $R_{\odot }$ and $M_{\odot }$ denote the solar radius and solar mass. Cancellation of the total mass is the special Roche-lobe-filling period-density relation combined with $R_2\propto M_2$.

Between eruptions the conservative binary mass transfer has $M$ constant and ${\dot{M}}_1=-{\dot{M}}_2$. Neglecting spin, writeIts logarithmic derivative givesDifferentiating the Roche lobe relation consequently givesFor $\dot{J}=0$, the Roche-lobe radius response exponent is ${\zeta }_L=2q-5/3$. If the donor follows its specified equilibrium radius sequence, ${\dot{R}}_2/R_2={\dot{M}}_2/M_2$. Its logarithmic overflow therefore changes bySince ${\dot{M}}_2<0$, $q<4/3$ gives negative feedback in this equilibrium-response model, whereas $q>4/3$ gives increasing overflow. For $q>4/3$, loss of donor mass shrinks its Roche lobe faster than its modeled stellar radius, increasing transfer and hence amplifying the original disturbance. The $q>4/3$ comparison is a formal extension; the initial system is specified to have $q<1$.

The word dynamically needs a qualification. Thermal equilibrium specifies ${\zeta }_{\mathrm{eq}}=1$, but actual dynamical stability of binary mass transfer depends on the adiabatic stellar radius response exponent ${\zeta }_{\mathrm{ad}}$, which need not equal one. Stability after a rapid small mass loss requires ${\zeta }_{\mathrm{ad}}>{\zeta }_L$, because $d\log M_2<0$. The equilibrium relation alone does not supply this adiabatic response. As a formal counterexample to deriving a universal dynamical threshold from it, a radius model with a dimensionless entropy label $S$, $R(M_2,S)=C{M}_2^2{e}^S$, can have equilibrium $S_{\mathrm{eq}}=S_0-\log (M_2/M_{\ast })$, giving $R_{\mathrm{eq}}\propto M_2$ but ${\zeta }_{\mathrm{ad}}=2$ at fixed $S$. At $q=3/2$, ${\zeta }_L=4/3<2$, so that model is dynamically stable despite the extrapolated $q>4/3$ condition. This shows the missing response hypothesis, not a claim that this toy radius law describes the given low-mass star.

A more relevant illustration is an ideal fully convective adiabatic stellar polytrope of index $3/2$. At fixed entropy its polytropic mass-radius relation is $R\propto M_2^{-1/3}$, giving ${\zeta }_{\mathrm{ad}}=-1/3$ and a dynamical critical ratio $q=2/3$ under the same Roche lobe approximation. For example, $q=0.9<4/3$ can be unstable dynamically even though the equilibrium comparison has negative feedback. The requested $4/3$ threshold is the intended result with instantaneous response $R_2\propto M_2$; it is not a universal dynamical criterion. The subsequent cycle calculation uses the supplied equilibrium sequence and presumes a physically stable, sufficiently slow transfer regime.

Two important angular-momentum-loss mechanisms are gravitational-wave emission from a binary system and magnetic braking of a binary star. Gravitational waves directly remove orbital energy and angular momentum. A magnetized stellar wind removes donor spin angular momentum with a large magnetic lever arm; tidal synchronization replenishes that spin from the orbit, so the net effect is orbital angular-momentum loss. Treating the wind mass loss as small compared with the Roche-transfer flow is consistent with the conservative approximation used here; otherwise the mass and angular-momentum equations must both be modified. Neglecting the stored spin does not eliminate the torque transmitted through it.

Such losses shrink the Roche lobe and offset the donor's tendency to move inside it. Maintaining contact on the equilibrium sequence requires ${\dot{R}}_L/R_L={\dot{M}}_2/M_2$, hence the binary mass-transfer contact equation becomesFor $q<4/3$ and $\dot{J}<0$, this drives ${\dot{M}}_2<0$ and maintains the modeled semi-detached binary.

