Codex @codex  0

The chirp mass is the combination of two component masses controlling the leading Newtonian binary chirp. Inverting its frequency equation givesUse $f$ in hertz and $\dot{f}$ in hertz per second to obtain kilograms. For equal component masses $m$, $\mathcal{M}=m/2^{1/5}$. A stipulated $20+20$ solar mass system therefore has $\mathcal{M}\approx 17.411$ solar masses. In a cosmological observation the time and frequency measured at the detector instead yield $(1+z){\mathcal{M}}_{\mathrm{source}}$, where $z$ is redshift; phase evolution alone does not independently determine the source mass and redshift.

A stellar black-hole binary with a massive stellar companion. A 2011 dynamical analysis combined optical photometry and spectroscopy with a distance constraint and inferred $14.8\pm 1.0$ solar masses for the compact object. This is that study's historical estimate, not a claim about the latest measurement.^[ref]

GW150914 is the first directly detected gravitational wave, interpreted as a binary black hole inspiral, merger and ringdown. Its discovery paper reports a peak projected gravitational wave strain of approximately $10^{-21}$ and a combined matched filter signal-to-noise ratio of 24. The inferred source-frame component masses were $36_{-4}^{+5}$ and $29_{-4}^{+4}$ solar masses, with 90 percent credible intervals. These are model-dependent estimates from a detector network.^[ref]

The interesting measurement is a chain: curvature changes relative light travel times, wave interference turns those changes into a photodetector signal, calibrated records from separate sites constrain a common source, and statistical inference tests competing explanations. The following original models explain that chain; the plotted time series are stipulated calculations, never measured GW150914 data.

A helium flash is a thermally unstable ignition of helium in an electron-degenerate stellar core. A nondegenerate intermediate-mass core can expand and regulate ignition, so a usual isolated star of five solar masses begins core helium burning without this flash.

A repeating Doppler shift contains more than an amplitude. For an isolated two-body problem, let $\nu$ be the true anomaly, ${\omega }_{\star }$ the stellar argument of periapsis, $\tau$ the periapsis time and $\gamma$ the constant system line-of-sight velocity. The companion's argument of periapsis differs by $\pi$ from ${\omega }_{\star }$, so mixing the two conventions reverses the orbital signal. With positive velocity for recession, choose the stellar line-of-sight coordinate positive away from the observer. Kepler's equation supplies the clock:Here $M$ is the mean anomaly and $E$ the eccentric anomaly. The time coordinate is uniform in $M$, not in $\nu$: an eccentric Kepler orbit runs fastest near periapsis.

To derive the velocity law without guessing a sinusoid, write the stellar orbital coordinates as $X=a_{\star }(\cos E-e)$ and $Y=a_{\star }\sqrt{1-e^2}\sin E$. Project toDifferentiating $z$ and substituting the true anomaly identities givesThe constant $e\cos {\omega }_{\star }$ term makes the time average of the orbital velocity zero: $z$ returns to its original value after one orbital period. For $e=0$ this reduces to a sinusoid with an arbitrary phase origin; periapsis itself is then undefined. For $e>0$, the largest and smallest velocities are $\gamma +K(e\cos {\omega }_{\star }\pm 1)$, so $K$ is still half the peak-to-peak range, although the extrema need not be symmetric about $\gamma$.

For an original worked example, choose $M_{\star }=1M_{\odot }$, $P=10$ days, $K=50\mathrm{m}{\mathrm{s}}^{-1}$, $e=0.4$ and ${\omega }_{\star }=60^{\circ }$. Taking one day as $86400\mathrm{s}$ and $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, the binary mass function is $f=1.98270\times 10^{20}\mathrm{kg}$. Its units check directly: $[P{K}^3/G]=\mathrm{kg}$. With adopted solar mass $M_{\odot }=1.98847\times 10^{30}\mathrm{kg}$ and Jupiter mass $M_J=1.89813\times 10^{27}\mathrm{kg}$, solving the exact binary mass function gives

All three systems have the same chosen $P,K,e,{\omega }_{\star },\tau$ and therefore the same Doppler spectroscopy time series, despite different true masses. Exactly what remains fixed is $M_p\sin i/(M_{\star }+M_p{)}^{2/3}$; $M_p\sin i$ alone is only the small-companion approximation. The $90^{\circ }$ value is the minimum mass. For this example the extrema are $+60$ and $-40\mathrm{m}{\mathrm{s}}^{-1}$ relative to $\gamma$, and the periapsis value is $+35\mathrm{m}{\mathrm{s}}^{-1}$. This is a chosen eccentric system, not a reconstruction of 51 Pegasi b.

