Re: cosmic analysis doc

[Elton Smith <elton@jlab.org>]

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
>
>