Planar Undulator Considerations

Transcription

1 LCC-0085 July 2002 Linear Collider Collaboration Teh Notes Planar Undulator Considerations John C. Sheard Stanford Linear Aelerator Center Stanford University Menlo Park, California Abstrat: This note onsists of informal working notes that doument an effort to understand the TESLA baseline, unolarized, undulator based ositron soure. This is the first ste in the design roess of an undulator based ositron system for the NLC. The exressions and methodologies develoed herein are used in subsequent memos that referene this note. In regards to the TESLA design, it is found that a 135 m long (versus 100 m length) undulator is onsistent with the erformane desritions in the TDR text. And while oeration of the TESLA system with a 250 GeV drive beam energy rovides a safety margin of a fator of 2 in e+ yield, oeration of the same system at 160 GeV results in a yield of about 0.6.

2 Planar Undulator Considerations J. C. Sheard rev. 0: 11/1/01 rev. 1: 11/19/01 rev. 2: 7/03/02 Referenes: 1. H. Wiedemann, Partile Aelerator Physis II, V.V. Balandin, et al., Conetual Design of a Positron Injetor for the TESLA Linear Collider, Mosow-Hamburg 2000, DESY-TESLA TESLA TDR II Aelerators, haters 4 and 5. Abstrat This note onsists of informal working notes that doument an effort to understand the TESLA baseline, unolarized, undulator based ositron soure. This is the first ste in the design roess of an undulator based ositron system for the NLC. The exressions and methodologies develoed herein are used in subsequent memos that referene this note. In regards to the TESLA design, it is found that a 135 m long (versus 100 m length) undulator is onsistent with the erformane desritions in the TDR text. And while oeration of the TESLA system with a 250 GeV drive beam energy rovides a safety margin of a fator of 2 in e+ yield, oeration of the same system at 160 GeV results in a yield of about 0.6. Question: How long is the undulator? Start with how many ositrons are required, N, and a multiliative safety margin, f>1: fn = N Y L Y (1) ee hυ uhυ ς where N e = number of eletrons, e Y hν = yield of eletrons to hotons er meter of undulator, L u = the length of the undulator in meters, hν Y = yield of hotons to ositrons leaving the target, and ς = the ature effiieny. Note that e Y hν, hν Y, and ς all deend in detail of the undulator arameters, the inident eletron energy, the sot size on the target, and details of the olletion and ature system. L u is extrated from (1) to give: fn L = u N Y Y (2) e e hυ hυ ς For the TESLA design, N=Ne and f =2 for E e = 250 GeV, the eletron energy. For the TESLA undulator, B 0 = 7.5 kg and λ = 1.42 m (age II-122) [3]. From Wiedemann [1], K = B ( kg) λ ( m) = (3) 0

3 The total radiated energy er eletron er meter is given as E: E ( GeV) K EeV = = ev m = MeV m (4) 2 2 e 7 ( ) / 22.3 / 2 λ ( m) The TESLA design reort says that on average 3 GeV is lost, this imlies an undulator 3GeV length of 134.8m 22.3 MeV / m =. This is at odds with the text of the reort whih indiates that the undulator is 100 m long. Similarly, the reort indiates that 135 kw of ower is generated by the wiggler. From (4) the average ower outut is given as P avg : P = qn EeV ( )/ m L = 135kW (5) avg e u 19 J Plugging in the numbers, q = , and ev MeV/m, then gives a value for L u of Ne 14 = / s and E=22.3 L u 135kW = = 135.2m qn EeV ( )/ m e (6) This is at least onsistent with the 3 GeV average loss. Next a quik hek of the utoff energy of the undulator first harmoni, E 10 : E 2 4πγ λ = hω = h (7) ( 1+ K 2) For λ = 1.42 m, E 10 = 28 MeV (27.96 MeV, atually) whih is essentially the same as in Table (II-120). So next figure out yields for this artiular undulator. The number of undulator hotons is found by equating the integral of "universal" undulator setrum, hsum(ww10)[note: see z:/ositrons/olarized ositrons/1149df.m and teslsa150_250wrks.m], to the total energy radiated to find the normalization onstant whih in turn is used with the integral of the hoton number setrum. So here goes (Matlab is used through out for the numerial results). P avg ω δω (8) e a d hsum a hsum N whih gives P a= δω avg Ne ( J / e ) hsum (9)

