D B. An Introduction to the Physics and Technology of e+e- Linear Colliders. Lecture 2: Main Linac. Peter (PT) Tenenbaum (SLAC)

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1 An Introduction to the Physics and Technology of e+e- Linear Colliders Lecture : Main Linac Peter (PT) Tenenbaum (SLAC) Nick Walker DESY DESY Summer Student Lecture USPAS Santa Barbara, CA, st June, 003 July 00 Maxwell s Equations in MKSA units D = ρ B = 0 D H = J + t B E = t Where: B= µ H D= ε E 1

2 Wave Equations for E and H Make use of a vector calculus identity: A A A ( ) ( ) Apply to the Electric field curl equation, and assume no charges or currents are present: = B ( E) ( E) E ero (no charges present) Reverse Order of Derivatives t E = B t ( ) Wave Equations for E and H () Replace B with µh, and replce curl of B via Maxwell s magnetic curl equation: E µ ( H) t D J = µ + t t = And similarly for the magnetic field: E µε t E = 0 H µε t H = 0

3 Free-Space Solution to the Wave Eqn Let us define the -axis parallel to the direction of propagation, and solve the wave equation as a superposition of travelling plane waves: E = E0 exp i t, H H0 exp t ( ω k) = i( ω k) Absence of boundaries (isotropic, homogeneous vacuum) requires that vectors E 0 and H 0 be constant for all time and space Applying the wave equation to the electric field yields: E µε E exp i ωt k t { 0 ( ) } = Free-Space Solution to the Wave Eqn () Since E 0 is constant, both the time derivative and Laplacian operate only on the complex exponential. After cancellation of constant factors, we find: ω ω 1 k = µεω = = k k µε i.e., a plane wave with a phase velocity and a group velocity = 1/ µε. In vacuum, then, a wave which propagates at v=c. 3

4 Free-Space Solution to the Wave Eqn (3) Go back and apply Maxwell s equations to this solution: E E = = ike0, exp i ( ωt k) = 0 k=0: trivial solution, no wave! One of these (k or E 0, ) must be ero! E 0, = 0: electric field accelerates beam transverse to direction of wave propagation! If beam is accelerated in x while wave moves in, then wave will first accelerate, then decelerate, the beam! No good for acceleration! Bounded Solution to the Wave Equation Next simplest solution: apply some kind of boundaries in x and y, so that non-ero x and y derivatives of the electric field can cancel derivative (ie, permits non-ero E 0, while still obeying Maxwell). Try a conducting pipe of radius b, oriented along axis: y b x 4

5 Circular Waveguide Solution Once again, E = E0 exp i t, H H0 exp t ( ω k) = i( ω k) This time vectors E 0 and H 0 are functions of transverse coordinates x and y (or r and θ) but not or t. Thus we can simplify some derivatives: = ik, = k = + iω, = ω t t Circular Waveguide Boundary Conditions At the boundary (r=b), the normal component of B and the tangential component of E are continuous. If the conductor is perfect, then within the conductor the electric and magnetic field are identically ero. Thus at r=b, H r, E, and E θ 0. Since E θ =0, the θ component of the magnetic curl equation must go to ero. In total, then: E = θ 0 E = 0 H r = 0 H = 0 r = b 5

6 Circular Waveguide: Longitudinal Fields Rewrite the component of the wave equation: E k E + µεω E = 0 Now we can:.cancel common complex exponent factor from E.Define k c µεω -k + = E k E 0, c 0, 0 ( ) cos( ) E = a J k r nθ + θ 0, n n c n n= 0 Applying Boundary Conditions 1. n is an integer because field is single-valued -- cos[n(θ+π)] = cos nθ iff n is integer.. To force E r=b, k c b = np, where np is the p th ero of J n. Note that k c >0 as a result, and ( ) cos( ) E = a J k r nθ + θ 0, np n c, np np p= 1 n= 0 k Note that: cnp, np = = µεω ω b cnp, µε b 1 np k So k=0 must correspond to a nonero ω: = Cutoff Frequency 6

