Orbital dynamics. Test particle in gravitational potential Cylindrical polar coordinates (r, φ, z) Newtonian dynamics

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1 Orbital dynamics Test particle in gravitational potential Cylindrical polar coordinates (r, φ, z) Newtonian dynamics Φ Assume: Φ = Φ(r, z) Φ(r, z) = Φ(r, z) axisymmetric symmetric Special case: Φ= GM(r 2 + z 2 ) 1/2 point-mass potential Keplerian orbits

2 Orbital dynamics Equation of motion r = Φ { r r φ 2 = Φ,r r φ + 2ṙ φ =0 z = Φ,z Specific energy Specific angular momentum Reduces to 2D problem Effective potential ε = 1 2 ṙ 2 + Φ = const h = r 2 φ = const r = Φ eff,r z = Φ eff,z Φ eff = Φ + h2 2r 2

3 Orbital dynamics Circular orbit in midplane (z = 0) 0=Φ eff,z (r, 0) by symmetry 0=Φ eff,r (r, 0) = Φ,r (r, 0) h2 r 3 ε = h2 + Φ(r, 0) 2r2 }defining h (r) ε (r) Important relation dε dr = h r 2 dh dr h2 r 3 +Φ,r(r, 0) dε dh = h r 2 = φ =Ω orbital angular velocity

4 Orbital dynamics Keplerian case Φ(r, 0) = GM r h =(GMr) 1/2 ε = GM 2r ( GM Ω = r 3 ) 1/2

5 Orbital dynamics Reminder of general Keplerian orbits r = GMr r 3 dh dt = d (r ṙ) =ṙ ṙ + r r = 0 dt Orbit is confined to plane h, so introduce polar coordinates (r, φ) : r r φ 2 = GM r 2 h = r 2 φ = const d Let r =1/u and note that : dt = φ d dφ = d hu2 dφ hu 2 d [ hu 2 d ( )] 1 h 2 u 3 = GMu 2 dφ dφ u d 2 u dφ 2 + u = GM h 2

6 Orbital dynamics d 2 u dφ 2 + u = GM h 2 General solution (with two arbitrary constants) u = GM h 2 [1 + e cos(φ ϖ)] Polar equation of conic section: r = λ 1+e cos(φ ϖ) e =0 0 <e<1 e =1 e>1 circle ellipse parabola hyperbola bound bound marginally unbound (ε < 0) unbound (ε > 0)

7 Orbital dynamics Perturbations (δr, δz) of circular orbits in midplane ( h fixed) δr = Ω 2 r δr δz = Ω 2 z δz Ω 2 r =Φ eff,rr(r, 0) Ω 2 z =Φ eff,zz(r, 0) [ Φ eff,rz(r, 0) = 0 by symmetry] Ω r usually called κ Ω z sometimes called µ (horizontal) epicyclic frequency vertical (epicyclic) frequency Orbit is stable if Ω 2 r > 0 (i.e. κ 2 > 0 ) and Ω 2 z > 0 i.e. if orbit is of minimum energy for given h

8 Orbital dynamics Now κ 2 =Φ,rr (r, 0) + 3h2 = d dr ( h 2 r 3 ) r 4 + 3h2 r 4 = 1 r 3 dh 2 dr = 4Ω 2 +2rΩ dω dr Ω 2 z =Φ,zz (r, 0)

9 Orbital dynamics Keplerian case κ =Ω z =Ω descending node apocentre pericentre ascending node κ =Ω Ω z =Ω eccentric Keplerian orbit inclined Keplerian orbit

10 Orbital dynamics Precession If κ Ω and/or Ω z Ω, describe as slowly precessing orbit Minimum r (pericentre) occurs at time intervals φ = 2πΩ κ ( ) Ω =2π κ 1 ( ) Ω =2π κ 1 Apsidal precession rate +2π mod 2π φ t =Ω κ t = 2π κ Similarly, nodal precession rate =Ω Ω z

