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Separate the center-of-mass motion from the relative motion. Two equal quark masses give reduced mass $\mu =m_q/2$. Replacing the hydrogen Coulomb strength by $a>0$ gives the Coulomb binding energies and length scaleThe factor of two in the reduced mass matters: using $m_q$ directly would double the binding energy. These are relative binding energies measured from the free quark–antiquark threshold; the rest energy $2m_q{c}^2$ is excluded.

Take overdots in this question to mean differentiation with respect to the orbiting particle's proper time $\tau$. Write $\mu =GM/c^2$, $x=r/\mu$ and $\ell =hc/(GM)$, with $h$ the specific angular momentum. Then the dimensionless timelike effective potential isThe last term is the relativistic correction to the Newtonian centrifugal-plus-gravitational potential. At large $r$ the two agree. Newtonian angular momentum gives an infinite inner centrifugal barrier for nonzero $h$; the relativistic correction lowers that barrier and permits plunge. The exterior problem starts at $r=2\mu$, where $U=-c^2/2$; extrapolating the potential towards $r=0$ is not a static exterior description.

Circular orbits obeyThere are two extrema when $h^2>12(GM/c{)}^2$: an inner maximum representing an unstable circular orbit and an outer minimum representing a stable one. At equality they merge at $r=6\mu$; below it neither exists. More explicitly, evaluate the curvature at a circular orbit:Timelike circular orbits require $r>3\mu$. They are radially stable for $r>6\mu$, so the marginally stable inner limit is $r_{\mathrm{ISCO}}=6GM/c^2$. The Newtonian potential instead has one stable minimum at $r=h^2/(GM)$, without an innermost stable circular orbit.

For the Schwarzschild circular-orbit local speed in the question, $\gamma$ is the Lorentz factor relative to a local Schwarzschild static observer, not $dt/d\tau$. That observer's clock obeys $dT=\sqrt{1-2\mu /r}dt$. Since the particle passes with local speed $v$, $rd\varphi /d\tau =\gamma v$. Circular normalization givesThus $v=c/2$ at the innermost stable circular orbit. Using Schwarzschild time in place of the local clock would give a different speed.

For the conserved particle energy measured at infinity, substitute $h^2$ in the circular energy equation. Setting $y=\mu /r$ givesTake the positive root for a future-directed exterior orbit. It approaches $m{c}^2-GMm/(2r)$ at large radius. The marginally bound circular Schwarzschild orbit solves $k=1$, giving $y(4y-1)=0$ and hence $r_{\mathrm{mb}}=4GM/c^2$ at finite radius. It is unstable; strictly bound circular orbits have $r>4\mu$, with $E<m{c}^2$. The phrase smallest bound orbit here denotes this marginal circular limit, not a prohibition on a bound-energy trajectory plunging below it.

At the innermost stable circular orbit,A radiatively efficient thin disc with a zero-torque inner boundary condition can therefore release about 5.7% of the accreted rest energy before the plunge. energy advection, stress at the inner edge and photon capture can alter the escaping luminosity, so this is a model efficiency rather than a universal emitted fraction.

For the Kerr ISCO spin limits of an equatorial prograde disc around a rapidly rotating Kerr black hole, the innermost stable circular orbit moves inward and its specific energy decreases. In the ideal extremal limit its Boyer-Lindquist radius tends to $\mu$, $E/(m{c}^2)\to 1/\sqrt{3}$, and the corresponding binding-energy efficiency tends to 42.3%. Retrograde rotation has the opposite trend: the extremal limit is $r_{\mathrm{ISCO}}=9\mu$, $E/(m{c}^2)=5/(3\sqrt{3})$, giving about 3.8%. The prograde limiting coordinate radius does not identify a massive orbit with the null horizon generator.^[ref]

Finally $\dot{t}=k/(1-2\mu /r)=1/\sqrt{1-3\mu /r}$. Dividing $\dot{\varphi }=h/r^2$ by this givesIts algebraic form is exactly Newtonian, but $r$ is the Schwarzschild spacetime areal radius and $t$ Schwarzschild time, normalized at infinity. Local clock rates, orbit stability and binding energy remain relativistic. With $G{M}_{\odot }\simeq 1.3271\times 10^{20}{\mathrm{m}}^3{\mathrm{s}}^{-2}$,The marginally bound circular-orbit frequency is about $0.404\mathrm{kHz}$. These Schwarzschild ISCO orbital frequencies and marginally bound frequencies are rotation frequencies per distant stationary observer's time, as intended by using $d\varphi /dt$. A static observer at finite radius $r_o$ instead divides both frequencies by $\sqrt{1-2\mu /r_o}$, leaving their ratio unchanged. Identifying an observed quasi-periodic oscillation with the innermost stable circular orbit orbital frequency is an additional astrophysical assumption.

A donor star can feed a compact companion through Roche-lobe overflow near the inner Lagrange point, or through capture of a stellar wind. The transferred gas usually carries angular momentum; when its circularization radius exceeds the accretor's effective inner radius it forms an accretion disc. Outward angular momentum transport permits inward mass flow, while dissipation converts gravitational binding energy into heat and radiation. For a compact object with a surface $L\sim GM\dot{M}/R$; for a black hole write $L=\eta \dot{M}{c}^2$, with efficiency determined by the flow and inner boundary. Optically thick disc regions emit a sum of thermal radiation spectrum components; a hot optically thin accretion corona can inverse-Compton scatter softer photons into a harder X-ray component. A black hole has no material surface on which to radiate a stellar boundary layer.

