We derive a general formula expressing the resistive wall impedance in the ultrarelativistic limit for a beam pipe of arbitrary cross section through the 'normal derivative' of its electric capacitance. An application to the case of rectangular cross secrion yields a closed form of expression of the corresponding longitudinal impedance in terms of elliptic integrals

Ruggiero, F

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EUROPEAN ORGANIZATION FOR NUCLEAR RESEARCHCERN - SL DivisionCERN-SL-95-95 (AP)Resistive Wall Impedanceas Derivative of the Electric Capacitancefor a Beam Pipe of Arbitrary Cross SectionFrancesco RuggieroABSTRACTWe derive a general formula expressing the resistive wall impedance in theultrarelativistic limit for a beam pipe of arbitrary cross section through the`normal derivative' of its electric capacitance. An application to the case ofrectangular cross section yields a closed form expression of the correspondinglongitudinal impedance in terms of elliptic integrals.Geneva, Switzerland7 November 19951 Introduction and summaryThe electromagnetic eld associated with an ultrarelativistic bunch of charged particlestravelling in a perfectly conducting pipe of arbitrary, but constant, cross section can bedetermined by solving a two-dimensional electrostatic problem. Specically, the electriceld E =  r?' is derivable from a scalar potential ' satisfying Poisson's equation in thetransverse plane, with equipotential boundary conditions at the metallic beam pipe. Indeed,thanks to the translation invariance of the pipe cross section, the electromagnetic eld isobtained by a Lorentz transformation of the purely electrostatic eld in the rest frame of thebunch. In the extreme relativistic limit, the bunch becomes innitely long in its rest frameand the electrostatic potential at a given point is only determined by the `slice' of beamcharge in the corresponding transverse plane.For a pipe of nite resistivity, the electromagnetic eld is no longer purely transverse andthe Fourier transform of the longitudinal electric eld on the beam axis, responsible for theparasitic loss, is associated with the longitudinal resistive wall impedance ZL. In the caseof a thick pipe with uniform resistivity , one can treat the eect of resistivity as a smallperturbation and assume that the transverse elds can be approximated by those obtainedfor a perfectly conducting pipe. Each Fourier component of the longitudinal wall current Iz,equal to the tangential magnetic eld Htand owing in the direction opposite to the beamcurrent Io, is therefore proportional to r?' at the metallic boundary. On the other hand,using Ohm's law, the longitudinal electric eld can be approximately written as Ez= ZwIz,where the wall surface impedance Zw= (1  i)= depends on the skin depth . The powerZLjIoj2lost by the beam is equal to the outgoing ux of the complex Poynting vectorE Hacross the pipe wall and the longitudinal impedance per unit length ZL=L can therefore beexpressed as a contour integral of EzIz= ZwjIzj2over the pipe perimeter [1]ZLL=ZwjIoj2IjHtj2d` = Zw"o2joj2Ijr?'j2d`: (1)Here "ois the permittivity of free space and o= Io=c the linear charge density of the beam.For a pencil beam located at r?= rbin the transverse plane, the electrostatic potential 'satises the two-dimensional Poisson equationr2?' =  o"o(r?  rb):Therefore the ratio  = '=ois real and depends only on the pipe geometry and on thebeam position. A similar expression holds also for the transverse (dipole) impedance ZT,provided the electrostatic problem is solved using a dipole source term [1].In the next section we show that the contour integral of (r?)2, required to computethe resistive wall impedance, is proportional to the `normal derivative' of the electrostaticenergy stored in the region between the beam and the surrounding pipe (see Fig. 1). Thiselectrostatic energy can be expressed through the specic capacitance C = C=L of the systembeam{pipe and, denoting by n the innitesimal outward displacement along the normal tothe pipe surface, we arrive at the following expression for the longitudinal resistive wallimpedance:ZLL= Zwn"oC: (2)2Note that the specic capacitance C has the same dimensions as the permittivity of freespace (i.e., farad/metre), so that the ratio "o=C is dimensionless and its normal derivativetimes the wall surface impedance yields an impedance per unit length1.In practice, to nd the resistive wall impedance one has rst to solve the two-dimensionalelectrostatic problem for a uniform beam with unit linear charge density and to compute thecorresponding electric potential dierence  = 1=C from the beam (assumed of innites-imal, but constant radius) to the pipe (assumed equipotential). The calculation is thenrepeated for a pipe of slightly larger dimensions, each surface element being displaced bya constant amount n along the normal to the surface: this yields