Lecture Notes on Topics in Accelerator Physics

328 Pages·2001·2.24 MB·English

by  Alex Chao

SLAC-PUB-9574
November 2002
Lecture Notes on
Topics in Accelerator Physics
Alex Chao
Stanford Linear Accelerator Center
These are lecture notes that cover a selection of topics, some of them under
current research, in accelerator physics. I try to derive the results from first
principles,althoughthestudentsareassumedtohaveanintroductoryknowledge
of the basics. The topics covered are:
1. Panofsky-Wenzel and Planar Wake Theorems
2. Echo Effect
3. Crystalline Beam
4. Fast Ion Instability
5. Lawson-Woodward Theorem and Laser Acceleration in Free Space
6. Spin Dynamics and Siberian Snakes
7. Symplectic Approximation of Maps
8. Truncated Power Series Algebra
9. Lie Algebra Technique for nonlinear Dynamics
Thepurposeoftheselecturesisnottoelaborate,buttopreparethestudents
so that they can do their own research. Each topic can be read independently
of the others.
ManyusefulcommentsandhelpatthelecturingfromGennadyStupakovof
SLAC and Jeff Holmes of ORNL are greatly appreciated.
*Work supported by Department of Energy Contract DE-AC03-76SF00515.
Stanford Linear Accelerator Center, Stanford University, Stanford, CA 94309, USA
1 Panofsky-Wenzel and Planar Wake Theorems
1.1 Concept of Wakefields
Toalargedegree,acceleratorphysicsandplasmaphysicsarequitesimilar. Both
involve nonlinear dynamics (single-particle effects) and collective instabilities
(multi-particle effects). However, there is an important difference:
beam self fields>external applied fields (plasma)
beam self fields(cid:1)external applied fields (accelerators)
This difference means perturbation techniques are applicable to accelerators
with
unperturbed motion = external fields,
perturbation = self fields, or “wakefields”
In fact, in accelerator physics, a first order perturbation often suffices.
Itisimportanttoappreciatethefactthatourinstabilityanalysisinacceler-
ators is based on the validity of this perturbation technique. In particular, the
concept of wakefields is based on the validity of this perturbation technique as
appliedtohighenergyaccelerators–thewords“highenergy”arecritical,aswe
will explain.
Consider a beam with distribution ψ. The dynamics of the evolution of ψ is
described by the Vlasov equation,
∂ψ p(cid:6) ∂ψ ∂ψ
+ · +f(cid:6)· =0
∂t m ∂(cid:6)q ∂p(cid:6)
(cid:6)v
f(cid:6)=e(E(cid:6) + ×B(cid:6)) (1.1)
c
In case the beam is intense, the EM fields contain two contributions,
E(cid:6) =E(cid:6) +E(cid:6)
ext wake
B(cid:6) =B(cid:6) +B(cid:6) (1.2)
ext wake
The wakefields (N is beam intensity)
(E(cid:6),B(cid:6)) ∝N
wake
(E(cid:6),B(cid:6)) (cid:1)(E(cid:6),B(cid:6)) (1.3)
wake ext
are determined by the Maxwell equations where the source terms ρ and(cid:6)j are
determined by the beam distribution ψ:
(cid:1) (cid:1)
ρ= d3p ψ, (cid:6)j = d3p(cid:6)vψ (1.4)
We therefore have the situation when the beam distribution is described
by the Vlasov equation whose force terms are given by the electromagnetic
1
fields, while the electromagnetic fields are described by the Maxwell equations
whose source terms are given by the beam distribution. It is clear that a full
treatment of the beam-wakefield system requires solving a coupled “Vlasov-
Maxwell equation”.
Most wakefields are generated by beam-structure interaction. Figure 1.1
shows no wakefields when the beam pipe is smooth and perfectly conducting,
while a structure causes wakefields to be generated.
Smooth Pipe Pipe with Structure
No Wakefields Wakefields
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8322A22
Figure 1.1: Wakefields are generated when the beam pipe is not smooth.
Beam-structure interaction is a difficult problem in general. Its solution
often involves numerical solution using particle-in-cell (PIC) codes. Applying
PIC codes is reasonable for small devices such as electron guns and klystrons,
but becomes impractical for large accelerators.