11 Mechanics of accretion Consider two particles in circular orbits in the midplane Can energy be released by an exchange of angular momentum? Total energy and angular momentum: E = E 1 + E 2 = m 1 ε 1 + m 2 ε 2 H = H 1 + H 2 = m 1 h 1 + m 2 h 2 In an infinitesimal exchange: dh de =de 1 +de 2 = m 1 Ω 1 dh 1 + m 2 Ω 2 dh 2 If dh =dh 1 +dh 2 = m 1 dh 1 + m 2 dh 2 dh =0 then de = (Ω 1 Ω 2 )dh 1 In practice dω/dr <0 Energy released by transferring angular momentum outwards

12 Mechanics of accretion Generalize argument to allow for exchange of mass: dm =dm 1 +dm 2 =0 dh =dh 1 +dh 2 =0 dh i = m i dh i + h i dm i de i = m i Ω i dh i + ε i dm i =Ω i dh i +(ε i h i Ω i )dm i de = (Ω 1 Ω 2 )dh 1 + [(ε 1 h 1 Ω 1 ) (ε 2 h 2 Ω 2 )]dm 1 In practice d(ε hω)/dr = h dω/dr >0 Energy released by transferring angular momentum outwards and mass inwards This is the physical basis of an accretion disc dh dm

13 Equations of astrophysical fluid dynamics Astrophysical fluid dynamics (AFD): Basic model: Newtonian gas dynamics: u t + u u = Φ 1 ρ p ρ + u ρ = ρ u t p + u p = γp u t Compressible Ideal (inviscid, adiabatic) Non-relativistic (Galilean-invariant) Lagrangian (material) derivative: D Dt = t + u u Φ ρ p γ velocity gravitational potential density pressure adiabatic exponent

14 Equations of astrophysical fluid dynamics Gravity: Non-self-gravitating fluid: Φ is prescribed (fixed / external potential) Self-gravitating fluid: Φ is determined (in part) from the density of the fluid: 2 Φ=4 πgρ

15 Equations of astrophysical fluid dynamics Extensions of the basic model: Viscosity: Usually extremely small May be needed to provide small-scale dissipation May be introduced to model turbulent transport u t = + 1 ρ T T =2µS + µ b ( u)i S = 1 2 [ u +( u) T ] 1 ( u)i 3 T µ µ b S I viscous stress tensor (shear) viscosity bulk viscosity shear tensor unit tensor kinematic viscosity ν = µ/ρ

16 Equations of astrophysical fluid dynamics Non-adiabatic effects: Thermal energy equation: Heating: ρt Ds Dt = H C Viscous: Cooling: T s H C H = T : u =2µS 2 + µ b ( u) 2 Radiative: C = F Diffusion approximation: (optically thick regions) F = 16σT 3 3κρ T σ κ temperature specific entropy heating / unit volume cooling / unit volume (non-adiabatic effects) Stefan-Boltzmann constant opacity (Rosseland mean)

17 Equations of astrophysical fluid dynamics Equation of state: p = p(ρ, T ) Ideal gas with radiation: p = p g + p r = p r kρt + 4σT 4 µ m m p 3c important at very high T k µ m m p c Thermal energy equation in dynamical variables: ( )( 1 ρt ds = dp γ ) 1p γ 3 1 ρ dρ ( )( 1 Dp γ 3 1 Dt γ ) 1p Dρ = H C ρ Dt Boltzmann constant mean molecular weight proton mass speed of light µ m =0.5 for fully ionized H, µ m =2 for molecular H, etc. For ideal gas of constant ratio of specific heats, γ 1 = γ 2 = γ 3 = γ

18 Equations of astrophysical fluid dynamics Extensions of the basic model: Magnetohydrodynamics (MHD) Radiation hydrodynamics (RHD) Relativistic formulations Kinetic theory / plasma physics Simplifications of the basic model: Incompressible fluid: u =0 Boussinesq / anelastic approximations Barotropic fluid: p = p(ρ)

19 Equations of astrophysical fluid dynamics Magnetohydrodynamics (MHD): Electrically conducting fluid (plasma, metal, weakly ionized gas) Pre-Maxwell equations (without displacement current): B t = E B =0 B = µ 0 J ( E equation not required) Galilean invariance: x = x vt t = t solenoidal constraint B = B B E J µ 0 E = E + v B J = J magnetic field electric field electric current density permeability of free space

