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Boost Analysis Status Matt Weaver MDI Meeting March 31, 2006.

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1 Boost Analysis Status Matt Weaver MDI Meeting March 31, 2006

2 Combining  2 y’(z) and dy’/dy(z) Fits angular spread at waists precisely determined asymptotic angular spread (z >>  ) poorly determined z scale factor Together, these fits will determine both HER and LER emittances Assuming equal  *s and waist locations

3 Monte Carlo Validation 14mm  * 10mm  * 6mm  *  y’ dy’/dy z (cm) mrad-cm mrad ≤ 1 week data 5 fit parameters  Y * y-waist z offset  H  L  2 YY’ detector error Fit  2 is a simple sum of the two  2 s No    y’ dy’/dy} covariance terms included

4  * (mm) waist z (mm)  H (nm)  L (nm) generated value Monte Carlo Fit Results combined fit  y’(z) fit only

5 Run 5 Fit Results  * (mm) waist z (mm)  H (nm)  L (nm) combined fit  y’(z) fit only

6 waist z (mm) z-centroid (mm) (Relative) waist offset anti-correlates with movement of z-centroid as expected if magnetic waist is static Waist Location

7 22 22 24 d.o.f. Combined Fit

8 waist z (mm) z-centroid (mm) (Relative) waist offset anti-correlates with movement of z-centroid as expected if magnetic waist is static Waist Location

9 waist z (mm) Waist Location (more)  2 YY’ (mrad-cm)  2 YY’ detector error should be constant, except for significant changes to detector (e.g. loss of SVT module) Any variation in the fit may indicate additional waist motion (HER vs LER) Ideally, this parameter will be replaced with a constant  2 YY’ and an additional waist offset.  2 YY’ (mrad-cm) Monte Carlo

10  Y (  m) Specific Luminosity  Y derived from combined-fit results shows good anti-correlation with specific luminosity

11 Luminous  Y (0) (  m) Comparison with Josh  * (mm) Don’t see much correlation. Error bars are complementary in scale. Run 5b may be an interesting comparison

12 HER Bunch Current (mA) LER Bunch Current (mA) Luminosity Other Machine Conditions

13 Run 4 Combined Fit Results  * (mm) waist z (mm)  H (nm)  L (nm)

14  Y (  m)  Y L (  m) 22 22 24 d.o.f.


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