Understand and Correct a Defective Lattice: Phasors

Pictures

Resonance: in blue Distortion

Resonances and Distortions

• A resonance is an abnormality of phase space.

• A distortion is an unacceptable deviation from linearity (or mid-plane symmetry) but the map is still a rotation in disguise.

• All these effects are the results of iterating a map!

We want to Understand MN

• Globally MsN=As◦RN◦As

-1 is impossible if resonances are present.

• Therefore, let us raise the map “manually” to a power “N”

• Phasors will emerge from that treatment!

Problem• M = M1◦ (I + C+…)

• M1 is the linear part of the map • C is the quadratic distortion of this map• “…” represents higher order terms which

we will temporarily neglect • Question: MN = M1◦ (I + C+…) ◦ … ◦ M1◦ (I + C+…)

Can we compute this map to first order in ?

Normalize what you understand already: Linear Part

A-1 ◦ MN ◦ A ={ A-1 ◦ M1◦ A ◦ A-1 ◦ (I + C+…) ◦A }N

= {R ◦ (I + Ĉ +…)}N

Where Ĉ = A-1C ◦ A We will symbolically denote the effect of A on C

as: Ĉ = AC

Example: C is a sextupole kick• C=(0,kx2)• A is a Courant-Snyder Matrix

• C ◦ A =(0,kx2)• Ĉ =A-1 C ◦ A =(0, kx2 )

MN to first order: Origin of Phasors

{R ◦ (I + Ĉ +…)}N =? = R◦(I + Ĉ +…) ◦R◦(I + Ĉ +…) ◦ … ◦ R◦(I + Ĉ +…) ◦R◦(I + Ĉ +…)

= R◦(I + Ĉ +…) ◦ R-1 ◦R2 ◦(I + Ĉ +…) ◦ R-2◦ R3 ◦(I + Ĉ +…) R-3 ◦ … ◦ RN ◦(I + Ĉ +…) ◦R-N ◦RN

= (I + (R1 Ĉ+ R2 Ĉ+ R3 Ĉ+ R4 Ĉ+ … + RN Ĉ ) + Order() ) ◦RN

• Therefore understanding Rk Ĉ is extremely important.

Phasors: Eigen-operators

• We will see that Ĉ is connected to a vector field and therefore in the Hamiltonian case to a simple Hamiltonian function.

• For the moment, let us examine the general case of Rk Ĉ where Ĉ is a vector function. (general = non Hamiltonian)

• We want to find the eigenvector field: R Ĉ = = R-1 Ĉ ◦ R = Ĉ

Resonance Basis

• h±=x±i p

or (x1new,x2

new)= (x1old+i x2

old , x1old - i x2

old )• This transformation can be written in matrix form as:

• The matrix R in that new basis is given by E=B-1RB

which is just

B -1= 1 i1 -i

E=B -1RB= e-i 00 e i where R= cos sin

-sin cos

Definition of Phasors

The n-turn map is:= (I + (E1 G + E2 G + E3 G + E4 G + … + EN G ) + O

rder() ) ◦ En

Ek G = Ek G ◦ Ek Eab = ab exp( (-1)a i)

In the new basis : Gi = Gi;nmx1nx2

m

Ek G = Gi;nm exp(ik{m-n-(-1)a} ) x1nx2

m

Summing up

• k=1,N Ek G |a = k=1,N Gi;nm exp(ik{m-n-(-1)a} ) x1n

x2m

= Gi;nm {1-exp(iNnm) }/{exp(-inm)-1} x1nx2

m

nm = {m-n-(-1)a}

Sextupole Order m+n=2 ; m-n-(-1)a =±3, ±1; Let us classify the cases.

Phasors at 1st Order in Sextupoles

GTotal=G1;20x12

-=k

+G1;11x1x2=k

+G1;02x223=k

G2;20x12-3=k

+G2;11x1x2-=k

+G2;02x22=k

Ga;mn= exp iN m-n- -1 a - 1 / exp -i m-n- -1 a -1

B -1A -1MNAB = 1 + G EN

If 3≈k2, then G1;02 and G2;20 can get real big!

