Lattice Modules
223
up to the first order of δ. As a result, we have
T x,xδ +
T a,aδ = 0,
which, combined with the previous result, leads to
T x,xδ = 0,
T a,aδ = 0.
In the original space,
T x,δ 2
T a,δ 2
= ˆ
A
−1
T x,δ 2
T a,δ 2
=
0
0
,
and
T x,xδ T x,aδ
T a,xδ T a,aδ
= ˆ
A
−1
T x,xδ
T x,aδ
T a,xδ
T a,aδ
ˆ
A
=
√
β x
0
−α x /
√
β x 1/
√
β x
0
T x,aδ
−
T x,aδ 0
1/
√
β x
0
α x /
√
β x
√
β x
=
α x β x
−γ x −α x
T x,aδ .
The Twiss parameters here are the periodic solution of the cell. In conclusion,
there is only one independent chromatic aberration in the x plane. The same
conclusion can be reached for the y plane following the same procedure. This
proves, from the global point of view, that only two second order chromatic
aberrations are independent.
Furthermore, it is easy to show that the remaining terms
T x,aδ and
T y,bδ are
simply the chromaticities. We can write the transfer matrix in the normalized
space as
ˆ
M x =
1
T x,aδ δ
−
T x,aδ δ
1
=
cos (2π)
s i n( 2 π) +
T x,aδ δ
− sin (2π) −
T x,aδ δ
cos (2π)
= 1
⎛
⎝
cos
2π +
T x,aδ δ
sin
2π +
T x,aδ δ
− sin
2π +
T x,aδ δ
cos
2π +
T x,aδ δ
⎞
⎠ ,
and we obtain
ξ x =
T x,aδ .
Similarly,
ˆ
M y = 1
⎛
⎝
cos
2π +
T y,bδ δ
sin
2π +
T y,bδ δ
− sin
2π +
T y,bδ δ
cos
2π +
T y,bδ δ
⎞
⎠ ,