Transverse-Longitudinal Coupling by Space Charge in Cyclotrons

0.2 MB·English

by  C. Baumgarten

Transverse-Longitudinal Coupling by Space Charge in Cyclotrons
C. Baumgarten1
Paul Scherrer Institute, Switzerlanda)
(Dated: 5 January 2012)
A method is presented that enables to compute the parameters of matched beams with space charge in
cyclotrons with emphasis on the effect of the transverse-longitudinal coupling. Equations describing the
transverse-longitudinal coupling and corresponding tune-shifts in first order are derived for the model of
an azimuthally symmetric cyclotron. The eigenellipsoid of the beam is calculated and the transfer matrix
is transformed into block-diagonal form. The influence of the slope of the phase curve on the transverse-
longitudinal coupling is accounted for. The results are generalized and numerical procedures for the case of
2 anazimuthallyvaryingfieldcyclotronarepresented. ThealgorithmisappliedtothePSIInjectorIIandRing
1
cyclotron and the results are compared to TRANSPORT.
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2
PACS numbers: 77.22.Jp,29.20.dg,41.85.Lc

n Keywords: cyclotrons, space charge effects, beam matching
a
J

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INTRODUCTION I. THE SIMPLIFIED CYCLOTRON MODEL

]
h Thesimplestrelativisticcyclotronmodelisbasedona
p There is continuous interest in the understanding of magneticfieldwithcylindricalandmid-planesymmetrie.
c- space charge effects in isochronous cyclotrons1–9. The The nominal orbital frequency of an ion beam coasting
c area is of special relevance for the conceptual designs in an isochronous cyclotron is
a of cyclotrons as energy-efficient drivers for accelerator
q q
s. driven systems (ADS)10,11. Nevertheless there is no self- ωo = B = B0γ, (1)
c consistent first order theory of matched bunched beams mγ mγ
i
s with space charge in cyclotrons up to date known to the where q is the charge and m the rest mass of the ions.
y author. Bertrand and Ricaud showed that it is possible TheaxialmagneticfieldisB =B(r)=B γ. Wedefinea
0
h to set up and solve linearized equations of motion for a cyclotron length unit a= c using the orbital frequency
p simple cyclotronmodel9, butthey didnot determine the ω . The particle velocity iωsov = ω r so that β = v/c =
[ o o
parametersofmatchedbeamsincyclotronsandtheydid r/a. Hence the relativistic factor γ can be written as

2 not generalize their results in order to find a numerical
v method to compute matched beams for sector-focused γ = 1 . (2)
6 cyclotrons. Information on the matched ellipsoid is of 1 (r/a)2
9 −
special interest for modern high-performance codes like
9 The axialmagnetic fieldpin dependence of the radius r is
1 OPAL, that enable to simulate high intensity beams in then given by
. the space charge dominated regime12,13.