20 Equations of astrophysical fluid dynamics Ohm s law: J = σe J = σ(e + u B) Combine with Maxwell: B t = E = (u B) in rest frame of conductor σ : electrical conductivity for conducting fluid with velocity u(x,t) ( J σ ) = (u B) (η B) magnetic diffusivity η = 1 µ 0 σ resistivity = (u B)+η 2 B η if uniform Induction equation : vector advection-diffusion equation ω cf. vorticity equation for t = (u ω)+ν 2 ω ω = u

21 Equations of astrophysical fluid dynamics Ideal MHD (perfect conductor: σ, η 0): B t = (u B) = B u u B B( u)+u( B) Magnetic field is frozen in to fluid: Field lines behave as material lines Magnetic flux through an open material surface is conserved Valid for large magnetic Reynolds number Rm = LU η cf. Re = LU ν (advection versus diffusion) Much easier to achieve on astrophysical scales

22 Equations of astrophysical fluid dynamics Lorentz force per unit volume F m = J B = 1 ( B) B µ 0 = 1 ( ) B 2 B B µ 0 2µ 0 curvature force: magnetic tension gradient of magnetic pressure T p m = B 2 m = B 2 (= magnetic energy density) µ 0 2µ 0 F m = M M = BB µ 0 B 2 2µ 0 I Maxwell stress tensor If B = B e z, M = p m p m T m p m

23 Equations of astrophysical fluid dynamics Lorentz force: Magnetic tension + frozen-in field Alfvén waves v a = ( Tm ρ ) 1/2 cf. elastic string v a =(µ 0 ρ) 1/2 B vector Alfvén velocity Magnetic pressure magnetoacoustic waves

24 Equations of astrophysical fluid dynamics Ideal MHD equations: u t + u u = Φ 1 ρ p + 1 ( B) B µ 0 ρ ρ + u ρ = ρ u t p + u p = γp u t B = (u B) t B =0 Or can expand out E and J eliminated Nonlinearities in equation of motion and induction equation

25 Equations of astrophysical fluid dynamics Total energy equation in ideal MHD: ] [ [ρ( 12 t u 2 +Φ+ e)+ B 2 + ρu( 1 2µ 2 u 2 +Φ+ w)+ E B ] 0 µ 0 =0 specific internal energy For ideal gas of constant γ : e = p (γ 1)ρ w = e + p ρ = γp (γ 1)ρ With self-gravity, Φ=Φ int +Φ ext contributes to the energy density and only specific enthalpy Poynting flux 1 2 Φ int +Φ ext E = u B

26 Equations of astrophysical fluid dynamics Forces as the divergences of a stress tensor: Equation of motion can be written ρ Du Dt = ρ Φ+ T Related to conservative form for momentum: (ρu)+ (ρuu T) = ρ Φ t Contributions to stress tensor T : pressure p I viscous 2µS + µ b ( u)i self-gravity gg 4πG + g 2 8πG I magnetic BB µ 0 B 2 2µ 0 I (check using Poisson s equation) 2 Φ=4 πgρ g = Φ also turbulent stresses from correlations of fluctuating fields

27 Evolution of an accretion disc Regulated by conservation of mass and angular momentum Mass conservation in 3D: ρ t + (ρu) =0 Momentum conservation in 3D: (ρu)+ (ρuu T) = ρ Φ t Φ is axisymmetric gravitational potential we considered for orbits (later, allow non-axisymmetric potential and tidal torque) T represents collective effects, including self-gravity

28 Evolution of an accretion disc Write in cylindrical polar coordinates Reduce from 3D to 1D Mass conservation: (r, φ, z) ρ t + 1 r r (rρu r)+ 1 r φ (ρu φ)+ z (ρu z)=0 Integrate r 2π 0 dφ dz over cylinder of radius r M t + F r =0 1D mass density Radial mass flux M(r, t) = F(r, t) = assuming no vertical mass loss or gain 2π 0 2π 0 ρr dφ dz ρu r r dφ dz