So, can we simplify it for our purpose? The answer is yes. For high energy
accelerators, this complication can be avoided due to two simplifying approx-
imations. These simplifications lead to the concepts of “wake function” and
“impedance”.
Rigid beam approximation The first simplification is the rigid beam ap-
proximation. Athighenergies,beammotionislittleaffectedduringthepassage
of a structure. This means one can calculate the wakefields assuming the beam
shapeisrigidanditsmotionisultrarelativisticwithv =c. Infact,weonlyneed
to calculate the wakefields generated by a “rigid cosmθ ring beam” as shown
in Fig.1.2, where m = 0 is monopole moment (net charge), m = 1 is dipole
moment, etc. Wakefield of a general beam can be obtained by superposition of
wakefields due to the ring beams with different m’s and different ring radii.
Impulse approximation The second simplification is the impulse approxi-
mation. First, let’s note that we don’t need to know the instantaneous E(cid:6) or B(cid:6)
separately. We need only to know f(cid:6). Second, for high energies, we don’t even
need the instantaneous f(cid:6). We only need the integrated impulse
(cid:1)
∞
∆p(cid:6)= dt f(cid:6) (1.5)
−∞
2
a
ν=c
ρ∝cosmθ
6–97
8322A19
Figure 1.2: An ultrarelativistic cosmθ ring beam going down the beam pipe in
the rigid-beam approximation.
where the integration over t is performed along the unperturbed trajectory of
the test charge e, holding D fixed. See Fig.1.3. The integration from −∞ to ∞
assumes the wakefield is localized to the neighborhood of the structure in the
beam pipe.
ν=c
e ν=c
D
6–97
8322A20
Figure 1.3: Configuration of a ring beam and a test charge that follows it.
The ring beam generates a wakefield. The test charge receives a wake-induced
impulse in the impulse approximation.
The instantaneous wakefields are complicated, but as we will soon see, ∆p(cid:6)
is much simpler and it is ∆p(cid:6) that we need! Mother nature has been very kind.
Thequantityc∆p(cid:6)issometimescalledthe“wakepotential”. Notethatalthough
the beam is considered to be rigid during the passage, the impulse will affect
the subsequent beam motion after the passage.
Reasoningalongthislineturnsouttobequitefruitful. Inthefollowing,Iwill
firstderiveatheorem(Panofsky-Wenzel),andthenconsidervariousapplications
along similar lines, including the introduction of another theorem called the
planar wake theorem. The Panofsky-Wenzel theorem is the basis of all beam
instability analyses in high energy accelerators.
3
1.2 Panofsky-Wenzel Theorem
We derive the theorem in this section. Details of the derivation are different
from the original paper. Maxwell equations read
∇·E(cid:6) =4πρ
1∂E(cid:6)
∇×B(cid:6) − =4πβρzˆ
c ∂t
∇·B(cid:6) =0
1∂B(cid:6)
∇×E(cid:6) + =0 (1.6)
c ∂t
where we have made the important rigid beam approximation(cid:6)j = ρ(cid:6)v and (cid:6)v =
βczˆ.
The Lorentz force, Eq.(1.1) is given by
f(cid:6)=e(E(cid:6) +βzˆ×B(cid:6)) (1.7)
which leads to
∇·f(cid:6) = e[4πρ+β∇·(zˆ×B(cid:6))]
= e[4πρ−βzˆ·∇×B(cid:6)]
(cid:2) (cid:3)
β ∂E(cid:6)
= e 4πρ− zˆ· −4πβ2ρ
c ∂t
eρ eβ∂E
= 4π − z (1.8)
γ2 c ∂t
(cid:2) (cid:3)
1∂B(cid:6)
∇×f(cid:6) = e − +β∇×(zˆ×B(cid:6))
c ∂t
(cid:4) (cid:5)
1 ∂ ∂
= −e +β B(cid:6) (1.9)
c∂t ∂z
As mentioned, we are only interested in the impulse given by Eq.(1.5). To
be more specific, we want to calculate the net kick received by a test charge e
which has a transverse position (x,y) and longitudinal position D relative to
the moving beam (see Fig.1.3). Both the beam and the test charge move with
(cid:6)v =βczˆ. Eq.(1.5) is then written more precisely as
(cid:1)
∞
∆p(cid:6)(x,y,D)= dt f(cid:6)(x,y,D+βct,t) (1.10)
−∞
Note that D <0 if the test charge is trailing the beam.
