USPAS Accelerator Physics 2013 Duke University

Transcription

USPAS Accelerator Physics 2013
Duke University
Chapter 9: RF Linear Accelerators and RF Cavities
Todd Satogata (Jefferson Lab) / satogata@jlab.org
Waldo MacKay (BNL, retired) / waldo@bnl.gov
Timofey Zolkin (FNAL/Chicago) / t.v.zolkin@gmail.com
http://www.toddsatogata.net/2013-USPAS
T. Satogata / January 2013
USPAS Accelerator Physics
1
2
RF Concepts and Design
§  Much of RF is really a review of graduate-level E&M
§  See, e.g., J.D. Jackson, “Classical Electrodynamics”
§  The beginning of this lecture is hopefully review
•  But it’s still important so we’ll go through it
•  Includes some comments about electromagnetic polarization
§  We’ll get to interesting applications later in the lecture
•  (or tomorrow)
§  Particularly important are cylindrical waveguides and
cylindrical RF cavities
•  Will find transverse boundary conditions are typically roots of
Bessel functions
•  TM (transverse magnetic) and TE (transverse electric) modes
•  RF concepts (shunt impedance, quality factor, resistive losses)
3
9.1: Maxwell’s Equations
Electric charge density
Electric current density
4
Constitutive Relations and Ohm’s Law
� = εE
� = εr ε0 E
�
D
ε0 = 8.85 × 10
−12
� : Electric displacement field
D
� : Electric field
E
C
N · m2
1
ε0 µ 0 = 2
c
ε : Permittivity
� = µH
� = µr µ0 H
�
B
µ0 = 4π × 10−7
� : Magnetic field
B
� : Magnetizing field
H
ε : Permeability
N · s2
C2
�
J� = σ E
σ : conductivity
5
Boundary Conditions
§  Boundary conditions on fields at the surface between two
media depend on the surface charge and current
densities:
§ 
E� and B⊥ are continuous
§ 
D⊥
changes by the surface charge density
ρs (scalar)
§ 
H�
changes by the surface current density
J�s
�1 − H
� 2 ) = J�s
n̂ × (H
§  C&M Chapter 9: no dielectric or magnetic materials
µ = µ0
� = �0
6
Wave Equations and Symmetry
§  Taking the curl of each curl equation and using the identity
� × (∇
� × E)
� = ∇(
� ∇
� · E)
� − ∇2 E
� = −∇2 E
�
∇
then gives us two identical wave equations
2�
�
E
∂
E
∂
2�
∇ E = µσ
+ µ� 2
∂t
∂t
2�
�
H
∂
H
∂
2�
∇ H = µσ
+ µ� 2
∂t
∂t
These linear wave equations reflect the deep symmetry
between electric and magnetic fields. Harmonic solutions:
� r, t) = Ψ̂(�r)e−iωt
Ψ(�
⇒
∇2 Ψ̂ = Γ2 Ψ̂
� = E,
� H
�
for Ψ
where Γ2 = (−µ�ω 2 + iµσω)
7
Conductivity and Skin Depth
§  For high conductivity, σ � ω� , charges move freely
enough to keep electric field lines perpendicular to surface
(RF oscillations are adiabatic vs movement of charges)
7
−1
−1
§  Copper: σ ≈ 6 × 10 Ω m
§  Conductivity condition holds for very high frequencies (1015 Hz)
§  For this condition
Γ≈
�
µωσ
(1 + i)
2
§  Inside the conductor, the fields drop off exponentially
Ψ̂(z) = Ψ(0) e−z/δ
skin depth
δ≡
�
e.g. for Copper
2
µωσ
8
Surface Resistance and Power Losses
§  There is still non-zero power loss for finite resistivity
§  Surface resistance: resistance to current flow per unit area
�
µω
1
surface resistance
Rs ≡
=
2σ
σδ
�
Rs
surface power loss
�Ploss � =
|H� |2 dS
2 S
§  This isn’t just applicable to RF cavities, but to transmission
lines, power lines, waveguides, etc – anywhere that
electromagnetic fields are interacting with a resistive media
§  Deriving the power loss is part of your homework!
§  We’ll talk more about transmission lines and waveguides
after one brief clarification…
9
Anomalous Skin Effect
§  Conductivity really depends on the frequency ω and mean
time between electron interactions τ (or inverse temperature)
σ0
σ(ω) =
(1 + iωτ )
§  Classical limit is ωτ → 0 , adiabatic field wrt electron interactions
� no longer applies since electrons
§  For non-classical limit, J� = σ E
see changing fields between single interactions