9
1+ε(r)
0 B(r)=B , (3)
Firstwedescribetheusualsimplifiedazimuthallysym- 0
11 metric cyclotron model (see for instance14 or15), but in- 1− ar22
clude the linearized space charge forces. The model ne- q
: where ε is a small distortion of the isochronism (ε
v glects possible effects of rf-acceleration and is restricted ≪
i 1). The radial field increase compensates the relativistic
X tothedescriptionofacoastingbeam. Amodifiedversion
mass increase and the cyclotron is strictly isochronous
ofTeng’smethodtodecouplelongitudinalandhorizontal
r for ε(r)=0. The field index n= r dB is given by
a motionwillbepresentedandappliedtothesimplifiedan- B dr
alytical model16,17. Later we describe how this method r dB dε
n= r +γ2 1, (4)
can be applied in an iterative numerical procedure to B dr ≈ dr −
determine the parameters of matched beams with space
so that
charge in sector-focused cyclotrons. Finally we present
dε
some results as computed for the cyclotrons of the PSI 1+n γ2+r . (5)
high intensity accelerator facility. ≈ dr
The phase shift per turn ∆φ dφ is (see for instance14:
≡ dn
∆φ = T dφ =T dφhf N dθ
dt dt − h dt
= 2ωπc (ωhf −(cid:16)Nhωc) (cid:17) (6)
= 2π (N ω N ω )
ωc h o− h c
a)Electronicmail: [email protected] = 2πNh ωωoc −1 ,
(cid:16) (cid:17)
2
where θ is the azimuthal angle, ω = θ˙ is the real or- Let S be the skew-symmetric matrix with S2 = 1
c
−
bital frequency and ω = N ω the frequency of the
hf h o 0 1 0 0
accelerating high frequency system operated at the har-
1 0 0 0
monic number Nh. The bending radius r of a particle S = −0 0 0 1 , (13)
with charge q in a magnetic field B is given by r = p
qB 0 0 1 0
and hence ω can be written as  − 
c  
so that – since M is sympletic – the following relation
ω = v = mcγβ = p
c r mγr mγr (7) holds:
= qB(r) = q B0γ(1+ε) =ω (1+ε)
mγ m γ o MT SM=S. (14)
so that
IfthereexistsaninvertibletransformationmatrixE and
ωo = 1 1 ε a diagonal matrix λ such that
ωc 1+ε ≈ − (8)
ε ∆φ = ωo 1 F=EλE 1, (15)
≈ −2πNh − ωc − −
Hence dε is proportionalto the ch(cid:16)angeof(cid:17)the phase shift thenits is easyto showthatthe followingrelationholds:
dr
per turn with radius. M=E exp(λs)E 1 =EΛE 1. (16)
− −
Thereferenceorbitcorrespondingtothekineticenergy
E =mc2(γ 1)=E (γ 1) is a circle with the radius E is the matrix of columnwise eigenvectors and λ is the
− 0 − (diagonal) matrix of the corresponding eigenvalues of F.
γ2 1 The (imaginary) eigenvalues of F are the betatron fre-
r(E)=a γ− . (9) quencies of the orbital motion. F and M have the same
p eigenvectors – and the eigenvalues of M are the expo-
We will use x = r −r(E) as the horizontal coordinate nentials of the eigenvalues of F. The eigenellipsoid σE
of a specific orbit relative to the reference orbit, y as
defined by
the axial deviation from the median plane and z as the
longitudinal coordinate. σE =MσEMT , (17)
can then be written as follows19:
II. SOME GENERAL REMARKS σE = E E−1S, (18)
− D
where is the diagonal matrix with the eigenvalues of
Fromtheassumptionofmidplanesymmetrieitfollows D
σ S –whicharetheemittances(apartfromfactors i).
E
that the axial coordinate y is not coupled neither to the ±
horizontal nor to the longitudinal coordinate. Therefore
it is sufficient to describe the coupling of horizontal and
III. THE EQUATIONS OF MOTION (EQOM)
longitudinalmotion. Sinceweassumeazimuthalsymme-
trie, the matched beam ellipse must also be azimuthally
The Hamiltonian is given by
symmetric, i.e. constant along the orbital length coordi-
nate s. If the matched beam is constant, then the forces H = x′2 + δ2 + kx−Kx x2 γ2Kz l2 hxδ. (19)
are constant - evenif space charge is takeninto account. 2 2γ2 2 − 2 −
The vector describing the (horizontal and longitudi-
where the canonicalcoordinatesare x and l and the mo-
nal) displacement of an orbit at position s relative to
menta aregivenby x andδ. The EQOM(in firstorder)
the reference orbit is x = (x,x,z,δ)T where x = dx ′
′ ′ ds are:
is the horizontal direction angle with the reference orbit
and δ = p−p0 is the momentum deviation. We use the x 0 1 0 0 x
fdoersmcraiblisemd bpo0yf lRin.eaTraHlmaamni1l8toinnicalnudsiynsgtemthseamsefothroidnstoafnAce. dds xl′ = −kx−+hKx 00 00 γh12  xl′  ,
Wolski to compute the matched beam matrix19. The  δ   0 0 Kzγ2 0   δ 
equations of motion (to first order) can be written as      (20)
where h = 1 is the inverse bending radius and k =
x(s)=Fx(s), (10) r x
′ h2(1+n) is the horizontallyrestoring force. K and K
x z
whereFis a constant4 4-matrix. The generalsolution representthe strength of the horizontaland longitudinal
is × space charge forces20:
x(s)=exp(Fs)x(0)=M(s)x(0), (11) Kx = (σxK+3σ(y1)−σfx)σz K3 = 20√5π3ε0qmIλc3β2γ3
twiahlerMeM(s)(sis)is the transfer matrix. The matrixexponen- KKyz == σ(σxKxKσ+3y3σfσ(y1z)−σfy)σz f ≈ √3σγxσσzy
(Fs)2 (Fs)3
M(s)=exp(Fs)=1+Fs+ + +... . (12) (21)
2! 3!
