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Page 1: LECTURE 2 Maxwell’s equations: Electric field Ert Brt Jrt ...dugan/USPAS/Lect2.pdfParticle trajectory equations of motion in accelerators 11/21/01 USPAS Lecture 2 2 Review of basic

11/21/01 USPAS Lecture 2 1

LECTURE 2

Review of basic electrodynamicsMagnetic guide fields used in accelerators

Particle trajectory equations of motion in accelerators

11/21/01 USPAS Lecture 2 2

Review of basic electrodynamicsMaxwell’s equations: Electric field

r rE r t( , ), magnetic field

r rB r t( , ),

charge density ρ( , )rr t , current density

r rJ r t( , )

r r r r

r r

r rr

r rr

r

r rr

r r

∇ • = => • =

∇ • =

∇ × = − ∂∂

=> • = − ∂∂

∇ × = + ∂∂

=>

• = +

∫∫

E E daQ

B

EBt

E dlBt

da

B JEt

B dl I

enclosed

S

SC

enclosed

ρε ε

µ µ ε

µ µ ε

0 0

0 0 0

0 0 0

0

Gauss' Law

Faraday' s Law

∂∂

•∫∫r

rEt

daSC

Ampere' s Law

11/21/01 USPAS Lecture 2 3

Electrodynamic potentials: r r rA r t V r t( , ), ( , )

r r r r rr

B A E VAt

= ∇ × = −∇ − ∂∂

Conservation of charge:r r∇ • + ∂

∂=J

0

Ohm’s Law:r rJ E= σ

Lorentz Force Law: force on charge er r r rF e E v B= + ×( )

The Lorentz force generated by the accelerator’s guide fielddetermines the trajectory of particles in the accelerator. Before we

discuss the trajectory equations, we should make some remarksabout the guide field.

11/21/01 USPAS Lecture 2 4

Magnetic guide fields used in accelerators

The simplest guide field for a circular accelerator is a uniformfield.

B0

e v

Pole

Pole

orbitρ

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11/21/01 USPAS Lecture 2 5

The Lorentz force provides the centripetal accelerationv evB

meBmv

Bp

20 0

0

1

10 2998

ρ ρ

ρ

= =

[ ] = [ ][ ]

,

.mT

GeV / c1

Using this relation, we sometimes measure momentum in units ofT-m:

Bpρ( )[ ] = [ ]

T - mGeV / c0 2998.

The (Bρ) product is called the magnetic rigidity.

More than a simple uniform bending field is required. Focusingfields are required to insure stability to small displacements from

the orbit.

11/21/01 USPAS Lecture 2 6

In a synchrotron, the guide field (including bending and focusingfields) is achieved by a series of separate magnets.

e

v

B0

B0

B0

B0

bending fields

focusing fields

11/21/01 USPAS Lecture 2 7

The focusing fields may be separate from the bending fields(“separated function” machine) or may be combined with the

bending fields (“combined function” machine).If the bending fields all have the same design value, the machine is

called isomagnetic.

Magnetic focusing fields:Optical analogy: Thin lens, focal length f

θx

f

θ = x

f

11/21/01 USPAS Lecture 2 8

Magnetic lens:

x

f

p

θ ∆p

L

Β

e

This linear dependence of field on position can be generated by aquadrupole magnet

∆ ∆p F t evBLv

eB Ly y= =

=

θ ≈ =∆pp

eB L

py

If BB

xx B xy

y=∂∂

= ′ , then θ ≈ ′ =eB xL

p

x

f

The field acts like a focusing lens offocal length

1f

eB Lp

B LB

≈ ′ = ′( )ρ

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11/21/01 USPAS Lecture 2 9

General description of the guide fields in the region of the particlebeam

For typical accelerator magnets, the magnet length is much largerthan the transverse dimensions of the field gap

(L>>G)

x

y

zL

GPole

poleB

The idealized fields extend only over the length L (“effectivelength”) and are independent of z and transverse to the beam.

Idealized fields ignore the fringe fields. (Note that this is does not

11/21/01 USPAS Lecture 2 10

apply to solenoids, and is a poor approximation for wigglers orundulators).

In the region of the beam,

r r∇ × =B 0 so

r rB = −∇ φr r

∇ • =B 0 so ∇ =2 0φ

The magnetic scalar potential is a solution to LaPlace’s equation.For the idealized fields, the magnetic potential is only a function of

x and y.In this case, the general solution to LaPlace’s equation in Cartesian

coordinates

11/21/01 USPAS Lecture 2 11

∂∂

+ ∂∂

=2

2

2

2 0φ φ

x y can be written as

φ( , ) Re ( )x y C x iymm

m= +=

∞∑

0where Cm is a (complex) constant determined by the boundary

conditions.

