Lecture Plan (1) • Review of particle equations of motion in 2D without space-charge - Courant-Snyder parameters, envelope equation

• Examples of space-charge • Particle and envelope equations with linear space-charge –Space-charge (Laslett) Tune shift –Image effects

• Envelope oscillations, resonances 2 Monday, 23 May 2011

Lecture Plan (2) • General particle equations under space-charge • Non-linear beams - rms beam sizes, rms emittance - rms envelope equations - evolution of rms emittance

• Examples of 2D distributions - KV, waterbag, Gaussian - concept of stationary distributions

• Beam halo – causes, measurement (kurtosis) 3 Monday, 23 May 2011

Lecture Plan (3) • Longitudinal space-charge • Self-consistent distributions - Hofmann-Pederson model

• Microwave instability • Acceleration cycle in a synchrotron • Bunch compression • Long and short bunches 4 Monday, 23 May 2011

Reading E.J.N. Wilson: Introduction to Accelerators S.Y. Lee: Accelerator Physics M. Reiser: Theory and Design of Charged Particle Beams D. Edwards & M. Syphers: An Introduction to the Physics of High Energy Accelerators M. Conte & W. MacKay: An Introduction to the Physics of Particle Accelerators R. Dilao & R. Alves-Pires: Nonlinear Dynamics in Particle Accelerators R. Davidson & Hong Qin: Physics of Intense Charged Particle Beams in High Energy Accelerators Monday, 23 May 2011

5

Notation and Basic Formulae

Rest mass

c = 2.99792458 × 108 m/sec v β = , v = βc c 1 1 γ=� =� 2 v2 1−β 1− 2 c m0

Relativistic mass

m = m0 γ

Momentum

p = mv = m0 γv = m0 γβc

Energy

E = mc2 = m0 γc2

Velocity of light Relative velocity Relativistic gamma

Kinetic energy

Note:

T = E − m0 c2 = m0 (γ − 1)c2

E2 2 2 2 = p + m 0c 2 c 6

Monday, 23 May 2011

Maxwell’s Equations E

Electric field

B

Magnetic flux density

ρ

Charge density

j

Current density

ρ ∇·E= �0 ∇ · B =0

∂B ∇∧E= − ∂t 1 ∂E ∇ ∧ B = µ0 j + 2 c ∂t

µ0

Permeability of free space, 4π × 10−7

�0

Permittivity of free space, 8.854 × 10−12

1 � 0 µ0 = 2 c

James Clerk Maxwell 7 Monday, 23 May 2011

Gauss’ Flux Theorem ρ � = ∇·E �0 Equivalent to Gauss’ Flux Theorem:

⇐⇒

ρ ∇·E= �0 ��� �� ��� 1 Q ∇ · E dV = E · dS = ρ dV = �0 �0 V S

The flux of electric field out of a closed region is proportional to the total electric charge Q enclosed within the surface.

8 Monday, 23 May 2011

Ampère’s Circuital Law 1 ∂E ∇ ∧ B = µ0 j + 2 c ∂t

⇐⇒

��

∇ ∧ B = µ0 j S

∇ ∧ B · dS =

�

B · dl = µ0

��

S

j · dS = µ0 I

For a straight line current: µ0 I Bθ = 2πr

9 Monday, 23 May 2011

Review of Simple Particle Dynamics Equation of motion in electromagnetic fields: � � dp Lorentz =q E+v∧B Force dt produces acceleration

produces bending

Total energy E = mc2 and E 2 = p2 c2 + m20 c4 dE dp 2 =⇒ E =c p· = qc2 p · E dt dt Energy change only from dE =⇒ = qv · E Electric fields dt 10 Monday, 23 May 2011

Motion in Constant Magnetic Field d (m0 γv) = qv ∧ B dt dv q =⇒ = v∧B dt m0 γ 2 v⊥ q =⇒ = v⊥ B ρ m0 γ =⇒ =⇒

m0 γv⊥ circular motion with radius ρ = qB v⊥ qB qB at an angular frequency ω = = = ρ m0 γ m

