Nuclear Instruments and Methods in Physics Research A 707 (2013) 54–57
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Nuclear Instruments and Methods in Physics Research A journal homepage: www.elsevier.com/locate/nima
Particle orbit in a radial-sector scaling FFAG Tae-Yeon Lee n, Hyungsuck Suh, Hee-Seock Lee Pohang Accelerator Laboratory, San 31, Hyoja-dong, Namgu, Pohang, Kyungbuk, Republic of Korea
a r t i c l e i n f o
a b s t r a c t
Article history: Received 11 July 2012 Received in revised form 15 December 2012 Accepted 15 December 2012 Available online 25 December 2012
The particle orbit in the bending magnet of a radial-sector FFAG accelerator is derived analytically in the approximation valid for a small bending angle. The orbit is described in terms of the distance from the machine center, r, and its variation is derived as a function of the bending angle. Also, the variation of the distance from the bending center, r, is derived as a function of the bending angle. The result shows how the ratio, r=r, can be made independent of the particle momentum, which is one of the conditions of the scaling property. & 2012 Elsevier B.V. All rights reserved.
Keywords: FFAG Orbit Momentum Scale
1. Introduction A fixed-field alternating gradient (FFAG) accelerator [1,2] is receiving attention recently as a suitable choice for a high energy, high current yet small sized proton accelerator, which can be used for an accelerator-driven nuclear reactor system. There is no unique type of FFAG accelerator but there are a few different types with different magnets and different operational principles, respectively. For example, some FFAG accelerators use sector-type magnets while others use spiral-type magnets. The central concept of FFAG accelerators is that the magnetic field is temporally fixed but spatially variable. In a synchrotron, the particle orbit virtually remains unchanged while the machine accelerates particles by increasing its magnetic field. This acceleration process of a synchrotron is not very fast and, thus, its repetition rate is limited to at most well below a hundred Hz. On the other hand, in an FFAG accelerator, as a particle gains energy from the RF section, its orbit radius increases. Unless the moving particle is relativistic, its revolution time changes and the RF frequency should also change. Because the RF change is much faster than the magnetic field change, an FFAG accelerator can achieve a much higher repetition rate and, hence, a much higher current proton beam than a synchrotron. Although this is a property shared by cyclotrons, a cyclotron has a limitation in its achievable beam energy as well as other differences compared to FFAG accelerators. Some of the differences are that the FFAG magnet lattice makes use of strong focusing by the distribution of alternating gradients, and the beam motion in an FFAG accelerator is practically periodic. Cyclotrons also
n
Corresponding author. Tel.: þ82 54 279 1461; fax: þ 82 54 279 1069. E-mail address:
[email protected] (T.-Y. Lee).
0168-9002/$ - see front matter & 2012 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.nima.2012.12.086
have gradients, but these gradients are mainly used to maintain the isochronous orbit as the beam energy goes relativistic and not for focusing, and the cyclotron orbit is more spiral. Hence, FFAG accelerators face the issue of betatron resonance, as in synchrotron accelerators. In fact, the most important issue in FFAG physics is how to solve this resonance problem. Of the different types of FFAG accelerators, the original and most traditional type of FFAG accelerator solves this problem by keeping the chromaticity to 0 during acceleration in order for the betatron tunes to remain unchanged during acceleration. While the zero-chromaticity is achieved by controlling sextupole magnets in a synchrotron, this FFAG accelerator does not have such dedicated magnets for chromaticity control but uses a kinematic method. One of the kinematic requirements is that the orbits of different beam energies have the same shape (in other words, orbits scale). Hence, this type of FFAG is called a scaling FFAG. Since most of the FFAG accelerators