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In the collisionless exosphere, an atom escapes if its outward trajectory has positive total mechanical energy. Neglect tides and stellar forces and use Newtonian gravity at exobase radius $r_e$:With thermal speed $v_{\mathrm{th}}=\sqrt{2k_B{T}_e/m}$, the Jeans escape parameter isA thermal distribution always has an escaping tail; Jeans escape is exponentially suppressed for ${\lambda }_e\gg 1$, with Jeans escape flux proportional to $(1+{\lambda }_e)e^{-{\lambda }_e}$. Efficient escape requires ${\lambda }_e$ of order a few or smaller, with an order-unity energetic estimate $k_B{T}_e\sim G{M}_{p}m/r_e$.

Assume a Neptune-like mass and radius, $r_e\simeq R_p$, and atomic hydrogen. Using the supplied rounded constants givesHenceUsing the mean kinetic energy $3k_{B}T/2$ instead gives an order-unity coefficient and $T\simeq 2.3\times 10^4\mathrm{K}$. An expanded exobase has weaker binding and lowers the estimate by $R_p/r_e$. A comet-like tail can also be shaped by radiation pressure and stellar-wind interactions; it does not by itself measure $T_e$ or prove that a hydrostatic Jeans model is valid. At ${\lambda }_e\sim 1$, $H/r_e\sim 1$ and hydrostatic equilibrium fails as a global description: substantial mass loss must usually be treated as hydrodynamic atmospheric escape.

The exobase is defined by mean free path $\ell \sim H$, not by a universal pressure. For a neutral hydrostatic gas with collision cross-section ${\sigma }_c$,For example, explicitly assuming ${\sigma }_c\sim 10^{-19}\text{–}10^{-20}{\mathrm{m}}^2$ gives $P_e\sim 2\times 10^{-7}\text{–}2\times 10^{-6}\mathrm{Pa}$, or $2\times 10^{-12}\text{–}2\times 10^{-11}\mathrm{bar}$. These are representative extremely dilute neutral-exobase pressures, with orders of magnitude varying with composition, cross-sections and expansion. The supplied constants contain no collision information, so they cannot uniquely determine an exobase pressure; ionization or a non-hydrostatic density profile changes this estimate.

Assume the planet and present-day Jupiter have the same mass, comparable composition and initial cooling normalization, and both can be described by the supplied power-law planetary cooling over the relevant ages. Take the age of present-day Jupiter to be $t_J\simeq 4.5\mathrm{Gyr}$ and the young planet's age to be $0.050\mathrm{Gyr}$. The normalization cancels:The 5-AU orbit is used to regard stellar irradiation as modest compared with a hot-Jupiter orbit; the ratio is specifically intrinsic cooling, not the sum of intrinsic and reradiated luminosity. Initial entropy and irradiation can change the assumed normalization, so this estimate is not independent of formation conditions.

Such a young self-luminous giant is a favourable target for near-infrared exoplanet direct imaging. The increased intrinsic luminosity improves its contrast with the host star, and a wide physical orbit is more readily separated on the sky than a hot-Jupiter orbit. Two essential observing considerations are:

Thus direct imaging of a nearby young system, with sufficient angular resolution and contrast, is the natural discovery method, subject to the actual stellar brightness and instrumental limits.

For a quasi-static self-gravitating object of fixed mass, negligible surface pressure, and no accretion, nuclear source or additional deep heating, conservation of energy gives the intrinsic planetary luminosityHere $U$ is internal energy, and $\Omega$ is negative gravitational potential energy. Gravitational contraction makes $\Omega$ more negative and releases power $L_{\mathrm{grav}}\equiv -d\Omega /dt>0$. Some of that release raises the internal energy; it cannot all be radiated while the object maintains hydrostatic equilibrium.

To quantify the split assume thermal gas pressure with constant specific-heat ratio $\gamma >4/3$ and negligible rotation or magnetic support. This excludes the dynamically unstable gas-supported regime. The stellar virial theorem givesIt follows thatFor a monatomic ideal gas, $\gamma =5/3$, so $L_{\mathrm{int}}=L_{\mathrm{grav}}/2$: half the released gravitational energy heats the gas and half escapes. If the density profile remains homologous, write $\Omega =-qG{M}^2/R$, with $q>0$ a fixed structure constant. Thenup to the structure and virial coefficients, where ${\tau }_{\mathrm{KH}}$ is the Kelvin-Helmholtz cooling time.

