Lecture 4 - Transverse Optics II
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Transcript of Lecture 4 - Transverse Optics II
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Lecture 4 - E. Wilson –- Slide 1
Lecture 4 - Transverse Optics II
ACCELERATOR PHYSICS
MT 2009
E. J. N. Wilson
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Lecture 4 - E. Wilson –- Slide 2
Contents of previous lecture - Transverse Optics I
Transverse coordinates Vertical Focusing Cosmotron Weak focusing in a synchrotron The “n-value” Gutter Transverse ellipse Cosmotron people Alternating gradients Equation of motion in transverse co-ordinates
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Lecture 4 - E. Wilson –- 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
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Lecture 4 - E. Wilson –- Slide 4
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) Physical meaning (H or V planes)
EnvelopeMaximum excursions
k 1B
dBz
dx
s ds
s
y s ˆ y / s
s
y s sin s 0
d 2 yds 2 + k s y 0
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Lecture 4 - E. Wilson –- Slide 5
Check Solution of Hill
Differentiatesubstituting
Necessary condition for solution to be true
so
Differentiate again
add both sides
w , = (s) + o
y 1
2 w (s) cos dds
w(s)sin
dds
1 (s)
1
w2(s)
y 1
2 w (s) cos 1w (s)
sin
y 1
2 w (s)cos w (s)w2(s)
sin w (s)w2(s)
sin
1
w3(s)cos
ky kw(s)cos
cancels to 0
must be zero 0
y (s) cos (s) + o
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Lecture 4 - E. Wilson –- Slide 6
Continue checking
The condition that these three coefficientssum to zero is a differential equation for
the envelope
y 1
2 w (s)cos w (s)w2(s)
sin w (s)w2(s)
sin
1
w3(s)cos
ky kw(s)cos
cancels to 0
must be zero 0
w (s) kw(s) 1
w3(s)0
12 1
42 k 2 1
alternatively
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Lecture 4 - E. Wilson –- Slide 7
All such linear motion from points 1 to 2 can be described by a matrix like:
To find elements first use notation We know Differentiate 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'1/2w' cos 0
w
1/2
sin 0
Twiss Matrix
y s2 y' s2
a b
c d
y s1 y' s1
M12
y s1 y' s1
.
w
y 1/2w cos 0
1
1
w2
0
/ 2
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Lecture 4 - E. Wilson –- Slide 8
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.
2 1
M12
w2
w1
cos w2w1' sin , w1w2 sin
1 w1w1
' w2w2'
w1w2 sin w1
'
w2
w2'
w1
cos ,
w1
w2 cos w1w2
' sin
w1 w2 w , w 1 w 2 w , = 2 1 2Q
M cos ww ' sin , w2 sin
1w2w '2
w2 sin , cos ww' sin
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Lecture 4 - E. Wilson –- Slide 9
Twiss concluded
Can be simplified if we define the “Twiss” parameters:
Giving the matrix for a ring (or period)
M cos ww ' sin , w2 sin
1w2w '2
w2 sin , cos ww' sin
w2 , = 12
, =1+ 2
w2 , =
12
, =1+ 2
M cos sin , sin
sin , cos sin
M
cos sin , sin sin , cos sin
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Lecture 4 - E. Wilson –- Slide 10
The lattice
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Lecture 4 - E. Wilson –- Slide 11
Beam sections
after,pct
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Lecture 4 - E. Wilson –- Slide 12
Physical meaning of Q and
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Lecture 4 - E. Wilson –- Slide 13
Smooth approximation
ds d
2R
2Q
RQ
tr Q
1tr
2 D R
D R
Q2
N 2Q
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Lecture 4 - E. Wilson –- Slide 14
Principal trajectories
0
00)(pp
sDysSysCsy
0
00)(pp
sDysSysCsy
10
01
00
00
SC
SC
0
0
D
D
D
D
pp
y
y
SC
SC
y
y
s
0
0100
pp
y
y
DSC
DSC
pp
y
y
s
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Lecture 4 - E. Wilson –- Slide 15
Effect of a drift length and a quadrupole
1
fx
x2
x2'
1 , 0
1 f , 1
x1
x1'
Quadrupole
Drift length
x2
x2'
1 , 0
kl , 1
x1
x1'
1
1
2
2
10
1
x
x
x
x
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Lecture 4 - E. Wilson –- Slide 16
Focusing in a sector magnet
Mx cos , sin
1
sin , cos
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Lecture 4 - E. Wilson –- Slide 17
The lattice (1% of SPS)
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Lecture 4 - E. Wilson –- Slide 18
Calculating the Twiss parameters
M
cos sin , sin sin , cos sin
a b
c d
THEORY COMPUTATION(multiply elements)
Real hard numbers
Solve to get Twiss parameters:
cos 1Tr M
2
cos 1
a d2
b / sin
a d
2sin c / sin
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Lecture 4 - E. Wilson –- Slide 19
Meaning of Twiss parameters
is either :» Emittance of a beam anywhere in the ring
» Courant and Snyder invariant fro one particle anywhere in the ring
22 )()(2)( ysyysys
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Lecture 4 - E. Wilson –- Slide 20
Example of Beam Size Calculation
Emittance at 10 GeV/c 20 mm.mrad 20 10 6 m.rad
ˆ = 108 m
46 10 6
46.10 3 m
= 46 mm.
108.20.10 6
0.43 10 6
0.43.10 3 rad
= 0.43 mrad.
20.10 6
108
x
x
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Lecture 4 - E. Wilson –- Slide 21
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Lecture 4 - E. Wilson –- Slide 22
Summary
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