Phys 954, Homework #2, Solutions
Physics 954
Homework Solutions, Homework #2
1) a)
The electric conductivity is
n e2
 e
m
i.e. for a given plasma density n the product m determines the conductivity. In principle, both
ions and electrons can conduct electric current. If e >> i the ion current may become
dominant. The collision frequency is
v
 e ,in  e,i  nn   coll ve ,i
e,in
Electron and ion currents are added in plasmas to make up the total current. Thus we treat both
currents as parallel. Because of the difference in mass the ion conductivity is usually much lower
than the electron conductivity, unless the electron collision frequency is much higher.
i) Comparing ions and electrons: ei = ie, thus we only need to check collisions with neutrals.
For ions the relevant velocity is usw, because the solar wind is a cold ion beam with uth << usw
while for electrons it is the thermal velocity with a typical temperature of 40 eV (500,000K).
 en  en uthe  en


 in  in usw  in
mikTe
2 10 16 1640  40

 0.4
me 2E sw 3 10 15 2  1000
The two collision frequencies are of the same order of magnitude, with the electron-neutral
frequency somewhat lower. But the mobility of the electrons is much higher than that of the ions
by √( me/mi ) << 1. Therefore, the electron conductivity dominates, even if the collision
frequencies are slightly different.
ii) Let us turn now to collisions of electrons with ions and let us compare this with the electron
neutral collisions at 1 AU. Both collisions slow down the electrons and thus act like “serial
resistors”. The Coulomb collision frequency is
u
nsw
kTe
 Coul  the 

4
2
Coul 5.9  10  T
me

1 30
5.910 4 5 10
 [m]
5 2
1.38 10 23  5 10 5
[m / sec]100[cm / m]
9.1 1031

 (1.9 57)108 sec 1
whereas the neutral collision frequency is:
 en  uthe  nISM   en  2.7108 cm / sec 0.1cm 3  210 16 cm
 5.510 9 sec 1
1
Phys 954, Homework #2, Solutions
At 1 AU:
Coul is the highest collision frequency and therefore is responsible for the
conductivity. (Always the highest frequency determines the conductivity in a serial circuit!)
2
rE
At 100 AU: nsw (100 AU) = nsw (1 AU) 
100 AU 2
invoking flux conservation and Vsw almost constant beyond 1AU. Variation of the electron
temperature with distance is also neglected for simplicity.
nsw (100 AU) = (1 – 30)  10-4 cm-3
Coul = (1.9 – 57)  10-12 sec-1
The e-i collision frequency is greatly reduced, whereas en remains constant because of the
constant ISM density. Clearly already beyond 3 – 10 AU the interaction with the neutral gas
dominates.
Using MKSA units:
nsw  e2
(1 30)  106 (1.6  1019 )2
A
  e (1AU)
 1.5 10 6
31
8
me  Coul
9.1 10  (1  30)  1.9 10
Vm
nsw  e 2
(1  30) 10 6 (1.6 10 19 )2
A
  e (100 AU)
104  (5  150)10 2
31
19
me en
9.1 10 5.5  10
Vm
2
The conductivity is decreasing with 1/r , because the neutral collisions are dominant and Nsw
proportional 1/r2. If Coulomb collisions were dominant, the conductivity would remain constant
because nsw appears in the numerator and the denominator.
b)
The magnetic diffusivity is

At 1 AU:
1
 o
1
Vm Am
m2


1.9
4 10 7  1.5 106 A Vs
s
The magnetic Reynolds number for the Earth's magnetosphere (lo = 10 Earthradii; 1 RE = 6.5 106
m) is (using Coul, Coulomb collisions are most important at 1 AU):
10  6.5 10 6  5 105
o  usw
Rm 

1.7 1013

1.9
i.e. a huge value.
Note: Rm does not depend on nsw, because nsw cancels in e for Coulomb collisions.
c)
The time it takes the solar wind to the heliospheric boundary is:
100( AU)  1.5 1011m
 sw 
310 7 sec  (1year)
5
5  10 m / sec
2
Phys 954, Homework #2, Solutions
With
B
1
B
2
 2 B
we get:
 B

L
t
With being the time constant for diffusion and L the typical scale length for variations in B.
This means that the size of a structure, which is destroyed by diffusion during the convection
with the solar wind to 100 AU, amounts to:
L   sw
We use en (neutral collisions) throughout, because they dominate beyond a few AU. Therefore,
 varies with distance we average. (using en from 1a)
100 r 2 dr
1
1
10 4
≈ (1.6 - 50) 107 m2/s

