Solution:
1. Without loss of generality, we use the Frenet-Serret coordinate system of Fig. 2.1 and
derive equation of motion for positively charged ions in the accelerator. It is easy to
modify the equations of motion for electrons.
(a) The coordinate of a particle is ~r = (ρ + x)x̂ + z ẑ. Using Eq. (2.6) and (2.8), we
obtain
dx̂
dx̂
ṡ
x̂˙ =
= ṡ
= ŝ,
dt
ds
ρ
dŝ
x̂
ŝ˙ = ṡ = ṡ −
ds
ρ
!
= −θ̇ x̂,
i.e. ~r˙ = ẋx̂ + (ρ + x)θ̇ŝ + ż ẑ, and
~¨r = (ẍx̂ + ẋθ̇ŝ) + ẋθ̇ŝ + (ρ + x)θ̈ŝ + (ρ + x)θ̇ŝ˙ + z̈ ẑ
= [ẍ − (ρ + x)θ̇ 2 ]x̂ + [2ẋθ̇ + (ρ + x)θ̈]ŝ + z̈ ẑ.
(b) From Newton’s law:
d~p
~ = evs ŝ × (Bx x̂ + Bz ẑ) = evs Bz x̂ − evs Bx ẑ,
= γm~¨r = e~v × B
dt
we obtain
ẍ − (ρ + x)θ̇ 2 =
evs Bz
v 2 Bz
= s ,
γm
Bρ
z̈ = −
evs Bx
v 2 Bx
=− s .
γm
Bρ
(c) Transformation from t to s for the independent coordinate with
ds
vs
vs
,
ṡ =
= ρθ̇ =
,
ρ+x
dt
1 + x/ρ
!
!
2
d
d
d d
d2
2 d
=
ṡ
ṡ
=
(
ṡ)
=
,
dt2
dt dt
ds ds
ds2
θ̇ =
we obtain
x
ρ+x
Bz
1+
x00 −
=
2
ρ
Bρ
ρ
!2
Bx
x
z 00 = −
1+
Bρ
ρ
,
!2
2. The solution for Hill’s equation y 00 + K(s)y = 0 with the constant K(s) = K is
y=

As + B

√

A cos( q
Ks + B)


K=0
K>0
A cosh( |K| + B) K < 0
K=0
√
K>0
y = −A
q K sin( qKs + B)


A |K| sinh( |K| + B) K < 0
0

A


√
where A, B are constants decided by the initial conditions y = y1 and y 0 = y10 at s = s1 .
Setting s1 = 0 and s2 = s, we obtain
6
• K = 0: We have B = y(s1 ) = y1 and A = y 0 (s1 ) = y10 , and
y1
1 s
=
y10
0 1
√
• K > 0: We have A cos B = y1 and −A K sin B = y10 , and
√
√
!
√1 sin
Ks y1 cos Ks
y2
K
√
√
√
=
y20
y10
− K sin Ks
cos Ks
y2
y20
q
• K < 0: We have A cosh B = y1 and A |K| sinh B = y10 , and
y2
y20
=

