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<div class="moz-cite-prefix">Hello Stepan,
<br>
Thank you for having read my note and for your prompt reply.
<br>
Let me start answering to your questions:
<br>
<br>
Il 25/03/2014 21:55, Stepan Stepanyan ha scritto:<br>
</div>
<blockquote cite="mid:5331ED26.20801@jlab.org" type="cite">
Hello Luca,<br>
<br>
Thanks for the work and the note. Sorry, meeting was cut short so
I could not<br>
ask my questions. So, here they are (I read not and will comment
on it) - <br>
- I am confused with your definition of \delat E and ECal
resolution. In the proposal <br>
we quoted sigma/E=4.5%/sqrt(E) for ECal resolution. This means
that the sigma will <br>
be sigma=4.5%*sqrt(E). This will end up for 1.5 GeV energy sigma ~
4.9 MeV, for <br>
2.2 GeV sigma~5.9, and for 6.6 GeV it is ~10 MeV. So, I am not
sure where this 30 MeV,<br>
20 MeV, and 6 MeV come from, and what they represent.<br>
</blockquote>
Yes, the formula I've been using for ECal resolution is:
<br>
sigmaE/E=4.5%/sqrt(E) (GeV)
<br>
which can be rewritten as:
<br>
sigmaE= 0.045 sqrt(E) (GeV)
<br>
<br>
I have double checked my code and I found a typo, so the numbers
that appear in the slides and the note are not exact.
<br>
I've repeated the calculations and I obtain the following numbers:
<br>
<br>
sigma E= 0.045*sqrt(1.5)GeV=0.045*1.2247 GeV=0.055 GeV =55 MeV @ 1.5
GeV
<br>
<br>
sigma E= 0.045*sqrt(2.2)GeV=0.045*1.4899 GeV=0.067 GeV =67 MeV @ 2.2
GeV
<br>
<br>
sigma E= 0.045*sqrt(6.6)GeV=0.045*2.569 GeV=0.115 GeV =115 MeV @ 6.6
GeV
<br>
<br>
which are even higher than previous ones.
<br>
<br>
If I use the formula:
<br>
<br>
sigma E= 0.04 sqrt(E) (GeV)
<br>
<br>
I obtain:
<br>
sigma E= 0.04*sqrt(1.5)GeV=0.04*1.2247 GeV=0.048 GeV =48 MeV @ 1.5
GeV
<br>
<br>
sigma E= 0.04*sqrt(2.2)GeV=0.04*1.4899 GeV=0.059 GeV =59 MeV @ 2.2
GeV
<br>
<br>
sigma E= 0.04*sqrt(6.6)GeV=0.04*2.569 GeV=0.103 GeV =103 MeV @ 6.6
at GeV
<br>
<br>
These are exactly ten times your numbers. What am I doing wrong?
<br>
<br>
<blockquote cite="mid:5331ED26.20801@jlab.org" type="cite"> <br>
- From your slides, Eq.(8) in the note is for 2.2 GeV and Eq.(9)
is for 6.6 GeV. Please <br>
mention that i the note<br>
</blockquote>
You are right. Done!
<blockquote cite="mid:5331ED26.20801@jlab.org" type="cite"> <br>
- on page 3, you have L=34, which I assume is the with for 23
modules and it includes<br>
thickness of wrapping, while H=6.5 cm, that is calculated without
taking into account <br>
vertical space between modules. It may not matter, but will be
better to be consistent.<br>
</blockquote>
I have information about the dimensions related to the width of the
calorimeter but not to the height,
<br>
so I decided to just consider 5 crystals without "vertical
wrapping"; but you're right, it is not consistent.
<br>
Where may I find the information about the exact vertical
dimensions?
<br>
As soon as I will access the value for effective height of the front
face of the Ecal, including wraps,
<br>
I'll be glad to adjust the calculations: it won't take much.
<br>
Meanwhile I've re-run the code considering no wrapping at all also
for width (5 rows of 23 crystals -> L=30 cm).
<br>
Results are in the attached file.
