Understand and Correct a Defective Lattice: Phasors

1. Understand and Correct a Defective Lattice: Phasors

2. Pictures Resonance: in blue Distortion

3. 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!

4. 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!

5. Problem • M = M1◦ (I + e 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 + eC+…) ◦ … ◦ M1◦ (I + eC+…) Can we compute this map to first order in e?

6. Normalize what you understand already: Linear Part A-1◦ MN ◦A ={ A-1◦ M1◦ A ◦ A-1◦ (I + eC+…) ◦A}N = {R ◦ (I + eĈ+…)}N Where Ĉ = A-1C ◦ A We will symbolically denote the effect of A on C as: Ĉ = AC

7. Example: C is a sextupole kick • C=(0,kx2) • A is a Courant-Snyder Matrix • C ◦ A=(0,kb x2) • Ĉ =A-1 C ◦ A=(0, kb3/2x2 )

8. MN to first order: Origin of Phasors {R ◦ (I + eĈ+…)}N =? = R◦(I + eĈ+…) ◦R◦(I + eĈ+…) ◦ … ◦ R◦(I + eĈ+…) ◦R◦(I + eĈ+…) = R◦(I + eĈ+…) ◦ R-1◦R2 ◦(I + eĈ+…) ◦ R-2◦R3◦(I + eĈ+…) R-3◦ …◦ RN◦(I + eĈ+…) ◦R-N◦RN = (I + e(R1Ĉ+ R2Ĉ+ R3Ĉ+ R4Ĉ+ … + RNĈ )+ Order(e2))◦RN • Therefore understanding RkĈ is extremely important.

9. 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 RkĈ where Ĉ is a vector function. (general = non Hamiltonian) • We want to find the eigenvector field: R Ĉ = = R-1 Ĉ◦ R =lĈ

10. Resonance Basis • h±=x±i p or (x1new,x2new)= (x1old+i x2old , x1old - i x2old ) • This transformation can be written in matrix form as: • The matrix R in that new basis is given by E=B-1RBwhich is just

11. Definition of Phasors The n-turn map is: = (I + e (E1G+ E2G+ E3G+ E4G+ … + ENG)+ Order(e2))◦ En EkG= E-k G◦ EkEab = dabexp( (-1)a im) In the new basis : Gi = S Gi;nmx1nx2m EkG = S Gi;nm exp(ik{m-n-(-1)a} m) x1nx2m

12. Summing up • Sk=1,NEkG |a = Sk=1,NS Gi;nm exp(ik{m-n-(-1)a} m) x1nx2m = S Gi;nm {1-exp(iNQnm) }/{exp(-iQnm)-1}x1nx2m Qnm = {m-n-(-1)a} m Sextupole Order m+n=2 ; m-n-(-1)a =±3, ±1; Let us classify the cases.

13. Phasors at 1st Order in Sextupoles 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.

14. 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-1contain secular terms which we reinterpret as being part of the rotation!

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

16. 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.

17. Making I+eG Hamiltonian • x=I+eG+… = exp(:eg:)I • Here :g:f is defined as [g,f] • Notice that x=exp(:eg:)I is an exact solution of dx/de = [x,-g], so x is exactly Hamiltonian • Furthermore g is an invariant of x!

18. Pictures: Go back to Problem Resonance: in blue Distortion

19. Start with Resonance • Globally Ms=As◦R◦As-1 is impossible. • Globally Ms=As◦N◦As-1 such that N=R2p/3◦N◦R2p/3-1 is possible.

20. One-Resonance Normalization

21. In our New Phasor Language • M = exp(:g:)R where g contains the non linear effects • Remove phasors not related to k3n: New map is N = Aexp(:g:)R A-1 = exp(:r:)R • Cube the map : N3= exp(:r3:)R3 • N3 is near the identity • Therefore N3 = exp(:3H:) • H is an invariant which potentially includes islands!

22. So H is to order dodecapole • H=-(dJ+aJ2+bJ3) +J3/2((A+CJ)cos(3f) + (B+DJ)sin(3f)) +J3(Ecos(6f) + Fsin(6f)) • Now for fun, we will correct a small lattice using sextupoles just to see if these tools make any sense!

23. Three corrections H=-(dJ+aJ2+bJ3)+J3/2((A+CJ)cos(3f)+(B+DJ)sin(3f))+J3(Ecos(6f)+Fsin(6f)) Standard Correction Methods • Correct Linear Terms A=B=0 • If not good enough, try A=B=C=D=0 • Correct tune around island: • d+2aJ0+3bJ02 =0 • A+CJ0=B+DJ0=0. • This can be done numerically or with H!

24. Next Lecture • We will look in details in the computer implementation