An explicit example is the wind-fed black-hole binary Cygnus X-1. Its massive stellar companion supplies gas, and the accretion flow produces strong X-ray emission.^[ref] Compact-object evidence comes from orbital dynamics rather than X-ray brightness alone. The single-lined binary mass function constrains the unseen mass; inclination, donor properties and distance refine it. Jerome A. Orosz and collaborators combined optical photometry, radial velocities, rotation and a distance constraint to obtain a black-hole mass of $14.8\pm 1.0M_{\odot }$ in their 2011 analysis. The inference depends on the stellar and orbital model, and that historical value should not be confused with a latest measurement.^[ref]

photon orbit and the Newtonian limit. Write $r_g=GM/c^2$ and $e=E/(m{c}^2)$. For a timelike circular Schwarzschild geodesic,The denominator tends to zero at $r=3r_g$, so the energy per unit rest mass diverges as a massive circular orbit approaches that radius. The corresponding local orbital speed measured by a static observer satisfies $v_{\mathrm{loc}}^2/c^2=r_g/(r-2r_g)$ and tends to one. This suggests a null circular orbit, but one must not simply insert $m=0$ into a massive-particle formula. For a photon, the radial effective potential is proportional to $(1-2r_g/r)/r^2$; its derivative vanishes at $r=3r_g$ and its second derivative there is negative. Thus the photon sphere is at $3GM/c^2$, and its circular null orbits are unstable.

At $r\gg r_g$, expansion givesThe second term is the total Newtonian circular-orbit energy, kinetic plus potential; the relativistic energy also includes rest energy. Here $E$ is the conserved energy referred to a stationary observer at infinity, not the local kinetic energy.

The stable inner edge. At fixed angular momentum, write $\ell =L/(mc)$. The timelike radial geodesic equation can be writtenThe circular condition $V_{\mathrm{eff}}^{\prime }=0$ gives ${\ell }^2=r_g{r}^2/(r-3r_g)$. Substituting into the second derivative taken at fixed $\ell$ givesTimelike circular orbits exist for $r>3r_g$; they are stable against radial perturbations for $r>6r_g$ and unstable for $3r_g<r<6r_g$. ThereforeThe boundary orbit is marginally stable. Differentiating $e(r)$ also gives a minimum there, consistently with this fixed-angular-momentum stability calculation.

accretion efficiency and black-hole spin. Matter supplied slowly from far away initially brings rest energy $d{m}_0{c}^2$. A radiatively efficient, zero-inner-torque thin disk releases the binding energy down to the innermost stable circular orbit; the remaining energy $e_{\mathrm{ISCO}}d{m}_0{c}^2$ is carried into the hole during the plunge. ThusHere $\eta$ denotes the efficiency called $ϵ$ in this question. The ideal relation assumes negligible additional dissipation during plunge and neglects photon capture and advection. A hot radiatively inefficient flow need not have this efficiency.

For a Kerr black hole, an equatorial prograde disk has a smaller innermost stable circular orbit and greater binding-energy efficiency than the Schwarzschild value; a retrograde disk has the opposite trend. The formal extremal prograde limits are $r_{\mathrm{ISCO}}\to r_g$ and $\eta \to 1-1/\sqrt{3}\simeq 0.423$. These are ideal limiting values, not a universal realized efficiency. photon capture and the accretion geometry affect the achievable black-hole spin and escaping radiation.

radiation force and the limiting luminosity. For isotropic luminosity $L$, the outward radiation momentum flux at radius $r$ is $L/(4\pi r^{2}c)$. In fully ionized hydrogen, Thomson scattering acts mainly on the electron, while electrostatic coupling makes the electron and proton move together. Neglecting the electron mass, one electron–proton pair experiencesEquating them yieldsThis Eddington luminosity uses electron-scattering opacity, spherical or isotropic force balance and negligible relativistic corrections at the force-comparison radius. Composition, other opacity, anisotropy and photon trapping can change the relevant limit.

Exponential growth and doubling. Distinguish rest-mass supply ${\dot{m}}_0$ from black-hole mass growth $\dot{M}$:At the Eddington limit,Using the printed scattering cross section gives $t_E\simeq 4.50\times 10^8\mathrm{yr}$. The Salpeter time $t_S$ is an e-folding time; the Eddington-limited doubling time isfor the Schwarzschild thin-disk efficiency. If one approximates $\dot{M}\simeq {\dot{m}}_0$, the factor $1/(1-\eta )$ is dropped and the estimate is about $1.79\times 10^7\mathrm{yr}$. The retained-energy version is the mass-conserving expression.

The efficiency bound. A growth factor $10^8$ requires $N=\log (10^8)=18.4207$ e-folds. Eddington-limited accretion is the fastest growth allowed by the adopted luminosity bound at fixed efficiency, so growth within a time $\Delta t$ requiresThis efficiency upper bound from Eddington-limited growth is an upper bound: higher efficiency radiates more of each supplied rest mass, and the luminosity ceiling therefore allows less mass to enter per unit time. A duty cycle below unity or sub-Eddington luminosity tightens it. The approximate convention $\dot{M}={\dot{m}}_0$ would instead give $\eta \lesssim 0.121$; both figures refer to the same effect, with different treatment of the radiated mass deficit. Mergers, super-Eddington growth or a different seed history would change the premises.

