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### Accelerator PhysicsTopic IAcceleration

Joseph Bisognano

Synchrotron Radiation Center

University of Wisconsin

J. J. Bisognano

Relativity

J. J. Bisognano

Maxwell’s Equations

J. J. Bisognano

Propagation in Conductors

J. J. Bisognano

Free Space Propagation

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Conductive Propagation

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Boundary Conditions

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AC Resistance

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Cylindrical Waveguides

- Assume a cylindrical system with axis z
- For the electric field we have
- And likewise for the magnetic field

J. J. Bisognano

Solving for Etangential

J. J. Bisognano

Maxwell’s equations then imply (k=/c)

J. J. Bisognano

All this implies that E0zandB0z tell it all with their equations

- For simple waveguides, there are separate solutions with one or other zero (TM or TE)
- For complicated geometries (periodic structures, dielectric boundaries), can be hybrid modes

J. J. Bisognano

a TE mode Example

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Circular Waveguide TEm,nModes

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Circular Waveguide TEm,nModes

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Circular Waveguide TMm,nModes

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Circular Waveguide Modes

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Energy Change of Wall Movement

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Lorentz Theorem

- Let and be two distinct solutions to Maxwell’s equations, but at the same frequency
- Start with the expression

J. J. Bisognano

Vector Arithmetic

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Using curl relations for non-tensor m, e one can show that expression is zero

- So, in particular, for waveguide junctions with an isotropic medium we have

S2

S3

S1

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S-matrix

- Let apamplitude of incident electric field normalize so that ap2 = 2(incident power) and bp2 = 2(scattered power)

J. J. Bisognano

Implication of Lorentz Theorem

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Unitarity of S-matrix

- Dissipated power P is given by
- For a lossless junction and arbitrary this implies

J. J. Bisognano

Symmetrical Two-Port Junction

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Power Flow

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Power Flow/cont.

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At Resonance

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Shunt Impedance

- Consider a cavity with a longitudinal electric field along the particle trajectory
- Following P. Wilson

z2

z1

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Shunt Impedance/cont

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Shunt Impedance/cont.

- Define
- where P is the power dissipated in the wall (the term)
- From the analysis of the coupling “b”
- where is the generator power

J. J. Bisognano

Beam Loading

- When a point charge is accelerated by a cavity, the loss of cavity field energy can be described by a charge induced field partially canceling the existing field
- By superposition, when a point charge crosses an empty cavity, a beam induced voltage appears
- To fully describe acceleration, we need to include this voltage
- Consider a cavity with an excitation V and a stored energy
- What is ?

J. J. Bisognano

Beam Loading/cont.

- Let a charge pass through the cavity inducing and experiencing on itself.
- Assume a relative phase
- Let charge be bend around for another pass after a phase delay of

J. J. Bisognano

Beam Loading/cont.

- With negligible loss
- But particle loses
- Since q is arbitrary, e =0 and

J. J. Bisognano

Beam Loading/cont.

- Note: we have same constant (R/Q) determining both required power and charge-cavity coupling

J. J. Bisognano

Summary of Beam Loading

- References: Microwave Circuits (Altman); HE Electron Linacs (Wilson, 1981 Fermilab Summer School)

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Vector Algebra

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E.g, assume q=j

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Summary of Scaling

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More Multicell Cavities

- Given a solution of a single cell cavity, one can consider the coupling of multiple cells in that mode by an expansion where
- the expansion coefficients give the strength of excitation of each cell in that mode
- coupling comes from perturbation of purely conductive boundary by holes communicating field between cells
- This is a recondite subject, with all sorts of dangers from “conditional” covergence of Fourier series; see Slater (and Gluckstern) for a complete picture of this

J. J. Bisognano

Periodic Structures

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Expansion Equations

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Holes

- But if there are holes in the cavity talking to neighboring cells, we have
- E.g., Bethe says

Tangential electric

field

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Coupled Equations

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Result of Floquet Theorem that solutions of differential equations with periodic coefficients have form of periodic function times exp(jb0z)

- Phase velocities less than c particle acceleration possible
- From Slater, RMP 20,473

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Floquet Theorem

- For example, for a disk loaded circular cylindrical structure, the TM01 is of the form

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Field Relations for Cylindrical Systems

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Field Relations for Cylindrical Systems

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Integrated Force at v=c

- Let be longitudinal force seen by a particle. Consider a trajectory z=vt, r=r0. The integrated force is then=
- Only q=/v contributes;i.e.bn=/v

J. J. Bisognano

E.g., a TM Mode

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Force on Relativistic Particle

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Panofsky Wenzel Theorem

- Pure TE mode doesn’t kick; pure TM mode, as in previous example, behaves to cancel denominator, so falls off as -2 ; hybrid modes don’t cancel denominator, so finite kick may obtain even when v=c

