M. LindroosNUFACT06 School Accelerator Physics Transverse motion Mats Lindroos

M. Lindroos NUFACT06 School

Accelerator PhysicsTransverse motion

Mats Lindroos

http://nufact06.physics.uci.edu/Images/PosterHigh.jpg


Co-ordinates



2 particles in a dipole field

Two particles in a dipole field, with different initial angles



The second particle oscillates around the first. This type of oscillation forms the basis of all transverse motion in an accelerator.

The horizontal displacement of the second particle with respect to the first



B

LBL

2

1

22sin

BLB

Effect of a uniform dipole field of length L and field BThe beam is deviated by an angle ,

B = magnetic rigidity.

Dipole



Magnetic Rigidity

The force evB on a charged particle moving with velocity v in a dipole field of strength B is equal to it’s mass multiplied by it’s acceleration towards the centre of it’s circular path.

e

p

e

mvB

mvevBF

2

where = radius of curvature of the path

remember p = momentum = mv

B is called the magnetic rigidity, and if we put in all the correct units we get :-

B [kG.m] = 33.356p (if p is in “GeV/c” )B [T.m] = 3.3356p (if p is in “GeV/c” )



Quadrupoles

A quadrupole has 4 poles, (2 North and 2 South) arranged symmetrically around the beam. No magnetic field along the central axis

Magnetic field

Bx

By



The “normalised” gradient, k is defined as )( 2m

B

K

On the x (horizontal) axis the field is vertical and is given by:-

By xOn the y (vertical) axis the field is horizontal and is given by:-

Bx y

The field gradient, K is defined as

dx

Byd 1Tm



This example is a Focusing Quadrupole (QF)It focuses the beam horizontally and defocuses verticallyRotating the poles by 90 degrees we get a Defocusing Quadrupole (QD)

Force on a particle

Fy

Fx



FODO Cell

We will study the FODO cell in detail during the tutorial!



As the particles move around the accelerator or storage ring, whenever their divergence (angle) causes them to stray too far from the central trajectory the quadrupoles focus then back towards the central trajectory.This is rather like a ball rolling around a circular gutter.



We characterize the position of the particle in this transverse motion by two things:- Position or displacement from central path, and angle with respect to central path.

x = displacement x’ = angle = dx/ds

This example is for a constant restoring force



These transverse oscillations are called Betatron Oscillations, and they exist in both horizontal and vertical planes. The number of such oscillations/turn is Qx or Qy. (Betatron Tune)

0)(2

2

xsKds

xd

(Hill’s Equation) describes this motion

If the restoring force (K) is constant in “s” then this is just SHM

Remember “s” is just longitudinal displacement around the ring

Hill’s equation



What happens to the motion of the ball in the circular gutter if we allow the shape of the gutter to vary?

The phase advance and the amplitude modulation of the oscillation are determined by the shape of the gutter.

The overall oscillation amplitude will depend on the initial conditions, i.e. how the motion of the ball started.

0)(2

2

xsKds

xd

K varies strongly with “s”….

Therefore we need to solve Hill’s equation for K varying as a function of “s”



0)(cos)(. ssx

0)(2

2

xsKds

xd

To solve it, try:

and 0 are constants, which depend on the initial conditions.(s) = the amplitude modulation due to the changing focusing strength.(s) = the phase advance, which also depends on focusing strength.

Hill’s equation



sin/cos/' x

This will give us:

cos.x

If we plot x .v. x’ as goes from 0 to 2 we get an ellipse, which is called a phase space ellipse

x’

x

/

.

.

/

The area of this ellipse is

NB This area does not depend on or

/



x’

x

x’

x .

.

/

/

As we move around the machine the shape of this ellipse will changeas changes under the influence of the quadrupoles

However the area of the ellipse () will not change

is called the transverse emittance and is determined by the initial beam conditions



x’

x

x’

x .

.