3
The eigenvalues of F and M are: if dε < 0, i.e. if the radial increase of the magnetic field
dr
is below the isochronous field increase, then the longitu-
λ = Diag(iΩ, iΩ,iω, iω) dinal focusing is strengthened.
− −
Λ = Diag(eiΩs,e−iΩs,eiωs,e−iωs) The matrix of eigenvectors E of the force matrix F is
ab ≡ Kkx−(KK2x−K+zh2γ2 k ) (22) 1 1 1 1
z x x
≡ −
Ω = a+√a2−b E= iiΩΩA −iiΩΩA iiωωB −iiωωB  . (24)
ω = pa−√a2−b.  Kzγ2A K−zγ2A Kzγ2B K−zγ2B 
p  
Note that b must be positive to give a real–valued fre-
The inverse matrix E 1 is given by:
quency ω and hence a longitudinally focused orbit. This −
is especially important, if we allow for a (small) field er-
B iB i 1
ror ε(r). In this case the change of the orbital frequency − Ω −Ω Kzγ2
as given by ddrε can play a significant role: E−1 = 2(A1 B) −AB −ΩiiBA Ωii Kz1γ12  , (25)
kx == hh22(1γ+2+n)r=dεh2=(1h+2γr2dd+rε +hγdε2−1) (23) −  A −iωAω −ωωi −−KKzz1γγ22 
dr dr  
b = Kz(cid:0)(Kx−hddεr(cid:1)) where
If Kx < hddrε, then the longitudinal focusing frequency A = Ω2+hKz
ω is imaginary and the longitudinal beam size increases B = h (26)
ω2+Kz
exponentially with s. This in fact is a surprising feature
ofthistypeofcouplingincombinationwithisochronism:
Longitudinal focusing requires a strong enough horizon- The transfer matrix M canbe computed by using eq.16
tally defocusing space charge force. On the other hand, and one obtains:
Ac BC As˜ BS S s˜ AB(C c)
− ω − Ω Ω − ω −
1 ΩBS ωAs˜ Ac BC C c AB(ωs˜ ΩS)
M=  − − − −  , (27)
A B AB(ΩS ωs˜) AB(c C) AC Bc AB(Bωs˜ AΩS)
−  c C− s˜ −S S − s˜ AC −Bc 
 − ω − Ω BΩ − Aω − 
 
where IV. THE EIGENELLIPSOID
The matrix is explicitely given by =
C = cos(Ωs) S = sin(Ωs) (28) Diag(iε1, iε1, Diε2,iε2)19. ThematchedeigenelDlipsoid
c = cos(ωs) s˜ = sin(ωs) − −
can then be calculated using eq. 18:
Bε1 + Aε2 0 0 ε1 + ε2
Ω ω Ω ω
1 0 Bε Ω+Aε ω ε1Ω+ε2ω 0
σE = B A  0 1ε1Ω+ε2ω 2 Aε1KΩz+γB2ε2ω 0  (29)
−  ε1 + ε2 Kz0γ2 Kz0γ2 ε1 + ε2 
 Ω ω BΩ Aω 
 
Thebeamdimensionsaregivenbythediagonalelements ial motion can be treated separately and one finds
of the matrix representing the eigenellipsoid:
ε
σ2 = y . (31)
σx2 = B−1A BΩεx + Aωεz (30) y h2νy2−Ky
σz2 = B−1A(cid:0)AεxKΩz+γB2εz(cid:1)ω , IfoneconsidersEQ.21tqogetherwithEQ.30andEQ.31,
(cid:16) (cid:17)
where ε = ε and ε = ε are identified with the hori- then it is obvious, that this result does not enable to
1 x 2 z
zontal and longitudinal emittance, respectively. The ax- start a straightforward calculation, since the beam sizes
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depend on the space charge forces and vice versa in an SymplecticRotation “Symplectic Boost”
