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Status Update

Chris RogersAnalysis PC

20th April 06

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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

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Recap

Last time, I tried to derive the first (M2) and second order transfer maps (M3) for a solenoid

Where a transfer map transports the phase space coordinates U U(z+dz) = M U(z)

dz

BB

B

BB

B

M

042

0

1002

200

4

02

10

1

200

0

020

0

2 dz

BB

B

BB

B

pM t

042

0

1002

200

4

02

10

)1(1

200

0

020

0

3

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2nd Moment Transport As before, 2nd moments are transported via <uiuj>fin = <MuiMuj>in

Formally for some pdf h(U) it can be shown that (Janaki & Rangarajan, Phys Rev E, Vol 59, 4577, 1999)

<uiuj>fin = integral(hfin(U) ui’ uj’ ) d2nU = integral( hin(U) (Mui’) (Muj’) ) Then in the linear approximation, i.e. using M2

<uiuj>fin = <M2ui M2uj> In the second order approximation, i.e. using M3 = M2 (1 + pt/0)

<uiuj>fin = <M2(1 + pt/0)ui M2(1 + pt/0)uj>

= <M2uiM2uj(1+pt/0)2> Find d/dz as a function of 2nd and 3rd moments

Then d2/dz2 as a function of 2nd, 3rd and 4th moments Assume beam is nearly Gaussian and apply Gaussian Joint Variable

Theorem <uvw> = 0 <tuvw>=<tu><vw> + <tv><uw> + <tw><uv> Assume <ptx>, <ptpx> are 0 (remember pt is just particle energy)

Then we get d/dz=0 and d2/dz2 as a function of 2nd moments This is a place to start, build up intuition/understanding

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Transverse Emittance Growth in Drift

For drift, consider 2d M2 such that M3 is a matrix (rather than a tensor in 6d) M3 =

Apply the transfer map to normalised emittance (recall px is normalised to p0) p0/m(<x2><px

2>-<xpx>2)1/2 -> p0/m(<x2> <px2> -<xpx>2

+2<px2><xpxpt>dz/0 - 2<xpx><px

2pt>dz/0)1/2

Use dn2/dz=2 dn/dz

Then dn/dz = p0/m(<px2><xpxpt>dz/0 - <xpx><px

2pt>dz/0)

Now consider an initially Gaussian beam so that 3rd moments are initially 0 <px

2pt> will always be 0 as momentum is constant in drift <xpxpt>=<M3px M3pt M3x>=<(pxpt)(x+pxdz+pxptdz/0)>

=<xpxpt>+<px2pt>dz + <px

2 pt2>dz

Again use <px2pt>=0

For a Gaussian beam, <px2pt

2>=<px2>< pt

2>+2<pxpt>2

Then d2n/dz2 is the change in dn2/dz; for <pxpt> = 0

d2n/dz2 = <px2>2<pt

2>/0 and n = (d2n/dz2) z2

10

)/1(1 0tpdz

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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) + p0/m <px

2>2<pt2>/0

2 z2, Remember that px, pt here is normalised to p0

Need 4th order for last few %? (~ 0.03% of 6)

5.90E-03

6.00E-03

6.10E-03

6.20E-03

6.30E-03

6.40E-03

6.50E-03

6.60E-03

0.00E+00 2.00E-01 4.00E-01 6.00E-01 8.00E-01 1.00E+00 1.20E+00

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4D vs 2D Why is 4D emittance growth the same as 2D? Non-linear terms appear in <qiqj> and <qi pj>

But if <px2pt>=0, non-linear term in <qi pj> is 0

For a general cylindrically symmetric beam (un-normalised) 4D emittance is (Penn Muc071)

(m/p0)4 n4 = |V| = (<x2><Px

2>-<xPx>2-<xPy>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/p0n)4/dz = d/dz([<xx> <pypy> - <xpy>2-<xPx>2]2)