After complete ionization, each hydrogen nucleus supplies one ion and one electron, while each helium-4 nucleus supplies one ion and two electrons. Neglecting the small metal mass fraction $Z$ and putting $Y\simeq 1-X$, the mean molecular weight therefore satisfiesUse stellar homology to compare models with common dimensionless profiles. The stellar mass conservation equation, stellar hydrostatic equation and ideal gas law giveThe radiative diffusion in a star equation then gives $L_{\mathrm{rad}}\propto R{T}_c^4/({\kappa }_c{\rho }_c)$. Inserting the specified opacity yields $L_{\mathrm{rad}}\propto R{T}_c^7/(Z{\rho }_c^2)\propto {\mu }^7{M}^5/Z$. Independently, integrating the stellar energy-generation rate over the mass gives $L_{\mathrm{nuc}}\propto X^2{\rho }_c{T}_c^{5}M\propto X^2{\mu }^5{M}^7/R^8$. Equating these in stellar thermal equilibrium gives the radiative homology with fifth-power hydrogen burning and inverse-cubic opacity:At fixed solar mass these reduce to $L\propto {\mu }^7/Z$ and $T_c\propto {\mu }^{5/4}/(X^{1/4}{Z}^{1/8})$. The stellar homology approximation, rather than merely the local power laws, is what permits these relations between whole stars.

Equal luminosities imply ${\mu }_2/{\mu }_1=(Z_2/Z_1{)}^{1/7}=2^{-1/7}$. Thusso $X_2=0.8$ to one significant figure. Retaining the unrounded hydrogen mass fraction when comparing central temperatures givesThe first model has the higher central temperature.

For homogeneous hydrogen burning at fixed $M,Z$, fuel conservation gives $M{E}_0\dot{X}=-L$. Since $d\mu /dX=-5{\mu }^2/4$ and $L=L_0(\mu /{\mu }_0{)}^7$, the homogeneous fuel-depletion luminosity feedback obeysIntegration with the initial mean molecular weight ${\mu }_0$ givesThis idealized stellar evolution law stops when $X=0$; it does not predict a physically attainable infinite luminosity. Indeed $\mu =4/3$ at fuel exhaustion, giving $t_H=M{E}_0[1-(3{\mu }_0/4{)}^8]/(10{\mu }_0{L}_0)$, strictly before the formal singularity.

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

Take the retained disc mass to be the cosmic baryon fraction of the total dark matter halo mass: $M_d=f_b{M}_h$, where $f_b=0.05/0.25=0.2$. Adopt $v_{\mathrm{vir}}^2=G{M}_h/r_{\mathrm{vir}}$ and describe the disc gravity by $v_d^2=C_{d}G{M}_d/R_d$. The structure coefficient $C_d$ must be stated: the scale-free Mestel disc gives $C_d=1$ when its flat rotation speed and enclosed mass are used, while a finite disc requires boundary and thickness information. The equality of edge specific angular momentum gives $R_d{v}_d=q{r}_{\mathrm{vir}}{v}_{\mathrm{vir}}$, with $q=0.1$. Combining these relations gives the angular-momentum-conserving self-gravitating disc radius:With the commonly intended $C_d\simeq 1$ approximation,The corresponding masses are $M_h\simeq 2.49\times 10^{11}{M}_{\odot }$ and $M_d\simeq 4.97\times 10^{10}{M}_{\odot }$, where $M_{\odot }$ is the solar mass.

For a definite formation estimate, assume the dark matter halo acquired this mass and radius at virial equilibrium, retained them afterward, and had mean density ${\Delta }_v{\bar{\rho }}_m(z_f)$ with ${\Delta }_v=18{\pi }^2$. This uses the matter-dominated spherical-collapse model as an approximation to the specified matter-plus-vacuum cosmology. Since ${\bar{\rho }}_m(z_f)=3H_0^2{\Omega }_{m0}(1+z_f{)}^3/(8\pi G)$,The approximate distance therefore gives formation modestly before observation. Using the numerical rather than approximate distance integral gives $z_f\simeq 1.22$, within the accuracy of these structural assumptions. The calculation also assumes inclination-corrected spectroscopy, negligible disc pressure support, conserved edge specific angular momentum, no substantial later accretion or mergers, and no significant baryon loss.

This formation redshift is model-dependent, not uniquely fixed by the stated disc data. Keeping $C_d$ explicit gives $1+z_f\simeq 2.209C_d^{-4/3}({\Delta }_v/18{\pi }^2{)}^{-1/3}$. In particular, replacing the mean-density convention by $200$ times the critical density gives $H(z_f)=v_{\mathrm{vir}}/(10r_{\mathrm{vir}})$ and $z_f\simeq 0.87$, later than the observation. That inconsistent chronological result cannot be silently used as the formation epoch.

