[Hps-ecal] ECal Meeting

Thu Mar 27 08:04:26 EDT 2014

Hello Stepan,
Thank you for having read my note and for your prompt reply.
Let me start answering to your questions:

Il 25/03/2014 21:55, Stepan Stepanyan ha scritto:
> Hello Luca,
>
> Thanks for the work and the note. Sorry, meeting was cut short so I 
> could not
> ask my questions. So, here they are (I read not and will comment on it) -
> - I am confused with your definition of \delat E and ECal resolution. 
> In the proposal
> we quoted sigma/E=4.5%/sqrt(E) for ECal resolution. This means that 
> the sigma will
> be sigma=4.5%*sqrt(E). This will end up for 1.5 GeV energy sigma ~ 4.9 
> MeV, for
> 2.2 GeV sigma~5.9, and for 6.6 GeV it is ~10 MeV. So, I am not sure 
> where this 30 MeV,
> 20 MeV, and 6 MeV come from, and what they represent.
Yes, the formula I've been using for ECal resolution is:
sigmaE/E=4.5%/sqrt(E) (GeV)
which can be rewritten as:
sigmaE= 0.045 sqrt(E) (GeV)

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.
I've repeated the calculations and I obtain the following numbers:

sigma E= 0.045*sqrt(1.5)GeV=0.045*1.2247 GeV=0.055 GeV =55 MeV @ 1.5 GeV

sigma E= 0.045*sqrt(2.2)GeV=0.045*1.4899 GeV=0.067 GeV =67 MeV @ 2.2 GeV

sigma E= 0.045*sqrt(6.6)GeV=0.045*2.569 GeV=0.115 GeV =115 MeV @ 6.6 GeV

which are even higher than previous ones.

If I use the formula:

sigma E= 0.04 sqrt(E) (GeV)

I obtain:
sigma E= 0.04*sqrt(1.5)GeV=0.04*1.2247 GeV=0.048 GeV =48 MeV @ 1.5 GeV

sigma E= 0.04*sqrt(2.2)GeV=0.04*1.4899 GeV=0.059 GeV =59 MeV @ 2.2 GeV

sigma E= 0.04*sqrt(6.6)GeV=0.04*2.569 GeV=0.103 GeV =103 MeV @ 6.6 at  GeV

These are exactly ten times your numbers. What am I doing wrong?

>
> - From your slides, Eq.(8) in the note is for 2.2 GeV and Eq.(9) is 
> for 6.6 GeV. Please
> mention that i the note
You are right. Done!
>
> - on page 3, you have L=34, which I assume is the with for 23 modules 
> and it includes
> thickness of wrapping, while H=6.5 cm, that is calculated without 
> taking into account
> vertical space between modules. It may not matter, but will be better 
> to be consistent.
I have information about the dimensions related to the width of the 
calorimeter but not to the height,
so I decided to just consider 5 crystals without "vertical wrapping"; 
but you're right, it is not consistent.
Where may I find the information about the exact vertical dimensions?
As soon as I will access the value for effective height of the front 
face of the Ecal, including wraps,
I'll be glad to adjust the calculations: it won't take much.
Meanwhile I've re-run the code considering no wrapping at all also for 
width (5 rows of 23 crystals -> L=30 cm).
Results are in the attached file.

> I do not understand what is h=4 cm, if it is distance from the x-z 
> plane to the bottom
> of the face of the ecal crystals it should be 2 cm, since the gap 
> between two ecal modules
> is 4 cm.
>
h is indeed the distance from the x-z plane. I made a mistake in 
inserting the dimension in my code.
I ran the code with the correct value for h=2 cm. You may find in in 
attachment. The new count rates
are larger, because of the smaller scattering angles corresponding to 
the various sectors, but remain
critical @ 6-6 GeV.

> - Table 2 and Fig. 3, this fiducial acceptances, they do not take into 
> account magnetic field,
> is not it.
>
It is true, neither the magnetic field or the presence of the SVT have 
been taken into account.
> I am attaching the note that we wrote with Takashi, will be 
> interesting to compare your numbers
> with numbers in the note, it seems your cross sections are much higher.
>
We started to compare my results with the ones in your note. We found 
that one main source
of difference is the useful Luminosity we have used:

1. We have considered the elastic Coulomb scattering process on the 
tungsten, for which
we understand the nucleus is considered as a whole.

The formula for the luminosity we used is:

L = (I_beam/q_e ) rho L N_av / A

where I_beam= 200 nA is the beam current, q_e=1.6 10(^-19)C is the 
electron charge,
rho=19.3 gm/cm2 is the tungsten density, N_av=6.022 19(^-23) is the 
Avogadro number,
L= 5 microns is the target length and A=183.35 is the atomic number.
We divided by A, since the volume density of scattering centers should 
include the number of
"nuclei" per unit of volume.
We obtain:
L = 39.5 micro barn (^-1) s(^-1) = 3.95 10(^31) cm-2 s-1

2. The formula you have used is very similar but you DO NOT DIVIDE per A,
since, to our understanding, you are considering the quasi-elastic 
scattering process
from the single nucleons.
You obtain:

L = 6360 micro barn (^-1) s(^-1)
which is approximately 183 times bigger.

Do you agree?
If we are both correct our first impression is that the reduced cross 
section of the inclusive-quasi-elastic
process is compensated by the different useful luminosity for the two 
processes.

At our first sight your calculations correspond to the ones in our first 
sector only, that is, between 15 and 60 mrad.

To make better comparison I would integrate my cross section in the 
theta bins listed in table II of your note and compare
final rates.

Best regards

Luca

> Regards, Stepan
>
> On 3/24/14 1:07 PM, Luca Colaneri wrote:
>> Il 24/03/2014 17:47, Raphaël Dupré ha scritto:
>>> Hello,
>>>
>>> I think we passed the maximum length of meeting due to the time change.
>>> So we will have to continue the discussion next week.
>>>
>>> Luca, could you send your slides around so everybody sees your results
>>> and conclusion?
>>>
>>> Sorry for the problem, next week Europe will be back on the usual
>>> schedule and this issue should not bother us again.
>>>
>>> Best,
>>>
>> Ok, no problem.
>> I wrote a report on what I did, you'll find it in the attachment 
>> along with the slides.
>> If you have any question, please write me.
>>
>>
>> my bests
>>
>> L.
>>
>>
>> _______________________________________________
>> Hps-ecal mailing list
>> Hps-ecal at jlab.org
>> https://mailman.jlab.org/mailman/listinfo/hps-ecal
>

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