Lattice Modules
213
Finally, let us look at chromaticities for FODO cells.
ξ = −
1
4π
β(s)k 0 (s)ds = −
1
4π
N
β max
f
−
β min
f
= −
1
4π
N
f
1 + sin(
μ
2
) − 1 + sin(
μ
2
)
2l
sin μ
= −
N
4π
4 sin(μ/2)
sin μ
2 sin(
μ
2
) = −
N
π
tan(
μ
2
) = −ν
tan(μ/2)
μ/2
, (ν =
N μ
2π
).
For the Fermilab Main Injector,
μ x =
π
2
, ν x = 26.425 =⇒ ξ x = −33.6,
μ y =
π
2
, ν y = 25.415 =⇒ ξ y = −32.4,
which is not far from the exact values (ξ x = −33.6 and ξ y = −33.9) showing
again the usefulness of the thin lens model. Without correction, and assuming
the momentum spread of ±1%, we have
Δν x = ξ x
Δp
p 0
= ±33.6 × 0.01 = ±0.336,
Δν y = ξ y
Δp
p 0
= ±32.4 × 0.01 = ±0.324.
Clearly, without chromaticity correction, the momentum acceptance of the
ring would be very small (probably below 0.1% due to the fact that the distance between ν x and the half-integer is only 0.075). To correct chromaticity, we place two sextupoles in each FODO cell, one next to the focusing
quadrupole and the other next to the defocusing quadrupole. Using the thin
lens model of the sextupoles and ignoring the distance between the sextupoles
and their adjacent quadrupoles, we obtain the total chromaticities from eqs.
(8.4) and (8.5)
ξ x = −
1
4π
β(s) [k x (s) − D x (s)k s (s)] ds
= −
1
4π
N
β max
1
f
− D max k sF
+ β min
−
1
f
− D min k sD
,
ξ y = −
1
4π
β(s) [k x (s) + D x (s)k s (s)] ds
= −
1
4π
N
β min
−
1
f
+ D max k sF
+ β max
1
f
+ D min k sD
,
where k sF and k sD are integrated strengths of the sextupoles next to the