Constant magnetic field gives uniform spiral about B with constant energy.

m0 γv p Bρ = = q q

Magnetic Rigidity 11

Monday, 23 May 2011

External Forces • Maxwell’s time-independent equations, no sources: � =0 ∇∧B

� = ∇φ =⇒ ∃φ such that B

� =0 ∇·B

=⇒

∇2 φ = 0,

• Simplest solutions, with z = x + iy

n=1 n=2 n=3

Scalar magnetic potential Laplace’s equation

(i =

φ ∝ x + iy φ ∝ (x2 − y 2 ) + 2ixy φ ∝ x(x2 − 3y 2 ) + iy(3x2 − y 2 )

√

−1) are

dipole quadrupole sextupole

• Then 12 Monday, 23 May 2011

Equations of Motion No bends

✻ x

design orbit ✑◗ ✑ ◗ y ✑ ◗ � ◗ ✑ ✰✑s ✑

x˙ 1 d n−1 � m0 (γ�v ) = q�v ∧ B = q y˙ ∧ Kn i z dt s˙ 0 =⇒

γ x˙ −is˙ d z n−1 γ y˙ = Knq s˙ m0 dt γ s˙ ix˙ − y˙

x˙ � y˙ Write x = , y = , and note in an accelerator x, ˙ y˙ � s˙ s˙ s˙ �

�

�

so that v = (x , y , 1)s, ˙

d d = s˙ x , y � 1 and dt ds �

�

13 Monday, 23 May 2011

d From above, (γ s) ˙ ∼0 Paraxial Equations dt 1 � � − 1 2 � � x˙ 2 + y˙ 2 + s˙ 2 − 2 �2 �2 2 γ = 1− ≈ 1 − β (x + y + 1) c2 2 −1 = (1 − β ) 2 + second order terms Ignoring second and higher velocity terms, equations of motion are � � � �� � −i x n−1 2 2 = Knqβc z m0 γβ c y �� 1 Field strength k =q × m0 γβc = Field strength/Bρ

Monday, 23 May 2011

Equations of Motion in Dipoles

�

�

��

��

2

x +κ x y ��

�

�

�

� � 1 x + 2x 0 = = ρ 0 �� y

x + 2ky + k y y �� − 2kx� − k � x

�

=

��

�

0 0

� 15

Monday, 23 May 2011

Mathieu-Hill Equations Paraxial equation of motion in periodic systems: x�� (s) + kx (s)x =

0

y �� (s) + ky (s)y

0

=

where s is distance along beam axis kx (s), ky (s) periodic focusing functions, k(s + L) = k(s)

George Hill

Floquet’s Theorem confirms two independent solutions: u = w(s)eiψ(s) ,

v = w(s)e−iψ(s)

The Wronskian is W (u, v) = uv � − vu� = −2iw2 ψ � = C, a constant dψ 1 � Choose C = −2i =⇒ = ψ = 2. ds w Then, substitute u or v into Mathieu-Hill equation: 16 Monday, 23 May 2011

u = (w + iwψ )e �

�

�

iψ

= (w + i/w)e , �

iψ

u�� = (w�� − iw� /w2 + iw� ψ � − ψ � /w)eiψ = (w�� − 1/w3 )eiψ u + ku = 0 ��

=⇒

1 w + kw − 3 = 0 w ��

Any solution of Matthieu-Hill is a linear combination of u, v, so set

x = A w(s) cos(ψ(s) + φ). A x d �x� = − 2 sin(ψ + φ) = A cos(ψ + φ), w ds w w �2 x2 � � � 2 =⇒ + wx − w x = A . 2 w Or:

where

Monday, 23 May 2011

βˆ = w2 ,

ˆ �2 = A2 γˆ x2 + 2α ˆ xx� + βx

1 ˆ� � α ˆ = −ww = − β , 2

2 1 1 + α ˆ γˆ = 2 + w�2 = w βˆ

17

Phase-Space Ellipse; Emittance � �2 ˆ γˆ (s)x2 + 2α(s)xx ˆ + β(s)x = A2

ˆγ − α Area of ellipse is πA2 (βˆ ˆ 2 ) = πA2

�2 ˆ γˆ x +2ˆ αxx + βx ≤ � is beam ellipse in x-x� phase space and � is called the beam emittance 2

