Äîêóìåíò âçÿò èç êýøà ïîèñêîâîé ìàøèíû. Àäðåñ îðèãèíàëüíîãî äîêóìåíòà : http://www.mso.anu.edu.au/~geoff/HEA/11_Accretion_Disks_II.pdf Äàòà èçìåíåíèÿ: Thu Jun 12 17:41:11 2003 Äàòà èíäåêñèðîâàíèÿ: Tue Oct 2 05:22:36 2012 Êîäèðîâêà: Ïîèñêîâûå ñëîâà: m 8

Accretion Disks II
1 The dynamical equations for accretion discs 1.1 Setting up the model

Thin accretion disc z r 2h(r )

Compact object
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The simplest accretion disc model to construct is that of the thin disc. We obtain a simplified set of equations by integrating over the z-dimension and by assuming that the flow is steady in the mean and that it is axisymmetric in the mean. Because of this approximation, we can regard our ensemble average as an azimuthal average, i.e. we average over independent regions around the annulus of an accretion disc. Assumptions In the following development of the theory of accretion discs, we assume that: · The disk is steady in the mean, i.e. time derivatives of the mean flow variables are zero. · The disk is axisymmetric, i.e. there is no dependence of the mean flow on the azimuthal angle
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· The disk is thin. This means that we can construct useful equations by vertical averaging. The validity of this assumption can be justified a posteriori · The velocity in the disk is dominated by the Keplerian veloc~ ~ ~ ity v = GM / R . In particular v « v . Again this can be
r

justified a posteriori · The disk does not have a substantial wind. Coordinates · In view of the geometry and the physical assumptions, we use cylindrical polar coordinates, ( r, , z ) . · Occasionally use the spherical radius, R .

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Mathematical approach · We develop the accretion disk equations by examining the statistically averaged equations for conservation of mass, momentum and energy in this axisymmetric coordinate system. · Integrated equations are obtained by integrating the equations over a disk height, i.e. between z = ­ h and z = h . 1.2 Continuity 1 ~ ~ ( v z ) + -- ( rv r ) = 0 r r z
(1.2-1)

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Multiply by 2 r and integrate over z :
h h ~ ~ 2 ( rv z ) dz + 2 ( rv r ) dz = 0 ­h z ­h r h d ~ ~ 2 r vz + ----- 2 ( rv r ) dz = 0 ­h ­ h dr h

(1.2-2)

We define the mass accretion rate by: = ­ 2 h rv dz ~ Ma r ­h
(1.2-3)

We have assumed that there is no wind from the top and bottom ~ surfaces of the disc. Therefore v z = 0 at z = ± h . Hence, d ----- M a = 0 M a = constant dr
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(1.2-4)
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i.e. the accretion rate is constant with radius. 1.3 Momentum All three components of the momentum equations GM p ~~ ­ --------( vi v j ) + v i v j = ­ ------ ­ x j x j x i xi R p GM x i = ­ ------ ­ ----------------x i R3 give us essential information.

(1.3-1)

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Since the accretion disc is confined to near z = 0 (i.e. h « r ) we treat the gravitational term in the following way. The spherical polar radius R = ( r 2 + z2 )1 / 2 r Therefore: GM ^ GM GM ---------, 0, GM z --------- R = --------- ( r, 0, z ) -------- r2 R2 R3 r3
(1.3-3)

at

z0

(1.3-2)

In the following we write the hydrodynamical equations in cylindrical polars. This can be accomplished in a number of ways 1.Use the Christoffel symbols for a cylindrical coordinate system to calculate the divergence terms etc. 2.Look up the equations in Landau & Lifshitz
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1.4 Radial momentum The radial momentum balance is given by: 1 ~ 2) + ( -- ( r v r r r z 1 + -- ( r v r 2 ) r r GM = ­ --------r2 ~ v ) ­ 1 v 2 v r ~ z -- ~ r + v r v z z p ­ ----r
(1.4-1)

