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 + 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 + e C+…) ◦ … ◦ M1◦ (I + e C+…) Can we compute this map to first order in e?

Normalize what you understand already: Linear Part A-1 ◦ MN ◦ A ={ A-1 ◦ M1◦ A ◦ A-1 ◦ (I + e C+…) ◦A }N = {R ◦ (I + e 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,kb x2) Ĉ =A-1 C ◦ A =(0, kb3/2 x2 )

MN to first order: Origin of Phasors {R ◦ (I + e Ĉ +…)}N =? = R◦(I + e Ĉ +…) ◦R◦(I + e Ĉ +…) ◦ … ◦ R◦(I + e Ĉ +…) ◦R◦(I + e Ĉ +…) = R◦(I + e Ĉ +…) ◦ R-1 ◦R2 ◦(I + e Ĉ +…) ◦ R-2◦ R3 ◦(I + e Ĉ +…) R-3 ◦ … ◦ RN ◦(I + e Ĉ +…) ◦R-N ◦RN = (I + e (R1 Ĉ+ R2 Ĉ+ R3 Ĉ+ R4 Ĉ+ … + RN Ĉ ) + Order(e2) ) ◦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 =l Ĉ

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-1RB which is just

Definition of Phasors The n-turn map is: = (I + e (E1 G + E2 G + E3 G + E4 G + … + EN G ) + Order(e2) ) ◦ En Ek G = E-k G ◦ Ek Eab = dab exp( (-1)a im) In the new basis : Gi = S Gi;nmx1nx2m Ek G = S Gi;nm exp(ik{m-n-(-1)a} m) x1nx2m

Sk=1,N Ek G |a = Sk=1,N S Gi;nm exp(ik{m-n-(-1)a} m) x1nx2m Summing up Sk=1,N Ek G |a = Sk=1,N S 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.

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.

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!

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

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+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!

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=R2p/3◦N◦R2p/3-1 is possible.

One-Resonance Normalization

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!

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!

Three corrections Standard Correction Methods 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!

Next Lecture We will look in details in the computer implementation