We estimate some parameters pertaining to the transition crossing of the main Injector. these include the nonadiabatic time, bunch length and bunch height at transition, the microwave growth across transition driven by a longitudinal impedance, and the parameters that govern the Umstätter's and Johnsen's effects.
At time Tc before and after transition, the bucket changes so rapidly that the bunch is not able to follow it. We call this region the nonadiabatic region. this characteristic time is given by1,2
E0 = 0.93827 GeV, rest energy of proton
gt = 20.4 = (1 - bt2)-1/2, transition gamma,
w0 = 566.78 kHz, angular revolution frequency,
h = 588, rf harmonic number,
Vrf = 2.78, rf voltage,
f = 37.6° rf phase,
is the rate of change of g at transition. We find Tc = 1.96 ms. If every particle crosses transition at exactly the same time, the evolution of the bunch can be computed easily. At transition, the bunch ellipse in the longitudinal phase space is tilted. The maximum rms bunch length (not at zero momentum offset) is given by
where G(1/3) = 2.678939 is the gamma function and the bunch area A = 0.4 eV-sec has been assumed. The maximum rms energy spread is
or sE/E = 3.45 × 10-3. Note that here 6psEst is not equal to the bunch area A because the ellipse is tilted.
The growth of microwave amplitudes across transition is unavoidable because, of a certain time interval, the frequence-slip parameter h = 1/gt2 - 1/g2 is too small to provided enough frequency spread for Landau damping. It has been shown by Courant and Synder1,2 as well as by Herrera3 that, it one assumes h/E to be a linear function of time in that interval, the invariant of the longitudinal phase space can be solved analytically in terms of Bessel function and Neumann function of order 2/3. A dispersion relation can be set up and the growth rate can then be solved numerically.4 If we further assume that Z/n is real and the bunches gaussian in shape, the problem can be solved approximately resulting in a handy formula.5 The total growth across transition is exp(Sb + Sa), where
represent the integrated growth rate ImDW before and after transition. The handy formula gives
where N is the number of particle of charge e per bunch, n is the harmonic of the microwave frequency, and F1 = 8.735 is a numerical constant. In the above, st is the rms time spread of the bunch at time -t0 when stability is lost before transition or at time t0 when stability is gained after transition. This time t0 and time spread st are found to be
Assuming that ( NZ/n ) 10 ohms, we obtain
where N is in 1010 and Z/n in ohms. An illustration is given in Table I. The growth of the microwave amplitude in the last row was computed by assuming a broad band centered at 1.5 GHz corresponding to n = 16629. One has to bear in mind that the actual growth is usually less than indicated because nonlinear effect may come in eventually to suppress that growth rate. The growth of the microwave amplitude will dilute the bunch area and lead to a growth of the bunch area. However, the relation between the two growths is not known.
The longitudinal space-charge force will help stability before transition but help instability after transition. It leads to the shortening of t0 before transition and lengthening of t0 after transition. Nevertheless, its contribution will not affect the estimation in Table I by very much.
|t0||1.52 ms||3.83 ms||6.57 ms|
|st||0.380 ns||0.478 ns||0.547 ns|
|3.11 × 10-5||1.57 × 10-4||4.04 × 10-4|
The transverse space-charge force will lower the betatron tune of those particles at that transverse edge of the bunch near the center, and therefore lower the transitiong. These particles will cross transition at a time earlier than the synchronous particle.6 This effect is called the Umstätter's effect. Roughly, the depression of gt is given by
is the linear particle density at the center of the bunch and
with a and b the half-width and half-height of the beam, rp = 1.535 × 10-18 m the classical proton's r, R = 528.30 m the ring radius, e1 = 0.172 the electrostatic image coefficient corresponding to rectangular 2" by 4" beam pipe and hv the half-height of the vacuum chamber. Assuming a normalized transverse emitance of the beta 20p mm mr and a minimum beta-function11.6 m, we get a = b = 6.74 mm and therefor e = 2.52 × 10-12. Using st = 0.380 ns and N = 5.10 × 1010 from Section II, we get for the maximum linear density l(0) = 16.0 × 1010. This leads to a maximum depression of Dgt = 0.00995. From Eq. (2.2), the rate of acceleration is t = 163.1. Therefore, some particles at the center of the bunch will cross transition at a time DT = Dg/ = 0.583 ms earlier than the synchronous particle. Since this time is much less than the adiabatic time Tc = 1.96 ms, Umstätter's effect should be negligible.
Each particle inside a bunch has momentum slightly different from the synchronous momentum p0. It travels along a different closed orbit and has a different momentum compaction factor. If the momentum deviation is Dp, its orbit length is given by
where L0 is the synchronous orbit length and a0 is the momentum compaction factor of the synchronous particle. Obviously, this off-momentum particle will have a different gt and crosses transition at a time DT earlier. It can be shown that DT, which is also called the nonlinear time, is given by7
where DE/E is approximately equal to the relative height of the bunch if every particle crosses transition at exactly the same time (or a1 = -1.5). From Section II, sE/E = 3.45 × 10-3 at transition, we have maximum DE/E ~ 8.45 × 10-3. This gives
If one assumes a perfect FODO-cell structure of phase advanced Ãc with two sets of sextupoles at, respectively, the F-quads and D-quads canceling a fraction of the natural chromaticity, a1 can be derived readily to give8
where sc = sin Ãc/2. If the sextupoles are not turned on, = 0. The Main Injector consists of 90°-cells. Therefore a1 = 25/23. Thus, we obtain maximum DT = 2.1 ms or maximum DT/Tc = 1.4. the growth of bunch area due to Johnsen's effect will be appreciable.9 On the other hand, if there is a complete cancellation of chromaticicy, = 1 and a1 = 1/23. Then, DT/Tc = 0.84. The growth in bunch area will be much less. The detail will be discussed elsewhere.