29 Evolution of an accretion disc Angular momentum conservation: (r, φ, z) t (ρu φ)+ 1 r 2 r [r2 (ρu r u φ T rφ )] + 1 r φ (ρu2 φ T φφ )+ z (ρu φu z T φz )=0 Integrate r 2 2π 0 dφ dz over cylinder of radius r Assume that ru φ = h(r) from orbital dynamics (examine this assumption later) t (Mh)+ (Fh + G) =0 r Internal torque G(r, t) = 2π 0 assuming no vertical loss or gain r 2 T rφ dφ dz

30 Evolution of an accretion disc Since h depends only on r : Eliminate M : M + F t r =0 M t h + (Fh + G) =0 r F dh dr + G r =0 M t Interpretation: = r [ (dh dr ) 1 G r which determines F angular momentum determines orbital radius angular momentum transport determines mass evolution ]

31 Evolution of an accretion disc More usual notation: M =2πrΣ Σ(r, t) F =2πrΣū r G = 2π νσr 3 dω dr ū r (r, t) ν(r, t) surface density mean radial velocity mean effective kinematic viscosity Equivalent to: Σ= ρ dz Σū r = = 1 2π 2π 0 dφ νσ = ρu r dz µ dz T rφ = µr dω dr

32 Evolution of an accretion disc Then obtain: Σ t = 1 r r [ (dh dr ) 1 r ( νσr 3 dω ) ] dr Keplerian disc: Ω r 3/2 h = r 2 Ω r 1/2 Σ t = 3 r r [ r 1/2 ] r (r1/2 νσ) diffusion equation for surface density

33 Evolution of an accretion disc Interpretation: angular momentum transport dω dr < 0 dh dr > 0 orbital shear viscous spreading

34 Analysis of the diffusion equation Inner boundary condition: Important because spreading ring will reach Depends on nature of central object r =0 in finite time Star with negligible magnetic field: Disc may extend down to stellar surface Star usually rotates at only a fraction of Keplerian rate: ( ) 1/2 GM Ω < R 3 (star is supported mainly by pressure) Angular velocity makes a rapid adjustment from Keplerian to stellar in a viscous boundary layer

35 Analysis of the diffusion equation dω Usual argument: at so there dr =0 r = r in ( R ) G =0

36 Analysis of the diffusion equation Black hole: Circular orbits are unstable close to event horizon Non-rotating black hole: Ω = ( GM r 3 ) 1/2 κ 2 =Ω 2 ( 1 6 GM c 2 r Orbits unstable for r< 6 GM c 2 =3r S ) Rapid inspiral from r ms ū r and Σ rapidly, so effectively G 0 at r in r ms

37 Analysis of the diffusion equation Star with significant magnetic field (e.g. dipole): magnetospheric radius r m polar accretion (observed) disc magnetospheric cavity G not necessarily 0 at r in r m Star can exert torque on disc Depends on magnetic coupling (complicated problem)

38 Analysis of the diffusion equation Mathematically, we may let r in 0 To allow mass flux at origin but no torque: F cst G 0 } as r 0 To allow torque at origin: G cst as r 0 For a Keplerian disc, F rσū r r 1/2 r (r1/2 νσ) G r 1/2 νσ First case (no torque): Second case (torque): νσ cst r 1/2 νσ cst as r 0 as r 0

39 Analysis of the diffusion equation Consideration of angular momentum transport processes and local vertical structure of disc (later) leads to ν = ν(r, Σ) (e.g. double power law) If If ν = ν(r) only, diffusion equation is linear ν = ν(r, Σ), diffusion equation is nonlinear Usually solve as initial-value problem or look at steady solutions Linear case can be treated using Green s function G(r, s, t) : Solution of diffusion equation with initial condition Σ(r, 0) = δ(r s) (i.e. very narrow ring) Then solution for any initial condition Σ(r, t) = 0 G(r, s, t)σ 0 (s)ds Σ 0 (r) is

40 Analysis of the diffusion equation Can calculate G(r, s, t) in terms of Bessel functions for any power law ν r p and any boundary conditions Classic example: ν = cst G =0 inner BC at r =0 no outer boundary G(r, s, t) = r 1/4 s 5/4 6 νt exp Becomes more asymmetrical as [ r2 + s 2 12 νt modified Bessel function ] ( rs ) I 1/4 6 νt increases (omit derivation) As t, all mass accreted by central object and all angular momentum carried to r = by negligible mass t (Lynden-Bell & Pringle 1974)

