112
An Introduction to Beam Physics
The second order equations then read
(x|xx)
x
2
i +(x|xa)
x i a i +(x|aa)
a
2
i = (a|xx)x
2
i +(a|xa)x i a i +(a|aa)a
2
i ,
(a|xx)
x
2
i +(a|xa)
x i a i +(a|aa)
a
2
i = −[(x|xx)x
2
i +(x|xa)x i a i +(x|aa)a
2
i ]
+ k[(x|x)
2 x
2
i +2(x|x)(x|a)x i a i +(x|a)
2 a
2
i ],
where the last line proportional to k is the inhomogeneous part
Q 2 , and using
eq. (4.39),
Q 2 =
Q 2x
Q 2a
=
0
k
cos
2 s · x
2
i + 2 cos s sin s · x i a i + sin
2 s · a
2
i
.
We make the ansatz
R 2 (s) = ˆ
L(s)
T (s) :
(x|xx)x
2
i + (x|xa)x i a i + (x|aa)a
2
i
(a|xx)x
2
i + (a|xa)x i a i + (a|aa)a
2
i
=
cos s sin s
− sin s cos s
T (s) .
From eq. (4.38),
T (s) =
T x
T a
=
s
0
ˆ
L
−1
Q 2 d¯ s =
s
0
cos ¯
s − sin ¯
s
sin ¯
s cos ¯
s
0
Q 2a
d¯ s,
so
T x =
s
0
(− sin ¯
s · Q 2a ) d¯ s
= k
s
0
− cos
2 ¯
s sin ¯
s · x
2
i − 2 cos ¯
s sin
2 ¯
s · x i a i − sin
3 ¯
s · a
2
i
d¯ s
= k
1
3
(cos
3 s − 1)x
2
i −
2
3
sin
3 s · x i a i +
cos s −
1
3
cos
3 s −
2
3
a
2
i
,
T a =
s
0
cos ¯
s · Q 2a d¯ s
= k
s
0
cos
3 ¯
s · x
2
i + 2 cos
2 ¯
s sin ¯
s · x i a i + cos ¯
s sin
2 ¯
s · a
2
i
d¯ s
= k
sin s −
1
3
sin
3 s
x
2
i −
2
3
cos
3 s − 1
x i a i +
1
3
sin
3 s · a
2
i
.
Then we obtain
R 2 (s) = ˆ
L (s)·
T (s) , which yields the second order elements