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In a weak applied field, magnetic susceptibility relates magnetization to magnetic field intensity, $M=\chi H$. Its sign, temperature dependence and nonlinear field dependence distinguish several mechanisms.

Diamagnetism is the orbital response opposing an applied magnetic field, with $\chi <0$. Closed-shell atoms and filled bands generally give a weak, nearly temperature-independent response. In a simple free electron metal, Landau diamagnetism contributes ${\chi }_L=-{\chi }_P/3$; the factor is specific to the isotropic parabolic-band model and is not universal in crystals.

Paramagnetism gives $\chi >0$. Localized magnetic moments align statistically with the field against thermal disorder. For independent moments of angular momentum $J$,which is the Curie law. Exchange interactions often lead approximately to the Curie–Weiss law, $\chi =C/(T-\Theta )$, sufficiently far above any ordering temperature. The sign of $\Theta$ suggests predominantly parallel or antiparallel exchange, though it is not a complete classification. Conduction electrons instead exhibit Pauli paramagnetism:for $g\simeq 2$ and a spin-summed density of states per unit volume. Only states close to the Fermi energy change occupation, making this response nearly temperature independent when $k_{B}T\ll E_F$.

Ferromagnetism has spontaneous parallel magnetic order below a Curie temperature. Exchange interaction, not simply classical dipole attraction, provides the dominant alignment mechanism. In the convention $H=-J_{\mathrm{ex}}{\sum }_{⟨ij⟩}{\mathbf{S}}_i\cdot {\mathbf{S}}_j$, positive $J_{\mathrm{ex}}$ favours alignment. Magnetic domains can make a ferromagnet's total magnetization vanish before an external field aligns its domains. Domain motion and anisotropy produce saturation, hysteresis, remanent magnetization and a reverse field measured by magnetic coercivity.

Antiferromagnetism has oppositely ordered sublattices below a Néel temperature; equal opposing moments give zero net magnetization while retaining a nonzero staggered order parameter. A susceptibility anomaly and magnetic neutron diffraction reveal this order even when bulk magnetization vanishes. Ferrimagnetism also has opposing sublattices but unequal contributions, giving a nonzero spontaneous net moment. Above their ordering temperatures both can become paramagnetic. Measurements with a magnetometer as a function of field and temperature, supplemented by magnetic diffraction, therefore distinguishes a compensated ordered material from an ordinary paramagnet.

For the untwisted quantum rotor, periodicity on the angle circle requires $\psi (\varphi +2\pi )=\psi (\varphi )$. The normalized energy eigenstates and their angular momentum areTaking the trace in this orthonormal basis therefore givesThe n=0 state is nondegenerate and each nonzero $|n|$ has two opposite-angular-momentum states. No flux or twisted angular boundary condition is present.

Use periodic plane waves on the ring. For $n\in \mathbb{Z}$,They are orthonormal on one circumference and complete by Fourier series. Thus $\boxed{E_k={\hslash }^2{k}^2/(2m{a}^2)}$, with degeneracy $\boxed{1\text{ for }k=0,2\text{ for }k\ge 1}$: the two signs of nonzero angular momentum have the same energy. Equivalently one may use normalized sine and cosine eigenfunctions for each positive $k$.

For the position measurement put $\theta =x/a$. Direct integration givesThe energy measurement requires projection onto whole degenerate eigenspaces, rather than just onto one traveling wave. The expansion coefficients areFor instance the imaginary integral vanishes by reflection about $\theta =\pi$, and the product-to-sum identity evaluates the real integral as $4/(1-4n^2)$. ConsequentlyEvery numerator is positive and no integer $k$ makes a denominator zero, so every energy has nonzero probability. Parseval's identity guarantees that these probabilities sum to one. The supplied state is a valid normalized periodic $L^2$ state; its derivative need not obey the operator-domain periodic condition for the spectral probabilities to be defined.

Expanding the total angular momentum and replacing transverse components by ladder operators givesThe particle exchange operator obeys $P_{12}^2=I$ and $P_{12}{s}_{1\alpha }{P}_{12}=s_{2\alpha }$, with the reverse relation for index $2$. These statements also apply to the ladder operators. Exchanging the indices preserves $s_1^2+s_2^2$ and $s_{1z}+s_{2z}$; it interchanges the last two terms in $S^2$. Operators on different tensor factors commute. HenceThe full particle exchange includes the spin coordinates as well as the spatial coordinates.