41 Analysis of the diffusion equation Easier example: suppose ν = Ar with A = const : ( ) 1/2 4r Let y = and g = r 1/2 νσ = Ar 3/2 Σ to obtain 3A g t = 2 g y 2 classical diffusion equation Spreading-ring solution with zero torque ( g =0) at centre: g t {exp [ 1/2 (y y 0) 2 ] exp [ (y + y 0) 2 ]} 4t 4t Thus G(r, s, t) = r 3/2 t 1/2 [ ] [ ] (3πA) exp (r + s) 2(rs) 1/2 sinh 1/2 3At 3At [ ( s ) ] 1/2 Remaining mass G(r, s, t) r dr = erf 3At 0

42 Analysis of the diffusion equation G(r, s, t) = r 3/2 t 1/2 (3πA) 1/2 exp [ ] (r + s) 3At sinh [ 2(rs) 1/2 3At ] t =0.1 s/a t =0.03 s/a t = s/a t =0.001 s/a t =0.003 s/a t =0.01 s/a r/s

43 Analysis of the diffusion equation G(r, s, t) = r 3/2 t 1/2 (3πA) 1/2 exp [ ] (r + s) 3At sinh [ 2(rs) 1/2 3At ] Gr mass / unit radius Gr 3/2 ang mom / unit radius r/s r/s t = {0.01, 0.03, 0.1, 0.3, 1} s/a

44 Analysis of the diffusion equation Nonlinear case with νσ r p Σ q and r in 0 : No intrinsic length-scale Special algebraic similarity solutions Generally attract solutions of initial-value problem But if q<0, viscous instability occurs (see later)

45 Analysis of the diffusion equation Steady discs F = cst = Ṁ Ṁ = mass accretion rate (but always neglect slow increase of M ) F dh dr + dg dr =0 Ṁh+ G = cst If G =0 at r = r in, solution is G = Ṁ(h h in) In Keplerian case (recall G = 2π νσr 3 dω ) dr νσ = Ṁ [ ( rin ) ] 1/2 1 3π r If we know ν(r, Σ), we can solve for Σ(r)

46 Analysis of the diffusion equation Other solutions: Non-accreting disc: Decretion disc: Note that F < 0 F > 0 requires F =0, G = cst dg dr > 0 } require torque from central object Energy balance in steady disc: Energy dissipation rate per unit volume = T rφ r dω dr = µ Energy emission rate per unit area (from each face of disc) = 1 ( 2 νσ r dω ) 2 dr Effective blackbody temperature T eff (r) given by σteff 4 = 9 8 νσω2 = 3GMṀ [ ( rin ) ] 1/2 8πr 3 1 r ( r dω ) 2 dr

47 Analysis of the diffusion equation Typical (theoretical) spectrum of emitted radiation: inner disc outer disc

48 Analysis of the diffusion equation Total luminosity of disc (r in < r < ) [ 3GMṀ ( rin ) ] 1/2 L disc = 8πr 3 1 r r in 2πr dr = 1 2 GMṀ r in = rate of release of orbital binding energy ( r in ) = 1 2 rate of release of potential energy (remaining kinetic energy released in boundary layer if Ω =0)

49 Vertical disc structure Vertical hydrostatic equilibrium Dominant balance (for non-self-gravitating gas disc) Expand potential 0= ρ Φ z p z Φ(r, z) = Φ(r, 0) Φ zz(r, 0)z 2 + So for a thin disc Φ z Ω2 zz Equation of vertical hydrostatic equilibrium p z ρω2 zz ( Ω z =Ω for Keplerian disc)

50 Vertical disc structure r H Disc supported against gravity by pressure in vertical direction Centrifugal support in radial direction

51 Vertical disc structure Order-of-magnitude estimates and time-scales Simple scaling arguments ( Important dimensionless parameter: H r For a thin disc: H may represent: ), omitting factors of order unity (angular semi-thickness, aspect ratio) H r 1 { true semi-thickness height of photosphere above midplane measure of density scale-height From vertical hydrostatic equilibrium: p H ρω2 H c s ΩH c s = ( ) 1/2 p ρ isothermal sound speed

