1. 1 Status Update Chris Rogers Analysis PC 20th April 06

2. 2 Overview Continuation/Update of talk given at last PC Try to explain emittance growth ito “non-linear beam optics” e.g. chromatic aberations Particles at different energies see different focusing This is the focus of my work atm Alternative explanation is emittance growth ito resonances Magnets do not focus at all at some energies I haven’t continued work on this yet A few corrections and some predictions for emittance growth due to chromatic aberrations

3. 3 Recap Last time, I tried to derive the first (M 2 ) and second order transfer maps (M 3 ) for a solenoid Where a transfer map transports the phase space coordinates U U(z+dz) = M U(z)

4. 4 2nd Moment Transport As before, 2nd moments are transported via fin = in Formally for some pdf h(U) it can be shown that (Janaki & Rangarajan, Phys Rev E, Vol 59, 4577, 1999) fin = integral(h fin (U) u i ’ u j ’ ) d 2n U = integral( h in (U) (Mu i ’) (Mu j ’) ) Then in the linear approximation, i.e. using M 2 fin = In the second order approximation, i.e. using M 3 = M 2 (1 + p t /  0 ) fin = = Find d  /dz as a function of 2nd and 3rd moments Then d 2  /dz 2 as a function of 2nd, 3rd and 4th moments Assume beam is nearly Gaussian and apply Gaussian Joint Variable Theorem = 0 = + + Assume, are 0 (remember p t is just particle energy) Then we get d  /dz=0 and d 2  /dz 2 as a function of 2nd moments This is a place to start, build up intuition/understanding

5. 5 Transverse Emittance Growth in Drift For drift, consider 2d M 2 such that M 3 is a matrix (rather than a tensor in 6d) M 3 = Apply the transfer map to normalised emittance (recall p x is normalised to p 0 ) p 0 /m  ( - 2 ) 1/2 -> p 0 /m  ( - 2 +2 dz/  0 - 2 dz/  0 ) 1/2 Use d  n 2 /dz=2 d  n /dz Then d  n /dz = p 0 /m  ( dz/  0 - dz/  0 ) Now consider an initially Gaussian beam so that 3 rd moments are initially 0 will always be 0 as momentum is constant in drift = = = + dz + dz Again use =0 For a Gaussian beam, = +2 2 Then d 2  n /dz 2 is the change in d  n 2 /dz; for = 0 d 2  n /dz 2 = 2 /  0 and  n = (d 2  n /dz 2 ) z 2

6. 6 Prediction vs Tracking ICOOL (4d) transverse emittance in blue Initially 6  beam transported through 1 metre long drift Initial conditions beta 333 with kinetic angular momentum for 4T field Predicted (2d) emittance in pink  n (z) =  n (0) + p 0 /m  2 /  0 2 z 2, Remember that p x, p t here is normalised to p 0 Need 4 th order for last few %? (~ 0.03% of 6  )

7. 7 4D vs 2D Why is 4D emittance growth the same as 2D? Non-linear terms appear in and But if =0, non-linear term in is 0 For a general cylindrically symmetric beam (un-normalised) 4D emittance is (Penn Muc071) (m/p 0 ) 4  n 4 = |V| = ( - 2 - 2 ) 2 |V| is the covariance matrix Where green term gives emittance growth Non-linear terms cannot produce a cylindrical asymmetry so other terms can be ignored Then emittance growth is 1/4 d(  n 4 )/dz d (m/p 0  n ) 4 /dz = d/dz([ - 2 - 2 ] 2 ) = 2 d/dz([ - 2 - 2 ] = 2 d/dz All constant terms disappear and we are left with the expression for 2D emittance

8. 8 Emittance Growth in constant B z Next step is to look at a constant B-field We don’t have to worry about canonical momentum vs kinetic momentum yet Canonical momentum p x c = p x k +A x /p 0 A x ~ a 0 yB z + a 1 yr 2 B z ’’ +…+a i yr 2i B z (2i) = yB z in constant field But this is just a shear in phase space i.e. emittance conserving Not the case with a changing B-field (B z ’’ != 0) So I can calculate emittance with canonical momenta and it will be the same as emittance calculated with kinetic momenta Exactly true only in a constant B z I don’t explicitly mention A y but that’s the same

9. 9 Kinetic coordinates Make the transformation p x -> p x - yB 0 /2 p x -> p x + xB 0 /2 Then the 2nd order transfer map simplifies to And we can get the emittance growth for a “nearly Gaussian” beam in a constant field to be d 2  n /dz 2 =p 0 /m [( +B 0 ) 2 - B 0 2 ] Reduces to the expression for drift when B 0 =0 Nearly Gaussian means obeys Gaussian Joint Variable Theorem

