Download Lecture 9 - Angular Momentum Transport o  

Lecture 9 - Angular Momentum Transport
o  Topics in today’s lecture:
o  Angular momentum transport
o  Magnetic braking
o  Stellar winds
o  Viscous Disk
Angular Momentum Transport
o 
Three processes have been proposed:
1.  Magnetic braking
2.  Stellar wind
3.  Gas viscosity
o 
Each theory attempts to explain why the Sun now contains the majority of matter
in the Solar System, and little of its AM.
1. Magnetic braking
o 
Material in the inner part of the disk must be
ionized. Fields are frozen if the plasma beta
value is large:
β=
PG
PB
where PG = nkT + ½ ρv2 and PB = B2 / 8 π.
o 
If PG > PB => β > 1 (high beta plasma; fields
frozen €
into plasma).
o 
If PG < PB => β < 1 (low beta plasma; fields
slip through plasma).
o 
For β > 1, the magnetic field of the Sun will
experience resistance as it drags through the
plasma, which will slow down the rotation of
the Sun.
PY4A01 Solar System Science
1. Magnetic braking (cont.)
o  Star’s magnetic field co-rotates the wind out to the Alfven Radius.
o  We know that the gas pressure of the wind is
PG = nkT + ½ ρ v2 = nkT + 1/2 nm v2
o  For v > 500 km s-1, 1/2 nmv2 >nkT => PG ~ 1/2 nm v2.
o  Equating the gas and magnetic pressures:
B2
= 1/2nmv 2
8π
o  As B = MB / r3 (MB is the magnetic dipole moment) gives
M B2 €
2
=
1/2nmv
8πr 6
€
Eqn. 1
1. Magnetic Braking (cont.)
o  Now from conservation of mass
dM
˙ = 4 πr 2 ρv = 4 πr 2 nmv
=M
dt
˙
M
=> nm =
4 πr 2v
o  Substituting this into Eqn. 1 and rearranging gives
€
" M B2 %1/ 4
rA = $
'
˙
M
v
#
&
Alfven Radius
o  Solar wind co-roates with the magnetic field for r<rA. For the Sun, rA ~ 5-50Rsun.
€
2. Stellar wind braking
o  Stellar wind literally blows the angular momentum problem away.
o  The total AM is
L = Msun vr
as v = rω (ω is the angular velocity) => L = Msun r2ω
o  The rate of loss of AM is
dL dM sun 2
=
rω
dt
dt
Eqn. 2
o  Particles lost from the star also carry away angular momentum. Given an initial
mass, rotation rate, and radius, we can thus calculate the rate of AM loss.
€
o  We know that solar-type star have strong T-Tauri winds in their youth, which
ultimately clear away residual gas and dust in the disk.
o  This theory also explains why more massive stars rotate faster: they do not have
proper winds like the Sun, and therefore are not as good at losing angular
momentum.
2. Stellar wind braking (cont.)
o  Eqn. 2 only applies to equatorial regions. The correct global form for distance
beyond the Alfven radius is:
dL 2 ˙ 2
Eqn. 3
= MrA ω
dt 3
where the 2/3 accounts for the fact that higher altitudes contribute less to AM
losses.
L 2 ˙ 2
= MrA ω
τD 3
€
o  Setting dL/dt ~ L/τD gives
o  As the AM of a spherically symmetric star is L = 2/5 Mr2ω we can then find an
estimate for the spin-down time
€
3 Mr 2
τD =
˙ rA2
5M
o  This gives the spin-down time in terms of the mass loss rate and the Alfven radius.
This predicts a spin down time of <1010 years for the Sun.
€
2. Stellar wind braking (cont.)
o  But how does the angular velocity change with time depend on the mass loss rate?
o  We know L = 2/5Mr2ω. This gives
# dM
dL
dω &
= 2 /5r 2 %ω
+M
(
$ dt
dt
dt '
o  Equating this with Eqn. 3:
2 /3
€
o  Rearranging gives
# dM
dM 2
dω &
rA ω = 2 /5r 2 %ω
+m
(
$ dt
dt
dt '
3/5r 2 M
€
dω
dM
= (rA2 − 3/5r 2 )ω
dt
dt
o  Since 3/5r2 << rA2, we can write:
€
rA2ω
€
dM
dω
= 3/5Mr 2
dt
dt
2. Stellar wind braking (cont.)
dω 5rA2
∫ω 0 ω = 3r 2
ω
o  Integrating then gives:
o  Therefore,
∫
dM
M0 M
M
# ω & 5rA2 # M &
ln% ( = 2 ln%
(
$ ω 0 ' 3r $ M 0 '
€
2
# M & 5rA / 3r
ω = ω 0%
(
M
$ 0'
€
2
where M and M0 are the final and initial masses, and ω and ω0 are the final and
initial AM.
