Stellar winds
Jiˇr´ı Krtiˇcka
Masaryk University
Introduction: why mass-loss?
Evidence for mass-loss: shells around stars
Abell 39 nebula
1
Evidence for mass-loss: shells around stars
nebula around WR 124
1
Evidence for mass-loss: shells around stars
nebula around Mira (o Cet)
1
Evidence for mass-loss: interstellar medium
NGC 3603 cluster
2
Evidence for mass-loss: heavy elements
After the period of primordial nucleosynthesys, the Universe was
composed mostly form H and He (with a tiny amount of heavier elements
like Li). Heavy elements (C, N, O, Fe, . . . ) were completely missing.
However, there are heavy elements around us. Where do they come
from? Heavier elements are synthethised during thermonuclear reactions
in the stellar interiors. How do they get into the interstellar medium? 3
How the mass can espape gravitational wells of stars?
Let us start with the momentum equation with gravity
ρ
∂v
∂t
+ ρv
∂v
∂r
= ρgw −
∂p
∂r
−
ρGM
r2
,
which can be simplified assuming stationary isothermal outflow (p = a2
ρ)
v
dv
dr
= gw −
a2
ρ
dρ
dr
−
GM
r2
,
where gw gives the force that drives the wind (force per unit of mass, i.e.,
the acceleration) and a is the isothermal sound speed.
We can integrate the equation from the stellar surface R∗ to infinity
∞
R∗
v
dv
dr
dr =
∞
R∗
gw dr −
∞
R∗
a2
ρ
dρ
dr
dr −
∞
R∗
GM
r2
dr
yielding
1
2
v2
∞ −
1
2
v2
0 =
∞
R∗
gw dr − a2
ln
ρ∞
ρ0
−
GM
R∗
.
4
Three ways
The individual terms in
1
2
v2
∞ −
1
2
v2
0 =
∞
R∗
gw dr − a2
ln
ρ∞
ρ0
−
GM
R∗
describe (from left to right) change of the kinetic energy (per unit of
mass), work of drivin forces, work of pressure force, and the potential
energy (per unit of mass).
There are three ways to initiate the outflow. Either the initial velocity is
larger than the escape speed vesc,
1
2
v2
0 ≥
GM
R∗
, v0 ≥ vesc =
2GM
R∗
, vesc = 620 kms−1 M
M⊙
R⊙
R
.
The initial kinetic energy of the flow should be larger than the absolute
value of the potential energy. This is fullfilled for explosive outflows like
supernovae or supernova impostors (e.g., η Car).
5
Three ways
The individual terms in
1
2
v2
∞ −
1
2
v2
0 =
∞
R∗
gw dr − a2
ln
ρ∞
ρ0
−
GM
R∗
describe (from left to right) change of the kinetic energy (per unit of
mass), work of drivin forces, work of pressure force, and the potential
energy (per unit of mass).
The other possibility is that the driving force is large enough
∞
R∗
gw dr ≥
GM
R∗
.
This is true for winds driven radiatively, either due to the absorption in
lines (in hot stars) or on dust particles (in luminous cool stars).
5
Three ways
The individual terms in
1
2
v2
∞ −
1
2
v2
0 =
∞
R∗
gw dr − a2
ln
ρ∞
ρ0
−
GM
R∗
describe (from left to right) change of the kinetic energy (per unit of
mass), work of drivin forces, work of pressure force, and the potential
energy (per unit of mass).
The last possibility is that the work done by pressure forces is large,
a2
ln
ρ0
ρ∞
≥
GM
R∗
.
Because ln(ρ0/ρ∞) is of the order of ten at most, this implies that
a ≈ vesc. This happens in coronal winds of cool main-sequence stars.
5
Coronal winds
Is there any evidence for the wind of our Sun?
two types of the comet tails (Biermann 1951)
6
Is there any evidence for the wind of our Sun?
aurorae
6
Is there any evidence for the wind of our Sun?
satellite observations
flux of particles streaming from our Sun (protons, electrons, He, . . . )
speed about ∼ 500 km s−1
number density (r = 1 a.u.) ∼ 107
particles m−3
mass-loss rate
˙M = 4πr2
ρv ≈ 2 × 10−14
M⊙ yr−1
6
Thermally driven wind
We have seen that the spherically symmetric atmosphere cannot be in
hydrostatic equilibrium. However, the root mean square of the total
velocity of particles vth = 3kT
mH
in the atmosphere with T = 6000 K is
about vth = 12 kms−1
, which is significantly lower than the escape speed
vesc = 620 kms−1
.
7
Thermally driven wind
The Sun has a large outer
atmosphere called corona. The
corona can be in optical light
observed only during the solar
eclipses or using satellites. The
detection of lines of highly ionized
atoms (Ca XII, Fe XIII, Ni XVI,. . . ,
”coronium”, Grotrian 1939, Edl´en
1942) shows that the temperature of the solar corona is about
105
− 106
K. Corresponding root mean square of the total velocity of the
order of 100 kms−1
is comparable with the escape speed. Consequently,
the thermal expansion of the solar corona is thought to be the source of
the solar wind (Parker 1958).
This coins the term coronal wind.
7
Parker model of the coronal wind
Let us assume that the coronal wind wind can be described as a
spherically symmetric, stationary, and isothermal outlow. Then the
corresponding hydrodynamical equations are
1
r2
d
dr
r2
ρv = 0,
ρv
dv
dr
= −a2 dρ
dr
−
ρGM
r2
.
The integration of the continuity equation gives the mass-loss rate
˙M ≡ 4πr2
ρv = const.
Inserting dρ/dr from the continuity equation into the equation of motion
gives ordinary differential equation for velocity
1
v
v2
− a2 dv
dr
=
2a2
r
−
GM
r2
,
which can be solved analytically.
8
Parker model of the coronal wind
We shall study the momentum equation
1
v
v2
− a2 dv
dr
=
2a2
r
−
GM
r2
in more detail.
At radius r = rc given by 2a2
/rc = GM/r2
c the right-hand side of
momentum equation is equal to zero. This implies either v = a or
dv/dr = 0.