For a diatomic gas with $\gamma =7/5$ the idealized fraction is $1/6$, already demonstrating that the half-factor is not universal. Partial electron degeneracy pressure, dissociation, changing structure, accretion and deep heating require the full energy equation. The general link between luminosity and Kelvin-Helmholtz contraction is $L_{\mathrm{int}}=-d(U+\Omega )/dt$, with any genuine source terms added explicitly.

Hot-Jupiter radius inflation is the excess radius of many irradiated gas giants relative to standard cooling models of the same mass, composition and age. A young gas giant starts with high entropy and a large radius, but ordinarily contracts through Kelvin-Helmholtz contraction. An old inflated object must either retain that thermal reservoir unusually well or receive power that affects the deep interior.

The first class is delayed cooling of an inflated giant planet. Two specific mechanisms are increased atmospheric opacity, which makes internal radiation escape less readily, and inhibited interior convection caused by composition gradients, potentially with layered transport. Both slow entropy loss. Stellar irradiation can also maintain an outer radiative blanket, but an appropriate irradiated boundary is already included in many baseline models and does not alone explain every extreme radius.

The second class is heating of an inflated giant planet. Two mechanisms are tidal heating from eccentricity, obliquity or other time-dependent tidal forcing, and Joule heating from wind-driven electric currents in an ionized atmosphere coupled to the magnetic field. Mechanical energy carried inward from atmospheric circulation is another candidate. Heating must be deposited sufficiently deeply and with sufficient power; superficial absorption that is promptly reradiated is not equivalent to deep interior heating. A circular synchronized isolated orbit does not automatically provide a persistent tidal source.

The classes can coexist and are tested through radius-age-irradiation trends, orbital properties and the required energy budget. Merely changing an observed transit altitude by a few atmospheric scale heights is generally distinct from inflating the bulk giant-planet interior.

At fixed mass, three broad controls of a super-Earth-sized body's observed radius are:

Bulk composition, envelope fraction/composition, and thermal/irradiation history are three independent controls. If mass is not fixed, the mass itself is an additional major variable. The label super-Earth does not ensure a rocky composition or an Earth-like atmosphere, and a radius alone does not determine which of these effects dominates.

Three research directions natural in the 2017 setting are:

Small-planet atmospheres, formation through composition, and atmospheric dynamics/variability provide three concrete emerging directions. The examples describe research aims as of the paper's date, rather than treating later discoveries or later operating missions as established in 2017.

Assume spherical fragments of the common mass density $\rho$, and let $n(D)=K{D}^{-\alpha }$ count fragments per unit diameter. Mass normalization of a power-law size distribution givesTheir geometric cross-sections are $\pi D^2/4$, hence the cross-section of a power-law fragment population is exactlyFor $3<\alpha <4$, large fragments dominate the mass integral and small fragments dominate the geometric cross-section integral. Neglecting the subdominant endpoints givesThe two actual error parameters are $(D_{\min }/D_{\max }{)}^{\alpha -3}$ and $(D_{\min }/D_{\max }{)}^{4-\alpha }$. They must both be small; the approximation is not uniform as $\alpha$ approaches either endpoint. At $\alpha =3$ or $4$ the corresponding integral is logarithmic, so the endpoint-dominated formula must be replaced by that integral.

Let $D_c$ be the parent comet's diameter, with the same mass density as the fragments, so $m=\pi \rho D_c^3/6$. The required geometric cross-section is ${\sigma }_{\mathrm{tot}}=4\pi a^2\sqrt{1-e^2}f$, using the corrected phase-mixed orbit interpretation.

Substitute this and the spherical mass into the endpoint-dominated fragment formula. The parent-body size from a fragment cross-section becomesThe common mass density cancels. The comet's radius is $D_c/2$. This expression treats both fragment cutoffs as specified independently and has the same endpoint-approximation restrictions as part (iii).

If a model instead ties the largest fragment to the parent through $D_{\max }=\eta D_c$, with fixed $0<\eta \le 1$, one must solve that dependence too:That extra assumption is not supplied by the PDF; setting $D_{\max }=D_c$ without saying so would change the requested size relation.

