Dear Didier,
I am running beam dynamics simulation for a periodic solenoid lattice. The .dat file is given below as
Repeat_ele 150 3
Lattice 3 0
SET_ADV 110
DRIFT 150 50 0 0 0
SOLENOID 100 1 50
DRIFT 150 50 0 0 0
END
1) While plotting Partran results, full current phase advance per period (Phase advances -> Beam in Chart page) is found to be greater than the zero current phase advance per period (Phase advances -> Structure in Chart page) as shown in attached figure "phase_adv.jpg".
As per my understanding this is inconsistent with the theory.
2) In addition, in the envelope results (Chart --> envelope) what does full current phase advance calculation means?
Kindly, help me with these.
I am attaching all relevant files for your reference.
Looking forward for your reply.
Thank you.
Beam phase advance calculation
Beam phase advance calculation
- Attachments
-
- Photo shows full current phase advance exceeding zero current phase advance (110 deg).
- phase_adv.JPG (72.51 KiB) Viewed 2668 times
-
- Solenoid.ini
- .ini file
- (30.88 KiB) Downloaded 218 times
Re: Beam phase advance calculation
Dear Pallavi,
(1) I fixed a issue in phase advance calculation in your specific conditions (Tracking with space-charge, very long structure and high transverse divergences with CW beam).
Please upgrade your code.
(2) I'm not sure to understand your question, but in manual you can find the phase advance definition according to selected option in 'Charts' page.
Regards,
Didier
(1) I fixed a issue in phase advance calculation in your specific conditions (Tracking with space-charge, very long structure and high transverse divergences with CW beam).
Please upgrade your code.
(2) I'm not sure to understand your question, but in manual you can find the phase advance definition according to selected option in 'Charts' page.
Regards,
Didier
Re: Beam phase advance calculation
Dear Didier,
Thank you for the prompt reply to the previous mail.
In case of solenoid channel (same .dat file as previous mail), I gave a run for zero beam current ( 0 mA Current in the "Main" page). If I plot "Qxy" in the "Chart" page, I expect the tune to correspond to the structure phase advance per period (or the zero current phase advance per period). However, in case of solenoid channels, this is not the case (Qxy plot is attached).
Also, I have checked for quadrupole channel (FODO), this works fine (i.e. 360*Qxy comes out to be zero current phase advance per period).
I request you to help me with this issue.
Looking forward to hearing from you.
Thank you,
--
Regards,
Pallavi
Thank you for the prompt reply to the previous mail.
In case of solenoid channel (same .dat file as previous mail), I gave a run for zero beam current ( 0 mA Current in the "Main" page). If I plot "Qxy" in the "Chart" page, I expect the tune to correspond to the structure phase advance per period (or the zero current phase advance per period). However, in case of solenoid channels, this is not the case (Qxy plot is attached).
Also, I have checked for quadrupole channel (FODO), this works fine (i.e. 360*Qxy comes out to be zero current phase advance per period).
I request you to help me with this issue.
Looking forward to hearing from you.
Thank you,
--
Regards,
Pallavi
- Attachments
-
- Qx-Qy plot for solenoid channel.
- Qx-Qy.png (13.87 KiB) Viewed 2607 times
Re: Beam phase advance calculation
Dear Pallavi,
I think that for the solenoid, it is a little more complex because of the X-Y' & Y-X' coupling. Bellow an example to illustrate the concequences on the trajectories of a particle.
Phase advance = 90
Input particle position: x=2 mm and y=0 mm
First image, quadrupole : Main Nux is easy to see.
Second image: solenoid: Nux is not easy to see and Nuy appears, because of the coupling.
One way to cancel this well known coupling is to replace the solenoid element by 2 solenoids of inverted sign and there the main harmonic is well mostly present, even if it's not perfect yet.
SOLENOID 50 1 150
SOLENOID 50 -1 150
Regards,
Didier
I think that for the solenoid, it is a little more complex because of the X-Y' & Y-X' coupling. Bellow an example to illustrate the concequences on the trajectories of a particle.
Phase advance = 90
Input particle position: x=2 mm and y=0 mm
First image, quadrupole : Main Nux is easy to see.
Second image: solenoid: Nux is not easy to see and Nuy appears, because of the coupling.
One way to cancel this well known coupling is to replace the solenoid element by 2 solenoids of inverted sign and there the main harmonic is well mostly present, even if it's not perfect yet.
SOLENOID 50 1 150
SOLENOID 50 -1 150
Regards,
Didier