1. Preliminery results from a 1 st study of misalignment.......and a fresh look at statistics David Forrest, University of Glasgow

2. Misalignment Study - Presented at CM21 http://www.cockcroft.ac.uk/events/MICE/CM21/presentations/Misalignment%20Study%20cm21%20june.ppt http://www.cockcroft.ac.uk/events/MICE/CM21/presentations/Misalignment%20Study%20cm21%20june.ppt - Misalignments were translational, rotational, and both combined. 10 mm, 3mrad and 10mm+3mrad respectively, considering X and Y separately - Step VI, 10,000 events - G4MICE and Grid (processing and storage) - Misaligning downstream tracker only, focussing on the relative misalignment between the two trackers

3. Example

4. Results 1.52.546810 Non-misaligned3.98 +/- 0.12-2.67 +/- 0.01-6.71 +/- 0.09-9.68 +/- 0.14-14.64 +/- 0.23-21.15 +/- 0.36 x 10mm Trans0.660.951.060.190.220.33 3mrad Rot1.851.561.420.550.210.81 Both1.390.650.680.370.440.21 y 10mm Trans1.020.520.260.330.29 3mrad Rot1.91.370.940.340.120.25 Both0.910.920.910.130.230.83 These are the fractional deviations from the non misaligned beam I confess this is a confusing slide. Take last column. With no misalignment we have a fractional drop in emittance of 21% with 0.36% “statistical error” (more later). The percentage errors due to misalignment are in the column below and all less than 1%, which is good news. I should rewrite this!

5. Check x10pi 3mm0.09 1mrad0.22 3mm+1mrad0.43 10 mm0.33 3mrad0.81 10mm+3mrad0.21 Expected that for small misalignments, the effect of misalignment is linear. (Hence studying unrealistic misalignments of 10mm for scaling purposes) To get manageable errors due to misalignment, based on previous slide, wanted 3mm, 1mrad tolerances (a third of what was studied) However this didn’t work out as planned. Assuming that the main reason is a need to separate the statistical error from the misalignment error, used this as an excuse to start looking at statistics (next slide).

6. Statistics (the story so far) – John Cobb The following is taken from an unpublished note by John, who has said he does not believe it to be true. However right now it’s the best we have by default. Let: Where α is the fractional change in momentum; N the number of events; ε eq the equilibrium emittance

7. Test: For a 10 pi beam, with α =-0.15, get K to be 0.1, if JHC formula is true. Running a fit, acquire σ = 7.59x10 -3 (for N=1000), σ = 1.95x10 -3 (for N=10,000) 500 1k event runs 50 10k event runs

8. “K” K = 0.2361 Slope is K

9. Future Work I aim to further this work in the second half of September. Prior to this I must attend local institution commitments and summer school programmes.