The vector potential has only one component and it is

A x y iC x iyz mm

m( , ) Re ( )= +

=

∞∑

0

Then

11/21/01 USPAS Lecture 2 12

Bx

mC x iy

By

imC x iy

x mm

m

y mm

m

= −∂∂

= − +

= −∂∂

= − +

=

∞−

=

∞−

φ

φ

Re ( )

Re ( )

1

1

1

1

These can be combined to give

B iB imC x iyy x mm

m+ = − +( )=

∞−∑

1

1.

This is the general multipole expansion for the two-dimensionalmagnetic field in a current-free region.

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11/21/01 USPAS Lecture 2 13

Since

− =−( )

∂∂

+ ∂∂

−= =

−= =

imCm

B

xi

B

xm

my

mx y

mx

mx y

11

1

10

1

10

!

we can write

B iBm

B iB x iyy xm m

m

m+ = + +( )=

∞∑ 1

0 !( ˜ )( ) ( )

where

11/21/01 USPAS Lecture 2 14

B B B B

B BB

xB B

Bx

B BB

xB B

B

x

y

x y

x

x y

y

x y

x

x y

( ) ( )

( ) ( )

( ) ( )

; ˜ ˜ ;

; ˜ ˜˜

; ˜ ˜

00

00

1

0

1

0

22

20

22

20

= =

= ′ =∂∂

= ′= ∂∂

= ′′ =∂∂

= ′′ = ∂∂

= = = =

= = = =

BB

xB

B

xm

my

mx y

mm

xm

x y

( ) ( ); ˜=∂∂

= ∂∂

= = = =0 0

Vector potential in this language:

A x ym

B iB x iyzm m m

m( , ) Re

!( ˜ )( )( ) ( )= − + +− −

=

∞∑ 1 1 1

1

11/21/01 USPAS Lecture 2 15

Let us write out the first few terms:

B B B x B yB

x y B xy

B B B x B yB

x y B xy

y

x

= + ′ − ′ + ′′ − − ′′ +

= + ′ + ′ + ′′ − + ′′ +

02 2

02 2

2

2

˜ ( ) ˜ ...

˜ ˜˜

( ) ...

The terms without the twiddle are called “normal” terms; the termswith the twiddle are called “skew” terms. The 0 coefficients arepure dipole terms, the linear are quadrupole terms, the quadratic

are sextupole terms.

11/21/01 USPAS Lecture 2 16

G

Integration path

Iron

NI

y

x

N

S

Pure dipole: NI turns/pole

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11/21/01 USPAS Lecture 2 17

Evaluate r rH dl Ienclosed• =∫ around the integration path shown.

For infinite permeability iron r

r

HB= →µ

0 inside the iron, so in the

gap

HNIG

B HNIG

BNI

G

= ⇒ = =

= −[ ][ ]

2 2

2 52

0 0µ µ

[ ] .TkA turns

mmThis dipole bends a positive particle moving in the z-direction to

the left.If the dipole is rotated clockwise by 90o about the z-axis, it

becomes a pure skew dipole.

11/21/01 USPAS Lecture 2 18

NS

NS

y

x

NI

R

Integration path

Iron

Pure quadrupole, NI turns/pole

11/21/01 USPAS Lecture 2 19

Evaluate r rH dl Ienclosed• =∫ around the integration path shown.

For infinite permeability iron r

r

HB= →µ

0 inside the iron, so in the

gapr rH dl B rdr

B RNI B

NI

R

BNI

R

R• = ′ = ′ = ⇒ ′ =

′ = −[ ][ ]

∫ ∫1

22

2 51

00

2

00 2

2

µ µµ

[ ] .Tm

A turns

mm

Quadrupole focal length fp

eB LBB L

≈′

=′

( )ρ

Note that (for a positive particle moving in the z-direction),this quadrupole is focusing in x, but defocusing in y.

If the quadrupole is rotated clockwise by 45o about the z-axis, itbecomes a pure skew quadrupole.

11/21/01 USPAS Lecture 2 20

R

N

S

N

S

N

S

NI

Integration path

Iron

PuresextupoleNI turns/pole

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11/21/01 USPAS Lecture 2 21

Evaluate r rH dl Ienclosed• =∫ around the integration path shown.

For infinite permeability iron r

r

HB= →µ

0 inside the iron, so in the

gap

r rH dl

Br dr

B RNI B

NI

R

BNI

R

R• = ′′ = ′′ = ⇒ =

′′ = −[ ][ ]

∫ ∫1

2 66

7540

0

20

3

00 3

3

µ µµ

[ ]Tm

A turns

mm

If the sextupole is rotated clockwise by 30o about the z-axis, itbecomes a pure skew sextupole.

