Table Of ContentBeam Coupling Impedances of Small Discontinuities
∗
Sergey S. Kurennoy
0
0 SNS Division, Los Alamos National Laboratory,
0 Los Alamos, NM 87545, USA
2
n
a
J
Abstract
7
2 Ageneralderivationofthebeamcouplingimpedancesproducedbysmalldiscontinuitiesonthewallof
thevacuumchamberofanacceleratorisreviewed. Acollectionofanalyticalformulasfortheimpedances
] of small obstacles is presented.
h
p
-
1 Introduction
c
c
a
A common tendency in design of modern accelerators is to minimize beam-chamber coupling impedances
.
s to avoid beam instabilities and reduce heating. Even contributions from tiny discontinuities like pumping
c
i holes have to be accounted for because of their large number. Numerical time-domain methods are rather
s
involvedand time-consuming for small obstacles,especially for those protruding inside the beam pipe. This
y
h makes analytical methods for calculating the impedances of small discontinuities very important.
p A discontinuity on the wall of a beam pipe — a hole, cavity, post, mask, etc. — is considered to be
[ small when its typical size is small compared to the size of the chamber cross section and the wavelength.
1 Analytical calculations of the coupling impedances of small obstacles are based on the Bethe theory of
v diffractionofelectromagneticwavesbyasmallholeinametalplane [1]. The method’sbasicideais thatthe
5 hole in the frequency rangewhere the wavelengthis largecomparedto the typical hole size, can be replaced
6
bytwoinduceddipoles,anelectricandamagneticone. Thedipolemagnitudesareproportionaltothebeam
0
fields at the hole location, with the coefficients called polarizabilities [2]. The fields diffracted by a hole into
1
0 the vacuum chamber can be found as those radiated by these effective electric and magnetic dipoles, and
0 integrating the fields gives us the coupling impedances. Following this path, the impedances produced by a
0 small hole on the round pipe were calculated first in [3].
/
s Since essentially the same idea works for any small obstacle, the method can be extended for arbitrary
c small discontinuities on the pipe with an arbitrary-shaped cross section. Analytical expressions for the
i
s couplingimpedances ofvarioussmalldiscontinuitiesonthe wallofacylindricalbeampipe withanarbitrary
y
single-connected cross section have been obtained in Refs. [3]-[9].
h
All the dependence on the discontinuity shape enters only through its polarizabilities. Therefore, the
p
: problem of calculating the impedance contribution from a given small discontinuity was reduced to finding
v
its electric and magnetic polarizabilities, which can be done by solving proper electro- or magnetostatic
i
X problem. Useful analytical results have been obtained for various axisymmetric obstacles (cavities and
r irises)[10],aswellasforholesandslots: circular[1]andelliptic [2]holeinazero-thicknesswall,circular[11]
a
and elliptic [12] hole in a thick wall, various slots (some results are compiled in [13]), and for a ring-shaped
cut [14]. The impedances of various protrusions (post, mask, etc) have been calculated in [15].
This text is organized as follows. In Sections 2 and 3 a general derivation of the coupling impedances
for a small discontinuity is given, and then in Sect. 4 the trapped modes are discussed. This part mostly
follows Ref. [9]. In Sect. 5 we collected for the reader convenience some practical formulas for the coupling
∗Lectures presented atthe USParticleAccelerator School inTucson, AZon January17-21, 2000, as apartof R.L.Gluck-
stern’scourse”AnalyticMethodsforCalculatingCouplingImpedances”
1
impedances of various small discontinuities. In a compact form, many of these formulas are included in the
handbook [16].
2 Fields
2.1 Beam Fields
Let us consider an infinite cylindricalpipe with an arbitrary cross section S and perfectly conducting walls.
The z axisis directed alongthe pipe axis,a hole (or another smalldiscontinuity like a post, cavityor mask)
is located at the point (~b,z =0), and its typical size h satisfies h b. To evaluate the coupling impedance
≪
one hasto calculate the fields inducedin the chamberby a givencurrent. Ifanultrarelativisticpoint charge
q moves parallel to the chamber axis with the transverse offset~s from the axis, the fields harmonics E~b,H~b
produced by this charge on the chamber wall without discontinuity would be
Eb(~s,z;ω) = Z Hb(~s,z;ω) (1)
ν 0 τ
= Z qeikz k−2e (~s) e (~b) ,
− 0 nm nm ∇ν nm
n,m
X
whereZ =120πOhmsistheimpedanceoffreespace,andk2 ,e (~r)areeigenvaluesandorthonormalized
0 nm nm
eigenfunctions (EFs) of the 2D boundary problem in S:
2+k2 e =0 ; e =0 . (2)
∇ nm nm nm ∂S
(cid:0) (cid:1) (cid:12)
Here ~ is the 2D gradient in plane S; k = ω/c; νˆ means an(cid:12)outward normal unit vector, τˆ is a unit
∇
vector tangent to the boundary ∂S of the chamber cross section S, and νˆ,τˆ,zˆ form a right-handed basis.
