The Effects of a Finite Beam Pipe on Resistive Wall Instabilities CBETA Tech Note 046

The Effects of a Finite Beam Pipe on Resistive Wall

Instabilities

J. Christie, Oxford Universityjonathan.christie@trinity.ox.ac.uk

August 6, 2019

Abstract

The effect of resistive wall instabilities on beam break-up is investigated for beam pipes with finite and infinitethickness walls. Analytic solutions for the electric field along the beam axis are determined for both finite andinfinite thickness walls using complex analysis. The longitudinal and transverse impedances are also determined;variations from the infinite wall case are observed when the wall thickness is of the order of the skin depth of thebeam pipe material.

1 Introduction

Resistive wall instabilities occur when charged particlestravel through a beam pipe made out of a metal with afinite conductivity, and can lead to beam break-up in lin-ear accelerators if not accounted for due to its long-rangenature [1].

Consider a beam of evenly spaced bunches of chargedparticles (called macroparticles) that are slightly offsetfrom the central axis of the beam pipe. The beam prop-agates through an infinitely long linear accelerator at ap-proximately the speed of light. As it does so, the elec-tromagnetic field of the first macroparticle propagates tothe beam pipe walls.

If the walls were perfectly conducting, the electromag-netic field would terminate at the surface of the beampipe wall and there would be no instability.

However, since the walls have some large but fi-nite conductivity, the electromagnetic field from themacroparticle penetrates slightly into the beam pipe wallitself. This leads to a residual electromagnetic field beingpresent within the walls themselves in order to satisfy thenecessary boundary conditions.

As a result, the next macroparticle in the beam has itstrajectory affected by the residual electromagnetic field,whilst also contributing to the total residual electromag-netic field, resulting in larger and larger deviations inmacroparticle trajectories and, eventually, beam break-up.

This instability can be demonstrated using a simplemodel [2]. A beam of nm ≈ 106 macroparticles sepa-rated by z ≈ 1 mm, each containing nb ≈ 109 electrons,initially propagates near the centre of a perfectly con-ducting, infinitely long pipe with infinitely thick walls.The beam has an initial offset xj ≈ 1 µm from the cen-

tral axis of the beam pipe. The beam then encountersa section of length L ≈ 1000 m with finite conductivityσ ≈ 2× 106 Ω−1m−1. The resulting net angular kick tomacroparticle i after passing through the resistive sectionin SI units is thus given by

∆x′i =nbe

2

mec2γ

1

nm

nm∑j=1

W1(zj − zi)xj , (1)

where γ ≈ 100 is the gamma factor of the 50 MeV elec-trons, e is the electronic charge, me is the rest mass ofthe electron, and c is the speed of light. W1(zj − zi),the transverse wake field, describes how the beam piperesponds to the dipole moment of the particle beam. Atlong distances behind the bunch, the transverse wakefield in SI units is given by

W1(z) ≈ 2L

πb3

√c

4πε0σ

1

|z| 12, (2)

where ε0 is the permittivity of free space, and b ≈ 5 cmis the radius of the beam pipe.

Figure 1: Net transverse angular kick received by eachbunch

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The Effects of a Finite Beam Pipe on Resistive Wall Instabilities CBETA Tech Note 046

e cc

r

s

b

z

t

a

Figure 2: Beam pipe layout [1]

Because the macroparticles are moving at approxi-mately the speed of light, causality requires that the wakefield generated by a given macroparticle is zero in frontof that particle. As a result, a given macroparticle i canonly be affected by the wake generated by the macropar-ticles in front of it.

Using this form of the long-range wake field, it isrelatively straightforward to simulate the above modeland determine how the net angular kick received byeach macroparticle varies with macroparticle number, asshown in Figure 1.

From the plot, it is clear that, at large bunch num-bers, the net angular kick received by each macroparticleis proportional to the square root of the macroparticlenumber, and hence proportional to the square root of dis-tance from the first macroparticle. As a result, such in-stabilities continue to grow as more macroparticles pass,until beam break-up occurs.

However, when deriving the form ofW1 in Equation 2,it was assumed that the pipe walls were infinitely thick.Whilst this is usually a good approximation for short-range wake effects, it is not necessarily good for long-range effects, especially if the pipe walls have a thicknesson the order of the skin depth of the metal at frequenciesof interest.