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We integrate 2 r times this equation with respect to z and obtain:
2 V h h d 2 dz + 2 r v v h ­ 2 -----~~ ~ - dz ----- 2 r v r rz ­h r ­h dr ­h h d 2 dz + 2 v v h + ----- 2 r v r rz ­h dr ­h h d GM h = ­ 2 --------- dz ­ ----- 2 r p dz ­h dr r 2 ­h

(1.4-2)

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We neglect all but the blue terms since: ~ ~ v r, v r « v ~ vz ( ±h ) = 0 ~2 p « v
(1.4-3)

v r v z ( ± h ) = 0 The last inequality is equivalent to:

~2 v p ~2 2 ~2 -- « v c s « v ----- » 1 2 cs

(1.4-4)

that is, the azimuthal speed is highly supersonic. This is justified a posteriori below.

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Surface density This important parameter is defined by
h ( r ) = dz ­h

(1.4-5)

The radial momentum balance equation therefore has the form: ~2 v GM 2 ( r ) ----- = 2 ( r ) --------r r2 ~ 2 = GM v --------r

(1.4-6)

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Keplerian speed The Keplerian speed (i.e. the speed of a particle in a circular orbit around the central object) is given by
2 v K GM 2 = GM ------ = --------- v K --------r r r2

(1.4-7)

Hence, the radial momentum equation tells us that ~ v = vK Disks in which this equation is valid are called Keplerian.
(1.4-8)

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1.5 Azimuthal momentum The -component of the momentum equations is: 1 2 ~~ ~ ---- (r v r v ) + ( v z r2r 1 2 + ---- [ r v r v ] r2r Multiply through by 2 r 2 : ~~ [ 2 r 2 vr v ] + r + [ 2 r 2 v r r ~~ [ 2 r 2 v v z v ] + [ 2 z ] z
(1.5-2)

~ vz ) + v v z = 0 z
(1.5-1)

r 2 v v z ] = 0
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Integrate over z : dh ~ ~ ( 2 r v r ) ( rv ) dz + 2 r 2 ­ h dr dh + ----- 2 r 2 v r v dz + 2 dr ­ h ~~ v vz h ­h r 2 v v z h = 0 ­h

(1.5-3)

~ Since v z = 0 and v v z ( ± h ) = 0 are both zero at the surface of the disk,
h ~ ) ( rv ) dz + h 2 r 2 v v dz d ­ h ( 2 r v r ~ r ­h -----

dr

(1.5-4)

=0

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~ The term rv is the z -component of angular momentum per unit mass. Write the first term in the brackets as:
h ~ ~ ~ rv â 2 r v r dz = ­ M a ( rv ) ­h

(1.5-5)

Integrate the above equation: rv + 2 r 2 h v v dz = Constant ­ Ma ~ r ­h
(1.5-6)

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r r0

Evaluate the constant using values at the innermost stable orbit. Also define:
h G r ( r ) = v r v d z ­h

(1.5-7)

Hence the azimuthal equation reads: ~ ­ M a rv ( r ) + 2 r 2 G r ( r )
2 ~ = ­ M a r 0 v ( r 0 ) + 2 r 0 Gr ( r 0 )

(1.5-8)

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Physical meaning of the angular momentum equation
h ~ ~ Consider ( 2 r v r ) ( rv ) dz ­h

z component of a ngular ~ rv = momentum per unit mass Flux of z component of angular ~ ~ ( v r ) â rv = momentum in r direction 2 rdz = Element of area Therefore: Flux of angular momentum h ~ ~ ( 2 r v r ) ( rv ) dz = ­h in r direction

(1.5-9) (1.5-10) (1.5-11)



(1.5-12)

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~ ­ M a rv ( r ) r
0

r

As we have seen this flux is negative and we have put it equal to ~ ­ M a rv , i.e. accretion implies that


there is a flux of angular momentum inwards.
~ ­ M a r 0 v ( r 0 )

Also, since M a is constant, the flux of angular momentum is larger at

~ larger r because rv ( r ) = ( GMr ) 1 / 2 This means that as a result of the accretion there is more angular momentum transported by accretion into the annulus between r and r 0 than is transported out at the smaller radius r = r 0 .
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Now consider:
2 h v v dz 2r r ­h