At the sonic point (v = a) either r = rc or dv/dr → ∞. At the sonic
point the sound speed is equal to half of the escape speed.
9
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 continuous solutions describing outflow (wind)
and inflow (accretion). There is one outflow solution that is supersonic at
large distances from the Sun (wind). Other outflow solutions are subsonic
(”breeze”). Observations show supersonic flow at the location of Earth. 10
Nature of coronal heating: unresoloved problem
The nature of coronal heating is one of the most important open
problems in astrophysics. It is most likley related to a deep hydrogen
convective zone in cool stars. The convection magnifies seed magnetic
fields. Sound waves generated by the convection interact with coronal
magnetic field and create MHD waves. Hybrid MHD waves or Alfv´en
waves possibly heat the corona.
11
Coronal winds: importance
Coronal winds of main-sequence stars are weak and do not influence
stellar evolution significantly. In earlier phases, they might be responsible
for evolution of interplanetary medium. Coronal winds are important for
the interaction of stars with exoplanets and for rotational braking of
main-sequence stars.
In cool gaints, the coronal winds are important for the mass-loss
(Cranmer & Saar 2011, Suzuki 2013). It is expected that our Sun will
lose a fraction of its mass by this mechanism (about 0.2).
12
Dust-driven winds
Evidence for wind in cool luminous stars
envelopes around stellar remnants
planetary nebula Abell 39
13
Evidence for wind in cool luminous stars
envelopes around stellar remnants
Cat’s Eye Nebula – NGC 6543 (HST)
13
Evidence for wind in cool luminous stars
envelopes around stellar remnants
nebula around Mira (o Cet)
13
Driving the wind of cool luminous stars
We have seen that there are observational indications of wind in cool
luminous stars (AGB stars, red supergiants). These winds are likely not
connected with coronae, due to missing strong X-ray emission and
chromospheric activity. On the other hand, these stars are luminous,
consequently radiative force is capable to drive a wind in these stars. As
a result of their low temperature, dust grains form in the envelopes of
cool luminous stars. Dust grain absorb the stelar radiation and accelerate
the wind giving rise to dust driven winds.
14
Dust driven wind equations
Again, let us assume that the dust driven winds can be described as a
spherically symmetric, stationary, and isothermal outlow. Then the
corresponding hydrodynamical equations are
1
r2
d
dr
r2
ρv = 0,
ρv
dv
dr
= −a2 dρ
dr
−
ρGM
r2
+ ρgrad,
where grad is the radiative force.
The integration of the continuity equation gives the mass-loss rate
˙M ≡ 4πr2
ρv = const.
Inserting dρ/dr from the continuity equation into the equation of motion
gives ordinary differential equation for velocity
1
v
v2
− a2 dv
dr
=
2a2
r
−
GM
r2
+ grad.
In cool dust driven winds, one can neglect 2a2
r ≪ GM
r2 . 15
Sonic point condition
From the momentum equation
1
v
v2
− a2 dv
dr
= grad −
GM
r2
follows that at the sonic point v = a the radiative force equals the gravity
(in magnitude)
grad =
GM
r2
.
The gravity is stronger than the radiative force in subsonic region v < a
grad <
GM
r2
,
while the radiation dominates in the supersonic region v > a
grad >
GM
r2
.
16
Radiative force due to the dust
The radiative force due to absorption of radiation on dust particles is
frad = ρgrad =
1
c
∞
0
χ(r, ν)F(r, ν) dν,
where χ(r, ν) is the absorption coefficient and F(r, ν) is the radiative
flux. Because κ(r, ν) = χ(r, ν)/ρ(r) varies only due to a change of the
dust fraction and F(r, ν)/F(r) varies mostly with frequency, the radiative
acceleration is
grad =
1
c
κ(r)F(r),
where the mean opacity is
κ(r) =
∞
0
κ(r, ν)
F(r, ν)
F(r)
dν
and F(r) is the integrated flux, F(r) = L/(4πr2
).
17
Eddington parameter due to dust
Introducing the ratio of the radiative and gravity acceleration,
Γd(r) =
κ(r)L
4πcGM
,
the momentum equation can be rewritten as
1
v
v2
− a2 dv
dr
= −
GM
r2
(1 − Γd(r)) .
There are three regions of the wind:
subsonic region, v < a, Γd(r) < 1,
sonic point, v = a, Γd(r) = 1 (formation of the dust),
supersonic region v > a, Γd(r) > 1.
18
Estimating the mass-loss rate
Let us start with the momentum equation
ρv
dv
dr
= −
dp
dr
−
ρGM
r2
+ Γd
ρGM
r2
and let us assume that dust starts to form close to the sonic point rc,
where Γd changes from Γd ≪ 1 to Γd ≫ 1. We shall multiply the
momentum equation by 4πr2
and integrate the equation from R∗ to ∞
∞
R∗
4πr2
ρv
dv
dr
dr +
∞
R∗
4πr2 dp
dr
+
ρGM
r2
dr =
∞
R∗
4πr2
Γd
ρGM
r2
dr.
The term in the first integral can be taken out assuming constant
mass-loss rate ˙M = 4πr2
ρv. The integral can be evaluated assuming
hydrostatic equilibrium in the atmosphere v(R∗) ≈ 0 and using wind
terminal velocity v∞ = v(r = ∞). Therefore,
∞
R∗
4πr2
ρv
dv
dr
dr = ˙M [v]
∞
R∗
≈ ˙Mv∞.
19
Estimating the mass-loss rate
Let us start with the momentum equation
ρv
dv
dr
= −
dp
dr
−
ρGM
r2
+ Γd
ρGM
r2
and let us assume that dust starts to form close to the sonic point rc,
where Γd changes from Γd ≪ 1 to Γd ≫ 1. We shall multiply the
momentum equation by 4πr2
and integrate the equation from R∗ to ∞
∞
R∗
4πr2
ρv
dv
dr
dr +
∞
R∗
4πr2 dp
dr
+
ρGM
r2
dr =
∞
R∗
4πr2
Γd
ρGM
r2
dr.
