Dynamics of astrophysical discs

Lecture 2

FDEPS, Kyoto

29.11.11

NASA - In Saturn's Shadow http://www.
nasa.gov/mission_pages/cassini/
multimedia/pia08329.html

Gordon Ogilvie
DAMTP, University of Cambridge
2011年12月2日金曜日

gio10@cam.ac.uk

Evolution of an accretion disc

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FDEPS, Kyoto© GIO 2011

● Regulated by conservation of mass and angular momentum● Mass conservation in 3D:∂ρ+ ∇ · (ρu) = 0∂t● 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

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Evolution of an accretion disc
● Write in cylindrical polar coordinates (r, φ, z)● Reduce from 3D to 1D● Mass conservation:1 ∂∂∂ρ 1 ∂+(rρur ) +(ρuφ ) +(ρuz ) = 0
∂t
r ∂r
r ∂φ
∂z

● Integrater

� ∞ � 2π−∞0

· dφ dzover cylinder of radius r

∂M ∂F
⇒
+
=0
∂t
∂r

assuming no vertical mass loss or gain
� ∞ � 2πρ r dφ dz● 1D mass density M(r, t) =● Radial mass flux F(r, t) =2011年12月2日金曜日−∞ 0� ∞ � 2π−∞0

ρur r dφ dz

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Evolution of an accretion disc
● Angular momentum conservation:(r, φ, z)

1 ∂ 2
1 ∂
∂
∂
2
(ρuφ ) + 2 [r (ρur uφ − Trφ )] +(ρuφ − Tφφ ) + (ρuφ uz − Tφz ) = 0
∂t
r ∂r
r ∂φ
∂z

● Integrater2

� ∞ � 2π−∞0

· dφ dz over cylinder of radius r● Assume that ruφ = h(r) from orbital dynamics(examine this assumption later)∂∂⇒(Mh) +(Fh + G) = 0
assuming no vertical loss or gain∂t∂r
� ∞ � 2π
2
G(r,
t)
=
−
r
Trφ dφ dz● Internal torque−∞2011年12月2日金曜日

0Evolution of an accretion disc
● Since h depends only on r :
∂M ∂F
+
=0
∂t
∂r
∂
∂M
h+
(Fh + G) = 0∂t∂r● Eliminate M :

dh ∂G
+
=0
F
which determines F
dr
∂r
�� �
�
−1
∂
∂M
∂G
dh
=
∂t
∂r
dr
∂r

● Interpretation:● angular momentum determines orbital radius● angular momentum transport determines mass evolution
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Evolution of an accretion disc
● More usual notation:
M = 2πrΣF = 2πrΣūr

G = −2πν̄Σr

Σ(r, t)3 dΩ

dr● Equivalent to:� ∞Σ=�ρ� dz−∞Σūr =

� ∞

−∞

ν̄Σ =

� ∞

−∞

2011年12月2日金曜日

surface densityūr (r, t) mean radial velocityν̄(r, t)mean effective
kinematic viscosity

1
�·� =
2π

� 2π
0

· dφ

�ρur � dz
�µ� dzdΩTrφ = µrdr2-06
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Evolution of an accretion disc
● Then obtain:
1 ∂
∂Σ
=
∂t
r ∂r

��

dh
dr

�−1

∂
∂r

�

−ν̄Σr

3 dΩ

dr

��

● Keplerian disc:Ω ∝ r−3/2

h = r2 Ω ∝ r1/2
�
�
3 ∂
∂Σ
1/2 ∂
=
r
(r1/2 ν̄Σ)
∂t
r ∂r
∂r

2011年12月2日金曜日

diffusion equation
for surface density

Evolution of an accretion disc

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● Interpretation:
angular momentum
transport

dΩ
<0
dr

orbital shear

dh
>0
dr

“viscous” spreading

2011年12月2日金曜日

Analysis of the diffusion equation

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● Inner boundary condition:● Important because spreading ring will reach r = 0 in finite time
● Depends on nature of central object● 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 tostellar in a viscous boundary layer

2011年12月2日金曜日

Analysis of the diffusion equation

2-09
FDEPS, Kyoto© GIO 2011dΩ
= 0 at r = rin (≈ R∗ ) so G = 0 there● Usual argument:dr

