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[Opal] Ks and Ksc frame

Chronological Thread 
  • From: Philippe Piot <piot AT>
  • To: opal AT
  • Subject: [Opal] Ks and Ksc frame
  • Date: Fri, 3 Jan 2020 05:24:10 -0600

Dear All,
  I would like to clarify how the Ksc is found (sorry these may be very naive questions).
1/ I do not understand the statement in the user manual Sec "... coincides with the mean position of the particles and the mean momentum is parallel to z". I understand Ksc origin coincides with the bunch center but I would expect a statement on the basis vectors orientation. I take it from the statement that Ksc is generally aligned with (x,y,z) if the beam is moving rectilinearly?
2/ In fact, and in general, it appears from Fig. 6.2 that Ksc is rotated so that its "longitudinal" basis vector aligns with the local tangent of the reference trajectory? Though I do not get why Ksc is not exactly on Ks as the flow diagram of sec. 6.6 seems to imply.

  Finally what happened in the case of a beam with a strong spatial correlation between two degrees of freedom? For instance, I have a flat beam with a strong correlation in (x,y) is the Ksc frame rotated accordingly to ensure the cartesian space-charge grid is optimally matched to the principal axis of the beam?

  Thank you very much. All the best,  -- Philippe.

Philippe Piot,
Northern Illinois University, Dept of Physics and
Northern Illinois Center for Accelerator & Detector Development
DeKalb, IL 60115, USA
Tel: 815 753 6473, Web:

Argonne National Laboratory, Advanced Photon Source
Accelerator System Division
Lemont, IL 60439, USA
Tel: 630 252 2415, Web: 

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