Coherent prograde accretion spins up the hole, moves the inner edge inward and increases $\eta$. For fixed $L/L_{\mathrm{Edd}}$ the mass-growth e-folding time then grows, so the logarithmic growth curve becomes less steep; one cannot simply retain the initial Schwarzschild efficiency throughout. Episodic misaligned accretion can have a different black-hole spin history. In general,

energy available to affect the galaxy. Let $M_g$ denote galaxy mass, keeping it distinct from black-hole mass. At constant efficiency, negligible seed mass and no merger contribution, the emitted energy in building $M_{\mathrm{BH}}=0.002M_g$ isFor $v=300\mathrm{km}{\mathrm{s}}^{-1}$ and the nonspinning efficiency this is approximately $121$. The common small-efficiency approximation gives $2000\eta \simeq 114$, and a fiducial $\eta =0.1$ would give about $222$ with retained mass included. These differences do not alter the conclusion: the available accretion energy substantially exceeds the stated galactic binding-energy estimate.

Only a coupling fraction of order one percent would then be energetically sufficient to affect the gas strongly. radiation, winds or jets can heat gas, drive outflows and suppress further cooling and star formation, providing active-galactic-nucleus feedback and possible self-regulation of black-hole growth. energy availability alone does not prove that the whole galaxy is expelled: most radiation can escape, jets can be collimated, heated gas can cool, and collisionless stars couple differently from gas.

mass from motion. A resolved Keplerian orbit gives $M=4{\pi }^2{a}^3/(G{P}_{\mathrm{orb}}^2)$ when the orbiting object has negligible mass. At the Galactic centre, repeated infrared astrometry and spectroscopy measure sky motion, acceleration and line-of-sight velocity of stars around Sagittarius A*. Resolving a compact enclosed mass and fitting complete orbits avoids reliance on a single projected velocity. Source confusion, reference-frame stability, distance, inclination and possible extended mass must be included in the fit.^[ref]

For an unresolved stellar-mass binary in a circular orbit, the donor's velocity semiamplitude $K_2$ and orbital period give the binary mass functionIt provides a lower mass bound; donor mass and orbital inclination are needed for an actual mass estimate. For an eccentric orbit, multiply the observational combination $P_{\mathrm{orb}}{K}_2^3/(2\pi G)$ by $(1-e^2{)}^{3/2}$ when evaluating the binary mass function. Resolved gas or maser rotation and stellar-dynamical models also measure masses, provided pressure support, noncircular motion and line-of-sight projection are modeled. Resolving the black-hole sphere of influence improves separation of central mass from the host galaxy.

For active nuclei, reverberation mapping measures a lag $\tau$ between variable continuum and broad-line response. With a line velocity scale $\Delta v$,The factor $f_{\mathrm{geom}}$ accounts for geometry and kinematics and is not known simply from the lag. Cadence, line responsivity, inclination and winds require control. Rapid variability demonstrates compactness but is not by itself a unique mass measurement.

black-hole spin from strong-field emission. Define $a_{\ast }=cJ/(G{M}^2)$. Large-radius Keplerian motion is principally a mass probe and usually insensitive to black-hole spin. In the Kerr model, the radius and binding energy of the innermost stable circular orbit vary with $a_{\ast }$ and orbital orientation, making inner-disk radiation useful.

black-hole continuum fitting infers an inner emitting radius from the disk's thermal spectrum and flux normalization, then identifies it with the innermost stable circular orbit in a relativistic thin-disk model. Independent mass, distance and inclination, together with atmospheric spectral hardening, absorption and contamination, are needed. Spectra should be selected in a disk-dominated regime where truncation and strong advection are limited. A 2011 Cygnus X-1 analysis combined such spectra with independently constrained binary parameters and reported a near-extreme black-hole spin conditional on black-hole spin–orbit alignment and its disk model.^[ref]

black-hole reflection spectroscopy instead models a relativistically broadened iron-line and reflection continuum: orbital Doppler shifts and beaming broaden the profile, while gravitational redshift contributes to the extended red wing. Its innermost emitting radius can constrain black-hole spin if reflection extends to the stable inner edge. Ionization, abundance, absorption, coronal illumination, disk thickness and possible truncation are competing parameters; a broad line is not a direct black-hole spin reading independent of these assumptions. Timing or precession models can add information but introduce their own dynamical identification.

mass is principally inferred from gravitating dynamics; black-hole spin is inferred from strong-field effects with a specified spacetime and accretion model. Agreement between independent methods is especially valuable because disk inner radius, inclination and radiation-transfer assumptions can otherwise mimic a change of black-hole spin.

An active galactic nucleus is a compact, exceptionally luminous central source powered chiefly by black-hole accretion rather than ordinary stellar fusion. Typical signatures include a broad spectral-energy distribution extending from radio to X-rays, rapid variability implying a small emitting region, strong optical or ultraviolet emission lines and, in some objects, radio jets and lobes. Broad permitted lines indicate rapidly moving dense gas; narrower forbidden lines can arise farther out. Obscuration and viewing orientation can hide the broad-line region, and many active nuclei are radio quiet, so no one signature is required in every object.