{ }
The differential operator is the scalar product = νˆ ~. The eigenvalues and EFs for circular and
ν ν
∇ ∇ ·∇
rectangular cross sections are given in the Appendix.
Let us introduce the following notation for the sum in Eq. (1)
e (~s)= k−2e (~s) e (~b) (3)
ν − g g ∇ν g
g
X
where g = n,m is a generalized index. From a physical viewpoint, this is just a normalized electrostatic
{ }
field produced at the hole location by a filament charge displaced from the chamber axis by distance ~s. It
satisfies the normalization condition
dl e (~s)=1 , (4)
ν
I∂S
where integrationgoesalongthe boundary∂S, which is a consequence ofthe Gauss law. It followsfromthe
fact that Eq. (3) gives the boundary value of ~e (~s) ~Φ(~r ~s), where Φ(~r ~s) is the Green function
ν
≡ −∇ − −
of boundary problem (2): 2Φ(~r ~s) = δ(~r ~s). For the symmetric case of an on-axis beam in a
∇ − − −
circularpipe ofradiusbfromEq.(4)immediately followse (0)=1/(2πb). Itcanalsobe derivedbydirectly
ν
summing up the series in Eq. (3) for this particular case.
2.2 Effective Dipoles and Polarizabilities
At distances l suchthat h l b, the fields radiatedby the hole into the pipe areequalto those produced
≪ ≪
by effective dipoles [1, 2]
P = χε Eh/2; M =(ψ Hh+ψ Hh)/2;
ν − 0 ν τ ττ τ τz z
M = (ψ Hh+ψ Hh)/2 , (5)
z zτ τ zz z
2
where superscript ’h’ means that the fields are taken at the hole. Polarizabilities ψ,χ are related to the
effective ones α ,α used in [2, 3] as α = χ/2 and α =ψ/2, so that for a circular hole of radius a in a
e m e m
−
thin wall ψ =8a3/3 and χ=4a3/3 [1]. In general, ψ is a symmetric 2D-tensor, which can be diagonalized.
If the hole is symmetric, and its symmetry axis is parallel to zˆ, the skew terms vanish, i.e. ψ = ψ = 0.
τz zτ
In a more general case of a non-zero tilt angle α between the major symmetry axis and zˆ,
ψ = ψ cos2α+ψ sin2α ,
ττ ⊥ k
ψ = ψ =(ψ ψ )sinαcosα , (6)
τz zτ k ⊥
−
ψ = ψ sin2α+ψ cos2α ,
zz ⊥ k
where ψ is the longitudinal magnetic susceptibility (for the external magnetic field along the major axis),
k
and ψ is the transverse one (the field is transverse to the major axis of the hole). When the effective
⊥
dipoles are obtained, e.g., by substituting beam fields (1) into Eqs. (5), one can calculate the fields in the
chamber as a sum of waveguide eigenmodes excited in the chamber by the dipoles, and find the impedance.
This approach has been carried out for a circular pipe in [3], and for an arbitrary chamber in [5]. The
polarizabilities for various types of small discontinuities are discussed in detail in Section 5.