Let ${\mu }_p=G{M}_p$, $A=\beta G{M}_{\star }/c$ and $\kappa =A/a_p^2$. A tightly bound circumplanetary orbit lies well inside the Hill sphere, so $a/a_p\ll 1$ and its orbital period is short compared with the planet's year. Treat the stellar flux and stellar direction $\hat{\mathbf{R}}$ as constant during one dust orbit. With stellar-frame velocity ${\mathbf{V}}_p+\mathbf{u}$, the velocity-dependent acceleration isThe terms independent of $\mathbf{u}$, including the leading static radiation pressure, do no net work on an unperturbed closed circular orbit. HenceFor a coplanar circular orbit, $u^2={\mu }_p/a$ and the component along the stellar direction has mean square $u^2/2$. Thus $\dot{a}=2a^2{\dot{\epsilon }}_p/{\mu }_p$ yields the Poynting–Robertson decay of a circumplanetary orbitThe coefficient three assumes coplanarity, which the PDF does not state. For orbital normal $\hat{\mathbf{k}}$, the general circular covariance is $⟨u_i{u}_j⟩=(u^2/2)({\delta }_{ij}-k_i{k}_j)$, givingAn orbital plane normal to the stellar radial direction has coefficient two during that orbit, a counterexample to an orientation-independent coefficient three. Averaging also over the planet's circular stellar orbit, with fixed dust-plane orbital inclination $I$ to it, gives coefficient $3-\frac{1}{2}{\sin }^{2}I$. The conservative forces must be weak enough for the assumed approximately circular, planet-bound orbit to persist; the small planet-to-star mass ratio alone does not ensure this for arbitrary grain $\beta$.

Let $\chi =v_{\mathrm{esc}}/v_k$, with $v_k=\sqrt{G{M}_{\star }/a_p}$. The Safronov number is $\Theta ={\chi }^2/2$. In the instantaneous planet frame let the asymptotic vectors be ${\mathbf{u}}_{\mathrm{in}}$ and ${\mathbf{u}}_{\mathrm{out}}$; their magnitudes agree. In the stellar frame the corresponding specific orbital energy change during a short encounter isThe quadratic relative-speed terms cancel. A comet with semi-major axis comparable to $a_p$ has stellar binding energy per unit mass of order $v_k^2/2$. Consequently $\chi$ measures whether a single strong encounter can change a substantial fraction of that binding.

Large escape-to-orbital-speed ratios favour ejection; small ratios favour collision or accretion during repeated encounters. For $\chi \gg 1$, a suitably oriented gravitational assist can eject the comet before it strikes the planet. For $\chi \ll 1$, most individual kicks are too weak; many close passages may be needed, providing repeated chances of collision.

The crossover is of order unity, not a sharp theorem about each orbit. The initial binding energy, encounter speed and geometry matter: an already nearly parabolic comet may be ejected by a low-$\chi$ planet, while an unfavourably directed kick can make a comet more tightly bound even when $\chi$ is large.

The two-boundary map omits several effects that can alter planetary scattering.

The initial semi-major axis, orbital eccentricity, orbital inclination and orbital phase determine whether encounters occur and their relative velocities. A strongly bound comet needs more energy to escape; a nearly parabolic one needs less. The planetary radius and mass density, the finite comet radius, and gravitational focusing determine collision probabilities. Tidal disruption, atmospheric gas drag, sublimation and physical fragmentation can destroy a body before a nominal point-particle scattering sequence is completed.

Other planets can hand a comet from one scatterer to another, eject it, or lift its periapsis clear of the original scatterer's orbit. Mean-motion resonances and secular perturbations can protect objects from encounters or correlate kicks, contradicting the independent random walk assumption. Planetary migration changes the encounter geometry over time.

Stellar flybys and a galactic tide can change distant comet periapses, allowing new encounters or detaching an object from the planetary region. Stellar mass evolution changes both binding and planetary orbits. Finally, the age and the supply rate of new comets determine whether the observed population is a residual one or continuously replenished. Thus the mass-radius map is a useful conditional classification, not a complete survival law.

