240
Chapter 4. Particle scattering
In the low energy limit, β 1 « 1. Keeping only the lowest order
terms in the Taylor series, this reduces to
m 1 m 2
T
�
1
β
2
= 2
1 .
(4.14)
m 1 + m 2
We now define a quantity called the reduced mass M , given by
m 1 m 2
M =
,
(4.15)
m 1 + m 2
from which we write
T
� =
1
2
M β 1
2 .
(4.16)
We conclude from this that the two-body scattering problem in
the CM frame is mathematically equivalent to a single particle of
mass M scattering about a fixed point, which coincides with the
center of momentum. This reduction of the scattering problem is
called the equivalent one-body problem.
We are now in a position to consider the final state after scattering. This is shown in the CM frame in Figure 4.2, where the
particle with rest mass m 1 has final momentum q
�
1 , and the particle with rest mass m 2 has final momentum q
�
2 . We assume total
energies ε
�
1 and ε
�
2 for the two particles after scattering, where
2
� 2
2
ε
�
= q + m
1
1
1
ε
� 2 = q
� 2 + m
2
(4.17)
2
2
2 .
The scattering angle in the CM frame is defined as θ
� . In later
sections, we will address the central scattering problem, namely,
calculation of the scattered intensity as a function of θ
� . Consequently, we assume θ
� to be known for now from the calculation
to come.
Since measurement is always performed in the lab frame, we must
express the relevant quantities there. This is shown in Figure 4.3,
where the incident particle with mass m 1 is assumed to scatter
through angle θ 1 , and the incident particle with mass m 2 is assumed to scatter through angle θ 2 . We wish to calculate the scat
Chapter 4. Particle scattering
In the low energy limit, β 1 « 1. Keeping only the lowest order
terms in the Taylor series, this reduces to
m 1 m 2
T
�
1
β
2
= 2
1 .
(4.14)
m 1 + m 2
We now define a quantity called the reduced mass M , given by
m 1 m 2
M =
,
(4.15)
m 1 + m 2
from which we write
T
� =
1
2
M β 1
2 .
(4.16)
We conclude from this that the two-body scattering problem in
the CM frame is mathematically equivalent to a single particle of
mass M scattering about a fixed point, which coincides with the
center of momentum. This reduction of the scattering problem is
called the equivalent one-body problem.
We are now in a position to consider the final state after scattering. This is shown in the CM frame in Figure 4.2, where the
particle with rest mass m 1 has final momentum q
�
1 , and the particle with rest mass m 2 has final momentum q
�
2 . We assume total
energies ε
�
1 and ε
�
2 for the two particles after scattering, where
2
� 2
2
ε
�
= q + m
1
1
1
ε
� 2 = q
� 2 + m
2
(4.17)
2
2
2 .
The scattering angle in the CM frame is defined as θ
� . In later
sections, we will address the central scattering problem, namely,
calculation of the scattered intensity as a function of θ
� . Consequently, we assume θ
� to be known for now from the calculation
to come.
Since measurement is always performed in the lab frame, we must
express the relevant quantities there. This is shown in Figure 4.3,
where the incident particle with mass m 1 is assumed to scatter
through angle θ 1 , and the incident particle with mass m 2 is assumed to scatter through angle θ 2 . We wish to calculate the scat