Hi Adrian,
the "real data" that is reconstructed is always radiated and has undergone
energy loss. However, reconstruction presently does not account for
eloss. So, even if the apparatus is perfect, W would not peak at the
proton mass but at some shifted value. This is in fact demonstrated by
your radiated MC (where the reconstruction does not care of eloss).
When the corrections are determined, you impose ideal kinematics, i.e. you
require W to peak at M_p. This corresponds to the unradiated case, hence
no wonder that the corrected spectrum resembles the unradiated MC.
Determining the correction WITHOUT accounting for eloss in the
reconstruction EFFECTIVELY convolutes the energy loss with the shifts due
to the apparatus to an EFFECTIVE correction. Therefore, in order to obtain
a PID-independent correction, one has to explicitly account for eloss in
the reconstruction BEFORE the actual corrections are determined.
For the later use of such "net" corrections, one would always have to
account for eloss separately.
See also the meeting minutes (to come up).
Best regards,
Michael
On Thu, 2 Mar 2006, Adrian T Sindile wrote:
> Hi Karen,
> Michael explained to me that because energy loss is not included in
> reconstruction of real data (while both MCs include energy loss) this is what
> he expected... I somehow thought it should look like the radiated MC, that's
> why I showed the wrong one yesterday, which although had a double peak was
> more or less around the red curve... but that was wrong, as Michael pointed
> out in the meeting and I verified I had a sign error.
>
> It is still not clear to me why it is expected for the W spectrum to look
> like the unradiated MC once I apply kinematic corrections. If anyone can shed
> some light on this for me, I would appreciate it...
>
> Thank you,
> Adrian
>
> On Thu, 2 Mar 2006, Karen Dow wrote:
>
>> Adrian,
>>
>> I agree, the peak makes sense in terms of the 2 sectors agreeing. But now,
>> why is the peak at the UNradiated location, and the width comparable to
>> UNradiated? I can't remember from all the discussions -- do the kinematic
>> corrections also compensate for energy loss and radiation?
>>
>> Karen
>>
>> Adrian T Sindile wrote:
>>
>>> Hi,
>>> Michael was right (obviously), I had a sign error (I was applying
>>> corrections in the wrong direction).
>>>
>>> Here are the plots again:
>>>
>>> http://einstein.unh.edu/~adrian/Wraw.gif
>>>
>>> http://einstein.unh.edu/~adrian/Wcorrected.gif
>>>
>>> There is no double peak, once I apply momentum corrections right.
>>> Sorry for the confusion,
>>> Adrian
>>>
>>> -------------------------------
>>> Adrian Sindile
>>> Research Assistant
>>> Nuclear Physics Group
>>> University of New Hampshire
>>> phone: (603)862-1217
>>> FAX: (603)862-2998
>>> email: asindile@unh.edu
>>> http://einstein.unh.edu/~adrian/
>>>
>>
>
> --
>
> -------------------------------
> Adrian Sindile
> Research Assistant
> Nuclear Physics Group
> University of New Hampshire
> phone: (603)862-1217
> FAX: (603)862-2998
> email: asindile@unh.edu
> http://einstein.unh.edu/~adrian/
>
>
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