Hermiticity of the Cartesian angular momentum components givesThe angular momentum commutation relations $[L_i,L_j]=i\hslash {ϵ}_{ijk}{L}_k$ implyFor example, $[L_z,L_x+i{L}_y]=i\hslash L_y+\hslash L_x=\hslash L_{+}$. The two terms in $[L_x^2+L_y^2+L_z^2,L_i]$ cancel in pairs, giving the first commutator. Also $[L_{+},L_{-}]=2\hslash L_z$.

The given angular momentum commutation relations imply $[L_3,L_1]=i\hslash L_2$ and $[L_3,L_2]=-i\hslash L_1$. By linearity,Similarly,For the remaining commutators, the commutator product rule gives, for each $j$,The last sum vanishes because the Levi-Civita symbol is antisymmetric under $i\leftrightarrow k$, while $L_i{L}_k+L_k{L}_i$ is symmetric. Taking $j=3$ and then linear combinations with $j=1,2$ proves

If $L^2\psi =\lambda {\hslash }^2\psi$ and $L_3\psi =m\hslash \psi$, these identities yieldOne must also prove that this candidate eigenstate is nonzero. Since the angular momentum operators are self-adjoint operators, $L_{+}^{\ast }=L_{-}$, and expanding the product givesThereforeThis is nonnegative and, by the assumed inequality, nonzero. Hence it is positive. The angular momentum raising operator really does produce a nonzero simultaneous eigenstate with the claimed eigenvalues.

To prove that raising terminates, let $v$ be any nonzero state reached after $r$ steps, with projection $q=m+r$. Positivity givesso $q^2\le \lambda$. Thus the sequence cannot remain nonzero for every integer $r\ge 0$. At its last nonzero state $v=(L_{+}{)}^k\psi$, one has $L_{+}v=0$, and the squared norm identity givesUnder the paper's initial inequality the first step is nonzero, so in fact $k\ge 1$. An angular momentum ladder endpoint must therefore be reached: bounded projection follows from positivity, and the final raising norm fixes the total angular momentum eigenvalue. Writing $j=m+k$, the complementary product $L_{+}{L}_{-}=L^2-L_3^2+\hslash L_3$ gives $\Vert L_{-}v{\Vert }^2=2j{\hslash }^2\Vert v{\Vert }^2$, also showing $j\ge 0$.

In a sufficiently tight binary star, the main sequence companion can fill its Roche lobe. Roche-lobe overflow then transfers gas through the inner Lagrange point towards the neutron star. The gas carries angular momentum, so it generally forms an accretion disk rather than falling radially onto the surface. Internal stresses carry angular momentum outwards and permit accretion inwards. Capture of a stellar wind is another possible supply mechanism if the companion does not fill its Roche lobe.

Near the neutron star, the rapidly increasing magnetic dipole field, with $B\propto r^{-3}$, becomes dynamically important when its magnetic pressure $B^2/(2{\mu }_0)$ is comparable to the inflow stresses. The accretion disk is truncated around a magnetospheric radius. Ionized material attaches to magnetic field lines and streams along them in magnetically channelled accretion, forming an accretion column above one or both magnetic poles. The Lorentz force impedes motion across the magnetic field, but does not forbid motion parallel to it. Gas can therefore reach the polar surface, with its gravitational potential energy released in a hot shock wave and radiation.

This description presumes the neutron star rotates slowly enough for material to reach it: sufficiently rapid rotation can centrifugally expel the inflow instead. A strong magnetic field by itself does not guarantee surface accretion.

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]

In a close binary star, the rotating effective gravitational potential defines two Roche lobes meeting at the inner Lagrange point. Stellar expansion or orbital shrinkage can bring the donor into Roche-lobe overflow. Gas then flows through the inner Lagrange point; its angular momentum can form an accretion disk around the companion. wind mass transfer in a binary star is also possible without Roche-lobe filling. For a circular, synchronized binary, the Eggleton Roche-lobe radius approximation isThis describes the lobe geometry, not the donor's dynamical response.^[ref]

conservative binary mass transfer has fixed total mass and orbital angular momentum, neglecting stellar spin. Using $J=M_d{M}_a\sqrt{Ga/(M_d+M_a)}$ givesSince ${\dot{M}}_d<0$, transfer from the more massive donor initially shrinks the orbit; after binary mass ratio reversal it expands. Stability depends on both stars and their changing lobes. Let ${\zeta }_{\ast }=d\ln R_d/d\ln M_d$ and ${\zeta }_L=d\ln R_L/d\ln M_d$. A small mass loss reduces the relative overfill if ${\zeta }_{\ast }>{\zeta }_L$. Use the adiabatic donor response for dynamical stability, and its thermal response for slower adjustment; an equilibrium mass-radius law alone cannot decide both.