4 Similar thinking leads to the ratio of the number of hotons radiated to the number of eletrons, e Y hν : e Y hυ aδω Pavg( W) Ne( e / s) hsum./ ww10 = hsum./ ω = L qe ( ev) L ( m) hsum u 10 u (10) Note, a lot of the hoton energy setrum, hsum(ww10), is shown below; ww10 is the hoton energy (frequeny) normalized by the utoff energy of the first harmoni, E 10 : ω ww10 = (10a) ω 10 wherein ω / 10 = E 10 h. hsum./ ww10 Also the quantity hsum first 4 harmonis deends only on K. For K=1, and summing over the hsum./ ww10 = hsum (11) So, for the L u = 135 m, K=1, λ = 1.42 m, and E 10 = 28 MeV, Yh υ = = 1.005( hotons / meter / eletron ) (12) 135 e One an also figure out the average hoton energy, h υ E avg, hsum υ E = E hsum./ ww10 h avg 10 (13) For the ase at hand, υ E = MeV (14) h avg And the rodut, as it should. Y L E = GeV (15) e hυ u hυ avg

5 In order to determine the yield of hotons to ositrons leaving the target, hν Y?, one needs to use EGS4. A version of EGS4 user ode (urtzs.mortran, urtz.data, num_k1_120.out, ww10_k1_120.out, and Xegs4run*) has been written (with signifiant hel and guidane from R. Nelson) whih generates ositrons from an inident hoton beam of arbitrary setrum. The ode also allows one to seify an inident rms gaussian beam size in x and y. The rograms all reside in /afs/sla.stanford.edu/u/ad/js/egsruns/yieldal/htnstrm in Unix land. Figures 3-8 follow the text. Figures 3 and 4 are satter lots of the x-x hase sae of ositrons oming out of the bak surfae of the onversion target. It should be noted that the emittane of the emitted ositrons are only a fator of 2-3 time smaller than the hase sae of onventionally rodued ositrons. This is in large art due to the fat that the beam size of the emitted ositrons is large beause of the 1.4 m thikness of the 0.4 r.l. length Ti-alloy target. Figure 5 illustrates the energy setrum of emitted ositrons for the onditions noted in the figure. Figures 6-8 show the setrum and transverse rofiles of the inident hoton beam. Figures 3-8 are made from files written by EGS4. Figure 1: Universal undulator energy setrum, hsum.

6 Figure 2: The universal hoton number setrum, hsum/(ww10) At 250 GeV through the undulator, E 10 = 28 MeV, σ = σ = 0.7, 100,000 hotons x y mm into 0.4 R.L. Ti target results in 5404 emitted ositrons from the exit fae, a yield of hν Y = 5.4%. Similarly, at 160 GeV through the undulator, E 10 = MeV, σ = σ = 1.09, 100,000 hotons into 0.4 R.L. Ti target results in 1750 emitted x y mm ositrons from the exit fae, a yield of hν Y = 1.75%. The INR reort [2] states that the ature effiieny into ε x = ε y = m is ς utting this all together, I find 25%. So, Ne f( Ee = 250 GeV) = eyhυluhυyς = = 1.83 N At 160 GeV, Ne f( Ee = 160 GeV) = eyhυluhυyς = = 0.59 N The ratio

7 f(160 GeV) f(250 GeV ) = = This is similar in value to simly ounting the number of ositrons within the 5-D hase sae uts whih are shown in the emitted ositron figures. From the figures, N (160) N (250) = = = On the fae of this, it looks as if TESLA has real roblems making ositrons at 160 GeV. To get u to a ature yield of f(160 GeV ) = 1, the ature effiieny needs to inrease by a fator of 1.7 from 0.25 to Seems tough; ς 25% also feels otimisti and is quoted at a ositron energy of 274 MeV [2], not through the full 5 GeV of aeleration and transort. [Note 11/19/01: Yuri Batygin indiated that ς 25% may be doable but is very sensitive to the relative hasing of the ature rf and also of the lina hases, at the several degree level.] Time to do our own flux onentrator simulations. Also an figure out the energy ross over oint. Figure 3: x- x hase sae distribution of ositrons exiting an 0.4 r.l. thik Ti-alloy onversion target. The inident hoton setrum is that of a K=1 undulator with E = MeV and σ = σ = x y mm

8 Figure 4: x- x hase sae distribution of ositrons exiting an 0.4 r.l. thik Ti-alloy onversion target. The inident hoton setrum is that of a K=1 undulator with E10 = MeV and σ = σ = x y mm

9 Figure 5.: Emitted ositron energy setrum for K=1 undulator with E 10 = 28 MeV, 0.4 r.l. thik Ti-alloy target. Figure 6.: Inident hoton energy setrum for K=1 undulator with E = MeV

10 Figure 7.: Inident hoton beam transverse distribution, horizontal lane. Figure 7.: Inident hoton beam transverse distribution, vertial lane.