For the Schwarzschild circular-ring energy shift, set $r_g=GM/c^2$ and let $E_0$ be the photon energy in the orbiting gas frame. A local static observer in Schwarzschild spacetime measures $E_{\mathrm{stat}}$. For gas receding along the line-of-sight component of its velocity, special-relativistic frequency transformation givesPropagation from that static frame to a distant stationary observer contributes $E_{\infty }=E_{\mathrm{stat}}\sqrt{1-2r_g/r}$. HenceThe factors describe gravitational redshift, transverse time dilation and the longitudinal Doppler shift. Here the emitter is on the maximally receding azimuth of a circular disc ring. A general azimuth has a smaller projected velocity, and a fully ray-traced image must use the photon direction at emission rather than equating that local direction with the distant inclination. The supplied expression is the direct-projection transfer model, not a complete light-bending calculation.

The innermost stable circular orbit is the marginal stable timelike circular geodesic separating stable circular motion from radial instability. In Schwarzschild spacetime,For example, the circular-orbit specific angular momentum satisfies $j_c^2=GM{r}^2/(r-3r_g)$; its radial stability changes where $\mathrm{d}{j}_c^2/\mathrm{d}r=0$, at $6r_g$. This is outside both the horizon at $2r_g$ and the circular photon orbit at $3r_g$.

The printed local orbital-speed relation gives ${\beta }^2=r_g/(r-2r_g)$. Combining the gravitational and transverse factors,At the ISCO, $\beta =1/2$, soThe Edge-on Schwarzschild ISCO redshift envelope is reached for an edge-on disc on its receding side:The approaching edge has $1+z=\sqrt{2}(1-\frac{1}{2}\sin i)$. For a fixed inclination below $90$ degrees the maximum is smaller; the quoted number is an upper envelope, not the value for every observer. Within the stated ring model the receding factor increases inward across stable orbits, so the inner edge sets that envelope. Emission from a plunging region, occultation and returning rays are outside this ISCO-emitting model.

The Kerr ISCO efficiency limits follow from equatorial circular orbits in the Kerr metric. Prograde motion has a smaller ISCO than the non-spinning case, while retrograde motion has a larger one. A zero-torque thin disc radiating the orbital binding energy before the gas plunges has efficiency $\eta =1-E_{\mathrm{ISCO}}$, where $E_{\mathrm{ISCO}}$ is energy per unit rest-mass energy at infinity. The ideal limiting values areThus more orbital binding energy is available in the prograde inner disc. The extremal values are formal geodesic limits: photon capture, inner stresses, outflows and finite accretion thickness change the efficiency of real flows. An ISCO inside the ergoregion is not a claim that material or radiation escapes from inside the horizon.

For accretion output relative to galactic binding energy, distinguish accreted rest mass from the mass retained by the black hole. At constant radiative efficiency, building a hole from a negligible seed to $M_{\mathrm{BH}}=\alpha M$ emitsA non-negligible seed replaces $\alpha M$ by the actual mass increase. Let the first use of ${\sigma }_v$ mean the total three-dimensional rms random speed. For an approximately virialized galaxy with negligible boundary term, $K=\frac{1}{2}M{\sigma }_v^2$ and $W=-2K=-M{\sigma }_v^2$. The ratio to the magnitude of its gravitational potential energy is thereforeIf “binding energy” means the net energy needed to unbind the virialized system, $|K+W|=M{\sigma }_v^2/2$, the ratio is twice as large. The widely used order-of-magnitude estimate drops the retained-mass correction and writes $\eta \alpha (c/{\sigma }_v{)}^2$; it should not be confused with a precisely fixed structural coefficient.

For the numerical parameters given,No efficiency is specified, so this function is the numerical result. For an explicitly chosen illustration $\eta =0.1$, it is about $111$, or $222$ relative to the net virial binding energy. The non-spinning zero-torque efficiency gives about $60.6$ relative to $|W|$; the ideal retrograde and prograde limits give about $39.2$ and $731$, respectively. Spin affects the ratio through the adopted efficiency and orbital orientation, not through a universal rule that a larger unsigned spin always means a larger efficiency.

Such large energy ratios make even a modest coupling of the output to galactic gas dynamically important. Radiation and winds can heat or eject gas and reduce star formation or later black-hole feeding. For the $\eta =0.1$ illustration a percent-level coupling is already comparable to $|W|$. Energy available is not energy necessarily deposited: cooling, escape directions, gas geometry, photon opacity and the duration of activity determine the actual feedback. It does not follow that the entire stellar galaxy must be disrupted.

For the swept-shell calculation, the problem now explicitly redefines ${\sigma }_v$ as a one-dimensional line-of-sight dispersion. The specified isothermal profile hasNeglecting central-hole gravity and the shell's self-gravity, the inward force due to the remaining interior component isIn a single-scattering momentum-driven model with the shell intercepting the central luminosity, $F_{\mathrm{rad}}=L/c$. At the hydrogen Eddington luminosity this becomes $4\pi G{M}_{\mathrm{BH}}{m}_p/{\sigma }_T$. Sustained outward acceleration against that weight requiresand henceThe radius cancels because the swept isothermal mass is proportional to $R$. This is the isothermal-galaxy momentum-balance shell threshold, with its fourth-power dispersion scaling.