Assuming that the pipe walls have some finite thick-ness t, different boundary conditions apply. This leadsto different expressions for the wake field and its Fouriertransform, known as the impedance.

This note investigates the effects that wall thick-ness has on the long-range wake fields and on the low-frequency impedances, potentially leading to a feasiblesystem with reduced resistive wall instabilities and hencea more stable beam.

2 Resistive Wall Wake Fields

2.1 Solving Maxwell’s Equations

The first step towards finding the electromagnetic fieldsand wake fields is to solve Maxwell’s equations in cylin-drical coordinates. Given that the initial distribution ofthe charge creating the wake field is unknown, we canwrite charge density ρ and current density ~j in terms oftheir multipole moments,

ρ =

∞∑m=0

ρm ,

~j =

∞∑m=0

~jm ,

ρm =Imδ(s− ct)δ(r − a) cos (mθ)

πam+1(1 + δm0),

~jm = cρms ,

(3)

where m is the multipole moment present, Im is the mthmoment of the beam, and δm0 = 1 if m = 0, 0 otherwise.

Figure 2 gives the layout of the beam pipe; an in-finitesimally thin ring charge with charge density ρ, cur-rent density~j, radius a, and angular dependence cos (mθ)propagates along the s-axis of the pipe at the speed oflight. The beam pipe has conductivity σ ≈ 1017 s−1,radius b ≈ 5 cm, and thickness t ≈ 1 mm. The electro-magnetic field experienced by a test charge at a distancez behind the ring charge in CGS units is thus given bythe equations

1

r

∂(rEr)

∂r+

1

r

∂Eθ∂θ

+∂Es∂s

= 4πρ ,

1

r

∂Bs∂θ− ∂Bθ

∂s− 1

c

∂Er∂t

=4π

cjr ,

2

The Effects of a Finite Beam Pipe on Resistive Wall Instabilities CBETA Tech Note 046

∂Br∂s− ∂Bs

∂r− 1

c

∂Eθ∂t

=4π

cjθ ,

1

r

∂(rBθ)

∂r− 1

r

∂Br∂θ− 1

c

∂Es∂t

=4π

cjs ,

1

r

∂(rBr)

∂r+

1

r

∂Bθ∂θ

+∂Bs∂s

= 0 ,

1

r

∂Es∂θ− ∂Eθ

∂s+

1

c

∂Br∂t

= 0 ,

∂Er∂s− ∂Es

∂r+

1

c

∂Bθ∂t

= 0 ,

1

r

∂(rEθ)

∂r− 1

r

∂Er∂θ

+1

c

∂Bs∂t

= 0 .

(4)

In order to make the equations easier to solve, wewrite the electromagnetic field components in terms oftheir Fourier transforms,

(Es, Er, Bθ) = cos (mθ)

∫ ∞−∞

dk

2πeikz(Es, Er, Bθ),

(Eθ, Bs, Br) = sin (mθ)

∫ ∞−∞

dk

2πeikz(Eθ, Bs, Br),

(5)

where the Fourier transformed components are functionsof k and r and are complex quantities.

Solving for the case of the monopole moment m = 0,the field components (Eθ, Bs, Br) = 0. The Fouriertransformed components can then be solved for the re-gion inside the pipe r < b,

Es = A(k),

Er = Bθ =

−ikA r

2 , r < a,

−ikA r2 + 2q

r , a < r < b.

(6)

In order to determine the form of A(k), we need tosolve for the region inside the pipe walls r > b. Insidethe metal walls,

ρ = 0,

~j = σ ~E,(7)

where the conductivity σ is assumed to be independentof k.

We now consider the cases where the pipe walls areinfinitely thick, and where the pipe walls have some finitethickness t.