Anomalous skin effect
Mean free path
Asymptotic value
Normal skin effect
Skin depth
Reuter and Sondheimer, Proc. Roy. Soc. A195 (1948), from Calatroni SRF’11 Electrodynamics Tutorial
10
Plane Wave Properties
� x, t) = E
�0 e
0 ⇒ E(�
i�
k·�
x−iωt
i�
�
�
0 ⇒ B(�x, t) = B0 e k·�x−iωt
Generally F (z, t) =
�
∞
A(ω)ei(ωt−k(ω)z) dω
2π
(wave number)
λrf
ω
vph =
(phase velocity)
k
dω
vg =
(group velocity)
dk
k=
−∞
The source-free divergence Maxwell equations imply
�k · E
� 0 = �k · B
�0 = 0
k
so the fields are both transverse to the direction �
(This will be the ẑ direction in a lot of what’s to come)
Faraday’s law also implies that both fields are spatially
transverse to each other
�k × E
�0
�0
�k
nn̂
×
E
�0 =
n̂ ≡
index of refraction
B
=
k
ω
c
11
Standing Waves
i�
�
�
�
µ�ω ]E = 0 ⇒ E(�x, t) = E0 e k·�x−iωt
2
i�
k·�
x−iωt
�
�
�
µ�ω ]B = 0 ⇒ B(�x, t) = B0 e
2
Note that Maxwell’s equations are linear, so any linear
combination of magnetic/electric fields is also a solution.
Thus a standing wave solution is also acceptable, where
there are two plane waves moving in opposite directions:
� x, t) = E
�0
E(�
� x, t) = B
�0
B(�
�
�
e
e
i�
k·�
x−iωt
i�
k·�
x−iωt
+e
+e
−i�
k·�
x−iωt
−i�
k·�
x−iωt
�
�
12
Polarization
i�
�
�
�
µ�ω ]E = 0 ⇒ E(�x, t) = E0 e k·�x−iωt
2
i�
k·�
x−iωt
�
�
�
µ�ω ]B = 0 ⇒ B(�x, t) = B0 e
2
� fields are transverse, they can still
� and B
§  As long as the E
have different transverse components.
§  So our description of these fields is also incomplete until we
specify the transverse components at all locations in space
�
§  This is equivalent to an uncertainty in phase of rotation of E
� around the wave vector �k .
and B
§  The identification of this transverse field coordinate basis
defines the polarization of the field.
13
Linear Polarization
� = 0 ⇒ E(�
� x, t) = E
� 0 ei�k·�x−iωt
[∇2 + µ�ω 2 ]E
i�
�
�
�
[∇ + µ�ω ]B = 0 ⇒ B(�x, t) = B0 e k·�x−iωt
2
2
� and B
� transverse directions are
§  So, for example, if E
constant and do not change through the plane wave, the
wave is said to be linearly polarized.
E/B field oscillations
are IN phase
Fields each stay in one plane
14
Circular Polarization
� x, t) = E
�0
E(�
� x, t) = B
�0
B(�
�
�
e
i�
k·�
x−iωt
e
i�
k·�
x−iωt
+e
i�
k·�
x−iωt+π/2
+e
i�
k·�
x−iωt+π/2
�
�
� and B
� transverse directions vary with time, they can
§  If E
appear as two plane waves traveling out of phase. This phase
difference is 90 degrees for circular polarization.
Two equal-amplitude linearly
polarized plane waves
90 degrees out of phase
15
9.2: Cylindrical Waveguides
§  Consider a cylindrical waveguide, radius a, in z direction
� = E(r,
� θ)ei(ωt−kg z)
E
§  kg can be imaginary for attenuation down the guide
§  We’ll find constraints on this cutoff wave number
§  Maxwell in cylindrical coordinates (math happens) gives
(and the same for Hz )
k≡
�ω �
c
(free space wave number)
kc2 ≡ k 2 − kg2
16
Transverse Electromagnetic (TEM) Modes
§  Various subsets of solutions are interesting
§  For example, the Ez , Hz = 0 nontrivial solutions require
kc2 = k 2 − kg2 = 0
§  The wave number of the guide matches that of free space
§  Wave propagation is similar to that of free space
§  This is a TEM (transverse electromagnetic) mode
•  Requires multiple separate conductors for separate potentials
or free space
•  Sometimes similar to polarization pictures we had before
Coaxial
cable
17
Transverse Magnetic (TM) Modes
§  Maxwell’s equations are linear so superposition applies
§ 
§ 
§ 
§ 
We can break all fields down to TE and TM modes
TM: Transverse magnetic (Hz=0, Ez≠0)
TE: Transverse electric (Ez=0, Hz≠0)
TM provides particle energy gain or loss in z direction!
Separate variables Ez (r, θ) = R(r)Θ(θ)
Boundary conditions Ez (r = a) = Eθ (r = a) = 0