/
100
o en 0 rE 2 rE
4  10 7 5  150  102 3
L   sw = √(3 107 sec*(1.6 - 50) 107 m2/s) = (2.2 - 12) 107 m
I.e. a structure with a size of 22,000 - 120,000 km that is imprinted into the solar wind is just
destroyed after traveling 100 AU with the solar wind. The diffusion is not very strong because
the conductivity is so high. Therefore, a disturbance of the size of the Earth’s magnetosphere
would only be barely destroyed by diffusion.
There are, however, other effects in the solar wind that can alter the magnetic field structure,
such as the increased wrapping of the field around the sun at larger distances and turbulence.
3
Phys 954, Homework #2, Solutions
2) Angular Momentum Transport
To compute the values in a), b) and c), we need to know the Alfvén distance rA, because this
determines the angular momentum transported away from the sun by the solar wind. Up to the
Alfvén point the wind is still forced into rigid rotation with the sun. This determines the
maximum angular momentum
L rA2
that can be transmitted by the magnetic field. Further out the field slips w.r.t. the solar wind
plasma. To estimate rA we use the condition that MA = 1 is reached at the Alfvén point:
The Alfvén point is reached where the magnetic field and the RAM pressure strike a balance.
o ur2
at rA
B2A
Furthermore, we use the following two conservation relations
Br2r2 = const. and  ur r2 = const.
to relate the values to the values measured at 1 AU.
r2
o  ur (r) E  E 2 ur (rA )
rA
1
where rE = 1 AU
r4
2
Br (rE ) 4
rA
This results in:
1  MA 
2
ra rE
Br (rE )
o(rE )ur (rA )/ur (rE ) ur (rE )
Let us use at 1 AU:
Br ~ 5nT
nsw  10 cm-3
usw  500 km/sec
We know: cs = 180 km < ur(rA) < ur(rE) = 500 km/sec. Now let us assume an intermediate
value of: ur(rA)  340 km/sec, i.e. (cs + ur(rE))/2. The result is not very sensitive to our choice,
because rA is roughly proportional to √u(rA). The actual value of u(rA) lies between our choice
and vsw at 1 AU, because the solar wind speed rises rapidly close to the sun and then becomes
almost constant. Using our choice:
510 9
rA = 215 Rs 
18.4 Rs
4 107 1.61027 107 0.68 5105
This value is higher than the 12 Rs given in Kirk et al. The result depends on our assumptions on
nsw, usw, and Br. The location of the Alfvén point varies significantly with these parameters.
Let us adopt 18.4 Rs = 0.085 AU = 1.28  107 km for our further computations. Then the
angular momentum L per unit mass is:
2
2  1.281010 
m2
2
L   rA 
 4.31014
28 243600
sec
4
Phys 954, Homework #2, Solutions
a)
The tangent of the aberration angle for the solar wind with respect to the radial direction,
i.e. coming directly from the sun is
u/ur
(equal to the angle for small angles)
2
2
M  L /(r  ) 1
M A2 (rA2 / r2 ) 1
where
u    r  A



r

MA 2 1
M A2 1
We have to compute the Alfvénic Mach number at 1 AU (using the values given above):
1/ MA (1AU) 
2
Br2
(510 9 ) 2

 0.005
 o  ur2
4 10 7 1.610 27 107 (5 105 )2
MA 2 200 M A 14
From the conservation of the angular momentum outside rA:
M 2 (rA2 / rE2 ) 1
2 1.5 1011 200  (18.4 / 215)2  1
u =   rE  A


 800 m/ s
M A2 1
2.510 6
200  1
u/ur  0.0016

0.1
This aberration is very small.
Br2
.
nswur3
u and thus the aberration is directed with the sun's rotation, i.e. from west to east.
It varies approximately proportional to
b)
The orbital velocity of Earth is
uEarth  30 km/sec
uEarth/usur  0.06  3.4o
This is a much greater effect. Because the Earth is moving west to east around the sun
(counterclockwise) like the sun's rotation this effect is opposite to the effect computed in a)
leading to a combined effect:  -3.3o.
c)
The total angular momentum flux density of the sun is:
sw