Now take $\delta m>0$ for the mass expelled from the accretor in one classical nova. The white dwarf loses $\delta M_1=-\delta m$, while the donor has $\delta M_2=0$ and the total mass change is $\delta M=-\delta m$. Isotropic ejection in the accretor rest frame supplies no mean kick and carries the accretor's specific orbital angular momentum. Its orbital radius is $a_1=a{M}_2/M$, soThis is the angular-momentum budget for isotropic re-emission from a binary star, without an additional spin or frictional torque on the ejecta. The general logarithmic change of $J=M_1{M}_2\sqrt{Ga/M}$ isUsing $M/M_1=1+q$ gives the first-order nova changesThe donor mass and its assumed equilibrium radius are unchanged during the eruption. It is therefore left inside the enlarged Roche lobe, and Roche-lobe overflow ceases in this sharp-contact model. An extended atmosphere, irradiation-induced donor expansion or additional ejecta torques would require a different treatment.

The stipulated circular orbit is an important idealization. Isotropic loss that is slow compared with the orbital period can preserve circularity. If it is truly impulsive compared with that period, an initially circular point-mass orbit instead acquires eccentricity. With unchanged velocity and position, its new specific orbital energy is ${\epsilon }^{\prime }=GM/(2a)-G(M-\delta m)/a=-G(M-2\delta m)/(2a)$ and its squared specific angular momentum stays $h^2=GMa$. The Kepler orbit energy and eccentricity relations then give $a^{\prime }=a(M-\delta m)/(M-2\delta m)$ and $e^{\prime }=\delta m/(M-\delta m)$. The latter still has the same first-order separation increase. The question assumes circularity; its first-order circular budget is consistent with adiabatic orbital mass loss or with subsequent circularization that changes the separation only at second order. Work symbolically with $\delta m$, since the printed numerical layer mass has no explicit unit.

Finally put ${\mathcal{L}}_J=-\dot{J}>0$, constant over both intervals as specified, and let $J,M,q$ denote their leading-order values over one small cycle. While detached, masses and donor radius are fixed, so ${\dot{R}}_L/R_L=2\dot{J}/J=-2{\mathcal{L}}_J/J$. The fractional gap $4\delta m/(3M)$ closes afterOnce contact resumes, the contact equation gives $-{\dot{M}}_2=3M_2{\mathcal{L}}_J/[J(4-3q)]$. Accumulation of the next layer takesDividing yields the detached-to-semidetached time ratioBoth times are first-order in $\delta m/M$. Corrections to either time are second-order in that small mass fraction, so the displayed ratio has relative corrections of order $\delta m/M$. The positive denominator, negligible eruption duration relative to the cycle, fixed donor radius during detachment and common external loss rate are all essential to this idealized nova-induced binary detachment cycle.

The linearized azimuthal Euler momentum equation isFor the time dependence $e^{-i\omega t}$ and $u_R^{\prime }=-i\omega {\xi }_R$, it givesThis expresses conservation of the displaced element's specific angular momentum. It applies directly to nonzero-frequency modes, with the zero-frequency limit taken in the displacement formulation.

The radial advective acceleration supplies $-2\Omega v_{\varphi }$, while the pressure force perturbation is $\delta \rho \nabla p/{\rho }^2-\nabla \delta p/\rho$. Eliminate $v_{\varphi }$ and retain the vertical equation. Under the Cowling approximation, $\delta \Phi =0$, soThe continuity equation gives $\delta \rho =-\nabla \cdot (\rho \mathbf{\xi })$. The Lagrangian adiabatic relation $\Delta p/p=\gamma \Delta \rho /\rho$, with $\Delta \rho =-\rho \nabla \cdot \mathbf{\xi }$, givesHere $\kappa$ is the radial epicyclic frequency. These formulas define the axisymmetric adiabatic displacement operator.

The pressure-square term in the Cowling energy principle for a rotating barotropic star is nonnegative, and ${\mathcal{N}}^2>0$ makes its stratification term nonnegative. Write $j=R^2\Omega$ for specific angular momentum. ThenFor a regular star reaching the rotation axis, $j(0)=0$. The given $dj/dR>0$ therefore makes $j>0$ for $R>0$, and consequently ${\kappa }^2>0$. Equivalently, with the usual nonnegative angular-velocity convention the sign follows directly. All three energy terms are nonnegative, soThis is stability within the Cowling approximation used throughout. The orientation-independent rotational condition is $d(j^2)/dR\ge 0$, the Rayleigh discriminant criterion. If a fluid region excludes the axis and negative angular velocity is allowed, $dj/dR>0$ alone needs the additional sign of $j$; the squared-angular-momentum criterion avoids that ambiguity.

