Resistive wall impedance and tune shift for a chamber with a finite thickness

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1 PHYSICAL REVIEW E Resistive wall impedance and tune shift for a chamber with a finite thickness Yoshihiro Shobuda and Kaoru Yokoya KEK High Energy Accelerator Research Organization - Oho Tsukuba Ibaraki Japan Received 6 January 00; revised manuscript received 0 May 00; published 6 November 00 Since the resistive wall impedance for a beam pipe of a nonround cross section depends on the coordinates of a witness particle the witness particle receives an incoherent tune shift. When the expression for the impedance of an infinitely thick chamber is applied to the calculation of this tune shift it becomes infinite. We have derived the resistive wall impedance for a chamber with a finite thickness and calculated the tune shift. There is no ambiguity in this expression for the tune shift because it is automatically finite. DOI: 0.03/PhysRevE PACS numbers: 9.0.Dh 4.85.Ja 4.85.Lc I. INTRODUCTION Resistive wall impedances have been studied in accelerator physics. Gluckstern van Zeijts and Zotter GZZ derived formulas of impedances for the case of elliptical and rectangular cross section when the position of the source particle and the witness particle are the same and the thickness of the beam pipe is infinite. Yokoya gave a computer algorithm to calculate the impedances for a general cross section when the thickness of the beam pipe is infinite. It is known that a beam pipe of nonround cross section causes an incoherent tune shift because the resistive wall impedance due to the source particle depends not only on the coordinates of the source particle but also on those of the witness particle. However this tune shift becomes infinite when the impedances for a chamber with an infinite thickness are applied to its calculation because the resistive wall wake function is proportional to /s where s is the distance between the source particle and the witness particle. In previous studies the tune shift was calculated by introducing an artificial cutoff 34. This kind of cutoff causes an ambiguity in the calculation of the tune shift. It is possible to avoid this kind of infinity by considering the thickness of the beam pipe. The electromagnetic fields leak out when the thickness of the beam pipe is finite. Since the skin depth of the beam pipe material is proportional to (/k) where k is a wave number this leak actually occurs on a large time scale. For this case an important parameter is the thickness of the beam pipe instead of the skin depth. It is important to consider the effect of the thickness of the beam pipe and to understand how the divergence of the tune shift can be resolved. In Sec. II we show how we include the effect of the thickness of the chamber to the resistive impedances. In Sec. III we explicitly show the impedances and the tune shift. In Sec. IV we apply our theory to KEKB electron ring and compare the experimental data with our theory. A summary is given in Sec. V. II. DERIVATION OF IMPEDANCES FOR A CHAMBER WITH A FINITE THICKNESS We assume throughout that the beam pipe is uniform longitudinally and the beam is ultrarelativistic. Since all of the field quantities which are effective on the wake field are proportional to exp jk(ctz) where c is the velocity of light we omit this factor to define the fields. We denote by E z the longitudinal component of the electric field generated by a source particle. Once the field on the inner surface of the pipe is known E z can be calculated by the Kirchhoff integral formula 5 E z r E z rgrre z r Grr nds where is the integral along the boundary surface the prime denotes quantities at the boundary n is an outwardly directed normal to the boundary surface and is the twodimensional gradient. The function G is the Green function satisfying Grrrr. Since the Kirchhoff integral is valid for any Green function