The second integral gives the hydrostatic equilibrium density distribution
for r < rc, while the pressure gradient becomes neglibible for r < rc.
Therefore,
∞
R∗
4πr2 dp
dr
+
ρGM
r2
dr =
rc
R∗
4πr2 dp
dr
+
ρGM
r2
dr+
+
∞
rc
4πr2 dp
dr
+
ρGM
r2
dr ≈ 4πGM
∞
rc
ρ dr.
19
Estimating the mass-loss rate
Let us start with the momentum equation
ρv
dv
dr
= −
dp
dr
−
ρGM
r2
+ Γd
ρGM
r2
and let us assume that dust starts to form close to the sonic point rc,
where Γd changes from Γd ≪ 1 to Γd ≫ 1. We shall multiply the
momentum equation by 4πr2
and integrate the equation from R∗ to ∞
∞
R∗
4πr2
ρv
dv
dr
dr +
∞
R∗
4πr2 dp
dr
+
ρGM
r2
dr =
∞
R∗
4πr2
Γd
ρGM
r2
dr.
The right-hand side integral can be evaluated using wind optical depth
τW =
∞
rc
κ(r)ρ dr,
∞
R∗
4πr2
Γd
ρGM
r2
dr = 4πGM
∞
rc
Γdρ dr =
L
c
∞
rc
κ(r)ρ dr = τW
L
c
.
19
Estimating the mass-loss rate
Putting all the terms together we derive
˙Mv∞ = 4πGM
∞
rc
ρ dr + τW
L
c
.
For Γd ≫ 1 the gravity term can be neglected with respect to the
radiative acceleration and we derive the mass-loss rate estimate
˙Mv∞ = τW
L
c
.
We see that by assuming multiple scattering (τW > 1) the mass-loss rate
can be significantly higher than the single-scattering limit ˙Mv∞ = L/c
(Gail a Sedlmayr 1986, Netzer a Elitzur 1993).
20
Problem with too high condensation radii
The condition of radiative equilibrium on dust particles
∞
0
κλ Bλ(T) dλ =
∞
0
κλ Jλ dλ
predicts too high dust temperature in the atmosphere. The temperature
is higher than the condensation temperature Tc. The condensation may
appar only at larger distances from the star, which are higher than the
condensation radius, r > rc. This is a problem for radiative driving.
silicate graphite amorphous carbon
Tc = 1500 K Tc = 1500 K Tc = 1500 K
Teff rc/R∗ rc/R∗ rc/R∗
3000 2.99 4.03 3.42
2500 1.85 2.34 2.12
2000 1.15 1.29 1.24
(Lamers a Cassinelli 1999)
21
Solution of the problem with too high condensation radii
Luminous cool stars pulsate, consequently, pulsations may transfer the
stellar matter to a large distances from the star.
trajectories of pulsating particles without radiative acceleration on dust
particles (Bowen 1988) 22
Solution of the problem with too high condensation radii
Luminous cool stars pulsate, consequently, pulsations may transfer the
stellar matter to a large distances from the star.
trajectories of pulsating particles with radiative acceleration on dust
particles (Bowen 1988) 22
Wind radial velocity
Neglecting the gas pressure term, the momentum equation
v
dv
dr
= (Γd(r) − 1)
GM
r2
can be integrated. Substituting v dv
dr = 1
2
d(v2
)
dr , we have for the radial wind
velocity
v2
(r) = v2
(rc) + 2GM
r
rc
Γd − 1
r′2 dr′
.
The wind speed at the critical point can me typically neglected.
Moreover, one can assume that the Eddingtin parameter is constant
Γd = const. This yelds for the wind velocity ”beta” velocity law in the
form of
v(r) = v∞ 1 −
rc
r
with wind terminal velocity
v∞ = 2GM (Γd − 1)/rc
propotional to he escape speed.
23
Minimum stellar luminosity to drive a wind
As a necessary condition for the wind existence, the radiative force should
overcome the gravity,
Γd > 1.
Inserting the formula for the Eddington parameter,
κL
4πcGM
> 1,
this gives the condition for the stellar luminosity,
L >
4πcGM
κ
.
Assuming a typical mean opacity κ ≈ 30 cm2
g−1
, the minimum stellar
luminosity to drive a wind is (in scaled quantities)
L > 400 L⊙
M
1 M⊙
.
This means, that evolved solar-mass stars with L > 400 L⊙ may drive a
dust driven wind.
24
Dust driven winds: importance
Dust driven winds appear during the AGB phase of low-mass stars with
initial mass 0.4 M⊙ M0 8 M⊙. A typical wind mass-loss rate due to
dust driven wind is of the order of 10−7
M⊙ yr−1
− 10−6
M⊙ yr−1
. Given
a typical duration of AGB phase (which is of the order of 10 × 106
yr),
low-mass stars lose a significant amount of their mass via dust driven
winds.
Part of the material lost during the AGB phase is ionized in a subsequent
evolutionary phase and form a planetary nebula. Dust driven winds carry
freshly synthethised s-process elements, consequently AGB stars are an
important source of elements heavier than iron.
25
Line-driven winds of hot stars
Evidence for wind in hot stars
shells in the surroundings of hot stars
nebula close to the star WR 124 (HST)
26
Evidence for wind in hot stars
the interstellar medium around hot stars
open cluster NGC 3603 (HST)
26
Evidence for wind in hot stars
P Cyg line profiles in UV
Fλ[ergcm
-2
s
-1
A
-1
]
λ [A]
emission part
absorption part
C IV HD 163758
0
1 × 10
-11
2 × 10
-11
3 × 10
-11
4 × 10
-11
5 × 10
-11
6 × 10
-11
7 × 10
-11
1500 1510 1520 1530 1540 1550 1560 1570 1580
HD 163758 (HST)
26
Evidenceforwindinhotstars
X-rayemission
FeXXIV
FeXXIV
FeXVII
FeXIX
FeXXIII
FeXXIII
FeXXIV
FeXXIVFeXXIV
FeXVIII
FeXVIII
NeIX
FeXXIII
FeXXII
X-rayspectrumθ1
OriC
(CHANDRA,Schulzetal.2003)
26
Evidence for wind in hot stars
Hα emission line
relativeflux
λ [Å]
α Cam
0.9
1
1.1
1.2
1.3
1.4
6520 6540 6560 6580 6600
α Cam, 2m telescope in Ondˇrejov (Kub´at 2003)
26
Radiative driving
Hot stars are very luminous, consequently the radiative force can be
suspected to drive the wind. In a spherically symmetric case, the
radiative force is
frad =
1
c
∞
0
χ(r, ν)F(r, ν) dν,
where χ(r, ν) is the absorption coefficient and F(r, ν) the radiative flux.