Fix $M_{\star },{\rho }_p,t_{\star }$ and an initial-orbit ratio $\kappa =a_0/a_p$; use $\kappa =1$ for the figure. The qualitative finite-age planetary scattering regimes have two useful boundaries.

First, the order-unity escape-to-orbital-speed boundary $\chi =1$ isIt slopes down with logarithmic slope $-3/2$. Above it a planet can give large fractional binding-energy changes, favouring planetary ejection of a comet; below it weak kicks and repeated collision opportunities favour accretion.

Second, equating the characteristic comet energy diffusion time to the age givesThis boundary slopes up with logarithmic slope $3/4$. Above it the many-encounter diffusion estimate fits within the available age; below it that estimate exceeds the age.

Above both boundaries, rapid ejection is favoured. Above the age boundary but below the escape-speed boundary, repeated encounters can act during the age while individual kicks remain weak, so collision or accretion is commonly favoured. Below the age boundary, the diffusion model predicts incomplete ejection; the escape-speed boundary still distinguishes strong from weak individual kicks.

These labels are statistical expectations under the stated encounter model. In particular, the given diffusion coefficient supplies no collision rate, so it does not prove that every object in the low-$\chi$ region is accreted within $t_{\star }$. Nor does a long ejection time exclude faster collisions or other loss mechanisms. Very massive planets with $N_{\mathrm{ej}}\lesssim 1$ require a few-encounter treatment rather than an extrapolation of the diffusion formula.

For a spherical planet of mass density ${\rho }_p$, $R_p=(3M_p/4\pi {\rho }_p{)}^{1/3}$. The point-comet surface escape velocity and circular Kepler orbit speed therefore giveRaise this equation to the power $3/2$ and solve for $M_p$:Thus the constant of proportionality is $\boxed{\sqrt{3/(32\pi )}}$, and the gravitational constant cancels. Dimensional checking gives mass on the right: $M_{\star }^{3/2}{a}_p^{-3/2}{\rho }_p^{-1/2}$ has that dimension.

This uses the planet's own escape velocity. If instead $\chi$ is defined at finite-comet contact, an additional factor $(1+R/R_p{)}^{3/2}$ appears on the right, and $R_p$ itself depends on $M_p$; the quoted pure-planet scaling is then only the $R/R_p\to 0$ limit.

Put $u=\lambda v$ and let $n$ now mean number per area. During time $T$, the two strips of impact parameters between $b$ and $b+db$, on opposite sides of the trajectory, have combined area $2uTdb$. ThusThe same straight-line impulse approximation gives a kick $2Gm/(bu)$. Summing its squared size over these encounters givesassuming $b_{\max }\gg b_{\min }$. The relevant change is now the small random speed $u$, not the full circular speed $v$. Setting $D_v{t}_{\mathrm{rel}}\sim u^2$ yieldsThe razor-thin disk relaxation integral is dominated by nearby encounters, rather than equally by logarithmic intervals. Because kicks cease to be weak near the cutoff, its numerical coefficient is an estimate.

Here the weak-deflection cutoff must use the relative speed: $b_{\min }\sim Gm/u^2=Gm/({\lambda }^2{v}^2)$, which is larger than the spherical cutoff. Thus $t_{\mathrm{rel}}\sim u/(Gmn)$. Using $n\sim N/R^2$ and $v^2\sim GNm/R$ from the virial theorem givesFor example $n=N/(\pi R^2)$ gives the order-one coefficient $\pi /8$ in this last substitution; the paper's approximate equality suppresses such structural factors. The cancellation of $N$ is the central result.

An unsoftened point-particle system confined to an infinitesimally thin plane is therefore not collisionless over orbital times for $\lambda \sim 0.1$ in this encounter model. At fixed total mass, increasing $N$ decreases $m$ and the cutoff together; the enhanced importance of close planar encounters cancels the usual benefit of increasing particle number. This is a singular zero-thickness limit. A fixed nonzero gravitational softening length or physical thickness restores a different $N$ dependence, so the conclusion is not a prohibition on collisionless numerical models of disks.

