Lecture 4 - Transverse Optics II · Lecture 4 - E. Wilson - 22 Oct 2010 –-Slide2 Contents of...
Transcript of Lecture 4 - Transverse Optics II · Lecture 4 - E. Wilson - 22 Oct 2010 –-Slide2 Contents of...
Lecture 4 - E. Wilson - 22 Oct 2010 –- Slide 1
Lecture 4 - Transverse Optics II
ACCELERATOR PHYSICS
MT 2010
E. J. N. Wilson
Lecture 4 - E. Wilson - 22 Oct 2010 –- Slide 2
Contents of previous lecture -Transverse Optics I
Transverse coordinatesVertical FocusingCosmotronWeak focusing in a synchrotron The “n-value”Gutter Transverse ellipseCosmotron peopleAlternating gradients Equation of motion in transverse co-
ordinates
Lecture 4 - E. Wilson - 22 Oct 2010 –- Slide 3
Lecture 4 - Transverse Optics II Contents
Equation of motion in transverse co-ordinates
Check Solution of Hill Twiss Matrix Solving for a ring The lattice Beam sections Physical meaning of Q and beta Smooth approximation
Lecture 4 - E. Wilson - 22 Oct 2010 –- Slide 4
Relativistic H of a charged particle in an electromagnetic field
Remember from special relativity:
i.e. the energy of a free particle
Add in the electromagnetic field» electrostatic energy» magnetic vector potential has same dimensions
as momentum
px =mvx
1− β2 , x
py =mvy
1− β 2 , y
pz =mvz
1− β 2 , z
H = px2c2 + py
2c2 + pz2c2 + m0
2c4
eφeA
H q,p,t( ) = eφ + c p − eA( )2 + m02c4[ ]
12
Lecture 4 - E. Wilson - 22 Oct 2010 –- Slide 5
Hamiltonian for a particle in an accelerator
Note this is not independent of q because
Montague (pp 39 – 48) does a lot of rigorous, clever but confusing things but in the end he just turns H inside out
is the new Hamiltonian with s as the independent variable instead of t (see M 48)
Wilson obtains
» assumes curvature is small» assumes» assumes magnet has no ends» assumes small angles» ignores y plane
Dividing by
Finally (W Equ 8)
A = A x, y, s( )
ps
φ = 0 Ax = Ay = 0px << ps
H =
px2
2− eAs
P = px2 + py
2 + pz2
H =
p2
2−
eP
As =′ x ( )2
2−
As
Bρ( )
H q,p,t( ) = eφ + c p − eA( )2 + m02c4[ ]
12
Lecture 4 - E. Wilson - 22 Oct 2010 –- Slide 6
Multipoles in the Hamiltonian
We said contains x (and y) dependance
We find out how by comparing the two expressions:
We find a series of multipoles:
For a quadrupole n=2 and:
H =
′ x ( )2
2−
As
Bρ( )As
As = Ann∑ xn
H =
′ x ( )2
2+
k(s)x2
2
By ( y= 0)= −
∂As
∂x= − nAnn∑ xn−1
By ( y= 0)=
1(n −1)!
∂ (n −1)By
∂x(n−1) xn −1
H =
′ x ( )2
2+
1Bρ( )n
∑ 1n!
∂ (n −1)By
∂x(n −1) xn
Lecture 4 - E. Wilson - 22 Oct 2010 –- Slide 7
Hill’s equation in one ( or two) lines
Hamilton’s equations give an equation of motion (remember independent coordinate is now s not t )
and:
′ ′ x + kx = 0
Ý p x = −
∂H∂x
⇒ d( ′ x )
ds= −
∂H∂x
∴ ′ ′ x = −
∂H∂x
= −kx
H =
′ x ( )2
2+
k(s)x2
2
Lecture 4 - E. Wilson - 22 Oct 2010 –- Slide 8
Other interesting forms
H =
p2
2+
k s( ) x2
2
H =
px2
2+
1Be( )n=0
∞
∑ 1n!
∂ n −1( )Bz∂x n −1( )
xn
eAsp
=1Be
1n!
∂ n −1( )Bz∂x n −1( )
x n∑
H ≈ −
eAsp
+p x
2
2
H ≈ −
eAsp
− 1− p x2( )1/ 2
In x plane
Small divergences:
Substitute a Taylor series:
Multipoles each have a term:
Bz z = 0( ) =
∂As∂x
= n An x n−1( )
Quadrupoles:
But:
Lecture 4 - E. Wilson - 22 Oct 2010 –- Slide 9
Equation of motion in transverse co-ordinates
Hill’s equation (linear-periodic coefficients)
– where at quadrupoles
– like restoring constant in harmonic motion Solution (e.g. Horizontal plane)
Condition
Property of machine Property of the particle (beam) ε Physical meaning (H or V planes)
EnvelopeMaximum excursions
k = − 1
Bρ( )dBzdx
β s( ) ϕ = ds
β s( )∫
y = εβ s( ) ′ ˆ y = ε / β s( )
ε β s( )
y = β s( ) ε sin φ s( ) + φ0[ ]
d 2 yds 2 + k s( )y = 0
Lecture 4 - E. Wilson - 22 Oct 2010 –- Slide 10
Check Solution of Hill
Differentiatesubstituting
Necessary condition for solution to be true
so
Differentiate again
add both sides
w = β , φ = φ(s) + φo
′ y = ε1
2 ′ w (s) cosφ −dφds
w(s)sin φ
dφds
=1
β (s)=
1w2(s)
′ y = ε1
2 ′ w (s) cosφ −1
w (s)sin φ
′ ′ y = ε1
2 ′ ′ w (s)cosφ −′ w (s)
w2(s)sin φ +
′ w (s)w2(s)
sin φ
−1
w3(s)cosφ
+ky +kw(s)cosφ
cancels to 0
must be zero 0
y = β(s)ε cos φ(s) + φo( )
Lecture 4 - E. Wilson - 22 Oct 2010 –- Slide 11
Continue checking
The condition that these three coefficientssum to zero is a differential equation for the envelope
′ ′ y = ε1
2 ′ ′ w (s)cosφ −′ w (s)
w2(s)sin φ +
′ w (s)w2(s)
sin φ
−1
w3(s)cosφ
+ky +kw(s)cosφ
cancels to 0
must be zero 0
′ ′ w (s) + kw(s) −1
w3(s)= 0
12
β ′ ′ β −14
′ β 2 + kβ 2 = 1
alternatively
Lecture 4 - E. Wilson - 22 Oct 2010 –- Slide 12
All such linear motion from points 1 to 2 can be described by a matrix like:
To find elements first use notationWe knowDifferentiate and remember
Trace two rays one starts “cosine” The other starts with “sine”We just plug in the “c” and “s” expression
for displacement an divergence at point 1 and the general solutions at point 2 on LHS
Matrix then yields four simultaneous equations with unknowns : a b c d which can be solved
y'= ε1/2w' cos ϕ + φ0( ) −
εw
1/2sin ϕ + φ0( )
Twiss Matrix
y s2( )y' s2( )
=
a bc d
y s1( )y' s1( )
= M12
y s1( )y' s1( )
.
