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the momentum components of k to zero, we find (2.291)
∞
∞
2π
τ ˜(k r ; 0) =
dρ 0 ρ 0
dz 0 ψ(ρ 0 , z 0 )
dφ 0 exp (−ik r · f r ),
0
−∞
0
(2.293)
where k r is the three-vector spatial part of the six-vector k. We
have set the three-vector momentum part k p to zero. This is equivalent to integrating over all momentum values in direct space.
Next we must find the spatial trajectory displacement f r at time
t. For the case in which the two particles are initially at rest, the
scattering reduces to the Kepler problem for zero angular momentum, where the particles fly apart along a line joining them. In
this case the solution to the Kepler problem reduces to
2
e
r
r 0
r 0
t
3
=
1 − + tanh
−1 1 − ,
(2.294)
πf 0 mr 0
r 0
r
r
where t is transit time, r 0 is initial separation, and r is separation
at time t. Substituting this into (2.293) we obtain
∞
∞
r
τ ˜(k ρ , 0; 0) = 2π
dρ 0 ρ 0
dz 0 ψ(ρ 0 , z 0 ) J 0
1
2
k ρ ρ 0
− 1 ,
0
−∞
r 0
(2.295)
where k ρ is the two-vector spatial part in the transverse (ρ, φ)
plane. We have made use of assumed axial symmetry and the
integral representation of the Bessel function J 0 (x) as
2π
J 0 (x) =
1
dφ e
−i x cos φ .
(2.296)
2π 0
We assume the particles are initially uniformly distributed over
a cylindrical volume with radius a and length L. It follows that
ψ 0 (r 0 ) is convolution of a cylindrical volume with itself, as follows:
⎡
⎤
1
2
ρ 0
ρ 0
ρ 0
2
|z 0 |
ψ 0 (r 0 ) =
· ⎣ cos
−1
−
1 −
⎦ · 1 −
,
πa 2 L π
2a
2a
2a
L
(2.297)
where ψ 0 is nonzero for
0 ≤ ρ 0 ≤ 2a, −L ≤ z 0 ≤ L,
(2.298)
108
Chapter 2. Geometrical optics
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the momentum components of k to zero, we find (2.291)
∞
∞
2π
τ ˜(k r ; 0) =
dρ 0 ρ 0
dz 0 ψ(ρ 0 , z 0 )
dφ 0 exp (−ik r · f r ),
0
−∞
0
(2.293)
where k r is the three-vector spatial part of the six-vector k. We
have set the three-vector momentum part k p to zero. This is equivalent to integrating over all momentum values in direct space.
Next we must find the spatial trajectory displacement f r at time
t. For the case in which the two particles are initially at rest, the
scattering reduces to the Kepler problem for zero angular momentum, where the particles fly apart along a line joining them. In
this case the solution to the Kepler problem reduces to
2
e
r
r 0
r 0
t
3
=
1 − + tanh
−1 1 − ,
(2.294)
πf 0 mr 0
r 0
r
r
where t is transit time, r 0 is initial separation, and r is separation
at time t. Substituting this into (2.293) we obtain
∞
∞
r
τ ˜(k ρ , 0; 0) = 2π
dρ 0 ρ 0
dz 0 ψ(ρ 0 , z 0 ) J 0
1
2
k ρ ρ 0
− 1 ,
0
−∞
r 0
(2.295)
where k ρ is the two-vector spatial part in the transverse (ρ, φ)
plane. We have made use of assumed axial symmetry and the
integral representation of the Bessel function J 0 (x) as
2π
J 0 (x) =
1
dφ e
−i x cos φ .
(2.296)
2π 0
We assume the particles are initially uniformly distributed over
a cylindrical volume with radius a and length L. It follows that
ψ 0 (r 0 ) is convolution of a cylindrical volume with itself, as follows:
⎡
⎤
1
2
ρ 0
ρ 0
ρ 0
2
|z 0 |
ψ 0 (r 0 ) =
· ⎣ cos
−1
−
1 −
⎦ · 1 −
,
πa 2 L π
2a
2a
2a
L
(2.297)
where ψ 0 is nonzero for
0 ≤ ρ 0 ≤ 2a, −L ≤ z 0 ≤ L,
(2.298)
108
Chapter 2. Geometrical optics