2.3 Radiated Fields
The radiated fields in the chamber can be expanded in a series in TM- and TE-eigenmodes [2] as
F~ = A+ F~(E)+θ(z)+A− F~(E)−θ( z) + (7)
nm nm nm nm −
Xnmh i
B+ F~(H)+θ(z)+B− F~(H)−θ( z) ,
nm nm nm nm −
Xnmh i
whereF~ meanseitherE~ orH~,superscripts’ ’denote wavesradiatedrespectivelyinthe positive (+,z >0)
±
or negative ( , z < 0) direction, and θ(z) is the Heaviside step function. The fields F(E) of n,m th
nm
− { }
TM-eigenmode in Eq. (7) are expressed [2] in terms of EFs (2)
E∓ = k2 e exp( Γ z) ; H∓ =0 ;
z nm nm ± nm z
E~∓ = Γ ~e exp( Γ z) ; (8)
t ± nm∇ nm ± nm
ik
H~∓ = zˆ ~e exp( Γ z) ,
t Z ×∇ nm ± nm
0
where propagationfactors Γ =(k2 k2)1/2 should be replacedby iβ with β =(k2 k2 )1/2 for
nm nm− − nm nm − nm
k >k . For givenvalues ofdipoles (5)the unknowncoefficients A± canbe found [3,5]using the Lorentz
nm nm
reciprocity theorem
A± =a M b P , (9)
nm nm τ ± nm ν
with
ikZ 1
a = 0 eh ; b = eh . (10)
nm −2Γ k2 ∇ν nm nm 2ε k2 ∇ν nm
nm nm 0 nm
The fields F(H) of the TE -eigenmode in Eq. (7) are
nm nm
H∓ = k′2 h exp( Γ′ z) ; E∓ =0 ;
z nm nm ± nm z
H~∓ = Γ′ ~h exp( Γ′ z) ; (11)
t ± nm∇ nm ± nm
E~∓ = ikZ zˆ ~h exp( Γ′ z) ,
t − 0 ×∇ nm ± nm
with propagation factors Γ′ =(k′2 k2)1/2 replaced by iβ′ = i(k2 k′2 )1/2 when k >k′ . Here
nm nm− − nm − − nm nm
EFs h satisfy the boundary problem (2) with the Neumann boundary condition h =0, and k′2
nm ∇ν nm|∂S nm
3
are corresponding eigenvalues, see in Appendix. The TE-mode excitation coefficients in the expansion (7)
for the radiated fields are
B± = c M +d P +q M , (12)
nm ± nm τ nm ν nm z
where
1 1
c = hh ; q = hh ;
nm 2k′2 ∇τ nm nm 2Γ′ nm
nm nm
ik
d = hh . (13)
nm −2Z ε Γ′ k′2 ∇τ nm
0 0 nm nm
2.4 Fields near Hole with Radiation Corrections
A more refined theory should take into account the reaction of radiated waves back on the hole. Adding
corrections to the beam fields (1) due to the radiated waves in the vicinity of the hole gives
Eb +ψ Σ′Z H +ψ Σ′Z H
E = ν zτ x 0 τ zz x 0 z, (14)
ν 1 χ(Σ Σ′)
− 1− 1
Hb+ψ (Σ Σ′)H
H = τ τz 2− 2 z, (15)
τ 1 ψ (Σ Σ′)
− ττ 2− 2
χΣ′E /Z +ψ Σ′H
H = x ν 0 zτ 3 τ, (16)
z 1 ψ Σ′
− zz 3
where (s= n,m is a generalized index)
{ }
1 Γ eh 2 k2 eh 2
Σ = s ∇ν s ; Σ = ∇ν s ;
1 4 k2 2 4 Γ k2
s (cid:0) s (cid:1) s (cid:0) s s(cid:1)
X X
k2 hh 2 1 Γ′ hh 2
Σ′ = ∇τ s ; Σ′ = s ∇τ s ;
1 4 Γ′k′2 2 4 k′2
s (cid:0) s s(cid:1) s (cid:0) s (cid:1)
X X
k hh hh 1 k′2 hh 2
Σ′ = i s∇τ s ; Σ′ = s s . (17)
x 4 Γ′ 3 4 Γ′
s s s (cid:0)s (cid:1)
X X
Since this consideration works at distances larger than h, one should restrict the summation in Eq. (17) to
the values of s= n,m such that k h 1 and k′h 1.
{ } s ≤ s ≤
3 Impedance
3.1 Longitudinal Impedance
The generalizedlongitudinalimpedance of the hole depends on the transverseoffsets from the chamber axis
~s of the leading particle and~t of the test particle, and is defined [6] as
1 ∞
Z(k;~s,~t)= dze−ikzE (~t,z;ω) , (18)
z
−q
Z−∞
wherethelongitudinalfieldE (~t,z;ω)istakenalongthetestparticlepath. Thedisplacementsfromthe axis
z
are assumed to be small, s b and t b. The impedance Z(k;~s,~t) includes higher multipole longitudinal
≪ ≪
impedances,andinthelimits,t 0givestheusualmonopoleoneZ(k)=Z(k;0,0). TocalculateE (~t,z;ω),
z
→
4
we use Eq.(7) with coefficients (9)and (12)inwhich the correctednear-holefields (14)-(16)are substituted
[a dependence on~s enters via beam fields (1)]. It yields
ikZ e (~s)e (~t)
Z(k;~s,~t)= 0 ν ν (19)
− 2 ×
ψ χ
ττ + ψ2 Σ′ ,
× 1 ψ (Σ Σ′) τz 3− 1 χ(Σ Σ′)
(cid:20) − ττ 2− 2 − 1− 1 (cid:21)
where e (~r) is defined above by Eq. (3). In practice, we are usually interested only in the monopole term
ν
Z(k) = Z(k;0,0), and will mostly use Eq. (19) with replacement e (~s)e (~t ) e˜2, where e˜ e (0).
