OD Calorimetry Study Update — Leakage Effect and relative calibration Jaewon Park University of Rochester MINERvA/Jupiter Group Meeting, May 31, 2006

2 Contents OD calibration factor varies a lot depending on ID track length To understand what cause calibration factor change too much, try some other OD geometry  New geom: 5.75—5.72—5.72—11.43cm  5.75cm x 25 layers  2.5cm x 25 layers I can study 5.75cm x 10 layers geom by throwing data from layer > 10 Resolution will be affected by granularity (sampling) and leakage How to recover leakage Relative calibration

3 Leakage is big problem Same color and symbol are used for two groups:  5.72cm x 5, 10, 15, 20, 25 layers  2.5cm x 5, 10, 15, 20, 25 layers For 5.72cm geom, more layers didn’t affect much on resolution without leakage effect. 2.5cm geom shows better resolution than 5.72cm geom For 5.72cm geom, leakage is almost eliminated when more than 10 layers are used. Uncalibrated resolution wasn’t affect much by increasing more layers

4 How to recover the leakage? If particle starts showing almost near last layer of OD, there will be big energy loss in that layer Hoping last layer energy is correlated with leaked energy, try to see correlation between them Also whole OD energy vs leaked energy It looks pretty similar with ID vs. OD energy plot

5 Raw data – energy loss in each layer (position) Between red and blue line: 1-5 layers energy loss Six events are shown

6 Relative calibration – r-dependence Theta=50deg Relative factor, α applied to OD α=0.55 gives minimum resolution Minimizing α-value slightly depends on ID track length  It’ll also depend on theta and phi

7 Relative calibration- z dependence Theta=0deg DS-HCAL

8 Relative calibration and leakage Relative calibration, i.e. applying 0.55 to OD decreases overall resolution but didn’t help recovering leakage

9 Relative calibration and leakage Relative calibration, i.e. applying 0.55 to DS- HCAL decreases overall resolution It make mean shift more negative unevenly

10 Relative calibration to 4 OD layers – r-dependence Theta=50deg Relative factor, α applied to OD Fix α=0.55 Introduce anonther parameter β Outer layer is weighted more, in linearly increasing way with factor of β

11 Summary Simplest algorithm uses energy loss in ID, and each OD layers  Cut that requires showing in ID enhanced resolution  Relative calibration enhanced resolution Recovering leakage energy using energy loss in ID and OD layers looks hard Geometric information i.e. tracks and show shape might be useful  Finding correlation between incident shower object and OD energy loss Should we try neural network after all?