Blondel, A

Fabre, Jean-Paul

Gilardoni, S S

Vassilopoulos, N

Blondel, A.

Fabre, J.P.

Gilardoni, S.

Vassilopoulos, N.

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A thin target scheme for the muon source

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A thin target scheme for the Muon sourceAlain Blondel a , Jean Paul Fabre b , Simone Gilardoni c,Nikolaos Vassilopoulos baLPNHE, Ecole Polytechnique, 91128 Palaiseau Cedex, FrancebCern, 1211, Geneva 23, SwitzerlandcINFN Milano and Politecnico di Milano, Milano, ItalyAbstractA pion production source using a thin target and a recirculating storage ring isconsidered. The potential advantages of the scheme are: i) the thin target couldallow a better collection eciency; ii) the proton beam dump is avoided; iii) lighttargets could be used. The disadvantages are: i) the high beam current stored in therecirculator, ii) constraints on timing. This scheme could be used for the NeutrinoFactory but not for Muon Colliders.The purpose of this paper is to perform a preliminary analysis of a thin targetand recirculator scheme for the pion source of a muon machine such as aNeutrino Factory. A sketch of this is shown in Figure 1. This is similar tostudies performed for antiproton sources (1).Protons incoming from the source (i.e. a linac or a synchrotron) interact with athin target and produce pions. Most protons do not interact and are capturedTargetProton recirculator + accumulatorRFProton dump directionHornPion tracksProton source(linac)Fig. 1. Schema for the recirculating scenarioPreprint submitted to Elsevier Preprint 4 November 1999in a recirculator which sends them back onto the target for several turns.There is no need for a dump, since all the protons produced by the injectorare eventually absorbed in the target. The target can thus be thin (at limitpoint-like) and composed of low-Z materials which have the advantage of notproducing heavy radioactive nuclei. This scheme could also be considered withan H− linac, the target acting in this case as a strip foil.This scheme leads to a number of questions: i) will multiple scattering in thetarget increase the emittance so much that cooling of the protons should benecessary? ii) the energy deposited in the target is very localised and couldbecome unbearable; iii) the accumulated current in the recirculator might betoo high, iv) with the protons recirculating, the time structure of the pionbeam acquires a structure at about one microsecond. Although these implica-tions will not nd here an answer in the context of a consistent scenario of aNeutrino Factory, they are addressed in turn.1 Multiple scatteringThe proton beam is spread by multiple scattering according to a characteristicangle 0:0 =13:6(GeV )cp(GeV )sxX0(1 + log terms) (1)where X0 is the radiation length for the given material (see ref. (2)), x isthe target thickness and p the momentum of the incoming protons. This willadd up to the angular spread of the incident proton beam, initial =q"in=where " is the emittance of the beam and  the optical focusing function atthe target.After one passage in the target the proton beam has a total angular spreadwhich is the quadratic sum of these two contributions.2;out ="in+ 02 (2)and the corresponding transverse emittance is:"out = "in + 02 (3)At each turn, the fraction of protons that will interact, produce secondaryparticles and disappear from the beam is a = xXI, where x is the thickness of2the target, and XI is the interaction length for the given material. A typicalproton will thus pass through the target XIxtimes. (e.g. for a target composedby x=1 cm of Beryllium which has XI= 40 cm a proton passes on average 40times), and the average emittance will then be "out = "in + 02  XIx.A similar result is obtained if one lls the recirculator continuously. Let usnow assume that the recirculation time is matched to the timing of the protoninjector, as would be the case if the target also acted as an H− strip foil. Atevery turn a number n0 of protons is added to the bunch, with an emittanceequal to "in. The remaining fraction (1− a) of the proton beam is assumed tobe recovered by the recirculation system. After innite number of passages,the total angular divergence of the beam tends to:"equil: =n0P10 "inpkn0P10 pk+n0P10 k20pkn0P10 pk(4)where n0pk is the number of particles remaining after k turns, with pk =(1− a)k. With little algebra equation 4 becomes"equil: = "in + 201− aa(5)Playing with the denition of 20 and a, the equilibrium beam emittance can beexpressed as a function of the interaction and radiation lengths of the targetmaterial:"equil: = "in + 20XIX0(6)where 20 =13:6cp; for a 2 GeV beam (kinetic energy) 0 = 5mrad.For a given material, the emittance is a function only of the optical function at the target. One can then calculate the value crit that corresponds to anylimiting emittance. As an example we give in the table below the values ofcrit for which "equil: = 2 "in for the linac design presented in ref. (5) wherethe beam emittance is given to be "in = 0:6 m.Mat. XI(g=cm2) X0(g=cm2) XIX0crit(cm)D2 