In the Lambda-CDM model, nearly Gaussian primordial density contrasts grow under gravity within an expanding universe containing cold dark matter, ordinary baryons and a cosmological constant. The cold dark matter is effectively collisionless and has negligible pressure on galactic scales. Before recombination, baryons are coupled to the photon fluid: radiation pressure and acoustic oscillations prevent their perturbations from behaving like pressureless matter. After recombination they can fall into the gravitational potentials already established by dark matter, subject to gas pressure and the Jeans mass.

For small density contrasts, evolution is linear. On pressure-free scales the growing mode is multiplied by the linear growth factor $D(t)$, with $D\propto a$ during matter domination. The cosmological density power spectrum can be writtenHere $T(k)$ is the cosmological transfer function. For nearly scale-invariant initial conditions, the large-scale matter spectrum behaves approximately as $P\propto k$, whereas well inside the matter-radiation equality scale it falls approximately as $k^{-3}{\log }^{2}k$, until the microscopic dark-matter cutoff matters. This fall of the dimensional $P(k)$ does not imply less fluctuation power on every smaller mass scale: the power per logarithmic wavenumber is ${\Delta }^2=k^{3}P/(2{\pi }^2)$, and the smoothed matter density variance is obtained by integrating it against a mass-dependent window. Over the relevant cold-dark-matter hierarchy, smaller mass windows generally have larger variance.

The hierarchical galaxy formation picture follows: fluctuations on small mass scales typically reach the nonlinear collapse threshold first, while larger objects assemble later through accretion and dark-matter halo mergers. It is a statistical ordering, not a claim that every small object precedes every large rare peak. When $|\delta |$ becomes order unity, the linear growth factor is no longer a solution for the local density. Collisionless dark matter develops multistream regions and bound dark matter halos; phase mixing and violent relaxation redistribute orbital energies, and virialized structures approximately obey the virial theorem.

Baryons have an additional nonlinear route. Infall and shocks convert bulk kinetic energy into thermal energy, with characteristic virial temperatureUnlike collisionless dark matter, the gas can lose this energy through radiative cooling. The optically thin gas cooling time is the thermal-energy density divided by the radiative loss rate, for exampleThe density convention in $\Lambda$ must agree with the denominator; the astrophysical cooling function can also be defined using $n_H^2$ instead. The cooling criterion for galaxy formation compares this time with the collapse or supply time. Rapidly cooling gas loses pressure support, contracts and can form stars. Slowly cooling gas remains in a hot atmosphere. Stable virial shocks are not obligatory in every low-mass system: gas can also arrive in cold streams and cool while being accreted.

Angular momentum prevents indefinite radial contraction. Tidal torque theory supplies an initial halo spin, and later mergers change it. If gas radiates energy while retaining much of its specific angular momentum, it settles into a rotationally supported galactic disk rather than reaching the centre. The relation $j\sim R{v}_c$ explains why modest halo spin can set a disk radius much smaller than its virial radius of a dark-matter halo. Torques, bars and gravitational encounters can transport angular momentum outwards and feed central concentrations; radiative cooling alone does not remove it.

Galaxy mergers alter stellar structure as well as assembling mass. A major galaxy merger can strongly disturb or destroy an existing galactic disk, randomizing stellar orbits and creating a spheroid through violent relaxation. A gas-rich galaxy merger also permits dissipation, inflow and a burst of star formation; gas left over or accreted afterwards can rebuild a galactic disk. Minor galaxy mergers add stars to outer components, thicken disks and grow bulges. Dry galaxy mergers add stellar mass and can increase size without much new star formation. Halo merging therefore does not imply instantaneous merging of its galaxies: satellite orbital decay requires Chandrasekhar dynamical friction and can take a substantial time.