The coefficient identifies the approximations. Self-gravity in a thin spherical shell expansion adds the shell's inward pressure-equivalent force $G{M}_{\mathrm{sh}}^2/(2R^2)$ and replaces $f(1-f)$ by $f(1-f/2)$. A shell intercepting only a fraction of the photon momentum needs a larger luminosity; Thomson-thin material does not automatically receive $L/c$. Conversely, multiple scattering or an energy-driven hot wind changes the force model. A shell already carrying sufficient outward kinetic energy need not satisfy this instantaneous acceleration condition everywhere to escape. The printed inequality is the stated quasistatic, fully coupled momentum-balance estimate, not an assumption-free lower bound for every possible shell trajectory.

Astrophysical jets are directed plasma outflows whose bulk energy and momentum remain concentrated near an axis over distances much larger than the launch region. They occur around accreting black holes and neutron stars and in forming stellar systems, with very different characteristic speeds and scales. An accretion reservoir supplies binding energy; a rotating black hole can additionally provide spin energy through electromagnetic stresses. An observed jet therefore does not uniquely identify either a black hole or a particular launch mechanism.

Magnetic fields couple the rotating engine to the outflow. Gas tied to suitable open field lines can be accelerated centrifugally from a disc; a large-scale poloidal field threading a rotating hole can support outward electromagnetic energy flux. Farther out, magnetic stresses, hoop stresses and external pressure can aid collimation. Pressure-driven acceleration, magnetic-to-kinetic conversion and entrainment depend on loading and environment. A simple stationary axisymmetric picture does not guarantee stability against kinks, mixing or disruption.

The emission commonly contains synchrotron radiation from relativistic electrons spiraling in magnetic fields. Its broad continuum and linear polarization constrain the particles and ordered field component; inverse Compton scattering can extend radiation to higher photon energies. Compact knots can be shocks or moving patterns rather than individual material parcels. Interaction with the surroundings creates shocks, hotspots, cocoons and radio lobes, transporting energy far beyond the central emission region.

Relativistic viewing effects are crucial. If a feature moves at speed $\beta c$ at angle $\theta$ toward the observer, its Doppler factor isA comoving emission frequency is observed at ${\nu }_{\mathrm{obs}}={\delta }_D{\nu }^{\prime }/(1+z_{\cos })$. For a discrete isotropically emitting moving feature with spectral index $F_{\nu }^{\prime }\propto {\nu }^{\prime -s}$, the flux enhancement scales as ${\delta }_D^{3+s}$ in the usual unresolved-blob approximation. Continuous-jet conventions give a different exponent, so apparent brightness is not a direct orientation-free jet-power measurement.

Light-travel delays also produce apparent superluminal motion. In a source time interval $\Delta t$, the transverse displacement is $\beta c\Delta t\sin \theta$, but the observer receives the two signals separated by $\Delta t(1-\beta \cos \theta )$. ThereforeThe maximum is $\Gamma \beta$, at $\cos \theta =\beta$. An apparent speed above $c$ violates no local speed limit. The receding jet is deboosted, so an intrinsically two-sided source may look one-sided; absorption and environmental differences can also contribute.

Jets feed back on their host systems by depositing mechanical energy, inflating cavities, accelerating particles and modifying surrounding gas. Radio luminosity measures only part of that budget; unseen ions, magnetic fields, radiative losses and work on the environment all matter. Connecting the compact engine to a large lobe therefore requires both relativistic emission physics and the dynamics of its surroundings.

Adopt the approximate mass-luminosity relation $L/L_{\odot }\simeq (M/M_{\odot }{)}^3$ and equal available fuel fractions. The main-sequence lifetime then scales as $M/L$, givingUsing an exponent $3.5$ instead gives about $5.4\times 10^7$ and $3.1\times 10^8$ years; these are model-dependent estimates. At $r=3r_s=6GM/c^2$ the Newtonian gravitational potential energy released from infinity is $GMm/r=m{c}^2/6$. Half remains as kinetic energy in a circular orbit, so the radiatable binding energy is $GMm/(2r)$:Using the whole potential energy loss would give $1/6$, which does not subtract the final kinetic energy. A fully relativistic calculation for a nonrotating black hole gives $1-\sqrt{8/9}\simeq 5.7\%$. When the surviving star exhausts hydrogen burning, it expands off the main sequence and can fill its Roche lobe. Binary mass transfer to the black hole can form an accretion disk and an X-ray binary; unstable binary mass transfer may instead produce a common envelope. This is conditional on the binary star remaining bound after the first remnant forms. The supplied black hole endpoint is an assumption, not a universal outcome for an isolated $8M_{\odot }$ star.