7 Physical Meaning of Cutoff Frequency ω > ω c : Real-valued k is possible, wave is complex exponential ω < ω c : k is imaginary, wave falls off exponentially with (can t propagate) -- evanescent wave! Calculate phase and group velocity of the wave: v gr ω 1 ω ωc = = < c k µε ω v ph ω 1 ωc = = + > c k µε k Since particle moves < c, wave will overtake particle -- no acceleration possible! TE and TM modes The longitudinal electric field vector is given by: ( ) cos( ) E = a J k r nθ + θ 0, np n c, np np p= 1 n= 0 A similar solution is available for the magnetic field vector: ( ) cos( ) H = f J k r uθ + θ 0, uv u c, uv uv v= 1 u= 0 In this case the BC s require that dh/dr 0, so different cutoff frequency from electric field. 7

8 TE and TM modes () Since J and J have different eroes, in general waves with nonero H 0, have different phase/group velocities and cutoff frequencies from waves with nonero E 0,. Thus, in general a wave with a given phase and group velocity cannot have both a longitudinal electric field and a longitudinal magnetic field! Waves with H 0, 0 are called TM (transverse magnetic) modes; waves with E 0, 0 are called TE (transverse electric) modes. Usually the modes are referred to with their index numbers, TE uv or TM np. Note: TM 01 mode has nonero E, E r, H θ components only! TE and TM Modes (3) Cutoff Frequencies for a few modes in circular waveguide: 8

9 So where are we? Considered waves in free space and in regular cylindrical waveguide Waves in free space are no good for acceleration electric field momentum vector Waves in cylindrical waveguide also no good can get electric field parallel to momentum vector, but electric field phase velocity > c, wave overtakes particle Can force modes in waveguide to work by separating beam from wave after < 1 wavelength requires addition of boundary conditions in!!!! How about perfectly-conducting end walls? Waves in a Conducting Hollow Cylinder In addition to existing BC s, we find:.at =0 and =L, E r and E θ 0 (E normal to any conducting boundary). At =0 and =L, H 0 (H tangential to any conducting boundary).dh θ /d = dh r /d = 0 at ends 9

10 TM 010 (Standing Wave) Solution If we consider a rightward (+)-propagating TM 01 wave and a leftwardpropagating wave with the same frequency and amplitude, the nonero field vectors become: ( c ) E = J k r cos( k)exp( iωt) 0,01 k Er = J' 0 ( kc,01r) sin( k)exp( iωt) kc,01 iωε Hθ = J' 0 ( kc,01r) cos( k)exp( iωt) k c,01 TM 010 () The =0 and =L boundary conditions are automatically satisfied iff: kl = jπ, j This wave is usually called a TM npj mode, and only certain discrete frequencies ω npj 1 np jπ = + µε b L are permitted in this mode. 10

11 Properties of TM npj Modes 1. Since J n (0) 0 iff n=0, n=0 modes are optimal for acceleration (otherwise you have to put your particle far from the center of the cavity!). The rightward-travelling component of the standing wave will always accelerate the beam (assuming an appropriate choice of phase w.r.t. the beam arrival time). 3. For most combinations of parameters, the leftward-travelling component of the standing wave will decelerate the beam if j>0, hence j=0 is optimal for real-world applications. 4. For a TM 0p0 mode cavity, if the field is given by E exp = E0 ( iωt) the maximum energy gain achievable by a particle is E 0 LT, where sin ( ψ / ) and ψ = the phase variation of the wave during the T ψ / particle s passage (transit angle) Stored Energy in a Cavity; R/Q The stored energy in an AC electromagnetic field is given by: * ( ) E D = ε E0J0 kc,0pr cos ωt 1 U = E D + B H ( * * ) * ωε E 0 B H = µ J' 0 ( kc,0pr) sin ωt kc,0 p To simplify the problem, choose t=0 so all energy is in electric field! 11

12 Stored Energy and R/Q () When we integrate the electric field energy over the volume, we find a relationship between the max acceleration V E 0 LT and the stored energy: V T L = U πε J ( ) b 1 0p If we introduce the impedance of the medium Z (µ/ε) (vacuum: 377Ω) and perform a few simplifications and substitutions, then: ω V T L R 0p0 = Z U b Q ( R over Q or R upon Q ) -- a purely geometric quantity which relates stored energy, accelerating voltage, and frequency Stored Energy and R/Q (3) With appropriate substitutions, we can rewrite R/Q: R sin ( ψ /) Z c Q = 0 p ψ / µε R/Q is maximied when ψ ~ 134. This is less useful than it sounds, but in general the stored energy can be minimied by optimiing the transit angle and increasing the frequency (recall that R/Q = V /ωu). Furthermore, since 0p increases monotonically with p, selection of p=1 will also minimie the stored energy required for a given voltage, and we will assume such a selection henceforth. 1

