Magnetospheric Physics - Homework Solutions, 01/31/2014
3. Debye Screening
(a) Demonstrate that the plasma definition heφi Λ = nλ3D 1.
D
m 2
v
2
E
= kB T implies quasi-neutrality, i.e.,
−1/2
(b) The plasma parameter has the basic dependence Λ ∝ n0 T 3/2 . While the dependence on
temperature is intuitively clear, the density dependence appears odd because lower densities seems
to imply fewer particles and less shielding. Why is intuition wrong and why does the plasma
parameter improve (increase) with decreasing density?
Solution:
3
(a) With φtyp ' e/(0 rtyp ) and n ' 1/rtyp
0 kB T
Λ=n
ne2
!3/2
= n−1/2
0 kB T
e2
!3/2

3/2
' rtyp 
0 23
D
m 2
v
2
2
e
E 3/2

D
'
m 2
v
2
E 3/2
heφi

1
3
decreasing density n implies a larger typical separation of particles.
(b) Because of n ' 1/rtyp
However, for a larger separation the potential energy becomes smaller (∼ 1/r), i.e., less important
compared to the thermal energy, such that the plasma approximation improves for constant average
kinetic energy.
4. Plasma properties
a) Assume a plasma density of 1 cm−3 , temperature equivalent to 1 keV, and a magnetic field of
20 nT which are typical for the near Earth magnetotail. Determine Debye radius and the plasma
parameter Λ.
b) Compute the electron plasma frequency, the collision frequency for electrons, and the thermal
velocity, and the mean free path of an electron based on these numbers. Can one neglect collisions
for these electrons for the length scales of the magnetosphere of 100 RE ? (1 RE =6400 km)
Solution: Note 1 eV = 1.6 · 10−19 J = kB T0 ,→ T0 = 1.6 · 10−19 /1.38 · 10−23 K = 1.16 · 104 K
Measuring thermal energy in equivalents of 1eV : ε = T̃ · 1.6 · 10−19 J (which yields an actual
temperature of T = T̃ T0 ). Furthermore measuring density in particles/cm3 through n = ñ106
In this notation ñ = 1 and T̃ = 1000. Below, λD and Λ are derived as general impression for input
in units of eV and particles/cm3 . This was not needed for the homework.
(a) Debye length:
0 kB T
ne2
λD =
!1/2
=
0 eT̃
ne2
8.85 · 10−12 T̃
106 · 1.6 · 10−19 ñ
=
T̃
= 7.44
ñ
!1/2
!1/2
8.85 · 10
m=
1.6
1/2
T̃
ñ
!1/2
m
!1/2
m = 235m
Plasma parameter: Note that even though we used as input particles/cm3 the Debye length has
units of m. To get the plasma parameter we can just evaluate Λ = nλ3D = 1.3 · 1013 . To obtain a
more general expression with n in particles/cm3 as above we need λD in cm:
0 kB T
Λ = n
ne2
!3/2
= ñ · 7.443 · 103·2
= 4.12 · 108
= 106 ñλ3D
T̃ 3/2
ñ3/2
T̃ 3/2
= 1.3 · 1013
ñ1/2
Sometimes the plasma parameter is defined as particles in a Debye sphere in which case one
multiplies the result with a factor of 4π/3 ' 10.
(b) Plasma frequency:
ωpe =
=
ne2
me 0
!1/2
1.62 · 103
9.1 · 8.85
=
106 · 1.62 · 10−2·19
9.1 × 10−31 · 8.85 · 10−12
!1/2
ñ1/2
!1/2
104 · ñ1/2 = 5.64 · 104 · ñ1/2 s−1
Collision frequency:
3/2
ωpe
5.64 · 104
ñ
8 T̃
=
ln Λ =
ln 4.12 · 10 1/2
4πΛ
4π · 4.12 · 108 T̃ 3/2
ñ
"
νei
ñ
−5
= 1.09 · 10
T̃ 3/2
"
8 T̃
ln 4.12 · 10
h
3/2
#
s−1
#
ñ1/2
s−1
i
= 1.09 · 10−10 101/2 ln 1.3 · 1013 s−1
= 3.44 · 10−10 · 30.2 s−1 = 1.04 · 10−8 s−1
Alternatively (with a more accurate evaluation of the velocity dependence of the collision cross
section)
r
π 1 ωpe
νc =
ln [12πΛ]
2 32π Λ
q
which gives an additional factor of π/2/8 = 0.157 and a factor of 12π in the Coulomb logarithm
ln [12πΛ] = ln [4.9 · 1014 ] = 33.8 to yield
νei = 5.4 · 10−11 · 33.8 s−1 = 1.8 · 10−9 s−1
Thermal velocity: There are different equally justified definitions but their use may depend on the
2
2
= 32 kB T = thermal
particular application. Definitions: (a) vth
= hv 2 i = 3kB T /m such that 21 mvth
2
= hvi2 i = kB T /m; (c) by
energy; (b) by the average along only a single cartesian direction vth
1/2
the mean of the magnitude vth = h|v|i = (8kB T /πm) or; (d) by the halfwidth of a Maxwell