Adopt the approximate mass-luminosity relation $L/L_{\odot }\simeq (M/M_{\odot }{)}^3$ and equal available fuel fractions. The main-sequence lifetime then scales as $M/L$, givingUsing an exponent $3.5$ instead gives about $5.4\times 10^7$ and $3.1\times 10^8$ years; these are model-dependent estimates. At $r=3r_s=6GM/c^2$ the Newtonian gravitational potential energy released from infinity is $GMm/r=m{c}^2/6$. Half remains as kinetic energy in a circular orbit, so the radiatable binding energy is $GMm/(2r)$:Using the whole potential energy loss would give $1/6$, which does not subtract the final kinetic energy. A fully relativistic calculation for a nonrotating black hole gives $1-\sqrt{8/9}\simeq 5.7\%$. When the surviving star exhausts hydrogen burning, it expands off the main sequence and can fill its Roche lobe. Binary mass transfer to the black hole can form an accretion disk and an X-ray binary; unstable binary mass transfer may instead produce a common envelope. This is conditional on the binary star remaining bound after the first remnant forms. The supplied black hole endpoint is an assumption, not a universal outcome for an isolated $8M_{\odot }$ star.

Assume dilute, nondegenerate ideal electrons, protons and neutral atoms in thermal and chemical equilibrium, with charge neutrality $n_p=n_e$. Set ${\mu }_H={\mu }_p+{\mu }_e$, using each ideal species' translational factor $n{\lambda }_{\mathrm{th}}^3/g$ and hydrogen ground-state binding energy $-R$. Neglect excited states and take the heavy proton/atom thermal-wavelength ratio as one. With electron spin and neutral-atom degeneracies treated in the convention of the source, the remaining factor isDifferent partition-degeneracy conventions must be propagated consistently. Screening, dense-plasma level dissolution and departures from local thermodynamic equilibrium lie outside this derivation.

Ensemble and applications. A grand canonical ensemble describes a system exchanging both energy and particles with a large reservoir. Its fixed control parameters are $T,V,\mu$, while microstates with different $N$ have probabilitiesWeak coupling allows the system's energy levels to remain meaningful; equilibrium fixes the reservoir temperature and chemical potential. The grand potential is $\Phi =-k_{B}T\ln \Xi =U-TS-\mu N$, with $N$ now the mean population. It generates entropy, pressure and mean particle number by differentiation. This ensemble is convenient for gases in contact with particle reservoirs, independently occupied quantum levels and surface adsorption; its particle-number fluctuations are part of the description, not an error in counting.

Classical single-level limit. For the single-level grand partition function in the dilute limit, put $x_k=e^{-\beta ({ϵ}_k-\mu )}$. Empty and singly occupied levels contribute $1$ and $x_k$. In the dilute classical limit $x_k\ll 1$, higher occupancies contribute only at order $x_k^2$ or beyond:This truncation does not impose the Pauli principle on a classical gas. A fermion state has exactly $1+x_k$, a boson state has $(1-x_k{)}^{-1}$, and Maxwell-Boltzmann occupancy weights give $e^{x_k}$; they all agree to this order. In particular, $\ln {\Xi }_k=x_k+O(x_k^2)$, with the classical Maxwell-Boltzmann result exactly $x_k$.

grand potential and chemical potential. With $z=e^{\beta \mu }$, summing the single-level logarithms and using the stated density of states givesSubstitute $ϵ=k_{B}T{x}^2$, for whichCombining this with the density-of-states prefactor gives $\ln \Xi =zV/{\lambda }^3$. ThusDifferentiating at fixed $T,V$,Similarly $p=-({\partial }_V\Phi {)}_{T,\mu }=k_{B}Tz/{\lambda }^3$, so $pV=N{k}_{B}T$. This is the grand-canonical chemical potential of a classical ideal gas. The condition $N{\lambda }^3/V\ll 1$ states the classical regime. The supplied density of states counts one internal state; extra internal degeneracies would have to be included consistently rather than inserted at only one stage.

One adsorption site. Take the free-vapour energy reference as zero and the binding energy as $-ϵ$. A site is either empty or singly occupied; the latter has vibrational partition function $z_s(T)$. Its adsorption-site grand partition function and mean occupation areThe chemical potential of an adsorbed species on equivalent sites follows by inverting the occupation ratioThe factor $1-\theta$ is the vacant-site probability. The binding reference and the energies included in $z_s$ must be chosen consistently, so that vibrational zero-point energy is not counted twice.

Isotherm. The vibrational factor in a Langmuir equilibrium constant is fixed by adsorption equilibrium, ${\mu }_s={\mu }_{\mathrm{vap}}$. Neglecting vapour internal degrees of freedom and writing its number density as $p/(k_{B}T)$ gives $e^{\beta {\mu }_{\mathrm{vap}}}=p{\lambda }^3/(k_{B}T)$. ThereforeThis is the Langmuir isotherm: initially linear, increasing monotonically and concave down, approaching unit coverage. Half coverage occurs at $p_{1/2}=1/b_s(T)$. The coefficient has inverse-pressure units. It assumes equivalent independent sites, at most one molecule per site and no lateral interaction or multilayer adsorption. The molecular mass appropriate to the vapour determines $\lambda$.

A real uptake measurement must separate adsorbed molecules from free gas in the apparatus. Langmuir's original mica experiment compared a mica-containing bulb with a similar blank and inferred uptake from calibrated gas volumes and pressures. His stated blank contribution and bath-temperature drift limited the inferred adsorption; a saturating curve alone does not establish ideal equivalent sites.^[ref]

An electron ejected following photon absorption. For a free residual ion with negligible recoil, the emitted electron's kinetic energy equals incident photon energy minus the ionization binding energy.