27
Radiative driving: free electrons
The free electrons are the most numerous particles in the wind. In a
nonrelativistic limit, the opacity due to the light scattering on free
electrons is χ(r, ν) = σThne(r), where σTh is the Thomson scattering
cross-section and ne(r) is the electron density. Because the σTh is
frequency independent, the integral can be easily evaluated as
frad =
σThne(r)L
4πr2c
,
where L = 4πr2 ∞
0
F(r, ν) dν is the stellar luminosity. This can be easily
compared with the force due to gravity fgrav = ρ(r)GM/r2
to give the
Eddington parameter
Γ ≡
frad
fgrav
=
σT
ne(r)
ρ(r) L
4πcGM
, in scaled quantities, Γ ≈ 10−5 L
1 L⊙
M
1 M⊙
−1
.
For a typical stars Γ is roughly constant and Γ < 1, therefore free
electrons are not suitable to drive a wind.
28
Radiative driving: line transitions
Line (bound-bound) transitions provide another viable source of the
radiative force. The opacity due to the line transitions is
χ(r, ν) =
πe2
mec
lines
ϕij (ν)gi fij
ni (r)
gi
−
nj(r)
gj
,
where ϕij (ν) is the line profile normalized over frequencies,
∞
0 ϕij (ν) dν = 1, fij is the oscillator strength, and ni (r), nj(r) are level
occupation number with statistical weights gi , gj.
The radiative force due to the line transitions is then
fline =
πe2
mec2
∞
0 line
gi fij
ni (r)
gi
−
nj(r)
gj
ϕij (ν)F(r, ν) dν.
This cannot be easily evaluated due to the dependence of flux on
frequency (lines may be optically thick). Therefore, F(r, ν) should be
derived from the radiative transfer equation. This introduces coupling
with hydrodynamical equations.
29
Radiative driving: optically thin lines
A crude estimate of the radiative force (maximum) can be obtained
assuming that the lines are optically thin. In such case the flux F(r, ν) is
constant over the frequencies corresponding to a given line yilding
f max
lines =
πe2
mec2
lines
gi fij
ni (r)
gi
−
nj(r)
gj
F(r, νij ),
where νij is the frequency of the line center. Again, we can compare the
radiative force with gravity, which gives the ratio
f max
lines
fgrav
= Γ
lines
σij
σTh
ni
ne
νij Lν(νij )
L
,
where we have neglected upper level population nj (r) ≪ ni (r) and where
σij =
πe2
fij
νij mec
and Lν(νij ) = 4πr2
F(r, νij ).
For lines of heavier elements σij /σTh ≈ 107
and therefore f max
line /fgrav is up
to 103
(Abbott 1982, Gayley 1995).
This enables the appearance of line driven winds. 30
Radiative driving: general line depths
In general, lines may be optically thick. Therefore, the dependence of the
line force on density and velocity (due to the Doppler efect) becomes
very complex. Moreover, the level populations depend on the radiative
field. This introduces intricate feedback that has to be resolved
numerically in general.
However, there exists an approximation named after Sobolev, which
allows to derive an approximate expression for the radiative force in
supersonic flows. This allows to obtain approximate solutions of wind
equations.
31
Sobolev approximation: the essence
Stellar wind velocity increases with frequency. This causes a Doppler shift
of the frequency at which the line absorbs the stellar radiation. In a
supersonic wind, the width of the line (gray area) is smaller than the shift.
radius
velocity
frequency
32
Sobolev approximation: going into details
velocity
frequency
radius
∆νD2
SL2
If the Doppler width of the line ∆νD is smaller than the Doppler shift due
to the radial expansion, then the radiative transfer equation has to be
solved only accross the Sobolev length
LS ≡
vth
dv
dr
= c
∆νD
νij
1
dv
dr
.
This is possible if the Sobolev length is smaller than the density scale
height H, LS = vth/ dv
dr ≪ H = ρ/ dρ
dr ≈ v/ dv
dr , i.e., for v ≫ vth. 33
Sobolev approximation: analytical approximation
Solution of the comoving frame (CMF) radiative transfer equation
provides a straightforward way to derive the Sobolev approximation. This
equation for spherically symmetric stationary flow reads
µ
∂
∂r
I(r, µ, ν) +
1 − µ2
r
∂
∂µ
I(r, µ, ν)−
νv(r)
cr
1 − µ2
+
µ2
r
v(r)
dv(r)
dr
∂
∂ν
I(r, µ, ν) =
= η(r, ν) − χ(r, ν)I(r, µ, ν).
When the spatial derivatives can be neglected (Sobolev approximation),
∂
∂r I(r, µ, ν) ∼ I(r,µ,ν)
r ≪ νv(r)
cr
∂
∂ν I(r, µ, ν) ∼ νv(r)
cr
I(r,µ,ν)
∆νD
, i.e, for
v(r) ≫ vth, the CMF radiative transfer equation is
−
νv(r)
cr
1 − µ2
+
µ2
r
v(r)
dv(r)
dr
∂
∂ν
I(r, µ, ν) =
= η(r, ν) − χ(r, ν)I(r, µ, ν).
