Transcript Physics of accelerators (JUAS 98 INTRODUCTION)

Transverse Dynamics – E. Wilson –
CERN – 16th September 2003
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The lattice calculated
Solution of Hill
Solution of Hill (conc)
Meaning of Twiss parameters
Liouville’s Theorem
Closed orbit of an ideal machine
Dispersion – from the “sine and cosine” trajectories
From “three by three” matrices
Making an orbit bump grow
Measuring the orbit
Overlapping beam bumps
Gradient errors
Resonance condition
Multipole field expansion
Taylor series expansion
Multipole field shapes
Correction of Chromaticity
Luminosity
Zeuten 2 - E. Wilson - 7/21/2015 - Slide 1
The lattice
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Solution of Hill

y   (s)  cos (s) + o 
w  ,
 =  (s) + o
Differentiate
substituting
d


y   w(s) cos  
w(s)sin  


ds
1

2
Necessary condition for solution to be true
d
1
1

 2
ds  (s) w (s)


1
y   w(s) cos  
sin  

w (s)

1
so

2
Differentiate again


w(s)
w(s)
y   w(s)cos

  2 sin   2 sin  

w (s)
w (s)

1
2

1
cos 
3
w (s)
and add to both sides
ky
kw(s)cos 
Zeuten 2 - E. Wilson - 7/21/2015 - Slide 3
cancels to 0
must be zero 0
Solution of Hill (conc)


w(s)
w(s)
y   w(s)cos

  2 sin   2 sin  

w (s)
w (s)

1
2
1
 3 cos 
w (s)
cancels to 0
ky
 The
kw(s)cos 
must be zero 0
condition that these three coefficients
sum to zero is a differential equation for
the envelope
1
w (s)
  kw(s)  3  0
w (s)
alternatively
1
1 2
   k 2  1
2
4
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Meaning of Twiss parameters
 is
either :
» Emittance of a beam anywhere in the ring
» Courant and Snyder invariant for one particle
anywhere in the ring
 ( s) y 2  2( s) yy  ( s) y2  
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Liouville’s Theorem
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“The area of a contour which encloses all the beam in
phase space is conserved”
This area = pis the “emittance”
It is the same all round the ring
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NOT TRUE:
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during acceleration
in an electron machine where synchrotron emission damps
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Closed orbit of an ideal machine
 In
general particles executing betatron
oscillations have a finite amplitude
 One particle will have zero amplitude and
follows an orbit which closes on itself
 In an ideal machine this passes down the axis
x 
x
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Closed orbit
Zero betatron
amplitude
Dispersion- reminder

Low momentum particle is bent more
It should spiral inwards but:
There is a displaced (inwards) closed orbit
Closer to axis in the D’s
Extra (outward) force balances extra bends

D(s) is the “dispersion function”
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x  D(s)
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Fig. cas 1.7-7.1C
p
p
Dispersion – from the “sine and
cosine” trajectories

The combination of displacement, divergence and
dispersion gives:
 x   C S  x 
p  D 
   
  
 
x

C

S

x

 s 
 s 0
p D

Expressed as a matrix
 x  C S D  x 

 


x'

C

S

D

x'





 




p ps  0 0 1 p ps0

It can be shown that:
s
s
1
1
D(s)  S(s) 
Ct dt  C(s) 
St dt
(t)
(t)
s0
s0

Fulfils the particular solution of Hill’s eqn. when
forced :
1
D(s)  K(s)D(s) 
(s)
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From “three by three” matrices

Adding momentum defect to horizontal divergence
and displacement vector–
 x  m11

 
x'
 m21



 
p p2  0

m12
m22
0
m13  x 


m23  x' 


1 p p1
Compute the ring as a product of small matrices and
then use:
m13  m12 D( s)
D( s ) 
1  m11
m13 m21  1  m11 m23
D' s 
1  m11 1  m22   m21 m12
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To find the dispersion vector at the starting point
Repeat for other points in the ring
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Making an orbit bump grow
DIPOLE
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As we slowly raise the current in a dipole:
The zero-amplitude betatron particle follows a
distorted orbit
The distorted orbit is CLOSED
It is still obeying Hill’s Equation
Except at the kink (dipole) it follows a betatron
oscillation.
Other particles with finite amplitudes oscillate about
this new closed orbit
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FNAL MEASUREMENT
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Historic measurement from FNAL main ring
Each bar is the position at a quadrupole
+/- 100 is width of vacuum chamber
Note mixture of 19th and 20th harmonic
The Q value was 19.25
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Overlapping beam bumps
 Each
colour shows a triad bump centred on a
beam position measurement.
 A computer calculates the superposition of
the currents in the dipoles and corrects the
whole orbit simultaneously
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Gradient errors
0
 1
m0  
 ,
k 0sds 1
1
0