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Analysis of the diffusion equation● Black hole:● Circular orbits are unstable close to event horizon

Non-rotating black hole:
�1/2
�
GM
Ω=
r3
�
�
6 GM
2
2
κ =Ω 1− 2c rOrbits unstable for

6 GMr<= 3 rS2c● Rapid inspiral from rms● ūr ↑ and Σ ↓ rapidly, so effectively G ≈ 0 at rin ≈ rms
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Analysis of the diffusion equation● Star with significant magnetic field (e.g. dipole):

magnetosphericradius rm

polar accretion(observed)

discmagnetosphericcavity● G not necessarily 0 at rin ≈ rm
● Star can exert torque on disc

● Depends on magnetic coupling (complicated problem)
2011年12月2日金曜日

Analysis of the diffusion equation● Mathematically, we may let rin → 0● To allow mass flux at origin but no torque:F → cstG→0}

as r → 0● To allow torque at origin:
G → cstas r → 0● For a Keplerian disc,F ∝ rΣūr ∝ r
G ∝ r1/2 ν̄Σ● First case (no torque):

1/2 ∂

∂r

(r1/2 ν̄Σ)ν̄Σ → cst as r → 0● Second case (torque): r1/2 ν̄Σ → cst as r → 0
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Analysis of the diffusion equation

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● Consideration of angular momentum transport processes
and local vertical structure of disc (later) leads toν̄ = ν̄(r, Σ)(e.g. double power law)

● If ν̄ = ν̄(r) only, diffusion equation is linear● If ν̄ = ν̄(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 Σ0 (r) is� ∞Σ(r, t) =G(r, s, t) Σ0 (s) ds
0

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Analysis of the diffusion equation● Can calculate G(r, s, t) in terms of Bessel functions for anypν̄∝rpower lawand any boundary conditions(Lynden-Bell & Pringle 1974)● Classic example:ν̄ = cst

G = 0 inner BC at r = 0

modified Bessel function

no outer boundaryG(r, s, t) =r

−1/4 5/4

s
6ν̄t

�

r +s
exp −
12ν̄t
2

2

�

I1/4

� rs �
6ν̄t(omit derivation)● Becomes more asymmetrical as t increases● As t → ∞, all mass accreted by central object
and all angular momentum carried to r = ∞ by negligible mass
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Analysis of the diffusion equation● Easier example: suppose ν̄ = Ar with A = const :� �1/24r● Let y =and g = r1/2 ν̄Σ = Ar3/2 Σ to obtain
3A
∂g
∂2g
=
∂t
∂y 2

classical diffusion equation● Spreading-ring solution with zero torque ( g = 0 ) at centre:
�
�
��
�
�
2
2
(y + y0 )(y − y0 )−1/2− exp −
g∝t
exp −
4t
4t
● Thus

�

�

�

�

r
(r + s)
t
2(rs)
sinh
G(r, s, t) =
exp −
1/2
3At
3At
(3πA)
�
��
� ∞
�
s 1/2
G(r, s, t) r dr = erf● Remaining mass ∝3At
0−3/2 −1/2

2011年12月2日金曜日

1/2

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Analysis of the diffusion equation
�

�

�

r
(r + s)
t
2(rs)
sinh
G(r, s, t) =
exp −
1/2
3At
3At
(3πA)
−3/2 −1/2

1/2

�

10

8

t = 0.0003 s/A
6t = 0.1 s/A
4t = 0.001 s/A

t = 0.03 s/A

t = 0.003 s/A
t = 0.01 s/A

20.0

2011年12月2日金曜日

0.5

1.0

1.5

2.0

r/s

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Analysis of the diffusion equation
�

�

�

r
(r + s)
t
2(rs)
sinh
G(r, s, t) =
exp −
1/2
3At
3At
(3πA)
−3/2 −1/2

1.0

1.0

0.8

0.8

0.6

0.6

0.4

0.4

0.2

0.2

0.5

1.0

1.5

2.0

2.5

3.0

0.0

0.5

r/s
t = {0.01, 0.03, 0.1, 0.3, 1} s/A
2011年12月2日金曜日

�

Gr3/2 ∝ ang mom / unit radiusGr ∝ mass / unit radius

0.0

1/2

1.0

1.5

2.0

2.5

3.0

r/s

Analysis of the diffusion equation
p q
ν̄Σ
∝
r
Σ and rin → 0 :● Nonlinear case with