Assume a time-independent spherical relative potential $\psi (r)$, with acceleration $\nabla \psi$. The relative energy and specific angular momentum areAlong an orbit, $\dot{E}=\nabla \psi \cdot \mathbf{v}-\mathbf{v}\cdot \nabla \psi =0$ and $\dot{\mathbf{\ell }}=\mathbf{r}\times \nabla \psi =0$, since a spherical gradient is radial. Thus both $E$ and $L$ are integrals of motion. Choose the escape level $\psi =0$ and take the bound galactic distribution function to vanish for $E\le 0$.

For an isotropic $f(E)$, integrate over a spherical coordinate system in velocity space:Differentiating gives ${\rho }^{\prime }(\psi )=2\pi \sqrt{2}{\int }_0^{\psi }f(E)/\sqrt{\psi -E}dE$. Convolve once more with the reciprocal-square-root kernel. Reversing the order of integration, under the usual integrability assumptions, usesTherefore ${\int }_0^{E_0}{\rho }^{\prime }(\psi )/\sqrt{E_0-\psi }d\psi =2\sqrt{2}{\pi }^2{\int }_0^{E_0}f(E)dE$. Differentiating proves Eddington inversion:Here ${\pi }^2$ is outside the square root, as in the original PDF; the TeX transcription wrongly places it inside. The assumptions include a locally integrable bound distribution and sufficient density regularity, with no extra unbound or boundary population. An inverted expression must additionally be nonnegative to be physical. Since the velocity measure and $f$ are invariant under all velocity rotations, $⟨v_r^2⟩=⟨v_{\theta }^2⟩=⟨v_{\varphi }^2⟩=⟨v^2⟩/3$ whenever these moments exist; mixed moments vanish.

For the half-anisotropic distribution function, write the polar velocity angle from the radial direction as $\eta$. Then $L=rv\sin \eta$, andThe apparent $1/L$ singularity is integrable because it cancels the factor $v\sin \eta$ in the velocity measure. Integrating $\eta \in [0,\pi ]$ and $\phi \in [0,2\pi ]$ givesThe density relation requires $r\rho \to 0$ at the escape boundary and $f_E\ge 0$ for a physical distribution. Consider a monotonically decreasing $\psi$. The largest allowed radius at fixed binding energy is the radial-orbit limit $r_E$ with $\psi (r_E)=E$; a nonzero-$L$ orbit normally has a smaller apocenter because of its centrifugal term. Changing from $\psi$ to $r$ givesThe numerator in the original PDF instead uses $\rho +d\rho /d\psi$. That is a genuine printed error: the chain rule for $d(r\rho )/d\psi$ requires $\rho +rd\rho /dr$. It is not legitimate to prove the printed expression as written. A concrete counterexample is the unit-parameter Hernquist model at $r=1$: ${\psi }^{\prime }=-1/4$, ${\rho }^{\prime }=-(5/2)\rho$ and $d\rho /d\psi =10\rho$. The printed quotient is $-44\rho$, whereas the correct $g$ is $6\rho >0$. The former would give a negative distribution.

The angular weight after cancellation is uniform in $\eta$, so the averages of ${\cos }^2\eta$ and ${\sin }^2\eta$ are both $1/2$. The azimuthal angle splits the tangential term equally. Hence $⟨v_r^2⟩=⟨v_{\theta }^2+v_{\varphi }^2⟩$ and each individual tangential moment is half the radial one. The velocity-anisotropy parameter is consequentlyThis is a radially biased constant-anisotropy distribution function, not an isotropic one or purely radial motion. It does not by itself prove dynamical stability; it preferentially weights small angular momentum while remaining integrable.

For the Hernquist model, use the Poisson equation for Newtonian gravity with the relative-potential sign, ${\nabla }^2\psi =-4\pi G\rho$. Since ${\psi }^{\prime }=-GM/(r+a{)}^2$,Its enclosed mass is $M{r}^2/(r+a{)}^2$, which tends to $M$. Expressing the augmented density in the relative potential gives $r\rho =a{\psi }^3/(2\pi G^3{M}^2)$. The preceding derivative therefore giveswith zero bound-model population outside the allowed energy domain. It is nonnegative and reproduces the density by direct integration. The Hernquist model has a central density cusp, so the formulas for the local $1/L$ distribution are interpreted at $r>0$. As a further check, the radial second moment obtained from the same integral is $\psi /4$ and each tangential second moment is $\psi /8$, consistent with $\beta =1/2$.

