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 v
with respect to the accreting object.
The object is able to accrete only
particles with negative total energy,
Ekin + Epot < 0.
Assuming spherically symmetric
gravitional 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
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36
10
37
10
38
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34
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estimatedLX[ergs
-1
]
observed LX [erg s
-1
]
10
33
10
34
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35
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36
10
37
10
38
10
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10
34
10
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10
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estimatedLX[ergs
-1
]
observed LX [erg s
-1
]
10
33
10
34
10
35
10
36
10
37
10
38
10
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10
34
10
35
10
36
10
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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
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10
37
10
38
estimatedLX[ergs
-1
]
observed LX [erg s
-1
]
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 accretion disk from 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, 1990, 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