158
An Introduction to Beam Physics
where ˆ
T
T
x = ˆ
T x , ˆ
T
T
y = ˆ
T y and ˆ
D
T = ˆ
D
−1 . Using the relation ˆ
D
T = ˆ
D
−1 , we
obtain immediately
ˆ
I cos ϕ ˆ
D
−1 sin ϕ
− ˆ
D sin ϕ ˆ
I cos ϕ
=
ˆ
I cos ϕ ˆ
D
T sin ϕ
− ˆ
D sin ϕ ˆ
I cos ϕ
=
ˆ
I cos ϕ − ˆ
D
−1 sin ϕ
ˆ
D sin ϕ
ˆ
I cos ϕ
T
.
As a result, we have
ˆ
T =
ˆ
I cos ϕ − ˆ
D
−1 sin ϕ
ˆ
D sin ϕ
ˆ
I cos ϕ
T ˆ
T x ˆ 0
ˆ 0 ˆ
T y
ˆ
I cos ϕ − ˆ
D
−1 sin ϕ
ˆ
D sin ϕ
ˆ
I cos ϕ
,
which means that the coordinate change that block diagonalizes ˆ
T in the
linear form also block diagonalizes ˆ
T in the bilinear form. The matrices ˆ
T x
and ˆ
T y can be decomposed the same way as eq. (6.6) and we obtain
ˆ
A
T
x
ˆ 0
ˆ 0 ˆ
A
T
y
ˆ
I cos ϕ ˆ
D
−1 sin ϕ
− ˆ
D sin ϕ ˆ
I cos ϕ
T
ˆ
T
ˆ
I cos ϕ ˆ
D
−1 sin ϕ
− ˆ
D sin ϕ ˆ
I cos ϕ
ˆ
A x ˆ 0
ˆ 0 ˆ
A y
=
ˆ
I ˆ 0
ˆ 0 ˆ
I
,
(6.19)
where
ˆ
A x,y =
β x,y
0
−α x,y /
β x,y 1/
β x,y
.
Similar to eq. (6.9), this equation shows that the matrix
ˆ
A 4 =
ˆ
I cos ϕ ˆ
D
−1 sin ϕ
− ˆ
D sin ϕ ˆ
I cos ϕ
ˆ
A x ˆ 0
ˆ 0 ˆ
A y
transforms the four-dimensional ellipse into two decoupled circles and the
transfer matrix can be written as
ˆ
M 4 =
ˆ
I cos ϕ 2 ˆ
D
−1
2 sin ϕ 2
− ˆ
D 2 sin ϕ 2 ˆ
I cos ϕ 2
ˆ
A x2 ˆ 0
ˆ 0 ˆ
A y2
ˆ
R x ˆ 0
ˆ 0 ˆ
R y
·
ˆ
A
−1
x1
ˆ 0
ˆ 0 ˆ
A
−1
y1
ˆ
I cos ϕ 1 − ˆ
D
−1
1 sin ϕ 1
ˆ
D 1 sin ϕ
ˆ
I cos ϕ 1
,
where
ˆ
R x,y =
cos φ x,y sin φ x,y
− sin φ x,y cos φ x,y
.