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By Alexander Wu Chao
Attracts realization to special theoretical and mathematical facets of beam instabilities. Introduces and analyzes numerous collective instability results for top power accelerators. makes use of an important quantity of versions and soluble examples as illustrations.
Read Online or Download Physics of Collective Beam Instabilities in High Energy Accelerators (Wiley Series in Beam Physics and Accelerator Technology) PDF
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Additional info for Physics of Collective Beam Instabilities in High Energy Accelerators (Wiley Series in Beam Physics and Accelerator Technology)
Sample text
Physically the force comes from an imbalance between the amounts of charge behind and in front of the test charge being considered. In case there are more charges in the front, A’ is positive, leading to a retarding force. In case there are more charges in back, the force is accelerating. 40) satisfy the Maxwell equations if A B Xb2/y2. For a bunched beam, this requires the bunch length I to be much longer than b/y. Show that the leading correction term to Eq. 36) is AB, rln-, e b a b Y a2 rln+ - r 2 2r’ where the two entries respectively.
As promised, the longitudinal force FS = eE, is proportional to A/y*. Physically the force comes from an imbalance between the amounts of charge behind and in front of the test charge being considered. In case there are more charges in the front, A’ is positive, leading to a retarding force. In case there are more charges in back, the force is accelerating. 40) satisfy the Maxwell equations if A B Xb2/y2. For a bunched beam, this requires the bunch length I to be much longer than b/y. Show that the leading correction term to Eq.
11) where the coefficient A is the same as that appearing in Eq. 3) to assure continuity of i,. From Eq. 12) ‘We assume 18% chromium, 8% nickel. 71f we do not assume IhI ZSPb-‘, g, will be written in terms of Bessel functions. This complication is not required for our purposes. 12). 13) k where a term -k/A in the denominator has been dropped because it is much smaller than the term ikb/2. , at r = b, because there is a surface charge on the wall pipe. ,, and B, to obtain the fields. The exact expressions will be given later, but first, investigate the region where IA/k 1 z=- I kbl.