Stable transfer can proceed on nuclear or thermal timescales, while dynamically unstable overflow may create a common envelope, orbital spiral-in and either envelope ejection or a merger. Escaping gas, gravitational radiation and magnetic braking make evolution nonconservative by changing orbital angular momentum. accretion onto a compact stellar object releases power of order $G{M}_a\dot{M}/R_a$, producing bright X-ray binaries; radiation-driven outflows can limit retained mass. Eclipses, radial velocities and disk emission constrain the geometry and masses, but inclination, irradiation and assumptions about mass or angular-momentum loss affect the inference. The essential feedback is the donor's radius relative to its evolving Roche lobe.

Corotation reaches the speed of light at $R_c=c/\Omega$. Extrapolating a dipole field gives $B_c\sim B_{\ast }(R_{\ast }/R_c{)}^3$ and magnetic energy density ${ϵ}_c\sim B_c^2/{\mu }_0$, suppressing order-one factors. An outward electromagnetic energy flux of scale ${ϵ}_{c}c$ through area of scale $R_c^2$ therefore givesThis assumes a dipole-dominated field tied to the rotating star and an effective outgoing Poynting flux near the light cylinder. The dipole extrapolation, angular pattern and surface-field convention affect the coefficient. A strictly aligned vacuum dipole would not radiate, whereas an oblique vacuum dipole does. A plasma-filled magnetosphere opens field lines and can spin down even an aligned rotator, so the light-cylinder estimate is a scaling argument, not an exact vacuum emission calculation or literal rigid motion at $c$. Force-free calculations explicitly demonstrate this distinction.^[ref]

The energy comes primarily from stellar rotation. With constant moment of inertia $I$, $E_{\mathrm{rot}}=I{\Omega }^2/2$, and conservation of energy gives $I\Omega \dot{\Omega }=-P$. ThusFor constant $K$, differentiating gives $\ddot{\Omega }=-3K{\Omega }^2\dot{\Omega }$, hence $n=\Omega \ddot{\Omega }/{\dot{\Omega }}^2=3$. Evolving field strength, inclination, moment of inertia or additional torques invalidate the constant-$K$ prediction; generally $n=3+(\Omega /\dot{\Omega })(\dot{K}/K)$ in this parametrization.

For radial inflow onto mass $M$, take $v_{\mathrm{ff}}\sim (GM/r{)}^{1/2}$ and $\rho \sim \dot{M}/(4\pi r^2{v}_{\mathrm{ff}})$. magnetic pressure decreases as $B_{\ast }^2{R}_{\ast }^6/({\mu }_0{r}^6)$, while ram pressure scales as $\rho v_{\mathrm{ff}}^2\propto \dot{M}(GM{)}^{1/2}{r}^{-5/2}$. Balancing them yieldsNumerical factors and disc geometry are not determined by this spherical estimate. The displayed proportionality holds $M$ fixed and requires a magnetospheric boundary outside the stellar surface.

In a disc, inflowing matter near $R_M$ has specific angular momentum approximately $\sqrt{GM{R}_M}$. If it is transferred to a slower star, the material torque is $N\sim \dot{M}\sqrt{GM{R}_M}$ and $I\dot{\Omega }\sim N>0$. The corotation radius is $R_{\mathrm{co}}=(GM/{\Omega }^2{)}^{1/3}$, distinct from the light-cylinder radius. As stellar spin approaches the inner disc's Keplerian frequency, net spin-up weakens. If $R_M$ lies outside corotation, the field rotates faster than the disc and can centrifugally expel matter in the propeller regime. A characteristic equilibrium isfor fixed mass and magnetic-coupling convention. A complete magnetic torque model can change the coefficient; a star also cannot rotate past its mass-shedding limit, of order $\sqrt{GM/R_{\ast }^3}$.

For opacity per unit mass $\kappa$, an isotropic radiative force balances gravity at $L_E=4\pi GMc/\kappa$. Efficient release of surface gravitational binding energy gives $L_{\mathrm{acc}}\sim GM\dot{M}/R_{\ast }$ and therefore the surface accretion Eddington rateThe radius dependence matters: it uses efficiency $\eta \sim GM/(R_{\ast }{c}^2)$, not an arbitrarily fixed reference efficiency. Substituting this maximal rate into the equilibrium period givesmass, opacity and geometry factors are held fixed. This is the classical radiatively efficient Eddington-limited scaling, not a universal upper bound on every mass supply rate: beaming, advection, outflows and opacity changes can alter it. For sufficiently small field, $R_M\lesssim R_{\ast }$ and the magnetic scaling must be replaced by surface-disc and mass-shedding physics.