52 Vertical disc structure (Effective) viscosity: Dimensional argument: [µ] =ML 1 T 1 = [ p Ω] Write µ = αp in terms of dimensionless viscosity α Ω ( alpha viscosity prescription, Shakura & Sunyaev 1973) (Mean) kinematic viscosity ν µ ρ αc2 s Ω αc sh Analogous to kinetic theory of molecular transport: ν vl for mean speed v and mean free path l Molecular viscosity usually negligible for astrophysical discs Similar estimate applies to effective eddy viscosity of turbulence of mean turbulent speed v and correlation length l Assuming that v c s (subsonic turbulence) and l H, expect that α 1

53 Vertical disc structure Stress in a Keplerian disc: T rφ = µr dω dr = 3 2 αp Alternative version of alpha prescription: T rφ = αp Whatever process gives rise to stress, scales with local pressure Reasonable assumption for local processes (see later)

54 Vertical disc structure Three important characteristic time-scales: Dynamical time-scale: t dyn 1 Ω (orbital motion) H c s (hydrostatic) Viscous time-scale: t visc r2 ν α 1 ( H r ) 2 t dyn (radial motion) Thermal time-scale: t th ph νσω 2 c2 s νω 2 H2 ν α 1 t dyn (thermal balance) For a thin disc with α< 1 : t dyn <t th t visc All three time-scales usually increase rapidly with r

55 Vertical disc structure Mach number of orbital motion: Ma rω c s ( H r ) 1 Characteristic radial velocity due to viscosity: ) ū r ν r α ( H r c s For a thin disc, ū r c s rω orbital motion: highly supersonic accretion flow: highly subsonic

56 Vertical disc structure Corrections to orbital motion: Contribution of radial pressure gradient: p r / ρrω 2 c2 s r 2 Ω 2 ( H r ) 2 Vertical variations of radial gravitational acceleration ( H r ) 2 u r Other terms, e.g. ρu r, are smaller r [ ( ) ] 2 H Conclude that u φ = rω 1+O r i.e. fluid velocity close to orbital velocity of test particle In general, thin-disc approximations involve fractional errors and a formal asymptotic treatment is possible O ( H r ) 2

57 Vertical disc structure Typical values of H/r : protoplanetary discs : binary stars : active galactic nuclei : planetary rings :

58 Vertical disc structure Barotropic models: Vertical hydrostatic equilibrium: Can be solved if pressure is a known function of density (physical arguments, or just for analytical convenience) Vertically isothermal model: p = c 2 s ρ Solution: p z = ρω2 zz ρ(r, z) =ρ 0 (r, 0) e z2 /2H 2 c s independent of z H = c s Ω z isothermal scale height

59 Vertical disc structure Polytropic model: p = Kρ 1+1/n K, n independent of z Introduce (pseudo-)enthalpy: dp w = =(n + 1)Kρ1/n ρ w z = Ω2 zz Solution: w = 1 2 Ω2 z(h 2 z 2 ) H = true semi-thickness ρ(r, z) =ρ(r, 0) p(r, z) =p(r, 0) (1 z2 H 2 (1 z2 H 2 ) n (vacuum above z = H ) ) n+1

60 Vertical disc structure Radiative models: Energy dissipated by viscosity carried away by radiation Radiative diffusion (optically thick disc): F = 16σT 3 3κρ T κ = Rosseland mean opacity Dominant balance in thermal energy equation (thin disc): ( 0=µ r dω ) 2 F z dr z Contributions from O(H/r) 2 F r and from radial advection are smaller by

61 Vertical disc structure Vertical structure of radiative Keplerian disc: p z = ρω2 z F z = 9 4 µω2 F = 16σT 3 3κρ together with: T z equation of state, e.g. viscosity prescription opacity function κ(ρ, T ) p = kρt + 4σT 4 µ m m p 3c boundary conditions, e.g. ρ = p = T =0 (or match to atmopsheric model) Analogous to radial structure of a star at (ideal gas + radiation) z = H

62 Vertical disc structure Opacity often approximated by a power law, e.g. for ionized discs: κ = const 0.33 cm 2 g 1 κ = C κ ρt 7/2 C κ cm 5 g 2 K 7/2 Thomson opacity electron scattering hotter regions Kramers opacity free-free / bound-free cooler regions Cooler discs: dust, molecules,...

