Optimum Serpentine Acceleration in Scaling FFAG Shane Koscielniak September 20, 2013
Abstract Serpentine acceleration is typified by fixed radio frequency, fixed magnetic field and a near (but not) isochronous lattice, radial motion of the orbit, and two or more reversals of the motion in RF phase. This was discovered in 2003 for linear non-scaling FFAGs in the relativistic regime. In 2012, Kyoto University School of Engineering showed that serpentine accelera- tion is possible also in scaling FFAGs and may span the non-relativistic to relativistic regime. As a function of two key parameters, field index and synchronous energy, this paper shows how to optimize the extraction en- ergy and the voltage per turn for the scaling case. Optimization is difficult, and typically leads to poor performance: either extreme voltage or small acceleration range. Nevertheless, designs with credible acceleration pa- rameters can be obtained; and indicative examples are presented herein. 1
Introduction In the scaling FFAG, the magnet field has the form: B z ( R, z = 0) = ( R/R 0 ) k where k > 0 is the field index. R 0 is a reference radius. The subscript s shall denote synchronous value. The general orbit radius is given by R/R s = ( P/P s ) α where α = 1 / (1 + k ) < 1 is solely a property of the lattice. It follows that revolution period T as a function of E, P is given by T/T s = ( E/E s )( P/P s ) ( − 1+ α ) = ( β s /β )[( βγ ) / ( β s γ s )] α . Here γ is the relativistic kinematic factor, E = E 0 γ and E 0 = m 0 c 2 is the rest mass energy. We define T ≡ T ( γ ) , T s ≡ T ( γ s ) and T t ≡ T ( γ t ) where E s = E 0 γ s is a synchronous energy and E t = E 0 γ t is the transition energy. One may eliminate β = v/c in favour of γ . 2
Orbit Revolution Period T � Tg 1.8 1.6 1.4 1.2 Γ s1 2 3 4 5 6 Period versus energy ( γ ) for α = 1/2 (blue), 1/4 (red), 1/8 (yellow), 1/16 (green). 3
Two Synchronous Energies The curves are ”U” or ”V”-shaped. γ ( T ) is a double valued function: to each value of T belongs two values of γ . Each curve have a minimum which defines the transition energy. Solving ∂ ( T/T s ) /∂γ = 0 , one finds γ t = 1 / √ α . For brevity, let γ s 1 ≡ γ 1 and γ s 2 ≡ γ 2 be two energies having the same revolution period; there is a continuum of such doublets. We shall adhere to the convention that γ 1 < γ t < γ 2 . A certain doublet is chosen to be the synchronous reference when we set the radio frequency (RF) to be co- periodic with the orbit period T ( γ 1 ) = T ( γ 2 ) . Once this is chosen E 1 , E 2 become fixed points of the motion. Both values of the synchronous E s are equally valid! 4
It is a little arbitrary, but we choose to work with the lower E s 1 because it exists in the narrow range 1 < γ s 1 < γ t . T � Tg 1.8 1.6 1.4 1.2 Γ s1 2 3 4 5 6 The general features of the T/T g curves are a very steep rise as γ → 1 , and a long slow ramp for γ ≫ γ t . When selecting reference doublets, this has the consequence that as γ 1 → 1 , so γ 2 → ∞ . Thus the range of acceleration is unbounded. But this range is illusory, and corresponds to a linac-like regime with prodigious voltage requirement. 5
Hamiltonian H ( E, P, φ ) ≡ − Eh + h ( PP s ) ( P/P s ) α E s (1 + α ) + eV cos φ . (1) 2 π Because of the FFAG scaling property, the Hamiltonian is invariant whether we use E 1 or E 2 for the synchronous energy. We set h times their common revolution frequency equal to the radio frequency. These two energies are either side of transition; so, during acceleration, the direction of phase slip for the entire beam reverses twice. 6
6 30 25 5 20 4 15 3 10 2 5 Case 1: Case 3: � 3 � 2 � 1 0 1 2 3 � 3 � 2 � 1 0 1 2 3 ( α, γ s ) = (0 . 09357 , 1 . 650) ( α, γ s ) = (0 . 1371 , 2 . 427) Phase space contours: energy ( γ ) versus RF phase ( φ ) . Serpentine acceleration in the S-shape channel between two RF buckets offset in energy can be greater than the range (bottom to top) within a single RF bucket.