34
CMF radiative transfer equation equation: one line
We shall solve the CMF radiative transfer equation for one line,
−
νv(r)
cr
1 − µ2
+
µ2
r
v(r)
dv(r)
dr
∂
∂ν
I(r, µ, ν) =
= χL(r)ϕij (ν) (SL(r) − I(r, µ, ν)) ,
where
χ(r, ν) = χL(r)ϕij (ν)
η(r, ν) = χL(r)SL(r)ϕij (ν)
where χL(r) =
πe2
mec
gi fij
ni (r)
gi
−
nj (r)
gj
.
35
Solving the CMF equation
We shall introduce a new variable
y =
∞
ν
dν′
ϕij (ν′
),
which is y = 0 for the incoming side of the line and y = 1 for the
outgoing side of the line.
This transforms the CMF radiative transfer equation
−
νv(r)
cr
1 − µ2
+
µ2
r
v(r)
dv(r)
dr
∂
∂ν
I(r, µ, ν) =
= χL(r)ϕij (ν) (SL(r) − I(r, µ, ν)) ,
into
νv(r)
cr
1 − µ2
+
µ2
r
v(r)
dv(r)
dr
∂
∂y
I(r, µ, y) =
= χL(r) (SL(r) − I(r, µ, y)) .
36
Integrating the CMF equation
The transformed CMF radiative transfer equation
νv(r)
cr
1 − µ2
+
µ2
r
v(r)
dv(r)
dr
∂
∂y
I(r, µ, y) =
= χL(r) (SL(r) − I(r, µ, y)) .
can be integrated assuming that the variables do not significantly vary
with r within the Sobolev ”resonance zone”. Replacing ν by
center-of-line frequency ν0, the integration gives
I(y) = Ic(µ) exp [−τ(µ)y] + SL {1 − exp [−τ(µ)y]} ,
where the Sobolev optical depth is given by
τ(µ) =
χL(r)cr
ν0v(r) 1 − µ2 + µ2r
v(r)
dv(r)
dr
and where we assumed the boundary condition I(y = 0) = Ic(µ). Clearly,
the Sobolev optical depth is inversely proportional to the velocity
gradient, τ ∼ dv
dr
−1
.
37
Calculating the radiative force
The radial component of the radiative force is
frad =
1
c
∞
0
χ(r, ν)F(r, ν) dν.
Inserting the expression for the specific intensity and opacity gives
frad =
2π
c
∞
0
dν χL(r)ϕij (ν)
1
−1
dµ µI(r, µ, ν).
Transforming the intregral using variable y yields.
frad =
2πχL(r)
c
1
0
dy
1
−1
dµ µI(r, µ, y).
Inserting the solution of CMF equation
frad =
2πχL(r)
c
1
0
dy
1
−1
dµ µ {Ic(µ) exp [−τ(µ)y] + SL {1 − exp [−τ(µ)y]}}
shows that there is no net contribution of the emission to the radiative
force, because the Sobolev optical depth is an even function of µ
(assuming that SL is isotropic in the CMF).
38
The radiative force
The Sobolev radiative force
frad =
2πχL(r)
c
1
0
dy
1
−1
dµ µIc(µ) exp [−τ(µ)y]
after the integration over µ is
frad =
2πχL(r)
c
1
−1
dµ µIc(µ)
1 − exp [−τ(µ)]
τ(µ)
and after inserting the expression for the optical depth we arrive at the
radiative force in the form of (with σ(r) = r
v(r)
dv(r)
dr − 1)
frad =
2πν0v(r)
rc2
1
−1
dµ µIc(µ) 1 + µ2
σ(r) ×
× 1 − exp −
χL(r)cr
ν0v(r) (1 + µ2σ(r))
,
which was derived by Sobolev (1957), Castor (1974), and Rybicki &
Hummer (1978).
39
Optically thin lines: checking the consistency
For optically thin lines τ(µ) ≪ 1 the radiative force
frad =
2πχL(r)
c
1
−1
dµ µIc(µ)
1 − exp [−τ(µ)]
τ(µ)
can be simplified using exp [−τ(µ)] ≈ 1 − τ(µ), which gives
1 − exp [−τ(µ)]
τ(µ)
≈ 1
and for the radiative force
frad =
2π
c
1
−1
dµ µIc(µ)χL(r) =
1
c
χL(r)F(r).
Therefore, the optically thin radiative force proportional to the radiative
flux F(r) and to opacity (density). The same result can be derived for
the static medium.
40
Optically thick lines
For optically thick lines τ(µ) ≫ 1 one can neglect the exp [−τ(µ)] term in
frad =
2πχL(r)
c
1
−1
dµ µIc(µ)
1 − exp [−τ(µ)]
τ(µ)
,
which, after inserting of the optical depth cancels out the opacity,
frad =
2πν0v(r)
rc2
1
−1
dµ µIc(µ) 1 + µ2 r
v(r)
dv(r)
dr
− 1 .
Neglecting the limb darkening one can assume that
Ic(µ) =
Ic = const., µ ≥ µ∗,
0, µ < µ∗
, where µ∗ = 1 −
R2
∗
r2
,
and the radiative force is
frad =
ν0v(r)F(r)
rc2
1 +
r
v(r)
dv(r)
dr
− 1 1 −
1
2
R2
∗
r2
,
where F = 2π
1
µ∗
dµ µIc = π
R2
∗
r2 Ic.
41
Optically thick lines
At large distances from the star r ≫ R∗ the optically thick radiative force
frad =
ν0v(r)F(r)
rc2
1 +
r
v(r)
dv(r)
dr
− 1 1 −
1
2
R2
∗
r2
is
frad ≈
ν0F(r)
c2
dv(r)
dr
.
Therefore, the radiative force is proportional to the radiative flux and to
the velocity gradient, but does not depend on the level populations or on
the density.
42
Solving the hydrodynamical equations
Because the optically thick lines significantly contribute to the radiative
force, one can solve stationary hydrodynamical equations
1
r2
d
dr
r2
ρv = 0
ρv
dv
dr
= −a2 dρ
dr
+ frad −
ρGM(1 − Γ)
r2
with optically thick line force only.
