CEA Codes forum

Dear Didier,

I was wondering how the implementation of a bending magnet would work when taking into account higher order multipole coefficients, e.g. up to n=6 (dodecapole). When using "QUAD" elements in the structure file, I can easily add information about the sextupole, octupole, ... gradient. However, I do not see something comparable for the "BEND".

Is there some workaround via the "MULTIPOLE" element in order to implement a bending magnet with higher order multipole components or what would be the most elegant way to do this?

Thanks a lot in advance for your answer.

Best
Marius

Dear Marius,

It depends on how you simulate your dipole. In fact, I didn't understand if you wanted to use a BEND element or a field map of your dipole.
If you use the BEND element, then cut them in half and write your multipolar QUAD element with zero length in the middle.
If you use a field map overlaying multipolar maps that you have to create yourself. The MULTIPOLAR element cannot be overlaid as it is, but it can help you by generating this type of map for you, which you can then use.

Regards,

Didier

Hi Didier,

Thanks a lot for your answer. I was trying to find a way how to use the "BEND" element with multipoles and your answer made it perfectly clear.

Thanks again and best
Marius

Dear Didier,

About the implementation of multipole coefficient in bending magnets using "bend" and thin "quad" elements.
I want to confirm if the following implementation is correct:

1) Bending magnet:

EDGE 45 4500 100 0.45 2.8 100 1
BEND -90 4500 0 100 1
EDGE 45 4500 100 0.45 2.8 100 1

2) Divide the bend in two and add a zero-length Quad

EDGE 22.5 4500 100 0.45 2.8 100 1
BEND -45 4500 0 100 1
EDGE 22.5 4500 100 0.45 2.8 100 1

Quad 0 0 100 0.0 1e-3 1e-3 0 0 45

EDGE 22.5 4500 100 0.45 2.8 100 1
BEND -45 4500 0 100 1
EDGE 22.5 4500 100 0.45 2.8 100 1

I have a doubt about the value of the Gradient of the zero-length Quadrupole.
Which value I should use?
I set zero, but I am not so sure if it is correct.

Best regards,

Bruce

Bruce,

Est zero is ok, but your syntax is not exact, the rigth one is following

EDGE 22.5 4500 100 0.45 2.8 100 1
BEND -45 4500 0 100 1
EDGE 22.5 4500 100 1e-20 1e-20 100 1

Quad 0 0 100 0.0 1e-3 1e-3 0 0 45

EDGE 22.5 4500 100 1e-20 1e-20 100 1
BEND -45 4500 0 100 1
EDGE 22.5 4500 100 0.45 2.8 100 1

This 1e-20 value set in EDGE is put here to cancel Fringe-field at the jonction (0 will give default value, 0.45/2.8)

Regards,

Didier

Dear Didier,

Thanks for your answer.
I have a couple of questions:

1) By using a zero-length quadrupole, according to the manual:
If L equals 0, the quadrupole is simulated as thin lens and all gradients have to be replaced by their integral (over longitudinal direction) values.

So, the G, G3,G4, G5 must be the integral over the dipole length, right?

2) If G (Magnetic field gradient) is set zero (as in my case) and GFR is no null, Are the multipole coefficients the relative components (u)?
I mentioned that because the manual says that

u ={10000x Gn XGFR^{N-2} (m) }/ G

And the relative multipole components are defined according to the magnetic field, which is zero in this case.

Thanks for your help,

Bruce

Dear Bruce,

1) yes
2) if G=0 and GFR=0, then all components will be set to 0

Regards,

Didier

Dear Didier,

Thank you very much for the explanation.
I was thinking that if G=0 and GFR=0, the high multipole coefficients are defined as Gn (i.e., Sextupole gradient (T/m^2) ).

Because my understanding is that if GFR is not null, the input coefficients are Un (relative)
If not (GFR =0), the input coefficients are Gn.

In that way, the definition of relative multipole is slightly different from the one that is usually reported (
Un =1e4* Gn/G.).

In TraceWin there is an extra GFR^{n-2} (m), Could you comment on that?

Thanks for your time.

Bruce

Dear Bruce,

Sorry, you're rigth, if G=0 & GFR=0 the the input coefficients are oviously Gn.

For GFR^{n-2}, A magnet designer will typically refer to higher order multipoles as a relative amount on a specified good field radius (GFR) in the so-called units (u, 1e-4). For a certain quad design, e.g. the relative sextupole component will be constant: u3 = 1e4 * G3*GFR / G,or in more general terms, un = 1e4 * Gn*(GFR)^(n-2) / G.

So honestly the explanation below is not from me, but from our specialists in magnetic elements and I applied it as is and at the time (2014) it seemed rather relevant to me and other more recent exchanges on the subject have rather confirmed this approach.

Regards,

Didier

Dear Didier,

Thank you very much for your help, Could you share the 2014 reference if you have?

best regards,

Bruce