252
An Introduction to Beam Physics
the similar relations for (l|y) and (l|b), we obtain
Δη
ph
2 (s) =
1
C
(l|x)ΔD x2 + (l|a)ΔD
x2 + (l|y)ΔD y2 + (l|b)ΔD
y2
= −
1
C
k s ds
D
2
x − D
2
y
2 sin (πν x )
{[(a|x)D x + (1 − (x|x)) D
x ] β x cos (πν x )
− [(1 − (a|a)) D x + (x|a)D
x ] (sin (πν x ) − α x cos (πν x ))}
+
1
C
k s dsD x D y
sin (πν y )
(b|y)D y + (1 − (y|y)) D
y
β y cos (πν y )
−
(1 − (b|b)) D y + (y|b)D
y
(sin (πν y ) − α y cos (πν y ))
.
Taking into account the fact that the vertical part in the curly brackets is the
same as that of the horizontal part, we immediately arrive at the final result,
which is
Δη
ph
2 =
1
C
C
0
k s (s)
D
3
x − 3D x D
2
y
ds.
10.3 Longitudinal Dynamics
Based on the previous two sections, we can construct the one turn map
with the RF cavity present. From eq. (10.2), we can write the general form
of energy gain
ΔK (r, t) = qV 0 (r) cos [φ (t)] ,
where φ(t) = ωt. For the TM 010 mode of a pillbox cavity, we have
V 0 (r) = E 0 J 0
x 01 r
R c
LT,
where L, instead of l, is used to represent the length of the cavity to avoid
confusion. Converting to the canonical coordinates, we have
φ (l) = φ 0 +
ω
κ
l,
and
δ f (r, l) =
K 0
K 0 + ΔK (0, 0)
δ i +
ΔK (r, l) − ΔK (0, 0)
K 0 + ΔK (0, 0)
=
K 0
K 0 + qE 0 LT cos (φ 0 )
δ i +
qE 0 LT
K 0 + qE 0 LT cos (φ 0 )
·
J 0
x 01 r
R c
cos
φ 0 +
ω
κ
l
− cos (φ 0 )
.