the `normal derivative'of the potential dierence, proportional to that of the pipe electric capacitance. With thehelp of this prescription, one avoids the contour integral of (r?)2and performs insteada simple derivative. Moreover, the numerical estimate of the capacitance (and of its nor-mal derivative) for complicated beam pipe cross sections can be improved using variationaltechniques.In section 3, we rst apply our result to the known case of a beam in a circular pipe andthen to the more complicated case of a centred beam in a rectangular pipe, for which only aseries expansion of the impedance is available [1]. Using Eq. (2), we arrive at a closed formexpression of the longitudinal impedance in terms of elliptic integrals.a)QoSoSSnb)Q SoSnFigure 1: (a) Two closed equipotential curves Soand S, the former enclosing the beamcharge Q and the latter representing the pipe cross section with unit normal vector n.(b) Normal variation of the beam pipe geometry: the vector n = n n has constant norm n.2 Normal variation of the electrostatic energyWe consider the electrostatic potential  corresponding to a uniform beam with unit linearcharge density travelling in a perfectly conducting pipe of arbitrary cross section. The charge1The dimensionless ratio "o=C is also equal to the ratio L=obetween the specic inductance L (of thetransmission line consisting of the beam and of its surrounding pipe) and the permeability oof free space [2].However, it is more natural to express the electrostatic energy in terms of capacitance.3density induced on the inner surface of the beam pipe is1LdQd`=  "oEand the electrostatic force on a surface element of length L and tangential extent d` isdF =12EdQ;where the factor 1=2 is due to the fact that the electric eld vanishes inside the conductor.Since E =  r? =  nr?, where n denotes the unit vector normal to the surface andoriented in the outward direction (see Fig. 1), we obtaindFd`=  12r?dQd`=  "oL2(r?)2n:The work required to modify the cross section of the beam pipe by an innitesimalamount n in the outward direction is2U =  Id`dFd` n ="oL2Id` (r?)2n:From the principle of energy conservation, the corresponding variation of the electrostaticenergy U for a uniform normal variation n is thus proportional to the contour integral ofthe square of the electric eld over the pipe perimeterUn="oL2Id` (r?)2:For a beam of unit linear charge density Q=L = 1, the specic electrostatic energyU = U=L stored in the region between the beam and the metallic pipe can be expressed interms of the electric capacitance per unit length C = C=L as follows:U = UL =12Q2C! U =12C:ThereforeId` (r?)2=1"on1C(3)and, recalling that  = '=ois real, from Eq. (1) we obtain expression (2) for the resistivewall impedance. Since the electrostatic energy is also given by U = Q=2, the inversespecic capacitance 1=C equals the electric potential dierence  between the beam andthe surrounding pipe.In this context, the eect of resistivity can be interpreted as a longitudinal friction force,proportional to the normal electrostatic force F on the pipe surface: the friction coecientdepends on the frequency ! through the skin depth  =q2=(!o), appearing in the wallsurface impedance Zw.2This work does not include the contribution of the force on the opposite charge induced on the outersurface of the beam pipe: such contribution vanishes for an innitely thick or for a grounded pipe.43 ExamplesAs a simple application of our result Eq. (2), we rst compute the longitudinal impedanceof a circular pipe with a centred beam and then consider a transverse beam oset. Finallywe discuss the more complicated case of a centred beam in a rectangular pipe.3.1 Circular pipeThe electrostatic potential  of a uniform pencil beam with unit linear charge density,travelling at the centre of a circular pipe of radius b, is(r) =  12"oln(r) for   r  b;where r is the radial distance from the pipe axis and  denotes the (innitesimal) beamradius. The potential dierence  from the beam to the pipe is therefore =1C= ()  (b) =12"oln b!and, since the normal derivative for a circular pipe corresponds to an ordinary derivativewith respect to the radius b, from Eq. (2) we immediately obtain the well known resultZLL= Zw@@b"oC=Zw2b: (4)We now consider the case of a beam oset a < b in the horizontal direction x and writethe potential (x; y) in rectangular coordinates, using an opposite image charge at x = b2=ato satisfy the equipotential boundary condition at x2+ y2= b2:(x; y) =  14"oln"(x  a)2+ y2(x  b2=a)2+ y2#:The potential dierence  from the beam, of innitesimal radius , to the pipe is =1C= (a+ ; 0)  (b; 0) =14"o8<:ln"ab2#  ln24 a+   b2=a!2359=;and we obtain@@b"oC=  12(1b+2b=aa+   b2=a):In the limit ! 0, Eq. (2) yieldsZLL=Zw2bb2+ a2b2  a2; (5)in agreement with the known result (see Ref. [3], Exercise 2.31 on p. 118). Taking the limita; b ! 