These are now quasi-secular terms and we must treat them with care: this is best done if we reveal the Hamiltonian nature of the map explicitly.

Phasors at 1st Order (continue)

If 3m≈k2p , then G1;02 and G2;20 can get real big!

These are now quasi-secular terms and we must treat them with care: this is best done if we reveal the Hamiltonian nature of the map explicitly.

Are secular terms a property of resonance? Answer: no! Tune shifts with amplitude are unavoidable in a nonlinear map! Therefore even maps which can be made into a rotation M=A◦R◦A-1 contain secular terms which we reinterpret as being part of the rotation!

GOctupole=G1;30x13-2=k

+G1;21x12x2secular

+G1;12x1x222=k

+G1;02x234=k

G2;30x13-4=k

+G2;21x12x2-2=k

+G2;12x1x22secular

+G2;03x232=k

Hamiltonian MethodsThe coefficients Gi;mn in the map I+G+… are not arbitrary. In fact, it can be shown easily (we will show that) that there are only two independent terms related to the =k phasors in the sextupole order map.

G=G1;20x12

-=k

+G1;11x1x2=k

+G1;02x223=k

G2;20x12-3=k

+G2;11x1x2-=k

+G2;02x22=k

Hamiltonian → G2;11 + 2G1;20= 0 and G1;11 + 2G2;02 = 0.

Hamiltonian → G1 = 2i ∂2 g and G2 = -2i ∂1 g .

Or G = [g,Identity] where [x,p]=1 and therefore [h+,h-]=-2i

From g to G

• g= g30 h+3+ g21 h+

2 h- + g12 h+ h-

2 + g03 h-3

• G1=[g,h+]= 2i{ g21 h+2 +2 g12 h+

h- + 3g03 h-2 }

• G2=[g,h-]= -2i{ 3g30 h+2 +2 g21 h+

h- + g12h-2 }

• G1;20 = 2i g21 G1;11 = 4i g12 G1;02 = 6i g03

• G2;20 = -6i g30 G2;11 = -4i g21 G2;02 = -2i g12

Therefore G2;11 + 2G1;20= 0 and G1;11 + 2G2;02 = 0.

Making I+G Hamiltonian

• =I+G+… = exp(:g:)I• Here :g:f is defined as [g,f]

• Notice that =exp(:g:)I is an exact solution of d/d-gso is exactly Hamiltonian

• Furthermore g is an invariant of

Pictures: Go back to Problem

Resonance: in blue Distortion

Start with Resonance• Globally Ms=As◦R◦As

-1 is impossible.

• Globally Ms=As◦N◦As-1 such that

N=R2/3◦N◦R2/3-1

is possible.

One-Resonance Normalization

In our New Phasor Language1. M = exp(:g:)R where g contains the non

linear effects2. Remove phasors not related to k3: New

map is N = Aexp(:g:)R A-1 = exp(:r:)R 3. Cube the map : N3 = exp(:r3:)R3

4. N3 is near the identity 5. Therefore N3 = exp(:3H:)6. H is an invariant which potentially includes

islands!

So H is to order dodecapole

• H=-(J+aJ2+bJ3)

+J3/2 ((A+CJ)cos(3) + (B+DJ)sin(3)) +J3 (Ecos(6) + Fsin(6)) • Now for fun, we will correct a small lattice usi

ng sextupoles just to see if these tools make any sense!

Three corrections

Standard Correction Methods1. Correct Linear Terms A=B=02. If not good enough, try A=B=C=D=03. Correct tune around island: • +2aJ0+3bJ0

2 =0

• A+CJ0=B+DJ0=0. • This can be done numerically or with H!

H=-(J+aJ2+bJ3)+J3/2 ((A+CJ)cos(3)+(B+DJ)sin(3))+J3 (Ecos(6)+Fsin(6))

Next Lecture

• We will look in details in the computer implementation