Synchrotron Motion
247
The phase slip factor is defined as
η
ph
≡ −
Δt
t 0 δ
=
1
v 0 t 0 δ
C
0
1 − (1+ hx δ )
1+
η 0
1 + η 0
δ
1+ 2
1 + η 0
2 + η 0
δ +
η 0
2 + η 0
δ 2 − a 2
δ
ds
= η
ph
1 + η
ph
2 δ + · · · .
(10.3)
Note that the variable δ is defined as ΔK/K 0 . In a pure magnetic system, momentum is the more natural variable since it scales linearly with the magnetic
field. To this end, we recall eq. (3.16) and have
p
p 0
2
=
η (2 + η)
η 0 (2 + η 0 )
.
Using the relations
p
p 0
= 1 +
Δp
p 0
≡ 1 + δ p , η = η 0 (1 + δ) ,
we obtain
(1 + δ p )
2 = (1 + δ)
1 +
η 0
2 + η 0
δ
.
After a little bit of algebraic manipulations, the exact functional relation
between δ and Δp/p 0 is obtained, which is
δ =
1 + η 0
η 0
−1 +
1 +
η 0 (2 + η 0 )
(1 + η 0 )
2
2δ p + δ 2
p
.
As a result, the phase slip factor can be written as
η
ph
≡ −
Δt
t 0 δ p
=
1
v 0 t 0 δ p
C
0
1 − (1+ hx δ )
1+
η 0 (2 + η 0 )
(1 + η 0 )
2
2δ p + δ 2
p
(1+ δ p )
2 − a 2
δ
ds
= η
ph
1 + η
ph
2 δ p + · · · .
(10.4)
Taylor expanding eq. (10.4) to the leading order and taking into account the
fact that
x δ = D
Δp
p 0
,
we obtain
η
ph
1 =
1
(1 + η 0 )
2 −
1
C
C
0
D (s)
ρ (s)
ds =
1
γ 2
0
−
1
C
C
0
D (s)
ρ (s)
ds.
247
The phase slip factor is defined as
η
ph
≡ −
Δt
t 0 δ
=
1
v 0 t 0 δ
C
0
1 − (1+ hx δ )
1+
η 0
1 + η 0
δ
1+ 2
1 + η 0
2 + η 0
δ +
η 0
2 + η 0
δ 2 − a 2
δ
ds
= η
ph
1 + η
ph
2 δ + · · · .
(10.3)
Note that the variable δ is defined as ΔK/K 0 . In a pure magnetic system, momentum is the more natural variable since it scales linearly with the magnetic
field. To this end, we recall eq. (3.16) and have
p
p 0
2
=
η (2 + η)
η 0 (2 + η 0 )
.
Using the relations
p
p 0
= 1 +
Δp
p 0
≡ 1 + δ p , η = η 0 (1 + δ) ,
we obtain
(1 + δ p )
2 = (1 + δ)
1 +
η 0
2 + η 0
δ
.
After a little bit of algebraic manipulations, the exact functional relation
between δ and Δp/p 0 is obtained, which is
δ =
1 + η 0
η 0
−1 +
1 +
η 0 (2 + η 0 )
(1 + η 0 )
2
2δ p + δ 2
p
.
As a result, the phase slip factor can be written as
η
ph
≡ −
Δt
t 0 δ p
=
1
v 0 t 0 δ p
C
0
1 − (1+ hx δ )
1+
η 0 (2 + η 0 )
(1 + η 0 )
2
2δ p + δ 2
p
(1+ δ p )
2 − a 2
δ
ds
= η
ph
1 + η
ph
2 δ p + · · · .
(10.4)
Taylor expanding eq. (10.4) to the leading order and taking into account the
fact that
x δ = D
Δp
p 0
,
we obtain
η
ph
1 =
1
(1 + η 0 )
2 −
1
C
C
0
D (s)
ρ (s)
ds =
1
γ 2
0
−
1
C
C
0
D (s)
ρ (s)
ds.
