Space Plasma Physics — Sample Solutions —

International University Bremen
Joachim Vogt
Space Plasma Physics
— Sample Solutions —
Sheet 3
Spring 2006
9. Graphical solution of the collisionless Boltzmann equation
(a) The phase space orbit of a particle with initial position z = 0 at time t = t0 , and initial
velocity v0 = v(t = t0 ) is given by
v(t) = v0 − g(t − t0 ) ,
g
z(t) = v0 (t − t0 ) − (t − t0 )2 ,
2
therefore,
π(t; t0 , v0 ) =
z(t)
v(t)
!
−g(t − t0 )2 /2
−g(t − t0 )
=
!
+ v0
t − t0
1
!
= π 0 (t; t0 ) + v0 π 1 (t; t0 ) .
Since this is the parametric representation of a straight line with parameter v0 , all particles
ejected at the same time t0 with different velocities v0 will stay connected through a straight
line at all times t.
(b) According to the collisionless Boltzmann equation, the particle distribution function is
passively advected along particle trajectories in phase space. The initial dust particle
distribution is located between the two trajectories with initial velocities vmin and vmax ,
and between the times t = 0 and t = τ . At t = 3τ , t = 5τ , t = 7τ , the distribution
must stay between the two curves but has advanced in time as suggested by the time tags
attached to the curves. (Note that since the velocity is a linear function of time, time tags
t = τ, 2τ, 3τ, . . . can be easily attached to the trajectories as they correspond to equally
spaced values on the velocity axis.) The resulting distribution functions are shown in the
upper four panels of figure 1.
(c) The particle number density n(z, t) is obtained by integration over the velocity variable v.
In this example one may infer the extent of the distribution in v direction from the plot of
the distribution function, and then use this information to sketch the density as it evolves
in time. The result should look like the lower four panels of figure 1.
(d) The two-dimensional phase space occupied by the dust particle distribution changes shape
as time evolves, however, the volume remains constant as a consequence of the collisionless
Boltzmann equation.
1
Figure 1: Evolution of the distribution function and the number density in problem 9.
Although not required for the solution of the problem, it is instructive and convenient to nor2 /g, f , and f v
malize the equations using vmin , vmin /g, vmin
∗
∗ min to scale velocity, time, position,
distribution function f , and number density n, respectively. This normalization is used in figure 1. The scaled particle trajectories can be written as follows
zˆ(tˆ) = vˆ0 tˆ − tˆ2 /2 ,
vˆ(tˆ) = vˆ0 − tˆ
(for t0 = 0).
10. Collisionless atmosphere
(a) Since the total energy E = E(r, v) is conserved along particle orbits, and all functions
f (r, v) = g(E) solve the collisionless Boltzmann equation, we have to find a function
g(E(z, v)) which gives f0 (v) at the lower boundary z = 0. The total energy is given by
E(z, v) =
mv 2
+ mgz ,
2
thus
mv 2
+ mgz
2
g(E) = g
2
!
.
At z = 0,
q
f (z = 0, v) = f0 (v) = f0
2E(z = 0, v)/m
.
Therefore, the function
s
f (z, v) = f0 