The atomic and molecular cooling thresholds for galaxy formation supply a lower characteristic scale. Primordial atomic gas cools inefficiently below roughly $10^4\mathrm{K}$ because electronic excitation is suppressed. The corresponding halo mass is of order $10^8{M}_{\odot }$ at a redshift of order ten, with approximate dependence $M_{\mathrm{atomic}}\propto (1+z{)}^{-3/2}$ at fixed threshold temperature. Molecular hydrogen can cool gas at hundreds of kelvin and permit smaller early objects, provided it forms and survives dissociating radiation. Metal-line cooling changes these thresholds after enrichment. Thus the atomic threshold is not an absolute minimum mass for all stellar systems.

At the other end, sufficiently massive dark matter halos have high virial temperatures and low-density hot gas. Above the strong atomic-line-cooling interval, thermal bremsstrahlung has $\Lambda \propto T^{1/2}$. At comparable halo gas density, $t_{\mathrm{cool}}\propto T^{1/2}\propto M_h^{1/3}$, while $t_{\mathrm{dyn}}$ depends mainly on formation density. Cooling therefore becomes less able to condense all the gas within the available time. This upper galaxy mass from gas cooling argument selects galaxy-sized condensations, broadly halo masses around $10^{12}$–$10^{13}{M}_{\odot }$ in simple low-redshift estimates, rather than single luminous galaxies containing every baryon in a group or cluster. Its numerical scale depends on epoch, metallicity and gas profile. Subsequent galaxy mergers can build larger stellar systems; cooling is not an absolute upper bound on their final mass.

Finally, a Press-Schechter halo mass function has many low-mass objects and a steep high-mass cutoff, while the luminosity function of galaxies also has a faint component and a bright cutoff, often described by a Schechter function. The two shapes are related through the halo-to-galaxy luminosity mapping, not by identifying luminosity with total halo mass. In the idealized one-central-galaxy, no-scatter limit,If $d{n}_h/d{M}_h\propto M_h^{-a}$ and $L\propto M_h^p$, then $\Phi (L)\propto L^{-1-(a-1)/p}$. A constant conversion gives similar shapes. Actual stellar feedback, inefficient low-temperature cooling and reionization suppress faint galaxies relative to small haloes, while long cooling times and active-galactic-nucleus feedback suppress luminosity at large halo masses. Satellites, scatter, stellar populations and dust further affect the correspondence. The halo hierarchy supplies the gravitational framework; cooling, angular momentum and feedback determine which parts become luminous galaxies.

Let $M_h,R_h$ describe the halo and let $v=v_{\mathrm{vir}}$. The cosmic baryon fraction is $f_b={\Omega }_b/{\Omega }_m=0.20$, so the specified settled fraction gives $M_d=0.10M_h$. The specific angular momentum constraint at the disk edge isTo infer an actual circular speed one must specify the disk's radial mass distribution: enclosed mass does not determine a disc rotation curve. Introduce a finite geometry coefficient $\kappa$ by $v_d^2=\kappa G{M}_d/R_d$. A rounded, centrally concentrated disk permits a monopole estimate $\kappa \simeq 1$ at its outer edge; it is an explicit approximation, not the spherical shell theorem applied exactly to a razor-thin disk. Ignore the force of the unsettled baryons as well as the dark matter within the disk. Using $v^2=G{M}_h/R_h$, the angular-momentum estimate of a self-gravitating galactic disk givesAn exact disk answer cannot be fixed by total disk mass alone; the dependence on $\kappa$ records that missing input.

For the numerical virial radius of a dark-matter halo, adopt mean density ${\Delta }_c=200$ times the critical density at the formation epoch. This conventional definition givesThe supplied expansion law gives $H(3)=286.49\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$, and hence $R_h=v/(10H)=97.74\mathrm{kpc}$. Taking $\kappa =1$ yieldsThe corresponding virial mass of a dark-matter halo is $M_h\simeq 1.78\times 10^{12}{M}_{\odot }$, and $M_d\simeq 1.78\times 10^{11}{M}_{\odot }$. Other overdensity conventions give $R_d\propto {\Delta }_c^{-1/2}$ at fixed $H,v,\kappa$.

