Accretion & disks The disks are everywhere Jiˇr´ı Krtiˇcka Masaryk University Bondi-Hoyle-Lyttleton accretion theory Bondi radius Let us assume that the medium (ISM, wind, . . . ) moves with velocity Ú with respect to the accreting object. The object is able to accrete only particles with negative total energy, Ekin + Epot < 0. Assuming spherically symmetric gravitaional field, 1 2 ρv2 < ρ GM r . RBondi v This gives condition for a limiting Bondi radius, from which a spherically symmetric object can accrete matter, as RBondi = 2GM v2 . 1 Mass-accretion rate from stellar wind in a binary The total amount of accreted matter per unit of time (mass-accretion rate) is given by matter swept by a circle with radius RBondi, ˙Macc = ˙M R2 Bondi 4D2 . Here D is the orbital separation. For a compact accreting object (white dwarf, neutron star, or a black hole) with radius R, the energy released by accretion (per unit of time) is LX = ˙Macc GM R . 10 33 10 34 10 35 10 36 10 37 10 38 10 33 10 34 10 35 10 36 10 37 10 38 estimatedLX[ergs -1 ] observed LX [erg s -1 ] 10 33 10 34 10 35 10 36 10 37 10 38 10 33 10 34 10 35 10 36 10 37 10 38 estimatedLX[ergs -1 ] observed LX [erg s -1 ] 10 33 10 34 10 35 10 36 10 37 10 38 10 33 10 34 10 35 10 36 10 37 10 38 estimatedLX[ergs -1 ] observed LX [erg s -1 ] 10 33 10 34 10 35 10 36 10 37 10 38 10 33 10 34 10 35 10 36 10 37 10 38 estimatedLX[ergs -1 ] observed LX [erg s -1 ] (Krtiˇcka et al. 2019) This process can explain X-ray luminosity of high-mass X-ray binaries. 2 An alternative view: spherically symmetric accretion Alternatively, accretion flow in a spherically symmetric case can be described by the same equations as that used for the coronal flow, 1 r2 d dr r2 ρv = 0, ρv dv dr = −a2 dρ dr − ρGM r2 . The integration of the continuity equation gives the mass-accretion rate ˙M ≡ −4πr2 ρv = const. Inserting the continuity equation, the momentum equation takes the form 1 v v2 a2 − 1 dv dr = 2 r − GM r2a2 . 3 Solution of the Parker equation 0 0.5 1 1.5 2 2.5 3 0 0.5 1 1.5 2 2.5 3 v/a r/rc There are two types of solutions describing outflow (wind) and inflow (accretion). There is one outflow solution that is subsonic at large distances from the star and supersonic close to the star corresponding to the accretion from stationary interstellar medium. Other inflow solution correspond to accretion of matter with non-zero velocity. 4 Analytical solution The momentum equation 1 v v2 a2 − 1 dv dr = 2 r − GM r2a2 can be easily integrated. To do so, it is convenient to introduce a critical point radius rc = GM 2a2 = 1 2 RBondi(v = a), using which the momentum equation reads (after integration) v a 2 − ln v a 2 = 4 ln r rc + 4 rc r + K, where K is an integration constant. Selecting the solution that smoothly passes throught the criticar point v(rc) = a, K = −3 and v a 2 − ln v a 2 = 4 ln r rc + 4 rc r − 3. 5 Mass-accretion rate For a subsonic flow v ≪ a at large distances from the star, r ≫ rc, the solution of the momentum equation can be simplified as − ln v a 2 = 4 ln r rc − 3, which gives for the velocity v = −a rc r 2 e 3 2 . Here we selected a negative root, which gives positive Bondi accretion rate ˙M ≡ −4πr2 ρ∞v = 4πe 3 2 r2 c ρ∞a = πe 3 2 (GM) 2 a3 ρ∞, where ρ∞ is the density of the medium at large distances from the star. The accretion rate is clearly proportional to T−3/2 . 