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An Introduction to Beam Physics
Plugging in the first two components of eq. (9.2), we arrive at
cos μ x sin μ x
− sin μ x cos μ x
T x,δ 2 δ
2
T a,δ 2 δ
2
=
T x,δ 2 δ
2
T a,δ 2 δ
2
and
T x,xδ δ
T x,aδ δ
T a,xδ δ
T a,aδ δ
cos μ x sin μ x
− sin μ x cos μ x
=
cos μ x sin μ x
− sin μ x cos μ x
T x,xδ δ
T x,aδ δ
T a,xδ δ
T a,aδ δ
.
For the pure chromatic terms, we have
1 − cos μ x − sin μ x
sin μ x
1 − cos μ x
T x,δ 2
T a,δ 2
=
0
0
.
Since, for n > 1, we have
det
1 − cos μ x − sin μ x
sin μ x
1 − cos μ x
= 2 (1 − cos μ x ) = 0,
we conclude that
T x,δ 2 = 0,
T a,δ 2 = 0, for n > 1.
For the mixed chromatic terms, we have
T x,xδ cos μ x −
T x,aδ sin μ x
T x,xδ sin μ x +
T x,aδ cos μ x
T a,xδ cos μ x −
T a,aδ sin μ x
T a,xδ sin μ x +
T a,aδ cos μ x
=
T x,xδ cos μ x +
T a,xδ sin μ x
T x,aδ cos μ x +
T a,aδ sin μ x
−
T x,xδ sin μ x +
T a,xδ cos μ x −
T x,aδ sin μ x +
T a,aδ cos μ x
.
For each component, we have
−
T x,aδ sin μ x =
T a,xδ sin μ x ,
T x,xδ sin μ x = T a,aδ sin μ x ,
−
T a,aδ sin μ x = −
T x,xδ sin μ x ,
T a,xδ sin μ x = −
T x,aδ sin μ x .
For n = 2, sin μ x = 0. We have
T x,xδ −
T a,aδ = 0,
T x,aδ +
T a,xδ = 0.
From symplectic symmetry, we have
det
1 +
T x,xδ δ
T x,aδ δ
T a,xδ δ
1 +
T a,aδ δ
= 1