3 Non-linear Dynamics in Accelerators
85
– c xx , c xy , c yy : detuning with amplitude
The coefficients are the various aberrations of the optics.
As a first example one can look at the effect of a single (thin) sextupole. The map
is:
M = e − : μJ x + μJ y +
1
2 α c δ 2 : e
: k(x 3 − 3xy 2 ) +
p 2
x +p 2
y
2(1+δ) :
(3.122)
we get for h eff (see e.g. [6, 7]):
h eff = μ x J x + μ y J y +
1
2 α c δ 2 − kD 3 δ 3 − 3kβ x J x Dδ + 3kβ y J y Dδ
Then it follows:
Q x (J x , J y , δ) =
1
2π
∂h eff
∂J x
=
1
2π
(μ x − 3kβ x Dδ)
(3.123)
Q y (J x , J y , δ) =
1
2π
∂h eff
∂J y
=
1
2π
(μ y + 3kβ y Dδ)
(3.124)
Since it was developed to first order only, there is no non-linear detuning with
amplitude.
As a second example one can use a linear rotation followed by an octupole, the
Hamiltonian is:
H =
μ
2
(x
2
+ p
2
x ) + δ(s − s 0 )
x 4
4
= μJ + δ(s − s 0 )
x 4
4
with : J =
(x 2 + p 2
x )
2
(3.125)
The first part of the Hamiltonian corresponds to the generator of a linear rotation
and the second part to the localized octupole.
The map, written in Lie representation becomes:
M = e (−
μ
2 : x 2 + p 2
x :) e :
x 4
4 : = e : −μJ : e :
x 4
4 : = R e :
x 4
4 :
(3.126)
The purpose is now to find a generator F for a transformation
e − : F : M e : F : = e − : F : e :
x 4
4 : e : F :
(3.127)
such that the exponents of the map depend only on J and not on x.