Phys 954, Solar Wind and Cosmic Rays
E. Möbius
Homework #4 Solutions
1) Pickup Ion Trajectories
a) For an interplanetary magnetic field that is perpendicular to vsw the ions are
injected exactly perpendicular to the field with vsw. Therefore, in the solar
wind frame they gyrate with the gyro radius rc about the IMF B with speed vsw.
In polar coordinates their velocity is then (vr, v) = (0, vsw). In Cartesian
coordinates (with pointing away from the sun) this turns into:
vx = - vsw . cos(ct)
vz = vsw . sin(ct) where c is the gyro frequency
z
vsw
x
B
In the rest frame (adding vsw) this turns into:
vx = vsw . (1 - cos(ct))
vz = vsw . sin(ct)
The trajectory is:
x(t) = xo + vsw . (1 - sin(ct)/c) z(t) = vsw . (1 - cos(ct)/c)
i.e the trajectory of a cycloid.
For vsw = 440 km/sec the energy of solar wind ions is 1 keV/nucleon. The
maximum velocity of pickup ions is 2 vsw, i.e. the maximum energy is 4
keV/nucleon. Since the mass of He is 4 the maximum energy of He pickup
ions is: 16 keV.
i)
The expansion of the Li cloud from the AMPTE IRM spacecraft along
with pickup ion trajectories is shown in the following sketch:
Li Cloud
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Phys 954, Solar Wind and Cosmic Rays
E. Möbius
Using the velocity relations from a) we get for the arrival direction  of the ions:
v
sin  t
1cos  t
  arctan( v zx )  arctan( 1cosc c t )  arctan( 1cos cc t )
Let us do the 2nd half of the time calculation first. Using the relations on the
pickup velocity from a) we see that the ions come at 45o w.r.t. the solar wind
when vx = vz = vsw, i.e. when t = π/2. and parallel to the solar wind when t = π.
The time of arrival of these ions is composed of the time it takes the cloud to
expand as much that Li ions are produced far enough away from the spacecraft to
fulfill these conditions and of their travel time along the cycloid to the spacecraft.
To compute the latter we need the gyro frequenc. The travel time is 1/4 and 1/2 of
the gyro period, respectively. (A = 7; Q = 1; B = 10 nT)
 ci = eB/2πm = 1.52 . 107 (Q/A).B[T] ≈ 0.022 Hz
1/4: 11.5 sec
1/2: 23 sec
To compute the distance of their origin from the spacecraft, i.e. from the point
where the cloud was released, we need their gyro radius:
The ions are created in the solar wind frame at vsw = 440 km/sec (from a). The
gyro radius is:
rci = vsw/ci = vsw/2πci
rci = 440/(2π . 0.022) ≈ 3600 km
The distance of the origin of the ions from the spacecraft is:
d(t)  x(t)2  z(t)2
with x( /4) = (π/2 - 1) rci
and x( /2) = π rci
i.e.:
d( / 4)  ( / 2  1)2  1 rci and
d( / 2)   2  4 rci
d(/4) = 1.15 rci = 4150 km and d(/2) = 3.72 rci = 13400 km
The expansion speed is for (T = 2000 K; A = 7) with mvth2/2 = 3kT
vthLi = (3kT/m)1/2 = 0.155 km/sec (T[K]/A)1/2 ≈ 2.68 km/sec
Therefore, the Li ions will arrive from 45o (with vx= vz = vsw; E = mvsw2 = 14
keV) after 1540 sec (≈ 25 min) and parallel to the solar wind after 5000 sec (≈1 h
20 min). The travel time of the ions on the cycloid is negligible compared with the
expansion of the cloud.
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Phys 954, Solar Wind and Cosmic Rays
E. Möbius
2) Interstellar Pickup Ions
a) We assume that the pickup ion distribution is an ideal gas for which the
adiabatic law holds:


 P 
 P 



const.