�

ˆ γˆ are Courant-Snyder parameters α ˆ , β, � ˆ Beam size (half-width) is a(s) = �β(s) � � � ds ds � Phase advance ψ = ψ ds = = 2 w βˆ 18 Monday, 23 May 2011

• � is a constant of the motion, independent of s ˆ γˆ determine the shape of the ellipse in x-x phase space • α ˆ , β, ˆ γˆ are periodic functions, so that the ellipse rotates and • For a ring α ˆ , β, shears with position s, while its area π� is conserved (no acceleration)

• Phase advance around � a ring gives the tune (number of oscillations per ds 1 revolution) Q = ˆ 2π β(s) • Beam envelope given by maximum value of x:

a = Amax w(s) =

√

� w(s)

Equation for w then gives the envelope equation: 1 �� w + kw − 3 = 0 w Monday, 23 May 2011

=⇒

2 � a�� + ka − 3 = 0 a

Space-Charge: simple idea 2D point charges q experience repulsive electrostatic force of magnitude

+q

+q

q2 Fe = . 2π�0 r

Coulomb repulsion

Particles moving with speed v equivalent to two current wires I = qv. Magnetostatic force between two current wires is attractive of magnitude µ0 I 2 µo q 2 v 2 v2 Fm = = = 2 Fe . Magnetic attraction 2πr 2πr c ! ! � � 2 v Combined force is a repulsive self-field 1 − 2 Fe . c For electrons travelling at or close to c, space-charge forces can be negligible. v For proton or ion machines, where = β ∼ 0.5, eﬀects are important. c Note: other factors come in, e.g. intensity. 20 Monday, 23 May 2011

Relativistic Transformation of Fields Consider a single particle, charge q, moving with velocity v. In rest-frame, electrostatic field E0 , and B0 = 0. In lab-frame, transformation equations are � � E⊥ = γ E0 ⊥ − v × B0 � � 1 B⊥ = γ B0 ⊥ + 2 v × E0 c

E� = E0 � B� = B0 �

Transverse Lorentz force is � � � � 1 F⊥ = q E⊥ + v × B = γq E0 ⊥ + 2 v × (v × E0 ) c � � v·v = γq 1 − 2 E0 ⊥ c 1 = q 2 E⊥ γ 21 Monday, 23 May 2011

Illustration of Space-Charge; Tune Shift Consider a 2D axisymmetric beam with charge density ρ(r) = qn(r). Electric field is radial and inside the beam is given by Gauss’ Flux Theorem. 1 {Flux of E through circle of radius r} = × {charge enclosed}. �0 � � r 1 1 =⇒ 2πrEr (r) = ρ dV = 2πrρ(r) dr �0 �0 0 � r q 1 =⇒ Er (r) = rn(r) dr �0 r 0 � Magnetic field is angular, from Amp`ere’s Law B · dl = µ0 ×{current flowing through loop} � =⇒

=⇒

2πrBθ Bθ

= µ0 =

r

βcρ(r)2πr dr 0 � r qβ 1 rn(r) dr c�0 r 0

22 Monday, 23 May 2011

Space charge force on a particle is �

�

�

r � q 1� F (r) = q Er − βcBθ = 1 − β2 rn(r) dr �0 r 0 � � 2 r Consider a Gaussian distribution n(r) = A exp − 2 2σ N where A = and N is the number of particles per unit length. 2 2πσ � � 2 2 � Nq 1 r � Then F (r) = 1 − exp − 2 2 2π�0 γ r 2σ 2