ν ν → ν ν ≡ ν
In deriving Eq. (19) we have neglected the coupling terms between E , H and H , cf. Eqs. (14)-(16),
ν τ z
which contribute to the third order of an expansion discussed below, and also have taken into account that
ψ =ψ .
τz zτ
For a smalldiscontinuity,polarizabilitiesψ,χ=O(h3), andthey aresmallcomparedto b3. If we expand
the impedance (19) in a perturbation series in polarizabilities, the first order gives
ikZ e˜2
Z (k)= 0 ν (ψ χ) , (20)
1 ττ
− 2 −
that is exactly the inductive impedance obtained in [5] for an arbitrary cross section of the chamber. For a
particular case of a circular pipe, from either direct summation in (1) or applying the Gauss law, one gets
e˜ =1/(2πb). Substituting that into Eq. (20) leads to a well-known result [3, 4]:
ν
ψ χ α +α
ττ e m
Z(k)= ikZ − = ikZ , (21)
− 0 8π2b2 − 0 4π2b2
where we recall that two definitions of the polarizabilities are related as α = χ/2 and α =ψ /2.
e m ττ
−
From a physical point of view, keeping only the first order term (20) corresponds to dropping out all
radiation corrections in Eqs. (14)-(16). These corrections first reveal themselves in the second order term
ikZ e˜2
Z (k)= 0 ν ψ2 (Σ Σ′)+ψ2 Σ′ (22)
2 − 2 ττ 2− 2 τz 3
(cid:2) + χ2(Σ′ Σ ) ,
1− 1
which at frequencies above the chamber cutoff has both a real and i(cid:3)maginary part. The real part of the
impedance is
k3Z e˜2 < k′2 hh 2
ReZ (k) = 0 ν ψ2 s s (23)
2 8 τz k2β′
( s (cid:0) s(cid:1)
X
< eh 2 < β′ hh 2
+ ψ2 ∇ν s + s ∇τ s
ττ β k2 k2k′2
" s (cid:0) s s(cid:1) s (cid:0) s (cid:1) #
X X
< β eh 2 < hh 2
+ χ2 s ∇ν s + ∇τ s ,
k2k2 β′k′2
" s (cid:0) s (cid:1) s (cid:0) s s(cid:1) #)
X X
where the sums include only a finite number of the eigenmodes propagating in the chamber at a given
frequency, i.e. those with k <k or k′ <k.
s s
The dependence of ReZ on frequency is rather complicated; it has sharp peaks near the cutoffs of all
propagating eigenmodes of the chamber, and increases in average with the frequency increase. Well above
the chamber cutoff, i.e. when kb 1 (but still kh 1 to justify the Bethe approach), this dependence can
≫ ≪
be derived as follows. If the waveguide cross section S is a simply connected region, the average number
n(k) of the eigenvalues k (or k′) which are less than k, for kb 1, is proportionalto k2 [17]:
s s ≫
S
n(k) k2+O(k) ,
≃ 4π
5
where S is the area of the cross section. Using this property, and taking into account that eh k eh,
∇ν s ∝ s s
and hh k′hh, we replace sums in the RHS of Eq.(23) by integrals as < kdk d n(k). It turns out
∇τ s ∝ s s s → dk
that all sums in Eq. (23) have the same asymptotic behavior, being linear in k, and as a result, ReZ k4.
P R ∝
Obtainingthe exactcoefficientinthis dependence seemsratherinvolvedfora generalS, butit canbe easily
done for a rectangular chamber, see in Appendix B. The result is
Z k4e˜2
ReZ = 0 ν(ψ2 +ψ2 +χ2) . (24)
12π ττ τz
Remarkably, the same answer (for ψ = 0) has been obtained in Ref. [3] simply by calculating the energy
τz
radiatedbythedipolesintoahalf-space. Thephysicalreasonforthiscoincidenceisclear: atfrequencieswell
abovethe cutoff the effective dipoles radiate into the waveguidethe same energy as into an open half-space.
Strictly speaking,the realpartofimpedance is non-zeroevenbelow the chambercutoff,due to radiation
outside. In the case of a thin wall, ReZ below the cutoff can be estimated by Eq. (24), and twice that for
high frequencies, kb 1. For a thick wall, the contribution of the radiationoutside to ReZ is still given by
≫
Eq. (24), but with the outside polarizabilities substituted, and it decreases exponentially with the thickness
increase [4].