63 Vertical disc structure Keplerian disc, alpha viscosity, gas pressure, Thomson opacity: p z = ρω2 z F z = 9 4 αpω F = 16σT 3 3κρ p = kρt µ m m p T z Σ= H H ρ dz

64 Vertical disc structure Order-of-magnitude treatment: p z = ρω2 z F z = 9 4 αpω F = 16σT 3 3κρ p = kρt µ m m p T z p H ρω2 H F H αpω F σt 3 κρ p kρt T H µ m m p Σ= H H ρ dz Σ ρh Algebraic solution: H α 1/6 Σ 1/3 Ω 5/6 ( µ m m p k ) 2/3 ( σ κ ) 1/6

65 Vertical disc structure Viscosity: ν αc s H αωh 2 α 4/3 Σ 2/3 Ω 2/3 ( µ m m p k Thus ν rσ 2/3 ) 4/3 ( σ κ ) 1/3 Full treatment: Use order-of-magnitude treatment as a dimensional analysis It defines characteristic units U H,U ρ,u p,u T, etc. Then write z = U H z, ρ = U ρ ρ( z), etc. to obtain a system of dimensionless ODEs with no free parameters Solve numerically Find ν = ArΣ 2/3 with a precise coefficient A Power-law constitutive relations give power-law viscosity

66 Vertical disc structure pressure temperature density radiative flux similar to polytropic models ρ (H 2 z 2 ) n p (H 2 z 2 ) n+1 T (H 2 z 2 )

67 Thermal and viscous stability Viscous stability: Nonlinear diffusion equation for Keplerian disc: Σ t = 3 [ r 1/2 ] r r r (r1/2 νσ) ν = ν(r, Σ) Linearize about any given solution Σ 0 (r, t) : Σ(r, t) =Σ 0 (r, t)+σ (r, t) Σ Σ 0 ( νσ) = ( νσ) Σ Σ = q νσ q = ln( νσ) ln Σ Linearized diffusion equation: Σ t = 3 r Unstable for r [ r 1/2 ] r (r1/2 q νσ ) q<0 : rapid growth on short length-scales

68 Thermal and viscous stability Thermal stability: So far, assumed a balance between heating and cooling: 9 4 νσω2 = H = C =2F + Relax this assumption, but assume that t dyn t th t visc Consider behaviour on the timescale t th : disc is hydrostatic surface density does not evolve α 1 so that By solving equations of vertical structure except thermal balance, can calculate H and C as functions of (Σ, νσ) In fact H depends only on νσ Equation of thermal balance defines a curve in the (Σ, νσ) plane

69 Thermal and viscous stability Infinitesimal perturbations: dh = dh d( νσ) d( νσ) C dc = Σ dσ + C ( νσ) d( νσ) Along the equilibrium curve, dh =dc and d( νσ) = q ν dσ dh d( νσ) = 1 q ν C Σ + C ( νσ) q = ln( νσ) ln Σ Thermal energy content of disc per unit area ph (Ω/α) νσ If some heat is added, νσ increases but Σ is fixed on t th Unstable if excess heating exceeds excess cooling, i.e. if dh d( νσ) > dc d( νσ) i.e. 1 q ν C Σ > 0 In practice C/ Σ < 0 (because, at fixed νσ, Σ 1/ ν 1/(αT )) so thermal instability occurs (like viscous instability) when q<0

70 Thermal and viscous stability Outbursts: Radiative disc with gas pressure and Thomson opacity has νσ rσ 5/3 and is viscously and thermally stable For cooler discs undergoing H ionization, instability can occur S curve and limit cycle hot cool for steady accretion: νσ = Ṁ [ ( rin ) ] 1/2 1 3π r Ṁ 3π Explains outbursts in many cataclysmic variables, X-ray binaries and other systems