08/06/2015 1 fcc-ee with kinks can we conserve polarized beams?

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08/06/2015 1 FCC-ee with kinks Can we conserve polarized beams?

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Page 1: 08/06/2015 1 FCC-ee with kinks Can we conserve polarized beams?

08/06/2015 1

FCC-ee with kinks

Can we conserve polarized beams?

Page 2: 08/06/2015 1 FCC-ee with kinks Can we conserve polarized beams?

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Transverse polarization both in the Z resonance region : 44 to 47 GeV beam energy, around half integer spin tunes) mZ , Z and in the WW thrshold region: 80 to 82 GeV also at half integer spin tunes) mW

is at the heart of the precision FCC-ee physics program.

In fact it is one of these things that makes it very special wrt to hadronic or linear colliders or even smaller e+e- colliders such as CEPC.

It needs to be used continuously for both the W and Z mass measurements.

A measurable polarization of a few percent is needed at both energies to proceed to spin matching to bring polarization level to ~10% to allow resonant depolarization.

Physics requirement

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It has been proposed to kink the FCC ring with kinks of 0.4 to 1.7 % to reduce somewhat the depth of the deepest pits and underground caverns.

The kink angles are of the order of 0.4% to 1.4%. At least two such kinks are necessary.

This 1%=10mrad seems small but this results in very large kicks for the spin, which undergoes a rotation with respect to the axis of the imposed trajectory rotation which is multiplied by the spin tune

=E(GeV)/0.44065 =

Given the spin tune of 103.5 at the Z peak and 182.5 at the W threshold,-- spin rotation resulting from a 0.4% kink is0.41 rad (23 degrees) at the Z energy0.73 rad (42 degrees) at the WW threshold-- spin rotation resulting from a 1.4% kink is1.45 rad (83 degrees) at the Z energy (almost flat!)2.56 rad (146 degrees) at the WW threshold (almost reversed!)

KINKS

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The high value of spin tune combined with the stochastic energy variationsof particles due to synchrotron radiation horizonal component of spin is destroyed in typically a few longitudinal damping times, in a way which is related to the energy spread.

Only the component of the spin that is parallel to the magnetic field is conserved

in case of a kink there is a continuous (twice per turn) change of the axis of themagnetic field which kicks the spin in the horizontal plane and would lead toimmediate depolarization if uncorrected.

it is essential to compensate locally the spin rotation due to the kink with a spin rotator. This compensation has to be effective for any beam particle -- within the energy acceptance and -- within the transverse emittance.

The KINK problem, an explanation with words:

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�⃗� �⃗�

in the magnet that causes the kink:small rotation of trajectory large rotation of spin, amplified by = 103.x

in the arc magnet:spin precession at rate of = 103.x turns of spin per turn of machine. This rotation will be augmented every turn by the the kink-induced rotation with a random phase (because of energy spread) rapid depolarization.

spin normally orient itself oppositeto magnetic field, for electrons.

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POSSIBLE CORRECTION:

1. Example of compensation of a known imperfection that creates depolarization.

2. an idea of a possible correction scheme

3. discussion, warning.

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Example: experimental solenoid compensation at LEP

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small spin rotator

Spin matching bumps were double-pi bumps

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Double pi bumps:vertical orbit and vertical dispersion cancel except inside bumps themselves.-- effect of errors?-- effect of dispersion inside bumps?

Bspin kink= 0.066 rad

𝜽𝒔

principle: equilibrium spin vector is in stable position (vertical in arcs)

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spin motion through the double pi bump:

Symmetry guarantees that spin returns to verticalat the end of the section.

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note that 66 mrad ‘spin kink’ (equivalent to 0.06% real kink) is enough to essentially kill polarization correction was necessary.

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solenoid spin compensation allowed us to go from 4% (resonant depolarization difficult) to 16% (resonant depolarization easy)

further correction needed for machine imperfection.

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examples of spin matching (I)

Solenoid spin matching : fully deterministic (know kick from exp. solenoids)Deterministic Harmonic spin matching measure orbit, decompose in harmonics, cancel components near to spin tune. NO FIDDLING AROUND. This worked very well at LEP-Z and should work even better at FCC-ee-Z if orbit is measured better.