So far, we have not assumed any detailed shape of the beam. Neither have
we assumed any information of the pipe boundary. We have kept β in the
derivation, postponing setting β =1 until later.
4
With Eq.(1.10), we have
(cid:1) ∞ (cid:6) (cid:7)
∇×∆p(cid:6) = dt ∇(cid:3)×f(cid:6)(x,y,z,t)
−∞(cid:1) (cid:8)(cid:4) (cid:5)z=D+βct (cid:9)
∞
1 ∂ ∂
= −e dt +β B(cid:6)(x,y,z,t)
c∂t ∂z
−∞ (cid:10) z=D+βct
e (cid:10)t=∞
= − B(cid:6)(x,y,D+βct,t)(cid:10) (1.11)
c t=−∞
In Eq.(1.11), ∇ refers to taking derivative with respect to coordinates (x,y,D),
while ∇(cid:3) refers to taking derivative with respect to coordinates (x,y,z). If the
wakefield B(cid:6) vanishes far away from the region of interest, we have
∇×∆p(cid:6)=(cid:6)0 (1.12)
which is the Panofsky-Wenzel theorem.
OnemightaskifthePanofsky-Wenzeltheoremexhaustsalltheusefulinfor-
mation contained in the Maxwell equations regarding the wake impulse. The
answer to this question is no. Panofsky-Wenzel theorem, in this sense, is nec-
essary but not sufficient. In particular, one observes that the Panofsky-Wenzel
theorem (1.12) makes use of Eq.(1.9), but not Eq.(1.8). Also, Eq.(1.12) implies
that ∆p(cid:6) can be written as the gradient of another quantity W, but it does not
saywhatW is. Ontheotherhand, MaxwellequationsallowexpressionsforW,
as illustrated in Exercises 4 and 5.
We introduce W by
∆p(cid:6)(x,y,D)=−e∇W(x,y,D) (1.13)
Equation (1.17) in Exercise 2 then requires the Laplace condition (whenβ =1)
∂2W ∂2W
∇2W =0 or + =0 (1.14)
⊥ ∂x2 ∂y2
Exercise 1Equation(1.12)isavectorequation. Onecandecompose
it into a component parallel to zˆand a component perpendicular to
zˆby taking zˆ· or zˆ× operations to it. Use these operations to show
∇·(zˆ×∆p(cid:6))=0 (1.15)
∂
∂D∆p(cid:6)⊥ =∇⊥∆pz (1.16)
Eq.(1.15) says something about the transverse components of ∆p(cid:6).
Eq.(1.16) says that the transverse gradient of the longitudinal wake
potentialisequaltothelongitudinalgradientofthetransversewake
potential.
Exercise 2 When β =1, show that
∇⊥·∆p(cid:6)⊥ =0 (1.17)
5
By setting β = 1, we have dropped the direct space-charge terms,
i.e. the 1-st term on the right-hand-side of Eq.(1.8).
Solution
(cid:1) ∞ (cid:6) (cid:7)
∇·∆p(cid:6) = dt ∇(cid:3)·f(cid:6)(x,y,z,t)
−∞(cid:1) (cid:4) (cid:5) z=D+ct(cid:1) (cid:4) (cid:5)
∞ ∞
e ∂E ∂E
= − dt z =e dt z
c ∂t ∂z
(cid:1) −∞(cid:4) (cid:5) z=D+ct −∞ z=D+ct
∞
∂f ∂∆p
= dt z = z (1.18)
∂z ∂D
−∞ z=D+ct
which proves (1.17).
Exercise 3 In Cartesian coordinates, Eq.(1.15) gives ∂∆px = ∂∆py,
∂y ∂x
while Eq.(1.17) gives ∂∆px + ∂∆py = 0 when β = 1. Combining
∂x ∂y
these equations give ( ∂2 + ∂2 )∆p =0 if β =1. It is clear that
∂x2 ∂y2 x,y
Panofsky-Wenzel theorem imposes strong conditions on ∆p(cid:6).