● No intrinsic length-scale● Special algebraic similarity solutions● Generally attract solutions of initial-value problem● But if q < 0 , viscous instability occurs (see later)

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Analysis of the diffusion equation● Example: ν̄ = Ar2 Σ● Two similarity solutions of the form
�a
��
�
R(t)
Σ = σ(t)−1 ,r ≤ R(t)r● Solution 1:

1
σ=
15At

a=1

R = Ct2/5(R − 5r)ūr = −
10t
r = R − ct

1/2

a = 5/4

1
σ=
15At

R = Ct1/2
r
ūr =
2t

(trajectories)● Accretion disc: almost all
mass accreted in finite time
(no torque at centre)
2011年12月2日金曜日● Solution 2:

r = ct1/2● Decretion disc: almost all
mass expelledrequires torque at centre)

Analysis of the diffusion equation

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● Steady discs
F = cst = −Ṁ

Ṁ = mass accretion rate
(but always neglect slow increase of M )

dh dG
+
=0
F
dr
dr

⇒

−Ṁ h + G = cst● If G = 0 at r = rin , solution is G = Ṁ (h − hin )3 dΩ● In Keplerian case (recall G = −2πν̄Σr
)
dr
�
�
�
�
rin 1/2
Ṁ
1−
ν̄Σ =
3π
r
● If we know ν̄(r, Σ) , we can solve for Σ(r)2011年12月2日金曜日

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Analysis of the diffusion equation● Other solutions:● Non-accreting disc:● Decretion disc:F = 0, G = cst

F >0

}

require torquefrom central object

dG>0● Note that F < 0 requires
dr● Energy balance in steady disc:

�

dΩ
dΩ
=µ r
● Energy dissipation rate per unit volume = Trφ rdrdr● Energy emission rate per unit area (from each face of disc)
�
�2
dΩ
1
= ν̄Σ r2
dr
● Effective blackbody temperature Teﬀ (r) given by
�
�
�
�
1/2
9
3GM
Ṁ
r
in
4
1−
σTeﬀ
= ν̄ΣΩ2 =3
88πrr
2011年12月2日金曜日

�2

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Analysis of the diffusion equation● Typical (theoretical) spectrum of emitted radiation:
inner discouter disc2011年12月2日金曜日

Analysis of the diffusion equation

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● Total luminosity of disc (rin < r < ∞)
�
�
� ∞
�
�
1 GM Ṁ
3GM Ṁ
rin 1/21−
2πr dr =Ldisc =
3
8πr
r
2 rin
rin
= rate of release of orbital binding energy (∞ → rin )1= rate of release of potential energy2(remaining kinetic energy released in boundary layer
if Ω∗ = 0 )2011年12月2日金曜日

Vertical disc structure

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● Vertical hydrostatic equilibrium● Dominant balance (for non-self-gravitating gas disc)∂Φ ∂p−
0 = −ρ
∂z∂z● Expand potential1Φ(r, z) = Φ (r, 0) + Φzz (r, 0)z 2 + · · ·2● So for a thin disc∂Φ( Ωz = Ω for Keplerian disc)≈ Ω2z z∂z
● Equation of vertical hydrostatic equilibrium∂p≈ −ρΩ2z z∂z

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Vertical disc structurer

H● Disc supported against gravity by pressure in vertical direction● Centrifugal support in radial direction

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Vertical disc structure● Order-of-magnitude estimates and time-scales● Simple scaling arguments ( ∼ ), omitting factors of order unity● Important dimensionless parameter:
Hr● For a thin disc:(angular semi-thickness, aspect ratio)H�1r● H may represent:{true semi-thickness
height of photosphere above midplanemeasure of density scale-height● From vertical hydrostatic equilibrium:p∼ ρΩ2 HH
2011年12月2日金曜日⇒

cs ∼ ΩH� �1/2pcs =ρ

isothermalsound speed

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Vertical disc structure● (Effective) viscosity:● Dimensional argument: [µ] = M L−1 T −1 =