1, with constant distance b   a = d from the beam to the pipe, the longitudinalimpedance per unit length becomes ZL=L = Zw=(2d): therefore the parasitic loss is thesame for a beam travelling in the centre of a circular pipe of radius b or parallel to an innitemetallic plane of equal resistivity at a distance d = b.53.2 Rectangular pipeWe now consider the case of a rectangular pipe with sides a and b and write the electrostaticpotential (z), using the complex notation z = x + iy, for a uniform pencil beam of unitcharge density, travelling at the centre zo= (a+ ib)=2 of the rectangle, as [4](z) =12"oRe(ln"sn2(Kz=a; k)  sn2(Kzo=a; k)sn2(Kz=a; k)  sn2(Kzo=a; k)#): (6)Here K = K(k) is the complete elliptic integral of the rst kind with modulus k, whilesn(u; k) denotes the Jacobian elliptic sine-amplitude, again with the same modulus k: thelatter depends on the ratio a=b between the sides of the rectangle and is implicitly denedbyK=K0= a=b; (7)where K0= K(p1   k2). The potential (6) is obtained by the conformal transformationw = sn2(Kz=a; k), which maps the inside of the rectangle in the z-plane into the upper halfof the w-plane. The electrostatic problem is then solved by adding an opposite image chargein the lower w-plane, such that the real w-axis be at zero potential. Therefore (z) vanisheson the pipe boundary and the potential dierence  between the beam, of innitesimalradius , and the rectangular pipe is given by the limit of (zo+ ) for ! 0. As shown inthe Appendix, neglecting the divergent self-potential proportional to ln(), this limit is =1C=14"oln abKK0!: (8)The innitesimal variation of "o=C can then be written"oC=14 aa+bb (KK0)KK0!and for a normal variation n of the rectangular pipe cross section we must requirea = b = 2n:Thereforen"oC=14 2a+2b 1KK0(KK0)n!: (9)In the following, we need the following identities [5]:EK0+ E0K  KK0=2; Legendre identity (10)dKdk=1kE1   k2 K!dK0dk=  k1  k2 E0k2 K0!; (11)where E = E(k) is the complete elliptic integral of the second kind and E0= E(p1   k2).6Using Eq. (7), we can now establish a relation between n and the variation k of theelliptic modulus. Indeedab=ba  abb2=b  ab22n = KK0:Since, from Eqs. (10) and (11),KK0=2K02kk(1   k2);it follows thatn =4K02b2b  akk(1  k2)and, using Eq. (7), Eq. (9) can be writtenn"oC=12"1a+1b 21a 1bk(1  k2)(KK0)k#:In the case of a square pipe of side 2d, i.e., for a = b = 2d, the term proportional to1=a   1=b vanishes and substituting this formula into Eq. (2) yields ZL=L = Zw=(2d). Wethus nd that, for a centred beam and for a given wall resistivity, a square pipe has thesame longitudinal impedance of the inscribed circular pipe. In the general case, using againEqs. (10) and (11), the derivative of KK0with respect to the modulus k can be expressedin closed form as follows(KK0)k=EK0  E0K + (2k2  1)KK0k(1   k2)and the longitudinal impedance per unit length becomesZLL=Zw21a+1b 21a 1bEK0  E0K + (2k2  1)KK0: (12)This formula is symmetric in a and b, since for an exchange of the sides of the rectangle k2istrasformed into 1 k2, K into K0and E into E0. The corresponding resistive loss, normalizedto that of the inscribed circular pipe, is plotted in Fig. 2: the loss is the same when the sidesof the rectangle are equal (square pipe) or when one of them becomes innite (two parallelplates). Indeed, for a ! 1, k = 1 and K is divergent, while K0, E and E0stay nite [5];therefore, the Legendre identity Eq. (10) implies that E0K = KK0and EK0= =2, so thatthe square braket equals 2=b, i.e., the inverse of the radius of the inscribed pipe. When theratio between the sides of the rectangle is around 1:33, the resistive loss reaches a minimumvalue about 6% lower than that corresponding to a square pipe.4 DiscussionThe result presented in this paper can be considered as a special example of the variationalformulation discussed in Ref. [6], where the tune shift due to a gradient perturbation in a71 2 3 4 50.940.950.960.970.980.991PrectPcirca=bFigure 2: Parasitic loss for a centred beam in a rectangular pipe, normalized to that for theinscribed circular pipe, as a function of the ratio a=b between the sides of the rectangle.circular accelerator was obtained by a rst order variation of a suitable action integral withrespect to the gradient perturbation. The variation with respect to the betatron functionvanishes by virtue of the corresponding Euler equation and the stationary value of the actionintegral coincides with the tune of the accelerator. In the present paper, we consider thevolume integral of the square of the electrostatic eld over the charge-free region between thebeam and its surrounding pipe. The stationary value of this action integral, for an equipo-tential boundary, corresponds to the stored electrostatic energy. For a normal variationof the boundary, the corresponding variation of the electrostatic energy is proportional tothe resistive wall impedance, while the variation due to the change of the