Write the dimensionless scale factor as $a=R/R_0$ and the density contrast as $\delta =(\rho -\bar{\rho })/\bar{\rho }$. Gravity amplifies overdensities while expansion and pressure oppose growth. In a sub-Hubble Newtonian single-fluid approximation, a comoving Fourier mode obeysFor pressureless matter in an Einstein-de Sitter background, the growing solution is $\delta \propto t^{2/3}\propto a$ and the decaying solution is $t^{-1}$. Jeans support suppresses growth when the pressure term dominates. The equation is not a substitute for relativistic perturbation theory on horizon-sized scales.

cold dark matter clusters without baryonic pressure support. Sub-horizon growth is much weaker during radiation domination, and baryons are initially tied to the photon fluid, producing acoustic oscillations rather than unrestricted collapse. Horizon entry, radiation-matter equality, acoustic evolution and diffusion therefore give a scale-dependent transfer function. The primordial curvature spectrum and the later matter-density spectrum are not the same function: evolution and the gravitational constraints relate them. Late dark-energy domination reduces the growth rate.

When $|\delta |$ approaches unity, linear theory fails. Collapse, virialization, mergers and tidal interactions form a hierarchy of haloes. Gas must cool and lose angular momentum to make dense stellar systems; feedback then changes its supply and distribution. Simulations combine collisionless gravitational dynamics with increasingly complex gas physics rather than extending the linear growing mode indefinitely. Galaxy clustering measures biased tracers of matter; gravitational lensing and CMB constraints provide complementary tests. Approximately adiabatic, nearly scale-invariant initial perturbations explain the coherent acoustic pattern and supply seeds for this later growth.^[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.

A radio pulsar is a rapidly rotating, magnetized neutron star whose emission beam periodically crosses the line of sight. Collapse of a massive stellar core can leave such a compact remnant. Angular-momentum conservation during contraction helps explain rapid rotation, while amplified magnetic fields and a plasma-filled magnetosphere enable particle acceleration and beamed emission. The pulses are not repeated stellar contractions. Not every newborn neutron star is seen as a radio pulsar, because emission, beaming and observational selection matter.

rotation supplies energy to radiation and the magnetized wind. With approximately constant moment of inertia $I$,An ideal fixed magnetic-dipole torque has $\dot{\Omega }=-K{\Omega }^3$. Its integration givesso the pulsar characteristic age ${\tau }_c=P/(2\dot{P})$ approximates the true age only for a short birth period, constant torque coefficient and braking index three. Real magnetospheres, field evolution and kinematic contributions to observed $\dot{P}$ complicate both age and inferred magnetic field.

As the star spins down, the available accelerating potential falls. If particle acceleration and pair production can no longer sustain the radio-emitting magnetosphere, the object crosses a model-dependent pulsar death line and ceases to be detectable as an ordinary radio pulsar. It still exists as a cooling neutron star; “death” refers to the emission state, not disappearance of the star. The death boundary is not a precisely universal spin period.

In a surviving binary, later Roche-lobe overflow from a companion can create an accretion disc. Transferred angular momentum spins the old neutron star up, and the accretion luminosity can produce an X-ray binary. Long accretion episodes can produce millisecond rotation, commonly accompanied by a smaller inferred external magnetic field. After mass transfer declines, the radio pulsar can switch on again as a recycled pulsar, often with a white dwarf companion. Other binary histories can leave a second neutron star, an ablated companion or an isolated recycled object; there is no single obligatory evolutionary sequence.

A direct observational bridge is the coherent approximately $2.49\mathrm{ms}$ X-ray pulsation discovered with RXTE in SAX J1808.4-3658, an accretion-powered binary. It demonstrated an accreting neutron star rotating at a rate appropriate to recycled millisecond radio pulsars.^[ref] Timing analysis must correct orbital Doppler shifts, telescope and solar-system timing, and instrumental aliases. Radio dispersion, sky coverage and beaming selection affect demographic inferences. The sequence is rotational pulsar spin-down, possible loss of radio activity, and possible accretion-driven recycling—not a necessary path for every neutron star.