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When all else fails, empirical spin matching: excite harmonics one by one to measure directly their effect on polarization and fit for pole in 4-D space. Here 8% polarization at 61 GeV.

examples of harmonic spin matching (II)

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from LEP to FCC-ee

1. Double pi bumps worked very well at LEP.

There was a particular relationship between vertical phase advance and spin tune. This has to be studied again before concluding that scheme can work for FCC-ee

2. the same principle (2 pi bumps) as for solenoids was used for harmonic and empirical spin compensation

3. and can be used in principle to spin-match small vertical orbit kicks (<1mrad) such as proposed for having FCC-ee in Butterfly shape -- this may or may not work for achieving a large (> 10mrad) orbit kick by decomposing it into several small ones

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top view

Artist view

side view

arc bends interspaced with vertical bends

vertical kicks interspaced with arc bends

Can one design kicks so that spin motion cancels?

The butterfly a single kick of 10mradis way too large to swallow ... break it up

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top view

side view

spin viewwrt trajectory

Case of a kink of 0.8% split in two pieces, for s= 103.5

Horizontal bendsVertical bends

kink =0.4%=4mrad 0.4%

spin = 414 mrad spin = spin = 414 mrad

what we would like:

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Comments on this simple scheme, assuming it can be realized in practice:

1. cancellation of spin precession in the vertical bend is ensured by precession in the interleaved horizontal bend

2. this cancellation will be affected by synchrotron radiation inside the system which changes one or two of the precession angles. given the large radius of FCC-ee expect this to be small. TBV

3. it will also become less precise if particle is off-momentum

4. it will work better if kinks are smaller, keeping even number of kinks -- depolarization scales as 2.

5. it will only work for a precise beam energy.

6. for small changes of energy (such as a few percent changes)it should be feasible to correct mismatch with a small double pi bump

7. this and simmilar schemes will not work at the WW energy if it works at the Z and vice versa.This is a fundamental difficulty.

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Investigation of practical implementation

Bastian in Washington.

one such cell bends by 0.4% if B rotted by 90o...This FODO cell bends by 4.08 mrad spin is rotated by 0.422 at Z, 0.745 at W pi rotation is realized with 7.44 such cells at vs = 103.5 (Z peak) pi rotation is realized with 4.22 such cells at vs = 182.5 (WW)

pi bump in vertical space is realized with 3 such cells

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AT THIS POINT:1. Kinks in the Arcs of FCC produce VERY LARGE spin rotations. -- a 1.7% kink at 80.5 GeV (WW) simply flips the spin.

2. Uncorrected, this would result in 0.0% polarization at all energies of interest.

3. There is an almost thinkable solution at the Z with splitting the kick bends in an even number of segments in such a way that the there is a pi spin rotation between them. (assuming the imperfect pi rotation in 8 cells can be fixed, e.g. switching off two of the 32 bends or decreasing sthem by 6% or...) 4. There is an almost thinkable, but different solution at the WW (assuming the imperfect pi spin rotation in 4 cells can be fixed, e.g. by increasing strength by 5% or....)

5. I have no clue at the moment on how to unify the spin compensation for Z and WW to keep the same tunnel/vac. chamber geometry. Perhaps using orbit correctors to complement the system ?

5’ I have not envisaged the effect on vertical emmittance. 08/06/2015

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AT THIS POINT:

6. to go one step further one would need to make some progress. -- implementation of FCC-ee lattice in the spin simulators -- start to play with the correctors to find the ‘elementary spin rotator’ (the equivalent of the pi bump of LEP) -- see if they can help with the problem. 7. In any case cannot help notice that --- for a given tunnel radius and given luminosity in FCC-eethe vertical kicks will require an increase of the number of magnetsand an increase of the strength of the horizontal bending magnetsthat will result in an increase of energy loss per turn which is as largeas 7.5% for a 1.4% kick. This will require more volts and more MW, for main ring and injector, thus many more MCHF, to be compared with the gain in the depth of the shafts. (and that for a given FCC-hh beam energy they will require an increase of the tunnel length and dipole magnets of 2.8%)

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