Sign of summation in second order RDTs wrt sextupoles

[blanco-garcia]

Dear all,
I have a doubt about the summation used in the calculation of the second order rdts with respect to sextupoles.

I see that in the elegant file 'twiss.cc' line 4997 the sign of the sum is written as

Code: Select all

termSign = SIGN(ed[iE].s - ed[jE].s);

which is negative for j>i, being j downstream wrt to i.

This termSign is the opposite to the Equation (46) by Chun-Xi Wang "Explicit formulas for 2nd-order driving terms due to sextupoles and chromatic effects of quadrupoles". ANL/APS/LS-330. March 10, 2012.

Is this difference a sign convention in Elegant ? or is there any other reason ?

Best regards,
o

[michael_borland]

I see what you mean. I had previously checked elegant against OPA and found good agreement, but need to look into this more.

--Michael

[michael_borland]

I compared the output from elegant to OPA again. If I change the sign of that term, the agreement is poor. Since I consider OPA to be the gold standard on these calculations, I think perhaps Chun-xi's note has an error.

Do you have other reasons that lead you to think elegant's calculations are wrong?

--Michael

[blanco-garcia]

Dear Michail,
thank you for checking the results. I don't have right now any other reason to think it is wrong.

Before starting this post I derived on my own the term in Eq. (8) and I got the same result that Chun-xi wrote on his paper, and on the way I confirmed the order of the sum is consistent. I will try to attach some of my notes in case they are of any interest.

By today I have confirmed that Eqs(8) to (24) are correct and all consistent with Bengtsson's article. This time I used a simpler derivation using Lie algebra from Chun-xi's first citation. It is the Eq.(54) in C.-x. Wang and A. Chao, “Notes on Lie algebraic analysis of achromats,” SLAC note AP-100 (1995).

I have not yet gone any further. I understand that these terms could be used to derive the tune shift with amplitude, but it will take me some time to get to that result while checking that all I use is consistent. Before taking that road I wanted to know if there was something obvious I was missing in how Elegant works.

Best regards,
o

[blanco-garcia]

Here are the notes I mentioned in the previous post.

o