11/21/01 USPAS Lecture 2 22

Iron

y

x

Gradient magnet (for combined function machines): provides botha bending field and a gradient (due to changing gap dimension).

11/21/01 USPAS Lecture 2 23

I

z

x

y

L

N turns

The transverse fringe fields at the solenoid ends are crucial to thesolenoid focusing action, so in this case the idealized fields mustinclude end fields. Maxwell’s equations allow the end fields to

have the form BBz

x BBz

yxz

yz= − ∂

∂= − ∂

∂12

12

Solenoid focal length:

fL

peB L

BBz z

=

= ( )

2 22 2ρ

Solenoid: produces an axialfield

BNILz = µ0

BNI

Lz[ ] .TkA turns

m= × −[ ]

[ ]−1 26 10 3

11/21/01 USPAS Lecture 2 24

Particle trajectory equations of motion in accelerators

Consider a particle of charge e, rest mass m0, momentumr rp m v= 0γ (γ 2

21

1

=−

vc

), moving under the action of the Lorentz

force:

The trajectory equation of motion is given by Newton’s Law:

dpdt

ddt

m v F e E v Br

r r r r r= ( ) = = + ×( )0γ

Solution is this equation is the trajectory rr t( )

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11/21/01 USPAS Lecture 2 25

Write rr t( ) in terms of the “reference trajectory”:

e

p

B0

B0

B0

B0

bending fields

focusing fields

0

Referencetrajectory

ρ

The “reference trajectory” rr t0( ) is the trajectory of a “reference

particle” of momentum rp0 that passes through the center of

symmetry of all the idealized magnetic guide fields. The referencetrajectory is an idealization.

11/21/01 USPAS Lecture 2 26

For a circular accelerator, this trajectory is closed (i.e., it repeatsitself exactly on every revolution)

Reference trajectory

rrrr0

δrr

Actual trajectory

r r rr t r t r t( ) ( ) ( )= +0 δ

We want to write the trajectory equation of motion in the “natural”coordinate system of the reference trajectory.

11/21/01 USPAS Lecture 2 27

From differential geometry: “natural” coordinate system of a curvein space

Reference trajectory

rr0

s

tdr

ds=

r0

ˆ ( )ˆ

n sdt

ds= −ρ

ˆ ˆ ˆb t n= × ˆ, ˆ ˆ

ˆ ˆ ˆ

b t n

b t n

and are unit vectors

= = =1

11/21/01 USPAS Lecture 2 28

For any space curve rr s0( ), where s is the arc length along the

curve, the vector tdrds

=r0 is a unit vector tangent to the curve. The

vector dtds

ˆ measures the rate at which the tangent vector changes,

which is inversely proportional to ρ( )s , the radius of curvature:dtds

ns

ˆ ˆ( )

= −ρ

, where n (called the principal normal to the curve) is a

unit vector normal to t . The vectors t , n, and ˆ ˆ ˆb t n= × form anorthogonal, right-handed coordinate system which moves (and may

rotate) as we progress along the space curve.

The change in the unit vectors with s is given by the

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11/21/01 USPAS Lecture 2 29

Frenet-Serret relations:dtds

n dnds

bt db

dsn

ˆ ˆ

ˆ ˆ ˆ

ˆˆ= − = + = −

ρτ

ρτ

where τ(s) is called the torsion.

For most accelerators, the reference orbit lies in a plane: thiscorresponds to τ(s)=0. For circular accelerators, since the orbit is

closed,ds

sdds

dsρ

θ π( )

= =∫∫ 2

Take the space curve to be the reference trajectory, and write thetrajectory equations using the Frenet-Serret co-ordinate system:

r r r rr r r r xn yb= + = + +0 0δ ˆ ˆ

11/21/01 USPAS Lecture 2 30

(δrr lies in x-y plane since coordinate system moves with the

particle)

rr r r

vdrdt

drdt

xn xdndt

yb sdrds

xn xsdnds

yb

st xnx

st yb stx

xn yb

= = + + + = + + +

= + + + = +

+ +

0 0

1

˙ ˆˆ

˙ ˆ ˙ ˙ ˆ ˙ˆ

˙ ˆ

˙ˆ ˙ ˆ ˙ˆ ˙ ˆ ˙ˆ ˙ ˆ ˙ ˆρ ρ

s

y

x

rr0

δrr

Reference trajectory

Actual trajectory

˙ ˙fdfdt

dsdt

dfds

sdfds

≡ = =

11/21/01 USPAS Lecture 2 31

Newton’s Law: dpdt

ddt

m v F e E v Br

r r r r r= ( ) = = + ×( )0γ

ddt

m vddt

m xnddt

m ybddt

m sx

t

nddt

m xm s x

bddt

m y tddt

m s

0 0 0 0

00

2

0 0

1

1

γ γ γ γρ

γ γρ ρ

γ γ

r[ ] = [ ] + [ ] + +

= [ ] − +

+ [ ] +

˙ ˆ ˙ ˆ ˙ ˆ

ˆ ˙˙ ˆ ˙ ˆ ˙ 11 0+

+

x m xsρ

γρ

˙˙

r rv B

n b t

x y sx

B B B

n yB sx

B b sx

B xB t xB yB

x y s

s y x s y x

× = +

= − +

+ +

+ −[ ]

