156
An Introduction to Beam Physics
generic symplectic matrix, all the conclusions hold after replacing cos μ 1 −
cos μ 2 with (tr ˆ
A − tr ˆ
B)/2. As a result, we have
cos (2ϕ) =
tr( ˆ
M − ˆ
N )
2 (cos μ 1 − cos μ 2 )
.
From the fourth equation of (6.16), on the other hand, we have
ˆ
n = −
ˆ
A ˆ
D
−1
− ˆ
D
−1 ˆ
B
sin ϕ cos ϕ =
ˆ
A ˆ
J ˆ
D
T ˆ
J − ˆ
J ˆ
D
T ˆ
J ˆ
B
sin ϕ cos ϕ,
and hence
ˆ
J ˆ
n
T ˆ
J
T = ˆ
J
ˆ
J
T ˆ
D ˆ
J ˆ
A
T
− ˆ
B
T ˆ
J
T ˆ
D ˆ
J
ˆ
J
T sin ϕ cos ϕ
= −
ˆ
D ˆ
A
−1
− ˆ
B
−1 ˆ
D
sin ϕ cos ϕ.
Adding the third equation of (6.16) on both sides, we obtain
ˆ
m + ˆ
J ˆ
n
T ˆ
J
T = −
ˆ
D
ˆ
A + ˆ
A
−1
−
ˆ
B + ˆ
B
−1
ˆ
D
sin ϕ cos ϕ
= −
ˆ
D(tr ˆ
A) − (tr ˆ
B) ˆ
D
sin ϕ cos ϕ
= − (cos μ 1 − cos μ 2 ) sin (2ϕ) ˆ
D.
(6.18)
Finally, we obtain
ˆ
D = −
ˆ
m + ˆ
J ˆ
n
T ˆ
J
T
(cos μ 1 − cos μ 2 ) sin (2ϕ)
.
Now, the only unknown quantity is cos μ 1 − cos μ 2 , which can be obtained
below. From eq. (6.18), we have
det
ˆ
m + ˆ
J ˆ
n
T ˆ
J
T
= (cos μ 1 − cos μ 2 )
2 sin
2 (2ϕ) .
Adding the square of eq. (6.17), we obtain
cos μ 1 − cos μ 2 =
1
2
tr( ˆ
M − ˆ
N )
1 +
det( ˆ
m + ˆ
J ˆ
n T ˆ
J T )
tr( ˆ
M − ˆ
N )/2
2 .
Now let us consider the case that ˆ
M 4 is symmetric, i.e., ˆ
M
T = ˆ
M , ˆ
N
T = ˆ
N
and ˆ
n
T = ˆ
m. As a result, we have
ˆ
D = −
ˆ
m + ˆ
J ˆ
m ˆ
J
T
(cos μ 1 − cos μ 2 ) sin (2ϕ)
,