§  Maxwell’s equations in cylindrical coordinates give
2
�
2
�
d R 1 dR
n
2
+
+ kc − 2 R = 0
2
dr
r dr
r
d2 Θ
2
+
n
Θ=0
2
dθ
Bessel equation
SHO equation
=> n integer, real
18
TM Mode Solutions
§  The radial equation has solutions of Bessel functions of
first Jn(kcr) and second Nn(kcr) kind
§  Toss Nn since they diverge at r=0
§  Boundary conditions at r=a require Jn(kca)=0
§  This gives a constraint on kc
Xnj is the jth nonzero root of Jn
�
�
Xnj
⇒ Ez (r, θ, t) = (C1 cos nθ + C2 sin nθ)Jn
r ei(ωt−kg z)
a
kc = Xnj /a
where
§  This mode of this field is commonly known as the TMnj mode
§  The first index corresponds to theta periodicity, while the
second corresponds to number of radial Bessel nodes
§  TM01 is the usual fundamental accelerating mode
19
TM Mode Visualizations
⇒ Ez (r, θ, t) = (C1 cos nθ + C2 sin nθ)Jn
�
�
Xnj
r ei(ωt−kg z)
a
§  Java app visualizations are also linked to the class website
§  For example, http://www.falstad.com/emwave2 (TM visualization)
•  Scroll down to Circular Modes 1 or 2 in first pulldown
20
Cutoff Frequency
§  It’s easy to see that there is a minimum frequency for our
waveguide that obeys the boundary conditions
§  Modes at lower frequencies will suffer resistance
§  Dispersion relation in terms of radial boundary condition zero
�
�2
� ω �2
Xnj
k 2 − kg2 =
− kg2 = kc2 =
c
a
§  A propagating wave must have real group velocity kg2 > 0
§  This gives the expected lower bound on the frequency
ω
Xnj
≥
c
a
⇒
cutoﬀ frequency ωc =
cXnj
a
§  Wavelength of lowest cutoff frequency for j=1 is
λc =
2π
≈ 2.61a
kc
21
Dispersion or Brillouin Diagram
Propagating frequencies
ω > ωc
αph > 45◦ , vph > c
•  Circular waveguides above cutoff have phase velocity >c
•  Cannot easily be used for particle acceleration over long distances
22
Iris-Loaded Waveguides
§  Solution: Add impedance by varying cylinder radius
§  This changes the dispersion condition by loading the waveguide
23
Cylindrical RF Cavities
§  What happens if we close two ends of a waveguide?
§  With the correct length corresponding to the longitudinal
wavelength, we can produce longitudinal standing waves
§  Can produce long-standing waves over a full linac
§  Modes have another index for longitudinal periodicity
TM010 mode
Simplified “tuna fish can”
RF cavity model
24
Boundary Conditions
§  Additional boundary conditions at end-cap conductors at
z=0 and z=l
25
TEnmj Modes
Ez = 0 everywhere
Longitudinal
periodicity
�
Xnj
is jth root of Jn�
fnmj
c
=
2π
�
�
�
Xnj
a
�2
+
� mπ �2
l
26
TMnmj Modes
Hz = 0 everywhere
Ez (r, θ, z) ∼ Jn (kc r)(C1 cos nθ + C2 sin nθ) cos
� mπz �
l
Longitudinal
periodicity
Jn (kc a) = 0
fnmj
c
=
2π
��
Xnj
a
�2
+
Xnj is the jth root of Jn
� mπ �2
l
TM010 is (again) the preferred acceleration mode
27
Equivalent Circuit
§  A TM010 cavity looks much like a
lumped LRC circuit
§  Ends are capacitive
§  Stored magnetic energy is inductive
§  Currents move over resistive walls
§  E, H fields 90 degrees out of phase
§  Stored energy
C&M 9.82
28
TM010 Average Power Loss
ends
sides
§  Compare to stored energy
§  Both vary like square of field, square of Bessel root
§  Dimensional area vs dimensional volume in stored energy
�0 2 2
U = E0 πa l[J1 (X01 )]2
2
29
Quality Factor
§  Ratio of average stored energy to average power lost (or
energy dissipated) during one RF cycle
§ 
§ 
§ 
§ 
How many cycles does it take to dissipate its energy?
High Q: nondissipative resonator
Copper RF: Q~103 to 106
SRF: Q up to 1011
Uω
Q=
�Ploss �
U (t) = U0 e−ω0 t/Q
Pillbox cavity
al
Q=
δ(a + l)
30
Niobium Cavity SRF “Q Slope” Problem
31

USPAS Accelerator Physics 2013 Duke University

Transcription

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