2.2 Infinite Pipe Wall Thickness

If the pipe walls are infinitely thick, we can solve for thefields inside the pipe walls using equations (4) and (7),as well as the conditions that (Es, Bθ) are continuous atr = b, and that (Es, Bθ) → 0 as r → ∞. This gives thesolution

A(k) =2q

b

(ikb

2− λ

k

)−1,

λ(k) = i

√−i4πσk

c,

(8)

where a term − kλ has been dropped since k

λ is much

smaller than λk or ikb

2 .The field component Es can now be determined by

performing the inverse Fourier transform on A(k),

Es =

∫ ∞−∞

dk

2πeikzA(k) . (9)

In order to calculate this integral, we need to find thelocation of the poles and branch cuts of A(k) [3]. It canbe shown that the poles of A(k) are located at

kp = α23 (±√

3− i) , (10)

where α is a constant given by

α =1

b

√2πσ

c≈ 106 m−

32 , (11)

and that A(k) has a branch cut along the negative imag-inary axis from the origin to infinity.

Figure 3: Plot of the integration contour and of the polesand branch cuts of A(k)

This allows us to calculate Es in the infinite wall case,

Es = −16q

b2

(1

3eu cos

(√3u)−√

2

π

∫ ∞0

x2eux2

x6 + 8dx

),

u = zα23 ≈ 104z ,

(12)using values of σ ≈ 1017 s−1 and b ≈ 5 cm. The fullderivation of the location of the poles and branch cuts ofA(k) and of Es can be found in Appendix A.

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The Effects of a Finite Beam Pipe on Resistive Wall Instabilities CBETA Tech Note 046

From Figure 4, we can see that Es is proportional toz

32 at long ranges in the infinite wall case. We shall now

investigate the long-range behaviour of Es in the finitewall case.

Figure 4: Long-range Es for infinite pipe walls

2.3 Finite Pipe Wall Thickness

If the pipe walls have finite thickness t instead, theboundary conditions are similar to those in Section 2.2,

except that (Es, Bθ) = 0 at r = b + t, instead of atr →∞. This leads to a different expression for A(k),

A(k) =2q

b

(ikb

2− λ

k

1 + e2iλt

1− e2iλt

)−1,

=2q

b2α23

(ik

k0+k0ik

√− ikk0

coth

(√− ikk1

))−1,

(13)

where k0 and k1 are constants given by

k0 = 2α23 = 2

(2πσ

cb2

) 13

≈ 2× 104 m−1 ,

k1 =1

2(bαt)2=

1

2t2c

2πσ≈ 2× 10−4 m−1 ,

(14)

using a wall thickness of t ≈ 1 mm and the same valuesof σ and b as in the infinite wall case.

(a) k1 → 0 (b) k1 = 0.001

(c) k1 = 0.025(d) k1 = 0.1

Figure 5: 3D contour plots of poles of A(k) for different values of k1

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The Effects of a Finite Beam Pipe on Resistive Wall Instabilities CBETA Tech Note 046

3D contour plots of A(k) for different values of k1 (af-ter setting k0 = 1m−1) are included in Figures 5(a) to(d).

Figure 5(a) shows that the infinite wall case (k1 → 0)has only 2 poles in the negative imaginary half-plane aswell as a branch cut along the negative imaginary axis.

As k1 increases to 0.001 in Figure 5(b), poles startto form along the negative imaginary axis. These polesspread further apart as k1 increases further, in Figures5(c) and (d).

From Equation 14, k1 ≈ 10−8 k0, which implies thata very large number of poles along the negative imagi-nary axis are very closely spaced together, similar to thek1 = 0.001 case.

In order to determine an analytic form of Es for thefinite wall case, we need to find the poles and branch cutsof A(k). To determine if branch cuts are present, we canexpand the coth term in Equation 13 about zero,

ik

k0+k0ik

√− ikk0

coth

(√− ikk1

)

≈ ik

k0+k0ik

√− ikk0

((√−k1ik

)+

1

3

(√− ikk1

)− ...

).

(15)Since there are no fractional powers of k present in thedenominator, A(k) does not have a branch cut in the fi-nite wall case. This is fundamentally different from theinfinite wall case, where a branch cut extends along thenegative imaginary axis. As a result, we expect to see adifferent analytic form for Es in the finite thickness wallcase.