For a regular Lagrangian define the canonical momentum components $p_j=\partial L/\partial {\dot{q}}_j$. Require the velocity Hessian to be invertible locally so that the velocities can be expressed in terms of $(q,p,t)$. The Legendre transform in mechanics definesThe explicit time argument cannot generally be omitted when the given Lagrangian depends on time. Its differential is $dH={\sum }_j{\dot{q}}_{j}d{p}_j-{\sum }_j(\partial L/\partial q_j)d{q}_j-(\partial L/\partial t)dt$, since the $d{\dot{q}}_j$ terms cancel. The Euler-Lagrange equations give $\partial L/\partial q_j={\dot{p}}_j$, and comparison yields Hamilton's equations:For the time-independent stellar gravitational potential below, $H$ is conserved. Its sign convention differs from the binding energy in question 2: here $E=H=v^2/2-\psi$ is the ordinary specific orbital energy.

The prescribed nondegenerate prolate spheroidal coordinates require $0<\alpha <\beta$; equality collapses the coordinate family. Their physical velocity components are $v_j=h_j{\dot{q}}_j$, and $h_{\varphi }=R$. Per unit stellar mass the Lagrangian is $L=\frac{1}{2}{\sum }_j{h}_j^2{\dot{q}}_j^2+\psi$, so $p_j=h_j^2{\dot{q}}_j=h_j{v}_j$. Its Hamiltonian isThis uses acceleration $+\nabla \psi$, as specified, rather than importing the opposite gravitational potential convention.

For the Stäckel potential, use the Hamilton-Jacobi equation with principal function $S=W_{\lambda }(\lambda )+W_{\mu }(\mu )+l\varphi -Et$. Axisymmetry makes $p_{\varphi }=l=R{v}_{\varphi }$ constant. Write $d=\lambda -\mu$ and $A(\tau )=(\tau +\alpha )(\tau +\beta )$. Multiplying $H=E$ by $d$ and usingseparates it as $F(\lambda )-E\lambda =F(\mu )-E\mu$, where $F(\tau )=2A(\tau )(W_{\tau }^{\prime }{)}^2-(\alpha -\beta )l^2/[2(\tau +\alpha )]-G(\tau )$. Each side depends on its own coordinate only and therefore equals a constant, chosen as $-I$. Here $G(\tau )$ is the generating function of the gravitational potential, not the gravitational constant. HenceThis proves the conserved axisymmetric Stäckel third integral. Eliminating $E$ between the two expressions and using $p_j=h_j{v}_j$ givesAt regular generic phase-space points the three separation constants $E,l,I$ are independent; special circular or degenerate orbits can reduce their local independence. Thus the model is an integrable Hamiltonian system.

For the intended regular, phase-mixed steady model, Jeans theorem permits a galactic distribution function $f=F(E,l,I)$. At any fixed point, $E$ and $I$ depend on $v_{\lambda }$ and $v_{\mu }$ only through their squares, and $l$ depends only on $v_{\varphi }$. Thus $f$ is even separately in $v_{\lambda }$ and $v_{\mu }$, whether or not it is even in $v_{\varphi }$. Odd integration proves velocity-ellipsoid alignment in a Stäckel potential:The argument allows azimuthal streaming: it does not require zero mean $v_{\varphi }$. The means in the two meridional directions vanish, so the off-diagonal velocity covariance also vanishes. Second moments must exist for this conclusion.

There is a hypothesis behind the printed universal wording. On a generic nonresonant regular invariant torus, a continuous steady distribution is constant along a dense orbit and hence depends only on the torus integrals; this justifies the phase-mixed form. Bare stationarity alone is not sufficient in a degenerate resonant gravitational potential with additional integrals. For example $H=\frac{1}{2}(v^2+{\omega }^2{r}^2)$ is a Stäckel Hamiltonian: choose $G(\tau )=-\frac{1}{2}{\omega }^2[{\tau }^2+(\alpha +\beta )\tau ]$, up to a constant vacuum shift. The additional conserved quantity $K_{xy}=v_x{v}_y+{\omega }^{2}xy$ gives a positive normalizable steady tracer $f\propto \exp (-aH+ϵK_{xy})$ for $a>|ϵ|>0$. Its local $⟨v_x{v}_y⟩=ϵ/(a^2-{ϵ}^2)\ne 0$. At azimuth zero and a regular point where the $\lambda$ direction has a radial component, this gives nonzero $⟨v_{\lambda }{v}_{\varphi }⟩$. This tracer need not itself source the harmonic gravitational potential. It exhibits why the stronger, ordinary phase-mixed three-integral assumption is necessary instead of claiming that every imaginable steady distribution has the asserted alignment.