Let a conducting sample and conducting analyzer be in good electrical contact, unbiased, at equilibrium, and without appreciable charging during illumination. Their Fermi levels then coincide. Choose that common Fermi level as zero. Define $E_B>0$ as the binding energy relative to this zero: removing an electron into the electron reservoir at the Fermi level costs $E_B$. It is not the cost of removing the electron all the way to the sample's vacuum level.

Write the sample and analyzer work functions as ${\varphi }_s$ and ${\varphi }_a$. Their local vacuum levels lie at ${\varphi }_s$ and ${\varphi }_a$ above the common Fermi level. Conservation of energy for an electron emitted without additional losses gives total outgoing energy $h\nu -E_B$, where $h$ is the Planck constant and $\nu$ the photon frequency. Immediately outside the sample its kinetic energy is $K_s=h\nu -E_B-{\varphi }_s$. The contact-related electric potential changes this toThus the sample work function cancels when the measured kinetic energy is referenced to the analyzer. Substituting ${\varphi }_s$ for ${\varphi }_a$ in the final conversion would mix two different energy references. An instrument's calibrated binding energy scale can incorporate its effective analyzer work function, so the correction need not appear as a separate operator-entered number.

As an original numerical example, stipulate $h\nu =1486.6\mathrm{eV}$, ${\varphi }_a=4.5\mathrm{eV}$ and a line at $E_B=284.0\mathrm{eV}$. The predicted analyzer kinetic energy is $1198.1\mathrm{eV}$. If ${\varphi }_s=5.0\mathrm{eV}$, the surface value is instead $K_s=1197.6\mathrm{eV}$; the contact-related increase is $0.5\mathrm{eV}$. A second line at $286.0\mathrm{eV}$ gives $1196.1\mathrm{eV}$ at the analyzer. A $+2.0\mathrm{eV}$ chemical shift in photoelectron spectroscopy therefore appears as a $-2.0\mathrm{eV}$ shift in measured kinetic energy. These are chosen values, not a measured carbon spectrum.

Charging breaks the assumed reference. If a uniformly charging sample rises by electric potential $V>0$ relative to the analyzer, an escaping electron loses an additional energy $eV$, with $e$ the positive elementary charge. Applying the unbiased conversion consequently returnsTwo features from the same stable potential retain their separation. Features from regions at different potentials acquire different offsets, which can mimic chemical differences or broaden a line. A conducting holder alone does not establish electrical contact through an insulating surface layer. Compare repeat scans, electrical contact and charge-compensation settings before interpreting an offset as a change in chemical bonds.

Measurement calibration is a separate issue: a known reference line tests the scale, while charge compensation controls the specimen's electric potential. In an original offset model $E_{B,\mathrm{read}}=E_{B,\mathrm{true}}+b$, one known line determines $b$. An unknown scale factor in $E_{B,\mathrm{read}}=a{E}_{B,\mathrm{true}}+b$ requires two distinct known lines to determine both $a$ and $b$. Re-reading a reference after a sample checks drift. Treating an unidentified contamination peak as a universally fixed reference would assume the chemical identity and electrical behavior that the measurement should test.

Choose two hypothetical environments A and B of one chemical element, with line centers ${\mu }_A=284\mathrm{eV}$ and ${\mu }_B=286\mathrm{eV}$. Model each line by a normalized Gaussian function with common standard deviation $\sigma$, and stipulate integrated count weights $w_A=3/4$, $w_B=1/4$. On the binding energy axis $E$, the normalized count density isAll energies in this expression use electronvolts. Each component's integrated area equals its count weight, and $\int I(E)dE=1$. There is no background, statistical counting fluctuation, or additional emission feature in this stipulated model. These choices deliberately isolate photoelectron peak overlap and resolution.

Direct integration, or the law of total expectation applied to the component label, gives the centroidThe centroid lies between the chemical environments and is not either component's binding energy. Expanding $(E-\bar{E}{)}^2$ within each component, the odd cross term integrates to zero. The resulting variance isFor $\sigma =0.25\mathrm{eV}$ the variance is $0.8125{\mathrm{eV}}^2$ and the two maxima are separated. For $\sigma =1.00\mathrm{eV}$ it is $1.7500{\mathrm{eV}}^2$ and there is one maximum. Each Gaussian's full width at half maximum is $2\sqrt{2\ln 2}\sigma$, approximately $0.589$ and $2.355\mathrm{eV}$ respectively. The chemical separation remains $2\mathrm{eV}$ in both cases. The number of visible maxima therefore does not equal the number of chemical components.

One interpretation of this change is additional independent Gaussian instrument blur. The convolution of probability densities adds the variances, so broadening $\sigma =0.25$ to $1$ requires response standard deviation $\sqrt{1-0.25^2}=\sqrt{15}/4\simeq 0.968246\mathrm{eV}$. A normalized response with zero mean preserves the centroid and integrated counts. Real line shapes and backgrounds can depart from this model, and separating hidden components is an inverse problem; a plausible fit does not alone establish uniqueness.