6 Accretion disks in a nutshell The need for angular momentum conservation The very existence of accretion disks stems from the angular momentum conservation. Assuming that a typical blob of interstellar medium with mass M = 1 M⊙ ≈ 1033 g, radius rISM = 0.1 pc ≈ 1017 cm, and velocity vISM = 1 kms−1 = 105 cm s−1 forms a star with about the same mass M = 1 M⊙ ≈ 1033 g and radius R∗ = 1 R⊙ ≈ 1011 cm, the need for the angular momentum conservation implies rotational velocities of the order of rISMvISM/R∗ = 1011 cm s−1 , which is significantly higher than the escape speed and in fact even higher than speed of the light. Therefore, during accretion, the matter has to get rid of angular momentum. As a result of universality of angular momentum conservation law, the accretion disks appear in different astrophysical environment including pre-main sequence stars, binaries with mass transfer, and centers of active galactic nuclei. 7 Mass and angular momentum transfer mass angular momentum 8 Basic hydrodynamics of the disks Assuming spherically symmetric outflow, the radial component of the momentum equation ∂vR ∂t + vR ∂vR ∂R + vφ R ∂vR ∂φ + vz ∂vR ∂z − v2 φ R = − 1 ρ ∂p ∂R + gR can be in a stationary state (∂vR /∂t = 0) neglecting the density gradient for a slow inflow (vR ≪ vφ) rewritten as gR + v2 φ R = 0, which in equatorial plane gR = −GM/R2 gives the Keplerian velocity vφ = GM R . Matter moves on Keplerian orbits in accretion disks. The angular velocity Ω = vφ/R ∼ R−3/2 decreases with radius, whereas magnitude of the angular momentum per unit of mass j = Rvφ ∼ R1/2 increases with radius. Consequently, matter losses angular momentum during accretion. 9 The need for anomalous viscosity The momentum transfer in the direction perpendicular to the flow (orbital) motion requires viscosity. As a result of viscosity, the particles moving on neigbouring orbits may interchange the angular momentum. The coefficient of viscosity ν is given as a product of the particle mean free path ℓ and the mean velocity v , ν ≈ ℓ v . The particle mean free path is given as ℓ = 1/(nσ), where n is particle number density and σ is the particle cross section. The viscous time, a characteristic time of the angular momentum transfer, is given by τ ≈ R2 /ν. For a characteristic radius R = 1 au ≈ 1013 cm with a typical particle number density n ≈ 1014 cm−3 and v ≈ 105 cm s−1 , for a molecular viscosity with σ ≈ 10−16 cm2 the viscous time τ ≈ 1019 s ≈ 1011 yr is too large to generate a meaningfull accretion. Therefore, there is a need for additional anomalous viscosity. 10 Accretion disks in a more detail Viscous disk equations: equation of continuity Integrating the equation of continuity in cylindrical coordinates ∂ρ ∂t + 1 R ∂ ∂R (RρvR ) + 1 R ∂ ∂φ (ρvφ) + ∂ ∂z (ρvz ) = 0 over the vertical variable (z) assuming axisymmetric flow gives ∂Σ ∂t + 1 R ∂ ∂R (RΣvR ) = 0, where Σ = ∞ −∞ ρ dz and where we assumed that vR does not depend on z and that the disk disk density is zero in the limit z → ±∞ (therefore ∞ −∞ ∂/∂z (ρvz ) = 0). 11 Viscous disk equations: z component of the equation of motion The z component of the equation of motion in cylindrical coordinates ∂vz ∂t + vR ∂vz ∂R + vφ R ∂vz ∂φ + vz ∂vz ∂z = − 1 ρ ∂p ∂z + gz gives with no vertical motion vz = 0 equation of hydrostatic equilibrium − ∂p ∂z + ρgz = 0. The z component of the gravity is using the first order Taylor expansion gz = − GM r2 z r = − GMz (R2 + z2)3/2 ≈ − GMz R3 . The equation of hydrostatic equilibrium can be integrated with p = a2 ρ and assuming (vertically) isothermal disk, ρ(z) = ρ0e− GMz2 2a2R3 = ρ0e− 1 2 z2 H2 , which gives the Gaussian density distribution with scale-height H = aR3/2 /(GM) = aR/vK. The mid-plane disk density is related to the vertically integrated density via Σ = √ 2πρ0H. 12 Viscous disk equations: R component of equation of motion Integrating the radial component of the equation of motion in cylindrical coordinates ∂vR ∂t + vR ∂vR ∂R + vφ R ∂vR ∂φ + vz ∂vR ∂z − v2 φ R = − 1 ρ ∂p ∂R + gR multiplied by ρ over the vertical variable (z) assuming axisymmetric flow gives ∂vR ∂t + vR ∂vR ∂R − v2 φ R = − 1 Σ ∂ ∂R ( ∞ −∞ p dz) + 1 Σ ∞ −∞ ρgR dz. Introducing vertically independent isothermal speed of sound a in the first integral and assuming Gaussian vertical density distribution in the second integral with the Taylor expansion of gR in terms of z gR = − GM r2 R r = − GMR (R2 + z2)3/2 ≈ − GM R2 1 − 3 2 z2 R2 gives after the integration over z the momentum equation ∂vR ∂t + vR ∂vR ∂R − v2 φ R = − 1 Σ ∂ ∂R (a2 Σ) − GM R2 + 3 2 a2 R . 