(1.5-13)

Flux of turbulent momentum density (1.5-14) v r v = in r direction Flux of z component of turbulent r v r v = angular momentum in r direction Therefore Turbulent flux of angular h 2 r r v r v d z = ­h momentum in r direction
(1.5-16) (1.5-15)

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It is this flux that balances the build up of angular momentum between r and r 0 . The equation: ~ ­ M a rv ( r ) + 2 r 2 G r ( r )
2 ~ = ­ M a r 0 v ( r 0 ) + 2 r 0 Gr ( r 0 )

(1.5-17)

tells us that Flux of angular Flux of angular momentum = (1.5-18) momentum through r 0 through r

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That is angular momentum is conserved in between r and r 0 . This is correct since there are no torques acting on the disk material. Note also that this analysis applies to any annulus, not just one involving the innermost stable orbit.
h We can now solve for G r = v r v dz : ­h

~ ~ r0 2 M a v r 0 v ( r 0 ) G r ( r ) = ---- G r ( r 0 ) + ------------- 1 ­ -------------------- r ~ 2r rv ( r ) ~ r0 2 r0 1 / 2 M a v = ---- G r ( r 0 ) + ------------- 1 ­ ---- r r 2r
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(1.5-19)

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The effect of the inner boundary condition decreases quite rapidly with r so that we often neglect it and take: ~ M a v r0 1 / 2 G r ( r ) = ------------- 1 ­ ---- r 2r
(1.5-20)

1.6 Vertical equilibrium The momentum equation in the z --direction reads: 1 ~~ -- [ r v r v z + v r v z ] + [ r r z 2 ] ­ GM --------= ­ [ p + vz z r2
High Energy Astrophysics: Accretion Disks II

2 vz ]

(1.6-1)

z
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We assume: ~ 1. There is no wind ( v z = 0 ). 2. The turbulent stresses are much less than the pressure ( v 2 « p ). This means that we can neglect the terms involving v r v z and v z 2 . Thus the equation for vertical equilibrium reduces to: GMz p = ­ ----------z r3 Isothermal disk In order to get a quantitative feel for the implications of this equation, let us assume that the disk is isothermal.
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(1.6-2)


We put ~ kT p = ---------µm p ~ With T = constant ~ kT 1 GM ---------- -- = ­ --------- z -µm p z r3 µ GM m p ----- = exp ­ -------------------- z 2 ~ c 2 kT r 3
(1.6-4) (1.6-3)

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This defines the disk scale height hs 2 kT 2 = -------------------- ---- = 2 ---------- hs µm r µ GM m p p 2 kT r 3 GM ­ 1 --------r
(1.6-5)

2 as 2 = 2 ------ = -------2 2 vK MK

Hence the condition that the disc be thin is equivalent to the condition that the Mach number of the Keplerian flow be supersonic.

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1.7 The energy equation In the following we take a section of an accretion disk and calculate the radiation emitted from the surface as a result of the dissipation within the disk.

Radiation z=h Dissipation z = ­h

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Rate of production of turbulent energy Knowing the Reynolds stress from the angular momentum equation, we can evaluate the local rate of production of turbulent energy. This is ~ ~ ­ v i v j s ij ­ 2 v r v s r where the r component of shear is given by: 1 ~ s r = -- r 2 The angular velocity is given by the Keplerian value = GM 1 / 2 --------r3
(1.7-3) (1.7-2) (1.7-1)

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The derivative of this quantity is: 3 3 ( GM ) 1 / 2 - = ­ -- ---------------------- = ­ -- --2r 2 r5 / 2
(1.7-4)

The production rate per unit volume of turbulent energy is therefore ~ ( r, z ) = ­ v i v j s ij ­ v r v r 3 = -- v r v 2
(1.7-5)

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The production rate of turbulent energy per unit area of the disk is
h 3 h ( r ) = ­ v r v r dz = ------- v r v dz ­h 2 ­h

