Estimates of local heating due to trapped modes in vacuum chamber

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1 Estimates of local heating due to trapped modes in vacuum chamber Gennady Stupakov SLAC National Accelerator Laboratory, Menlo Park, CA CERN, April 29, 2016

2 2 Motivation The motivation for this analysis is to try to estimate heating of the vacuum chamber due to possible trapped modes in the machine. The biggest danger is for the high-current scenario, with the current I = 1.45 A in bunches (how does the filling pattern affect the heating?). We derive how the heating scales with the parameters of the trapped mode and the beam. We model the central part of the IP region as a cylindrical beryllium pipe of radius a = 2 cm and length l = 50 cm.

3 3 Modes can be trapped in unexpected locations Trapped modes can be generated by various protrusions in the vacuum chamber (BPMs, flanges, bellows, etc). They can even be generated even by collimators! From Ref. 1 : thin iris collimator Flat collimator 1 S. Heifets et al. Review of impedance issues for B-factory, SLAC-PUB (2004).

4 Power deposited by the beam 4 Power deposited by the beam into an element with the longitudinal impedance Z(ω) P = 1 T0 2 ρ(nω 0 ) 2 Re Z(nω 0 ) where n= ρ(ω) = C 0 ρ(z)e iωz/c dz ρ(z) is the charge per unit length, C is the circumference of the ring, T 0 = C/c is the revolution period, ω 0 = 2π/T 0. For FCC, with the circumference of C = 100 km, we have ω 0 = 2π 3.0 khz and the sampling in the equation for P goes every 3 khz. Assuming a mode with characteristic frequency of ω r 2π 5 GHz, and the quality factor Q = we find that the resonance width ω r /Q is much larger than ω 0. We can use P 1 πt 0 0 dω ρ(ω) 2 Re Z(ω)

5 5 Z(ω) for trapped mode For a trapped mode we take the resonant impedance in the form Z(ω) = R/Q Q 1 + i(ω r /ω ω/ω r ) where ω r is the resonant frequency, R is the shunt impedance and Q is the quality factor. The loss factor for the resonant impedance is κ = Rω r /2Q. Consider a mode in a pipe of radius a = 2 cm and length l = 50 cm. The most dangerous modes are of type TM 0n0 ; for these modes an approximate value of R/Q is given by 2 R Q Z 2 sin 2 (j 0n l/2a) 0 j 0n j 0n l/2a where j 0n is the nth root of the Bessel function J 0, Z 0 = 377 Ohm. For TM 010 ω r = j 01c a = 2π 5.74 GHz 2 D. H. Whittum, Techniques and Concepts of High-Energy Physics, NATO Sci. Ser. C 534, 387 (1999).

6 6 Z(ω) for trapped mode For the parameters chosen above we obtain R Q 10 Ω For the quality factor due to resistive wall losses, for TM 010 there is a formula σ 1 Q = f(ghz) σ Cu 1 + a/l where σ is the wall conductivity and σ Cu = S/m is the copper conductivity at room temperature. For beryllium, σ Be = S/m and Q The width of the resonance is approximately ω ω r Q 0.4 MHz 3 D. H. Whittum, SLAC-PUB-8026 (1998).

7 Beam resonances Assume M 1 bunches in the ring distributed with the distance between the bunches s b. Note that M can be smaller than C/s b which means that the number of bunches in the ring is smaller than the harmonic number of the machine: this means a gap in the fill of length C Ms b. Assuming short bunches (σ z 2 4 mm): M 1 ρ(z) = Ne δ(s ns b ) n=0 with N the number of particles in the bunch, gives ρ(ω) 2 = N 2 e 2 sin2 (Ms b ω/2c) sin 2 (s b ω/2c) MN 2 e 2 2πc s b n= δ (ω nω) 7 ρ(ω) 2 is periodic function with the period Ω = 2πc/s b which is equal to the RF frequency if s b is equal to the distance between the RF buckets. The width of each peak of this function is Ω/M ω 0.

8 Resonant heating In the worst-case scenario one of these delta functions hits the resonant frequency ω r. This happens when ω r = nω + ω (ω r is close to an integer of the RF frequency). We then find for the heating power P 2MN2 e 2 R s c/s b T (2Q ω/ω r ) 2 = 2NeI c s b ( ) Rs Q 1 Q 1 + (2Q ω/ω r ) 2 For similar trapped modes, this heating is about the same as in B-factories. For the parameters calculated above Even for Q = 100 we get 6 kw. P max = 880 kw

9 9 Two-bunch excitation of trapped modes in IP There are two bunches of opposite charges, N and N +, colliding at IP. Assuming that the collision occurs at the center of the pipe, we find that different longitudinal modes TM 01n j 2 01 ω r = c a 2 + π2 n 2 l 2 are excited with different amplitudes, if compared with the case of one beam 4. The amplitude A of the even modes (n = 2m) is proportional to A N + + N, and for odd modes (n = 2m + l), A N + N. Therefore, the power deposited in even modes scales as P (N + + N ) 2 ; it increases in comparison with a single beam case. 4 S. Heifets et al. Review of impedance issues for B-factory, SLAC-PUB (2004).

10 0 Detuning of the trapped modes Trapped modes due to small local expansions of the vacuum chamber sit near the cutoff frequency of the pipe. For a round pipe there are dangerous radii corresponding to cutoffs of TM 01 : a = cj 01 2πnf RF where n is an integer. For f RF = 400 MHz we find the radii that should be avoided: 3.59 cm (3.2 GHz), 3.19 cm (3.6 GHz), 2.87 cm (4 GHz),... For a more complicated shape of the pipe it makes sense to calculate the cutoff frequencies of the lowest modes using an EM code. What is the cutoff frequency for this cross section?

11 11 Can nonuniform filling help with detuning? Assume q identical trains with q gaps: where P = q2 T 2 0 n= ρ(ω) = ρ(qnω 0 ) 2 Re Z(qnω 0 ) C/q 0 ρ(z)e iωz/c dz The sampling now occurs at qω 0 frequencies, but this is still a small number, khz for q = Even high-q resonances will not be missed.

12 12 Randomization of bunches in the ring? Consider the case of M = bunches in buckets. Make a random distribution of bunches in the ring (should be correlated with the other ring). Calculate the beam spectrum ρ(ω) 2 and compare with the case of one train and one gap. Sample of random distribution of the bunches.

13 Randomization of bunches in the ring? Fine structure of the resonant peak in ρ(ω) 2 with ω = nω. Blue a single train with a gap, red a random distribution of the bunches over RF buckets. The area under the red peak is 4.25 times smaller than in the case of the bunch train. Hence the heating will be suppressed by this factor. ρ ω ω ω ω 13 There may be some filling patterns that make the heating smaller more work needed.

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