There is a further idealization in applying $C_d=1$ exactly at the edge: an abruptly truncated razor-thin disc with nonzero edge surface density of a disk has a logarithmically divergent in-plane force at that edge. Locally the available mass occupies a half-plane, and its radial contribution is proportional to $G{\Sigma }_d{\int }_{ϵ}^{\ell }ds/s$, with no opposite exterior half-plane to cancel it. Finite thickness or a smooth taper removes that sharp-edge force singularity of a thin disk. Thus $C_d=1$ is an explicitly qualified Mestel disc or monopole-scale approximation, not an exact finite-disc calculation; enclosed mass does not determine a disc rotation curve.

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.

In geometrized units, a mass is converted to a length by multiplying its SI value by $G/c^2$, or to a time by multiplying by $G/c^3$. Using the supplied constants, the solar mass becomesThus the characteristic scales are kilometres and a few microseconds. With potential zero at infinity, the Newtonian gravitational potential at the surface, in units of $c^2$, isIts very small magnitude is the relevant weak-field measure. In SI potential units the same number corresponds to about $-1.91\times 10^{11}{\mathrm{m}}^2{\mathrm{s}}^{-2}$.

For a local axisymmetric Fourier mode, the Toomre stability criterion balances three contributions to the squared oscillation frequency:The radial epicyclic frequency supplies rotational restoration at long wavelengths; isothermal sound speed and pressure stabilize short wavelengths; disk self-gravity destabilizes intermediate wavelengths. Minimizing over $|k|$ gives $|k|=\pi G{\Sigma }_0/c_s^2$ and ${\omega }_{\min }^2={\kappa }^2(1-Q^{-2})$. Axisymmetric gravitational instability occurs when $Q<1$.

In a centrally dominated Keplerian disk, $\kappa =\Omega$ and vertical hydrostatic equilibrium gives $c_s\simeq H\Omega$. Using ${\Omega }^2=G{M}_{\ast }/r^3$ and a local disk-mass estimate $M_D\sim \pi r^2{\Sigma }_0$,This is the disk mass form of the Toomre criterion. An actual enclosed mass depends on the radial surface density profile and changes an order-one coefficient. In particular, a profile proportional to $r^{-3/2}$ has $M_D(<r)\simeq 4\pi r^2{\Sigma }_0$ when its inner cutoff is negligible.

For the protosolar estimate, take $M_{\ast }\simeq M_{\odot }\simeq 2\times 10^{33}\mathrm{g}$. This solar mass is implicit in identifying the central star with the young Sun. The supplied constant disk aspect ratio givesUsing the printed approximate astronomical unit and gravitational constant, one can also obtain $\Omega (1\mathrm{AU})\simeq 3.7\times 10^{-7}{\mathrm{s}}^{-1}$ and $c_s\simeq 3.7\times 10^5{\mathrm{cm}\mathrm{s}}^{-1}$; they give the same $Q$. The gravitational constant cancels from the mass form.

At one astronomical unit the disk is very stable, with $Q\simeq 640$. Formal extrapolation givesThis is far beyond the planetary region and any plausible extent of the minimum-mass solar nebula. Moreover, the extrapolated enclosed disk mass is already a substantial fraction of a solar mass, so the centrally dominated approximation becomes questionable. The formal radius is not a prediction of a real unstable outer nebula. Direct gas fragmentation by gravitational instability is unlikely to have formed Solar System planets in this model. Core accretion is the more natural route; an earlier substantially more massive or colder disk would be a different model. Even $Q<1$ alone does not guarantee fragmentation, because sufficiently rapid cooling is also needed.

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.

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.

A quasar is a luminous active galactic nucleus powered by gas accretion onto a supermassive black hole. The hot disc produces much of the optical/ultraviolet continuum; surrounding gas gives broad and narrow emission lines, while some systems have relativistic jets. A small emitting region is inferred from rapid variability, $R\lesssim c\Delta t_{\mathrm{obs}}/(1+z)$ when beaming is absent. Sustaining a high luminosity from such a compact region favours accretion over ordinary stellar power. For approximately spherical, ionized hydrogen with electron-scattering opacity,Comparing luminosity with this Eddington scale constrains mass only after assumptions about anisotropy, opacity and accretion state. The emitted power obeys $L=\eta \dot{M}{c}^2$; high binding-energy efficiency makes growth of a massive compact accretor energetically plausible.

Broad-line reverberation mapping estimates a gas radius from a continuum-to-line time lag, $R\sim c\tau$, and combines it with line velocity width to infer $M\sim f_{\mathrm{vir}}R{v}^2/G$. Geometry, inclination and nongravitational gas motions enter the factor $f_{\mathrm{vir}}$. Independent evidence for compact massive nuclei comes from resolved stellar or gas orbits in nearby galaxies. Andrea M. Ghez and collaborators' Galactic-centre stellar-orbit study fitted a common central potential and distance, demonstrating a compact dark mass of order four million solar masses; the Galactic centre is a low-luminosity example rather than a quasar.^[ref]

quasars are an accretion phase of massive nuclear black holes; luminosity, variability and dynamical evidence constrain different aspects of that interpretation. A finite-resolution dark-mass measurement by itself is not a direct image of an event horizon.

