Chapter 3 Accretion Disks

Chapter 3

Accretion disks

Accretion of matter onto compact objects is perhaps the only mechanism that gives sufficient energy to feed the most luminous objects in the Universe. Examples can be found in different classes of narrow binary systems and nuclei of active galaxies and quasars. How much energy can be produces by accretion, is shown by the following. The amount of energy ∆E gained by accretion of a mass m at a star of mass M and with radius R is ∆E = GMm/R. For a typical neutron star we have M = 1 M⊙ and R = 10 km and thus ∆E is about 1.3 1016 J per swallowed kg of matter! Most of this energy is released again as electromagnetic× radiation. For comparison: burning wood gives 1.6 107 J/kg, and nuclear fusion (the hydrogen cycle) gives ∆E 0.007mc2 or 6 ×1014 J/kg. It is clear that the more compact a source is (for larger≃ M/R), the× more efficient is the energy conversion process. Therefore in particular accretion on neutron stars and black holes is very important. For white 4 dwarfs (M = 1 M⊙, R = 10 km) nuclear fusion is 25–50 times more efficient than accretion. But even in white dwarfs accretion plays a dominant role: the time scale for nuclear fusion is so short (nova outburst) that during the main part of the life of the white dwarf accretion is the most important energy source. Binaries with accretion onto a white dwarf are called cataclysmic variables. They are found frequently in our Galaxy. In general we can distinguish accretion onto two types of objects: accretion onto Galactic objects (black holes, neutron stars, white dwarfs) of 1 M⊙, and accretion ∼ 6 9 onto extragalactic supermassive black holes (nuclei of galaxies) of 10 –10 M⊙. The luminosity that is released around an accreting source cannot be unlimited: with increasing luminosity the radiation pressure increases, until this is so large that all matter that wants to fall onto the compact object is blown away again. The radiation pressure works primarily on electrons (Thomson scattering): for protons scattering is a factor of me/mp smaller. Because of the electrostatic interaction between electrons and protons however, the protons must follow the electrons. The radiation force on a pair (electron + proton) is therefore given by σTS/c, where S is the incoming flux (W m−2). The total luminosity is L =4πr2S, and therefore the net force F onto an electron-proton pair is given by

(GMmp LσT/4πc) F = − (3.1) r2 If that quantity becomes zero, the maximum luminosity is reached. This deﬁnes

90 the so-called Eddington limit:

4πGMmpc 31 LEdd = = 1.3 10 (M/M⊙) W (3.2) σT × The Eddington luminosity therefore increases with increasing mass. For M = 9 40 13 10 M⊙ one gets the typical luminosity of a bright quasar: 10 W (or 3 10 L⊙). The actual luminosity L of an accreting source can be expressed in the× amount of matter that is being accreted per unit of time. For a neutron star or white dwarf we have

L = GMM/R˙ (3.3) This can be written also as

26 L =1.3 10 M˙ 13M0/R7 W (3.4) × or

29 L =1.3 10 M˙ 13M0/R4 W (3.5) × 13 where M˙ 13 is the amount of accretion in units of 10 kg/s, M0 the mass of the 0 7 star in units of 10 M⊙ and R7 the radius in 10 m. (3.4) is characteristic for accretion onto white dwarfs, (3.5) is more relevant for neutron stars. The left hand side of both equations gives a reasonable impression of the typical luminosity of both types of objects. For black holes, on the other hand, we can write

2 L =2ηGMM/R˙ S = ηMc˙ (3.6) 2 where the Schwarzschild radius (Rs = 2GM/c ). Further, η is the eﬃciency of the energy release. Later we will show that η is usually of the order of 0.1 (compare to η =0.15 for accretion onto a neutron star and η =0.007 for hydrogen fusion). For 8 a black hole of 10 M⊙ radiating at the Eddington limit, one ﬁnds a Schwarzschild radius of 3 1011 m (2 astronomical units – an Earth orbit) and an amount of × 23 accretion of 1.4 10 kg/s or 2.3 M⊙/year. A bright quasar therefore must swallow × the equivalent of a few stars per year in order to maintain its luminosity! It is also possible to say something about the characteristic spectrum of the source. If all radiation is emitted as black body radiation, Stefan-Boltzmann’s law is applicable:

2 4 L =4πR σTb (3.7) −8 −2 −4 where σ is the constant of Stefan–Boltzmann (σ 5.67 10 W m K ). Tb is the temperature of the black body. Therefore we have≃ ×

1 2 4 Tb =(L/4πR σ) (3.8)

We can also deﬁne the temperature Tth that the matter would have if all potential energy would be converted into thermal energy. Per electron–proton pair the gained potential energy is GMmp/R (note that me mp) and the thermal energy is 2 3 kT , yielding ≪ × 2

91 Tth = GMmp/3kR (3.9) If the accretion ﬂow is optically thick, the radiation will reach thermal equilib- rium with the matter and we have Trad = Tb. However, if the accretion ﬂow is optically thin, then all radiation can escape and we have Trad = Tth. In general we therefore have:

Tb < Trad < Tth (3.10) ∼ ∼ We now approximate for the emitted photons E kTrad. For a white dwarf, 8 ≃ neutron star and 10 M⊙ black hole at the Eddington limit we therefore ﬁnd using (3.4), (3.5), and (3.2), respectively: white dwarf: 3 eV < E < 50 keV neutron star: 0.6 keV ∼< E ∼< 50 MeV AGN: 20 eV ∼< E ∼< 50 MeV ∼ ∼ 3.2 Gas dynamics

Matter falling onto a compact object must obey a number of boundary conditions. We discuss these conservation laws in some detail. First, there is mass conservation: ∂ρ + ρv = 0 (3.11) ∂t ∇· Further we have conservation of momentum: ∂v ρ + ρ(v )v = p + f (3.12) ∂t ·∇ −∇ The term with ρ(v )v represents the convection of matter. This terms allows for stationary solutions·∇ with v = 0 for those situations where ∂/∂t = 0. The force term with f can be composed6 of diﬀerent components: a gravity term f = ρg, − magnetic forces j B or viscosity. As the third conservation law we have the energy equation: ×

∂ 1 2 1 2 ( ρv + ρǫ)+ [( ρv + ρǫ + p)v]= f v Frad q (3.13) ∂t 2 ∇· 2 · −∇· −∇· Here ǫ is the internal energy per unit mass, and is given by 3 ǫ = kT/µm (3.14) 2 H where µ is the so-called mean molecular weight. It should be noted that the designation ”mean molecular weight” is misleading, but that word is used widely in the community, unfortunately. In a plasma, we have in general few molecules. More accurately, µ is the mean mass per particle expressed in atomic mass units, so that one can write the mass density ρ (in kg s−1) as

ρ = µmHntot, (3.15)

92 with ntot the total number of particles (ions and electrons). For a neutral plasma with cosmic abundances, we have ntot = 1.10nH and µ = 1.30; for a fully ionised plasma with cosmic abundances, ntot =2.30nH and µ =0.62. The left hand side of (3.13) is very similar to the continuity equation (3.11), with 1 2 as a diﬀerence that the conserved quantity now is not given by ρ but by 2 ρv + ρǫ. The last term within the square brackets is the force exerted by the pressure. The term with Frad is the gradient of the radiation pressure. This term generally depends in a complicated way on the speciﬁc intensity of the radiation Iν. In some special cases, however, it can be determined quite easily: for optically thin matter we have

∞

τ 1 : Frad = ǫνdν (3.16) ≪ ∇· Z 0 −3 −1 where ǫν is the emissivity (W m Hz ). For this one can take for example a cosmic plasma as treated before in these lectures. In the other extreme case the optical depth is very large (like in the interior of a star). In that case Frad approaches the black body ﬂux and we can use the Rosseland approximation:

16σ 3 τ 1 : Frad = T T (3.17) ≫ −3κRρ ∇

Here σ is again the universal radiation constant of Stefan–Boltzmann and κR is the Rosseland opacity. For Bremsstrahlung we have (Kramers law):

21 −7/2 2 −1 κR =6.6 10 ρT m kg (3.18) × For pure Thomson scattering we have

2 −1 κR = κes =0.040 m kg (3.19) Note that in (3.18) ρ is expressed in kg m−3 and T in K. Finally the term with q in (3.13) is the heat conduction coeﬃcient: for classical, unsaturated heat ﬂow one∇· can approximate this well with:

q 10−17T 5/2 T W m−3 (3.20) ≃ ∇ Finally we have of course the ideal gas law:

p = ρkT/µmH (3.21) An important illustrative example is given by stationary ﬂows. In that case all terms with ∂/∂t in the conservation laws can be ignored. We ignore here also the radiation pressure and heat conduction. It then follows that:

ρv = 0 (3.22) ∇·

ρ(v )v = p + f (3.23) ·∇ −∇ 1 [( v2 + ǫ + p/ρ) ρv]= f v (3.24) ∇· 2 · We now substitute (3.22) into (3.24) and ﬁnd

93 1 f v = ρv ( v2 + ǫ + p/ρ) (3.25) · ·∇ 2 On the other hand, it follows from (3.23) that (check this): 1 f v = ρv ( v2)+ v p (3.26) · ·∇ 2 ·∇ Combination of (3.25) with (3.26) immediately gives:

ρv (ǫ + p/ρ)= v p (3.27) ·∇ ·∇ We can write this as 1 v [ ǫ + p ] = 0 (3.28) · ∇ ∇ρ It follows that along a stream line dǫ + pd(1/ρ) = 0. If we substitute into this the deﬁnition of ǫ (3.14) as well as the ideal gas law (3.21), then it follows that 3 2 dT + ρT d(1/ρ) = 0, or also

pρ−5/3 =constant (3.29) This equation describes an adiabatic ﬂow. A somewhat more general form of (3.29) is

pρ−γ =constant (3.30) The constant γ is the so-called adiabatic index. The same description is also useful if we want to describe an isothermal ﬂow. In that case the full energy equation is replaced by T = constant. Formally this can be described by the ideal gas law as pρ = constant, which is of the same form as (3.30) with now γ = 1. Finally we repeat here the deﬁnition of the sound speed cs:

2 cs = γp/ρ (3.31) −1 4 Numerically cs 10√T4 km s where T4 is the temperature in 10 K, for those ≃ −1 cases where the plasma is mostly neutral, or cs 15√T4 km s for fully ionised plasma (in CIE, the transition occurs roughly at≃ a temperature of 14 000 K, where half of the hydrogen atoms are ionised).

3.3 Stationary, spherical symmetric accretion.

In this section we treat the most simple accretion problem. We assume spherical symmetry and consider the stationary accretion of matter onto a compact source at the centre of our coordinate system. We assume that for r the gas velocity → ∞ is zero (an undisturbed medium) while T and ρ remain ﬁnite for r . Mass conservation for the stationary case (3.22) now becomes in spherical coordinates:→ ∞ 1 d (r2ρv) = 0 (3.32) r2 dr Integration of this equation immediately gives

94 M˙ = 4πr2ρv (3.33) − We assume that only gravity contributes to the external force f in (3.23), so that dv 1 dp GM v + + = 0 (3.34) dr ρ dr r2 For the energy equation we can take now (see previous section) the polytropic relation

p = Kργ (3.35) with K a given constant. With this we can describe both adiabatic (γ = 5/3) and isothermal (γ = 1) flow. The differential equations are non-linear. It is therefore useful to look first qualitatively to these equations. In order to do that, we re-write the basic set of equations. First we have

1 dp 1 dp dρ 1 γp dρ c2 dρ = = = s (3.36) ρ dr ρ dρ dr ρ ρ dr ρ dr But from the continuity equation (3.32) it follows by diﬀerentiation that 1 dρ 1 d = (vr2) (3.37) ρ dr −vr2 dr so that we ﬁnd for (3.34):

d 1 c2 d GM ( v2) s (vr2)+ = 0 (3.38) dr 2 − vr2 dr r2 After some rearrangement we get

1 c2 d GM 2c2r (1 s ) (v2)= (1 s ) (3.39) 2 − v2 dr − r2 − GM This equation is suﬃcient to obtain a qualitative picture of the solution. Note that for r , T and thereby also cs remains ﬁnite, and therefore for r the right hand→∞ side is always positive. →∞ Further, we demand that for r also v 0 (accretion only occurs close 2 → ∞ cs → around the black hole). This makes 1 v2 < 0. Using this, it follows that for d 2 − r we have dr (v ) < 0. →∞Both boundary conditions imply that for r we have v2

