Basic Plasma Physics - Homework, 3/2/2012 1. Debye shielding ...

Basic Plasma Physics - Homework, 3/2/2012
1. Debye shielding:
Consider a test charge q t and compute the net charge of the Debye shielding cloud (Yukawa potential) as
a function of radial distance measured in units of the Debye length.
Compute the net charge of the shielding cloud for the Yukawa potential.
with
Poisson’s equations:
Φ D =
q (
t
4πɛ 0 r exp − r )
λ D
λ −2
D = n 0e 2
ɛ 0 k B
( 1
T e
+ 1 T i
)
∇ 2 Φ = − 1 q t δ(r) + 1 Φ(r) = − 1 ρ
ɛ 0 λ 2 c
D ɛ 0
(1)
This yields for the charge density:
ρ c = q t δ(r) − ɛ 0
Φ(r)
λ 2 D
= q t δ(r) − q t
4πλ 2 D
Integration over total space yields the total charge:
(
1
r exp − r )
λ D
Q net = q t − q t
4πλ 2 D
= q t − q t
λ 2 D
= q t − q t
λ 2 D
= q t − q t
λ 2 D
∫ 2π ∫ π ∫ R
0
∫ R
0
[ ] R
= q t + 1
λ D
0
r exp
[
−λ D r exp
[
−λ D R exp
0
(
− r
(
1
r exp − r )
r 2 dr sin θdθdφ
λ D
)
dr
λ D
(
− r
exp
(− R )
λ D
)
(
− λ 2 D exp − r )] R
λ D λ D 0
(− R )
− λ 2 D exp
(− R ) ]
+ λ 2 D
λ D λ D
where the integral over r is solved through integration by parts.
Note, an alternative derivation of the charge density can be obtained by computing ∇ 2 Φ D as indicated in
class (with the radial electric field given by

E r = − ∂Φ ( )
(
r
∂r = q t
+ 1
λ D 4πɛ 0 r exp − r )
2 λ D
Q net =
∫ R
sphere
ρ c d 3 r = ɛ 0
∫ R
sphere
∇ · Ed 3 r
∫ R
= ɛ 0 E · d 2 s = 4πR 2 ɛ 0 E r (R)
surface
( ) R
= q t + 1 exp
(− R )
λ D λ D
Note be care full to consider that ∇(1/r) = r/r 3 and ∇ · (r/r 3 ) = 4πδ(r).
2. Plasma properties:
(a) Calculate the electron thermal speed, Debye length, and the plasma parameter for
• a tokamak plasma with T e = 10 8 K, n 0 =10 19 m −3
• the tail magnetosphere with T e = 10 7 K, n 0 =10 6 m −3
• the ionosphere with T e = 10 3 K, n 0 =10 12 m −3
• the solar atmosphere with T e = 10 4 K, n 0 =10 20 m −3
with k B = 1.38 · 10 −23 JK −1 , ɛ 0 = 8.85 · 10 −12 F/m, m e = 9.11 · 10 −31 kg, and e = 1.60 · 10 −19 C
Solution: Electron thermal speed:
Electron Debye length:
( ) 1/2
kB T e
v the =
= 3.9 · 10 3 Te
1/2 m/s
m e
λ De =
=
( ) 1/2
ɛ0 k B T e
n 0 e 2
( 8.85 · 10
−12 · 1.38 · 10 −23 ) 1/2 (Te
1.6 2 · 10 −38
n 0
) 1/2
m
= 69.1 ·
( Te
n 0
) 1/2
m
Electron plasma parameter:
Λ e = n 0 λ 3 De = 3.3 · 10 5 T 3/2
e
n 1/2
0

T e / K n 0 / m −3 v the / m/s λ De / m Λ e
Tokamak 10 8 10 19 3.9 · 10 7 2.2 · 10 −4 1.04 · 10 8
Magnetotail 10 7 10 6 1.2 · 10 7 220 1.04 · 10 13
Ionosphere 10 3 10 12 1.2 · 10 5 2.2 · 10 −3 1.04 · 10 4
Solar atmosphere 10 4 10 20 3.9 · 10 5 6.9 · 10 −7 33
(b) Compute collision frequency and mean free path for these plasmas.
Solution: Plasma frequency:
ω pe =
(
n0 e 2
m e ɛ 0
) 1/2
(2)
ω pe =
Collision frequency:
(
n0 e 2
m e ɛ 0
) 1/2
= 1.60 · 10 −19 (
n 0
) 1/2
1.60 · 10 −19
=
9.11 · 10 −31 8.85 · 10 −12 2.84 · 10
−21
n1/2
0 = 56.34n 1/2
0
Mean free path:
√ π 1 ω p
ν c =
ln [12πΛ]
2 32π Λ
= 1.247 · 10 −2 ω p
ln (37.7Λ)
Λ
L c = v the /ν c
=
√ ( ) 1/2 (me )
√
2
π 32π kB T e ɛ 1/2 0 Λ 2
m e n 0 e 2 ln [12πΛ] = π 32π λ DeΛ
ln [12πΛ]
n 0 / m −3 Λ e v the / m/s ω p / s −1 ν c / s −1 ln (37.7Λ) L c / m
Tokamak 10 19 1.04 · 10 8 3.9 · 10 7 1.78 · 10 11 4.71 · 10 2 22.09 8.28 · 10 4
Magnetotail 10 6 1.04 · 10 13 1.2 · 10 7 5.63 · 10 4 2.27 · 10 −9 33.60 5.29 · 10 15
Ionosphere 10 12 1.04 · 10 4 1.2 · 10 5 5.63 · 10 7 8.69 · 10 2 12.88 1.38 · 10 2
Solar Atmosph. 10 20 33 3.9 · 10 5 5.63 · 10 11 1.52 · 10 9 7.13 2.5 · 10 −4

3. Plasma Definition
Can a fully ionized plasma be maintained at temperatures of T e = 100 K (Hint: derive a condition for the
density in relation to the temperature). How important is recombination for such a cold plasma
Solution:
Plasma definition:
Taking the square
Λ = nλ 3 D = n −1/2 (
ɛ0 k B T e
e 2 ) 3/2
≫ 1
n −1 (
ɛ0 k B T e
e 2 ) 3
= n −1 ( 8.85 · 10
−12 · 1.38 · 10 −23 · 10 2
1.6 2 · 10 −38 ) 3 [
m
−3 ]
= n −1 ( 4.8 · 10 5) 3 [ m −3] = n −1 · 1.1 · 10 17 [ m −3]
or taking number density as particle per cm −3 √ n ≪ 3.3 · 10
5
which implies that n could be as small as 10 7 or even 10 8 cm −3 to satisfy the plasma definition, i.e., that
the potential energy is much smaller than the kinetic energy of the charged particles.
While this is true there is the additional aspect of how long such a plasma could be maintained. Even
tho a sufficiently low density gas of charged particles can satisfy the plasma definition, recombination
caused by collisions can remove the the charged particles. For long life time of such a plasma it would
be necessary to have a very low level of collisions that could result in recombinations. In other words
the collision frequency must be sufficiently low and with a comparatively small value of the plasma
parameter of for instance 10 2 the collision frequency is still quite large and would lead to relatively fast
recombination leading to a loss of this plasma within a short time frame.