As always, the continuity equation gives the mass-loss rate
˙M ≡ 4πr2
ρv = const.
43
Solving the equation of motion
Neglecting the gas pressure term, the equation of motion reads
v
dv
dr
=
frad
ρ
−
GM(1 − Γ)
r2
.
Inserting the optically thick line force, the equation of motion can be
rewritten as
v −
ν0Lν
4πr2ρc2
1 −
1
2
R2
∗
r2
dv
dr
=
ν0v(r)Lν
8πρc2r3
−
GM(1 − Γ)
r2
,
where the monochromatic luminosity is Lν = 4πr2
F(r). This equation
has a critical point, which at large distances from the star is given by
v −
ν0Lν
4πr2ρc2
= 0.
This provides the mass-loss rate estimate
˙M ≡ 4πr2
ρv(r) =
ν0Lν
c2
≈
L
c2
.
The wind mass-loss rate is proportional to the equivalent photon
mass-loss rate. 44
Example: α Cam
The wind nass-loss rate estimate
˙M ≈
L
c2
was derived for one optically thick line. Assuming now that the mass-loss
is due to Nthick optically thick lines, the mass-loss rate prediction reads
˙M ≈ NthickL/c2
.
The NLTE calculations for the stars α Cam (Teff = 30 900K,
R∗ = 27.6 R⊙, and M = 43 M⊙) give Nthick ≈ 1000. This gives the
mass-loss rate prediction ˙M ≈ 4 × 10−5
M⊙ yr−1
not far from a more
detailed calculation, which gives 1.5 × 10−6
M⊙ yr−1
(Krtiˇcka & Kub´at
2008).
45
CAK theory
We have seen that the optically thin line force is
frad =
1
c
χL(r)F(r),
while the optically thick line force is
frad =
ν0F(r)
c2
dv
dr
.
This gives a possibility to introduce the Sobolev optical depth τS = χL(r)c
ν0
dv
dr
,
which enables us to rewrite the radiative force in a unified form
frad =
1
c
χL(r)F(r) τ−1
S
α
,
where α = 0 for optically thin line and α = 1 for optically thick line.
46
CAK theory
We derived the radiative force accounting for optically thick lines only.
However, in reality the wind is driven by a mixture of optically thick and
thin lines. Therefore, Castor, Abbott & Klein (1975) introduced the
radiative force in the form of
frad = k
σThneL
4πr2c
1
σThnevth
dv
dr
α
,
where k, α are constants (force multipliers) derived from a given line list
using NLTE calculations. Here σTh is the Thomson scattering
cross-section, ne is the electron number density, and vth is hydrogen
thermal speed (for T = Teff).
Gayley (1995) introduced a more physically motivated paremeter ¯Q,
which scales the line force in terms of optically thin line force,
frad =
1
1 − α
κeρF ¯Q
c
dv/dr
ρc ¯Qκe
α
.
47
Solving the momentum with CAK radiative force
Neglecting the gas pressure term, the momentum equation with CAK line
force reads after some manipulation
r2
v
dv
dr
= k
σThL
4πc
ne
ρ
ρ
ne
4πr2
v
σTh
˙Mvth
dv
dr
α
− GM(1 − Γ).
We will express the velocity in terms of the escape speed
w ≡
v2
v2
esc
, where v2
esc =
2GM(1 − Γ)
R∗
.
Introducing a new variable
x ≡ 1 −
R∗
r
the momentum equation can be rewritten as algebraic equation
1 + w′
= C (w′
)
α
,
where w′
≡ dw
dx and C is some uggly constant that depends on the
mass-loss rate.
48
Solving for the mass-loss rate
The algebraic equation
1 + w′
= C (w′
)
α
has zero, one or two solutions depending on the value of C (or ˙M).
0
1
2
3
4
5
0 1 2 3 4 5
C (w′)
1+w′
w′
0
1
2
3
4
5
0 1 2 3 4 5
C (w′)α
1+w′
w′
0
1
2
3
4
5
0 1 2 3 4 5
C (w′)α
1+w′
w′
49
Solving for the mass-loss rate and terminal velocity
Because C is inversely proportional to ˙M, the maximum mass-loss rate
appears for the minimum C, where both curves are tangent. This gives
w′
c =
α
1 − α
,
Cc =
(1 − α)α−1
αα
.
The second equation gives the wind mass-loss rate and the first one
(after the integration) the velocity profile
w =
α
1 − α
x ⇒ v = v∞ 1 −
R∗
r
1/2
,
in the form of the β-velocity law with β = 1/2 and wind terminal
velocity
v∞ = vesc
α
1 − α
proportional to the escape speed. Because vesc is of order of 100 kms−1
,
hot star winds are supersonic. For α Cam vesc = 620 kms−1
(α = 0.61)
prediction v∞ = 780 kms−1
is close to the empirical 1500 ± 200 kms−1
. 50
Wind instabilities I.: The ansatz
The Sobolev approximation gives reliable prediction of wind structure.
Therefore, it is reasonable to assume that is also provides a sound basis
for the study of instabilities. We shall start with time-dependent
hydrodynamical equations
∂ρ
∂t
+
1
r2
∂
∂r
r2
ρv = 0
ρ
∂v
∂t
+ ρv
∂v
∂r
= −a2 ∂ρ
∂r
+ frad −
ρGM(1 − Γ)
r2
We will study the instabilities in the comoving fluid-frame and assume
small perturbations δρ of stationary solution desctibed by ρ0 and v0:
ρ = ρ0 + δρ,
v = v0 + δv, v0 = 0.
51
Wind instabilities I.: Wave equation
Inserting the perturbation into hydrodynamical equations and neglecting
second order term we derive equations for perturbations δρ and δv
∂δρ
∂t
+ ρ0
∂δv
∂r
= 0,
ρ0
∂δv
∂t
= −a2 ∂δρ
∂r
+ δfrad,
where δfrad = ρ0g′
rad ∂δv/∂r and g′
rad ≡ ∂grad/∂ (dv/dr). Combining the
radial derivative of the equation of continuity with the temporal
derivative of the momentum equation we arrive into the wave equation in
the form of
∂2
δv
∂t2
= a2 ∂2
δv
∂r2
+ g′
rad
∂2
δv
∂t∂r
.