�p�
Ω

αp
● Write µ =in terms of dimensionless viscosity αΩ(“alpha viscosity prescription”, Shakura & Sunyaev 1973)

αc2sµ∼ αcs H● (Mean) kinematic viscosity ν̄ ∼ ∼ρΩ● Analogous to kinetic theory of molecular transport:
ν ∼ v� for mean speed v and mean free path �● Molecular viscosity usually negligible for astrophysical discs● Similar estimate applies to effective “eddy viscosity” of turbulenceof mean turbulent speed v and correlation length �● Assuming that v � cs (subsonic turbulence) and � � H ,expect that α � 1
2011年12月2日金曜日

Vertical disc structure

2-28
FDEPS, Kyoto© GIO 20113dΩ
= − αp● Stress in a Keplerian disc: Trφ = µr
dr2

● Alternative version of alpha prescription: Trφ = −αp● Whatever process gives rise to stress, scales with local pressure● Reasonable assumption for local processes (see later)

2011年12月2日金曜日

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Vertical disc structure● Three important characteristic time-scales:● Dynamical time-scale:1tdyn ∼Ω(orbital motion)● Viscous time-scale:
2

r
∼ α−1
tvisc ∼
ν̄

�

H
r

�−2

tdyn

H
∼
(hydrostatic)
cs(radial motion)● Thermal time-scale:
pH
c2s
H2
−1
∼
α
tth ∼
∼
∼
tdyn
2
2
ν̄ΣΩ
ν̄Ω
ν̄

● For a thin disc with α < 1 :tdyn < tth � tvisc(thermal balance)● All three time-scales usually increase rapidly with r

2011年12月2日金曜日

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Vertical disc structure● Mach number of orbital motion:
rΩ
Ma ∼
∼
cs

�

H
r

�−1

● Characteristic radial velocity due to viscosity:
� �
ν̄
H
cs
|ūr | ∼ ∼ αrr● For a thin disc,|ūr | � cs � rΩ

orbital motion: highly supersonicaccretion flow: highly subsonic

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Vertical disc structure● Corrections to orbital motion:● Contribution of radial pressure gradient:
� �2
�
2
cs
H
∂p
2
ρrΩ ∼ 2 2 ∼∂rr Ωr● Vertical variations of radial gravitational acceleration ∼∂ur● Other terms, e.g. ρur, are smaller∂r�� �2 �
H● Conclude that uφ = rΩ 1 + O
r

�

H
r

i.e. fluid velocity close to orbital velocity of test particle● In general, thin-disc approximations involve fractional errors O
and a formal asymptotic treatment is possible
2011年12月2日金曜日

�2

�

H
r

�2

Vertical disc structure
Typical values of H/r :● protoplanetary discs : 0.05 – 0.1● binary stars : 0.01 – 0.02
● active galactic nuclei : 0.001● planetary rings : 0.0000001

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Vertical disc structure● Barotropic models:● Vertical hydrostatic equilibrium:∂p= −ρΩ2z z∂z● Can be solved if pressure is a known function of density(physical arguments, or just for analytical convenience)● Vertically isothermal model:2p = cs ρ

cs independent of z

● Solution:−z 2 /2H 2

ρ(r, z) = ρ0 (r, 0) e

cs
H=
Ωz

2011年12月2日金曜日

isothermal scale height

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Vertical disc structure

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● Polytropic model:
p = Kρ1+1/n

K, n independent of z

● Introduce (pseudo-)enthalpy:
�dp= (n + 1)Kρ1/nw=
ρ∂w= −Ω2z z∂z
1 2 2
H = true semi-thickness● Solution: w = Ωz (H − z 2 )
2
�
�n
2
z
ρ(r, z) = ρ(r, 0) 1 − 2(vacuum above z = H )
H
�n+1
�
2
z
p(r, z) = p(r, 0) 1 − 2H
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Vertical disc structure● Radiative models:● Energy dissipated by viscosity carried away by radiation● Radiative diffusion (optically thick disc):
16σT 3∇TF =−3κρκ = Rosseland mean opacity● Dominant balance in thermal energy equation (thin disc):
�2
�
dΩ
∂Fz
0=µ r
−
dr
∂z
● Contributions from Fr and from radial advection are smaller byO(H/r)2