For betatron oscillations at angular frequency w, equation of particle motion is 2

2

d r F (r) Nq 1 2 +ω r = = 2 dt m0 γ 2π�0 m0 γ 3 r

�

� � r 1 − exp − 2 2σ �

2

23 Monday, 23 May 2011

Consider the Gaussian beam transported round a ring of mean radius R and let φ be the azimuthal angle round the ring. Then βc dt = R dφ and the equation of motion becomes � � � 2 2 2 2 � d r Nq R 1 r 2 +Q r = 1 − exp − 2 2 2 3 2 dφ 2π�0 m0 β γ c r 2σ 2

2N r0 R 1 = β2γ3 r

�

� � r 1 − exp − 2 , 2σ �

2

q2 where r0 = is the classical radius and Q is the tune. 2 4π�0 m0 c � � �� � � �� 2 2 1 r 1 r Now 1 − exp − 2 = 1 − 1 − 2 + ... r 2σ r 2σ r = + ... 2 2σ

24 Monday, 23 May 2011

Equation of motion becomes:

�

d r N r0 R 2 + Q − 2 2 3 2 dφ σ β γ 2

2

�

r=0

Equivalent to Q → Q + ∆Q, where 2 N r R 0 Q2 − 2 2 3 σ β γ

=

(Q + ∆Q)2

≈ Q2 + 2Q∆Q

Space-charge tune shift

=⇒

N r0 R2 ∆Q = − 2Qσ 2 β 2 γ 3

Most pronounced a low energies when β 2 γ 3 is small, particularly for bright, intense beams.

25 Monday, 23 May 2011

Equations of Motion with Space-Charge Consider a paraxial beam with uniform circular cross-section of radius a. Gauss’ Flux Theorem gives �

1 2πrEr = × {charge enclosed} = �0 N qr Er r 2 2π�0 a Er = Nq r 2π�0 r

ρπr2 /�0 ρπa2 /�0

ra

�a

Er

>a

Within the beam, space-charge forces are: x Ex = E r , r

y Ey = Er r

=⇒

Nq E= (x, y). 2 2π�0 a 26

Monday, 23 May 2011

Equations of Motion are: x + k(s)x = ��

For round beam, need axisymmetric focusing, kx=ky=k

or:

y �� + k(s)y

=

�

� K x�� + k(s) − 2 x = a � � K y �� + k(s) − 2 y = a

I is the beam current

q Ex 3 2 2 m0 γ β c q Ey 3 2 2 m0 γ β c 0 0

I 2 K= I0 β 3 γ 3

perveance

I = N qβc

4π�0 m0 c3 1 m0 c 2 I0 is the characteristic current I0 = ≈ q 30 q � 17 kA for electrons ≈ 31(A/Z) MA for ions Corresponding envelope equation is: 2 � K �� a + ka − 3 − = 0. a a 27 Monday, 23 May 2011

Incoherent Tune Shift Assume an unbunched beam, uniform density, circular cross-section in a ring, mean radius R. x�� + (k(s) + ksc (s)) x = 0 K I 2 1 2Ir0 1 where ksc = − 2 = − =− 3 3 2 3 3 2 a I0 β γ a qβ γ c a � � 1 1 ∆Qx = Kx (s)βx (s) ds = ksc (s)βx (s) ds 4π 4π 1 ∆Qx = − 4π

Monday, 23 May 2011

�

2πR 0

2r0 I βx (s) r0 RI ds = − 3 3 3 3 2 qβ γ c a qβ γ c

∆Qx,y

r0 RI =− 3 3 , qβ γ c�x,y

∆Qx,y

r0 N =− 2πβ 2 γ 3 �x,y

�

βx (s) a2 (s)

�

1 = �x

N qβc with I = 2πR 28

∆Qx,y

r0 N =− 2πβ 2 γ 3 �x,y

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