Dear Sasha,

Thank you for the quick comments.

Alexander Lanyov wrote:

>
>>>sect.3.2, para 1: figure 2, figure 3, figure 4 --> figures 2-4
>>>
>>>      
>>>
>>See my answer to Alan.
>>    
>>
>
>If you can send me LaTeX files, I can try to look how this can be done.
>  
>
Has been solved already.

>>>page 13, last para: Where PDG has the value (89.1+-4.1)% ?
>>>      
>>>
>>Page 200, last paragraph.
>>    
>>
>
>PDG quotes separate values for different channels.
>Why did you choose the sum of D0, D+ and D+s?
>Your value for D*+ is more precise than for D+s.
>Why not choose the sum of all four states, or calculate sigma(cc) for each
>channel separately and then combine them?
>The last option could give the most precise result.
>
>  
>
Our approach was that if we measured all open c ground state cross sections, we would have no error from the extrapolation to the full ccbar cross section. No we measure only a part of it (89%), but this fraction we want keep as high as we can, and there we sum up all channel that we measure (D* does not qualify, because it is not ground state, decays to D0).
Of course we could do it by calculating the ccbar cross section as a weighted average of cross section extrapolated from each of the channels, but this implies that:
- we assume that the fraction measured in e+e- are the same as in pA
- we understand how to separate the correlated and uncorrelated errors in the fractions


>  
>
>>>sect.5.3: You cite several times PDG [24], but I wasn't able to find
>>>         these numbers in PDG. Could you please tell where to look?
>>>      
>>>
>>PDG: page 200.
>>    
>>
>
>Taking the data from PDG page 200, I get
>  sigma(D+s)/(sigma(D0)+sigma(D+)) = 0.080/(0.565+0.246) = 0.10 +- 0.02
>which is different from your PDG value (0.13 +- 0.04).
>
>  
>
In the paper the value we actually use the value of 0.10 (p 15) , the 0.13 is in the note, has not been updated yet (was taken from the old reference by Gladilin).

>  
>
>>>page 14, last para: If you fit <pt> and beta, you get chi2/ndf=0.82
>>>                   but if you fix beta to 6, you get chi2/ndf=0.81.
>>>                   But then the first result is not a global minimum
>>>                   of chi2/ndf!
>>>
>>>      
>>>
>>This is a likelihood fit result which is not necessarily at the chi2
>>minimum. Chi2 is in this case only of informative nature. Note, however,
>>that we have by mistake quoted the Neumann chi2 returned by paw, and not
>>the more appropriate chi2 from the likelihood.
>>    
>>
>
>Are you going to change your description? (in the last draft I still see
>the same discussion).
>
>  
>
I do not think we have to change the description. In the present version we just quote chi2, and we note that the fit is a likelihood fit.

Note that we have also modified comparison to others in the leading/non-leading section. We only quote other experiments (and give their values ans xF ranges), and skip a comparison to Pythia.

Looking forward to hearing from you and Alan
Peter