= 2 d/dz([<xx> <pypy> - <xpy>2 -<xPx>2]

= 2<pypy> d/dz<xx>

All constant terms disappear and we are left with the expression for 2D emittance

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Emittance Growth in constant Bz

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 px

c = pxk+Ax/p0

Ax ~ a0yBz + a1yr2Bz’’ +…+ai yr2iBz(2i) = yBz in constant field

But this is just a shear in phase space i.e. emittance conserving Not the case with a changing B-field (Bz’’ != 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 Bz

I don’t explicitly mention Ay but that’s the same

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Kinetic coordinates

Make the transformation px -> px - yB0/2 px -> px + xB0/2

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 d2n/dz2=p0/m [(<px

2>+B0<xpy>)2 - B0<xpx>2]<pt2>

Reduces to the expression for drift when B0=0 Nearly Gaussian means obeys Gaussian Joint Variable Theorem dz

B

BpM t

000

1000

000

0010

)1(1

0

03

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Explicit expression for matched beam

Quote from Penn Muc-071 and use p0=pz, B0=2, =0 <xpx> = 0 <x2> = nm/p0 <px

2> = nm/p0 = nm/p0 (1/+ k2 )

<xpx> = - nm/p0 (Remember I use px normalised to p0)

Substitute into expression on previous slide d2n/dz2=p0/(m) (1-2 2)2

Finally note that for a beam with ’=0, ’’=0, =1 This comes from the differential equation for

Then d2n/dz2=0

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6.00E-03

6.01E-03

6.01E-03

6.02E-03

6.02E-03

6.03E-03

6.03E-03

6.04E-03

6.04E-03

6.05E-03

6.05E-03

0.00E+00 2.00E-01 4.00E-01 6.00E-01 8.00E-01 1.00E+00 1.20E+00

Matched Beam - L=0

ICOOL Emittance

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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

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Full Channel - emittance growth

Hoped emittance growth ~ <E2> (nb <E2>=) May be worth repeating with Gaussian distribution in energy

I think that <xpxpt> etc are still 0 with rectangular distribution Not sure about <xpxpt

2> etc

0

0.5

1

1.5

2

2.5

3

3.5

4

0 100 200 300 400 500 600 700 800

<E2>

4D

tra

nsv

ers

e e

mit

tan

ce c

hange

Periodic lattice

MICE SFoFoPalmer 1

Palmer 2

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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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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

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Physical Model842 430 30 40

230

15

150 630

No Detector Apertures

No Detector Apertures

No Detector Apertures

No Detector Apertures

All materials are copper

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200 MeV/c, L=0, <pz2>=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

radiu

s

z Px

x

MICE acceptance at diffuserRadius of MICE acceptance vs z

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200 MeV/c, L=spread, <pz2>=0

Repeat but introduce a spread in canonical angular momentum

radiu

s

z Px

x

MICE acceptance at diffuserRadius of MICE acceptance vs z

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200 MeV/c, L=0, <pz2>=spread

Introduce a spread in momentum well into the resonance region, no canonical angular momentum

radiu

s

z Px

x

MICE acceptance at diffuserRadius of MICE acceptance vs z

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(Recap) Beam Optics & Emittance Definition of linear beam optics:

Say we transport a beam from zin to zfin

Define an operator M s.t. U(zfin) = M U(zin) and M is called a transfer map In the linear approximation the elements of U(zfin) are a linear combination of

the elements of U(zin) e.g. x(zf) = m00 x(zin) + m01 y(zin) + m02 px(zin) + m03 py(zin) where mij are

constants Then M can be written as a matrix with elements mij such that ui(zf) =

jmijuj(zi)

2nd Moment Transport: Say we have a bunch with second moment particlesui (zfin) uj(zfin)/n Then at some point zfin, moments are particles(imik ui(zin)) (jmjk uj(zin))/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

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Non-linear beam optics Expand Hamiltonian as a polynomial series