The locations of the poles along the negative imagi-nary axis are determined by making the substitution

k = −ix2k0 (16)

into A(k), where x is a real number, so that A(x) is givenby

A(x) =2q

b2α23

(x2 +

1

xcot (px)

)−1, (17)

where p is a constant given by

p =

√k0k1

. (18)

Using the values of k0 and k1 from Equation 14,p ≈ 104 >> 1. Poles of A(x) occur when the denom-inator is equal to 0, corresponding to solutions of theequation

x3 + cot (px) = 0 ,

=⇒ x3 = − cot (px) ,

=⇒ x−3 = − tan (px) .

(19)

0 1 2 3 4x

0

0.5

1

1.5

2

y

Figure 6: Plot of y = x3 + cot (px) for p = 10

In the limit where x→ 0,

cot (px) ≈ 0 , =⇒ x ≈(n− 1

2 )π

p(n integer) , (20)

while in the opposite limit x→∞,

tan (px) ≈ 0 , =⇒ x ≈ nπ

p(n integer) . (21)

Using these limits, we can say that the poles of A(x) arelocated at

xpn =(n+ f(n))π

p, (22)

where f(n) varies from f(n) = − 12 at n = 1 to f(n) = 0

as n→∞.

In terms of k = −ix2k0,

kpn = −i(

(n+ f(n))π

p

)2

k0 . (23)

This makes finding the analytic form of Es much sim-pler; instead of needing to integrate along a branch cutas in the infinite wall case, we can use Cauchy’s Residuetheorem at each pole of A(k) to determine the complexintegral.

We can write A(x) as

A(x) =2qx

b2α23 (x3 + cot (px))

=g(x)

h(x), (24)

where g(x) and h(x) are

g(x) =2qx

b2α23

,

h(x) = x3 + cot (px) .

(25)

Thus, the residue of A(k) at k = kpn is

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The Effects of a Finite Beam Pipe on Resistive Wall Instabilities CBETA Tech Note 046

Res(A(k), kpn) =g(kpn)dhdk |k=kpn

,

=g(kpn)

dhdx |x=xpn

dxdk |x=xpn

,

=g(xpn)

h′(xpn)

dk

dx

∣∣∣x=xpn

.

(26)

Evaluating the expressions for g(x) and h(x) at x =xpn gives

Res(A(k), kpn) =2qxpn(−2ik0xpn)

b2α23 (3x2pn − p csc2 (pxpn))

,

=−4ik0qx

2pn

b2α23 (3x2pn − p(1 + cot2 (pxpn)))

,

=−4i(2α

23 )qx2pn

b2α23 (3x2pn − p(1 + x6pn))

,

=−8iqx2pn

b2(3x2pn − p(1 + x6pn)).

(27)

Using Cauchy’s Residue theorem the contribution to Esfrom the poles along the negative imaginary axis is

Es = −2πi

∞∑n=1

Res(A(k), kpn) ,

=8q

b2

∞∑n=1

x2pnezk0x

2pn

p(1 + x6pn)− 3x2pn.

(28)

Thus, including the contribution from the other two polesin the negative imaginary half-plane, the total electricfield along the s-axis is given by

Es ≈−16q

b2

(1

3eu cos

(√3u)− ...

− 1

2

8q

b2

∞∑n=1

x2pnezk0x

2pn

p(1 + x6pn)− 3x2pn

).

(29)

Whilst the first term is the same as in the infinite wallcase due to the poles at kp = α

23 (±√

3 − i) not moving,the branch cut integral has now been replaced by an in-finite sum of the residues of A(k) at every pole along theimaginary axis.

This makes it difficult to determine the long-rangebehaviour of Es in the finite wall case, since we needto include every term in the sum to obtain an accurateanalytic form of Es.

For the case of p = 1×104 (typical of most real-worldexamples), finding an approximate form of the sum is

difficult since the poles are not evenly spaced. As a re-sult, it is not possible to approximate the entire sum asan integral using the Midpoint Rule.

However, we can consider the case where p is a muchsmaller value, such as p = 10. Figure 6 shows that thepoles of A(x) become approximately evenly spaced afterthe first p poles, so we can approximate the sum by writ-ing out the explicit form of the residue of the first fewpoles, then approximating the rest of the sum with theMidpoint Rule.

