• from Vivek Jain

Hi,

here are some questions on your study to estimate ccbar contamination
(http://www-d0.fnal.gov/~bgv/talk-cc.ppt)

1) slide 5: According to this plot, there are approx. 10000 "signal" events
in 2 pb^-1. Does this number agree with the charm cross-section?

N = L*sigma*Branching Rat*Epsilon_(accept+reco)

L=2 pb^-1, N=10000, Br_rat = 0.68*0.04 = 0.027%

==> sigma*epsilon = 185 nb = 0.2 micro-barn.

Now, ppbar->bbbar cross-section is 150 microbarns, so the ccbar
cross-section must be higher.

In any case, to sigma >= 150 micro-barns, epsilon <= 0.13%. Does this sound
right?

this sounds ok however there are two problems to get exact numbers 1) to calculate any cross section for this sample we run into a usual thing that the semileptonic sample has been recorded by the single muon triggers which were prescaled, differently for each prescale set. in addition there is a trigger efficiency contribution to the total efficiency. so we never attempt to do this and in fact it's not needed for the result. another uncertainty is that most of the bkg to our sample comes from the gluon splitting which is not well known and that's why we are trying to estimate it from the data with minimum input from MC. we estimate it as a relative contribution to the sample composition so the xsection issues are useful only as sanity checks but not for the final result

2) slide 7: top plot. what are the three plots?

those are reconstructed D*(->pi muKX)D0 events in the muDo sample from the lifetime ratio analysis. all these events correspond to the plot on the previous page after cut dm<0.165. the red curve is D*(->pi muK) in the muD0 sample with right sign combination of pi and mu. the blue curve is the same but with wrong sign combination. the difference between them (excess of one over the other) is interpreted as the D*(->pi muKX)D0 contribution to the original muD* sample. magenta is sum of the two combinations. The comparison of the number of events in D0->Kpi peak for correct and wrong sign combinations shows that the (K mu pi) events give contribution mainly into the background but not into (D0->Kpi) peak.

3) what goes into converting 595+-206 to 12+-4%?

this is basically projection of 595 events to the muD0 sample:

From PDG:

1. Fragmentation c->D* anything is 0.255, 2. D*->D0 pi: 0.68 3. D0->K mu X: 0.032-0.065

It gives c->K mu pi X: 0.0087 (assuming D0->K mu X: 0.05) c->l X: 0.096 efficiency to reconstruct D*->pi muK = 50% (without muon eff) total muD0 candidates = 100000 fraction = 595 * 0.096/0.0087 /0.5 /100000 ~ 13% (Gennadi may have used slightly different numbers so he got 12%)

4) slide 11. how do you get 2.8+-0.2%. Is that the number of mu+D*+ in the
last bin divided by the number of mu+D*- in that bin?

The plot on this page was made with a special selection of events. We did not require the vertex between mu and D0 and removed some other cuts. That is why if you take the ratio of events in this last bin, it will give about 7%.

This quoted number (2.8%) was obtained from our working sample, with all selections applied as in the main analysis. Due to the requirement of muon and D0 making the same vertex and some additional cuts, we have this reduction of ccbar contribution.

5) slide 12: In the diagram that you have drawn, what happens if the (red)
muon is on the left side of the blue bbar quark? Won't the quasi B vertex be
in front of the PV? Why do you assume that a large fraction of bbbar events
will have negative decay length?

we found that these events populate equally negative and first positive bins - so it's symmetric which makes sense. see, page 28 here http://d0server1.fnal.gov/users/burdin/talks/sb20040813.pdf

6) On slide 14, how do you get 10%?

10% is difference between data (points) and estimation without ccbar contribution (blue hist).

> Hi Guennadi, Andrei and Sergey,
>
> I am looking through the new note and have a couple of quick questions.
> Will read it more carefully later.
>
> 1) On page 10, at the end of section 7.2, you say that the ratio of
> (c->D*)/(c->D0) was taken to be the same as the observed ratio of D* and
> D0 events.
> Where do you use this ratio? Is this an assumption you made? How did you
> arrive at this assumption? Also, does the denominator include all
> possible feed-downs, i.e., c->D*->D0, c->D**->D0
>
> The reason I ask is that from spin arguments one expects that c->D* is
> three times larger than c->D0 (direct production of D0 from c).
> Basically, there are three helicity states available to the D*.

