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Keep $A=\beta G{M}_{\star }/c$ constant. A circular circumstellar orbit stays circular in the secular drag approximation, and $da/dt=-2A/a$ integrates to $a^2=a_p^2-4At$. The inspiral time under Poynting–Robertson drag to the stellar surface isThe final expression treats the star as a point, or assumes $R_{\star }\ll a_p$.

For the coplanar circumplanetary orbit used in the preceding result, $da/dt=-3Aa/a_p^2$, so $a=a_0\exp (-3At/a_p^2)$. The inspiral time of circumplanetary dust to the planet's surface, for $a_0>R_p$, isThus the timescales have the same dependence on stellar flux and $\beta$, but planetary arrival contains a logarithm of the initial planetocentric radius. It is not always shorter: for a point star, $t_{\mathrm{cp}}<t_{\mathrm{cs}}$ only if $a_0/R_p<e^{3/4}$. A constant tilted-orbit average replaces three by its appropriate orientation coefficient. Sublimation or other removal can terminate the evolution before either idealized arrival time.

For a quasistatic spherical star with fixed total mass, the stellar structure equations in radius areHere $s$ is the specific entropy of a moving mass shell, and the entropy term accounts for release or storage of thermal and gravitational energy. In thermal equilibrium it vanishes. An equation of state, opacity, energy-generation law and composition evolution close the equations. The absence of mass loss fixes the outer mass coordinate; it does not require the star to be in thermal equilibrium.

The radiative temperature gradient needed to carry the full luminosity isHere the actual temperature gradient is $\nabla =d\log T/d\log P$. For uniform composition the Schwarzschild criterion says that a radiative layer is stable if ${\nabla }_{\mathrm{rad}}<{\nabla }_{\mathrm{ad}}$, with marginal stability at equality, where the adiabatic temperature gradient is $(\partial \log T/\partial \log P{)}_s$. Use $\nabla ={\nabla }_{\mathrm{rad}}$ there. If ${\nabla }_{\mathrm{rad}}>{\nabla }_{\mathrm{ad}}$, convection carries some of the flux. Efficient convection gives $\nabla \simeq {\nabla }_{\mathrm{ad}}$; inefficient surface convection needs a transport prescription and can be superadiabatic. For composition gradients, use the Ledoux criterion, with threshold ${\nabla }_{\mathrm{ad}}+\varphi {\nabla }_{{\mu }_{\mathrm{mol}}}/{\delta }_{\mathrm{th}}$ instead. Here ${\nabla }_{{\mu }_{\mathrm{mol}}}=d\log {\mu }_{\mathrm{mol}}/d\log P$, $\varphi =(\partial \log \rho /\partial \log {\mu }_{\mathrm{mol}}{)}_{P,T}$ and ${\delta }_{\mathrm{th}}=-(\partial \log \rho /\partial \log T{)}_{P,{\mu }_{\mathrm{mol}}}$, using the mean molecular weight ${\mu }_{\mathrm{mol}}$.

Above the thin burning shell, take constant luminosity $L$, constant mean molecular weight ${\mu }_{\mathrm{mol}}$, ideal-gas pressure $P=\mathcal{R}T\rho$ with $\mathcal{R}=k_B/({\mu }_{\mathrm{mol}}{m}_u)$, and the Kramers opacity law $\kappa ={\kappa }_0\rho T^{-7/2}$. Then the temperature equation isTo implement the printed assumption that $P,\rho$ and $M_r$ are all power laws while retaining mass conservation, write $M_r\propto r^m$, $\rho \propto r^b$, $P\propto r^p$ and $T\propto r^d$. The mass equation gives $m=b+3$; the ideal gas law gives $d=p-b$; hydrostatic equilibrium gives $d=m-1$. The diffusion equation gives $d-1=2b-2-13d/2$. Solving this linear system gives the self-gravitating Kramers power-law envelope:The coefficients can also be matched consistently. If $r_c$ is the core boundary and $M_c$ its enclosed mass,Radiative transfer fixes the constant luminosity throughwhose right-hand side is independent of radius for these exponents. Thus this is a solution of all four envelope equations, not just a dimensional estimate.

Matching the boundary temperature to the isothermal core and extrapolating the idealized power law to the specified photosphere givesFor reference $M_R/M_c=(R/r_c{)}^{1/11}=2500^{1/10}\simeq 2.19$, so the envelope mass is not negligible in this particular solution. The given $M_c\simeq M_{\odot }$ fixes the core-radius normalization $r_c=(11/42)G{M}_c/(\mathcal{R}10^7\mathrm{K})$; a numerical absolute radius needs the unspecified molecular weight. The exponent ratio also gives $\nabla =d/p=5/21<2/5$, so the fully ionized monatomic version of this idealization is convectively stable.