that are in operation are scaling-type, it is important to study characteristics of the scaling FFAG accelerators. In a synchrotron, there is only one orbit and the orbit remains unchanged, and the radius of curvature, r, is also constant. Thus, it is trivial to describe the orbit mathematically. On the other hand, in a scaling FFAG accelerator, the radius of curvature as well as the curvature center varies along the orbit. Therefore, in a scaling FFAG accelerator, the natural reference coordinate is not r but r, the distance from the machine center. ! ! ! ! Note that r and r are related by r ¼ r þ a , where a is the position from the machine center to the center of r. Hence, the FFAG magnetic field depends on r as in B ¼ B0 ðr=r m Þk where k is a constant and rm is a reference position, and it is a sector magnet with the angle of 2f with respect to r as shown in Fig. 1. Note that the magnetic field has a gradient in the r-direction, which provides the horizontal focusing, and horizontal betatron motion
T.-Y. Lee et al. / Nuclear Instruments and Methods in Physics Research A 707 (2013) 54–57
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position. The thick line AD denotes the orbit. Suppose that when the particle is at point Bn, which has a coordinate ðr ¼ r n , f ¼ fn Þ in ! the reference frame O, its radius of curvature vector is r n that is located at Cn and rotated angle y ¼ yn from the magnet center line OA as shown in Fig. 1. The orbit is the trace of the end point of the ! ! ! vector r ðyÞ. As can be seen in Fig. 1, r and r point to different directions, respectively, in general except when y is 0. The ! starting point of r varies depending on y and a is its distance from the machine center O. When the particle moves to Bn þ 1 by ! d r , which is depicted in a great exaggeration in Fig. 1, yn is ! ! increased by the amount dy and r n is changed to r n þ 1 . In other words, the angle between C n þ 1 A and C n þ 1 Bn þ 1 is
yn þ 1 ¼ yn þ dy:
ð2Þ !
!
The direction of d y is perpendicular to r n . From the triangle OC n Bn , we see that rn is computed as r n ¼ ½a2n þ r2n 2an rn cosðpyn Þ1=2 ¼ ða2n þ r2n þ 2an rn cos yn Þ1=2 : ð3Þ Similarly, from the triangle OC n þ 1 Bn þ 1 , r n þ 1 is given by r n þ 1 ¼ ½a2n þ 1 þ r2n þ 1 þ 2an þ 1 rn þ 1 cosðyn þ dyÞ1=2 :
ð4Þ
Expanding cosðyn þ dyÞ and using the approximations valid up to dy, cos dy 1,
sin dy dy,
ð5Þ
r n þ 1 is approximated, up to dy, as r n þ 1 ¼ ½a2n þ 1 þ r2n þ 1 þ 2an þ 1 rn þ 1 ðcos yn sin yn dyÞ1=2 : Fig. 1. A half of the F-magnet. AD is the particle orbit and 2y is the bending angle. 2 f is the angle of the F-magnet with respect to the machine center O.
occurs in the r-direction not in the r-direction. This magnet is called F-magnet. The necessary horizontal defocusing is provided by a reversed magnet that bends the beam in the negative (opposite) direction. This is called D-magnet. Obviously, Fmagnets bend the beam 3601, and a typical scaling FFAG accelerator is composed of D-F-D triplets. Both r and r vary on the particle orbit, but, if orbits are to scale during acceleration, they should not vary arbitrarily. Their relative position should remain unchanged during acceleration. Hence, r=r should be independent of the particle momentum during acceleration, if the orbits are to scale. However, this is not automatically satisfied by the above FFAG magnetic field only. A scaling FFAG should be designed in detail to keep r=r momentum-independent [2], and it is not trivial. The difficulty of dealing with FFAG is that it does not allow the analytic approach easily. The analytic form the FFAG particle orbit is still unknown and it is not easy even to prove by calculation that r=r can really be made momentum-independent, although it is proved by currently operating FFAG machines. In this paper, we compute the particle orbit in the F-magnet of a scaling radial sector FFAG in the approximation valid for a relatively small bending angle, and the variation of r and r on the orbit. These analytic results are used to derive the relation between the bending angle y and magnet angle f. It is also shown that r=r can be made independent of p, the particle momentum.