10. 10 Explicit expression for matched beam Quote from Penn Muc-071 and use p 0 =p z, B 0 =2 ,  =0 = 0 =  n m/p 0  =  n m/p 0  =  n m/p 0 (1/  +  k 2 ) = -  n m/p 0  (Remember I use p x normalised to p 0 ) Substitute into expression on previous slide d 2  n /dz 2 =p 0 /(  m) (1-  2  2 ) 2 Finally note that for a beam with  ’=0,  ’’=0,  =1 This comes from the differential equation for  Then d 2  n /dz 2 =0

11. 11 Matched Beam - L=0 ICOOL Emittance

12. 12 Full Channel - field map Consider two field maps Periodic focusing lattice MICE lattice Fire muons from -2750 to +2750 Use beam gaussian in transverse with beta=420 at 200 MeV/c Rectangular distribution in energy Also use Palmer’s beam files

13. 13 Full Channel - emittance growth Hoped emittance growth ~ (nb =) May be worth repeating with Gaussian distribution in energy I think that etc are still 0 with rectangular distribution Not sure about etc 4D transverse emittance change Periodic lattice MICE SFoFo Palmer 1 Palmer 2

14. 14 Conclusions Starting to understand emittance growth Can predict emittance growth in drift Starting to work with emittance growth in fields Be nice to get  n (  ) What if we drop the “nearly Gaussian” approximation? Need to understand longitudinal emittance growth My aim is to show longitudinal emittance growth due to energy straggling Need to pick up work on TOF/ longitudinal emittance resolution again This is the other plot I want in the TOF justification note I also have some work on scraping/MICE acceptance that tells us about detector apertures and so on

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16. 16 Scraping An aim of MICE is to measure the phase space accepted by the SFoFo channel So detector acceptance must be larger than SFoFo acceptance Any particles which get through the SFoFo must make it through the detectors also This defines the size required for the detectors Get the required size for the tracker window at the Helium/Vacuum interface and the diffuser diameter I only consider the 200 MeV/c lattice Fill phase space with muons Fire muons through SFoFo aperture Look at radius of muons that survive

17. 17 Physical Model 842 4303040 230 15 150630 No Detector Apertures All materials are copper

18. 18 200 MeV/c, L=0, =0 Radius of accepted phase space vs z Detector/diffuser radius must be larger than this radius I filled 2d transverse phase space All muons have no net canonical angular momentum All muons have 200 MeV/c momentum Detector radius has to be larger than the black line radius zPx x MICE acceptance at diffuserRadius of MICE acceptance vs z

19. 19 200 MeV/c, L=spread, =0 Repeat but introduce a spread in canonical angular momentum radius z Px x MICE acceptance at diffuserRadius of MICE acceptance vs z

20. 20 200 MeV/c, L=0, =spread Introduce a spread in momentum well into the resonance region, no canonical angular momentum radius zPx x MICE acceptance at diffuser Radius of MICE acceptance vs z

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22. 22 (Recap) Beam Optics & Emittance Definition of linear beam optics: Say we transport a beam from z in to z fin Define an operator M s.t. U(z fin ) = M U(z in ) and M is called a transfer map In the linear approximation the elements of U(z fin ) are a linear combination of the elements of U(z in ) e.g. x(z f ) = m 00 x(z in ) + m 01 y(z in ) + m 02 p x (z in ) + m 03 p y (z in ) where m ij are constants Then M can be written as a matrix with elements m ij such that u i (z f ) =  j m ij u j (z i ) 2nd Moment Transport: Say we have a bunch with second moment  particles u i (z fin ) u j (z fin )/n Then at some point z fin, moments are  particles (  i m ik u i (z in )) (  j m jk u j (z in ))/n But this is just a linear combination of input 2nd moments Emittance conservation: It can be shown that, in the linear approximation, so long as M is symplectic, emittance is conserved (Dragt, Neri, Rangarajan; PRA, Vol. 45, 2572, 1992) Symplectic means “Obeys Hamilton’s equations of motion” Sufficient condition for phase space volume conservation