A different common approximation neglects the envelope's self-gravity and sets $M_r\simeq M_c$ in the force equation. It cannot obey the exact mass equation with a nonzero density and exactly constant $M_r$. Under that additional approximation, the same calculation instead givesThis is the constant-core-mass power-law opacity radiative envelope limit, not the full power-law solution with changing enclosed mass. Stating which approximation is used resolves the otherwise different numerical radius ratios.

A compact nearly isothermal core, a thin luminosity-producing shell and a much larger cool envelope describe a shell-burning red giant. The full power-law profile is a deliberately simple model. The Kramers opacity law and an ionized ideal gas are not reliable at a $4000\mathrm{K}$ photosphere; partial ionization, other opacity sources and a convective envelope usually modify real cool giants. The quoted radius ratios are extrapolations within the stipulated model.

The stellar relaxation time is the time on which accumulated discrete gravitational encounters change a typical star's velocity by an amount comparable to its original velocity. It concerns two-body relaxation, rather than the much faster orbital evolution in a smooth gravitational potential. A collisionless stellar system requires this time to greatly exceed the duration being studied.

In the straight-line impulse approximation, an encounter at relative speed $u$ has transverse acceleration $Gmb/(b^2+u^2{t}^2{)}^{3/2}$. ThusA small-angle gravitational encounter needs $\Delta v_{\perp }\ll u$. The transition to order-one deflections is therefore $b\sim Gm/u^2$, ignoring factors such as the equal-mass relative deflection. For the spherical estimate $u\sim v$,This is the lower cutoff of the weak-scattering estimate; closer encounters actually occur and require strong-scattering treatment.

The number of encounters in time $T$ with impact parameters in $[b,b+db]$ is $2\pi bdbnuT$. Independent random transverse directions make mean kicks cancel while their variances add. ConsequentlyEach logarithmic interval contributes equally, giving the Coulomb logarithm in stellar dynamics. Defining a deflection time by variance $v^2$ and setting $u=v$ gives $v^3/(8\pi G^2{m}^{2}n\log \Lambda )$. The paper instead uses an order-one normalization three quarters of this estimate:Both have the same physical scaling. The specified characteristic speed, the word “comparable”, the velocity-distribution average and strong-encounter cutoff do not fix that numerical coefficient uniquely. The crude equal-speed impulse calculation must not be claimed to determine $3/(32\pi )$ exactly.

For a self-gravitating, approximately virialized system, the virial theorem gives $v^2\sim GNm/R$. With $n=3N/(4\pi R^3)$, $\Lambda \sim N$ and the stellar crossing time $t_{\mathrm{dyn}}=R/v$, substitution into the paper's convention givesAn externally dominated gravitational potential or a different structural constant changes this substitution. For an N-body simulation lasting $T=K{t}_{\mathrm{dyn}}$, a tolerable fractional velocity-squared diffusion $\epsilon \ll 1$ requires $T/t_{\mathrm{rel}}\lesssim \epsilon$, henceThere is no unique smallest $N$ without $K$, the error tolerance and the force prescription. For scale, equality at $t_{\mathrm{rel}}/t_{\mathrm{dyn}}=100$ occurs near $N=7.1\times 10^3$; $N=10^5$ gives about $1.1\times 10^3$ crossing times. A calculation lasting 100 crossing times therefore needs substantially more than the first threshold to have negligible relaxation. Gravitational softening can raise the effective cutoff and reduce artificial scattering, but it also sets the spatial force resolution.

For finite-thickness disk relaxation, assume a well-separated range $b_{\min }\ll z_0\ll R$ and approximately virialized self-gravity. The three factors have distinct origins. Replacing the characteristic encounter speed by $u=\lambda v$ shortens the diffusion time by ${\lambda }^3$: the variance-production rate has one inverse power of speed, and the target random velocity squared has two powers. Concentrating the same number of stars into volume of order $R^2{z}_0$, rather than $R^3$, raises the number volume density by $R/z_0$ and shortens the time by $z_0/R$. Finally, only separations below the thickness sample a three-dimensional encounter geometry. The logarithmic upper cutoff becomes $z_0$, and therefore the inverse-logarithm time acquires $\log (R/b_{\min })/\log (z_0/b_{\min })$. More distant encounters have the planar geometry and provide a convergent, non-logarithmic correction. A smaller Coulomb logarithm in stellar dynamics actually partly offsets the first two shortenings.