The geometry follows by multiplying the coordinate-defining equation and evaluating at $\tau =-\alpha ,-\beta$:Constant $\lambda$ surfaces are prolate spheroids, with equatorial semiaxis $\sqrt{\lambda +\alpha }$ and polar semiaxis $\sqrt{\lambda +\beta }$. Constant interior $\mu$ surfaces are two-sheeted hyperboloids. Their common foci lie at $z=\pm \sqrt{\beta -\alpha }$. The orbit volume in an axisymmetric Stäckel potential is determined byIts allowed coordinate intervals have turning boundaries or coordinate endpoints. A typical bound orbit with $l\ne 0$ oscillates between inner and outer spheroids, oscillates vertically between hyperboloidal limits, and circulates about the symmetry axis with $\dot{\varphi }=l/R^2$. The equatorial fold $\mu =-\beta$ joins the positive and negative $z$ branches; it is not necessarily a physical turning point in $z$.

For three incommensurate frequencies, the regular trajectory densely samples its projected allowed volume, with nonuniform occupation determined by its speeds. It does not follow one coordinate surface, and it need not close. Resonant orbits may be closed or lie in a smaller projected set; $l=0$ confines azimuthal motion to a meridional plane (with the usual axis coordinate qualification), and circular or equatorial limits are also lower-dimensional. Smooth Stäckel separation removes the generic chaotic behavior of nonintegrable potentials; it does not assert that every orbit fills a three-dimensional volume.

For this stellar population, let $m=M/M_{\odot }$ denote initial mass in units of the solar mass, and let $\tau =t/\mathrm{Gyr}$ denote present stellar age. Constant formation of equal numbers of stars per unit time makes the stellar age a uniform distribution on $[0,10]$ Gyr, so $Y=\tau /10$ is uniform on $[0,1]$. The normalized initial mass function has probability density function $f(m)=0.1m^{-2}$ for $m>0.1$, since ${\int }_{0.1}^{\infty }{m}^{-2}dm=10$. ConsequentlyThese formulae have the stated age and mass domains; outside them the relevant cumulative fractions saturate at zero or one. The probability integral transform makes $X$ uniform on $[0,1]$: $m=0.1/X$ gives $f(m)|dm/dX|=1$. The time-independent initial mass function and constant number formation rate give independence of $X$ and $Y$. All fractions here count objects, including white dwarfs, using the stipulated stellar evolution law.

A red giant hasThus the mass boundaries are $m=9/\tau$ and $m=10/\tau$ for positive $\tau$. Equivalently, for $0.9<m<1$ the red giant region runs from $9/m$ to the age cap $10$, and for $m\ge 1$ it runs from $9/m$ to $10/m$. There are no red giants with $m\le 0.9$. Boundaries have zero probability and their endpoint convention does not affect the fractions. In the uniform $(X,Y)$ square the red giant region is $9X<Y<10X$, with triangle vertices $(0,0)$, $(1/10,1)$ and $(1/9,1)$. The white dwarf region is the triangle $0<X<Y/10$.

At fixed $Y=y$, define the conditional probabilities of a red giant, white dwarf and main sequence star by $g(y)$, $w(y)$ and $s(y)$. Their interval widths areIntegration over the uniform distribution of age gives the individual-star fractionsThe systems form a coeval binary population: the two binary star components have the same age. Their masses are independent, so $(X_1,X_2,Y)$ has uniform probability density function on $[0,1{]}^3$ and their states have conditional independence given $Y$. Unconditional independence of their states would be incorrect: older systems make both evolved states more likely. The law of total probability now givesThe subtraction removes the double counting of systems containing two red giants.

