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Changes between Initial Version and Version 1 of Ticket #26


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Timestamp:
Sep 2, 2008, 11:11:29 AM (16 years ago)
Author:
Frank Krauss
Comment:

the answer is that Pythia takes the LO plain cross section - in contrast we compose our total cross section of the various entries, such that

sigma_tot = sigma_0 * weight_0 + sigma_1 * weight_1 ...

where sigma_i is the cross section for Z+i jets (with kt-jet measure as specified) and weight_i is the average Sudakov rejection weight on this configuration. In principle, both have the same formal accuracy, but they will probably always differ by up to 20-30 %. Depending on jet definitions, process etc., Sherpa's cross section then may be larger or smaller than the ordinary LO cross section.

yes, you're right, raising Q_cut lowers sigma_i and sigma_0 becomes the LO xsec, ~sigma(pythia) - there may be residual differences due to slightly different scale choices in the factorisation scale ... . But this is minor, conceptually you're right.

Concerning Leif's remark, well, let me put it that way: I know that for, e.g., inclusive Z production the NLO-LO K-factor is about 1.3, give or take. We found that in some setup our xsec is smaller than he LO one, so we move away from NLO ... . I do not take this as a sign for our xsec being "improved LO", closer to NLO. So, in my opinion, it's probably best to think of the xsec's as having LO accuracy.

From a more technical side, let me argue on orders: Typically CKKW is good for logs (as is the parton shower), but typically, NLO xsecs are dominated by finite terms (since most of the time they're quite inclusive - there are no logs then). CKKW has no well-defined way to get hold of these finite terms, and that's why I do not believe to produce anyting like an NLO xsec.

On the other hand, shapes are typically dominated by logs (erspecially jet pt's etc.) - therefore, CKKW can be expected to perform reasonably well there.

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  • Ticket #26

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  • Ticket #26 – Description

    initial v1  
    1313
    1414(Follow up question)
    15 Is it correct that if I raise Q_cut up to, say s^0.5 /2, I shoud get
    16 sigma_i ~ 0 for i > 0 and sigma_0 ~ sigma_pythia?
     15Is it correct that if I raise Q_cut up to, say s^0.5/2^, I shoud get sigma_i ~ 0 for i > 0 and sigma_0 ~ sigma_pythia?
    1716   
    1817About the formal accuracy, during the school in Debrecen Leif Lonnblad