algebraically complicated way. But with the additional Tr(M−N)/2 cos(2φ)∆ cosh(2ψ)∆
assumptionof a sphericalbeam, it is possible to derive a Dt sin(2φ)2∆2 −sinh(2ψ)2∆2
4th order equation for the beam size as will be shown in D −m+S2nTS2T m+S2nTS2T
∆sin(2φ) ∆sinh(2ψ)
Sec. VII. A M−D−1m tan(φ) M−D−1m tanh(ψ)
B N+Dn tan(φ) N −Dn tanh(ψ)
TABLEI. Comparison ofthemethod of symplecticrotation
V. DECOUPLING LONGITUDINAL AND TRANSVERSE
with the symplectic “Lorentz boost”. The difference ∆ is
MOTION defined by ∆≡cos(µ )−cos(µ ). The matrix S is defined
1 2 2
0 1
If sectors are considered, then the azimuthal symme- by S2 = −1 0 and Dt is defined by Dt = 2Det(m)+
trie is broken and hence the force terms in the EQOM Tr(nm).(cid:18) (cid:19)
andconsequentlythe beamellipsoiddepend onthe posi-
tion s. The vertical beam dynamics can still be treated
separately, but is has to be taken into account that the tains16,17:
periodic change of the vertical beam size influences the
space charge factors Kx and Kz. The design orbit usu- M = cos2(φ)+D−1 D sin2(φ)
A B
allyhastobecomputednumericallyasdescribedbyGor- N = cos2(φ)+D D 1 sin2(φ)
don21. Ifthishasbeendone,itispossibletocomputethe m = B(D D)Asin(−φ) cos(φ) (35)
transfer matrix for known starting conditions, since the n = −( DA−1B D 1 ) sin(φ) cos(φ)
− −
equations of motion are known and can be integrated. − A − B
The problem is to find the correct beam dimensions for The rotation angle φ can be computed using
agivenbeamcurrentandgivenemittancessuchthatthe
beam is matched, i.e. such that EQ. 17 is fulfilled. 1Tr(M N) = cos(2φ)(cos(µ ) cos(µ ))
2 − 1 − 2
Teng and Edwards described a parametrization for 2Det(m)+Tr(nm) = sin2(2φ)(cos(µ ) cos(µ ))2,
1 2
−
coupled motion in two and more dimensions that allows (36)
to find the decoupling matrix16,17. They called their If we apply this method to the transfer matrix com-
method symplectic rotation. We will give a brief sum- puted according to EQ. 16, we obtain:
mary of the method and apply it to the problem of de-
coupling longitudinal and transverse motion. 1Tr(M N) = B+A(cos(Ωs) cos(ωs))
Giventwo2 2symplecticmatrices and thatform 2Det(m2)+Tr(−nm) = B−A4AB (cos(Ω−s) cos(ωs))2,
a block-diagon×al(i.e. decoupled) transAfer maBtrix T0: −(B−A)2 − (37)
Comparison yields:
0
T0 = A0
(cid:18) B (cid:19) cos(2φ) = B+A
α β B A (38)
(µ ) = cos(µ )1+sin(µ ) 1 1 (32) sin2(2φ) = −4AB
A 1 1 1 (cid:18)−γ1 −α1 (cid:19) −(B−A)2
α β
(µ ) = cos(µ )1+sin(µ ) 2 2 , From EQ. 26 we find that A> 0 and B > 0 and B > A
B 2 2 2 (cid:18)−γ2 −α2 (cid:19) sinceΩ2 >ω2,sothatthemethodfailsinthecaseunder
study. Themethodofsymplecticrotationisnotgenerally
where α , β and γ are the familiar twiss-parameters.