Exercise 4 Use Maxwell equations and the Lorentz force equation
to show that (cid:1)
∞ (cid:10)
W = dt(φ−A )(cid:10) (1.19)
z x,y,D+ct,t
−∞
Solution With E(cid:6) =−∇φ− 1∂A(cid:11) and B(cid:6) =∇×A(cid:6), we have
c ∂t
ddp(cid:6)t =e(E(cid:6) +zˆ×B(cid:6))=−e∇φ− ec∂∂A(cid:6)t +e∇⊥Az−e∂∂A(cid:6)z⊥
We then use dA(cid:11) = ∂A(cid:11) +((cid:6)v·∇)A(cid:6) = ∂A(cid:11) +c∂A(cid:11), and integrate over t,
dt ∂t ∂t ∂z
to obtain (cid:1)
∞
∆p(cid:6)=−e dt(∇φ−∇A ) (1.20)
z
−∞
Eq.(1.19) then follows.
Exercise 5ShowthatonecancastthePanofsky-Wenzeltheoremin
terms of relativity 4-vectors as
(cid:1)
∞
∂W 1
∆pα =−e where W = dτAβu (1.21)
∂x c β
α −∞
where τ is the proper time.
Solution The 4-vectors are xα = (ct,x,y,z), uα = (γc,γ(cid:6)v), Aα =
(φ,A(cid:6)), pα = muα = (E/c,p(cid:6)). Start with the 4-vector equation of
motion for a relativistic particle,
(cid:4) (cid:5)
dpα duα e ∂Aβ ∂Aα
=m =− − u
dτ dτ c ∂x ∂x β
(cid:1) α β (cid:1)
e ∞ ∂Aβ e ∞ ∂Aα
=⇒ ∆pα =m∆uα =− dτ u + dτ u (1.22)
c ∂x β c ∂x β
−∞ α −∞ β
6
The second term o(cid:11)n the RHS can be writ(cid:10)ten as, using the fact that
u dτ = dx , −e ∞ dx ∂Aα = −eAα(cid:10)∞ = 0. Eq.(1.21) then
β β c −∞ β ∂xβ c −∞
follows. This result is consistent with Eq.(1.19).
It should be pointed out that Eqs.(1.19) and (1.21) are formal
expressions. ExplicitexpressionsofW stillrequiressolvingMaxwell
equations for Aα, which is the difficult part. Indeed, the power
of Panofsky-Wenzel theorem lies in the fact that one can obtain so
muchresultwithoutresortingtosolvingMaxwellequationsindetail.
Exercise 6 As an illustration of decomposing a general beam dis-
tribution into a superposition of cosmθ-ring components, consider
a point charge q located at r =r and θ =θ , moving with velocity
0 0
(cid:6)v =czˆ. Perform the cosmθ-ring decomposition for this beam.
Exercise 7 Panofsky-Wenzel theorem applies when the impulse is
caused by a Lorentz force. Consider a particle with magnetic mo-
mentµ(cid:6),whichexperiencesaStern-GerlachforceinsteadofaLorentz
force. Does the Panofsky-Wenzel theorem apply to its motion?
1.3 Cylindrically Symmetric Pipe
In cylindrical coordinates, Eq.(1.15) gives
∇·[zˆ×(∆p rˆ+∆p θˆ)]=0
r θ
∂ ∂
=⇒ (r∆p )= ∆p (1.23)
∂r θ ∂θ r
Eq.(1.16) gives
(cid:12) (cid:13)
∂ ∂ θˆ ∂
(∆p rˆ+∆p θˆ)= rˆ + ∆p
∂D r θ ∂r r∂θ z
(cid:14)
∂ ∆p = ∂ ∆p
=⇒ ∂D r ∂r z (1.24)
∂ ∆p = 1 ∂ ∆p
∂D θ r∂θ z
Eq.(1.17) gives
1 ∂ 1 ∂
(r∆p )+ ∆p =0
r∂r r r∂θ θ
∂ ∂
=⇒ (r∆p )=− ∆p (β =1) (1.25)
∂r r ∂θ θ
Equations(1.23-1.25)aresurprisinglysimple. Theydonotcontainanybeam
sourceterms. Exactshapeordistributionofthebeamdoesnotmatter. Neither
do they depend on the boundary conditions. The boundary can be perfectly
conducting or resistive metal, or it can be dielectric. The boundary does not
have to be a sharply defined surface; it can for example be a gradually fading
7
plasma surface. The boundary also does not have to consist of a single piece.
TheonlyinputsneededforthePanofsky-WenzeltheoremaretheMaxwellequa-
tions in free space and the rigid-beam and the impulse approximations.