The real part of the impedance is related to the power P scattered by the hole into the beam pipe as
ReZ =2P/q2. Theseenergyconsiderationscanbeusedasanalternativewayfortheimpedancecalculation.
The radiated power is
P = A2P(E)+B2P(H) ,
s s s s
Xs h i
wherewesumoverallpropagatingmodesinbothdirections,andP meansthetime-averagedpowerradiated
s
in sth eigenmode:
P(E) =kβ k2/(2Z ) and P(H) =Z kβ′k′2/2 .
s s s 0 s 0 s s
Substituting beam fields (1) into Eqs. (9)-(13) for the coefficients A and B and performing calculations
s s
givesusexactlytheresult(23). SuchanalternativederivationoftherealparthasbeencarriedoutinRef.[8]
for a circular pipe with a symmetric untilted hole (ψ = 0). Our result (23) coincides, in this particular
τz
case, with that of Reference [8]. It is appropriate to mention also that in this case at high frequencies the
series has been summed approximately [8] using asymptotic expressions for roots of the Bessel functions,
and the result, of course, agrees with Eq. (24).
One should note that the additional ψ2 -term in Eq. (23) is important in some particular cases. For
τz
example, this skew term gives a leading contribution to ReZ for a long and slightly tilted slot, because ψ
τz
can be much larger than ψ in this case, since ψ ψ , cf. Eqs. (6).
ττ k ⊥
≫
3.2 Transverse Impedance
WewillmakeuseoftheexpressionforthegeneralizedlongitudinalimpedanceZ(k;~s,~t),Eq.(19). According
tothePanofsky-Wenzeltheorem,thetransverseimpedancecanbederivedasZ~ (k;~s,~t)= ~Z(k;~s,~t)/(ks),
⊥
∇
see, e.g., [6] for details. This way leads to the expression
iZ edip(~s)~e (~t)
Z~ (k;~s,~t)= 0 ν ∇ ν (25)
⊥
− 2s ×
ψ χ
ττ +ψ2 Σ′ ,
× 1 ψ (Σ Σ′) τz 3− 1 χ(Σ Σ′)
(cid:20) − ττ 2− 2 − 1− 1 (cid:21)
where edip(~s)=~s ~e (~s).
ν ·∇ ν
Going to the limit s t 0, we get the usual dipole transverse impedance
→ →
1
Z~ (k)= iZ (d2 +d2)~a cos(ϕ ϕ ) (26)
⊥ − 02 x y d b− d ×
ψ χ
ττ +ψ2 Σ′ .
× 1 ψ (Σ Σ′) τz 3− 1 χ(Σ Σ′)
(cid:20) − ττ 2− 2 − 1− 1 (cid:21)
6
Herex,yarethehorizontalandverticalcoordinatesinthechambercrosssection;d ∂ e (0),d ∂ e (0);
x x ν y y ν
≡ ≡
ϕ =ϕ =ϕ istheazimuthalangleofthebeampositioninthecross-sectionplane;~a =~a cosϕ +~a sinϕ
b s t d x d y d
is a unit vector inthis plane indirectionϕ , whichis defined by conditions cosϕ =d / d2 +d2, sinϕ =
d d x x y d
d / d2 +d2. It is seen from Eq. (26) that the angle ϕ shows the direction of the traqnsverse-impedance
y x y d
vectqorZ~ and,therefore,ofthebeam-deflectingforce. Moreover,thevalueofZ ismaximalwhenthebeam
⊥ ⊥
is deflected along this direction and vanishes when the beam offset is perpendicular to it.
The equation(26)includes the correctionsdue towavesradiatedbythe holeintothe chamberinexactly
the same way as Eq. (19) for the longitudinal impedance. If we expand it in a series in the polarizabilities,
the first order of the square brackets in (26) gives (ψ χ), and the resulting inductive impedance is [5]:
ττ
−
1
Z~ (k)= iZ (d2 +d2)~a cos(ϕ ϕ )(ψ χ) . (27)
⊥ − 02 x y d b− d ττ −
For a circular pipe, d = cosϕ /(πb2) and d = sinϕ /(πb2), where ϕ is the azimuthal position of
x h y h h
the discontinuity (hole). As a result, ϕ = ϕ , and the deflecting force is directed toward (or opposite to)
d h
the hole. Note that in axisymmetric structures the beam-deflecting force is directed along the beam offset;
the presence of an obstacle obviously breaks this symmetry. For a general cross section, the direction of
the deflecting force depends on the hole position in a complicated way, see [5] for rectangular and elliptic
chambers.