GM 11 rs = 2 7.5 10 (M/M⊙)/T4 m (3.40) 2cs(rs) ≃ ×

At r = rs the right hand side of (3.39) is zero and therefore also the left hand side should be zero, thus at that point there are two possibilities:

2 2 v (rs)= cs (rs) (3.41)

95 or d (v2(r ))=0. (3.42) dr s Further inwards, the right hand side becomes negative. Because we expect that d 2 also close to the compact object dr (v ) < 0, it follows immediately that in the 2 2 innermost region v >cs , implying that in that region the ﬂow is supersonic. Using this it is possible to sketch a qualitative picture of the solution, see Fig. 3.1.

Figure 3.1: Spherical adiabatic gas flow in the gravity field of a compact source. For v < 0 these are accretion flows, for v > 0 stellar winds or breezes. The two trans-sonic solutions 1 and 2 divide the other solutions into the families 3–6.

There are two solutions for which cs = v at rs, these are called the trans-sonic solutions; rs is the sonic point of the solution. The solutions 3 and 4 are always subsonic or supersonic. Types 5 and 6 are multiple valued and do not cover the full radial domain; therefore we exclude them, although in situations where shocks occur they can describe a part of the correct solution. Only solution 1 is consistent with our boundary conditions (supersonic for small r, subsonic for large r). We will now solve these equations explicitly. From (3.35) it follows that ρ = ρ(p) and therefore after integration of (3.34) it follows that 1 v2 + dp/ρ GM/r =constant (3.43) 2 Z − γ−1 γ−1 2 Substitution of dp = Kγρ dρ gives, with Kγρ = γp/ρ = cs : v2 c2 GM + s =constant (3.44) 2 γ 1 − r − (provided γ = 1). Because lim v2 = 0 it follows that the constant must be equal 6 r→∞ 2 2 2 to c ( )/(γ 1). By application of (3.44) for r = rs, where v = c and rs is given s ∞ − s by (3.40), it follows after simple algebra that

2 cs(rs)= cs( ) (3.45) ∞ r5 3γ − From (3.33) it now follows that

96 ˙ 2 M =4πrs ρ(rs)cs(rs) (3.46) because M˙ is independent from r. Now using c2 ργ−1 we ﬁnd the relation s ∼ between ρ(rs) and ρ( ) as a function of cs(rs) and cs( ). Substitution into (3.46) gives ∞ ∞ 5 3γ πG2M 2ρ( ) 2 − M˙ = ∞ [ ]2γ 2 (3.47) c3( ) 5 3γ − s ∞ − Note that the factor with γ depends only weakly on γ; for γ 5/3 it approaches 1, for γ 1 it approaches e3/2 4.48. For γ = 1.4 (which→ appears to be a → ≃ reasonable value for the ISM) the factor is 2.5. If we further choose T = 104 K and ρ( )=10−21 kg m−3, which corresponds to a density of about 1 cm−3, then it ∞ follows that cs 10 km/s and therefore ≃ ˙ 8 M 2 −3 M 1.4 10 ( ) ρ( )−21cs( )4 kg/s (3.48) ≃ × M⊙ ∞ ∞ Even if we drop this amount of matter on a neutron star, it then follows from (3.5) that the typical luminosity cannot be larger than 2 1024 W, which is much smaller than the observed luminosity of many neutron stars.× At a typical distance of 1 kpc this luminosity would be completely unobservable. We conclude that spherical accretion never gives suﬃcient energy to explain the brightest galactic X-ray sources. Given the scaling M˙ M 2 in (3.47) the prospective that spherical accretion could ∼ power supermassive black holes in active galactic nuclei looks favourable. However, as we will see later, in that situation there are a number of complications that make spherical accretion unlikely.

3.4 Accretion in binary systems

The importance of accretion as a source of energy was ﬁrst discovered in the study of stellar binary systems, in particular X-ray binaries. In binary systems and then in particular in eclipsing systems one can obtain a lot of information on masses, sizes and geometry of the source. The importance of accretion is even more important if we consider that the majority of stars are a member of binary systems. In some evolutionary stages mass transfer between the stars can play a role. Angular momentum plays an important role in the accretion process. In many cases the matter cannot reach the star before losing most of its angular momentum. This leads to the formation of accretion disks, that we will treat in the next section in more detail. These disks constitute very eﬃcient ways to transform gravitational energy into radiation energy. This makes disks very attractive for for example active galactic nuclei. In the last case the prove for the existence of disks is somewhat harder; but for stellar binary systems there is an overwhelming amount of observational evidence for the existence of disks. There are two important reasons why binary stars sometimes can transfer matter:

1. One of the stars can increase its size during its evolution, or the distance between the stars can shrink, until the gravity of the companion object can remove the outer layers of the atmosphere (Roche-lobe overﬂow).

