Re: cosmic analysis doc

["Christina Kourkoumeli" <hkourkou@phys.uoa.gr>]

Dear Elton,
I totally agree with your calculations,they are more detailed than mine.
The reason I didn't comment more on this is that I was trying to get an
approximate value from the sigma and on the other hand I do not understand
the mean of the plot, namely the total time taken to travel the
8-13cm.This should be, as you say only the time taken for light to arive
from top to bottom(everything else cancels).So why is it 1.5nsec, and not
0.5nsec or so??Is there some absolute shift?
                            Cheers,
                              Christine
> Hi Christina,
>
> Thanks for your note. I took a brief look at it and would like to ask you
> to clarify the conclusion paragraph. My interpretation of your plots leads
> to slightly different numbers, but please comment. As I understand it you
> start from the measured sigma in Figure 14:
>
> sigma(15WC - 3WC) = 0.65 ns. Assuming the same time resolution for both 15
> and 3, we get
>
> sigma_measured(15WC)=sigma(3WC)=sigma(15WC-3WC)/sqrt(2) = 0.46 ns.
>
> This would correspond to the time resolution for one cell with an average
> energy deposition given by a cosmic-ray traversing one cell. The
> equivalent energy for minimum ionizing in the direction of the beam is
> about 170 MeV, and vertically the desposited energy is 2/3(170) = 113 MeV.
> The energy in one cell is 113MeV/3 ~ 40 MeV.
>
> Now there are corrections to this value due to fluctuations not associated
> with the timing measurement itself: 1) size of the source (although this
> should partially cancell out when taking the difference between 15 and 3)
> and 2) path length variations in the cosmic-ray tracks between #15 and #3.
> Let's take each one of these in turn (as my naive analysis differs from
> your numbers):
>
> 1. Size of the source. The paddles have a width of 9 cm, so a uniform
> distribution over 9cm gives a sigma=9cm/sqrt(12) = 2.6 cm. Using your
> number for the velocity in the barrel of 14.5 cm/ns, this contribution
> gives 2.6cm/14.5cm/ns = sigma_source=180 ps.
>
> 2. Path length variations. Here, I take your numbers of 8cm for direct
> path and 13 cm max for inclined paths (if I understand them correctly). I
> could take a naive estimation for the sigma as (13-8cm)/sqrt(12)=1.44 cm
> as above. Alternatively one can compute the sigma assuming a uniform
> distribution of cosmic rays in phase space dN/dcosthe = constant (this
> actually overestimates the variation):
>
> sigma_path^2 = <p^2> - <p>^2, where p = 8cm/costhe. If I calculated the
> averages correctly, I get the following:
>
> sigma_path = 8 cm sqrt(1.613 - 1.582) = 8 cm (0.176) = 1.4 cm, similar to
> above.
>
> Now, for cosmic rays, I think we should assume 30 cm/ns (velocity of
> light) since it does not relate to signal propagation in the barrel.
> Therefore, we get
>
> sigma_path = 1.4cm/30cm/ns = 47 ps.
>
> In summary, I would conclude the following:
>
> sigma(15WC) = sigma_measured(15WC) - 180 - 47 (in quadrature), which gives
> sigma(15WC) = 420 ps.
>
> In order to compare with Blake's analysis, we can plug in 40 MeV into his
> parametrization:
>
> sigma_t = 64/sqrt(0.040) + 134ps = 454 ps. I would consider this fairly
> good agreement if this interpretation is correct.
>
> Please comment on this reasoning and/or explain your conclusions in a
> little more detail.
>
> Thanks, Elton.
>
>
> Elton Smith
> Jefferson Lab MS 12H5
> 12000 Jefferson Ave
> Suite # 16
> Newport News, VA 23606
> elton@jlab.org
> (757) 269-7625
>
> On Tue, 3 Jul 2007, Christina Kourkoumeli wrote:
>
>> Dear all,
>> I posted doc-836 which studies the time info for the cosmic run of the
>> BCAL at Jlab.
>>                  Cheers,Christine
>>
>>
>
>