52
Wind instabilities I.: Searching for the solution
We shall search for the solution of the wave equation
∂2
δv
∂t2
= a2 ∂2
δv
∂r2
+ g′
rad
∂2
δv
∂t∂r
in the form of travelling waves, δv ∼ exp [i (ωt − kr)], which yields the
dispersion relation
ω2
+ g′
radωk − a2
k2
= 0,
which has a solution
ω
k
= −
1
2
g′
rad ±
1
4
g′2
rad + a2
1/2
.
With zero radiative force we obtain ordinary sound waves ω/k = ±a once
again as it should be. A general case gives a new type of waves –
radiative-acoustic (Abbott) waves (Abbott 1980, Feldmeier et al. 2008)
with downstream (+) and upstream (−) mode.
53
Wind instabilities I.: Understanding the critical point
A very special point is the critical point, at which the radial wind velocity
equals to the speed of (upstream) Abbott waves; they move in opposite
direction, therefore from vc = −ω/k the critical point conditions is
vc −
1
2
g′
rad −
1
4
g′2
rad + a2
1/2
= 0.
The siginificance of the critical point stems from the fact that no
information can travel from the regions with v > vc towards the stellar
surface. In that sense the critical surface resembles the even horizon of a
black hole (Feldmeier & Shloshman 2000). This also means that the
mass-loss rate is determined there.
As a bottom line, we have not found any instability of hot-star winds.
The hot star winds should be perfectly stable, as follows, for example,
from hydrodynamical simulations of Votruba et al. (2007).
54
Wind instabilities II.
Our stability analysis showed that the wind should be stable. However,
the question is what causes the occurrence of X-rays? Is there anything
wrong with our stability analysis? The problem is that the Sobolev
approximation is not valid for small-scale perturbations!
55
Wind instabilities II. – The origin of instability
ν0
I = I0 exp(-τ∫ν
∞
ϕ(ν)dν)
The plot shows the radiative transfer in the comoving frame. As the
unabsorbed photon comes from blue, it hits the rezonance zone where it
can be absorbed. This results in the decrease of the light intensity.
56
Wind instabilities II. – The origin of instability
ν0
I = I0 exp(-τ∫ν
∞
ϕ(ν)dν)
ϕ(ν)
While the intensity in a given line decreases with decreasing frequency,
the absorption line profile ϕ(ν) is symmetric accross the laboratory
frequency of a given line ν0 in the comoving frame.
57
Wind instabilities II. – The origin of instability
ν0
I = I0 exp(-τ∫ν
∞
ϕ(ν)dν)
ϕ(ν)
grad
Therefore, the line force, which is given by a product of opacity and flux
comes mostly from the blue part of a given line.
58
Wind instabilities II. – The origin of instability
ν0
I = I0 exp(-τ∫ν
∞
ϕ(ν)dν)
ϕ(ν)
ϕ(ν-δν)
grad
grad+δgrad
The line force can significantly change after introducing a small change
of the velocity. This is the essence of the line-driven wind instability
(Lucy & Solomon 1970, Owocki et al. 1984, Feldmeier et al. 1997).
59
Wind instabilities II. – Line force perturbation
To describe the instability, we shall evaluate the perturbation of the
radiative acceleration
grad =
2π
cρ
∞
0
dν χL(r)ϕij (ν)
1
−1
dµ µI(r, µ, ν)
assuming optically thin perturbation. This means, we shall perturb just
the line profile and not the intensity,
δgrad =
2π
cρ
∞
0
dν χL(r)δϕij (ν)
1
−1
dµ µI(r, µ, ν).
The perturbed line profile due to velocity perturbation is
δϕij (ν) =
∂ϕij (ν)
∂ν0
δν0 =
∂ϕij (ν)
∂ν0
ν0
δv
c
.
In short, the radiative force perturbation is
δgrad = Ωδv,
where the positive quantity Ω need not to be written down explicitly.
60
Wind instabilities II. – Wave equation
The equations for perturbations δρ and δv are the same as before,
∂δρ
∂t
+ ρ0
∂δv
∂r
= 0,
ρ0
∂δv
∂t
= −a2 ∂δρ
∂r
+ δfrad.
Combining these two gives the wave equation
∂2
δv
∂t2
= a2 ∂2
δv
∂r2
+ Ω
∂δv
∂t
.
As usual, we shall seek the solution in the form of travelling waves
δv ∼ exp [i (ωt − kr)], which gives the dispersion relation
ω2
+ iΩω − a2
k2
= 0.
This has a solution, which is
ω = −
1
2
iΩ ± −
1
4
Ω2
+ a2
k2
1/2
.
61
Wind instabilities II. – The dispersion relation
The dispersion relation
ω = −
1
2
iΩ ± −
1
4
Ω2
+ a2
k2
1/2
takes in the case of negligible gas pressure Ω2
≫ a2
k2
the simple form of
ω = −iΩ.
Therefore, the wave amplitude varies as (Ω > 0)
δv ∼ exp (iωt) = exp (Ωt) .
This leads to a strong instability of the radiative driving (Lucy &
Solomon 1970, MacGregor et al. 1979, Carlberg 1980, Owocki et al.
1984, Feldmeier et al. 1997).
62
Wind instabilities II. – Simulations
We have seen that line driven wind is stable for large scale perturbations,
while it becomes unstable for small-scale perturbations. This enables us
to derive basic wind properties from global models, while the small-scale
structure may affect some observables. This is referred to as clumping.
However, our instability analysis is
linear only and hydrodynamical
simulations are necessary to
describe the instability in detail
(Owocki et al. 1988, Feldmeier et
al. 1997, Runacres & Owocki
2002). 0.0
0.5
1.0
1.5
2.0
2.5
1 2 3 4 5 6 7
v(r)/1000kms-1
r / R∗
63
Hot star winds: micro-view
We have seen that the stellar wind of hot stars is accelerated due to the
scattering of radiation in lines and on free electrons. However, mostly
heavy elements such as carbon, nitrogen, silicon, and iron contribute to
the radiative force. How does it work on a micro-level?
