Elegant users forum

Hi everybody,

for demonstration of synchrotron radiation effects I tracked a bunch of electrons with large initial energy spread and observed the longitudinal radiation damping. Then I determined the damping time from an exponential fit, but the value is 40-50% smaller than expected.

Here is what I did:
- use CSBEND dipoles with use_rad_dist=1
- define a bunched_beam: gaussian with 50 particles, sigma_dp larger than equilibrium
- define a watch element in parameter mode to get energy spread Sdelta
- track 2-3 times the expected damping time using Pelegant (2-3e4 turns)
- plot Sdelta as a function of Pass (or Pass times revolution time, here 548e-9 s)
- fit f(x)=a*exp(-x/tau)+offset via a,tau and offset
-> tau is only 50-60% of the expected value (taudelta from .twi output)
Find the files attached.

It seems not to be a problem of statistics or the fit. I tried with more particles and different seeds. The fit errors are about 1%. Any ideas?

Thanks in advance!
Jan

Jan,

The problem is that the energy spread and bunch length don't decay exponentially. Rather, the square of these quantities decays exponentially. The expression is See the attachment for an corrected comparison. The agreement is very good.

By the way, USE_RAD_DIST=1 is rarely needed. It just makes it slower.

--Michael

Michael,

thank you very much for your reply. I do not understand why exponential decay of the trajectories leads to your expression. Can you give me a reference for the expression?

Thanks a lot!
Jan

Jan,

I don't know of a reference that has this result in this form, but a good starting point is Matthew Sand's old storage ring paper SLAC-Report 121. In this paper, he gives expressions for the quantum excitation and damping terms. The QE term adds a fixed change in Sdelta^2 each turn, while the damping term fights this. So the differential equation is something like where Z = Sdelta^2. At equilibrium, dZ/dt=0 and Z = SdeltaEq^2, which allows us to conclude that QE=2*SdeltaEq^2/tau. With that, we can perform the integration and get the equation I gave.

--Michael