13 Extending to Finite Conductivity What happens when there is some small resistance to the surface of the cavity? In this case the electric and magnetic fields in the conductor are nonero. Since the tangential magnetic field is continuous at the boundary, the tangential field no longer vanishes thereat. After a great deal of mathematics, the ohmic losses (product of the current density and the electric field in the conductor) can be expressed in terms of the tangential magnetic field at the surface of the cavity: dp da = µω c σ µ c permeability of conductor ( µ 0 ) H S σ conductivity of conductor (1/Ωm) Finite Conductivity () We can usefully rewrite the power dissipation equation: dp = R S ( ω) H S da R R S µ cω σ υ [ GH] [ ] where R S = frequency-dependent conductivity ( surface resistance ) 5 TC S [ Ω] 9 10 exp α SC + RRES T K T Normal conductors Superconductors α SC ~1.9, T C = 9. K (Niobium), R RES ~ 10-8 Ω 13

14 Wall Q of a Cavity If we return to our right-circular cylinder example and integrate the ohmic losses over the endcaps and the barrel, we arrive (after some Bessel function tricks): π ER 0 S Pcav = J 1 ( 01) b( L+ b) P, U both ~ E Z P ~ U Define: Q W ωu 01ZL = P R ( L+ b) S Wall Q, fractional loss of stored energy into walls in π x RF period. Shunt Impedance Now that we have power loss per unit energy, and voltage per unit energy, we can compute relationship between available input power and achievable voltage (voltage at which input power = wall losses). This relationship has ohmic units, hence is called Shunt Impedance: R cav = Qw Q V = P Z = R c µε c S R ( ψ ) sin / Treat the cavity like a resistor in phys lab! P = V /R (careful of factor of -- AC power)! ψ µε 01 14

Motivation for multi-celled RF Cavities Consider a single-celled RF cavity with shunt impedance R. A voltage V can be maintained if power P = V /R is supplied to the cavity.

15 Motivation for multi-celled RF Cavities Consider a single-celled RF cavity with shunt impedance R. A voltage V can be maintained if power P = V /R is supplied to the cavity. Alternately, if n cavities are available, each one can supply a voltage of V/n at a cost in power of n(v/n) /R = 1/n of the power needed to get there in a single cavity. Disadvantage: less stored energy in each cavity means that the beam current that we can accelerate is also reduced. Also, we need more space for the accelerating elements. However, for 100 GeV and a single 1 GH cavity, the input power is around 1.5 x watts. California s peak consumption is around 5 x watts. Apparently multi-celled cavities are a necessity if a voltage of 100 GV is to be achieved in a single pass. RF Accelerating Structures Consider an infinite cylindrical waveguide (radius b) divided into RF cavities by plates with thickness h and -spacing d. In order to allow the beam to pass thru, each plate (or disc) has a hole (iris) of radius a. h 15

16 RF Accelerating Structures () What do we know about the electric field in an infinite periodic structure? Floquet s theorem tells us: E r,, d, t E r,,, t exp d ik ( θ + ) = ( θ ) ( α + ) Also, as a 0, expect the solution to go asymptotically to single-cavity form. E, H E, H Let us define: 1 1 Fields in absence of endcap holes Fields with endcap holes included * * And consider the da ( E1 H E H1) expression: (Yuk) RF Accelerating Structures (3) * * * * ( ) ( ) * * * * ( 1) 1 ( ) ( 1) 1 ( ) d A E H E H = dvol E H E H = dvol H E E H + E H H E H 1 µ = iµω1 H1, etc. t After a bunch of such substitutions, collection of terms, and asserting that perturbed and unperturbed fields are ~ equal: * * * * da E 1 H E H 1 = i ω ω dvol µ H 1 H 1 + εe 1 E 1, or ( ) ( ) ( ) 1 * * i d A ( E ) 1 H E H 1 ( ω ω ) = 1 U 1 16