satisfying Eq. we choose the function with its value equal to zero at the boundary. By using this Green function we can calculate the longitudinal impedance when we only know E z at the boundary. When E z is obtained the transverse force F is calculated using the Panofsky-Wenzel theorem 6 F xy jk E z xy. Thus it is necessary to know E z at the inner surface of the chamber for the case of a finite chamber thickness. The tangential azimuthal magnetic component (H t ) is calculated by Z 0 H t E n xy xyz 0 rr where E n is the normal component of the electric field Z 0 is the impedance of free space and r is the transverse coordinate of the source particle. In order to obtain E z at the boundary we must find the relation between E z and H t along the boundary when the thickness of the chamber is finite. We consider the case when the radius of the beam pipe is sufficiently larger than the skin depth. In order to find the relation between E z and H t at the boundary it is sufficient to X/00/665/056506/$ The American Physical Society

2 Y. SHOBUDA AND K. YOKOYA PHYSICAL REVIEW E which give the following relations: E z 0kZ 0 j GH t0 3 G e jd/ jd/ e e jd/ jd/ e 4 in normal applications where k 0. For d Eq. 4 reproduces the relation I. FIG.. A wall with its thickness d. The beam runs in region consider the one-dimensional problem. Consider a wall with thickness d see Fig.. We call (x0) region I (0x d) region II and (dx) region III. We assume that the beam runs in region I. The beam creates fields on the inner surface of the wall (x0) which are written as E z (0) and H t (0). In region II Maxwell equations are written as follows: jkh t x jkc 0 E n x E n x 0 jk x 0 E z x0 E zx jke x n xjkc 0 H t x H t x jkc x 0 E z x. According to Eq. 6 E z (x) must satisfy E z x x j E zx0 where 0 /k and 0 / 0 c ( m for a copper chamber at room temperature. The solutions for region II are H t x j E z xc e jx/ c e jx/ The boundary conditions are as follows: jk 0 jkz 0 c e jx/ c e jx/. 9 E z 0c c 0c e jd/ c e jd/ H t 0 j 0 jk 0 jkz 0 c c E z 0kZ 0 j H t0 5 which is already known. We expect that Eq. 3 which is obtained in the one-dimensional case is almost satisfied for the general beam-pipe case when the radius of the beam pipe is sufficiently larger than the skin depth but the thickness of the chamber d may be small. In order to confirm that this expectation is plausible we can consider a simple example for the two-dimension case where the cross section of the pipe is round the radius of the pipe is b 0 and the thickness of the chamber is d). This problem was already solved by Chao 6. The boundary conditions also give Eq. 3 when b 0 and k 0. These situations suggest that Eq. 3 is almost satisfied for the general beam-pipe case when the skin depth is sufficiently smaller than the radius of the beam pipe. III. WAKE FUNCTIONS AND TUNE SHIFT According to Eqs. and 3 we find that E z for a chamber with a finite thickness is that with an infinite thickness multiplied by the G factor because H t does not depend on the properties of wall materials. We obtain E z xykj e jd/ jd/ e Z 0 e jd/ e jd/ b 3 D x x xd y y yd xy x y b D 0 j e jd/ jd/ e Z 0 F xy j e jd/ e jd/ b D 3 D D xx D y y D xy y x 6 7 where b is a typical vertical size of the cross section of the chamber; the analytic form of the D s is given in the Appendix for elliptic and rectangular cases. (D is relevant to the tune shift. D x D y and D xy are equal to W x /x s W y /y s and W x /x w in Fig. 8 of Ref. respectively. We should notice that D xy vanishes when the cross section of the chamber is round