The first 10 solutions of x3 + cot (px) = 0 for p = 10are located at

xp ≈ 0.1575, 0.4824, 0.8387, 1.2047, 1.5443,

1.8698, 2.1896, 2.5069, 2.8230, 3.1384...(30)

Then, the 11th pole and onwards are located at

xp11 ≈11π

p,

xp12 ≈12π

p,

xp13 ≈13π

p,

(31)

with a spacing

∆x ≈ π

p. (32)

Suppose we had intervals of width ∆x = πp starting from

x = 21π2p ,

23π2p , ..., which give midpoints of x = 11π

p , 12πp , ...and so forth. We can write the sum as

∞∑n=1

x2pnezk0x

2pn

p(1 + x6pn)− 3x2pn≈

10∑n=1

x2pnezk0x

2pn

p(1 + x6pn)− 3x2pn

+

∫ ∞21π2p

px2pezk0x

2p

π(p(1 + x6p)− 3x2p)dxp .

(33)

The electric field is then given by

Es ≈−16q

b2

(1

3eu cos

(√3u)− ...

− 1

2

10∑n=1

x2pnezk0x

2pn

p(1 + x6pn)− 3x2pn− ...

− 1

2

∫ ∞21π2p

px2pe2ux2

p

π(p(1 + x6p)− 3x2p)dxp

).

(34)

6

The Effects of a Finite Beam Pipe on Resistive Wall Instabilities CBETA Tech Note 046

Figure 7: Long-range electric field in both the infiniteand finite wall case for p = 10

Figure 8: Long-range electric field in both the infiniteand finite wall case for p = 30

Figure 7 shows that, in the finite thickness wall case,Es transitions from obeying a z

32 power law relation to

exponentially decaying as z becomes more negative. Thisis different from the infinite wall case, where it obeys az

32 power law relation for all zα

23 < −1, which suggests

that finite thickness walls reduce long-range wake effectsexperienced by charged particles, leading to greater beamstability.

Following the same procedure for p = 30, Figure 8shows that the point where Es transitions from obeyinga power law to exponential decay occurs at a more nega-tive z value than for p = 10. This suggests that Es wouldbegin to exponentially decay at large values of z for thecase where p = 104, meaning that the long-range wakeeffects are significantly reduced compared to that of theinfinite wall thickness case.

However, more work is required to determine thepoint where this transition occurs for large values of p,and to determine the accuracy of the integral approxi-mation to the sum.

3 Impedance

Given that it is difficult to determine the long-range be-haviour of the wake fields in terms of position, we caninstead examine the Fourier transform of the wake fields,also known as the impedance, in order to determine theirfrequency response.

The longitudinal impedance Z‖m(ω) and the trans-

verse impedance Z⊥m(ω) are related to A(k) by

Z‖m(ω)

L=ω

c

Z⊥m(ω)

L= − 1

ImcAm(k) , (35)

where ω = ck, Am(k) is the k-dependent coefficient of Escaused by the mth moment of the charge, and L is thelength of the beam pipe section that contains the wakefield.

A0(k) is given by Equations 8 and 13,

A0(k) =

2qb

(ikb2 −

λk

)−1, infinite wall,

2qb

(ikb2 −

λk

1+e2iλt

1−e2iλt

)−1, finite wall .

(36)

Solving Maxwell’s equations for the m ≥ 1 moments sim-ilarly gives expressions for Am(k) when m ≥ 1,

Am(k) =

4Imb2m+1

(ikbm+1 −

λk −

imkb

)−1,

4Imb2m+1

(ikbm+1 −

λk

1+e2iλt

1−e2iλt −imkb

)−1,

(37)

where the upper expression is for the infinite wall caseand the lower expression is for the finite wall case.

In this example, we focus on the m = 0 longitudinalimpedance and the m = 1 transverse impedance sincethey are the dominant terms for resistive wall effects.Solving for the relevant impedances gives

Z‖0 (ω)

L=

2bc

(λcω −

iωb2c

)−1,

2bc

(λcω

1+e2iλt

1−e2iλt −iωb2c

)−1,

(38)

Z⊥1 (ω)

L=

4ωb3

(λcω + ic

ωb −iωb2c

)−1,

4ωb3

(λcω

1+e2iλt

1−e2iλt + icωb −

iωb2c

)−1,

(39)

where the upper expressions are for the infinite wall case,and the lower expressions are for the finite wall case.