ˆ ˆ ˆ

˙ ˙ ˙

ˆ ˙ ˙ ˆ ˙ ˙ ˆ ˙ ˙

1

1 1

ρ

ρ ρ

11/21/01 USPAS Lecture 2 32

Three trajectory equations:

ddt

m xm s x

yeB sx

eB eE

ddt

m y sx

eB xeB eE

ddt

m sx m xs

s y x

x s y

00

2

0

00

1 1

1

1

γ γρ ρ ρ

γρ

γρ

γρ

˙˙

˙ ˙

˙ ˙ ˙

˙˙˙

˙

[ ] = +

+ − +

+

[ ] = +

− +

+

= − + xxeB yeB eEy x s− +˙

Write derivatives in terms of arc length s along the reference curveas the independent variable:

(with ′ ≡fdfds

)

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11/21/01 USPAS Lecture 2 33

dfdt

dfds

s f s= = ′˙ ˙

and introduce l=the path length of the particle along the orbit

ds

dlReferencetrajectory

Trajectory

dldt

vdlds

s l s lvs

= = = ′ ⇒ ′ =˙ ˙ ˙

11/21/01 USPAS Lecture 2 34

vl

dds

mvl

xm v

lx

yvl

eBvl

xeB eE

vl

dds

mvl

yvl

xeB x

vl

eB eE

v

s y x

x s y

′ ′′

=′

+

+ ′′

−′

+

+

′ ′′

=′

+

− ′′

+

00

2

0

1 1

1

γ γρ ρ ρ

γρ

lldds

mvl

x m x vl

xvl

eB yvl

eB eEy x s00

21γ

ργρ′

+

= − ′

+ ′

′− ′

′+

Divide by vl′

, use p m v= 0γ , and expand LHS:

11/21/01 USPAS Lecture 2 35

pl

x xdds

pl

pl

xy eB

xeB eE

lv

pl

y ydds

pl

xeB x eB eE

lv

x dds

p

s y x

x s y

′′′ + ′

=′

+

+ ′ − +

+ ′

′′′ + ′

= +

− ′ + ′

+

ρ ρ ρ

ρ

ρ

1 1

1

1ll

pl

dds

x pxl

x eB y eB eElvy x s

+′

+

= − ′′

+ ′ − ′ + ′1

ρ ρ

Let the electric field be zero: the particle energy is then constant.

Divide by pl

dds l

l

l′ ′

= − ′′′

and use1

2

11/21/01 USPAS Lecture 2 36

′′ − ′ ′′′

= +

+ ′ ′ − +

′′ − ′ ′′′

= ′ +

− ′ ′

+

′′′

− +

= ′ − ′

x xll

xy l

eBp

xl

eB

p

y yll

lx eB

px l

eBp

x ll

dds

x x

s y

x s

11 1

1

1 1

ρ ρ ρ

ρ

ρ ρ ρxx l

eB

py l

eBp

y x′ + ′ ′

From the relation for the velocity,

v s Kx x y sx

x y2 2 2 2 2 22

2 21 1= +( ) + + = +

+ ′ + ′

˙ ˙ ˙ ˙ρ

Solve for s

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11/21/01 USPAS Lecture 2 37

˙ ˙

sv

xx y

lvs

xx y=

+

+ ′ + ′

′ = = +

+ ′ + ′

1

12

2 2

22 2

ρ

ρand

Approximation:“paraxial” motion=> the derivatives ′ ′ <<x y2 2 1, so

′ ≈ + ′ ′′′

′ ′′′

lx

xll

yll

and we neglect and terms in the trajectory

equations(These neglected terms are called “kinematic terms”)

Typical values of

( ) cos

max

max

x s xs

xdxds

x

′ = ≈ ≈ ≈ −

β

β1010

10 3mmm

11/21/01 USPAS Lecture 2 38

Trajectory equations in the paraxial approximation:

′′ = +

− +

+ ′ +

′′ = +

− ′ +

xx x eB

py

eBp

x

yx eB

px

eBp

x

y s

x s

11 1 1

1 1

2

2

ρ ρ ρ ρ

ρ ρ

′ = +

lx