For Roche-lobe overflow in the stipulated stellar thermal equilibrium, $R_2=R_L$. Using the mass-radius relation and Kepler's third law,The total mass cancels, givingHere $R_{\odot }$ is the solar radius. This is the Roche-lobe-filling period-density relation specialized to $R_2\propto M_2$.

Neglecting spin, the angular momentum of a circular orbit is $J=M_1{M}_2\sqrt{Ga/M}$. Between classical novae the transfer conserves total mass, so $\dot{M}=0$ and ${\dot{M}}_1=-{\dot{M}}_2$. Its logarithmic derivative givesIn particular, for conservation of angular momentum,The source PDF has $\dot{J}=0$ here; the TeX's $J=0$ is a transcription defect. For $q>4/3$, the Roche-lobe radius response exponent exceeds the equilibrium stellar radius response exponent $1$. Since ${\dot{M}}_2<0$, the Roche lobe then shrinks faster than the donor star; the overfill increases and drives more transfer, giving positive feedback. Equality at $q=4/3$ is marginal in this linear response test.

Two qualifications matter. First, $q>4/3$ is a formal extrapolation outside the stated $q<1$ range, and the supplied Roche lobe approximation need not remain accurate there. Second, true dynamical stability of binary mass transfer compares ${\zeta }_L$ with the adiabatic stellar radius response exponent ${\zeta }_{\mathrm{ad}}$, not ${\zeta }_{\mathrm{eq}}=1$. The stellar thermal equilibrium argument supplies the displayed feedback threshold under the imposed radius law, rather than a necessary and sufficient physical dynamical threshold. For example, the polytropic mass-radius relation for a fully convective adiabatic stellar polytrope with index $3/2$ gives ${\zeta }_{\mathrm{ad}}=-1/3$; within this same Roche lobe approximation its dynamical threshold is $q=2/3$. Thus $q<4/3$ by itself does not guarantee dynamical stability.

Gravitational-wave emission from a binary system removes orbital energy and angular momentum. Magnetic braking of a binary star provides another important loss: a magnetized stellar wind carries away donor spin, while tidal locking couples that spin to the orbit. Tides alone redistribute angular momentum and are not an external sink. Matter expelled from the system can also carry orbital angular momentum, although appreciable continuous mass loss would require nonconservative modifications of the preceding equations. Setting ${\dot{R}}_L/R_L={\dot{R}}_2/R_2$ gives the binary mass-transfer contact equationFor $q<4/3$, a negative external $\dot{J}$ therefore sustains negative ${\dot{M}}_2$ along the stipulated stellar thermal equilibrium contact sequence, provided the system is otherwise stable and can remain thermally relaxed.

During the classical nova, the ejecta carry the white dwarf's specific angular momentum, not zero angular momentum. With ${\Omega }^2=GM/a^3$ and the accretor's distance from the centre of mass $a_1=a{M}_2/M$, isotropic escape givesThe slow ejection relative to the orbital period permits the adiabatic circular orbit approximation; the donor star does not transfer appreciable additional mass during it. Take $\delta M_1=-\delta m$, $\delta M_2=0$ and $\delta M=-\delta m$. Differentiating $J=M_1{M}_2\sqrt{Ga/M}$ givesSince $M/M_1=1+q$, this yieldsThese are first-order formulae with errors of order $(\delta m/M{)}^2$. The orbital widening agrees with Jeans-mode mass loss. The donor star's radius is unchanged to this order by the assumed ejection, so its Roche lobe expands away from it, causing nova-induced binary detachment. This idealization neglects changes in the donor from irradiation or interaction with ejecta; those are not supplied in the model.

Let $\Gamma =-\dot{J}>0$ be the fixed external loss rate. During the detached binary phase both component masses are fixed, and ${\dot{R}}_L/R_L=2\dot{J}/J=-2\Gamma /J$. Closing the fractional gap $4\delta m/(3M)$ takesIn the subsequent semidetached binary, the binary mass-transfer contact equation gives $-{\dot{M}}_2=3M_2\Gamma /[J(4-3q)]$. Accumulating a fresh layer of mass $\delta m$ therefore takesThroughout these first-order estimates $J,M,q$ can be evaluated at the start of the cycle: their fractional changes during it are of order $\delta m/M$. The limit $\Gamma =0$ is excluded, since no finite reconnection time follows without an external shrinkage mechanism.