2011年12月2日金曜日

Vertical disc structure

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● Vertical structure of radiative Keplerian disc:∂p= −ρΩ2 z
∂z
9 2
∂F
= µΩ
∂z
4
16σT 3 ∂TF =−3κρ ∂ztogether with:

kρT
4σT 4
+
● equation of state, e.g. p =(ideal gas + radiation)µm mp
3c● viscosity prescription● opacity function κ(ρ, T )● boundary conditions, e.g. ρ = p = T = 0 at z = H(or match to atmopsheric model)● Analogous to radial structure of a star
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Vertical disc structure● Opacity often approximated by a power law, e.g. for ionized discs:

κ = const ≈ 0.33 cm2 g−1

κ = Cκ ρT−7/2

Cκ ≈ 4.5 × 1024 cm5 g−2 K7/2

● Cooler discs: dust, molecules, ...

2011年12月2日金曜日

Thomson opacity
electron scattering
hotter regions
Kramers opacity
free-free / bound-freecooler regions

Vertical disc structure

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● Keplerian disc, alpha viscosity, gas pressure, Thomson opacity:∂p= −ρΩ2 z
∂z
9
∂F
= αpΩ
∂z
4
16σT 3 ∂TF =−3κρ ∂z

kρT
p=
µm mp
� H
ρ dz
Σ=
−H

2011年12月2日金曜日

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Vertical disc structure● Order-of-magnitude treatment:∂p= −ρΩ2 z∂zp∼ ρΩ2 H
H

9
∂F
= αpΩ
∂z
4
16σT 3 ∂TF =−3κρ ∂zF∼ αpΩH
σT 3 T
F ∼
κρ H

kρT
p=
µm mp
� H
ρ dz
Σ=

kρT
p∼
µm mp

−H

Σ ∼ ρH● Algebraic solution:� µ m �−2/3 � σ �−1/6m p
H ∼ α1/6 Σ1/3 Ω−5/6kκ
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Vertical disc structure

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● Viscosity:
� µ m �−4/3 � σ �−1/3m pν̄ ∼ αcs H ∼ αΩH 2 ∼ α4/3 Σ2/3 Ω−2/3kκ● Thus ν̄ ∝ rΣ2/3● Full treatment:● Use order-of-magnitude treatment as a dimensional analysis● It defines characteristic units UH , Uρ , Up , UT , etc.
● Then write z = UH z̃,ρ = Uρ ρ̃(z̃), etc. to obtain a system ofdimensionless ODEs with no free parameters● Solve numerically
2/3
ν̄
=
ArΣ
● Findwith a precise coefficient A● Power-law constitutive relations give power-law viscosity
2011年12月2日金曜日

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Vertical disc structure
pressuredensitysimilar topolytropic modelsρ ∝ (H 2 − z 2 )np ∝ (H 2 − z 2 )n+1
temperature

2011年12月2日金曜日

radiative flux

T ∝ (H 2 − z 2 )Vertical disc structure

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● A curious result:● Thomson opacity, radiation pressure-dominated (near black holes)∂p= −ρΩ2 z
∂z
9 2
∂F
= µΩ
∂z
4
16σT 3 ∂TF =−3κρ ∂z

4σT 4
p=
3c

● A solution is possible for only one value of the effective viscosity:4c≈ 4.0 × 1010 g cm−1 s−1
µ=
9κ
−3
−1 −1
µ
≈
9.0
×
10
g
cm
s )
(cf. water:
2011年12月2日金曜日

