NNPDF / hawaiian_vrap

vrap with pineappl
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Fix assumptions about quark PDFs #25

Closed cschwan closed 1 year ago

cschwan commented 1 year ago

This fixes the following assumptions,

and

for all neutral current related processes

scarlehoff commented 1 year ago

Could you fix also #24 since you are at it?

cschwan commented 1 year ago

I'm not quite sure I understand the piso branches; in there both up and down PDFs are replaced with (up + down) / 2 and the corresponding combination for the anti quarks, but not for strange, charm or bottom. Is this correct? @Radonirinaunimi might know whether this is a correct model of the hadron (which in our cases I think is a deuteron, @scarlehoff is that always true?).

I also fixed the ppbar if-branches, although we don't seem to use this in any dataset. However, since it sometimes overlaps with with pp and piso only partially fixing it is probably worse. The W+/W- branch I didn't touch.

cschwan commented 1 year ago

Could you fix also #24 since you are at it?

Done.

Radonirinaunimi commented 1 year ago

I'm not quite sure I understand the piso branches; in there both up and down PDFs are replaced with (up + down) / 2 and the corresponding combination for the anti quarks, but not for strange, charm or bottom. Is this correct? @Radonirinaunimi might know whether this is a correct model of the hadron (which in our cases I think is a deuteron, @scarlehoff is that always true?).

@cschwan From the branch name at least I would deduce that it is constructing a free isoscalar version of the proton. Isoscalarifying a nucleon basically means that each PDF flavour is replaced by: $$f^{\rm iso}_i (x, Q^2) = \left( \frac{ f^{p}_i (x, Q^2) + f^{n}_i (x, Q^2) }{2} \right)$$ where $f^p$ and $f^n$ are respectively the proton and neutron PDF. The neutron PDF can be extracted from the proton-bound using isospin symmetry which basically boils down to the following relations: $$u^n (x, Q^2) = d^p (x, Q^2) \, d^n (x, Q^2) = u^p (x, Q^2)~\text{(same relations hold for anti-u and anti-d)}$$ and $f^n_i (x, Q^2) = f^p_i (x, Q^2)$ for other flavours. If this is what is implemented, then everything is correct.

cschwan commented 1 year ago

@Radonirinaunimi thank you for this great explanation! That's indeed what's being implemented.

scarlehoff commented 1 year ago

I'm merging this. Once I've finished with the E906 problem and check the k-factors (I need to do that with the bugged version for a good comparison) I will release a vrap 1.4 (and update pinefarm accordingly)