cosh
q
q
|K|s
|K| sinh
q
√1
|K|
|K|s
sinh
cosh
q
q
|K|s
|K|s


y1
y10
The transfer matrix M (s2 |s1 ) can be obtained easily from the above equations,
as
q
shown in the exercise. In thin lens approximation with l → 0, we have |K| l → 0,
and Kl → 1/f , i.e.
√
√
√
1
lim √ sin K l = 0, lim K sin K l = 1/f ;
l→0
l→0
l→0
K
q
q
q
q
1
lim cosh |K|l = 1, lim q
sinh |K| l = 0, lim |K| sinh |K| l = −1/f,
l→0
l→0
l→0
|K|
lim cos
√
Kl = 1,
where f > 0 for focusing quadrupole and f < 0 for defocusing quadrupole. The
1
0
transfer matrix of a quadrupole in thin lens approximation is M =
.
−1/f 1
3. Effects of Dipole on betatron motion:
(a) Using Kx = 1/ρ2 and Kz = 0 for a sector dipole, we obtain
Mx =
cos ρl
− 1ρ sin
ρ sin ρl
cos ρl
l
ρ
!
cos θ
− 1ρ sin θ
=
ρ sin θ
cos θ
!
,
Mz =
1 l
0 1
(b) Horizontal Edge Focusing: When the edge angle δ > 0, the particle with x > 0
will experience less bending field than the reference particle in the horizontal
direction, the resulting effect is defocusing. The total
length without field of
R
the x > 0 particle is x tan δ, and the field integral is Bz ds = −Bx tan δ. The
focusing strength is3
1 ∂ (Bx tan δ)
tan δ
1
=−
=−
.
fx
Bρ
∂x
ρ
3
Since
R
R
(∂Bz /∂x)ds = −B tan δ = −( dB) tan δ, or
Z
Z
∂Bz
∂Bz
1
1
tan δ
tan δ
=−
ds = −
ds = −
.
fx
Bρ
∂x
Bρ
∂s
ρ
Bz ds = −Bx tan δ, we find
R
7
∂Br
∂z
Vertical Edge Focusing:4 Using Maxwell’s equation:
field Br obeys
∂
∂z
Z
Br ds =
∂
∂x
Z
=
∂Bz
,
∂x
the radial
Bz ds = B tan δ.
Since it is defocusing in x direction, the vertical focal length becomes
tan δ
1
1 ∂ Z
Br ds =
=
.
fz
Bρ ∂z
ρ
4. See Lecture Note
5. See Lecture Note
6. Differentiating β 00 + 2Kβ − 2γ = 0, we obtain β 000 + 4β 0 K + 2βK 0 = 0. The solutions
for piecewise constant K are
• Drift space: K = 0, β = a + bs + cs2 .
√
√
+
b
sin
2
Ksq+ c.
• Focusing quadrapole: K = constant > 0, β = a cos 2 Ks
q
• Defocusing quadrapole: K = constant < 0, β = a cosh 2 |K|s+b sinh 2 |K|s+c.
(a) Using the initial conditions at s = s0 :
β = β0 ,
β00 = −2α0 ,
β000 = −2Kβ0 +
2
(1 + α02 ) = −2Kβ0 + 2γ0 ,
β0
we obtain
• Drift space: a = β0 ,
b = β00 = −2α0 ,
c = 12 β000 = γ0 .
β = β0 − 2α0 s + γ0 s2 = γ0 (s − s∗ )2 + 1/γ0 ,
where s∗ = α0 /γ0 .
• Focusing quadrapole:
β0
γ0
β000
=
−
,
a=−
4K
2
2K
β00
α0
√
b=
= −√ ,
2 K
K
c = β0 − a =
β0
γ0
+
.
2
2K
• Defocusing quadrapole:
a=
β0
γ0
+
2
2K
α0
b = −q
|K|
c = β0 − a =
γ0
β0
−
2
2K
4
Because of the fringe field, the
H particle trajectory
R off mid-plane will experience a longitudinal field, B s .
~ · d~` = B0 z +
Using Ampere’s law, we obtain B
fringe Bs ds = 0. The longitudinal component Bk of the
fringe field of a dipole with
an
edge
angle
δ
can
be
divided
into two components Bs and Bx that are related
R
R
by Bx = Bs tan δ. Thus fringe Bx ds = fringe Bs tan δds = B0 z tan δ. The focal length becomes fz = ρ/ tan δ.
8
(b) In a drift space with a symmetry condition at s = s∗ , we have