64
Hot star winds: micro-view
Typical volume with: 1000 H ions
65
Hot star winds: micro-view
Typical volume with: 1000 H ions
radiative acceleration due to the line
absorption can be in most cases
neglected
radiative acceleration due to the
free-free processes also negligible
σp ≪ σe
65
Hot star winds: micro-view
Typical volume with: 1000 H ions + 100 He ions
65
Hot star winds: micro-view
Typical volume with: 1000 H ions + 100 He ions
radiative acceleration due to the line
absorption can be in most cases
neglected
radiative acceleration due to the
free-free processes also negligible
65
Hot star winds: micro-view
Typical volume with: 1000 H ions + 100 He ions + 1200 e−
65
Hot star winds: micro-view
Typical volume with: 1000 H ions + 100 He ions + 1200 e−
Γ = ge/ggrav ≈ 0.1 for many OB stars
⇒ significant contribution to the
radiative acceleration
65
Hot star winds: micro-view
Typical volume with: 1000 H ions + 100 He ions + 1200 e−
+ 2 metals
65
Hot star winds: micro-view
Typical volume with: 1000 H ions + 100 He ions + 1200 e−
+ 2 metals
maximum radiative acceleration due to
the lines gmax
line ≈ 1000 ggrav (Gayley
1995) ⇒ crucial contribution to the
radiative acceleration
65
Hot star winds: micro-view
Typical volume with: 1000 H ions + 100 He ions + 1200 e−
+ 2 metals
65
How can this work?
In order to accelerate the stellar wind, two efficient processes are
necessary.
A process which transfers momentum from radiative field to heavier ions.
This has to be efficitent in such a way that only a very small part of the
radiative energy is used to heat the wind. Therefore, the frequency of
absorbed and emitted photon should be precisely the same. Such
absorption is called scattering in astrophysics. We have seen that both
light scattering on free electrons and in spectral lines are exactly such
processes.
However, since most of radiative momentum is received by heavy
elements, a second process is necessary, which transfers momentum from
heavier ions to the bulk flow (H and He).
66
How to transfer momentum?
Stellar wind of hot stars is ionised. Therefore, Coulomb collisions are
efficient to transfer momentum from heavier elements to the passive
component. In such case the frictional force on passive component (p)
due to heavy ions (i) is
fpi = ρpgpi = npni
4πq2
pq2
i
kTip
ln Λ G(xip)
vi − vp
|vi − vp|
,
where np, ni are number densities of components, vi, vp are their radial
velocities, and qp, qi their charges. The frictional force depends on the
velocity difference between the wind componets via so-called
Chandrasekhar function.
xip =
|vi − vp|
αip
α2
ip =
2k (miTp + mpTi)
mimp 0.00
0.05
0.10
0.15
0.20
0.25
0 1 2 3 4 5
G(xip)
xip
Chandrasekhar function G(xip)
67
Momentum transfer efficiency: efficient transfer
The frictional force depends on the product of densities of individual
components. Consequently, to transfer a given amount of momentum at
high densities, just a small velocity difference is needed. In such a case
xip ≪ 1, the transfer of momentum from heavier ions to hydrogen and
helium is efficient and one-component models are sufficient in such case.
Moreove, such winds are stable, a decrease of velocity difference leads to
a stronger frictional force, which in turn leads to stronger coupling and
decrease of the velocity difference.
Such winds are typically found at
high densities or at high mass-loss
rates. Therefore, wind of luminous
OB stars and WR stars can treated
as one-component flow neglecting
all multicomponent effects.
0.00
0.05
0.10
0.15
0.20
0.25
0 1 2 3 4 5
G(xip)
xip
Chandrasekhar function G(xip)
68
Momentum transfer efficiency: multicomponent effects
For lower wind densities (or lower wind mass-loss rates) the momentum
transfer between heavy ions and hydrogen and helium becomes
inefficient. For xip 1 part of transfered energy goes to heating, which
leads to frictional heating of the wind.
For even lower densities xip > 1 and a new type of instability appears.
The Chandrasekhar function is a decreasing function of velocity
difference. This means that small positive perturbation of velocity
difference leads to decrease of the frictional force, which in turn allows
for even larger velocity separation. This leads to runaway of heavy ions or
to decoupling instability.
To model winds at low densities,
multicomponent approach is
needed. Such winds are typically
found in main-sequence B stars or
in stars at low metallicity. 0.00
0.05
0.10
0.15
0.20
0.25
0 1 2 3 4 5
G(xip)
xip
Chandrasekhar function G(xip)
69
Line-driven winds: importance
Despite the fact that massive stars are short lived (106
yr), with large
mass-loss rate (of the order of 10−6
M⊙ yr−1
) the massive stars may lose
a significant fraction of their mass on their way from main sequence to
the neutron star or a black hole. Amount of the mass-loss is one of the
factors that determine the nature of the final evolutionary stage.
Hot star winds are important also for the dynamics of the interstellar
medium. Moreover, cosmic-ray particles are generated at the boundary
between hot star wind bubble and the interstellar medium.
Line-driven winds appear in O and early B main-sequence stars, in OBA
supergiants, in hot subdwarfs and central stars of planetary nebulae and
in Wolf-Rayet stars. Accretion disks around supermassive black holes
may also be sources of line-driven wind.
70
Suggested reading
J. Castor: Radiation hydrodynamics
H. J. G. L. M. Lamers, & J. P. Cassinelli: Introduction to Stellar Winds
D. Mihalas & B. W. Mihalas: Foundations of Radiation Hydrodynamics
J. Puls, J. S. Vink, F. Najarro: Mass loss from hot massive stars
S. Owocki: Stellar wind mechanisms and instabilities
F. H. Shu: The physics of astrophysics: II. Hydrodynamics
71