€
o  For a particular mass loss rate,
mass loss.
€
M = M0 −
dM
τ M , where τωm is the mass duration of
dt
3. Gas viscosity
o  As well as the AM transfer of
angular momentum from star to the
disk, must also transport AM within
the disk.
o  One way we can do this is by
convection: numerical models
shows turbulent eddies can develop,
which can transport AM.
o  Gravitational interactions between
clumps of material in the disk can
cause angular momentum to be
transported outwards, while gas and
dust move inwards.
3. AM Transport in disk?
o 
The total energy associated with an orbit is: E = KE + PE = 1/2 mv2 - GMm/r.
o 
For a circular orbit, mv2/r = GMm/r2 => v2 = GM/r.
o 
o 
GM GMm
−
Therefore
r
r
GMm
=> E = −
2r
GM
The angular momentum is given by L = mvr = m
r
r1
r2
r
r
=> L = m GM
r
€
o 
m1
E = 1/2m
m2
Lynden-Bell and Pringle (1974) considered how AM could be transported in a disk. Take two
bodies in circular orbits with masses m1 and m2. The total energy and AM of the system are:
€
GM # m1 m2 &
% +
(
2 $ r1
r2 '
r
r
L = GM (m1 1 + m2 2 )
r1
r2
E =−
€
Eqn. 4
Eqn. 5
3. AM transport in disk
o 
Now suppose the orbits are perturbed slightly, while conserving L, Eqns.4
and 5 become:
GM $ m1
m2 '
ΔE =
δ
r
+
δ r2 )
Eqn.6
&
1
2 % r12
r2 2 (
ΔL = GM (m1 δ r1 + m2 δr2 ) = 0
o 
Eqn. 7 implies that
δr2 = −
€
o 
m1
r2
δr1
r1
Substituting this into Eqn. 6 gives ΔE =
€
o 
m1
m2
Eqn.7
GM % m1
m % m
'' 2 δ r1 + 22 ' − 1
2 & r1
r2 & m2
GMm1δ r1 % r12 r2 (
'1−
*
2r12 & r12 r1 )
3/2
%
(
GMm1δ r1 '% r1 (
=> ΔE = −
' * −1**
2
'
2r1
&& r2 )
)
=
r1
r2 ((
δr1 **
r1 )*)
r2
m2
€
If r1 < r2 we can reduce the energy of the system if Δr1 <0 i.e. moving m1 closer to the center. Thus the
energy can be reduced while conserving AM by moving the initially closer body in and moving the
€
initially outer body further out => all the mass is concentrated near the inner parts of the solar system,
while the AM occupies the outer parts. See Planetary Science by Cole and Woolfson for further details.
Properties of the Solar System
1. 
2. 
3. 
4. 
5. 
6. 
7. 
8. 
9. 
10. 
11. 
12. 
All the planets' orbits lie roughly in the ecliptic plane.
The Sun's rotational equator lies nearly in this plane.
Planetary orbits are slightly elliptical, very nearly circular.
The planets revolve in a west-to-east direction. The Sun rotates in the same west-to-east
direction.
The planets differ in composition. Their composition varies roughly with distance from
the Sun: dense, metal-rich planets are in the inner part and giant, hydrogen-rich planets
are in the outer part.
Meteorites differ in chemical and geologic properties from the planets and the Moon.
The Sun and most of the planets rotate in the same west-to-east direction. Their
obliquity (the tilt of their rotation axes with respect to their orbits) are small. Uranus and
Venus are exceptions.
The rotation rates of the planets and asteroids are similar-5 to 15 hours.
The planet distances from the Sun obey Bode's law.
Planet-satellite systems resemble the solar system.
The Oort Cloud and Kuiper Belt of comets.
The planets contain about 99% of the solar system's angular momentum but the Sun
contains over 99% of the solar system's mass.