β = β ∗,
β 0 = 0,
β 00 = 2/β ∗ .
Using previous result, β = a + bs + cs2 , we obtain
c=
1
β 00
= ∗,
2
β
Thus β = β ∗ +
b = −2cs∗ = −
1
(s
β∗
2s∗
β∗
a = β ∗ s − bs∗ − cs∗2 = β ∗ +
s∗2
.
β∗
− s∗ )2 , and
α = (s − s∗ )/β ∗ ,
γ = (1 + α2 )/β = 1/β ∗ = constant.
The phase advance from the IP to the high β quad becomes
ψ=
Z
s
s∗
ds
= β∗
1
∗
∗
2
β + β ∗ (s − s )
Z
s−s∗
0
du
s − s∗
= arctan
∗2
2
β +u
β∗
!
→
π
,
2
where we use the fact that s − s∗ β ∗ .
(c) The transfer matrices are related through similarity transformation, Eq. (2.46),
with
M (s2 ) = M (s2 |s1 )M (s1 )[M (s2 |s1 )]−1 .
Now the transfer matrix can be expressed in Courant-Snyder parametrization,
Eq. (2.50),
α2
M (s2 ) = I cos Φ +
−γ2
α1
M (s1 ) = I cos Φ +
−γ1
β2
sin Φ = I cos Φ + J2 sin Φ
−α2
β1
sin Φ = I cos Φ + J1 sin Φ.
−α1
Thus J2 = M (s2 |s1 )J1 [M (s2 |s1 )]−1 , i.e.
α2
−γ2
β2
−α2
=
M11
M21
M12
M22
·
α1
−γ1
β1
−α1
·
M22
−M21
−M12
M11
.
Then α2 , β2 and γ2 are related to α1 , β1 , and γ1 by the transfer matrix. Substituting appropriate transfer matrices, we find
• Drift space: β = β0 − 2α0 s + γ0 s2
• Focusing quadrapole:
√
√
√
√
α0
γ0
β = β0 cos2 Ks − 2 √ cos Ks sin Ks + sin2 Ks
K
K
!
!
√
√
β0
β0
α0
γ0
γ0
=
cos 2 Ks − √ sin 2 Ks +
−
+
2
2K
2
2K
K
9
• Focusing quadrapole:
β = β0 cosh2
q
q
q
q
α0
γ0
|K|s − 2 q
cosh |K|s sinh |K|s − sinh2 |K|s
K
|K|
!
q
q
β0
γ0
γ0
α0
β0
+
−
cosh 2 |K|s − q
sinh 2 |K|s +
2
2K
2
2K
|K|
=
!
.
Note that β(s) oscillates twice the betatron frequency.
7. In dealing with the mismatch problem, it is much easier to transform the phase space
coordinates into the normalized phase space coordinates where the phase space ellipse
is a circle. A mismatched injection ellipse is not a circle. The semi-major and minor
axes can be determined easily.
(a) With the normalized coordinates (Y, P ),
Y
P
=B
−1
y
y0
y
y0
Y
=B
P
,
where
B
−1
=
√
1/ √β
α/ β
√0
β
,
the matched ellipse is transformed to a circle Y 2 +P 2 = , and the injection ellipse
to
aY 2 + 2bY P + cP 2 = ,
where
a=
(α1 β − αβ1 )2
1
β
β
+
= + b2 ,
β1
ββ1
c β1
b=
α1 β − αβ1
,
β
c=
β1
β
and ac = 1 + b2 .
(b) The major and minor axes of the injection ellipse can be obtained by a coordinate
rotation to obtain AỸ 2 + C P̃ 2 = , where A, C are the roots of the equation
u2 − (a + c)u + 1 = 0, or
s
(a + c)
a+c 2
A, C =
−1
±
2
2
The mis-match factor Fmm is then
"
#
1
a+c
1 β 2 + β12 (α1 β − β1 α)2
= (γ1 β + β1 γ − 2α1 α)
Fmm =
=
+
2
2
ββ1
ββ1
2
The major and minor axes of the ellipse F+ , F− are
q
q
−1/2
1/2
1
2 −1
2 −1
= Fmm ∓ Fmm
= Fmm ± Fmm
C, A
In a perfect linear ring lattice, the injection ellipse rotates at twice the tune and
the beam envelope oscillates at a frequency twice the tune.
(c) Because of nonlinear betatron detuning, the mismatched phase space ellipse is
decohered, and the phase space area of the injection beam is increased by a factor
of
F± = √
F+2 = Fmm +
10
q
2 −1
Fmm