3.1 Skin Depth

In order to get an idea of the frequency range at whichlong-range wake effects start to dominate, consider thefrequency and skin depth to which the first pole of A(k)corresponds.

The skin depth δs of a material is a measure of howfar electromagnetic waves penetrate into the material be-fore they are attenuated. δs is related to the quantity λfrom Equation 8 by

δs =1

Im(λ)=

c√2πσ|ω|

, (40)

which shows that materials have a larger skin depth forlower frequency waves.

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The Effects of a Finite Beam Pipe on Resistive Wall Instabilities CBETA Tech Note 046

For δs << t, only a very small difference betweenthe infinite and finite wall wake field is expected as theelectromagnetic waves will be almost entirely attenuatedby the time they have travelled a distance t through thewall.

However, when δs ≈ t, a large proportion of the ini-tial electromagnetic wave will permeate through the fi-nite pipe wall, potentially leading to decreased long-rangewake effects.

Given a wall thickness of t ≈ 1 mm, conductivityσ ≈ 1017s−1, and pipe radius b ≈ 5 cm as in Section 2,the frequency at which δs ≈ t is given by

ω ≈ 1.4× 105 rad s−1 . (41)

Thus, if the first pole of A(k) corresponds to a frequencyof order 105 Hz, we expect the finite thickness of the pipewalls to have an effect on the long-range wake field.

From Equation 20, the first pole of A(k) occurs at

x ≈ π

2p, (42)

where p = 1×104 as is the case in most real-world situa-tions. Using k = −ix2k0 from Equation 16 and |ω| = c|k|gives

ω ≈ 1.5× 105 rad s−1 . (43)

This corresponds almost perfectly to the frequencyat which skin depth effects become important, suggestingthat wall thickness could have an effect on the long-rangebehaviour of the wake fields.

3.2 Longitudinal Impedance

Now that the frequency at which long-range wake effectsstart to dominate has been calculated, we can examinehow wall thickness affects the longitudinal and transverseimpedances.

Equation 38 shows that the low frequency behaviour

of Z‖0 for a pipe with infinitely thick walls is proportional

to |ω| 12 ,

Z‖0 (ω)

L≈ 1

bc

√|ω|2πσ

(1− i) . (44)

Figures 9(a) to (d) shows that the low-frequencyimpedance obeys different power laws for the finite andinfinite wall cases.

Figures 9(c) and 9(d) show that, at low frequencies,the finite wall impedance has a real and imaginary partproportional to |ω|2 and −|ω| respectively, in comparison

to the infinite wall behaviour of |ω| 12 and −|ω| 12 .As the frequency increases beyond 105 Hz, the fi-

nite wall impedance tends towards the infinite wallimpedance.

(a) Frequency response of the real part of thelongitudinal impedance

(b) Frequency response of the imaginary part of thelongitudinal impedance

(c) Plot of the real part of the longitudinal impedance (d) Plot of the imaginary part of the longitudinal impedance

Figure 9: Plots of the real and imaginary parts of Z‖0

8

The Effects of a Finite Beam Pipe on Resistive Wall Instabilities CBETA Tech Note 046

(a) Frequency response of the real part of thetransverse impedance

(b) Frequency response of the imaginary part of thetransverse impedance

(c) Plot of the real part of the transverse impedance (d) Plot of the imaginary part of the transverse impedance

Figure 10: Plots of the real and imaginary parts of Z⊥1

This corresponds to the skin depth becoming muchsmaller than the wall thickness, so that the finite thick-ness pipe wall is essentially infinitely thick to electromag-netic radiation.

It can also be seen that, for frequencies ω . 104 Hz,the total impedance is smaller in the finite wall case, lead-ing to smaller long-range wake effects than in infinite pipewalls.