17 RF Accelerating Structures (4) i d A E H E H ( ω ω 1) = U 1 * da ( E ) 1 H = 0 H 1 E E1 Except in the vicinity of the hole * * ( ) 1 1 Electric field is normal to the surface so parallel do the differential area vector Is known at all points from the TM 010 solution In principle, it s possible to calculate the frequency of the system (w.r.t. the unperturbed TM 010 frequency) after a lot of arithmetic and calculus ω Accelerating Structure Dispersion Relation a 1 1 exp / cos 3 01 = + 01 b µε 3π J1 ( 01) b d ( h a) ( k d) 3 01 vph = 1+ 1 exp 01h/ a cos kd kb 3π J µε 1 ( 01) b d a a ( ) ( ) 3 01 vgr = exp 01h/ a sin kd b µε 3π J1 ( 01) b ( ) ( ) What does it mean? Consider a system with b=10 cm:.regular waveguide, or.cavity with L = 10 cm, or.accelerator Structure with d = 10 cm, a = 5 cm 17

18 Disc Loaded Wave Guides v ph = c v gr = c 1.33 GH k d = 158 Finite-Length Disc-Loaded Waveguides Previous estimates of the dispersion relation relied upon the assumption that the structure is essentially infinite. In real life, a structure with N cell cells acts like an N cell -coupled oscillator system, with N cell oscillation modes. The modes are equally spaced in k, as a consequence the frequencies are closely-spaced at the upper and lower limits 18

19 v gr, v ph and k d What is the physical meaning of k d in the preceding equations? If we return to Floquet s theorem and assume a simple sinusoidal variation in time, then E( r, θ, + d)exp( iωt) = E( r, θ, )exp( ikd)exp( iωt) Er (, θ, + d)exp[ i( ωt kd)] = Er (, θ, )exp( iωt) So a particle with velocity ω/k encounters the same phase of RF in each cell the phase velocity ω/k is the velocity of the synchronous phase. At a given time t, then, k d is the phase difference between neighboring cells. Combine those facts, and find k d is synonymous with the transit angle ψ (though usually called the phase advance per cell in this context). Finally, we assert that v gr is the velocity of energy propagation thru the structure. k d values greater than π So far, nothing we have talked about precludes a value of k d > π. Since the dispersion relation is sinusoidal, if we allow k d > π then the relation becomes multi-valued (many values of k correspond to each value of ω) All k s yield same v gr (with varying signs) Only one has v ph = c. What s up with this? 19

20 Space Harmonics Remember that E() is a stepwise function (ie, E() is constant over 1 cell, then steps to a new value in the next one ) We can Fourier-decompose it into components with different wavelengths -- but they all oscillate with the same frequency The various Fourier components have different phase velocities! Standard interpretation:.for a given ω, the corresponding values of k d the various Fourier components of the accelerating field (space harmonics).only the space harmonic at the beam velocity provides net acceleration.the effective shunt impedance is reduced because some fraction of RF power (~0-35%) supports non-synchronous harmonics Travelling-Wave Accelerators 0

21 Power Flow through a TW DLWG P 0 (W) For now, neglect the beam U () (J/m) P() (W) U (J) p w () (W/m) Power Flow () Conserve Energy: Net power flow must balance net change in stored energy, or: du '( ) + p ( ) dp( ) 0 w + = dt d In steady state: Definition of du '( ) = 0 dt Definition group velocity: P( ) = U'( ) vgr ( ) of Q: pw ( ) ωu = Q '( ) w dp ( ) U ( ) '( ) P ( p ) w ω ω = = = d Q v Q ( ) w gr w 1

22 Constant Impedance Structure If v gr is a constant along the structure, then: where ( ) = 0 α0 P P exp( ) α 0 ω Qv w gr Introduce the gradient G 0 dv/d and the normalied shunt impedance r l = dr struc /d to find ( ) G = α rp exp( α ) 0 0 l 0 0 Constant Impedance Structure () Define τ α 0 L (Field e-folding factor per structure) V rlp Total voltage: l 0 = 1 e τ τ τ Filling Time: t f L = = v gr Q w ω τ Advantage: All cells are identical (make a illion copies of 1 design) Disadvantage: Gradient much higher at front end of structure (bad for gradient-limited structures)