m  
 .
k 0s  k sds 1
cos  0   0 sin  0 ,
0 sin  0

M0 s  
 .
cos  0  0 sin  0 
   0 sin  0 ,
M s  mm01M0 .

1
0
mm01  
 .
 ks1ds 1

cos 0  0 sin  0 ,
 0 sin  0

M  
 .
k sdscos  0  0 sin 0    sin  0 , k sds 0 sin  0  cos  0   0 sin  0 
T r M / 2  cos     sin 0 
2 p Q    
 s  k s ds
2
.
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sin  0
 0sksds
2
Q 
1
 sks ds .
4p 
Resonance condition
nQ  p ,
QH  mQV  p
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Multipole field expansion (polar)
Scalar potential  (r,  ) obeys Laplace
 2  2
1  2  1    
r
  0
or 2
2 
2  0
2 
x
y
r 
r  r   r 
whose solution is

    n r n sin n
n1
Example of an octupole whose potential
oscillates like sin 4around the circle
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Taylor series expansion

    n r n sin n
n1
Field in polar coordinates:

1 
Br  
, B 
r
r 
Br   nnr n1 sinn ,
B   nnr n1 cosn
To get vertical field
Bz  Br sin   B cos
  n nrn1 cos cosn  sin  sin n 
 n nr n1 cos n  1    n nxn1 (when y  0)
Taylor series of multipoles
Bz   o   2 2x   3 3x2   4  4x3 .......
1 Bz 1  2 Bz 1  3 Bz
 Bo 


.....
2
3
1!  x 2!  x
3! x
Dip. Quad Sext
Oct upole
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Fig. cas 1.2c
Multipole field shapes
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Chromaticity- reminder



The Q is determined by the lattice quadrupoles whose
strength is:
1 dBz
1
k

 B  dx
p
Differentiating:
Remember from gradient error analysis
k
p

.
k
p

Giving by substitution
1
Q 
 sks ds .

4p
Q’ is the chromaticity

“Natural” chromaticity
1
1
p


s

k

s

ds



s

k

s

ds
.






4p
4p
p
p
Q  Q 
p
Q 
Q   
1
4p

s k s  ds  1.3Q
N.B. Old books say
Zeuten 2 - E. Wilson - 7/21/2015 - Slide 19

p dQ Q 

Q dp
Q
Correction of Chromaticity



Parabolic field of a 6 pole is really a gradient which
rises linearly with x
If x is the product of momentum error and dispersion
The effect of all this extra focusing cancels
chromaticity
k 

B" D p
.
 B  p
Because gradient is opposite in v plane we must have
two sets of opposite polarity at F and D quads where
betas are different
 1 B"s sD sdsdp
Q   
 .
 B
4p
 p
Zeuten 2 - E. Wilson - 7/21/2015 - Slide 20
Luminosity
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
Imagine a blue particle colliding with a beam of cross
section area - A

N
Probability of collision is
A
For N particles in both beams

A
 N2
Suppose they meet f times per second at the revolution
frequency
Event rate
f rev
c

2 pR
f rev N 2

A
Make big
e.g. 10 25
Make small
LUMINOSITY
Zeuten 2 - E. Wilson - 7/21/2015 - Slide 21
 1030 to 1034 cm-2 s-1
Summary
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The lattice calculated
Solution of Hill
Solution of Hill (conc)
Meaning of Twiss parameters
Liouville’s Theorem
Closed orbit of an ideal machine
Dispersion – from the “sine and cosine” trajectories
From “three by three” matrices
Making an orbit bump grow
Measuring the orbit
Overlapping beam bumps
Gradient errors
Resonance condition
Multipole field expansion
Taylor series expansion
Multipole field shapes
Correction of Chromaticity
Luminosity
Zeuten 2 - E. Wilson - 7/21/2015 - Slide 22