electric potentialvanishes by virtue of the Laplace equation. It should be noted, however, that in generalthe variation of the potential is not zero on the original boundary (as one usually assumeswhen deriving the corresponding Euler equation). Nevertheless, the potential perturbationis harmonic and preserves the ux of the electric eld across the boundary; this is enough toprove that the associated rst order variation of the action integral vanishes. It is remarkablethat such complicated derivation is not necessary, if one makes use of the principle of energyconservation.We would like to stress that our starting equation (1) is only valid for relatively highfrequencies, typically above a few MHz, corresponding to skin depths much smaller thanboth the pipe thickness and its local radius of curvature. On the other hand, the perturbativetreatment of the wall resistivity3requires that the frequency be not too high: for example!=2  1012Hz for a cylindrical aluminum pipe, with 5 cm radius, at room temperature (see3The elegant derivation of Eq. (1) presented in Ref. [1] and based on the Lorentz reciprocity theorem maygive the impression of an exact result, with the only approximation introduced by the so-called Leontovichboundary condition, relating the longitudinal electric eld to the tangential magnetic eld via the wallsurface impedance Zw. However, the gradients appearing in the Poynting theorem and required to convertthe longitudinal impedance into a surface integral can be considered as transverse gradients only for aperfectly conducting pipe, while this is only approximately true for a pipe with wall losses. Such additionalapproximation is implicitly used in Ref. [1].8Ref. [3], p. 73). The range of validity of these approximations and thus of our result Eq. (2) isusually wide enough to yield accurate estimates of the parasitic loss in particle accelerators.These considerations can be extended to the case of anomalous skin eect (when the surfaceimpedance Zwhas a dierent dependence on frequency [7]) and, to some extent, also to thecase of non-uniform resistivity along the pipe perimeter. In the latter case, however, evenfor an innitely thick pipe the perturbative treatment of the wall resistivity breaks downat low frequency, when the induced currents tend to redistribute themselves among regionswith dierent resistivity following the path of least dissipation.References[1] R.L. Gluckstern, J. van Zeijts and B. Zotter, Coupling impedance of beam pipes ofgeneral cross section, Phys. Rev. E, 47, 1993.[2] J.D. Jackson, Classical Electrodynamics, (Wiley, New York, 1975), p. 262.[3] A.W. Chao, Physics of Collective Beam Instabilities in High Energy Accelerators, (Wi-ley, New York, 1993).[4] P.M. Morse and H. Feshbach, Methods of theoretical physics, (McGraw{Hill, New York,1953), Part II, p. 1252.[5] I.S. Gradshteyn and I.M. Ryzhik, Table of Integrals, Series, and Products, (AcademicPress, New York, 1980), p. 907 and p. 914.[6] R.L. Gluckstern and F. Ruggiero, Variational formulation for linear optics in a periodicfocussing system, Part. Accel. 39, 125{136 (1992).[7] W. Chou and F. Ruggiero, Anomalous Skin Eect and Resistive Wall Heating, CERN,LHC Project Note 2 (1995).AppendixIn this appendix we shall prove that, neglecting a divergent term proportional to ln(),the limit for  ! 0 of the potential (zo+ ), dened by Eq. (6), is given by Eq. (8).Indeed, replacing z = zo= (a+ ib)=2 in the numerator of expression (6) and expanding thedenominator to rst order in , we can write(zo+ ) =12"oRe8>><>>:ln26664sn2(u; k)  sn2(u; k)@sn2(u; k)@uKa377759>>=>>;;whereu =Kazo=K + iK02as a consequence of Eq. (7). Therefore(zo+ ) =12"o26664lnaK  ln() + ln0BBB@sn2(u; k)  sn2(u; k)@sn2(u; k)@u1CCCA377759and, since the last term in square brakets can be shown to be zero, neglecting the termproportional to ln() (which does not contribute to the normal derivative of ) yields =12"olnaK=14"oln abKK0!;where we have used again Eq. (7) to obtain an expression symmetric in a and b. This resultcoincides with Eq. (8).To prove that the last term in square brakets is zero, we start from the identities [5]@sn(u; k)@u= cn(u; k) dn(u; k) =q1  sn2(u; k)q1   k2sn2(u; k);sn(u; k) =p1 + k + ip1   kp2k:Here cn(u; k) and dn(u; k) denote the Jacobian elliptic cosine-amplitude and delta amplitude,respectively, and the rst of these identities is generally true, while the second holds as aconsequence of our choice of a centred beam, corresponding to the special value of u =(K + iK0)=2. Thensn2(u; k)  sn2(u; k) =2ikp1  k2;@sn2(u; k)@u=  2ikp1   k2k + ip1  k2and, since the elliptic modulus k ranges from 0 and 1, the ratio of these two complexquantities has a norm equal to unity: therefore its logarithm is zero.10

Resistive wall impedance as derivative of the electric capacitance for a beam pipe of arbitrary cross section

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