23 Constant Gradient Structure Goal: Create a structure such that, when wall losses are taken into account, gradient is (approximately) independent of length dp( ) ω U '( ) ωp( ) = w = = d Q v Q How? Recall: p ( ) If Q ~ constant and r l ~ constant, constant U constant gradient ( ) w gr w So dp/d, U, P()/v gr () must all be constant! Constant Gradient Structure () Let us define again τ such that P() = P 0 (1-e -τ ). Since dp/d is now a constant, dp P0 τ ωp( ) = ( 1 e ) = d L v ( ) Q P() decreases linearly with, therefore v gr () must do the same: v gr ( ) w gr τ ( 1 ) ω L e = τ Q 1 e w l ( ) V = rlp e τ 0 1 t f L = = v gr Q w ω τ 3

24 Standing Wave Accelerator Structures What happens if we supply input power to the input coupler and the output coupler of a TW structure? Input coupler power synchronous with beam, accelerates Output coupler power has phase velocity in opposite direction Backwards power accelerates in some cells, decelerates in others :: Backwards power is wasted, shunt impedance reduced by 50% Standing Wave Accelerator Structures () Exception: Structure that operates in π mode (180 /cell phase advance) In π mode, forwards and backwards waves have same phase velocity Naïve shunt impedance calculation is correct! Note that group velocity = 0 for π mode -- how s it get energy into the structure? 4

25 Filling a Standing-Wave Structure Recall that structure with N cell cells has N cell normal modes; each mode has its own resonant frequency ω, wave number k, all have about the same Q. When power in the π-mode frequency ω π hits the input coupler, all the other modes act like resonators driven off-frequency (they are!) Those other modes transport the power from the input coupler to the cells. For all practical purposes, it s valid to calculate as though input power at ω π fills the π-mode instantaneously and uniformly (ie, don t worry about causality!). Filling a Standing-Wave Structure () Want to trap a lot of energy in the structure, so the TW-style input coupler is no good (lets too much energy out). Use a coupler which lets almost no power through. Side-effect: lets almost no power in! Takes a long time to fill the structure. In a filled cavity with no input power, the ratio of power lost out the coupler to power lost into the walls β c (coupling coefficient) When the cavity is included, the effective ( loaded ) Q is lower: Q L Qw = 1 + β c 5

26 Filling a Standing-Wave Structure (3) Define the characteristic time of a standing wave structure: t c Q L Assuming a constant-amplitude of incoming RF (E in ), we find: βc t/ t ( 1 c Amplitude of emitted wave (RF power Ee = Ein e ) 1+ β leaking back out input coupler) c βc t/ t E = E ( 1 e c ) E 1+ β out in in c β Uc t tcpin e 1+ β c t/ t () = ( 1 ) c / R β c () ( 1 t t c Vt = e ) ωtp c in Q 1+ β c c Amplitue of emitted + reflected wave (destructive interference) Stored Energy Accelerating Voltage ω Superconducting Structures When would you use superconducting materials to make an accelerating structure? In a TW structure, input power is lost to the walls, the output coupler, and the beam. SC eliminates the first term but the others remain. For a structure with τ=0.6 (typical), about 70% of the RF power is lost into the walls. Recouping this is a nice idea but maybe not one that justifies the complexity and effort. Now consider a standing-wave structure: for 1 GH and Q w ~ 10 4, the e- folding time for stored energy is about 10 microseconds. For Q w ~ 10 9 the e-folding time is about 1 second. Because there s no output coupler, a SC SW structure is better than a normalconducting one by a factor of 10 5, not 3. 6

CERN Accelerator School. RF Cavities. Erk Jensen CERN BE-RF

CERN Accelerator School. RF Cavities. Erk Jensen CERN BE-RF CERN Accelerator School RF Cavities Erk Jensen CERN BE-RF CERN Accelerator School, Varna 010 - "Introduction to Accelerator Physics" What is a cavity? 3-Sept-010 CAS Varna/Bulgaria 010- RF Cavities Lorentz