J. J. Bisognano

Superconductivity

- Basic mechanism
- Condensation of charge carriers into Cooper pairs, coupled by lattice vibrations
- Bandgap arises, limiting response to small perturbations (e.g., scattering)
- No DC resistance
- At temperatures above 0 K, some of the Cooper pairs are “ionized”
- But for DC, these ionized pairs are “shorted out” and bulk resistance remains zero

J. J. Bisognano

RF Superconductivity

- But pairs exhibit some inertia to changing electromagnetic fields, and there are some residual AC fields (sort of a reactance)
- These residual fields can act on the ionized, normal conducting carriers and cause dissipation
- But it’s very small at microwave frequencies (getting worse as f2)
- At 1.3 GHz, copper has Rs~10 milli-ohm
- At 1.3 GHz, niobium has Rs~800 nano-ohm at 4.2 K
- At 1.3 GHz, niobium has Rs~15 nano-ohm at 2 K
- Q’s of 1010 vs. 104

J. J. Bisognano

CEBAF Cavity Assembly

J. J. Bisognano

Cavity Specifications

- Frequency 1497 MHz
- Nominal length 0.5 meters
- Gradient >5 MeV/m
- Accel Current 200mA 5 passes
- Number cells 5
- R/Q 480 ohms
- Nominal Q0 2.4·109
- Loaded QL 6.6 ·106

J. J. Bisognano

What’s QL ?

- Run in linac mode, essentially on crest, j = 0
- In storage rings, j 0 for longitudinal focusing
- Wall losses V2/R=(2.5 · 106volts)2 /(480W ·2.4·109 ) are 5.4 watts
- Power to beam: (200mA 5 passes)·(2.5 · 106 volts) is 2500 watts

J. J. Bisognano

Would Copper Work

- Typical Q is now 104
- Wall losses (2.5 · 106volts)2 /(480W ·104 ) are now 1.3 MW vs. beam power of 2500 watts
- Some optimizations could yields “2’s” of improvement
- More importantly, SRF losses are at 2 K, which requires cryogenic refrigeration.
- Efficiencies are order 10-3
- So, 5 watts at 2 K is 5 kW at room temperature, but still factor of 100 to the good

J. J. Bisognano

Higher Order Modes

- There are higher order modes, which can be excited by the beam
- These can generate wall losses, and fields can act on beam to generate destructive collective effects
- First question is whether wall losses are large or small compared to fundamental wall losses

J. J. Bisognano

Loss Factor for HOMs

- Bunch spectrum extends out to
- For a typical 1 ps linac bunch,
- For for a 1.5 GHz fundamental, there are many tens of longitudinal HOMs for the beam to couple; coupling is weaker than to fundamental because of more rapid temporal and spatial variation
- From codes,

J. J. Bisognano

Power Estimates

- For 1 mA CEBAF 5-pass beam, with 0.5 pC/superbunch, we have only10mW of loss
- But in, say FEL application, with 100 pC bunches at 5mA, we have 10 watts in the wall, more than the power dissipated by fundamental!
- So extraction of HOM power is issue for high current applications with short bunches

J. J. Bisognano

Couplers and Kicks

- Waveguide couplers break cylindrical symmetry
- Result is that the nominal TM01mode now has TE content and m0; hybridized
- By Panofsky-Wenzel, introduces steering and skew quad fields that require compensation

J. J. Bisognano

Homework: Topic I

- From Accelerator Physics, S.Y. Lee
- Reading: Chapter 3, VIII, p. 352 & onward
- Problems:3.8.1 (really table 3.10), 3.8.2, 3.8.4
- The next generation CEBAF cavity can achieve 20 MV/m gradients with a 7 cell structure. a) Assuming R/Q is 7/5 higher, what would be the heat load generated per cavity if the Q is unchanged? b) How much higher a Q is necessary to main the heat load at the levels from the first generation cavity discussed in the lecture?
- Consider a single cell RF system operating at 500 MHz with an effective length of 0.3 meters, which is to be operated at a gradient of 2 MV/m with beam current of 100 mA. Assume the R/Q of the cavity is 100 ohms and that the Q from resistive wall losses is 40,000.

A) Calculate the optimal coupling coefficient for powering the cavity with the 100 a mA beam passing through it.

B) Calculate the power necessary with 100 mA beam to power the cavity.

C) Calculate the power received by the beam and the power dissipated in wall losses.

D) Describe what will happen to the power requirements and reflected power from the cavity if the beam is lost, but the feedback system attempts to maintain the 2 MV/meter gradient.

- A CEBAF problem: Use the nominal specs given in lecture
- a) For a 1 mA at 5 MV/m calculate total power and reflected power on resonance and 10 degrees off resonance with bunch on crest.
- b) If this beam is allowed to pass for a second time through the cavity for energy recovery at 170 degrees off accelerating crest, simultaneously with the first pass beam on crest, calculate the total and reflected power when the cavity is tuned on resonance.

J. J. Bisognano

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