H=H2+H3+H4+… where Hn is a sum of nth order polynomials in phase space coordinates ui

Then the transfer map is given by a Lie algebra M = … exp(:f4:) exp(:f3:) exp(:f2:)

Here :f:g = [f,g] = ( ( f/ qi)( / pi) - ( f/ pi)( / qi) ) g exp(:f:) = 1 + :f:/1! + :f::f:/2! + :f::f::f:/3! + …

And fi are functions of (Hi, Hi-1 … H2)

fi are derived in e.g. Dragt, Forest, J. Math. Phys. Vol 24, 2734, 1983 in terms of the Hamiltonian terms Hi for “non-resonant H”

For a solenoid the Hi 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 B0

Or try Dragt, Numerical third-order transfer map for solenoid, NIM A Vol298, 441-459 1990 but none explicitly calculate f3, 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)

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Application to Solenoids - leading order

Use U = (Q,,P,Pt;z) and Q = (x/l, y/l); P = (px/p0, py/p0), = t/l, P = pt/cp0

H2(U,z) = P2/(2l) - B0(QxP).zu/l + B02Q2/(2l) + Pt

2/2(l)2

H3 (U,z) = Pt H2/

H4 (U,z) = … B0 = eBz/2p0

zu is the unit vector in the z direction H2 gives a matrix transfer map, M2

Use f2 = -H2 dz M2 = exp(:f2:) = 1+:f2:+:f2::f2:/2+…

:f2: = i{[(B02qi- B0(qi

uxP).zu) / pi] - [(pi-B0Qxpiu.zu) / qi]}dz/l

+ pt/(00) dz/l / :f2::f2: = 0 in limit dz->0 Remember if U is the phase space vector, Ufin=M2 Uin, with uj/ ui = ij

Ignoring the cross terms, this reduces to the usual transfer matrix for a thin lens with focusing strength (eBz/2p0)2

B0(qiuxP).zu/2 term gives canonical angular momentum conservation

Pi - B0Qxpiu.zu/2 is kinetic momentum

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Next to leading order

f3 is given by f3 = - H3(M2 U, z)dz H3(M2U,z)dz = Pt H2(M2 U, z)/0 dz = Pt H2(U, z)dz in limit dz->0 Then :f3: = :Pt H2:dz = Pt :H2:dz/0 + H2 :Pt:dz/0 = Pt :f2: /0 + H2 dz /0

d/d Again :f3:n = 0 in limit dz -> 0

The transfer map to 3rd order is M3= exp(:f3:) exp(:f2:)=(1+:f3:+…)(1+:f2:+…) =1+:f2:+:f3: in limit dz->0

In transverse phase space the transfer map becomes M3 = (1 + (M2-1)pt/0) where 1 is the (4x4) identity matrix

In longitudinal phase space the transfer map becomes pt

fin = ptin

fin = in + pt/(00) dz + (pt2/(0

2 0) + H2/0)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

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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: pt

fin = ptin

fin = in + ptin/(00) dz + ( (pt

in)2 /(02

0) + H2in/0)dz

Then in lim dz -> 0 (is this right? Only true if variables are independent?) <ptpt> = const <pt>fin = <pt>in + <ptpt>in/(00) dz + < (pt

2/(02

0) + H2/0)>dz <>fin = <>in + 2<pt>in/(00) dz + 2< (pt

2/(02

0) + H2/0)>dz Longitudinal emittance (squared) is given by

fin2= <ptpt>in (<>in + 2<pt>in/(00) dz) -

( (<pt>in)2 +2 <pt>in <ptpt>in/(00) dz ) +

2(<ptpt>in - <pt>in)< (pt2/(0

2 0) + H2/0)>dz

= in2 + 2(<ptpt>in - <pt>in)< (pt

2/(02

0) + H2/0)>dz Growth term looks at least related to amplitude momentum correlation Need to check against tracking to fix/test algebra

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