w = β
y = ε1/2w cos ϕ + φ0( )
ϕ =
1β
=1
w2
φ = 0φ = π / 2
Lecture 4 - E. Wilson - 22 Oct 2010 –- Slide 13
Twiss Matrix (continued)
Writing The matrix elements are
Above is the general case but to simplify we consider points which are separated by only one PERIOD and for which
The “period” matrix is then
If you have difficulty with the concept of a period just think of a single turn.
φ = φ2 − φ1
M12 =
w2w1
cos ϕ − w2w1' sin ϕ , w1w2 sin ϕ
−1 + w1w1
' w2w2'
w1w2 sin ϕ −
w1'
w2−
w2'
w1
cos ϕ , w1
w2 cos ϕ + w1w2
' sin ϕ
w1 = w2 = w , ′ w 1 = ′ w 2 = ′ w , µ = φ2 − φ1 = 2πQ
M = cos µ − ww ' sin µ , w2 sin µ
−1 + w2w '2
w2 sin µ , cos µ + ww' sin µ
Lecture 4 - E. Wilson - 22 Oct 2010 –- Slide 14
Twiss concluded
Can be simplified if we define the “Twiss” parameters:
Giving the matrix for a ring (or period)
M = cos µ − ww ' sin µ , w2 sin µ
−1 + w2w '2
w2 sin µ , cos µ + ww' sin µ
β = w2 , α = −12
′ β , γ = 1+ α 2
β
M = cos µ + α sin µ , β sin µ
−γ sin µ, cos µ − α sin µ
Lecture 4 - E. Wilson - 22 Oct 2010 –- Slide 18
Smooth approximation
dsβ∫ = dφ∫
2πRβ
= 2πQ
∴β = RQ
γ tr ≈ Q1
γ tr2 =
D R
∴ D = RQ2
Nµ = 2πQ
Lecture 4 - E. Wilson - 22 Oct 2010 –- Slide 19
Principal trajectories
( ) ( ) ( )0
00)(ppsDysSysCsy ∆
+′+=
( ) ( ) ( )0
00)(ppsDysSysCsy ∆′+′′+′=′
=
′′ 10
01
00
00
SCSC
=
′ 0
0DD
′
+
′
=
′′
′ D
Dpp
yy
SCSC
yy
s
∆
0
0100
′=
′′′
′
ppyy
DSCDSC
ppyy
s ∆∆
Lecture 4 - E. Wilson - 22 Oct 2010 –- Slide 20
Effect of a drift length and a quadrupole
θ =
1f
⋅ x
x2x2
'
=
1 , 0−1 f , 1
x1x1
'
Quadrupole
Drift length
x2
x2'
=
1 , 0−kl , 1
x1
x1'
′
=
′ 1
1
2
2
101
xx
xx
Lecture 4 - E. Wilson - 22 Oct 2010 –- Slide 21
Focusing in a sector magnet
Mx = cos θ , ρ sin θ
−1ρ
sin θ, cos θ
Lecture 4 - E. Wilson - 22 Oct 2010 –- Slide 23
Calculating the Twiss parameters
M =
cos µ + α sin µ , β sin µ− γ sin µ, cos µ − α sin µ
=
a bc d
THEORY COMPUTATION(multiply elements)
Real hard numbers
Solve to get Twiss parameters:
µ = cos−1 Tr M2
= cos−1 a + d
2
β = b / sin µ
α =a − d
2sin µγ = −c / sin µ
Lecture 4 - E. Wilson - 22 Oct 2010 –- Slide 24
Meaning of Twiss parameters
ε is either :» Emittance of a beam anywhere in the ring» Courant and Snyder invariant fro one particle
anywhere in the ring
ε=′β+′α+γ 22 )()(2)( ysyysys
Lecture 4 - E. Wilson - 22 Oct 2010 –- Slide 25
Example of Beam Size Calculation
Emittance at 10 GeV/cε = 20π mm.mrad = 20π ×10−6 m.radˆ β = 108 m
= 46 10−6
= 46.10−3 m= 46 mm.
εβ = 108.20.10−6
= 0.43 10−6
= 0.43.10−3 rad= 0.43 mrad.
εβ = 20.10−6
108
x
′ x