13 Viscous disk equations: φ component of equation of motion The Keplerian disk motion (vφ ∼ R−1/2 ) results in a radial shear. As a result of radial shear, fluid elements exchange angular momentum leading to a net flux of mass inwards and agular momentum outwards. Such angular momentum flux perpendicular to the mass flow is desribed by viscosity. However, the φ component of equation of motion in cylindrical coordinates, ∂vφ ∂t + vR ∂vφ ∂R + vφ R ∂vφ ∂φ + vz ∂vφ ∂z + vR vφ R = − 1 ρR ∂p ∂φ + gφ, which assuming axisymmetric flow and gφ = 0 reads as ∂vφ ∂t + vR R ∂(Rvφ) ∂R = 0, was derived assuming inviscid flow and therefore does not contain any viscosity. 14 Introducing anomalous viscosity The corresponding component of stress tensor τRφ = µR ∂Ω ∂R is parameterized assumimg proportionality to pressure via α parameter of viscosity (Shakura & Sunyaev 1973) as τRφ = αp. The viscosity is assumed to stem mostly from the magnetorotational instability for which α ≈ 0.01. The vertically integrated momentum equation is then ∂vφ ∂t + vR R ∂(Rvφ) ∂R + α ΣR2 ∂ ∂R R2 ∞ −∞ p dz = 0, or, with µ = αaρH (assuming eddies with length-scale H and speed a) ∂vφ ∂t + vR R ∂(Rvφ) ∂R = α ΣR2 ∂ ∂R a2 Σ R4 vφ ∂ vφ R ∂R . Apparently, angular momentum transfer is zero for rigidly rotating disks. 15 Stationary disks In stationary (∂/∂t = 0) disk the continuity equation reads 1 R d dR (RΣvR ) = 0, which has a solution 2πRΣvR = ˙M = const. giving the disk accretion rate. The orbital velocity dominates the radial component of equation of motion vφ ≫ a ≫ vR , therefore the equation again predicts the Keplerian velocity distribution close to the star vφ = GM R . 16 Stationary disks: example of solution Equations for stationary disks 1 R d(RΣvR ) dR = 0, vR dvR dR + 1 Σ d(a2 Σ) dR + GM R2 = v2 φ + 3 2 a2 R , vR d(Rvφ) dR = α ΣR d dR a2 Σ R4 vφ d vφ R dR predict nearly Keplerian rotation close to the star, decrease of the accretion velocity and of the angular momentum ˙J = ˙MvφR towards the 10 -5 10 -4 10 -3 10 -2 10 -1 1 10 1 10 100 1000 r / Req vr / a vφ / vK J . / J . K(Req) Rcrit α~ = 1 α~ = 0.1 α~ = 0.01 star. The disks are subsonic (vR < a) up to the critical radius Rcrit. Multiplication of the azimuthal component of the momentum equation by 2πR Σ gives for Keplerian rotation the accretion rate ˙M = 3παa2 ΣR/vφ. 17 Disk evolution Assuming the Keplerian rotation, the φ-component of the momentum equation yields equation for disk radial velocity vR = − 3α ΣRvφ ∂ ∂R (R2 a2 Σ). Inserting this expression into the continuity equation gives ∂Σ ∂t = 3α R ∂ ∂R 1 vφ ∂ ∂R (R2 a2 Σ) , which is a diffusion equation for Σ. From this equation the viscous time-scale is ∂Σ ∂t ≈ Σ τvis ≈ αa2 Σ Rvφ , or τvis = Rvφ αa2 = 1 αΩ R H 2 . The viscous time-scale is of the order of years for stellar disks (α ≈ 1). However, this equation is typically used to derive α from observations. 18 Thermodynamics of the disk Disk structural equations The vertical structure of the optically thick disk can be described by a similar structural equations as the equations used for the modelling of stellar structure or stellar atmospheres, i.e., hydrostatic equilibrium equation ∂p ∂z = −ρ GMz R3 , heat transport equation F = −4acT3 3κρ ∂T ∂z , ∇ ≡ d ln T d ln p < ∇add, Fconv, ∇ > ∇add, and the equation describing the frictional generation of heat ∂F ∂z = µ R dΩ dR 2 . These equations permit to get the thermal structure of the disk and to predict disk spectra solving the radiation transfer equation. 