i i i applicable and has to be extended. A workaround solu-
Then a generalsymplectic transfer matrix M of coupled
tion was found by replacing trigonometic by hyperbolic
motion can be written as
functions1.
M n
M= =RT R 1. (33) I cosh(ψ) D 1 sinh(ψ)
m N 0 − R = −
(cid:18) (cid:19) D sinh(ψ) I cosh(ψ)
(cid:18) (cid:19) (39)
wsyhmerpeleMcti,cNro,tmatioanndmnatarrixe.2T×en2gmsuagtrgiecsetsedantdoRwriistethRe R−1 = (cid:18)−IDcossinhh(ψ(ψ)) −DI−c1osshin(hψ()ψ)(cid:19) ,
in the form
with Det(D) = 1, i.e. D is not symplectic – but R is
−
I cos(φ) D 1 sin(φ) symplectic. Tab.Icomparestheformulasofthesymplec-
R = −
D sin(φ) I cos(φ) ticrotationandofsymplectic“Lorentzboost”. Applying
(cid:18)− (cid:19) (34)
I cos(φ) D 1 sin(φ)
R−1 = D sin(φ) −I−cos(φ) .
(cid:18) (cid:19)
where D is a symplectic 2 2 transfer matrix that de- 1 Formallythis extension is a rotation about an imaginary angle,
scribes the structure of th×e coupling. Then one ob- similartoaLorentzboostinMinkowskispace.
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this method to the case under study then gives: VI. THE ITERATION PROCESS AND EXAMPLES
sinh(2ψ) = 2√AB cosh(2ψ) = B+A We have shown for the symmetric analytical exam-
B A B A
tanh(2ψ) = 2√−AB sinh(ψ) = − A ple that a modified version of the method of Teng and
B+A B−A Edwards enables to bring the transfer matrix in block-
cosh(ψ) = B tanh(ψ) = qA , diagonal form according to EQ. 32. The diagonalization
B A B
q − q (40) of the block-diagonal matrices is given by EQ. 43 and
which can be solved. The matrices and are then hence the matrix of eigenvectors E and the matched
given by: A B beam ellipsoid can be constructed.
In order to take advantage of this method for the case
cos(Ωs) sin(Ωs)/Ω of sectored cyclotrons, the method has to be applied it-
=
A Ω sin(Ωs) cos(Ωs) eratively. The goal is to compute the properties of a
(cid:18)− cos(ωs) ABω(cid:19)sin(ωs) matched beam. Assumed that the beam emittances and
B = sin(ωs)/(ABω) − cos(ωs) (41) the current are given as boundary conditions, one pro-
(cid:18) 0 √AB (cid:19) ceeds as follows:
D =
1/√AB 0 1. Provide an initial guess of the beam dimensions
(cid:18) (cid:19)
σ (s), σ (s) and σ (s).
x y z
The signs in suggest a negative β - but this can be
2
B
compensated by using either a negative emittance or a 2. Compute the driving terms of the space charge
negative frequency ω, since the transfer matrix is invari- forces K (s), K (s) and K (s).
x y z
ant against a change of the sign of ω, while σ is not.
