�
�
�
�
Letting r → ∞ after the scattering has taken place, we obtain
2HL 2
0 = 1 + 1 +
cos (ϑ − θ 0 ).
(4.46)
M κ 2
It is evident from the figure that
2(θ 0 − ϑ) + ϑ = π,
(4.47)
from which it follows that
ϑ π
cos(ϑ − θ 0 ) = cos
−
= sin(ϑ/2).
(4.48)
2
2
The conserved angular momentum L is given at any given point
along the trajectory by
L = r × p.
(4.49)
Considering the incident particle far from the scattering center,
we write
√
b
L = lim r p 0 · = b 2M H,
(4.50)
r→∞
r
√
where p 0 = 2M H is the initial momentum. Substituting this
into (4.46), we find
4H
2 b
2 −1/2
sin(ϑ/2) = − 1 + κ 2
.
(4.51)
Solving for the impact parameter b, this leads to
κ
b = 2H
cot(ϑ/2).
(4.52)
Differentiating, we obtain
db
dϑ
= −
κ
4H
csc
2 (ϑ/2).
(4.53)
Substituting (4.52, 4.53) into (4.43), we obtain the result for the
differential cross section as
κ
2
σ(ϑ) =
.
(4.54)
16 H 2 sin
4 (ϑ/2)
251
4.2. Scattering cross section and classical scattering
�
�
�
Letting r → ∞ after the scattering has taken place, we obtain
2HL 2
0 = 1 + 1 +
cos (ϑ − θ 0 ).
(4.46)
M κ 2
It is evident from the figure that
2(θ 0 − ϑ) + ϑ = π,
(4.47)
from which it follows that
ϑ π
cos(ϑ − θ 0 ) = cos
−
= sin(ϑ/2).
(4.48)
2
2
The conserved angular momentum L is given at any given point
along the trajectory by
L = r × p.
(4.49)
Considering the incident particle far from the scattering center,
we write
√
b
L = lim r p 0 · = b 2M H,
(4.50)
r→∞
r
√
where p 0 = 2M H is the initial momentum. Substituting this
into (4.46), we find
4H
2 b
2 −1/2
sin(ϑ/2) = − 1 + κ 2
.
(4.51)
Solving for the impact parameter b, this leads to
κ
b = 2H
cot(ϑ/2).
(4.52)
Differentiating, we obtain
db
dϑ
= −
κ
4H
csc
2 (ϑ/2).
(4.53)
Substituting (4.52, 4.53) into (4.43), we obtain the result for the
differential cross section as
κ
2
σ(ϑ) =
.
(4.54)
16 H 2 sin
4 (ϑ/2)
251
4.2. Scattering cross section and classical scattering