Interpret the wavelength separation as an observed-frame local mean near the redshift in question. Since Lyman-alpha absorption appears at ${\lambda }_{\mathrm{obs}}={\lambda }_{\alpha }(1+z)$, the incidence isAssume one counted absorption system per intercepted disk, no unrelated forest systems or missed absorbers, unity neutral covering fraction, and a locally slowly varying population. Let ${\sigma }_{\mathrm{abs}}$ be the proper interception cross-section and $n_c$ the comoving number density. The proper density is $n_p=n_c(1+z{)}^3$, and the proper line element along the light path is $d{l}_p=cdz/[H(z)(1+z)]$. Thus the absorber incidence and comoving number density relation isFor a thin circular disk, the projected area is $\pi R_d^2|\cos i|$. Isotropically oriented normals have $⟨|\cos i|⟩=1/2$, so the random-orientation absorbing-disk cross-section is ${\sigma }_{\mathrm{abs}}=\pi R_d^2/2$. ConsequentlyIf all disks are taken face-on instead, $n_c\simeq 0.398{\mathrm{Mpc}}^{-3}$. Generally the random-orientation result scales as ${\kappa }^2({\Delta }_c/200)$ and is divided by the neutral covering fraction. The finite mean redshift spacing is sizable, so a precision inference would integrate the incidence over the actual redshift interval rather than identify it with one local value.

There is also a halo abundance mass-budget bound on this formal result. The present mean matter density for these parameters is ${\bar{\rho }}_{m,0}={\Omega }_{m}3H_0^2/(8\pi G)\simeq 3.40\times 10^{10}{M}_{\odot }{\mathrm{Mpc}}^{-3}$. Distinct haloes of this mass cannot have $n_c{M}_h>{\bar{\rho }}_{m,0}$, even if every matter particle belonged to them. Yet the inferred random-orientation population hasEven the face-on estimate exceeds this bound by about twenty-one. The numerical incidence result is conditional; the supplied population assumptions are not cosmologically consistent under this standard virial and compact-disk estimate. A larger neutral-gas absorption radius, a different absorber population, or different physical inputs are needed. For example, random orientations would require an absorbing radius at least about $63\mathrm{kpc}$ merely to reach the all-matter upper bound, much larger than the calculated centrifugal radius. This check does not change the algebraic answer, but prevents treating it as a realizable population.

Specific orbital energy, orbital eccentricity and the distribution. Let $\mu =G{M}_{\star }$, $v_K=\sqrt{\mu /a}$ and $c=\cos \theta$. Choose positive $\sin \theta$ to mean an outward radial kick. At release the tangential and radial velocities are $v_t=v_K(1+\gamma c)$ and $v_r=\gamma v_K\sin \theta$. The specific orbital energy and specific angular momentum areUsing ${\epsilon }^{\prime }=-\mu /(2a^{\prime })$ and $e^{\prime 2}=1+2{\epsilon }^{\prime }{h}^{\prime 2}/{\mu }^2$ therefore givesThe denominator must be positive for an elliptic Kepler orbit. A vanishing denominator describes a parabolic Kepler orbit; negative $a^{\prime }$ is the signed semi-major axis of a hyperbolic Kepler orbit.

Differentiating the squared orbital eccentricity givesIn the small-kick regime $0<\gamma <1$, $1+\gamma c>0$. Thus the minimum is at $c=-\gamma /3$, withThe entire expression, including the ${\gamma }^4/27$ term, lies under the square root in the PDF. This minimum is not a universal statement for arbitrary kick size. For example $\gamma =2$, $c=-1$ reverses the circular velocity and gives a retrograde circular orbit with $e^{\prime }=0$; the displayed stationary value is then not the global minimum.

For $0<\gamma <\sqrt{2}-1$, every kick direction gives a bound prograde Kepler orbit. The two endpoint maxima, one at each end of the accessible curve, areThe positive endpoint is the global maximum. There is one interior minimum, at the location just found. Uniform $\theta$ gives the kick-orbit distributionBoth signs of the radial kick occupy the same $e^{\prime }$–$a^{\prime }/a$ curve; they have opposite apsidal orientations. The distribution is concentrated near its endpoints, with integrable square-root singularities. For larger kicks retain the parametric curve and its bound portion $c<(1-{\gamma }^2)/(2\gamma )$; an unbound branch begins at $e^{\prime }=1$, $a^{\prime }/a\to \pm \infty$. In particular, a sketch of a wholly elliptic population presupposes the bound-kick restriction above.