How the location of quantum excitation is chosen in dipole?

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zhilong
Posts: 13
Joined: 12 Oct 2016, 07:19

How the location of quantum excitation is chosen in dipole?

Post by zhilong » 05 Apr 2019, 09:48

Hi Michael,

I am using elegant to do some single particle tracking for ring recently. When I use the element CSBEND for tracking, I wanna to know how it works for quantum excitation?

I read the source code for CSBEND first, and I find that:

1, if we set the flag "distributionBased=0", then it seems like there is no increase effects for the horizontal direction but only damp effects even though the ISR is set to 1. And from my simulation, there will be one or two events that the energy of emitted photon is lower than 0, I don't know whether that is a bug.

2, When we set the flag "distributionBased=1", then where is the photon emitted? it is at the begin of the kick part or the end of the kick part, or just a random location inside the kick part? It looks like at the begin from the source code. And as I know, the excitation of a photon will induce a horizontal kick of electron by dx=D(s)*u/Es and dxp=Dp(s)*u/Es (the increasing effects), here D(s) and Dp(s) are the dispersion function at the location of photon emitted. So the location will directly affect the horizontal emittance increase. But I don't see this effects in the AddRadiationKick function inside the CSBEND element. So how the elegant deal with this quantum excitation effects?

Thanks so much!
Zhilong

michael_borland
Posts: 1927
Joined: 19 May 2008, 09:33
Location: Argonne National Laboratory
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Re: How the location of quantum excitation is chosen in dipole?

Post by michael_borland » 05 Apr 2019, 12:23

Zhilong,

I'm not sure understand your first question. I think you are saying that you don't see quantum excitation when you only track one particle. By default, you should. Please verify that you haven't changed the ISR1PART parameter on the CSBEND elements. It should have the default value of 1. If the problem persists, please upload some files that show the issue.

Here's an example of tracking 10 particles for one turn each, showing that they have different momentum offsets because of ISR.
CdeltaVs-s.png
Also attached are the files I used to do this run.

For the second question, the response is the same regardless of whether distribution-based or gaussian-based radiation calculations are performed: the effect on the emittance originates in *subsequent* transport through bending systems. One does not need to explicitly put it into the implementation of radiation kicks. So, for example, if a photon is emitted by one electron in the middle of a dipole, the electron will start to diverge from the other electrons by virtue of getting a larger deflection in the downstream half of the dipole. This is confirmed by long-term multi-particle tracking with elegant, where we get the emittances we expected.

--Michael
Attachments
CdeltaExample.zip
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zhilong
Posts: 13
Joined: 12 Oct 2016, 07:19

Re: How the location of quantum excitation is chosen in dipole?

Post by zhilong » 06 Apr 2019, 07:16

Hi Michael,

Thanks so much for your kind response, and sorry about the bad expression.

For question 1, when set the "DistributionBased=0", I also see the ISR effects is on. I misunderstood that the horizontal kick is off in this setting, and from your answer to question 2, I know that the horizontal kick because of ISR is on now. Another issue is that I observed that there will be some minus energy photon emitted, which means the electron will get a energy increase because of ISR, for example the red line in your picture. I thought it is not a physics process.

For question 2, I wanna to check that the exact location of the photon emitted by set ISR=1, because the location is random in real physics emission process. If we set the location in the middle by default, then maybe the flag "n_kicks" will affect our simulation result by my opinion. Is it correct?

Best regards,

Zhilong

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