3.3 Transverse Impedance

In the case of the transverse impedance, Equation 39 canbe used to show that the low frequency behaviour of Z⊥1for a pipe with infinitely thick walls is proportional to|ω|− 1

2 ,

Z⊥1 (ω)

L≈ 1

b3

√2

πσ|ω|(1− i) . (45)

Figures 10(a) to (d) show that the finite wall caseobeys different power laws to the infinite wall case at lowfrequencies, and that the finite wall case tends towardsthe infinite wall case as frequency increases, similar tothe longitudinal impedance.

However, in this case, Figures 10(c) and 10(d) showthat the finite wall impedance has a real part propor-tional to |ω| and a constant imaginary part, in compar-

ison to the infinite wall behaviour of |ω|− 12 and −|ω|− 1

2

for the real and imaginary parts respectively.

Once again, the total impedance is smaller in the fi-nite wall case for frequencies ω . 104 Hz. Though furtherwork is needed to obtain the relevant wake fields throughinverse Fourier transforming the impedance expressions,we can see that the long-range, low-frequency wake is re-duced when a beam pipe with a finite thickness of about1 mm is used.

4 Conclusion

Using a beam pipe with finite thickness walls resultsin the long-range wake fields created by the monopolecharge distribution decaying exponentially, as opposedto following a |z| 32 power law as in the infinite wall case.A similar procedure to that used in Section 2.1 could beused to determine the other field components for highercharge multipole moments.

For the m = 0 component of Es, the Fourier trans-form of Es in the finite wall case has no branch cut alongthe negative imaginary axis like in the infinite wall case,but instead has multiple poles along the negative imag-inary axis. This allows Cauchy’s Residue theorem to beused to obtain an analytic expression for Es. Whilstthis expression can be approximated for small values ofp, further work is needed to determine a suitable wayof approximating the infinite sum term for the large pvalues that typically occur in most linear accelerators.

The low-frequency longitudinal and transverseimpedances of a beam pipe with finite thickness walls

9

The Effects of a Finite Beam Pipe on Resistive Wall Instabilities CBETA Tech Note 046

also obey different power laws than for a beam pipe withinfinitely thick walls. For frequencies less than about104 Hz, both impedances for the finite wall case aresmaller than for the infinite wall case, suggesting thatthe long-range wake field from the impedances is also re-duced. This could be further investigated by examiningthe long-range behaviour of the wake field expressionsthat are obtained through Inverse Fourier transformingthe impedance.

References

[1] A. W. Chao, Physics of collective beam instabilitiesin high energy accelerators. New York, NY: Wiley,1993.

[2] S. Peggs and V. Mane, “KRAKEN, a numericalmodel of RHIC impedances,” in Proceedings ParticleAccelerator Conference, pp. 3137–3139, May 1995.

[3] K. L. F. Bane and M. L. Sands, “The Short-RangeResistive Wall Wakefields,” June 1995.

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The Effects of a Finite Beam Pipe on Resistive Wall Instabilities CBETA Tech Note 046

A Solving Es in the Infinite WallThickness Case

A.1 Finding Poles and Branch Cuts ofA(k)

Inside the pipe region (r < b, where b is the radius of theinner region of the pipe),

Es = A , (46)

where

A =2q

b

(ikb

2− λ

k

)−1,

λ = i

√−i4πσk

c.

(47)

The function A has poles when its denominator iszero,

ikb

2− λ

k= 0 ,

orikb

2− i

k

√−i4πσk

c= 0 ,

=⇒ ik = − 1

ik

2

b

√−i4πσk

c,

=⇒ (−ik)32 = −2

b

√4πσ

c,

= −2√

2α ,

(48)

where α = 1b

√2πσc . Solving for (−ik),

(−ik)3 = 8α2 ,

= 8α2(e0, e−i2π, ei2π) .

=⇒ (−ik) = 2α23 (e0, e−i

2π3 , ei

2π3 ) .

(49)

However, for (−ik) = 2α23 ,

(−ik)32 + 2

√2α = 2

√2α+ 2

√2α 6= 0 . (50)

For (−ik) = 2α23 e±i

2π3 ,

(−ik)32 + 2

√2α = 2

√2αe±iπ + 2

√2α = 0 . (51)

Thus, the two poles of A are located at

(−ik) = 2α23 ei

2π3 ,

=⇒ k = 2α23 e−i

5π6 ,

= α23 (−√

3− i) ,

(52)

and

(−ik) = 2α23 e−i

2π3 .