We are now ready to consider a cosmθ ring beam with(cid:6)v =czˆas in Fig.1.3.
Eqs.(1.23-1.25)canbesolved,anditfollowsthatthereexistsafunctionW (D)
m
such that
c∆p(cid:6)⊥ = −eImWm(D)mrm−1(rˆcosmθ−θˆsinmθ)
c∆p = −eI W(cid:3) (D)rmcosmθ (1.26)
z m m
where I is the m-th multipole moment of the ring beam. The reader should
m
try to derive Eq.(1.26) from Eqs.(1.23-1.25). On the other hand, the derivation
can be recovered as a special case when we discuss Eqs.(1.28-1.33). Eq.(1.26)
can also be derived by solvng Eq.(1.14) to obtain
W =I W (D)rmcosmθ (1.27)
m m
The solution (1.26) contains explicit dependences of r and θ. The depen-
dence on D is through the wake function W (D) which can be obtained only
m
if boundary conditions are introduced. The fact that we can go so far without
much details shows the power of this line of study.
When the beam pipe is cylindrically symmetric, each m-multipole compo-
nent of the beam excites a wake pattern according to (1.26). Different m’s do
not mix.
I assume the reader is familiar with the properties of the wake function and
its Fourier transform, known as the impedance. Suffice it to remind the reader
that the rigid-beam and the impulse approximations have led to a drastically
simpler Vlasov system (instead of a Vlasov-Maxwell system) to solve, and as
a result one obtains a large amount of analytical results without resorting to
PICcodes. BelowwelookforotherapplicationsalongthelineofthePanofsky-
Wenzel analysis.
1.4 Cylindrically Symmetric Pipe With a Central Con-
ductor
Considernowacylindricallysymmetricbeampipeexceptthatthistimethereis
aperfectconductoratthecenter. Thepipethereforehasacoaxial-likegeometry,
as sketched in Fig.1.4. The beam has(cid:6)v =czˆand is necessarily ring shaped and
goes around the central conductor. The beam can again be decomposed into a
superposition of cosmθ ring beams. Other than being cylindrically symmetric,
the geometries of the pipe and the central conductor are arbitrary.
Everything from Maxwell equations (1.6) to the Panofsky-Wenzel theorem,
Eqs.(1.12-1.25) still hold. But Eq.(1.26) needs to be changed. In obtaining
Eq.(1.26) as solution to Eqs.(1.23-1.25), we have applied the condition that the
solution must be well-behaved at the pipe center r =0. Indeed, when r →0 in
8
(a)
ν=c
Central
ρ∝cosmθ Conductor
Beam Pipe
(b)
ν=c
Central
Conductor
6–97
8322A21
Figure 1.4: When there is a central conductor in the beam pipe.
Eq.(1.26), the wake potentials are well behaved.1
Butwithacentralconductor,thisconditiondoesnothavetoholdandthere
are new modes which can be excited by the beam. As we will see, the general
form of the wakefield will then be quite different from Eq.(1.26). This can be
of concern because one of the techniques often used to measure the wakefield
is to run a thin conducting wire down the pipe structure. The concern is then
whether the thin wire has profoundly perturbed the wakefield.
With a cosmθ beam, even with a central conductor, we can write
∆p =∆p¯ cosmθ, ∆p =∆p¯ sinmθ, ∆p =∆p¯ cosmθ (1.28)
r r θ θ z z
where ∆p¯ , ∆p¯ , ∆p¯ are functions of r and D. Substituting into Eqs.(1.23-
r θ z
1.25), we find
∂ ∂ ∂
(r∆p¯ )=−m∆p¯ , ∆p¯ = ∆p¯
∂r θ r ∂D r ∂r z
∂ m ∂
∆p¯ =− ∆p¯ , (r∆p¯ )=−m∆p¯ (1.29)
∂D θ r z ∂r r θ
It follows that, by eliminating ∆p¯ and ∆p¯ ,
r θ
(cid:4) (cid:5)
∂ ∂∆p¯
r r z −m2∆p¯ =0 (1.30)
∂r ∂r z
1Theonlyquestionablecaseiswhenm=0,therm−1-dependenceof∆p(cid:3)⊥seemsdivergent.
Butwhenm=0,theentire∆p(cid:3)⊥ vanishesbecause∆p(cid:3)⊥∝m.
9