1
Transport theory
Course 16209
Prof. Dr. H. Ruhl and P. Böhl
Exercise Sheet 1
May 05 , 2016
Problem 1: We know that the probability density
C(x1 , . . . , xN , p1 , . . . , pN , t) =
N
Y
δ 3 (xj − xj (t))δ 3 (pj − pj (t))
j=1
for a system of N classical point particles satisfies the Liouville-equation
N
N
X ∂C
∂C X ∂C
+
+
=0
vj
ṗj
∂t
∂xj j=1 ∂pj
j=1
Consider a force-free, non-interacting system of N point-particles where the initial positions x0,j and momenta p0,j
are given by an arbitrary function f (x0,1 , . . . , x0,N , p0,1 , . . . p0,N ).
Show that the averaged probability function
Z Y
N
ρN (x1 , . . . , xN , p1 , . . . , pN , t) =
d3 p0,j d3 x0,j C(x1 , . . . , xN , p1 , . . . , pN , t)f (x0,1 , . . . , x0,N , p0,1 , . . . , p0,N )
j=1
also satisfies the Liouville-equation. Hint: Integrate the equations of motion to obtain an explicit relation between
the coordinates and momenta (xj (t), pj (t)) and the initial conditions (x0,j , p0,j ).
Solution problem 1: For brevity we define X = (x1 , . . . , xN ) and P = (p1 , . . . , pN ). Since the system is force-free,
we have
dP
=0
dt
dX
=0
dt
→
P(t) = P0
→
X(t) = V0 t + X0
Now
Z
ρN (X, P) =
d3N P0 d3N X0 f (X0 , P0 )δ 3N (X − (V0 t + X0 ))δ 3N (P − P0 ))
= f (X − V0 t, P0 )
The Liouville equation in this case is given by:
∂C
∂C
+ V0
=0
∂t
∂X
For ρN we have:
∂ρN
= −f (X − V0 t, P0 )V0
∂t
and therefore
∂ρN
∂ρN
+ V0
=0
∂t
∂X
2
Problem 2: Consider the reduced s = 1 equation for hard spheres
Z
Z
∂t f (1) (~x1 , p~1 , t) + ~v1 · ∂~x1 f (1) (~x1 , p~1 , t) = d3 p2 dS12 ~n12 · (~v1 − ~v2 ) f (2) (~x1 , p~1 , ~x2 , p~2 , t) .
The integration
spheres.
R
dS12 is over the surface that fulfills the condition |~x1 − ~x2 | = σ, where σ is the diameter of the hard
Show that in the limit N → ∞, σ → 0, and 4πσ 3 N/3V , where V is the box volume, it is obtained
∂t f (1) (~x1 , p~1 , t) + ~v1 · ∂~x1 f (1) (~x1 , p~1 , t)
Z σ
Z
Z 2π
h
i
0
0
3
dbb f (2) (~x1 , p~1 , ~x2 , p~2 , t) − f (2) (~x1 , p~1 , ~x2 , p~2 , t) .
dφ
= d p2 |~v1 − ~v2 |
0
0
R
Solution problem 2: We divide the surface integration into an integration over two hemispheres S + dS12 with
12
R
~n12 · (~v1 − ~v2 ) > 0 and S − dS12 with ~n12 · (~v1 − ~v2 ) < 0 and obtain
12
Z
Z
(1)
(1)
3
∂t f (~x1 , p~1 , t) + ~v1 · ∂~x1 f (~x1 , p~1 , t) = d p2
dS12 |~n12 · (~v1 − ~v2 )| f (2) (~x1 , p~1 , ~x2 , p~2 , t)
+
S12
Z
−
d3 p2
Z
−
S12
dS12 |~n12 · (~v1 − ~v2 )| f (2) (~x1 , p~1 , ~x2 , p~2 , t) .
−
+
The points S12
are in-collisions points and points S12
out-collision points. We obtain further
Z
dS12 ~n12 · (~v1 − ~v2 ) f (2) (~x1 , p~1 , ~x2 , p~2 , t)
−
S12
= σ 2 |~v1 − ~v2 |
Z
2π
Z
π
dφ
dθ sin θ cos θ
0
π/2
×f (2) (~x1 , p~1 , ~x1 − σ ~n12 , p~2 , t)
Z 2π
Z π/2
= −σ 2 |~v1 − ~v2 |
dφ
dθ sin θ cos θ
0
0
×f
Z
+
S12
(2)
(~x1 , p~1 , ~x1 − σ ~n12 , p~2 , t) ,
dS12 ~n12 · (~v1 − ~v2 ) f (2) (~x1 , p~1 , ~x2 , p~2 , t)
= σ 2 |~v1 − ~v2 |
Z
2π
Z
π/2
dφ
0
×f
dθ sin θ cos θ
0
(2)
(~x1 , p~1 − ~n12 [~n12 · (~
p1 − p~2 )] , ~x1 − σ ~n12 , p~2 + ~n12 [~n12 · (~
p1 − p~2 )] , t)
With the help of b = σ sin θ, where σ denotes the impact parameter, we obtain dθ sin θ cos θσ 2 = dbb. In the limit
N → ∞, σ → 0, and 4πσ 3 N/3V , where V is the box volume, this leads to
∂t f (1) (~x1 , p~1 , t) + ~v1 · ∂~x1 f (1) (~x1 , p~1 , t)
Z
Z 2π
Z σ
h
i
0
0
= d3 p2 |~v1 − ~v2 |
dφ
dbb f (2) (~x1 , p~1 , ~x2 , p~2 , t) − f (2) (~x1 , p~1 , ~x2 , p~2 , t) .
0
0