E 3. Compute the one-turntransfermatrixM(s)forall
The transformation matrices are explicitely given by:
azimuthal angles.
1 0 0 A
4. Compute the eigenvectors of M(s) by diagonali-
0 1 1/B 0
R = B zation.
B A  0 A 1 0 
−
q 1/B 0 0 1
  5. Use the beam emittances to obtain the eigenellip-
  (42)
soid σ according to EQ. 18.
E
1 0 0 A
−
0 1 1/B 0 6. Take the beam sizes from σ and go back to step
R−1 = BBA  0 A −1 0  . 2, if beam sizes changed signEificantly compared to
− −
q 1/B 0 0 1 the previous iteration.
 − 
 
Note: R and R 1 are symplectic. To make the method Optionally one can start the iteration with a reduced
−
complete, we give the formula to split 2 2 matrices of beam current and increase it either during the iteration
the form of or : × process or use the result of the reduced beam current
A B as an initial guess for higher currents. The (speed of)
c = cosµ convergence strongly depends on the assumptions about
s = sinµ beam current and emittance. In the following examples,
c s/K e iµ 0 the process converged typically after less than 20 itera-
Ks c = R2 −0 eiµ R2−1 tions.
(cid:18)− (cid:19) (cid:18) (cid:19)
i/K i/K (43)
R2 = 1 −1
(cid:18) iK/2 1/2(cid:19) A. Example: PSI Injector II
R2−1 = −iK/2 1/2
(cid:18) (cid:19) The PSI high intensity proton accelerator facility23,24
(HIPA)consistsofaCockcroft-Waltonpre-accelerator,a
The matched beam ellipsoid can be computed for given four-sector injector cyclotron (Injector II, 72MeV) and
emittances accordingto EQ.18assoonasthe diagonali- the eight sector Ring cyclotron (590MeV). In routine
zation of the transfer matrix M is known. operation the beam current is 2.2mA.
Thefactthatwehavetoextendthedecouplingmethod The Injector II cyclotron can be modelled reasonably
of Teng and Edwards raises the question, whether the wellbyahardedgeapproximationofthesectormagnets.
description of decoupling is now complete or if there are Thishastheadvantagethatthecomputedmatchedbeam
other cases that require further modifications or exten- parameters can be compared to TRANSPORT25–27. We
sions. Theanswertothisquestionrequiresapropertwo- have computed the matched beam parameters for Injec-
dimensionalextensionofthe Courant-Snydertheory and torII.Theresultshavethenbeusedasinputparameters
a complete survey of all possible symplectic transforma- for TRANSPORT (with space charge). The results are
tions as presented in an accompanying paper22. shown in Fig. 1.
6
10 INJECTOR II, 0.842 MeV, 2.2 (mA), (ex,ey,ez)=(2.15,2.15,2.15) (p mm mrad) sthtaertoirnbgitroandeiuaslsRoeoob(θt0a)i.nsBaepsirdeecsisethvealpuuerfeorgeεo=me1tryωoof
8 − ω
6 2 sz (mm) for each EO21, which then allows to compute dε. In or-
dr
4 2 sy (mm) der to obtain the force matrix according to EQ. 20, the
2 following additional calculations have been done:
0
-2 1. Inverse bending radius h= 1 = B(r,θ)
--64 2 sx (mm) ρ B0p
-8 2. Local field index
SM1 SM2 SM3 SM4
-10
0 0.5 1 1.5 2 2.5 ρ dB B dB dB
s (m) n= = 0 p p .
