transverse dynamics 37
Substituting this to Eq. 2.19 gives us the Hill’s equation for
an off-energy particle:
(
)
d 2 x
1
1 Δp
− K(s) − (s) x =
(2.46)
ds 2
ρ 2
ρ(s) p 0
The answer to the above equation can be found using the following form:
dp
x = x 0 +
D
(2.47)
p
where D is the dispersion function and x 0 describes betatron
oscillation around the dispersive orbit as shown in Fig. 2.19.
'
FIGURE 2.19
Dispersion.
Substituting this to Eq. 2.46 will yield the following equation, which governs the evolution of the dispersion function:
d 2 D
1
(
1
− K(s)
ds 2
−
D =
(2.48)
ρ 2 (s)
)
ρ(s)
Dispersion function D can also be expressed in terms of
the principal trajectories as
D(s) = S(s)
s
s
C(t)
S(t)
dt − C(s)
dt
(2.49)
ρ(t)
ρ(t)
s 0
s 0
s
s
C(t)
S(t)
D
' (s) = S
' (s)
dt − C
' (s)
dt
(2.50)
ρ(t)
ρ(t)
s 0
s 0
We have assumed in this section that all bending occurs in
a horizontal plane and therefore only a horizontal dispersion
function is nonzero. This may not be the case in instances of
vertically bending magnets or coupling, discussed below.
As dispersion affects the space location of the reference
orbit for off-energy particles, it thus also affects the orbit’s
path length. Defining C as circumference or orbit path length
in curvilinear coordinates, the deviation of the path length
can be shown to be given by
dp D(s)
ΔC =
ds
(2.51)
p
ρ(s)