97 2. One of the stars can have a strong stellar wind; a part of that wind can be captured by the gravity of the companion. We focus here in particular on Roche–lobe overﬂow. The basis for that physics was laid by the 19th century astronomer Roche. Assume that there are two massive objects (stars) orbiting each other in a circular orbit. A circular orbit is usually a reasonable approximation for narrow binary systems, because the tidal forces tend to circularise the orbits. We further assume that the stars are small compared to their separation (point mass approximation). The system is characterised by the masses M1 and M2 (we shall express them in solar mass units) and the orbital period P . That period depends through Kepler’s law on the distance a between the stars:

2 3 2 4π a = G(M1 + M2)P (3.49)

If q is the mass ratio q = M2/M1, then numerically we have:

11 1/3 a =1.5 10 [M1(1 + q)Pyear] m × 9 1/3 a =2.9 10 [M1(1 + q)Pday] m × 8 1/3 a =3.5 10 [M1(1 + q)Phour] m × The gas stream is described by conservation laws as discussed in Section 3.2. It is handy to use a coordinate system that rotates together with the rotation of the binary, with angular velocity ω. Through this, however, centrifugal and Coriolis forces enter the equation of motion. These therefore become: ∂v 1 +(v )v = ΦR 2ω v p (3.50) ∂t ·∇ −∇ − × − ρ∇ The angular frequency ω is given by Kepler’s law:

1 2 M1(1 + q) / ω = 3 (3.51) h a i The term with ω v is the Coriolis force per unit mass. The potential ΦR × contains both gravity and centrifugal forces; this so-called Roche potential is given by:

GM1 GM2 1 2 ΦR = (ω r) (3.52) − r r1 − r r2 − 2 × | − | | − | where the positions of the stars are r1 and r2. An example of the resulting potential is given in Fig. 3.2. At large distance the potential is spherical, and looks similar to that of a single point mass at CM. Also very close around each star we find circular equipotentials. The 8–shaped line defines the Roche lobes of both stars. The Roche lobes touch each other in the innermost Lagrangian point L1; this point is a saddle–point and therefore at that location it is very easy for matter of one star that has arrived here to jump to the other star. The surface of both stars also forms an equipotential of ΦR: that follows immediately from (3.50), because in the rotating system v = 0, and therefore the pressure is only a function of ΦR. Mass transfer occurs when one of the stars (one usually calls this the secondary) fills its Roche lobe; the receiving star is usually called the primary. Note that the stellar surface of the secondary is strongly

98 Figure 3.2: Section in the orbital plane of a binary with q = 0.25. Drawn are the equipotential planes of ΦR. The centre of mass is indicated by CM. The equipotentials are labelled according to increasing ΦR. deformed; this can give rise to ellipsoidal variations in the light curve of narrow binaries, also in systems that do not yet fill their Roche lobe completely. When both starts fill their Roche lobe, we have a so-called contact binary (for example W UMa stars). In order to find the characteristic radius R2 of the secondary component, we 4 3 define this as V = 3 πR2 where V is the volume of the Roche lobe. Numerical calculation (Paczyński formulae) yields

0.5 q < 20 : R2 =0.38+0.20 log q (3.53) ≤ a 1/3 R2 q 0

By substituting q 1/q one obtains similarly the ratio R1/a. The distance b1 of the ﬁrst Lagrange point→ to the centre of the primary is given by the approximation of Plavec & Kratochvil: b 1 =0.500 0.227 log q (3.55) a − exercise 3.1. Demonstrate that for q < 0.5 it follows from (3.54) that the average density of the secondary only depends upon the period P of the binary:

5 1.1 10 3 q < 0.5:ρ ¯2 = ×2 kg m (3.56) Phr

In binaries with periods of hours therefore in particular the lower main sequence stars (¯ρ 103 – 105 kg m−3) can ﬁll their Roche lobe. For such stars (amongst which the∼ Sun) the following mass–radius relation holds approximately:

99 M/M⊙ = R/R⊙ (3.57) Substitution into the deﬁnition of the mean density then gives

1400 3 ρ¯ = 2 kg m (3.58) M2

where M2 is expressed in units of M⊙. (3.56) and (3.58) together lead to a period–mass relation

M2 0.11 Phr (3.59) ≃ and a period–radius relation

7 R2 8 10 Phr m. (3.60) ≃ × The above relations appear to work well for the secondary component in cataclysmic variables (accretion onto a white dwarf). The mass transfer in general leads to a changing value of q and also to a change in P and a (because of the transfer of angular momentum). It is therefore not trivial that the mass transfer will continue. We consider here in some more detail this problem. Assume that the total mass is conserved (all the mass lost of star 2 is given to star 1), that the total angular momentum is conserved, and that the last quantity is the sum of the angular momenta of both stars (point masses). From these conditions it follows that

M1(1 + q)=constant (3.61) and

2 2 M1a1ω + M2a2ω =constant (3.62)

where a1 = qa/(1 + q) and a2 = a/(1 + q) are the distances of both stars to the centre of gravity. From (3.62) it follows that

2 M1a q/P (1 + q)=constant (3.63) 2 4/3 Using Kepler’s law and (3.61) it follows that a P and M1 1/(1+ q) and therefore ∼ ∼

(1 + q)6 1 P 3 3 (3.64) ∼ q ∼ (M1M2) and therefore using Kepler’s law

(1 + q)4 1 a 2 2 (3.65) ∼ q ∼ (M1M2) We see immediately that P and a reach a minimum for q = 1 (equal masses). exercise 3.2. Demonstrate using (3.54) that when the most massive star ﬁlls its Roche lobe, both a and R2 decrease, so that the process will continue until the star shrinks due to evolutionary eﬀects or until q = 1 is reached.

100 When star 2 is the less massive component, then R2 will increase due to the mass transfer, and only expansion due to evolution or the loss of angular momentum can cause continued mass transfer. This last case happens with cataclysmic variables, where the lighter secondary fills its Roche lobe despite the fact that the evolutionary time scale is usually larger than the Hubble time. Roche lobe overflow will lead to the formation of an accretion disk in most binary systems. That works as follows. The velocity vr with which the matter flows over L1 towards the primary will be of the order of the sound speed at the secondary, and therefore never much larger that a few times 10 km s−1 (normal stellar atmospheres are cooler than 105 K). Inside the Roche lobe of the primary the equipotentials are approximately circular (see Fig. 3.2). In this co-moving frame there is always a typical tangential velocity component vφ = b1ω where b1 is the distance of L1 to the primary. Using (3.49) and the numerical expressions below that equation it then follows that

1/3 −1 v 100[M1(1 + q)/Pday] km s (3.66) φ ≃ where we have chosen b1 0.5a (cf. (3.55) for q not too extreme). We therefore ≃ ﬁnd a mainly tangential motion (vr vφ) in an approximately circular orbit. For a given orbital angular momentum matter≪ in a circular orbit has the lowest energy (compare also our chapter on orbits around black holes). Therefore dissipation in the rotating gas will lead to circularisation. If during this circularisation process not too much angular momentum is lost, we ﬁnd for the circularisation radius Rcirc

2 GM1 vφ(Rcirc)= (3.67) Rcirc with

2 Rcircvφ(Rcirc)= b1ω (3.68) Substitution of (3.55) gives after some algebra: R b circ =(1+ q)( 1 )4 =(1+ q)(0.500 0.227 log q)4 (3.69) a a −

We can easy show that (except when q < 0.05) usually Rcirc is 2–3 time smaller than the Roche lobe of the primary; for example,∼ for q = 1 we have Rcirc = 0.125a and R1 = 0.38a. The captured material will therefore follow a circular orbit that is clearly within the Roche lobe of the primary (except of course when the stellar radius R∗ of the primary is larger than Rcirc). Substitution of Kepler’s law gives an equivalent for (3.69):