The transverse impedance of a discontinuity on the wall of a circular pipe has a simple form [3]:
ψ χ α +α
Z~ (k)= iZ ττ − ~a cos(ϕ ϕ )= iZ m e ~a cos(ϕ ϕ ) , (28)
⊥circ − 0 2π2b4 h b− d − 0 π2b4 h b− d
where~a is a unit vector in the direction from the chamber axis to the discontinuity, orthogonalto zˆ.
h
For M (M 3) holes uniformly spaced in one cross section of a circular beam pipe, a vector sum of M
≥
expressions (28) gives the transverse impedance as
α +α M
Z~ (k)= iZ m e ~a , (29)
⊥circM − 0 π2b4 2 b
where~a =~s/~s isaunitvectorinthedirectionofthebeamtransverseoffset. Onecanseethatthedeflecting
b
| |
force is now directed along the beam offset, i.e. some kind of the axial symmetry restoration occurs. The
maximal value of Z for M holes which are uniformly spaced in one cross-section is only M/2 times larger
⊥
than that for M = 1. Moreover, the well-known empirical relation Z = (2/b2k)Z, which is justified only
⊥
for axisymmetric structures, holds in this case also.
The second order term in Eq. (26) includes ReZ , cf. Sect. 3.1 for the longitudinal impedance.
⊥
4 Trapped Modes
So far we consideredthe perturbation expansionof Eq. (19)implicitly assuming that correctionterms O(ψ)
andO(χ)inthedenominatorsofitsright-handside(RHS)aresmallcomparedto1. Undercertainconditions
this assumption is incorrect, and it leads to some non-perturbative results. Indeed, at frequencies slightly
below the chamber cut-offs, 0 < k k k (or the same with replacement k k′), a single term in
s − ≪ s s → s
sums Σ′, Σ , or Σ′ becomes very large, due to very small Γ =(k2 k2)1/2 (or Γ′) in its denominator, and
1 2 3 s s− s
then the “corrections” ψΣ or χΣ can be of the order of 1. As a result, one of the denominators of the RHS
of Eqs. (19) can vanish, which corresponds to a resonance of the coupling impedance. On the other hand,
vanishingdenominatorsinEqs.(14)-(16)meantheexistenceofnon-perturbativeeigenmodesofthechamber
with a hole, since non-trivial solutions E,H =0 exist even for vanishing external (beam) fields Eb,Hb =0.
6
These eigenmodes are nothing but the trapped modes studied in [18] for a circular waveguide with a small
discontinuity. In our approach, one can easily derive parameters of trapped modes for waveguides with an
arbitrary cross section.
7
4.1 Frequency Shifts
Let us for brevity restrict ourselves to the case ψ = 0 and consider Eq. (15) in more detail. For Hb = 0
τz
we have
k2 eh 2
H 1 ψ ∇ν s +... =0 , (30)
τ " − ττ 4(cid:0)Γsks2(cid:1) #
where s nm is the generalizedindex, and...meanallother terms ofthe seriesΣ ,Σ′. Atfrequency Ω
≡{ } 2 2 s
slightly below the cutoff frequency ω =k c of the TM -mode, the fraction in Eq. (30) is large due to small
s s s
Γ in its denominator, and one can neglect the other terms. Then the condition for a non-trivial solution
s
H =0 to exist is
τ
6 Γ 1ψ eh 2 . (31)
s ≃ 4 ττ ∇ν s
In other words, there is a solution of the homogeneou(cid:0)s, i.e.,(cid:1)without external currents, Maxwell equations
for the chamber with the hole, having the frequency Ω <ω — the sth trapped TM-mode. When Eq. (31)
s s
is satisfied, the series (7) is obviously dominated by the single term A FE; hence, the fields of the trapped
s s
mode have the form [cf. Eq. (8)]
= k2e exp( Γ z ) ; =0 ;
Ez s s − s| | Hz
~ = sgn(z)Γ ~e exp( Γ z ) ; (32)
t s s s
E ∇ − | |
Z ~ = ikzˆ ~e exp( Γ z ) ,
0 t s s
H ×∇ − | |
up to some arbitrary amplitude. Strictly speaking, these expressions are valid at distances z >b from the
| |
discontinuity. Typically, ψ = O(h3) and eh = O(1/b), and, as a result, Γ b 1. It follows that the
ττ ∇ν s s ≪
field of the trapped mode extends along the vacuum chamber overthe distance 1/Γ , largecomparedto the
s
chamber transverse dimension b.
