Hello Aaron,
I think I can answer these two.
On Thu, 30 Aug 2001, Aaron Joseph Maschinot wrote:
>
> Hello:
>
> I have some questions regarding the class TBLRecon, for anyone who happens
> to know.
>
> 1) There are multiple member functions of this class that perform fits of
> the data (e.g. WCfit1, WCfit2, and WCminuitDit). Which one is "the best"?
>
WCfit1 is the method I generally use, it is fast, and gives
a reasonable fit most of the time.
WCfit2 is an idea for a fitter which has not been followed up
on. Maybe someday.
WCminuitFit is a robust fitter, but it is also MUCH slower then
WCfit1. I can imagine a situation where if WCfit1
failed, one could try WCminuitFit.
> 2) To identify a scattered electron, you need six pieces of info: three
> vertex coordinates (x0, y0, z0) as well as the three spherical coordinates
> of that electron's initial momentum vector (p, theta, phi). When you
> reconstruct the electron's path, it is only possible to determine p,
> theta, and phi. information regarding x0 and y0 is gone due to the fact
> that the target has a nonzero rho value. however, it is still possible to
> detemine a corrected zc value. My question is: is this zc value as
> determined by the TBLRecon class the z coordinate of the point of closest
> approach to the z-axis of the reconstructed track? Or is there some other
> definition being used in the class. I tried to look at the actual code,
> but it's hard to decipher due to a lack of comments and the fact that i'm
> not sure what exact calculations were done, anyway.
It is true that the vertex might be described by (x0, y0, z0),
but in general a tracks vertex can at best be described by
(d0, z0), where d0 is the distance of closes approach to the
z-axis (rho). So in general there are five parameters to
fit (p, theta, phi, z0, d0). However, in the fitter I have
assumed that d0 -> 0 because the vertex is not constrained
by the target, rather it is constrained by the intersection
of the target and the beam line.
The best fit we could hope for if we fit d0 is about
sigma(d0) ~= 3*sigma(z0)
for the same reason that sigma(phi) ~= sigma(theta). That
means we could only fit d0 to 1 cm, but we know the beam to
a micron.
So I have assumed that we know d0 best from the beam
position. This also means that we only have four parameters
to fit (p, theta, phi, z0) which is much faster then four
parameters.
>
> Thank you,
>
> Aaron
>
Tim Smith
P.S. Ask more questions!
____________________________________________________________________
Timothy Paul Smith Research Scientist
MIT Bates Lab tim_smith@mit.edu
21 Manning Rd. tel: (617) 253-9207
Middleton, MA 01949 fax: (617) 253-9599
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