Thus, using the same cutoff for this comparison,For actual point particles the disk cutoff is $Gm/(\lambda v{)}^2$, whereas the spherical one is $Gm/v^2$; one must use the appropriate cutoff in each logarithm rather than silently identifying them. For a softened calculation it is instead of order the larger of the weak-deflection scale and the softening length. The formula is not valid when $z_0$ approaches that cutoff.

Write $h=z_0/R$ and $q=K/\epsilon$, with simulation duration $K{t}_{\mathrm{dyn}}$ and allowed fractional diffusion $\epsilon$. Combining the paper's spherical normalization and its comparison formula gives the leading criterionThe spherical comparison logarithm cancels when its numerical convention and a common comparison cutoff are used; order-one shape factors remain approximate. In an unsoftened disk with $v^2\sim GNm/R$, $z_0/b_{\min }\sim h{\lambda }^{2}N$. Hence the implicit requirement is $N/\log (h{\lambda }^{2}N)\gtrsim 8q/({\lambda }^{3}h)$, in the regime $h{\lambda }^{2}N\gg 1$. For $\lambda =h=0.1$, the stellar relaxation time is of order $10^{-4}$ of the spherical value apart from logarithms; $N\sim 10^8$ gives a time of order 100 crossing times in this estimate. Negligible relaxation over that duration requires an additional margin. A disk generally needs far more particles than a round galaxy for the same relaxation tolerance; no universal minimum follows without duration, thickness and softening. Collective spiral or bending responses are a separate source of evolution even in a collisionless stellar system.

Use the same specified homogeneous Newtonian background and Jeans swindle, but describe stars by a mass-normalized galactic distribution function. The Collisionless Boltzmann equation is ${\partial }_{t}f+\mathbf{v}\cdot {\nabla }_{x}f-\nabla \varphi \cdot {\nabla }_{v}f=0$. Let $f=f_0(\mathbf{v})+f_1{e}^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$ and similarly perturb the gravitational potential. To first order,Eliminating $f_1$ and the nonzero gravitational potential amplitude gives the collisionless Jeans dispersion relationFor growing modes $\mathrm{Im}\omega >0$ there is no real-velocity pole. Real and damped frequencies require a causal contour prescription or analytic continuation; one must not simply integrate through a pole without specifying it.

Take $b>0$ in the Cauchy velocity distribution. Its normalization is $\int f_0{d}^{3}v={\rho }_0$. Orient $\mathbf{k}$ along the $u=v_z$ axis and integrate over the perpendicular velocities:At the marginal mode $\omega =0$, $F^{\prime }(u)/u=-2{\rho }_{0}b/[\pi (u^2+b^2{)}^2]$, with its finite limiting value at zero. The provided Gamma function integral at power two givesConsequentlyTo check that this really separates growing modes, put $c=\omega /k$ in the upper half-plane. Residue integration, or integration by parts followed by the Cauchy resolvent, gives $\int F^{\prime }(u)/(u-c)du={\rho }_0/(c+ib{)}^2$. The upper-half-plane solution has $\omega =i(\sqrt{4\pi G{\rho }_0}-kb)$, which grows exactly for $0<k<k_J$. At the threshold it approaches the marginal mode continuously. Although $b$ plays the threshold role of a velocity scale, this distribution's second moment diverges: the radial integrand for $\int v^2{f}_0{d}^{3}v$ tends to a nonzero constant at large speed. It is therefore incorrect to identify $b^2$ with a finite velocity dispersion or to infer this result by substituting an rms speed into a fluid formula.

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.

A star of declination $\delta =\theta$ crosses the zenith. Its direction rotates around the celestial pole at ${\Omega }_{\mathrm{sid}}$, but its actual angular velocity on the celestial sphere is ${\Omega }_{\mathrm{sid}}\cos \theta$. Consequently the small-angle crossing-gap estimate becomesFor clarity, the factor $\cos \theta$ is the radius of the star's daily circle on a unit celestial sphere; it does not modify the sidereal hour angle rate.

