104
An Introduction to Beam Physics
4.4.2.4 The Solenoid with Hard Edge Fringe Fields
We now study an idealized long solenoid with vanishing field outside and
with a constant interior field of
B s (s) = B 0 ,
B x = B y = 0.
We begin the discussion with the observation that the form of the equation
of motion (4.28) entails that the quantities
Z and
C vary continuously even
when passing through a hard edge fringe field. In fact, different from the
situation for z and c, there are no delta functions appearing which led to the
discontinuities in eqs. (4.22) and (4.23). So in both the entrance and the exit
of the hard edge fringe field, instead of eqs. (4.22) and (4.23), we simply have
Z f =
Z i ,
C f =
C i .
(4.29)
Assuming that the beginning edge of the solenoid is located at s = 0, we
have that
θ(s) = −
B 0
2χ m0
s,
θ
=
dθ
ds
= −
B 0
2χ m0
(constant).
The angular frequency θ
is now constant on the inside, and we use the abbreviation
ω = −
B 0
2χ m0
(constant).
Thus we obtain a simple harmonic oscillator solution for
Z. Using the initial
conditions
Z 0 and
C 0 , we have
Z(s) = cos(ωs)
Z 0 +
1
ω
sin(ωs)
C 0 ,
C(s) = −ω sin(ωs)
Z 0 + cos(ωs)
C 0 .
(4.30)
Now this solution has to be expressed in terms of the original coordinates z
and c.
We begin by observing that outside the beginning of the solenoid, we simply
have
Z i = z i ,
C i = c i .
(4.31)
Next, because of eqs. (4.29), even just after entering the solenoid, we have
Z i = z i ,
C i = c i . In the solenoid itself, the quantities
Z and
C change according to eqs. (4.30) until the end of the solenoid is reached, where they have
the values
Z(L) and
C(L). When exiting the solenoid,
Z and
C again remain
unchanged because of eqs. (4.29). In the outside region, there is no field left,
so in eqs. (4.24) and (4.26) we have θ
= 0, thus the transformation equations
simplify to
z(L) = ˆ
R(ωL) ·
Z(L),
, c(L) = ˆ
R(ωL) ·
C(L).
(4.32)
An Introduction to Beam Physics
4.4.2.4 The Solenoid with Hard Edge Fringe Fields
We now study an idealized long solenoid with vanishing field outside and
with a constant interior field of
B s (s) = B 0 ,
B x = B y = 0.
We begin the discussion with the observation that the form of the equation
of motion (4.28) entails that the quantities
Z and
C vary continuously even
when passing through a hard edge fringe field. In fact, different from the
situation for z and c, there are no delta functions appearing which led to the
discontinuities in eqs. (4.22) and (4.23). So in both the entrance and the exit
of the hard edge fringe field, instead of eqs. (4.22) and (4.23), we simply have
Z f =
Z i ,
C f =
C i .
(4.29)
Assuming that the beginning edge of the solenoid is located at s = 0, we
have that
θ(s) = −
B 0
2χ m0
s,
θ
=
dθ
ds
= −
B 0
2χ m0
(constant).
The angular frequency θ
is now constant on the inside, and we use the abbreviation
ω = −
B 0
2χ m0
(constant).
Thus we obtain a simple harmonic oscillator solution for
Z. Using the initial
conditions
Z 0 and
C 0 , we have
Z(s) = cos(ωs)
Z 0 +
1
ω
sin(ωs)
C 0 ,
C(s) = −ω sin(ωs)
Z 0 + cos(ωs)
C 0 .
(4.30)
Now this solution has to be expressed in terms of the original coordinates z
and c.
We begin by observing that outside the beginning of the solenoid, we simply
have
Z i = z i ,
C i = c i .
(4.31)
Next, because of eqs. (4.29), even just after entering the solenoid, we have
Z i = z i ,
C i = c i . In the solenoid itself, the quantities
Z and
C change according to eqs. (4.30) until the end of the solenoid is reached, where they have
the values
Z(L) and
C(L). When exiting the solenoid,
Z and
C again remain
unchanged because of eqs. (4.29). In the outside region, there is no field left,
so in eqs. (4.24) and (4.26) we have θ
= 0, thus the transformation equations
simplify to
z(L) = ˆ
R(ωL) ·
Z(L),
, c(L) = ˆ
R(ωL) ·
C(L).
(4.32)