 
5 / 3 
with
P
 kT
m
T
2 / 3
we get
 const.
(I)
The total flux of pickup ions as transported with the solar wind is conserved and
vsw = const. Therefore:
4r2 pickup  const.
~
1
r2
(II)
The temperature of the pickup ion distribution is determined by the velocity of the
shell distribution.
v2 ~ T
(III)
The pickup ions are injected to fill a shell around the solar wind with radius vsw at
the location of their ionization r. Inserting II and III into I:
v2 r4/3 = const.
or
vsw2 r4/3 = v2 robservation4/3
2 / 3
r
  observati on 

Vsw
r

q.e.d.
(IV)
b) Along the axis of inflow towards the sun the neutral gas density of the
interstellar gas is reduced along the way according to
dN(r )
   ion (r) N(r)
dt
Assuming a constant inflow velocity VISM = vb (neglecting the influence of solar
gravity at first) we can write:
 dr  Vism dt
(gradient inward)
The ionization rate varies with the solar UV intensity as
 ion (r)  ionE rE 2 / r2
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Phys 954, Solar Wind and Cosmic Rays
E. Möbius
where ion E the ionization rate of rE = 1 AU.
2
dN(r )  ion E  rE dr

 2
N(r)
VISM
r
integrated:
 io nE rE 2
N(r)
ln

No
VISM r

N(r) No e
 ion E rE2
VISM r
or

 No e

r
i.e. the density falls off exponentially with the typical penetration distance
 = ionE  rE2/VISM, the gas approaches the sun.
c) With gravitation we need to look at this in a different way:
i) Again the problem simplifies greatly along the inflow axis, because here b = 0.
We use:
 ionE  rE 2 drdt
 ionE rE 2 dr
dN(r )


 

N(r)
r2
dr
r2
v
To determine v(r) we use energy conservation:
m 2
m 2
mMsG
2
v 
vISM 
or v 2  v 2ISM (1 
)
2
2
r
r
where  = VISM2/MsG. So:
ion E  rE 2
 ionE  rE 2 d
dN(r )
dr
 



N(r)
VISMr 2
VISM

1  2 / r
where   1  2 /  r
Integration yields:
 ion E rE 2


again with 
N(r)  N o e VISM
1 2 / r
There is a relatively simple extension for trajectories that pass the sun, i.e. with
collision parameter b ≠ 0.
ii) To simplify the task we only consider survival to ionization, not gravitational
focusing of the flow:
 r 2
  r 2 ddt
dN(r, )
  ionE 2 E  dt   ion E 2 E 
N(r, )
r
r
d
Now we can use conservation of angular momentum
r 2 d / dt  rvsin   bVISM
to replace the time dependence and to turn the time integral into an integral over
.
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Phys 954, Solar Wind and Cosmic Rays
 r 2
dN(r, )
  ionE E  d
N(r, )
VISMb
or
E. Möbius

N(r, ) No e
 ion E rE 2
VISM b

 No

 
e b
The same parameter =ionE  rE2/VISM appears as scale length, but now in
comparison with the impact parameter b.
d) Now let us assume we know N(r). Then the production rate of pickup ions is:
r 2
S(r) = ion E  E2  N(r)
r
S(r) decreases with the ionization rate. However, the density of pickup ions,
which have been created closer to the sun, also falls off as 1/r2 during the transport
with the solar wind due to the radial expansion. Therefore, we have to normalize
each spherical shell in the velocity distribution function f(v(r)) for this transport
effect by multiplying: S(r)  r2/rE2 if we observe the distribution at re = 1 AU.
3/ 2
 v 
3 ion E rE
Thus:
(V)
f (v)
N(r)  
8 Vsw
VSW 
does not contain rE2/r2 any more!
3/ 2
. Vsw 
We only have to note that r(v) = rE
according to (IV).
 v 
3/ 2
  ion AE rE Vsw 3/ 2  
3 ion E rE
v