3 RESISTIVE WALL IMPEDANCE AND TUNE SHIFT FOR... PHYSICAL REVIEW E condition b i.e. sb / 0 ). Here we divide W (s) into the part proportional to the coordinate of the source particle (x or y ) and the part proportional to the coordinate of the witness particle (x or y) and write W s W xsx W sx. W y sy W sy 3 W xy (s) cause a coherent tune shift and W (s) an incoherent tune shift. Since the equations of motion are FIG.. The s dependence of the transverse wake function W where W 0 cz 0 0 /(b 3 d)d. corresponds to Eq. and ---toeq.. Here we take the inverse Fourier transformation of F to obtain the transverse wake function W per unit length W s dke jks c D j j Z 0 b 3 e jd/ jd/ e e. jd/ jd/ e 8 The integrand in Eq. 8 does not have a cut but simple poles at kk n j 0 n 4d n By closing the contour by the upper semicircle for s0 W s cz 0 b D j 3 n0 Thus we obtain Residue kk n. 0 W s cz 0 0 b 3 d n0 e ( 0 s/d )(n/) D. Since the summation over n can be replaced by integration for the case 0 s/d i.e. d) we can reproduce W s cz 0 0 b 3 s D. d x n s ds d y n s ds K x sx n s mk E W x nmknct x m sklw nm knct x n s K y sy n s mk E W y nmknct y m sklw nm knct y n s 4 5 where K xy are focusing forces C is the charge per bunch E is the electron or positron energy ev ct is the distance between bunches L is the circumference of the ring and N is the total number of bunches. The coherent tune shift () ( cohxy ) for th mode and the incoherent tune shift ( incxy ) are given by () cohxy L xy 4 k sin () xy () sin xy N L xy 4 n0 Z 0 I 0 E E W xykle jk () xy e j () xy (/N) ct 0 b 3 D xy k 0 e ( 0 kl/d )(n/) () jk xy d Comparing the asymptotic region of s in both Eqs. and we find that Eq. deviates from /s in the region sd /( 0 ) and the wake function becomes exponentially damped see Fig.. This is consistent with our other sin () xy sin () e j () xy (/N) xy N 6

4 Y. SHOBUDA AND K. YOKOYA PHYSICAL REVIEW E incxy L xy 4 k E W kct L xy Z 0 I 0 ct 0 b 3 E d D xy n0 e ( 0 ct/d )(n/) 7 where () xy xy xy are tunes I 0 A is the total current xy are average values of functions around the ring and the upper lower sign corresponds to incx ( incy ). Since () xy 0 L/d and 0 ct/d in normal applications Eqs. 6 and 7 become () cohxy L xy Z 0 I 0 L 0 4 E b 3 L D xy k k cos () xy k () k j sin xy incxy D xy L xy 4 Z 0 I 0 E b 3 d 8 9 where we have replaced the summation over n by an integral in Eq. 6 and used n0 /(n/) / in Eq. 7. Here we should notice that the coherent tune shift is automatically finite even in the limit d and does not depend on d even if we use Eq. 6. On the other hand the incoherent tune shift is infinite for d. Thus we usually calculate the incoherent tune shift introducing a cutoff (N )in the number of bunches when we use Eq.. The tune shift is then written as incxy D xy L xy 4 Z 0 I 0 E N b 3 j 0ct. j 30 We can evaluate the effective cutoff. According to Eqs. 9 and 30 we obtain ctn d 4 0 ct max. 3 Equation 3 can be interpreted in terms of the skin depth i.e. one should get k min from dk min 3 and define ct max /k min. IV. APPLICATION Let us compare experimental results at KEKB electron ring with our theory. In KEKB parameters are given by x y xy.0 m L306 m E8 0 9 ev d60 3 m a50 3 m and b5 0 3 m(a and b are defined in the Appendix. In KEKB Ieiri measured the tune shift and got d x / meas 0.06/A and d y / meas 0.037/A 7. Recently a more precise measurement was done 8. The tune-shift dependence was fitted by x meas.340i b I y meas 7.76I b I where I b A is the bunch current. Following Zimmermann 4 let us separate the tune shift into the coherent part and the incoherent part as follows: d x meas d y meas 0.5 d cohr d incx d cohr d incy. 36 Here d cohr / is the coherent tune shift for a round chamber. The numerical factors come from those for the resistive wake Fig. 8 of Ref. for (ab)/(ab)0.35). We expect a similar ratio with some uncertainties even if the relevant wake is not of resistive wall type. From the data we obtain d cohr /A d incxy /A. 37 Using our theory T max.9 ms. According to Eqs. 8 and 9 the tune shift is given by d cohx /A d cohy /A d incxy /A. 38 We cannot explain the coherent tune shift by a resistive wake. We should consider the other sources of impedances. For the incoherent tune shift our value is about four times smaller than the data Eq. 37. This