3 vK M r0 1 / 2 3 = ------- G r ( r ) = -------------------- 1 ­ ---- r 4r 2

(1.7-6)

where we have used the result from the analysis of the angular momentum equation: ~ r0 1 / 2 M a v G r ( r ) = ------------- 1 ­ ---- r 2r
(1.7-7)

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Now GM 1 / 2 GM 1 / 2 GM ----------------vK = = --------r r3 r2 so that
h ~ ( r ) = ­ 2 v r v s r dz ­h

(1.7-8)

r0 1 / 2 3 GM M a = --------------------- 1 ­ ---- r 4r3

(1.7-9)

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Radiative flux We assume that the production of turbulence per unit area is equal to the dissipation into heat per unit area and that this heat is radiated in a quasi steady state away from the surface of the disk. The luminosity of the disk emitted between radii r 1 and r 2 is

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3 GM M a L disc ( r 1, r 2 ) = --------------------4 3 GM M a = --------------------2r0 3 GM M a = --------------------2



r0 1 / 2 r2 1 ---- 1 ­ ---- 2 r dr r r1 r 3
(1.7-10)



r0 1 / 2 r2 r0 2 r ---- ---- 1 ­ ---- dr r r1 r 0 1 2 ---- 1 ­ -- r1 3 r 0 1 / 2 r 0 1 / 2 1 2 ---- -­ ---- 1 ­ -- ---- r 1 r2 3 r 2



The total luminosity emitted by the disc is obtained by integrating between r 0 and : GM M a L = L disc ( r 0, ) = ----------------2r0
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(1.7-11)
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Relationship to overall energetics Consider the energy of a parcel of gas with a mass, m , at radius r . This is: 1 2 GMm 1 GM ­ GM E b = -- mv K ­ ------------- = m -- --------- --------2 r 2 r r 1 GMm = ­ -- ------------2r

(1.7-12)

This gas has been accreted, essentially from r = so that this represents the energy that has been lost by this parcel of gas during the time that it has spiralled in from r = to r .

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By the time the gas has reached the innermost stable orbit, the energy lost per unit mass is: GM --------2r0
(1.7-13)

The mass accreted per unit time is M a . Hence, the total power associated with the accretion is GM M a GM M a â --------- = ----------------2r0 2r0
(1.7-14)

Hence the total luminosity of the disk is the power associated with the energy lost per unit mass at radius r 0 times the mass flux.
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The gravitational radius With black holes or neutron stars in mind, we express the radius in units of the gravitational radius GM r g = --------c2 In terms of the gravitational radius, the luminosity is GM M a GM M a M ac2 L = ----------------- = --------------------------- = ---------------------2r0 2rg(r0 / rg) 2(r0 / rg)
(1.7-16) (1.7-15)

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For a spherical black hole, the innermost stable orbit is at 6 gravitational radii. If our Newtonian treatment were to be adequate at such radii L 0.083 M a c 2
(1.7-17)

i.e. approximately 8% of the infalling rest mass energy is converted into radiation. The exact answer from the general relativ istic treatment is L 0.057 M c 2 . That is, 5.7% of the mass
a

energy is converted into luminosity. For a rotating black hole the figure goes up to 42%. Typically, for order of magnitude purposes, we assume that black holes are 10% efficient. Neutron stars are actually slightly more efficient than spherical black holes because the kinetic energy of the infalling material is converted into radiation at a shock on the surface.
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2 Accretion disk temperature 2.1 Use of Stefan's law for an optically thick disk If we assume that the disc is optically thick (true in a large number of cases) then the dissipation per unit area appearing in radiation from each side of the disc is r0 1 / 2 3 GM M a 1 -- ( r ) = --------------------- 1 ­ ---- r 2 8r3 The temperature of the disk surface is given by 3c6 T 4 = ------------4G2 Ma ------M2 r0 1 / 2 r ­3 ---1 ­ ---- r r g
(2.1-2) (2.1-1)