Theexistenceofthetrappedmodesinacircularwaveguidewithasmallholewasfirstprovedin[18],and
conditions similarto Eq.(31)for this particularcase wereobtainedin [18,19], using the Lorentzreciprocity
theorem. From the general approach presented here for the waveguide with an arbitrary cross section,
their existence follows in a natural way. Moreover, in such a derivation, the physical mechanism of this
phenomenon becomes quite clear: a tangential magnetic field induces a magnetic moment on the hole, and
the induced magnetic moment supports this field if the resonance condition (31) is satisfied, so that the
mode can exist even without an externalsource. One should also note that the induced electric moment P
ν
is negligible for the trapped TM-mode, since P =O(Γ b)M , as follows from Eq. (32).
ν s τ
The equation (31) gives the frequency shift ∆ω ω Ω of the trapped sth TM-mode down from the
s s s
≡ −
cutoff ω
s
∆ωs 1 ψ2 eh 4 . (33)
ω ≃ 32k2 ττ ∇ν s
s s
(cid:0) (cid:1)
In the case of a small hole this frequency shift is very small, and for the trapped mode (32) to exist, the
widthoftheresonanceshouldbesmallerthan∆ω . Contributionstothe resonancewidthcomefromenergy
s
dissipationinthewaveguidewallduetoitsfiniteconductivity,andfromenergyradiationinsidethewaveguide
and outside, through the hole. Radiation escaping through the hole is easy to estimate [18], and for a thick
wall it is exponentially small, e.g., [4]. The damping rate due to a finite conductivity is γ =P/(2W), where
P is the time-averagedpower dissipation and W is the total field energy in the trapped mode, which yields
γ δ
s = dl( e )2 , (34)
ω 4k2 ∇ν s
s s I
where δ is the skin-depth at frequency Ω , and the integration is along the boundary ∂S. The evaluation
s
of the radiation into the lower waveguide modes propagating in the chamber at given frequency Ω is also
s
straightforward [8], if one makes use of the coefficients of mode excitation by effective dipoles on the hole,
Eqs.(9)-(13). Thecorrespondingdampingrateγ =O(ψ3)issmallcomparedto∆ω . Forinstance,ifthere
R s
8
is only one TE -mode with the frequency below that for the lowest TM -mode, like in a circular waveguide
p s
(H has a lower cutoff than E ),
11 01
γR = ψττβp′ hh 2 , (35)
∆ω k′2 ∇ν s
s p
(cid:0) (cid:1)
where β′ (k2 k′2)1/2 because k k .
p ≃ s − p ≃ s
Onecaneasilyseethatdenominator[1 χ(Σ Σ′)]inEq.(14)doesnotvanishbecausesingulartermsin
− 1− 1
Σ′ havea“wrong”sign. However,duetothecouplingbetweenE andH ,anon-trivialsolutionE ,H =0
1 ν z ν z 6
of simultaneous equations (14) and (16) can exist, even when Eb =0. The corresponding condition has the
form
Γ′ 1 ψ k′2 hh 2 χ hh 2 , (36)
nm ≃ 4 zz nm nm − ∇τ nm
which gives the frequency of the trappedhTE -m(cid:0)ode, p(cid:1)rovide(cid:0)d the RH(cid:1)Siof Eq. (36) is positive.
nm
4.2 Impedance
The trapped mode (32) gives a resonance contribution to the longitudinal coupling impedance at ω Ω
s
≈
2iΩ γ R
s s s
Z (ω)= , (37)
s ω2 (Ω iγ )2
s s
− −
where the shunt impedance R can be calculated as that for a cavity with given eigenmodes, e.g. [6],
s
2
σδ dzexp( iΩ z/c) (z)
s z
R = − E . (38)
s ds 2
(cid:12)R Sw |Hτ| (cid:12)
(cid:12) (cid:12)
R
The integral in the denominator is taken over the inner wall surface, and we assume here that the power
lossesduetoitsfiniteconductivitydominate. IntegratinginthenumeratoroneshouldincludeallTM-modes
generatedby the effective magnetic moment on the hole using Eqs.(9)-(13), in spite of a large amplitude of
onlythetrappedTM mode. WhileallotheramplitudesaresuppressedbyfactorΓ b 1,theircontributions
s s
≪
arecomparableto thatfromTM , because this integrationproduces the factor Γ b for anyTM mode. The
s q q
integral in the denominator is obviously dominated by TM . Performing calculations yields
s
Z e˜2ψ3 k eh 4
R = 0 ν ττ s ∇ν s , (39)
s 8δ dl( e )2
(cid:0)ν s (cid:1)
∇
where e˜ =e (0) is defined by Eq. (3). H
ν ν
Results for a particular shape of the chamber crosssection can be obtained fromthe equations above by
substituting the corresponding eigenfunctions (see Appendix).