2E(z, v) 
m
is a solution to the collisionless Boltzmann equation and satisfies the boundary condition.
(b) If the boundary distribution is of Maxwell-Boltzmann type
m
2πkB T
f (z = 0, v) = n0
3/2
!
m|v|2
exp −
2kB T
,
then
f (z, v) = n0
m
2πkB T
3/2
mgz
= n0 exp −
kB T
= n(z)
3/2
m
2πkB T
!
m|v|2 + mgz
exp −
2kB T
m
2πkB T
3/2
m|v|2
exp −
2kB T
m|v|2
exp −
2kB T
!
!
.
The density profile n(z) is given by
n(z) = n0 exp −
mgz
kB T
= n0 e−z/H
(barometric law) with the scale height H = kB T /(mg).
11. Krook collision term
(a) The number density is obtained from the distribution function through integration over
velocity space:
Z
n0 =
"
v∗
Z
3
2
f0 (|v|) d v = 4πf∗
v dv = 4πf∗
0
v3
3
#v=v∗
=
v=0
4πv∗3
f∗ .
3
The bulk velocity hvi is identically zero because
f∗
hvx i = hv sin ϑ cos ϕi =
n0
2π
Z
cos ϕ dϕ
0
Z
{z
=0
Z
sin ϕ dϕ
0
f∗
hvz i = hv cos ϑi =
n0
Z
2π
{z
Z
dϕ
0
π
Z
2
sin ϑ dϑ
0
=0
π
v 3 dv = 0 ,
}
Z
0
{z
=[− cos 2ϑ/4]π
0 =0
3
v∗
0
sin ϑ cos ϑ dϑ
|0
v 3 dv = 0 ,
0
}
2π
|
v∗
Z
2
sin ϑ dϑ
0
|
f∗
hvy i = hv sin ϑ sin ϕi =
n0
π
Z
}
v∗
v 3 dv = 0 .
Here ϕ and ϑ denote azimuth and polar angle in velocity space, respectively.
The scalar pressure is
4πf∗
mn0 D 2 E
mn0 2
= m
hvx i + hvy2 i + hvz2 i =
v
3
3
3
" #v=v∗
5
5
4πf∗ v
4πf∗ v∗
mn0 2
= m
= m
=
v .
3
5 v=0
3 5
5 ∗
Z
p0 =
v∗
v 4 dv
0
This yields for the kinetic temperature
T0 =
mv∗2
p0
=
.
n0 kB
5kB
(b) The equilibrium distribution of Maxwell-Boltzmann type
f∞ (|v|) = n∞
m
2πkB T∞
3/2
m|v − vb,∞ |2
exp −
2kB T∞
!
has the parameters n∞ (number density), vb (bulk velocity), and T∞ (temperature). Since
the system is spatially homogeneous, particle number conservation implies constant number density, thus
4πv∗3
n∞ = n0 =
f∗ .
3
Momentum conservation requires in this case n0 vb,0 = n∞ vb,∞ , hence vb,∞ = 0. Energy
conservation gives
mv∗2
m
5
T∞ = T0 =
⇒
=
5kB
2kB T∞
2v∗2
which, finally, yields for the equilibrium distribution
5
f∞ (|v|) =
3
r
10
5|v|2
f∗ exp − 2
π
2v∗
!
.
(c) Since the equilibrium distribution f∞ in the equation
f − f∞
∂f
= −
∂t
τ
does not depend on time t, we can define
δf = f − f∞
and write
δf
∂δf
= −
.
∂t
τ
This first-order linear differential equation has the solution
δf (v, t) = δf (v, t = 0) e−t/τ ,
thus
f (v, t) = f∞ (v) + δf (v, t) = f∞ + [f0 (v) − f∞ (v)] e−t/τ
= f0 (v) e−t/τ + f∞ (v) 1 − e−t/τ
4
.
12. Incompressibility in phase space (∗)
(a) The divergence in phase space
˜ ·Φ =
∇
∂
·Φ =
∂π
∂/∂r
∂/∂v
!
·Φ =
∂
∂ F(r, v, t)
·v +
·
∂v
m
|∂r{z }
=0
=
1 ∂Fy
1 ∂Fz
1 ∂Fx
+
+
m ∂vx
m ∂vy
m ∂vz
(note that we are looking at the non-relativistic case) reduces to partial derivatives of
the force components with respect to the components of velocity. Hence force-independent
force fields yield zero phase flow divergence.
(b) In cartesian components the Lorentz force is given by




Fx
vy Bz − vz By




q  Fy  = F = qv × B = q  vz Bx − vx Bz  .
Fz
vx By − vy Bx
Since none of Fi ’s depends on the corresponding velocity component vi (i = x, y, z),
∂Fi /∂vi = 0, and the phase flow divergence of the Lorentz force is identically zero.
For linear friction F = −kv (k = const) we obtain
∂Fx
∂Fy
∂Fz
=
=
= −k
∂vx
∂vy
∂vz
and thus
˜ · Φ = −3 k .
∇
m
5