> > Our idea was that both in the semileptonic B-decay and in ccbar > > events it is essentially the same process, i.e. the fragmentation > > of c-quark to D-meson. That is why we expect to a good approximation, > > that the number of (c->D*)/(c->D0) both in the semileptonic B-decays > > and in the c->D process should be the same. > > BTW, this assumption, > > within the experimental errors, is confirmed by the experimental > > data. > > > > We take the experimental ratio N(D*)/N(D0) both for B->mu D X and > > for c->D. This ratio automatically takes into account all ineficiencies > > and c->D**. > > > > The difference between c->D and B->D is that the lifetime distribution > > should be the same for c->D* and c->D0, while it is different > > for B->mu D* and B->mu D0. This is taken into account in our > > fitting procedure. > >

c->D* implies that c->D*+ and c->D*0 happens equally. So, c->D* leads to > 0.85 D0/D* and 0.15 D+/D*, and c->D0 is 100% D0. Is this what you meant > when you say that the lifetime distribution is the same for c->D* and c->D0?

Saying that the lifetime distribution is the same for c->D* and c->D0, I mean a very simple thing. We measure the lifetime as the distance to the vertex of muon and D0. Both for c->D* and c->D0, this distance should have exactly the same distribution, because D* decays immediately in the production point and D0 in both cases flies the same distance. There is a small decrease of momentum for decay D*->D0, but it is really a second order effect.

> 2) In table 3, a_3 is significantly shifted from 0. Is this the mean of
> the third Gaussian? Where is this bias coming from?
>
> vivek

also there was a discussion on the trigger bias which can be found at these pages dated September 2004 http://listserv.fnal.gov/archives/d0-run2eb-006.html. Some final plots of the muon pt follow

The plot in:

http://www-d0.fnal.gov/~bgv/d0_private/dt-pap/muonPtData.ps

shows the muon pt spectrum in data from decays B->mu D* X.

The plots in:

http://www-d0.fnal.gov/~bgv/d0_private/dt-pap/muonPt.ps

shows the muon pt spectrum in data from decays B->mu D* X.

The plots in:

http://www-d0.fnal.gov/~bgv/d0_private/dt-pap/muonPt.ps

shows muon pt distribution in MC for different subprocesses:

Page 1, upper plot: comparison of muon pt for B->mu D** X -> mu D0 X (points with errors) with muon pt for B->mu D0 nu (histogram)

Page 1, lower plot: comparison of muon pt for B->mu D** X -> mu D* X (points with errors) with muon pt for B->mu D0 nu (histogram)

Page 2, upper plot: comparison of muon pt for B->mu D* nu (points with errors) with muon pt for B->mu D0 nu (histogram)

All these plots were obtained after all our selections.

In general, these plots confirm the statements and plots of Andrei. The muon spectrum in data is quite different from the muon spectrum in MC due to the trigger conditions. However, the muon spectrum for different semileptonic decay channels of B (B->D0 mu nu, B->D* mu nu, B->D** mu nu) does not differ too much, which gives a reasonably small variation of the relative efficiency between channels. We estimated the systematic error in the realtive efficiency (coefficients Cy) at 10% level. And for our measurement, only the relative change of efficiency in different channels is important.

• from Kazu Hanagaki

Result of analyzing the new data ONLY.
The fit of the new data only gives:
dt/t0 = 0.061 +- 0.023
As a reminder, the fit of the old data (Moriond sample) gives:
dt/t0 = 0.093 +- 0.021
The results are consistent, the difference between them is 0.032 +- 0.031

In chi2 fit we use the ratio of events in each bin: N(D*)/N(D0). For the error of this ratio, we use the errors sqrt(N(D*)) and sqrt(N(D0)) and the error propagation. This approach is correct in the limit of the large numbers, otherwise there is some small bias. It gives the final bias in our toy Monte Carlo. We verified, that this bias decreases with increase of the number of events in the sample.

We subtracted this bias (0.00047) from our result. We added the systematic error 0.00047 to our systematic error.

Yes, it is a full MC result. Our MC was generated with 1.07
and we measured 1.084, it agrees with the measured value within
MC statistical error.

You can see the pull distributions for different input values
of the lifetime ratio in:

For t+/t0 - 1 = 0.060:
http://www-d0.fnal.gov/~bgv/d0_private/dt-pap/dtpull060.eps

For t+/t0 - 1 = 0.085:
http://www-d0.fnal.gov/~bgv/d0_private/dt-pap/dtpull085.eps

For t+/t0 - 1 = 0.095:
http://www-d0.fnal.gov/~bgv/d0_private/dt-pap/dtpull095.eps

In words, the pull distribution has sigma=1 and bias does not
change with the change of input value of lifetime difference.

also some follow-up discussion

Will do it. It was correctly mentioned before, that we applied the cut Pt>0.5 in our Ks study and the cut P(D0)>5 GeV for our lifetime analysis, which effectively gives the Pt spectrum of slow pion starting from about 0.4 GeV and peaking at 0.5 GeV. Hopefully, these momentum distributions are quite close. But we will provide the plots, which you request.

the pt(D0)>5 Gev cut makes the slow pion more energetic. so both cases are comparable.
You can find the corresponding plot for Ks in

http://www-d0.fnal.gov/~bgv/d0_private/dt-pap/pt-pi.ps