2. Computation of the orbit Fig. 1 shows the right half of a F-magnet that has the magnetic field of the form k r , ð1Þ BðrÞ ¼ B0 rm where r is the position from the machine center, rm is some reference position, and B0 is the magnetic field at the reference
ð6Þ
To obtain r n þ 1 , an þ 1 , and rn þ 1 , in the approximation valid up to dy, we use iteration. At the first stage, we set an þ 1 an and rn þ 1 rn , which makes it possible to express rn þ 1 simply in terms of dy. Then, at the next stage, we use r n þ 1 to obtain an þ 1 and rn þ 1 . Hence, using Eq. (6) in Eq. (4), we obtain first, up to dy, 1=2 2an rn sin y an rn sin y dy ¼ r n 1 dy : ð7Þ r n þ 1 ¼ r n 1 2 2 rn rn Then, the magnetic field Bn þ 1 is given by, up to dy, k rn þ 1 k an rn sin yn Bn þ 1 ¼ Bn ¼ Bn 1 d y rn r2 n kan rn sin yn ¼ Bn 1 dy : r 2n
ð8Þ
Since Br ¼ p=e where p is the particle momentum and so
y-independent, rn þ 1 is determined by Bn þ 1 as
rn þ 1 ¼ rn 1 þ
kan rn sin yn dy : 2 rn
ð9Þ
! ! Since dy points perpendicular to r , the increase of r should be compensated by the decrease of a. From the geometry of Fig. 1, the relation should be like this, an þ 1 an ¼ ðrn rn þ 1 Þcos yn : Hence, an þ 1 must be given by kr2 sin yn cos yn an þ 1 ¼ an 1 n dy : 2 rn
ð10Þ
ð11Þ
To derive r, a, and r in terms of y, note that rn þ 1 is given by kan r2n sin yn dy r 2n 2 kan1 rn1 sin yn1 kan r2n sin yn ¼ rn1 þ dy þ dy 2 r 2n r n1 ^ n X am r2m sin ym dy, ¼ r0 þk r 2m m¼0
rn þ 1 ¼ rn þ
ð12Þ
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T.-Y. Lee et al. / Nuclear Instruments and Methods in Physics Research A 707 (2013) 54–57
and similar results for an þ 1 and r n þ 1 . Hence, eventually, the three quantities are given by integrals. Z y aðyÞrðyÞ2 rðyÞ ¼ r0 þ k sin ydy, rðyÞ2 0 Z y 2 k aðyÞrðyÞ sin 2ydy, aðyÞ ¼ a0 2 0 rðyÞ2 Z y aðyÞrðyÞ sin ydy, ð13Þ rðyÞ ¼ r 0 rðyÞ 0 where the subscript 0 denotes the value at y ¼ 0. At this point, we need further iteration. First, we replace aðyÞ, rðyÞ, and rðyÞ by a0, r0 , and r0 and integrate over y. We obtain the lowest order result as
rðyÞ ¼ r0 ½1 þ kðm0 1ÞX,
k aðyÞ ¼ a0 1 ð1 þ cos yÞX , 2 rðyÞ ¼ r 0 ½1ðm0 1ÞX,
ð14Þ
where m0 ¼ r 0 =r0 and X is defined as XðyÞ ¼
1cos y
m20
:
ð15Þ
When y, the half-bending angle of a F-magnet, is far less than 1, cos y is very close to 1. Also, using the fact that m0 is greater than 1, we see that X is very small, in this approximation, XðyÞ
y2 2m20
51:
ð16Þ
Therefore, the result was and will be expanded in terms of X. The next step will be to use the results of Eq. (14) in the integration of Eq. (13) in the approximation of Eq. (16). After the integration, we find that k rðyÞ ¼ r0 1þ kðm0 1ÞX þkðm0 1Þ ðk þ 1Þðm0 1Þ ð2 þ cos yÞ X 2 : 6 ð17Þ Similarly, aðyÞ is obtained as k aðyÞ ¼ a0 1 ð1 þ cos yÞX 2 k 1 k ðk þ 1Þðm0 1Þð1 þ 2 cos yÞ ð1 þ cos yÞ2 X 2 , 2 3 8
ð18Þ
and the beam orbit rðyÞ is obtained as ðm 1Þ k ðk þ1Þðm0 1Þ ð2 þcos yÞ X 2 : rðyÞ ¼ r 0 1ðm0 1ÞX 0 2 3 ð19Þ
3. Discussion and conclusion When k¼0, the magnetic field is constant and r and a are reduced to r0 and a0, and r describes nothing but a circle of radius r0 . Fortunately, the exact formula of the circular orbit is given by r 0 ðyÞ ¼ ½r20 þa20 þ 2r0 a0 cos y1=2 ¼ r 0 r 0 ðm0 1ÞXr 0 ðm0 1Þ2
X2 þ , 2 ð20Þ
where the 2nd equality was obtained by expanding in terms of X and the three terms exactly agree with those obtained from Eq. (19) with k ¼0. When k is finite, r increases and a decreases, while r deviates from a circle. However, the deviation is only the high order terms beginning with X2,
Dr ¼ r0 ½16 ð2 þ cos yÞ12ðm0 1Þkðm0 1ÞX 2 þ ,
ð21Þ
which is very small. Hence, we see that the beam orbit in a scaling FFAG is close to a circular shape, or equivalently, a circle is a fairly
Fig. 2. Orbits expressed as a function of y. The solid line was obtained from the Zgoubi code and the dotted lines was obtained from the analytic formula of this paper. The dashed line is the circular orbit obtained when the magnetic field is constant (k ¼0). In this example, the discrepancies between the three lines are small.