One should note that typically the peak value R of the impedance resonance due to one small hole is
s
rathersmallexceptforthelimitofaperfectlyconductingwall,δ 0—indeed,R (h/b)9b/δ,andh b.
s
→ ∝ ≪
However, for many not-so-far separated holes, the resulting impedance can be much larger. The trapped
modes for many discontinuities on a circular waveguidehas been studied in Ref. [19], and the results canbe
readily transferred to the considered case of an arbitrary shape of the chamber cross section. In particular,
it was demonstratedthat the resonanceimpedance in the extreme case can be as large as N3 times that for
a single discontinuity, where N is the number of discontinuities. It strongly depends on the distribution of
discontinuities, or on the distance between them if a regular array is considered.
After the trapped modes in beam pipes with small holes were predicted theoretically [18, 19], their
existence was proved by experiments with perforated waveguides at CERN [20].
9
5 Analytical Formulae for Some Small Discontinuities
For reader convenience, in this section we collected analytical expressions for the coupling impedances of
various small discontinuities. The expressions give the inductive part of the impedance and work well at
frequencies below the chamber cutoff, and, in many cases, even at much higher frequencies. However, there
can also exist resonances of the real part at frequencies near the cutoff for holes and cavities due to the
trappedmodes, aswasshownin[18, 19], andthe realpartofthe impedance due to the energyradiatedinto
the beam pipe should be taken into account at frequencies above the cutoff, see [7, 9, 21].
It is worth noting that both the longitudinal and transverse impedance are proportional to the same
combination of polarizabilities, α +α , for any cross section of the beam pipe. Here we use the effective
e m
polarizabilities α ,α as defined in [2]; they are related to the magnetic susceptibility ψ and the electric
e m
polarizabilityχ of anobstacle as α = χ/2 andα =ψ/2. The realpart of the impedance is proportional
e m
−
to α2+α2 , and is usually small compared to the reactance at frequencies below the chamber cutoff. While
e m
the impedances below are written for a round pipe, more results for the other chamber cross sections, the
impedance dependence on the obstacle position on the wall and on the beam position can be found in
[5, 7, 9, 21, 22].
The longitudinal impedance of a small obstacle on the wall of a cylindrical beam pipe with a circular
transverse cross section of radius R is simply [3] (up to notations, it is the same Eq. (21) above)
α +α
e m
Z(k)= ikZ , (40)
− 0 4π2R2
where Z = 120π Ohms is the impedance of free space, k = ω/c is the wave number, and α ,α are the
0 e m
electric and magnetic polarizabilities of the discontinuity. The polarizabilities depend on the obstacle shape
and size.
The transverse dipole impedance of the discontinuity for this case is
α +α
Z~ (ω)= iZ m e~a cos(ϕ ϕ ) , (41)
⊥ − 0 π2R4 h h− b
where~a is the unit vectordirectedto the obstacle inthe chambertransversecrosssectioncontainingit, ϕ
h h
and ϕ are azimuthal angles of the obstacle and beam in this cross section.
b
5.1 Holes and Slots
For a circular hole with radius a in a thin wall, when thickness t a, the polarizabilities are α =
m
≪
4a3/3, α = 2a3/3, so that the impedance Eq. (40) takes a simple form
e
−
a3
Z(k)= ikZ , (42)
− 06π2R2
and similarly for Eq. (41). For the hole in a thick wall, t a, the sum (α +α ) = 2a3/3 should be
m e
≥
multiplied by a factor 0.56, see [4, 11]. There are also analytical expressions for polarizabilities of elliptic
holes in a thin wall [2], and paper [12] gives thickness corrections for this case. Surprisingly, the thickness
factor for (α +α ) exhibits only a weak dependence on ellipse eccentricity ε, changing its limiting value
m e
for the thick wall from 0.56 for ε=0 to 0.59 for ε=0.99.
For a longitudinal slot of length l andwidth w, w/l 1, in a thin wall, useful approximationshavebeen
≤
obtained [13]: for a rectangular slot
α +α =w3(0.1814 0.0344w/l);
m e
−
and for a rounded end slot
α +α =w3(0.1334 0.0500w/l);
m e
−
substituting of which into Eqs. (40)-(41) gives the impedances of slots. Figure 1 compares impedances for
different shapes of pumping holes.
10