19 Some general relations Integration of the equation for ∂F/∂z from the disk midplane upwards and assuming Keplerian disk rotation gives for the flux from the surface and effective disk temperature F0 = σT4 eff = 9 8 GM R2 αa2 Σ vK . The accretion leads to heating, which can lead to the emission of high-energy photons from the disk (UV, X-ray). 20 Application to cataclysmic variables Cataclysmic variables are binary systems in which a white dwarf accretes matter through an accretionm disk form a late-type main sequence companion, which fills its Roche lobe. These systems show semi-regular outbursts in which the brightness rises by several magnitudes. 7 8 9 10 11 12 13 14 0 200 400 600 800 1000 1200 1400 mv[mag] JD-2448700 Light curve of SS Cyg derived from AAVSO database 21 Outburst in cataclysmic variables Disk structural equations allow to obtain relation between the disk column density Σ and the accretion rate. In the region of partial ionization of hydrogen and helium this relation is double valued and allows two solutions connected by an unstable branch. This leads to cyclical behaviour (Meyer & Meyer-Hofmeister 1982). 22 Self-gravitating accretion disks The disk gravity In many cases, the mass of the disk is comparable to the mass of the central object (e.g., AGN disks and disks around young stellar objects). In such a case, the gravitational potential Φ in the equation of motion is given as a sum of two terms corresponding to the contribution of the central object and the disk, Φ = Φc + Φd. For geometrically thin disk, the potential can be obtained from the Poission’s equation in the form of ∇2 Φd = 4πGΣδ(z), where δ(z) is the Dirac δ-function. 23 Disk orbital velocity Again, assuming spherically symmetric inflow, the radial component of the momentum equation ∂vR ∂t + vR ∂vR ∂R + vφ R ∂vR ∂φ + vz ∂vR ∂z − v2 φ R = − 1 ρ ∂p ∂R − ∂Φ ∂R can be in a stationary state (∂vR /∂t = 0) neglecting the density gradient for a slow inflow (vR ≪ vφ) rewritten as v2 φ R = ∂Φ ∂R . For self-graviting disks with Σ = Σ0R0/R (Mestel 1963) the Poission’s equations gives (Lodato 2007) ∂Φd ∂R = 2πGΣ. In this case the orbital velocity is non-Keplerian, vφ = 1 + 2πΣ0R0 M RvK. This implies vφ = const. for strongly self-graviting disks. 24 Disk vertical structure For radially homogeneous slab we have the Poission’s equation ∂2 Φd ∂z2 = 4πGρ, which after integration from −z to z gives ∂Φd(z) ∂z = 2πGΣ(z), where Σ(z) = z −z ρ(z′ ) dz′ . Therefore, the hydrostatic equilibrium equation in the vertical direction is − a2 ρ ∂ρ ∂z = 2πGΣ(z). This equation has a solution ρ(z) = ρ0 cosh2 (z/Hsg) , Hsg = a2 πGΣ . 25 When the self-gravity becomes important? The vertical component of the gravitational field produced by the disk is of the order of 2πGΣ ≈ GMdisk R2 , while taking into account the projection effect the vertical component of the gravitational field of the central object is of the order of GM R2 H R . Therefore, the self-gravity of the disk becomes important already when Mdisk M ≈ H R , which is typically much smaller than 1. Therefore, the disk self-gravity may become important even if the disk mass is smaller than the mass of the central object. 26 Suggested reading F. Meyer: Reviews in Modern Astronomy, 3, 1 A. R. King, J. E. Pringle, & M. Livio, 2007, MNRAS, 376, 1740 D. Mihalas & B. W. Mihalas: Foundations of Radiation Hydrodynamics G. Lodato: Rivista del Nuovo Cimento, 30, 293 A. T. Okazaki: Bright Emissaries: Be Stars as Messengers of Star-Disk Physics, 2016, 3 F. H. Shu: The physics of astrophysics: II. Hydrodynamics 27