On Coherent Electron Cooling Page: 4 of 7
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We obtain
6n(z,T) = koZeoz+( +/)pT f ikoqz ex3 ( (/3 - . 2 (10)
The integral in (10) is easily computed
jdqeioz2 exp T --)3]) P 6 i -3nj (11)
(F 31/2-i
We see that for a given T (the undulator length) 6n(z, T) has a Gaussian distribution over
z. The maximal value of 5rn(z, T) is achieved at the point where the argument of the
exponential function in (11) is equal to zero, koz = T/3. Introducing the standard power
gain length Lg
Lg-1 = 2/3pk, (12)
we replace pT = l,/2/ 3Lg and obtain
31/4
max &n(z,T) = koZp Lgelu/2Lg(3
The longitudinal electric field 6E (z, T) generated by the density perturbation 6n(z, T) is
found from the 1D Poisson equation. This equation is trivially solved if one remembers that
6n (z, T) has a fast oscillating factor eiko, in it, hence
4ifre 4ir Ze 31/4 L
max 5E (z, T) = max lon(z, T) = p elu/2L14)
We can write the result (14) as the initial field Eo multiplied by an amplification factor G,
max E (z, T) = GEo, where
G =2 p-celu/2Lg. (15)
This factor G can be related to the (amplitude) FEL amplification factor GFEL. The latter is
usually defined as a ratio of the final (exit) amplitude of a sinusoidal density perturbation at
the resonant frequency (q = 0) to its initial value; in our notation GFEL = 16nq(T)/fnq(0) q-o-
Using (5) we find
GFEL = S1Hq(T)q0 = eu/2L6)4
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Stupakov, Gennady & Zolotorev, M.S. On Coherent Electron Cooling, article, March 8, 2013; United States. (https://digital.library.unt.edu/ark:/67531/metadc846172/m1/4/: accessed March 28, 2024), University of North Texas Libraries, UNT Digital Library, https://digital.library.unt.edu; crediting UNT Libraries Government Documents Department.