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Steady alpha discs● Steady Keplerian disc:
f Ṁ
,
ν̄Σ =
3π

f =1−

� r �1/2
inr● Combine with treatment of vertical structure, e.g. Thomson opacityand gas pressure:� µ m �−2/3 � σ �−1/6m p
H ∼ α1/6 Σ1/3 Ω−5/6kκ
� µ m �−4/3 � σ �−1/3m pν̄ ∼ αcs H ∼ αΩH 2 ∼ α4/3 Σ2/3 Ω−2/3kκ● Deduce that (insensitive to poorly known value of α )� µ m �−2/5 � σ �−1/10
m p−1/10 −7/10
1/5
H∼α
Ω
(f Ṁ )kκ
● Very slightly flared:H∝ f 1/5 r1/20
r
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Thermal and viscous stability● Viscous stability:● Nonlinear diffusion equation for Keplerian disc:
�
�
3 ∂
∂Σ
1/2 ∂
ν̄ = ν̄(r, Σ)=r(r1/2 ν̄Σ)
∂t
r ∂r
∂r
● Linearize about any given solution Σ0 (r, t) :Σ(r, t) = Σ0 (r, t) + Σ� (r, t)|Σ� | � Σ0∂(ν̄Σ) �Σ = qν̄Σ�(ν̄Σ) =
∂Σ∂ ln(ν̄Σ)
q=
∂ lnΣ

�

● Linearized diffusion equation:
�
�
�
3 ∂
∂Σ
1/2 ∂
=
r
(r1/2 qν̄Σ� )
∂t
r ∂r
∂r

● Unstable for q < 0 : rapid growth on short length-scales
2011年12月2日金曜日

Thermal and viscous stability

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● Thermal stability:● So far, assumed a balance between heating and cooling:9ν̄ΣΩ2 = H = C = 2F +4
● Relax this assumption, but assume that α � 1 so thattdyn � tth � tvisc● Consider behaviour on the timescale tth :● disc is hydrostatic● surface density does not evolve● By solving equations of vertical structure except thermal balance,
can calculate H and C as functions of (Σ, ν̄Σ)
● In fact H depends only on ν̄Σ● Equation of thermal balance defines a curve in the (Σ, ν̄Σ) plane2011年12月2日金曜日

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Thermal and viscous stability● Infinitesimal perturbations:dHd(ν̄Σ)dH =d(ν̄Σ)∂C∂CdΣ+d(ν̄Σ)dC =∂Σ∂(ν̄Σ)● Along the equilibrium curve, dH = dC and d(ν̄Σ) = qν̄ dΣ
1 ∂C
∂C
dH
=
+
⇒
d(ν̄Σ)qν̄ ∂Σ ∂(ν̄Σ)∂ ln(ν̄Σ)q=∂ lnΣ

● Thermal energy content of disc per unit area ∼ pH ∼ (Ω/α)ν̄Σ● If some heat is added, ν̄Σ increases but Σ is fixed on tth● Unstable if excess heating exceeds excess cooling, i.e. if
1 ∂C
dC
dH
>0
>
i.e.
qν̄ ∂Σ
d(ν̄Σ)d(ν̄Σ)● In practice ∂C/∂Σ < 0 (because, at fixed ν̄Σ , Σ ∝ 1/ν̄ ∝ 1/(αT ) )so thermal instability occurs (like viscous instability) when q < 0
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Thermal and viscous stability● Outbursts:● Radiative disc with gas pressure and Thomson opacity hasν̄Σ ∝ rΣ5/3 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:
�
�
�
�
Ṁ
rin 1/2
Ṁ
≈
1−
ν̄Σ =
3π
r
3π

● Explains outbursts in many cataclysmic variables, X-ray binaries
and other systems
2011年12月2日金曜日

Observations: cataclysmic variable stars

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SS Cygni
dwarf nova
V magnitude(range 12-8)
1900-2010
aavso.org
Also UV andsoft X-rays

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Observations: X-ray binary stars (1960s-)
GRO J1655-40 (“Nova Sco 1994”)

1996 97
Cyg X-1

2011年12月2日金曜日

98

99

2000

01

02

03

04

05

06

07

08

reference• NASA - In Saturn's Shadow

http://www.nasa.gov/mission_pages/cassini/multimedia/pia08329.ht
ml• http://aavso.org• Lynden-Bell, D. and Pringle, J. E., 1974: The evolution ofviscous discs and the origin of the nebular variables., Mon.
Not. Roy. Astr. Soc., 168, 603-637• Shakura, N. I. and Sunyaev, R. A., 1973: Black holes inbinary systems: Observational appearance., A & A, 24, 337355