discrepancy is significant even if we take into account the uncertainty of the coherent part. In our case the effect of the radiation damping is negligible because the damping time 46 ms is much larger than T max. The reason of the discrepancy may be other sources of impedances the effects of materials outside the chamber. Since the observation showed that the tune shift is proportional to the total current rather than to the bunch current the responsible wake has to be a long range wake but does not seem to be of narrow resonance type. Further the nonround cross section must be related to this wake because the sign of the tune shift is opposite for x and y. We

5 RESISTIVE WALL IMPEDANCE AND TUNE SHIFT FOR... think that case is impossible therefore we should consider the possibility of because the fields leak out of the chamber on a large time scale. The next step will be to consider the effects of these components outside the chamber. V. SUMMARY Resistive wall impedances with a finite thickness can be obtained by multiplying the G factor by impedances with infinite thickness under the approximation that the size of the chamber is sufficiently larger than a skin depth. This transverse wake function becomes exponentially damped for s d / 0 rather than falling like /s. The expression for an incoherent tune shift is automatically finite by using this impedance. Thus there is no ambiguity to evaluate the tune shift. The reason why the divergence of the tune shift is resolved is that the physically important parameter is the thickness of the beam pipe instead of the skin depth for a PHYSICAL REVIEW E long time scale. By reducing the thickness of the chamber we can reduce the tune shift linearly. In this paper we also try to explain the incoherent tune shift which is observed in KEKB. Our value is four times smaller than the experimental data. This situation suggests that we have to consider the effect of components that exist outside the chamber: for example magnets. The next step will be to consider the effects of these components to the tune shift. ACKNOWLEDGMENT The authors thank Dr. T. Ieiri Dr. R. Nagaoka Dr. F. Zimmermann Dr. K. Oide Dr. T. Nakamura and Dr. K. Sato for discussions. One of the authors Y.S. gives thanks for the financial support of Japan Society of the Promotion of Science. APPENDIX: D FUNCTIONS Here we denote the size of a cross section of a chamber as follows. The upper value holds for the case where the cross section is elliptic its major axis is a and minor axis is b; the lower value holds for the case where the cross section is rectangular its boundaries are given by xa and yb: sinh u 0 dv 0 D 0 G0u0 F 0 b a modd 0 v sinh u 0 sin v cosh ma b a b for elliptic case for cosh mb a rectangular case sinh 3 u 0 dv 4 0 D x Gxu0 F x b a 3 8 x v sinh u 0 sin v m modd sinh ma b sinh 3 u 0 dv 4 0 D y Gyu0 F y b a 3 8 y v sinh u 0 sin v b modd 3 m a 3 sinh ma b for elliptic case b meven 3 m a 3 cosh mb a for elliptic case m meven cosh mb a for for rectangular case rectangular case 0 v xy v dv 4 0 sinh D xy sinh3u0 u 0 sin v m 8 modd 3 cosh ma b for elliptic case b 3 m a 3 cosh mb for rectangular case a A where

6 Y. SHOBUDA AND K. YOKOYA PHYSICAL REVIEW E cos mv 0 v m m cosh mu 0 cosmv x v m m m coshmu 0 sinmv y v m m m sinhmu 0 xy v8 m m m cos mv cosh mu 0 a f cosh u 0 b f sinhu 0 for elliptic case. A Functions G 0xy (u 0 ) F 0xy () and 0xy (v) are calculated by GZZ. R. L. Gluckstern J. van Zeijts and B. Zotter Phys. Rev. E K. Yokoya Part. Accel S. Heifets Report No. SLAC-PUB unpublished. 4 F. Zimmermann Commissioning report 00 unpublished. 5 See e.g. J. K. Jackson Classical Electrodynamics Wiley New York 974 p A. W. Chao Physics of Collective Beam Instabilities in High Energy Accelerators Wiley New York T. Ieiri et al. inproceeding of the Second Asian Particle Accelerator Conference Beijing edited by S. Fang et al. IHEP Beijing 00. p T. Ieiri private communication