<br>
<br>
<br>
<br>
<blockquote cite="mid:5331ED26.20801@jlab.org" type="cite"> I do not
understand what is h=4 cm, if it is distance from the x-z plane to
the bottom<br>
of the face of the ecal crystals it should be 2 cm, since the gap
between two ecal modules<br>
is 4 cm.<br>
<br>
</blockquote>
h is indeed the distance from the x-z plane. I made a mistake in
inserting the dimension in my code.
<br>
I ran the code with the correct value for h=2 cm. You may find in in
attachment. The new count rates
<br>
are larger, because of the smaller scattering angles corresponding
to the various sectors, but remain
<br>
critical @ 6-6 GeV.
<br>
<br>
<blockquote cite="mid:5331ED26.20801@jlab.org" type="cite"> - Table
2 and Fig. 3, this fiducial acceptances, they do not take into
account magnetic field,<br>
is not it.<br>
<br>
</blockquote>
It is true, neither the magnetic field or the presence of the SVT
have been taken into account.
<blockquote cite="mid:5331ED26.20801@jlab.org" type="cite"> I am
attaching the note that we wrote with Takashi, will be interesting
to compare your numbers <br>
with numbers in the note, it seems your cross sections are much
higher.<br>
<br>
</blockquote>
We started to compare my results with the ones in your note. We
found that one main source
<br>
of difference is the useful Luminosity we have used:
<br>
<br>
1. We have considered the elastic Coulomb scattering process on the
tungsten, for which
<br>
we understand the nucleus is considered as a whole.
<br>
<br>
The formula for the luminosity we used is:
<br>
<br>
L = (I_beam/q_e ) rho L N_av / A
<br>
<br>
where I_beam= 200 nA is the beam current, q_e=1.6 10(^-19)C is the
electron charge,
<br>
rho=19.3 gm/cm2 is the tungsten density, N_av=6.022 19(^-23) is the
Avogadro number,
<br>
L= 5 microns is the target length and A=183.35 is the atomic number.
<br>
We divided by A, since the volume density of scattering centers
should include the number of
<br>
"nuclei" per unit of volume.
<br>
We obtain:
<br>
L = 39.5 micro barn (^-1) s(^-1) = 3.95 10(^31) cm-2 s-1
<br>
<br>
2. The formula you have used is very similar but you DO NOT DIVIDE
per A,
<br>
since, to our understanding, you are considering the quasi-elastic
scattering process
<br>
from the single nucleons.
<br>
You obtain:
<br>
<br>
L = 6360 micro barn (^-1) s(^-1)
<br>
which is approximately 183 times bigger.
<br>
<br>
Do you agree?
<br>
If we are both correct our first impression is that the reduced
cross section of the inclusive-quasi-elastic
<br>
process is compensated by the different useful luminosity for the
two processes.
<br>
<br>
At our first sight your calculations correspond to the ones in our
first sector only, that is, between 15 and 60 mrad.
<br>
<br>
To make better comparison I would integrate my cross section in the
theta bins listed in table II of your note and compare
<br>
final rates.
<br>
<br>
Best regards
<br>
<br>
Luca
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<blockquote cite="mid:5331ED26.20801@jlab.org" type="cite"> Regards,
Stepan<br>
<br>
<div class="moz-cite-prefix">On 3/24/14 1:07 PM, Luca Colaneri
wrote:<br>
</div>
<blockquote cite="mid:53306657.4090906@roma2.infn.it" type="cite">Il
24/03/2014 17:47, Raphaël Dupré ha scritto: <br>
<blockquote type="cite">Hello, <br>
<br>
I think we passed the maximum length of meeting due to the
time change. <br>
So we will have to continue the discussion next week. <br>
<br>
Luca, could you send your slides around so everybody sees your
results <br>
and conclusion? <br>
<br>
Sorry for the problem, next week Europe will be back on the
usual <br>
schedule and this issue should not bother us again. <br>
<br>
Best, <br>
<br>
</blockquote>
Ok, no problem. <br>
I wrote a report on what I did, you'll find it in the attachment
along with the slides. <br>
If you have any question, please write me. <br>
<br>
<br>
my bests <br>
<br>
L. <br>
<br>
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