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Hence, 1 r0 1 / 2 1 / 4 6 1 / 4 M a / 4 r ­ 3 / 4 3c T = ----------------1 ­ ---- (2.1-3) ------------- ---- r 4 G 2 M 1 / 2 r g The constant
19 3c6 1 / 4 ----------------= 2.88 â10 SI units 4 G 2

(2.1-4)

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Typical parameters for a galactic mass black hole are M a = 10 14 Kg/s and M = 3 M O giving L = 7.5 â10 W r0 1 / 2 1 / 4 7 r ­3 / 4 1 ­ ---- °K T = 3.7 â10 ---r r g
(2.1-5)
29

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Typical parameters for an extragalactic supermassive black hole in an AGN are M a = 0.1 M O /yr and M = 10 8 M O giving: L 4.7 â10
37

W
(2.1-6)

r0 1 / 2 1 / 4 5 r ­3 / 4 1 ­ ---- °K T 5.7 â10 ---r r g

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The emission for a galactic mass black hole therefore peaks in the X-ray; that from a supermassive black hole peaks in the UV. This the reason proposed for the UV bumps shown in the spectra of AGN at the left.

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3 The Eddington Luminosity 3.1 Derivation for spherically symmetric accretion The luminosity of an accreting object cannot increase indefinitely. There is a fundamental limit, known as the Eddington limit, which limits accretion by the radiation force on the accreting matter.
Electron Radiation field

An electron in a radiation field feels a force proportional to the momentum flux density, P rad , of the radiation field. This is given by F = T P rad
(3.1-1)
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(Intuitively, one can think Force = Pressure â Area .)

of

this

expression

as

T is the Thomson cross-section given by 8 T = ----- ------------------------- 3 4 m c 2 0e e2 2
2 8r0 ­ 29 = ----------- = 6.65 â10 m 2 (3.1-2) 3

where the electron radius,
­ 15 e2 r 0 = ------------------------- = 2.818 â10 m 4 0 m e c 2

(3.1-3)

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Calculation of momentum flux density The momentum flux density per unit frequency is given by: 1 P = -c and the flux density is F =
source

source



I cos2 d

(3.1-4)

I cos d

(3.1-5)

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For distances large compared to the dimensions of the source
r Source

F I I F P ---- = -----c c

(3.1-6)

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For an isotropic emitter the total luminosity is spread over the radius of a sphere at the distance r . Hence, L F â 4 r 2 = L F = ----------4r2 L L P = -------------- P rad = -------------4r2c 4r2c

(3.1-7)

where P rad and L both involve integrations over the frequency, . Hence, the force on an electron is L T F = -------------4r2c
High Energy Astrophysics: Accretion Disks II

(3.1-8)

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Now consider a hydrogen plasma consisting of electrons and protons in the gravitational field of an object of mass M . Consider the nett outward force exerted on each electron-proton pair. The radiation force is primarily exerted on the electron but the gravitational force is primarily exerted on the proton. However, the two cannot move apart since this would result in a large charge separation. The nett force on the electron-proton pair is: L T GM m p F nett = -------------- ­ ----------------4r2c r2
(3.1-9)

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If the above nett force is greater than zero, accretion cannot occur. Thus, for accretion, L T GM m p -------------- ­ ----------------- < r2 4r2c 4 GM m p c 31 L < --------------------------- = 1.3 â10 T The parameter 4 GMcm p L edd = -------------------------T
(3.1-11)

0
(3.1-10)

M ------------------------- W Solar mass

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is known as the Eddington luminosity. Although this limit has been derived here for the case of spherical accretion, this limit is an important benchmark in all accretion scenarios. This limit was originally derived by Eddington in the context of stars. For compact objects of order a solar mass in size, the Eddington luminosity is of order 10 31 W . For black holes of order 10 9 solar masses, the Eddington luminosity is of order 10 40 W . It is therefore not surprising that these luminosities represent the upper limits of what is normally observed in these environments.