=⇒ k = 2α23 e−i

13π6 ,

= α23 (√

3− i) ,

(53)

About the origin, replacing (−ik) with reiθ

ikb

2+

1

ik

√−i4πσk

c=−breiθ

2− 1

re−iθ

√4πσ

creiθ ,

=−breiθ

2− 1√

re−i

θ2

√4πσ

c,

=

br2 + b

√2αi√r, θ = π ,

br2 −

b√2αi√r, θ = −π .

(54)

This implies that the values of A at θ = ±π are different,which means that the origin is a branch point and thebranch cut is along the negative imaginary axis.

A.2 Solving the Integral

The electric field Es(z) is obtained by performing theinverse Fourier transform of A(k),

Es =

∫ ∞−∞

dk

2πeikz

2q

b(ikb2 + 1

ik

√4πσc (−ik)

) . (55)

Let (−ik) = s. The Laurent expansion of(ikb2 −

ik

√−i 4πσkc

)−1about s = 2α

23 ei

2π3 = s+,

1

ikb2 + 1

ik

√−i 4πσkc

=−1

sb2 + 1

s

√4πσsc

,

=−1

s−12

(b2s

32 +√

2bα) ,

=−s 1

2

b2 (s+ + (s− s+))

32 +√

2bα,

≈ s12

√2bα(

1 + 3(s−s+)2s+

+ ...− 1) ,

≈ 2s12 s+

3√

2bα(−ik − s+),

=2s

12 s+i

3√

2bα(k − s+i).

(56)

Thus, the residue at k = α23 (−√

3− i) (i.e. at the point

s = s+ = 2α23 ei

2π3 ) is given by

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The Effects of a Finite Beam Pipe on Resistive Wall Instabilities CBETA Tech Note 046

Res

(k = α

23 (−√

3− i))

=2is

32+

3√

2bα,

=2i(−2

√2α)

3√

2bα,

=−4i

3b.

(57)

Similarly, the Laurent expansion about the point s =

2α23 ei

−2π3 = s− is

1

ikb2 + 1

ik

√−i 4πσkc

=2s

12 s−i

3√

2bα(k − s−i).

=⇒ Res

(k = α

23 (√

3− i))

=2is

32−

3√

2bα,

=−4i

3b.

(58)

Using Cauchy’s residue theorem (since the integrationcontour is clockwise, the integral around one of the polesis (−2πi) times the residue of that pole),

Es = −2πi

(q

bπ

(−4i

3b(ezα

23 (1−

√3) + ezα

23 (1+

√3))))

+ [Branch cut integral] ,

= −16q

3b2eu cos

(√3u)

+ [Branch cut integral] ,

(59)

where u = zα23 .

Transforming coordinates from k to r using (−ik) =re±iπ,

k = ire±iπ

=⇒ dk = ie±iπdr .(60)

Performing the branch cut integral and substituting k,

I =

∫ 0

−∞

ieiπ

2πe−zre

iπ 2q

b(b2

(r + 2

√2√rαi))dr

+

∫ ∞0

ie−iπdr

2πe−zre

−iπ 2q

b(b2

(r − 2

√2√rαi))dr ,

= − 2q

b2π

(−∫ ∞0

iezrr − 2

√2√rαi

r2 + 8α2

r

dr

+

∫ ∞0

iezrr + 2

√2√rαi

r2 + 8α2

r

dr

),

= −8√

2q

b2π

(∫ ∞0

ezrαi2

r52 + 8α2√

r

dr

).

(61)

Using the substitution r = x2α23 =⇒ dr = 2xα

23 dx,

I =8√

2q

b2π

(∫ ∞0

ex2zα

23 α

x5α53 + 8α2

xα13

2xα23 dx

),

=16√

2q

b2π

(∫ ∞0

x2eux2

x6 + 8dx

),

(62)

where u = zα23 as before. Thus, Es is given by

Es = −16q

b2

(1

3eu cos

(√3u)−√

2

π

∫ ∞0

x2eux2

x6 + 8dx

).

(63)

Equation 63 gives the behaviour of Es at both small andlarge |z|.

12