Acceleration Range The range is the sum of three phase space arcs: (i) from the injection energy E i to the first synchronous energy E 1 ; (ii) a path between E 1 and E 2 ; (iii) from the second synchronous energy to the extraction energy E x . The extraction energy is obtained by equating H ( E x , P x , π ) = H ( E 2 , P 2 , 0) , writing E x = E 2 + δE x , and solving for the increment 2 V/πh δE 2 x ≈ +1 /E 2 − E 2 /P 2 2 (1 − α ) The injection energy is obtained by equating H ( E i , P i , 0) = H ( E 1 , P 1 , π ) , writing E i = E 1 − δE i , and solving for the increment 2 V/πh δE 2 i ≈ − 1 /E 1 + E 1 /P 2 1 (1 − α ) 7
The energy range of the machine is ∆ E = ( E x − E i ) ≈ ( E 2 − E 1 ) + δE i + δE x ∼ 2 E 2 which is expressible solely in terms of E 1 , E 2 . But E 2 is expressible in terms of E 1 : E 2 ( E 1 ) is the solution of T ( E 1 ) = T ( E 2 ) . Hence there is an expression for the energy range in terms of E 1 , V, α . √ √ Typically δE i ≪ δE x : δE i ∼ P 1 c 2 and δE x ∼ E 2 2 . Typically, ∆ E = ( E x − E i ) ∼ 2 E 2 . 8
The total acceleration range ∆ E (blue) and the contribution from the fixed points ( E 2 − E 1 ) (red) for a particular α as a function of γ s 1 . The quantities are normalized by the transition energy. As γ s → 1 the range becomes unbounded; and as γ s → γ t the range shrinks to zero. �Γ � Γ �Γ � Γ 20 10 8 15 6 10 4 5 2 Case 1. Case 2. Γ 1 Γ 1 1.0 1.5 2.0 2.5 3.0 1.0 1.5 2.0 2.5 �Γ � Γ �Γ � Γ 3.0 8 2.5 6 2.0 1.5 4 1.0 2 0.5 Case 5. Case 4. Γ 1 Γ 1 1.2 1.4 1.6 1.8 1.05 1.10 1.15 1.20 1.25 1.30 9
Minimum Voltage The condition to connect the two fixed points E 1 and E 2 by a phase space path of zero width is obtained by equating the two Hamiltonians H ( E 1 , P 1 , π ) = H ( E 2 , P 2 , 0) and solving for voltage per turn: πh = ( E 2 − E 1 ) + ( E 2 P 2 1 − E 1 P 2 eV 0 2 ) . (2) E 1 E 2 (1 + α ) Evidently, one prefers low harmonic number. Eliminating the momenta leads to = πh ( γ 2 − γ 1 )( γ 1 γ 2 α − 1) eV . (3) E 0 γ 1 γ 2 (1 + α ) This is a very significant relation. 10
linac and ring like regimes If αγ 1 γ 2 ≫ 1 this corresponds to acceleration in a linac-like regime (case 1) in which ∆ γ/γ t ≫ 1 and eV 0 /E 0 → ( γ 2 − γ 1 ) απh/ (1 + α ) This is a very few turn acceleration regime, and there is little point employ- ing an FFAG ring unless the particles are very short lived. The required voltage is prodigious: order the rest mass energy per turn; this may be acceptable for leptons (e.g. 0.5 MeV for e) but not for hadrons (e.g. 1 GeV for p). Contrastingly, if αγ 1 γ 2 → 1 then V → 0 . In principle, this implies ∆ E/eV → ∞ ; but ∆ γ/γ t → 0 . This corresponds to acceleration in a ring-like regime (case 3), with tiny voltage and many turns but with a small range. 11
Minimum voltage = πh ( γ 2 − γ 1 )( γ 1 γ 2 α − 1) eV . (4) E 0 γ 1 γ 2 (1 + α ) By fine tuning of parameters, this feature may be exploited to give a limited multi-turn acceleration (cases 2,4,5). αγ 1 γ 2 = 1 has the single solution is γ 1 γ 2 = γ t . For all other values such that T ( γ 1 ) = T ( γ 2 ) , αγ 1 γ 2 > 1 and rises progressively rapidly because γ 2 increases more quickly than γ 1 falls. Clearly, it is an advantage to use small α . 12
Optimization Our task would appear to be to maximize the acceleration range for a given value of the voltage per turn V . Figure shows the normalized range ∆ E/E t (red), voltage eV 0 /E 0 (yel- low), and ∆ E/eV 0 (blue) which is roughly the number of turns, and as function of γ s 1 . While ∆ E/eV rises, the acceleration range falls dramati- cally; the voltage per turn falls even more precipitously. These behaviours are common to all values of α . � E � eV, �Γ � Γ ,eV � Eo � E � eV, �Γ � Γ ,eV � Eo 20 10 8 15 6 10 4 5 2 Case 1. Case 2. 0 Γ 1 0 Γ 1 1.0 1.5 2.0 2.5 3.0 1.0 1.5 2.0 2.5 13
� E � eV, �Γ � Γ ,eV � Eo � E � eV, �Γ � Γ ,eV � Eo 10 4 8 3 6 2 4 1 2 Case 3. Case 5. 0 Γ 1 0 Γ 1 1.0 1.5 2.0 2.5 1.00 1.05 1.10 1.15 1.20 1.25 1.30 The minimum voltage per turn is essentially the product of range and a quantity that diminishes as γ s 1 → γ t . This has two consequences for the combination ∆ E/eV 0 : (i) it is independent of range; and (ii) it rises as the range diminishes. Contrary to expectations, ∆ E/eV is not a suitable figure of merit upon which to base optimization. So we must apply to ∆ E and eV 0 directly as the basis for optimization. 14
We know that γ s 1 → 1 (large range, large voltage, few turns) and γ s 1 → γ t (small range, tiny voltage, many turns) are both poor choices for the synchronous energy. But one may speculate that useful working points exit between these ex- tremes. Our approach is to take combinations [ γ 1 , γ 2 ] which satisfy T ( γ 1 ) = T ( γ 2 ) exactly, and roughly satisfy γ 1 γ 2 ≈ γ 2 t . The optimization amounts to scanning α, γ s 1 .
Normalized range (left) and required voltage (right) as function of α, γ s 1 . Range of α = [0 . 1 , 0 . 5] . Figure shows that maximizing the energy range and minimizing the voltage are contradictory efforts. Thus one must choose, for given index α , either the range and accept the voltage, or place a limit on voltage per turn and accept the energy range. 15
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