good approximation to the actual orbit. Fig. 2 below compares the circular orbit with the simulated orbit. Using mðyÞ to denote rðyÞ=rðyÞ and using Eqs. (17) and (19), we obtain 1 mðyÞ ¼ m0 1ðk þ 1Þðm0 1ÞXðk þ1Þðm0 1Þ ðk þ Þðm0 1Þ 2 k ð2 þ cos yÞ X 2 : ð22Þ 6 Since the terms in the square bracket are p-independent, this result clearly shows that m is p-independent only when m0 ¼ mð0Þ is p-independent. Hence, it is m0 that should be made pindependent by careful design and study to construct a scaling FFAG accelerator. Finally, from the following relation coming from the triangle OBC,
rn sin yn ¼ rn sin fn ,
ð23Þ
we see that f and y are related by [4] sin y ¼ mðyÞ: sin f
ð24Þ
Solving this equation for y, it will be given only by f, m0 , and k. Hence, we see another reason why m0 must be independent of p in a scaling FFAG accelerator, because otherwise the bending angle y will vary depending upon the momentum, which is the violation of the scaling rule. You can immediately see from Eq. (24) that the crudest zeroth order approximation is simply
y m0 f:
ð25Þ
To find out how accurate this analytic formula of rðyÞ is, it is compared with the computer output obtained by the code Zgoubi [3]. Fig. 2 compares the formula and the Zgoubi result for the 150 MeV scaling FFAG in Japan [5]. It is composed of 12 cells and each cell bends protons 301. Each cell consists of a D-F-D triplet and the F-magnet has the angle of 2f ¼ 5:121 and has the k-value of 7.6. Eq. (24) gives y ¼ 0:397 rad (22.741). As shown in Fig. 2, the two results agree perfectly well at a low y and the discrepancy grows as y grows. However, the discrepancy is still only 0.16% at y ¼ 0:397 rad. The 3rd line in Fig. 2 is the circular
T.-Y. Lee et al. / Nuclear Instruments and Methods in Physics Research A 707 (2013) 54–57
orbit obtained when k¼ 0. In this example, it is fairly close to the two orbits. In conclusion, the particle orbit in the F-magnet of a radialsector FFAG was derived analytically and its scaling property was examined. The derived analytic formula of the orbit agrees quite well with the result of the computer code, Zgoubi. The circular orbit with a constant radius of curvature is a fairly good approximation. It was shown that r=r can be p-independent if it is at y ¼ 0. Acknowledgement Authors thank Prof. Mori and Dr. Lagrange for useful discussions and information. This work was supported by National
57
Research Foundation of Korea under Grant number 20120008353.
References [1] T. Ohkawa, Proceedings of Annual Meeting of JPS, 1953. [2] E.R. Symon, D.W. Kerst, L.W. Jones, L.J. Laslett, K.M. Terwilliger, Physical Review 103 (1956) 1837. [3] D. Garetta, F. Meot, J. C. Faivre, S. Valero, /https://oraweb.cern.ch/pls/hhh/ code_website.disp_code?code_name=ZgoubiS. [4] S. Machida, Y. Mori, R. Ueno, Proceedings of EPAC 2000, Vienna, Austria, 2000, p. 557. [5] S. Machida, et al., Proceedings of the 2003 Particle Accelerator Conference, Portland, Oregon, USA, 2003, p. 3452.