Finally, count weights describe detection. In a simple response model, let $n_i$ denote the number of emitting sites sampled in environment $i$, $s_i$ its effective photoionization cross section, and $T_i$ the probability that its emitted electron contributes to the selected measured line. ThenThe common factor $C$ includes exposure. Here $T_i$ includes escape from the sample, analyzer transmission and detection. Count weights equal site fractions only if the relevant response factors are equal or corrected. They are not automatically molecular mole fractions; different molecules can supply different numbers of the selected emitting sites. In an original example, equal site numbers with $s_A{T}_A=3s_B{T}_B$ give the plotted $3:1$ count ratio despite a $1:1$ site ratio.

For emission angle $\theta$ measured from the surface normal, a free outgoing photoelectron has parallel wavevector $k_{\parallel }=\sqrt{2m_e{E}_{\mathrm{kin}}}\sin \theta /\hslash$. Translation symmetry parallel to a clean periodic surface conserves this component modulo a surface reciprocal lattice vector, after accounting for photon momentum when needed. With initial binding energy $E_B$ below the Fermi level, its kinetic energy immediately outside the sample is $E_{\mathrm{kin}}=h\nu -{\varphi }_s-E_B$, where ${\varphi }_s$ is the sample work function. This sample-vacuum energy is distinct from the analyzer-referenced energy when contact potentials are present. Perpendicular crystal momentum generally needs a final-state model, whereas a strictly two-dimensional band has no perpendicular dispersion to reconstruct.

For a hydrostatic monatomic ideal gas with negligible external pressure, $2U+{\Omega }_f=0$, where gravitational potential energy is negative. The final total energy is therefore ${\Omega }_f/2$, giving the displayed radiated energy when no mass or other energy carrier escapes. The common $-{\Omega }_f/2$ result additionally takes the initial total energy to be negligible. Initial rest does not remove initial self-gravitational binding energy.

In an optically thin plasma, radiative recombination removes the kinetic energy of captured electrons. A radiative recombination photon can also carry the binding energy released when a free electron becomes bound; that part is stored ionization energy, rather than newly removed thermal energy. The kinetic contribution per event is the recombination-weighted electron energy, which need not be $3k_{B}T/2$.

Equality with mean photoelectron heating requires two balances. If photoionization is the only ionization and heating process, and radiative recombination the only recombination and cooling process, photoionization equilibrium equates their event rates. Separately imposing thermal equilibrium then equates the mean heating energy per ionization to the mean kinetic cooling energy per recombination. Photoionization equilibrium alone does not do this: at the same balanced event rate, unequal energies can still heat or cool the gas. Additional atomic line cooling, collisional ionization or trapping of emitted photons changes the energy accounting.

For binding energy $\epsilon >0$, match the inner reduced radial wave $\sin (Kr)$ to an outer decaying exponential. The resulting equation is $K\cot (Ka)=-\chi$, with $K^2=2m(V_0-\epsilon )/{\hslash }^2$ and ${\chi }^2=2m\epsilon /{\hslash }^2$. A bound level reaches zero energy when $\kappa a=(n+1/2)\pi$. These are divergences of the spherical square-well scattering length. The exact zero-energy threshold wave is not square-integrable; a normalizable bound state appears on the more attractive side.

X-ray fluorescence is Characteristic X-ray emission following excitation by incident photons. In a simple initially unexcited chemical atom, photoionization can eject an inner electron if the incident photon supplies its binding energy. An outer electron can then fill the vacancy, releasing an emitted photon with the difference of shell binding energies. Nonradiative relaxation competes with this route, so one absorbed incident photon need not yield one escaping characteristic photon.

Choose a separate stipulated electron shell model with values $B_K=10.0$, $B_L=1.5$ and $B_M=0.3$, all in $\mathrm{keV}$. Ignore subshell splitting, recoil and changes of binding energy during relaxation. The possible shell differences areThese labels denote simplified shell groups, not resolved subshell electronic transitions. With $hc=1.23984\mathrm{keV}\mathrm{nm}$, their calculated wavelengths are respectively $0.145864$, $0.127819$ and $1.03320\mathrm{nm}$. An incident $9.8\mathrm{keV}$ photon exceeds both K-line energies but cannot directly create the K vacancy in this initially unexcited model. A $12.0\mathrm{keV}$ photon can, leaving approximately $2.0\mathrm{keV}$ of photoelectron kinetic energy. This comparison fixes neither the excitation probability nor the radiative branching probability.

For a fixed homogeneous absorber in a narrow-beam geometry, define the X-ray spectral attenuation operator on L1 space over a bounded positive energy interval. A physical input $f(E)\ge 0$ is an integrable function giving incident photon rate density. Let its linear attenuation coefficient $\mu (E)\ge 0$ be a measurable function, with $x\ge 0$. The transmitted uncollided density isHere $w(E)\ge 0$ is a fixed bounded detector response described by a measurable function; background has been subtracted. This is a positive linear operator on L1 space. Since $0\le e^{-\mu (E)x}\le 1$, it cannot increase the integrated absolute incident rate. Multiplication of exponentials proves $A_{x+y}=A_x{A}_y$ for successive layers of the same material. Ignore secondary radiation entering the detector, saturation and changes to the source. Discrete spectral lines use the finite-sum version below instead of an ordinary rate density.