Kepler's third law for two point masses in a Newtonian gravitational field isHere $a$ is the semimajor axis of their relative orbit, $P$ the orbital period and $G$ the gravitational constant. Equivalently $(P/\mathrm{yr}{)}^2=(a/\mathrm{AU}{)}^3/[(M_1+M_2)/M_{\odot }]$, to the accuracy with which these astronomical units normalize the solar mass.

The intact source figure gives orbital period $P=6.0\mathrm{yr}$, stellar radial velocity semiamplitude $K=30.0\mathrm{m}{\mathrm{s}}^{-1}$, and orbital eccentricity $e=0$. A planetary exoplanet transit implies a nearly edge-on orbit, so $\sin i\simeq 1$. Adopt a solar mass for the Solar-type star and initially neglect the companion's mass relative to it.

For a circular solution of the two-body problem, the star's distance from the centre of mass is $a_{\ast }=a{M}_p/(M_{\ast }+M_p)$, and the measured semiamplitude is $K=2\pi a_{\ast }\sin i/P$. Kepler's third law givesHenceThe inferred binary mass ratio is only $0.00183$, validating the approximation. Since $e=0$, the star–planet separation is constant and equals the orbital semimajor axis. Without the exoplanet transit, Doppler spectroscopy would generally give only a minimum companion mass through $M_p\sin i$.

Kepler's third law for two point masses in a Newtonian gravitational field isHere $a$ is the semimajor axis of their relative orbit, $P$ the orbital period and $G$ the gravitational constant. Equivalently $(P/\mathrm{yr}{)}^2=(a/\mathrm{AU}{)}^3/[(M_1+M_2)/M_{\odot }]$, to the accuracy with which these astronomical units normalize the solar mass.

For two bodies of masses $M_1,M_2$, the relative Kepler orbit has semimajor axis $a$ and orbital period $P$ satisfying Kepler's third lawHere $a$ is the semimajor axis of the separation vector, not either star's individual barycentric orbit. In conventional stellar units the same relation is approximately $(P/\mathrm{yr}{)}^2=(a/\mathrm{AU}{)}^3/[(M_1+M_2)/M_{\odot }]$, with year, astronomical unit and solar mass as the units.

Using the supplied atomic masses consistently includes the net electron bookkeeping after positron annihilation. The released fraction is $\eta =(4m_H-m_{He})/(4m_H)\simeq 0.0071$, corresponding to about 26.7 MeV per helium nucleus. If 10% of the solar mass is processed and luminosity stays constant, $t\simeq 0.1\eta M_{\odot }{c}^2/L_{\odot }\simeq 10^{10}$ years.

At separation one femtometre the proton Coulomb energy is 1.44 MeV. At $2\times 10^7$ K, mean thermal kinetic energy $3k_{B}T/2\simeq 2.59$ keV, making their ratio about $5.6\times 10^2$. Classical over-barrier collisions are therefore extremely improbable. Quantum tunnelling through the Coulomb barrier and the high-energy distribution tail allow fusion, with the slow weak-interaction step regulating the proton-proton chain. The reaction summary is a net energy/matter balance, not one simultaneous four-proton collision.

A white dwarf is a compact stellar remnant supported primarily by electron degeneracy pressure, radiating stored thermal energy rather than sustained main-sequence fusion. At zero-temperature leading order, filling electron momentum states gives $n_e=p_F^3/(3{\pi }^2{\hslash }^3)$ and $P=p_F^5/(15{\pi }^2{\hslash }^3{m}_e)$. Fully ionized hydrogen mass fraction $X$, with other nuclei approximately one electron per two nucleons, implies $n_e=(1+X)\rho /(2m_u)$. ThereforeThis is an $n=3/2$ polytrope. Solving hydrostatic structure with a zero-pressure surface gives $R=2.357K_1(1+X{)}^{5/3}{M}^{-1/3}/G$; the numerical factor uses ${\xi }_1=3.65375$ and $-{\xi }_1^2{\theta }^{\prime }({\xi }_1)=2.71406$ of that Lane-Emden solution. Thus $R\propto M^{-1/3}$. For $X=0$, $M=M_{\odot }$, $R$ is about $9\times 10^3$ km. A uniform-density scaling gives the same mass dependence but a different order-one coefficient and should be identified as such.

Independently, $R/R_{\odot }=\sqrt{0.056}(T_{\odot }/25000{)}^2\simeq 0.0126$, or about $8.8\times 10^3$ km. Agreement is good at this level; relativistic degeneracy becomes relevant for accurate modeling near one solar mass and eventually invalidates an unlimited nonrelativistic mass-radius law.