23. 23 Non-linear beam optics Expand Hamiltonian as a polynomial series H=H 2 +H 3 +H 4 +… where H n is a sum of n th order polynomials in phase space coordinates u i Then the transfer map is given by a Lie algebra M = … exp(:f 4 :) exp(:f 3 :) exp(:f 2 :) Here :f:g = [f,g] = ( (  f/  q i )(  /  p i ) - (  f/  p i )(  /  q i ) ) g exp(:f:) = 1 + :f:/1! + :f::f:/2! + :f::f::f:/3! + … And f i are functions of (H i, H i-1 … H 2 ) f i are derived in e.g. Dragt, Forest, J. Math. Phys. Vol 24, 2734, 1983 in terms of the Hamiltonian terms H i for “non-resonant H” For a solenoid the H i are given in e.g. Parsa, PAC 1993, “Effects of the Third Order Transfer Maps and Solenoid on a High Brightness Beam” as a function of B 0 Or try Dragt, Numerical third-order transfer map for solenoid, NIM A Vol298, 441-459 1990 but none explicitly calculate f 3, etc “Second order effects are purely chromatic aberrations” Alternative Taylor expansion treatment exists E.g. NIM A 2004, Vol 519, 162–174, Makino, Berz, Johnstone, Errede (uses COSY Infinity)

24. 24 Application to Solenoids - leading order Use U = (Q, ,P,P t ;z) and Q = (x/l, y/l); P = (p x /p 0, p y /p 0 ),  =  t/l, P  =  p t /cp 0 H 2 (U,z) = P 2 /(2l) - B 0 (QxP).z u /l + B 0 2 Q 2 /(2l) + P t 2 /2(     l) 2 H 3 (U,z) = P t H 2 /   H 4 (U,z) = … B 0 = eB z /2p 0 z u is the unit vector in the z direction H 2 gives a matrix transfer map, M 2 Use f 2 = -H 2 dz M 2 = exp(:f 2 :) = 1+:f 2 :+:f 2 ::f 2 :/2+… :f 2 : =  i {[ (B 0 2 q i - B 0 (q i u xP).z u )  /  p i ] - [(p i -B 0 Qxp i u.z u )  /  q i ]}dz/l + p t /(  0  0 ) dz/l  /   :f 2 ::f 2 : = 0 in limit dz->0 Remember if U is the phase space vector, U fin =M 2 U in, with  u j /  u i =  ij Ignoring the cross terms, this reduces to the usual transfer matrix for a thin lens with focusing strength (eB z /2p 0 ) 2 B 0 (q i u xP).z u /2 term gives canonical angular momentum conservation P i - B 0 Qxp i u.z u /2 is kinetic momentum

25. 25 Next to leading order f 3 is given by f 3 = - H 3 (M 2 U, z)dz H 3 (M 2 U,z)dz = P t H 2 (M 2 U, z)/  0 dz = P t H 2 (U, z)dz in limit dz->0 Then :f 3 : = :P t H 2 :dz = P t :H 2 :dz/  0 + H 2 :P t :dz/  0 = P t :f 2 : /  0 + H 2 dz /  0 d/d  Again :f 3 : n = 0 in limit dz -> 0 The transfer map to 3rd order is M 3 = exp(:f 3 :) exp(:f 2 :)=(1+:f 3 :+…)(1+:f 2 :+…) =1+:f 2 :+:f 3 : in limit dz->0 In transverse phase space the transfer map becomes M 3 = (1 + (M 2 -1)p t /  0 ) where 1 is the (4x4) identity matrix In longitudinal phase space the transfer map becomes p t fin = p t in  fin =  in + p t /(  0  0 ) dz + (p t 2 /(  0 2  0 ) + H 2 /  0 )dz Longitudinal and transverse phase space are now coupled It may be necessary to go to 4th/5th order to get good agreement with tracking

26. 26 Longitudinal Emittance Growth This was all triggered by a desire to see emittance growth from energy straggling so need to understand longitudinal emittance growth Use: p t fin = p t in  fin =  in + p t in /(  0  0 ) dz + ( (p t in ) 2 /(  0 2  0 ) + H 2 in /  0 )dz Then in lim dz -> 0 (is this right? Only true if variables are independent?) = const fin = in + in /(  0  0 ) dz + dz fin = in + 2 in /(  0  0 ) dz + 2 dz Longitudinal emittance (squared) is given by  fin 2  = in ( in + 2 in /(  0  0 ) dz) - ( ( in ) 2 +2 in in /(  0  0 ) dz ) + 2( in - in ) dz =  in 2 + 2( in - in ) dz Growth term looks at least related to amplitude momentum correlation Need to check against tracking to fix/test algebra

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