The estimate assumes the required slew is nearly $180^{\circ }$. A more exact ideal symmetric reacquisition calculation is possible. Let the endpoints have hour angles $-H,+H$ and suppose $0<\theta <\pi /2$, $H$ small enough that both endpoints are above the astronomical horizon. The horizontal direction components are $n=\sin \theta \cos \theta (1-\cos H)$ and $w=\pm \cos \theta \sin H$. Their shortest azimuth separation isThe minimum ideal gap satisfies $2H{\Omega }_{\mathrm{az}}/{\Omega }_{\mathrm{sid}}=\Delta \eta (H)$; its angular separation is ${\varphi }_{\mathrm{full}}=2\arcsin (\cos \theta \sin H)$. Expanding for a fast drive gives the boxed expression. At an equatorial site $\Delta \eta =\pi$ exactly. At a geographic pole the direction with $\delta =\theta$ is stationary, so the crossing argument is inapplicable and the limit is zero. A full near-zenith rate-limited footprint still requires specifying the trajectory and drive model.

Write $T_{\mathrm{sid}}$ for one sidereal day in minutes and ${\Omega }_{\mathrm{sid}}=2\pi /T_{\mathrm{sid}}$. An equatorial star at an equatorial site rises along a vertical great circle, passes through the zenith, and descends on the opposite side. Its azimuth changes by $\pi$ at the crossing. The restricted altitude travel prevents following it continuously by rotating over the zenith.

The maximum azimuth rate is ${\Omega }_{\mathrm{az}}=2\pi \omega$ radians per minute, so the shortest half-turn takes $\Delta t=\pi /{\Omega }_{\mathrm{az}}=1/(2\omega )$ minutes. During that lost-track interval the star moves throughFor $T_{\mathrm{sid}}\simeq 1436.07$ minutes this is approximately $0.12534^{\circ }/\omega$; using a 24-hour day gives $0.125^{\circ }/\omega$. If the quoted zenith blind spot size means the angular radius of a symmetrically placed crossing gap, it is ${\varphi }_{\mathrm{full}}/2$, approximately $0.06267^{\circ }/\omega$.

This derives the ideal crossing-gap size with instantaneous acceleration and an available altitude drive. It does not uniquely define a circular forbidden region for all nearby trajectories. For example, at an equatorial site a star of small nonzero declination $\delta$ has maximum azimuth rate ${\Omega }_{\mathrm{sid}}|\cot \delta |$ at transit: its minimum offset is constrained by $|\tan \delta |\ge {\Omega }_{\mathrm{sid}}/{\Omega }_{\mathrm{az}}$. That rate contour and the half-turn crossing gap are different definitions of a zenith blind spot. Actual tracking footprints also depend on acceleration and the reacquisition strategy, as discussed in the telescope designers' tracking discussion.

Use the ideal detection convention of one collected Electron per detected photon; otherwise replace photon totals by detected Electron totals using the quantum efficiency. Let $A$ be the summed measurement on the star patch and $B^{\prime }$ the independent background-only patch. After additive offsets are removed, their means are $Q+B$ and $B$. Independent Poisson distribution counts have variance equal to their mean, and a single read of $n$ detector pixels contributes $n{r}^2$ to each patch's variance. ThereforeThe two factors of two arise from the independently measured background and the second patch's read noise, not from doubling the stellar signal. The formula assumes equal background means, independent detector pixels and patch measurements, one read per patch, and negligible dark current or other noise. Several exposure reads or correlated detector pixels require their own variance and covariance terms.

For a circular Kepler orbit, $e=0$ remains an exact solution of the averaged Poynting–Robertson drag equations, and $\dot{a}=-2A/a$. Integrating gives $a^2(t)=a_0^2-4At$, so the point-star inspiral time under Poynting–Robertson drag isFor a star of radius $R_{\star }$, contact occurs instead at $(a_0^2-R_{\star }^2)/(4A)$ within this model. Sublimation or other forces can remove a real grain earlier.

To first order in orbital eccentricity, the semi-minor axis is $a+O(e^2)$, while the ellipse's center is displaced from the star by $-a\mathbf{e}$. Thus each orbit is a circle of radius $a$ about $-az$, in complex spatial coordinates.

Write $z=z_f+e_p{e}^{i{\varpi }_p}$ at one secular epoch. The centers of the individual circles lie on a circle of radius $a{e}_p$ about $C_f=-a{z}_f$. Their union is the proper-eccentricity annulusIts full radial width is $2a{e}_p$ and its center is displaced by $a{e}_f$ toward the forced apoapsis direction. Relative to the star, if the forced longitude of periapsis is zero,The middle panel of the preceding figure shows this construction. For fully sampled orbital phases and proper phases, the annulus is occupied throughout, but its density need not be uniform; turning points in the proper radial excursion give integrable edge enhancements. The construction requires $e_f,e_p\ll 1$ and is an instantaneous secular geometry.

