ARB Technical Note - raft - 1/7/97 Geometrical and RF Considerations for All Beam Collisions via Crab- Crossing Frank Zimmermann and avid H Whittum In this note we sketch the geometry for a crab-crossing scheme that achieves all beam collisions from a matrix linac, without temporal or spatial beam combining RF scalings are noted and appear not to present any challenges Clearly a more difficult idea should be found Geometry The matrix linac and one final focus scheme have been described in previous notes Here we consider the idea sketched in Fig1, of colliding beams at a crossing angle, θ The idea is that each bunch from N 5 parallel linacs, can be made to collide with each and every bunch of the opposing N 5 parallel linacs epth of focus, chromatic effects, and the effect of detector solenoidal field are issues; here we consider the zeroth order concerns: what angle is required for the matrix linac scheme, and what are rf requirements result for the crab cavities? nd Collision B L S θ θ 1st Collision FIGURE 1 Geometry for collision at crossing half-angle θ The geometrical relationships determining the required crossing angle are as follows: sin π θ = B S, (complementary angles) S = B +, (Theorem due to Pythagoras) L+ B=, (group delay in matrix transmission line) β g
Combining the first two relations, one finds S = sin θ, B= cot ( θ ) ( ) The third relation gives then the length travelled by the second bunch, to its collision point 1 L = cot ( θ) β g In order for the opposing bunch to arrive on time, we must have S = L, and thus we find 1 β g cot( θ)= csc ( θ), or tanθ = β g The current conceptual design for the matrix scheme uses β g, corresponding to θ 11 3 197rad For the sake of definiteness we fill in the other parameters, λ λ 8 164 mm, 4 β g 5 5 λ λ 41 8 mm 4 The duration of one pulse of the N 5 parallel beam lines corresponds to N βg 6 5λ 15λ 5 41 mm 7 14 ns RF Considerations To accomodate the crossing angle without loss of luminosity, one would like to insure head on collisions This is to say that one would like the head of the beam entering from the left, in Fig 1, to pass through the head and the tail of the beam entering from the right This is illustrated in Fig, although not to scale (we are interested in beams of transverse extent σx σy nm, and bunch lengths long in comparison σ z 3 µ m)
θ FIGURE (a) When beams cross at an angle with no head-to-tail offset, the head ot the first bunch fails to collide with the tail of the second bunch θ FIGURE (b) When beams cross at an angle with the proper head-to-tail offset, the head ot the first bunch collides with the head and the tail of the second bunch, and all beam slices in between To accomplish head-on collisions, we must rotate the beam in the crossing plane We can see this as follows Let us suppose the head of bunch #1 follows the trajectory r h1 = t( cos θ,sinθ ), while the tail follows at approximately the same angle, but at an offset r = t T cos θ,sinθ r, ( )( )+ t1 t1 and we offset the time coordinate by T, to be determined For the oncoming bunch, bunch #, we have r h = t( cos θ,sinθ ), and r t = ( t T) ( cos θ,sinθ )+ r t Evidently the two heads collide at r r h1 = h = t= To insure that the tail of bunch #1 collides with the head of bunch #, we must have, for some t, r = t T cos θ,sinθ r r t cos θ,sinθ, ( )( )+ = = ( ) t1 t1 h
and we define this time to be t=t, so that r t1 = T( cos θ,sinθ ) Similarly then tail # should collide with head #1, r = t T cos θ,sinθ r r t cos θ,sinθ, ( )( )+ = = ( ) t t h1 and r t = T( cos θ,sinθ ) Thus the tail trajectories are parameterized according to r t1 = ( t T)( cos θ,sinθ)+ T( cos θ,sinθ)= ([ t T] cos θ, tsinθ ), r = ( t T) ( cos θ,sinθ)+ r = t+ T cos θ, tsinθ t t ([ ] ) Evidently the tails collide at t = T Treating T as a parameter of beam slices, one can see that each segment of the beam from the left, encounters each segment of the beam from the right, colliding at a point on the dashed centerline in Fig (b), at a point displaced vertically in the figure by an amount depending on the displacement of the slice from the beam head The offset r ( T)= T( cos θ,sin θ) ( T, Tθ ) amounts to a bunch rotation in the plane of the collision, and to accomplish this, we may present to the beam, a time varying deflection, x = α T, followed by transport (eg, a drift) x = R1x = R1α T For this to work we must require that x Tθ, and this determines the time-varying kick in terms of the crossing angle, θ α R1 For example, for a m drift, and a mrad crossing angle, one has α 1 1 5 cm 1 At the same time we may relate α to RF parameters according to 1 1 1 1 ε α = = = = T x px px e V T mcγ mc γ t mc γ x mc γ x Considering a standing-wave cavity, this voltage gradient may be related to power dissipated using the relations
R V x = / ωu, P diss ω Q U V/ x = = Q R / Q w w [ ] Thus, in terms of the transverse shunt impedance R Qw R / Q, V/ x = R P diss Conversely we may express power dissipated in terms of our kick parameter α, = [ ] P diss = Eα, R where E is the full beam voltage, 5 TV Equivalent voltage for our example is Eα 5 MV/cm Naturally we become interested in transverse shunt impedance at this point Since the voltage required is looking a bit large, high gradient may be required, and thus short-wavelength, planar structures would be of interest 1 Let us consider the case of a rectangular geometry consisting of a pillbox of transverse dimensions a b, and length L In this case, T 8Lβ x, β ab where the transit angle factor is ( ) sin θ / T =, ( θ / ) the transit angle θ = β L, and wavenumbers are βx = π / a,βy = π / b, β = βx + βy We may simplify this to read, x sin ( θ / ) 16 β ( ) ab θ / β Maximum occurs for θ=13356, corresponding to Q 1 1 11 6 ab, 1+ a / b and for a 11 6 λ = b = λ/ (with λ the free-space wavelength) this may be expressed as 1 For the record, for a cylindrical pillbox of radius R, transit angle θ, sin ( θ / ) 39 1 ( θ / ) R avid H Whittum, ipole Mode eflection in the Subharmonic rive Experiment, ARB Technical Note 17
Wall quality factor takes the form 1 δ β β x y β = + + Q β a b L, w with the skin-depth δ = µ m/ f ( GHz) For our choice of aspect ratios this is λ Q w 181 δ The result for shunt impedance (for a single cavity) is then R Q R = w Z f 1 3/ 1 ( ) GHz MΩ / cm λδ If we couple N cavities together to be powered by a single feed, we may expect not quite N times the shunt impedance of a single cell (N/ if we do badly in the design) As an example, for a singe cell driven at 856 MHz, we find R 6 MΩ / cm If 1 cells were coupled into one single-feed standing wave structure, and derating the shunt impedance by a factor of, we would have 6MΩ/ cm, and 5 MV/cm P diss 1 1 1 MΩ/ cm MW, (S-Band) comfortably less than the output of a SLE d SLAC 545 This scheme is actually a bit wasteful since the fill time for the structure would be 16 µs ( Q w 15 1 4 ), much longer than the train of charge to be manipulated with the device At W-Band (9139 GHz) we find R 1 1 MΩ/ cm for one cell; for 1 cells, derated by a factor of two, we have 11 GΩ / cm Peak power required is then 6 kw, and the fill time is 9 ns ( Q w 6 1 3 ), a good figure for a beam with this structure More Work The effect of synchrotron radiation in the course of the deflection by the cavities remains to be checked; it is a concern due to the large kick angle Aperture considerations and other wakefields sold separately Not valid after last date stamped Use only as directed o not ingest