2
2
= 2kB T /m. Since all contain the factor
) which yieldsqvth
distribution, i.e, f ∼ exp (−v 2 /vth
kB T /m the result below uses definition (b) vth = kB T /m.
vthe =
kB
me
!1/2
Te1/2
=
1.38 · 10−23
9.1 · 10−31
!1/2
Te1/2 m s−1 = 3.9 · 103 Te1/2 m s−1 = 1.3 · 107 m s−1
with Te = 1.16 · 107 K. Using temperature measured in eV kB Te = eT̃e :
vthe =
eT̃e
me
!1/2
=
1.6 · 10−19
9.1 · 10−31
!1/2
T̃e1/2 m s−1 = 4.2 · 105 T̃e1/2 m s−1 = 1.3 · 107 m s−1
Mean free path:
vthe
1.3 · 107
=
m = 1.25 · 1015 m = 1.95 · 108 RE
νei
1.04 · 10−8
or for the more accurate collision frequency
1.3 · 107
=
m = 7.2 · 1015 m = 1.1 · 109 RE
1.8 · 10−9
Lc =
Lc
The mean free path is many order of magnitude larger than any dimension associated with the
Earth’s magnetosphere. Therefore the system is highly collisionless.
5. Plasma energy
(a) For the plasma in problem 4, determine the temperature in degrees Kelvin, and the thermal and
magnetic energy density. How do these compare?
3
. For the sake of simplicity
(b) Express the energy densities in kW hours/m3 and kW hours /RE
assume that the plasma sheet is represented by a cylinder with 15 RE radius and 100 RE length.
How long could a power plant with an output of 1000 MW operate on the thermal energy stored in
the plasma?
Solution:
(a) With the note in Problem 4: A temperature of 1 eV ,→ T0 = 1.16 · 104 K
Thermal energy density:
3
εth = nkB T = 1.5 · 106 · 1.38 · 10−23 · 1.16 · 107 m−3 JK −1 K = 2.4 · 10−10 J/m3
2
Magnetic energy density:
εB = B 2 /2µ0 =
22 · 10−2·8
m−3 J/m3 = 1.6 · 10−10 J/m3
2 · 4 · π · 10−7
Thermal and magnetic energy density are comparable.
(b) Power plant part:
1.6 · 10−10
kW hours/m3
ε = 2.4 · 10 J/m = 3
10 · 3600
= 6.7 · 10−17 kW hours/m3
−10
3
3
3
• In kW hours /RE
: ε = 6.7 · 10−17 · 6.43 · 1018 = 1.7 · 104 kW hours /RE
.
• Energy in 15 RE radius and 100 RE length cylinder: W = π · 152 · 102 · 1.7 · 104 = 1.2 · 109
kW hours
• Operation time for 1000 MW power plant t = W/106 = 1200 hours≈ 50 days
6. Collisionless Boltzmann equation
(a) Show that any distribution function f (x, v, t) = F (H, Py ) with H = mv2 /2 + qφ (r) and
Py = mvy +qAy solves the steady state (∂/∂t = 0) collisionless Boltzmann equation for ∂/∂y = 0.
(b) Consider a distribution function of the form F (H) = c0 exp (−H/kB T ). Show that the plasma
density is given by n (r) = n0 exp (−qφ/kB T ) and express c0 in terms of n0 and kB T .
Solution:
(a) Collisionless Boltzmann equation
∂f
q
+ v · ∇f + (E + v × B) · ∇v f = 0
∂t
m
Individual terms:
∂f
∂F (H, Py )
=
=0
∂t
∂t
∇f terms:
∂F ∂Py
∂F ∂H
∇φ +
v · ∇f = v ·
∇Ay
∂H ∂φ
∂Py ∂Ay
∂F
∂F
= −q
v·E+q
v · ∇Ay
∂H
∂Py
!
and ∇v f terms:
"
#
q
q
∂F
∂F
(E + v × B) · ∇v f =
(E + v × B) ·
∇v H +
∇ v Py
m
m
∂H
∂Py
"
#
q
m ∂F
∂F
2
=
(E + v × B) ·
∇v v + m
∇v vy
m
2 ∂H
∂Py
∂F
∂F
= q
E·v+q
(E + v × B) · ey
∂H
∂Py
Note (E + v × B)·ey = vz Bx −vx Bz with Bx = ∂y Az −∂z Ay = −∂z Ay and Bz = ∂x Ay −∂y Ax =
∂x Ay . Further Ey = −∂y φ = 0 because ∂y = 0. Therefore
(E + v × B) · ey = −vz ∂z Ay − vx ∂x Ay = −v · ∇Ay
such that the sum of the ∇f terms and the ∇v f terms is 0!
(b) With F (H) = c0 exp (−H/kB T ) the density is given by
n (r) =
Z +∞
−∞
3
d vf (r, v, t) = c0
qφ
= c0 exp −
kB T
!Z
qφ
= c0 exp −
kB T
!Z
+∞
−∞
+∞
−∞
Z +∞
−∞
d3 v exp (−H/kB T )
mv2
d v exp −
2kB T
!
3
!
!
!
vy2 Z +∞
vx2 Z +∞
vz2
dvx exp − 2
dvy exp − 2
dvz exp − 2
vth −∞
vth −∞
vth
!#3
+∞
v2
qφ
dv exp − 2
= c0 exp −
kB T
vth
−∞
! Z
3
+∞
qφ
3
c0 exp −
= vth
dṽ exp −ṽ 2
kB T
−∞
! "Z
!
qφ
=
exp −
=
kB T
2
with
vth
= 2kB T /m
and
3
π 3/2 vth
c0
!3/2
2πkB T
qφ
c0 exp −
m
kB T
ṽ = v/vth
where we used the indefinite integral
I0 =
Z +∞
dṽ exp −ṽ 2 =
−∞
in comparison with n (r) = n0 exp (−qφ/kB T ) we find
m
c0 =
2πkB T
3/2
n0
√
π
!