B dx B2 dr θ− rdθ r
(cid:18) (cid:19)
FIG. 1. Transport (re-)computation of a matched beam in
InjectorIIforacirculatingbeamof2.2mAatlowenergy. The
3. Path length element ∆s=√r2+r2∆θ.
boxes indicate theposition of the sector magnets SM1..SM4. ′
Thehorizontalbeamwidthisplottedasanegativevalue. The
4. Horizontal(vertical)focusingk = 1+n (k = n).
longitudinal beam size is plotted as 2σz. x ρ2 y −ρ2
5. h(E,θ), n(E,θ), γ(E) and the space charge forces
urn (deg) 12 Phase shift per turn df 590 MeV Ring mKaxtr,ixKFy., Kz are then used to compose the force
/t
0 6. Theexponentialseriesallowstocomputethetrans-
Df
fer matrix for a (short) interval [θ...θ+∆θ]:
x,y1.75
n 1.5 Horizontal Tune nx M=exp(F ∆s)=1+F ∆s+ F2 ∆s2+... (44)
1.25 2!
1
0.75 Vertical Tune ny (Typically 3...4 terms are sufficient.)
z 0.1
n0.075 7. Multiplication of all matrices yields the one-turn
0.05 Longitudinal Tune nz transfer matrix M.
0.025
8. Diagonalize M and compute eigenellipse σ .
0 E
100 200 300 400 500 600
9. The average beam sizes are obtained from the
E (MeV)
eigenellipsoid and used to recompute the force co-
FIG. 2. Top: Phase shift per turn for the ring cyclotron efficients.
computed from the measured magnetic field map. Middle:
10. Theconvergenceisthencheckedandeitherthenext
Horizontal and vertical betatron-tunes of the Ring-cyclotron
as computed by EO-code (gray solid lines), the transfer ma- iteration starts with step 5) or is stopped.
trix method over the EO (black solid lines) and with space
Fig.2showssomeresultsofthematchedbeamwithspace
charge(dashedlines). Bottom: Longitudinalbetatrontuneas
inducedbytransverse–longitudinalcoupling. Theregionwith chargeforthePSIringcyclotron,thephaseshiftperturn
anegativeslopeofthephaseshiftisindicatedbyverticallines. vs. energyinthe upper, the transversaltunes inthe sec-
Without further field trimming, the longitudinal focusing is ond and the axial tune in the last graph. The region
suppressedbytheterm−hdε andthelongitudinalbeamsize where the slope of the phase shift is negative has been
dr
mayincrease. Thecalculationhasbeenperformedforabeam marked by vertical lines. In this region the longitudi-
currentof2.2mAand(εx,εy,εz)=(1.5,2.5,0.5)πmmmrad. nal tune due to space charge forces apparently becomes
imaginary.
B. Example: PSI Ring Cyclotron
VII. SPHERICAL SYMMETRY
The PSI ring cyclotron is a separated sector
isochronous cyclotron with 8 sectors. But since the field The first item in the described algorithm is to provide
ofthemagnetsdoesnotfalloffassharpasitdoesinInjec- aninitialguess ofthe beamdimensions. In the following
tor II, a hard edge approximationdoes not work equally we show how to obtain a possible choice of the initial
well. Furthermore the sectors are much more spiralled. guess for “nearly” spherical beams. If the horizontal-
Forthe sakeofprecisionweusedthemeasuredfieldmap longitudinal coupling is strong enough, the beam will
B(r,θ) to compute the equilibrium orbits (EO) on a ra- usually be approximately circular in this plane29. The
dial grid28. The radius R (θ) of the equilibrium orbit axial size depends mainly on the vertical emittance and
eo
(EO) is then given as a function of the angle and of the tune.