4/3 4 2/3 Rcirc 4(1 + q) [0.500 0.227 log q] P R⊙ (3.70) ≃ − day We therefore see that for narrow binaries the circularisation radius is of the order of a solar radius. When in a narrow binary the primary is for example a main sequence star, the matter of the secondary will fall directly on the surface of the primary. This happens probably in some Algol–systems with periods of a few days. For compact objects, such as white dwarfs and neutron stars we ﬁnd on the other hand that R∗ < Rcirc. By continuing accretion an annulus of matter at Rcirc will be formed. For suﬃciently

101 high densities a part of the kinetic energy will be dissipated and transformed into thermal energy; by radiation losses this energy decreases again; the energy loss can only be compensated by a smaller orbital radius, but that implies loss of angular momentum. In general the loss time for angular momentum is much longer than the cooling time or the orbital period. The gas will therefore lose as much as energy as is possible for the given angular momentum. Therefore it will spiral slowly inwards through many almost circular orbits. Such a conﬁguration is known as an accretion disk. The matter that has fallen inwards has lost angular momentum: because of the conservation of angular momentum, there should also be a net transport of angular momentum outwards. The original annulus at Rcirc will spread both towards smaller and larger radii. When the matter reaches the stellar surface R∗, it has a binding energy of GM/2R∗. Because the gas starts at large distance from the star with negligible binding energy, the total luminosity of the disk in a stationary situation must be given by

GMM˙ 1 Ldisk = = Lacc (3.71) 2R∗ 2

where Lacc is the total accretion luminosity (3.3). The other half of the luminosity must be released therefore close to the surface of the star (at the stellar surface the matter must be stopped: the kinetic energy of the Keplerian motion will be lost). On the other hand, the speciﬁc angular momentum R2Ω R1/2 and with ∼ R∗ Rcirc it follows that the material must loose almost all of its original angular≪ momentum. At the outer edge of the disk some process must remove this angular momentum. Tidal forces of the secondary on the accretion disk could for example feed this angular momentum back into the orbital motion of the binary.

3.4.1 Wind accretion We conclude this section with some considerations about wind accretion. In this case we consider an early type O or B star with a neutron star or black hole in a narrow orbit around it. Examples are some bright sources like Cen X-3, SMC X-1, Vela X-1 and the black hole candidate Cyg X-1. The stellar wind of the O or B star −6 −5 is very intense: M˙ is typically 10 – 10 M⊙/year, while it is also supersonic:

vw(r) vesc(RE)= 2GME/RE (3.72) ≃ p where ME and RE are the mass and radius of the OB star. This wind velocity is typically a few thousand km/s. When the orbital velocity of the neutron star is vn, then the relative velocity of the wind with respect to the neutron star is typically 2 2 of the order of vrel v + v . ≃ n w p −1 exercise 3.3. Show that for binaries with periods of days vn is typically 200 km s , so that vrel vw is a good approximation. ≃ Wind particles for which the kinetic energy is smaller than the potential energy with respect to the compact object will be captured. This deﬁnes the accretion radius:

2GMn racc 2 (3.73) ≃ vrel

102 We now can derive which fraction of the stellar wind is captured by the neutron star. Let M˙ w be the total mas loss of the OB star, and M˙ the amount of accretion onto the neutron star. Then we have

M˙ πr2 GM 2 1 M 2 R 2 acc = n n E (3.74) ˙ 2 2 Mw ≃ 4πa avw ≃ 4ME a −4 −3 For typical parameters one finds M/˙ M˙ w 10 –10 . We see that accretion ∼ through a stellar wind is not very efficient, compared to Roche lobe overflow where almost all the lost stellar material falls onto the primary. However, because for OB stars the stellar wind is so strong, there still is something to see: we find typically 36 38 Lacc 10 –10 W. Whether a disk is formed is hard to say in general. Compared ≃ 4 to Roche lobe overflow Rcirc is typically a factor of (racc/a) smaller, and is close to 10 km, but this depends strongly upon the precise parameters and the specific solution for the stellar wind. We will not elaborate that here, but only give the expression fro Rcirc in this case:

3 4 Rcirc Mn (Mn + ME) RE = 4 4 (3.75) a 16λ (a)ME a 2 2 where λ(a) is given by vw = λ(r)vesc(RE), and is of order unity.

103