54 186 0.43 5.6H2O 85 36 2.36 1Be 75 65 1.15 2.1Pb( Hg) 194 6.4 30.3 0.083The best is clearly to chose a material with a short interaction length (XI)and a long radiation length (X0). The target thickness could then as long asa few centimetres and remain shorter than crit.2 Energy lossAnother constraint for the target choice is given by the energy deposition inthe target which is, as a rst approximation, due to ionization and can becomputed using dEdxfor the given material. The energy deposited by secondaryparticles is presumably very small, since only a small fraction of the protonsinteract, and the target is too thin for showers to develop. For one proton, theenergy loss at each passage is given by xdEdx. The recirculator must thereforebe equipped with a modest amount of re-accelerating devices (typically a fewMeV).Since the average total path that every proton eventually undergoes in the tar-get is one interaction length, the total energy loss is given by Ploss =dEdxXI foreach proton. This is calculated in the following table and should be comparedwith the proton energy. For heavy nuclei such as lead the loss is around 10%of the incoming power while for light materials this is kept around 5%. Thisis clearly smaller than for a total absorption target, where a larger fraction(20-30%) of the incident power is absorbed in the target itself.Mat. dEdx(MeV )XI % of incoming beam energy for 2 GeV protonsD2 112 5.5%H2O 169 8.5%Be 119 6%Pb 217 11%From this point of view again, low-Z materials show a non-negligible advantageover heavy materials. Moreover, the above power loss is computed for a protonbeam energy of 2 GeV but, if one considered higher energies (8 GeV to 32GeV), for a given total beam power, the power loss in the target would beaccordingly reduced.43 IntensitySince each proton will remain on average in the recirculator for XIxturns, thebeam intensity will be correspondingly increased. We give below the typicalbeam intensity to be considered, starting from 6  106 protons (i.e. 10 mA),for a target length equal to crit.Mat. (g=cm3) XI(cm) Iequil: = I0 XI=crit (A)D2 0.17 325 0.58H2O 1 85 0.85Be 1.85 40.5 0.19Pb 11.4 16.8 2.1Here beryllium which oers high density and low Z at the same time is theclear winner of the materials considered.4 Particle CaptureIt could be an advantage to handle a point-like source. In particular a point-like source can be placed exactly in the focal point of an optical capture devicesuch as a magnetic horn. A study in progress (3) to simulate a horn for theNeutrino Factory taking experience from the NGS studies (4) has alreadyreached a performance similar to that obtained with a solenoid capture canbe envisaged.5 Time StructureThe favoured operation mode for the Neutrino Factory that emerged fromNUFACT99 in Lyon would be pulsed operation with a repetition rate of 10-50 Hz. Pulsed devices being operational for several hundredth of microseconds,and the recirculation time being of less or equal to a microsecond (recirculatorring of less than 300 meters circumference), there is plenty of time for theprotons to recirculate from one or several initial bunches of protons injectedin the system. If one starts from one bunch or train of bunches of protons,the recirculator generates then a succession of bunches (or trains) with a timestructure equal to the recirculation time and with exponentially decreasingintensity Ik = Iinitiale−k:x=XI .5Provided that the muon storage ring does not have a length that coincideswith the recirculating time, it should be possible to store the successive muonbunches in the muon decay ring, provided an RF injection scheme is consid-ered. This recirculator scheme could therefore be acceptable and oer someadvantages for the Neutrino Factory. For Muon Colliders, evidently, the dilu-tion of intensity over so many bunches would result in unacceptable loss ofluminosity.6 ConclusionsThe possibility of recirculating the protons in a storage ring coupled with athin target had been investigated. A thin target composed by low-Z materiallike D2; H2O or Be seems the best choice. The advantages of having a capturesystem like a magnetic horn, in particular for a thin target scheme, are stillunder investigation. The timing of the system seems at rst look acceptablefor a Neutrino Factory, but not for a Muon Collider.References[1] N. Mokhov and A. Van Ginneken, Increasing anti-proton yields via re-circulating beam targeting, FERMILAB-FN-621 (1994).[2] Particle Data Group, Euro. Phys. Journal C3 (1998) 76[3] A. Ball, A. Blondel, S. Gilardoni, N. Vassilopoulos, Preliminary mag-netic horn study for the collection scheme of a Neutrino Factory, toappear in the proceedings of NUFACT99, Lyon 1999.[4] The CERN Neutrino beam to Gran Sasso (NGS), Addendum to re-port CERN 98-02, INFN/AE-98/05, CERN- SL/99-034(DI), INFN/AE-99/05[5] M. Vretenar in Prospective study of Muon Storage Ring at Cern, B.Autin, A. Blondel and J. Ellis eds, CERN 99-02 (1999) 8.6

A thin target scheme for the muon source

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