107  1.6  10-27
ur
5105

L
4.31014
= 3.4 kg
m2
/ (m2 sec)
sec
We have to integrate this over the surface of a sphere at 1 AU
assuming that the solar wind is isotropic
noting that the lever arm scales like cos, where  is the solar latitude.
5
at 1 AU.
Phys 954, Homework #2, Solutions
L( ) L cos2 
/ 2
The total change of angular momentum: dL/dt   swur  L 2 
0
cos  d  r 2 r
2
 2 rE 2  swur L  4.91023
Over 5 billion years:
L 4.91023  510 9  365 243600  7.710 40
kgm
sec
2
kgm2
sec
The angular momentum of the sun
2
2
2
42 kgm
Lsun   Ms Rs  110
5
sec
Assuming rigid rotation and a homogenous mass distribution! Of course the angular momentum
is smaller, because the density increases towards the center of the sun.
According to this estimate the sun was only 8% faster in the beginning. Models on the
formation of the solar system give a much higher angular momentum. Regular solar wind does
not solve the angular momentum problem of the sun. It only contributes a small portion.
6
Phys 954, Homework #2, Solutions
3) a)
(I)
We start with (I.5_4) from the Lecture Notes
1  o k u1  0
mass conservation
(II)
 o o u1  B1 (Bo k )k (B1 Bo )
momentum flow
(III)
B1 (Bo k )u1  Bo (k u1 )
Faraday's and "Ohm's" Law
(IV)
kB1  0
First we note in (I) that k u1  0 means also 1 ≠ 0.
Second we note in (II) that k perpendicular to Bo (propagation perpendicular to Bo) means k. Bo
= 0 and thus u1 || k.
Using III) we substitute B1in II).
(IIa)
o o 2 u1  Bo k u1  Bo k u1
 
 B k  kB k u  B  B ku 
2
o
o
1
o
o
1
Let us evaluate first whether there is a component of u1 || Bo . : 0 ≠ u 1 Bo : ?
 
(IIa.1) (IIa) Bo : o  2 Bou1  0
The first bracket is equal to the second bracket on right hand side of (IIa) and thus u1 Bo  0
Let us now evaluate ku 1 :
i.e. the longitudinal component of the wave.




2
2
2
(IIa.2) (IIa) k : o o (k u1 )  Bo k k u1  k Bo k u1  Bo k  k  Bo k u1 Bo  Bo k u1 




0
0
noting that the two terms in the first bracket cancel, we get:
       
  
 
 
o  o 2 (k u1 )  k 2 Bo 2 k u1
We have already treated k u1  0
which is a pure transverse mode with the Dispersion
Relation of the transverse (shear Alfvén) waves. The k vector varies with the cos of the angle
with Bo. Now we get a similar Dispersion Relation:
2
2 B o
2

 VA
k 2  o o
which represents a compressional Alfvén wave, since k u1  0
As can be seen, compressional Alfvén waves propagate in all directions with the same speed. An
exception is k || Bo ; here it cannot be compressional because Bo u 1  0
Here the wave is purely transverse!! But the combined set of waves propagates with equal speed
in all directions.
b)
For the hot plasma case we have to add -  o k P1 P1  on the R.H.S. of II)
With the adiabatic equation of state P1 = cs2 1 and with I) this becomes:
c 2
2
 o k P1   o k cs 1   o o k s k u1
 
7
Phys 954, Homework #2, Solutions
Adding this term to (IIa) and multiplying by Bo the new equation (IIa.1) reads:
o  o 2 Bo u1  o  o Bo k cs 2 k u1
(IIb.1)
   
 
Now the right hand side is not 0.
This keeps one more term in (IIa.2), and after adding the pressure term to (IIa.2) we get:
(IIb.2)
oo 2 (k u1 ) k 2 (Bok)(u1Bo ) k 2Bo2(k u1 )   oo c s2 k 2 (k u1)
Note: The first term on the r.h.s. is not 0 any more.
The equations then turn into:
(IIb.1)
 2 (Bou1 )cs 2(Bok )(k u1 )
(IIb.2)
and ( oo  k 2Bo2   oo cs2 k 2 )(k u1 )   k 2(Bok )(Bo u1)
ku 1  0  Bo u 1 is the trivial solution of this case, or the ordinary transverse Alfvén wave, which
also exists for the warm plasma.
After inserting (IIb.1) into (IIb.2) we retrieve the relation for the compressional mode:

 
  ku 
 2  o o 2  k 2 Bo 2   o oc s2 k 2  k u1   k 2 cs 2 Bo k
For k u1  0 we get the Dispersion Relation:
B2
B2
 4  2 k 2 o  cS 2 2 k 2  k 2 cs2 o cos2  k 2  0
o o
or
4  2
k4

k2
VA 2 
2
k2
o o
c s2  cs 2 VA 2 cos2   0
2
This is a bi-quadratic equation, which has 2 solutions for 2 .
k
2
2
2 2
2
2
2 2
2
2 2

V

c






V

c
V

c



s
s
s

2 2 A
  A
  A
 c 2 V 2 cos2 
k 2 
 2   2  s a
k
2
2
2
4
4

VA  cs  V 2 c 2  VA Cs  c 2 V 2 cos2 
s
A
 2  A s
4
with 2 positive solutions, the fast and the slow magnetosonic wave.
8
2
1