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3.2 The Eddington accretion rate Consider now the case where the luminosity of the central source is actually derived from accretion. As we have seen for an accretion disc, we can represent the total luminosity in the form: L = M ac2
(3.2-1)

where 0.1 for a black hole. Hence, in order to satisfy the Eddington constraint: 4 GM m p c M a c 2 < --------------------------T 4 GM m p M a < ­ 1 -----------------------c T
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(3.2-2)

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The parameter 4 GM m p 14 M M edd = ------------------------ = 1.4 â10 --------------- kg s ­ 1 M c T solar
­9 M = 2.2 â10 --------------- M solar yr ­ 1 M solar

(3.2-3)

Thus, 10 14 kg s ­ 1 is the maximum sort of accretion rate that one expects in a solar mass sized object. For an object with a size of order 10 9 solar masses, one expects accretion rates up to about a solar mass per year.

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4 The viscosity prescription ­ the parameter 4.1 General approach A large number of the relationships we have derived have been without a specific prescription for the Reynolds stress. All of our results have been expressed in terms of the accretion rate. This is typical of what we often do when modelling turbulent flows, e.g. jets. However, to derive other information e.g. the inflow ve~ locity V we need to be more prescriptive about the turbulent
r

model.

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As we indicated earlier the closure relations for turbulence are difficult to obtain. Shakura & Sunyaev in their classic paper on accretion discs lumped all of the unknowns into one simple equation: v r v P Hence v 2 v 2 --------------- -----2 P cs
(4.1-2) (4.1-1)

Normally, in a turbulent flow, the turbulent velocity is less than the sound speed. There are 2 reasons for this: 1. In a turbulent flow sound waves are emitted and the rate of emission goes up as the 8th power of the Mach number of the
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turbulence. This means that the emission of these waves would be a very effective dampener of supersonic turbulence. 2.Supersonic turbulence would very quickly form shocks. Therefore, we usually assume that <1 and frequently we find that «1
(4.1-4) (4.1-3)

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4.2 Use of the ­parameter to obtain the inflow velocity We already have: r0 1 / 2 2 v v dz = rM v 1 ­ ---- 2r aK r r ­ and using the prescription r0 1 / 2 h p 2 r 2 -- dz = rM a v K 1 ­ ---- r ­h That is, r0 1 / 2 ~ h kT 2 r 2 ---------- dz = rM a v K 1 ­ ---- r ­ h µ m p
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(4.2-1)

(4.2-2)

(4.2-3)
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Assuming isothermality: r0 1 / 2 ~ kT 2 ---------- = rM v 1 ­ ---- 2r aK µm r p r0 1 / 2 kT ­ 1 2 r = ­ 1 ---------M a v K 1 ­ ---- µm r p Now the mass accretion rate is = ­ 2 r h v dz = ­ 2 rv ~ ~ Ma r ­h r Ma ~ v r = ­ -----------2r
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(4.2-4)

(4.2-5)

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Therefore: r 0 1 / 2 ­1 kT ~ = ---------- v ­ 1 1 ­ ---- -K vr µm r p
2 as r 0 1 / 2 ­1 = ----- 1 ­ ---- r vK

(4.2-6)

r 0 1 / 2 ­1 ­ = a s M K1 1 ­ ---- r Hence, for < 1 the inflow velocity is much less than the sound speed.

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5 The effect of magnetic fields The influence of magnetic fields on accretion disc dynamics is very much a current research topic. Let me just note the equation for the angular momentum of the disc (the --component of the momentum equations) which reads
12 d~ M a rv + -- r B r B ­ 2 r 2 v r v dz ­ 2 dr

(5.0-1)

=0 The term B r B are related to the flux of the z --component of angular momentum in the radial direction by the magnetic field.

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It is important to consider the magnetic field since it is impossible to make an unmagnetised Keplerian disc turbulent. However, it has been shown by Balbus and Hawley that a weak magnetic field produces instability. Hence, the present focus on using magnetic fields to transport angular momentum (cf. the case for winds). The B r and B components which are responsible for this are supposed to be generated by the Balbus-Hawley instabilities. If this is the case then the dissipation in the disc will be through reconnection of magnetic fields. Much of the analysis that we have carried out here for unmagnetised disks carries through for magnetised disks.

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