Assume an isolated star of initial mass five solar masses of approximately solar stellar metallicity. Rotation, convective overshooting and mass loss change numerical ages and the extent of a blue loop, so the ages below are estimates. Take zero age at the zero-age main sequence; the pre-main-sequence star phase adds a much shorter contraction time.

The original schematic Hertzsprung-Russell diagram marks the stages discussed below. The strip denotes the approximate Cepheid instability strip; the track is illustrative, rather than a computed stellar model.

The main sequence lasts roughly $8\times 10^7$–$10^8$ years. CNO cycle hydrogen burning is concentrated in a mixed convective core. Its declining hydrogen mass fraction and rising helium mass fraction are nearly uniform there, while the retreating core leaves a composition gradient. The unprocessed envelope remains hydrogen rich. At the terminal-age main sequence, core hydrogen is exhausted and a hydrogen-burning shell takes over.

The inert helium core grows through shell burning. The Schönberg-Chandrasekhar limit is approximatelyBeyond this limit an isothermal nondegenerate core cannot remain in thermal equilibrium with its envelope. Core contraction and envelope expansion carry the star across the Hertzsprung gap. An initial slower shell-burning interval can precede the rapid crossing; the crossing itself is governed by Kelvin-Helmholtz contraction, with a representative $G{M}^2/(RL)\sim 10^5$–$10^6$ years. First dredge-up then mixes hydrogen-processed material into the convective envelope, lowering its hydrogen mass fraction and increasing helium and nitrogen, while leaving a composition discontinuity where the envelope later retreats.

At an age still of order $10^8$ years, core helium burning begins quietly: the core is nondegenerate, so there is no helium flash. The Triple-alpha process and subsequent alpha capture build a carbon-oxygen core. Core helium burning lasts roughly $1.6\times 10^7$–$2.2\times 10^7$ years in representative solar-composition models. Pols's stellar-evolution notes illustrate the substantial dependence of these lifetimes on convective overshooting.

During a blue loop the envelope contracts and effective temperature increases, before the star returns redward. The loop depends on core and envelope structure and the hydrogen discontinuity left by first dredge-up. A loop reaching the Cepheid instability strip produces two further crossings, blueward and redward, in addition to the earlier rapid crossing of the Hertzsprung gap. Cepheid variables pulsate through the opacity mechanism, involving helium ionization. Neither loop extent nor all three strip crossings are guaranteed for every composition or mixing prescription. The hydrogen-burning shell continues moving outward in enclosed mass while central helium is depleted. Lattanzio's five-solar-mass tutorial shows these composition changes.

After helium exhaustion, at an age roughly $1.0\times 10^8$–$1.2\times 10^8$ years, the star ascends the Asymptotic giant branch. An inert carbon-oxygen core is surrounded by helium-rich material and a hydrogen-rich envelope. Second dredge-up mixes helium and hydrogen burning products into the envelope and reduces the hydrogen-exhausted core mass. Subsequently a helium-burning shell and hydrogen-burning shell alternate in importance. A thermal pulse of an asymptotic-giant-branch star causes intershell convection, expansion and temporary quenching of hydrogen burning; third dredge-up can carry carbon and slow neutron-capture process products outward. Between pulses hydrogen burning rebuilds the helium layer. These mechanisms are illustrated in Lattanzio's AGB tutorial.

The following original stellar composition profile sketches distinguish hydrogen exhaustion from helium exhaustion; their mass boundaries are illustrative.

The final giant phases add at most a few million years at this level of accuracy. Strong stellar winds remove the envelope; the hot remnant can illuminate a planetary nebula and then cool as a carbon-oxygen white dwarf. A typical remnant is of order one solar mass. The ordinary isolated $5M_{\odot }$ case does not reach sustained carbon burning and core collapse. Thus the usual endpoint is a carbon-oxygen white dwarf, at a total age of order $10^8$ years, with its subsequent cooling age added separately.

For a common semi-major axis $a$ and proper eccentricity $e_p$, randomly distributed proper longitudes of periapsis give an annulus of radii $a(1-e_p)$ and $a(1+e_p)$ about the point displaced from the star by minus $a$ times the forced eccentricity vector, to first order in orbital eccentricity.

A starburst galaxy undergoes an episode of rapid star formation. An instantaneous-burst idealization places all newly formed stars at one stellar age, so their present evolutionary-state fractions follow from the initial mass function at that fixed age rather than from an average over formation times.