/

/

The projection of this ellipse onto the “x” axis gives the physical transverse beam size

Therefore the variation of(s) around the ring will tell us how the transverse beam size will vary



To be rigorous we should define the emittance slightly differently. • Observe all the particles at single position on one turn and measure both their position and angle• A large number of points on our phase space plot, each corresponding to a pair of x,x’ values for each particle.

The emittance is the area of the ellipse which contains all (or certain percentage) of the points or particles

beam

x’

x

emittance



This conservation of emittance is an illustration of Liouville’s Theorem.

Liouville’s theorem states that if x is the transverse position and px is the transverse mometum, then for a group of particles:-

... const

xx cmp ...0'.xx

.. constdxpx

Transverse velocity

NB =v/c, =m/m0

constant

emittance

..'... constdxxcmo

is the normalised emittance and it is conserved even during longitudinal acceleratione is only conserved if there is no longitudinal acceleration

Lorenz factors!



Matrix formalism

Represent the particles transverse position and angle by a column matrix .

'x

x

As we move around the ring will vary under the influence of the dipoles, quadrupoles and empty (drift) spaces. Express this modification in terms of a TRANSPORT MATRIX (M). If we know x1 and x1’ at some point s1, After the next element in the accelerator ring at s2 :-

'x

x

)'(

)(

)'(

)(

)'(

)(

1

1

1

1

2

2

sx

sx

dc

ba

sx

sxM

sx

sx



Drift space - No magnetic fields, length = L

'10

1

'

'0'

'

1

1

2

2

12

112

x

xL

x

x

xx

Lxxx

x2 = x1 + L.x1’

L

x1’

x1



Remember By = K.x and the deflection due to the magnetic field is

xB

LK

B

LBy.

'1

01

'

''

0

1

1

2

2

112

12

x

x

B

LKx

x

xxB

LKx

xx

(Provided L is small)

x2’

x1 x2

x1’

deflectionQuadrupole of length, L



KL

BDefine focal length of the quadrupole as f =

'1

01

'1

1

2

2

x

x

B

LKx

x

Define f as positive for a focusing effect then:-

'1

101

'1

1

2

2

x

x

fx

x



Multiply together our drift space and our quadrupole matrices to form Transport Matrices, to describe larger sections of our ringThese matrices will move our particles from one point (x(s1),x’(s1)) on our phase space plot to another (x(s2),x’(s2))

sin/cos/' x cos.x

)('

)(

)('

)(

1

1

2

2

sx

sx

dc

ba

sx

sx

The elements of this matrix are fixed by the elements through which the particles pass from s1 to s2.However we can also express (x,x’) as solutions to Hill’s equation:-



Substitute for x(s1), x(s1)’, x(s2) and x(s2)’ into this matrix equation. E.g:-

sin/cos/cos.)cos(.)( 2 bbasx

sin/cos/' x

cos.x

)('

)(

)('

)(

1

1

2

2

sx

sx

dc

ba

sx

sx

)cos(. x

)sin(/)cos(/' x

Assume that our transport matrix describes a complete turn around the machine

12ss Therefore

And = the change in betatron phase over a complete turn



2

2

1

'2'

These special functions are called TWISS PARAMETERS

Remember is the total betatron phase advance over one complete turn.

sincossin

sinsincos

Now our transport matrix is:-

2

Q

Number of betatron oscillations/turn



It is also interesting to note that (s) and Q or are related

2

Q

s

ds

o

sQ

sds

sd

2

1

1

Where = over a complete turn

But

Over one complete turn

Increasing the focusing strength decreases the size of the beam envelope () and increases Q and vice versa.



What happens if we change the focusing strength slightly?

Tune correction

dsdk

dsdk

dQh

dQv

F

D

HFHD

VFVD

4

1

4

14

1

4

1

This matrix relates the change in the tune to the change in strength of the quadrupoles.We can invert this matrix to calculate change in quadrupole field needed for a given change in tune


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M. LindroosNUFACT06 School Accelerator Physics Transverse motion Mats Lindroos