7
From EQ. 22 and EQ. 26 one quickly derives: Analyzing EQ. 50 one finds that
Ω2ω2 = K (K +h2γ2 k )
z x x lim x(α) = 1
−
ΩΩ22+ωω22 == kx(−k KxK−K+zK )2 4h2γ2K (45) lαim→0 x(α) = α1/3, (55)
x x z z α
−1 = − γ2hKz − →∞
B−A p√(kx−Kx+Kz)2−4h2γ2Kz so that the function x(α) = (1+α)1/3 is a reasonable
approximation. The maximalrelativedeviationofabout
With the assumption of an isochronous cyclotron (k =
x
h2γ2) and spherical symmetrie, i.e. 0.035 appears at α 3/2. Therefore we suggest to use
− ≈
the approximations
σ = σ =σ =σ γ
x y z
→ε = Kεx==Kεyx==εKz y =Kz = K3σ33γ (46) σ ≈(cid:26)σσ00((11++αα4)−1/3α322) ffoorr 05/≤2≤α≤α 5/2 . (56)
one finds
Usingthenormalizedemittanceε¯=βγε,thewavelength
Ω2ω2 = K2 λ = c = 2πc , the cyclotron radius a = c = r and
Ω2+ω2 = kx 2K the reνrlaftionωεoNch2 = 1 , the values of K , αωa0nd σβ can
Ω2 ω2 = k−2 4Kk (47) 0 µ0 3 0
− x− x be written as follows:
1 = K
B−A ph√1−4K/kx K = 3qIλ
The beam size σ then yields using EQ. 30: 3 = 20√53πqεµ00mIac3β2γ3
σ4 == ε24(εΩ(2Bω)ω2,+(BA−ΩA)2)2 (48) α == 1100√√55qqmmµµ00ccIIββa22√γγ332Nεγ3hr/a2N3/h2 (57)
kx−4K 5√10mcγNh ε¯
so that one obtains σ = 2rε = √2a(cid:0)βγ(cid:1)ε = √2aε¯.
0 γ γ γ
4K r2 4ε2r2 q
σ4 3 σ =0. (49) EQ. 56 and 57 describe the matched beam sizes for a
− 3γ − γ2
sphericalbeaminaperfectlyisochronouscyclotron. The
Ifonedefinesσ = 2rε,thenEQ.49canbewrittenas real beam sizes will usually differ from these values since
0 γ the spherical symmetrie requires that the vertical tune
q
x4 αx 1=0, (50) roughly equals the horizontal tune, while in most cy-
− − clotrons the vertical tune is below the horizontal tune.
where Neverthelessthe equationsderivedforthespecialcaseof
a spherical beam can be used as starting conditions for
x = σ
α = σ40K3r2 = K3√2γr >0. (51) the described iterative matching procedure.
3γσ03 3ε3/2
A polynom of 4th order has 4 solutions, namely the so-
VIII. SUMMARY
lutions of EQ. 50 are
1 12α A method has been developed that allows to compute
x= √Y Y , (52)
261/3 ± s± √Y − ! the parameters of a matched beam with space charge in
cyclotronsinlinearapproximationforgivenbeamcurrent
where the abbreviations used are defined by andknownemittances. As anexample, a matchedbeam
in the PSI INJECTOR II cyclotron has been computed.
Y = 21/3Z 831/3 0 The result has been used as input to TRANSPORT and
− Z ≥ (53)
Z = (9α2+√3√256+27α2)1/3 >7681/6. it has been shown that the beam is matched.
Furthermore it has been shown that the longitudinal
The minus sign in the square root of EQ. 52 belongs to
focusing as induced by the space charge forces depends
a pair of complex conjugate solutions and may not be
on the isochronism of the cyclotron. In case of the PSI
used. Starting from low values of α, Y monotonically
Ring cyclotron, the negative slope of the phase shift per
increases starting from zero so that we have to take the
turnmayreduceorevendestroythelongitudinalfocusing
plus infrontofthe squarerootas wellinorderto obtain
effect. The results are important for the design of high
positive solutions also for small values of α, so that
intensity cyclotrons. If beam emittance and current are
chosenaccordingly,then the space charge induced longi-
1 12α
x= √Y + Y . (54) tudinal focusing allows to replace flat-top cavities by ac-
261/3 s√Y − ! celerating cavities and thus increase the energy gain per
8
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portant to minimize beam losses and activation of com- (1996).
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