The Einstein tensor and its contractions areTo match the supplied Ricci tensor components, use the Riemann curvature tensor conventionThis convention also fixes the sign used for geodesic deviation above. In vacuum, tracing the Einstein field equations gives $-R+4\Lambda =0$. Substitution therefore yields $R_{\mu \nu }=\Lambda g_{\mu \nu }$.

Differentiate the angular field equation to obtainConsequently $R_{00}=\Lambda A=\Lambda g_{00}$ and $R_{11}=-\Lambda /A=\Lambda g_{11}$. The other angular component follows from $R_{33}={\sin }^2\theta R_{22}$. Thus the single angular equation implies all the displayed vacuum equations wherever this coordinate chart is regular.

Integrating $(rA{)}^{\prime }=1-\Lambda r^2$ gives $A=1+C/r-\Lambda r^2/3$. Match $g_{tt}\simeq 1+2\Phi$ with the Newtonian gravitational potential $\Phi =-GM/r$ in the weak-field overlap region to identify $C=-2GM$. Hence the Schwarzschild-de Sitter spacetime hasNegative $\Lambda$ instead gives the Schwarzschild--anti-de Sitter black hole case. This matching fixes the normalization locally; a spacetime with nonzero $\Lambda$ is not asymptotically flat.

For timelike geodesics in the orbital plane, take proper time as parameter and use $L=(A{\dot{t}}^2-{\dot{r}}^2/A-r^2{\dot{\varphi }}^2)/2$. The cyclic coordinates give $A\dot{t}=k$ and $r^2\dot{\varphi }=h$; the angular canonical momentum is $-h$ in this signature. The four-velocity normalization $2L=1$ then givesHere $k$ is conserved Killing energy per unit rest mass in the chosen time normalization, and $h$ is conserved angular momentum per unit rest mass. There is no general interpretation of $k$ as energy measured at an asymptotically flat infinity when $\Lambda \ne 0$.

For radial motion $h=0$, differentiation gives $2\dot{r}\ddot{r}=-A^{\prime }\dot{r}$. The radial Euler-Lagrange equation extends the resulting expression through a turning point where division by $\dot{r}$ would fail. ThusThis is the proper-time second derivative of the areal radius, not a nonzero four-acceleration of the freely falling particle. Positive $\Lambda$ contributes an outward term; negative $\Lambda$ reinforces inward attraction. Restoring units in the slow-motion, weak-field Newtonian limit givesThe extra term describes the gravitational effect of a cosmological constant, not drag or a contact force. The approximation requires $GM/(r{c}^2)\ll 1$, $|\Lambda |r^2\ll 1$ and speed much less than $c$.

Put $f=1-2\mu /r$ and let dots now denote an affine parameter derivative. The null geodesic lagrangian in the orbital plane is $L=(f{c}^2{\dot{t}}^2-{\dot{r}}^2/f-r^2{\dot{\varphi }}^2)/2$. The time and azimuthal Euler-Lagrange equations integrate once; the null normalization givesFor completeness, the differential radial geodesic equation isthe remaining differential equations are $\ddot{t}+(f^{\prime }/f)\dot{t}\dot{r}=0$ and $\ddot{\varphi }+2\dot{r}\dot{\varphi }/r=0$. Spherical symmetry permits the equatorial choice.

Eliminating $\dot{t},\dot{\varphi }$ yields ${\dot{r}}^2+h^{2}f/r^2=c^2{k}^2$. For $h\ne 0$, divide by $h^2$:The impact parameter is unchanged by a common rescaling of the affine parameter. At infinity it is $c$ times angular momentum divided by energy. Radial photons have $h=0$ and obey the undivided equation ${\dot{r}}^2=c^2{k}^2$.

The effective potential is zero at $r=2\mu$, positive outside, and tends to zero as $r\to \infty$. Its derivative is $V_{\mathrm{eff}}^{\prime }=2(3\mu -r)/r^4$. ThusThe circular null geodesic lies at a maximum, since $V_{\mathrm{eff}}^{\prime \prime }(3\mu )=-2/(81{\mu }^4)<0$. It is unstable: a small radial displacement away from the maximum permits motion away from the orbit. For incoming photons from infinity in the exterior of a black hole, an outer turning point exists when $b>b_{\mathrm{crit}}$; for $b<b_{\mathrm{crit}}$ the photon can cross the event horizon. For a material central body, the vacuum circular orbit is relevant only if $3\mu$ lies outside its surface.