Dear Peter,

Thank you for the answers.
I will look your new draft starting from Monday.

Please find some quick comments below.


On Fri, 15 Jun 2007, Peter Krizan 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.



>>> >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.



>>> >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).



>>> >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).


Have a nice weekend,

Sasha.