Therefore:
f (v) 
No exp 
 
8 Vsw4
 VISM  v  
 Vsw 
For the following problem the neutral density at 1 AU is already given, so that the
r dependence in N need not be used.
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Phys 954, Solar Wind and Cosmic Rays
E. Möbius
3) Size of the Heliosphere
a) In the interstellar gas ran direction the ram pressure balance between solar
wind and interstellar plasma is: (III.2-1).
mp nsw(r) Vsw2 = mp nplasma VISM2
VISM = b is the bulk velocity of the interstellar gas and plasma. Due to flux
conservation the solar wind density nsw(r) varies as
2
r
nsw(r) = nsw E E2
where
rE = 1 AU
r
Thus
rE2
nsw E 2 VSW 2  n plasm a VISM 2
r
n SW E Vsw
r  rE

n plasm a VIsM
r 1 AU 
5
440

 394 AU
0.01 25
b) Including the magnetic field in the pressure balance we get:
Note that we have to include the tension of the magnetic field, because the
heliosphere is moving through the field. Therefore, the B-field effect doubles. (I
don't subtract points for this subtlety.)
2
r 2
B
m p nSWE E2 Vsw 2  m p np lasma VISM2 
r
o
r  rE 
1 AU
m pnSWE VSW2
m pnp lasmaVISM 2  B2 / o
1.6102 7 510 6 (4.410 5 )2
1.610 2 7 10 4  (2.5 10 4 )2  (3 10 1 0)2 / 4 10 7
138 AU
The neutral gas would be the largest contributor to the pressure, if all of it would
be converted to ions, and then lead to:
m p nSWE  VSW2
r  rE 
m p (np lasma  nneutra l )VISM2  B2 /  o
1 AU
1.6102 7 5106 (4.4105 )2
1.610 2 7  (104  105 )  (2.5 10 4 )2  (3 10 1 0) 2 / 4 10 7
 92 AU
However, neutral gas generally flows through the heliosphere without interaction.
It only contributes to the interaction with the solar wind after pickup ions have
been created. This happens much further inside, where the solar wind density is
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Phys 954, Solar Wind and Cosmic Rays
E. Möbius
much higher. Thus the effect of neutrals on the size of the heliosphere is much
smaller. Therefore, adding neutral gas is not a valid assumption.
c) If the heliosphere is pushed to be smaller than 1.5 AU we can use both plasma
and neutral density in the pressure balance, because the neutral gas is now
ionized at the distance of the boundary and thus contributes fully to the
pressure balance. To estimate the density of an interstellar gas cloud we go
back to the first relation used in a). We neglect magnetic pressure, assuming
that the typical magnetic field strength is of the order as used above. We had
seen that under current conditions the magnetic field pressure is comparable
with the plasma pressure. Therefore, it will be negligible for a dense cloud that
makes the heliosphere so small. Furthermore, we only compute the total
density (plasma + gas) of the interstellar cloud, because both components now
react in the same way.
For a heliosphere of 1 AU we solve the relation for the interstellar density:
r2 V 2
nISM  nsw(1AU ) E2 SW 2
r VISM
440 2
)  1550 cm 3
25
This would be a fairly high-density cloud, but at distances that bring them
across our solar system within ≈ 50,000 years there are clouds with densities
of several 100 cm-3. Therefore, such a scenario is not unreasonable.
nISM  5 cm 3 (
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Phys 954, Solar Wind and Cosmic Rays
E. Möbius
4) Gravitational Interaction and Collisions
a) The typical distance for gravitational interaction of the sun (i.e. Ekin = Egrav) is:
2GMs 2  6.67  1011  2 10 30
rgrav 

 4.31011 m  2.8 AU
Vb 2
(2.5  104 )2
The mean free path for neutral-neutral collisions is:
1
1
17
15
3
coll 

 10 cm  10 m  6.610 AU
No  
0.110 16
coll
 510 1
3
rgrav
Therefore, the neutral gas flow through the solar system is collisionless.
b) Charge exchange of neutral interstellar gas atoms by the solar wind
Hsw  H 

Hsw H 
produces neutral particles, which travel with solar wind speed. This is a neutral
solar wind component. The neutral solar wind cannot be stopped by
magnetospheres or at the heliospheric boundary, because it does not feel magnetic
fields. It needs collisions to be stopped, and as shown in a) the mean free path for
collisions is much larger than the heliosphere.
Because of the large mean free path it can flow out of the solar system. The most
likely fate is charge exchange with the interstellar plasma, but this will modify the
interstellar plasma for distances larger than the heliosphere itself.
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