Is Sin 2 2    At certain values of  m 2 and integrated p.o.t., MINOS can achieve a tighter limit on sin 2 2  than Super-K. This implies that we.

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Presentation transcript:

Is Sin 2 2    At certain values of  m 2 and integrated p.o.t., MINOS can achieve a tighter limit on sin 2 2  than Super-K. This implies that we could exclude sin 2 2  at <90% C.L if the parameters are favourable. Why is this interesting? –Implications for theoretical models? –Ambiguity in measurement of sin 2 2   if sin 2 2   <1 (U   is double-valued) D.A. Petyt Dec’ 03

Sin 2 2  m 2 =0.002 eV 2   is minimised w.r.t.  m 2. C.L. intervals are drawn assuming 1 d.o.f.

Fits along the SK 90% CL contour See for more fitshttp://

Discrimination as a function of  m 2 and p.o.t.  C.L. discrimination requires more than 7.4e20 p.o.t. Assuming   surface scales as N, can calculate integrated p.o.t. required to exclude sin 2 2  =1 at 68,90,99% C.L.

Sin 2 2  discrimination, 7.4e20 p.o.t.

Sin 2 2  discrimination, 16e20 p.o.t.

Sin 2 2  discrimination, 25e20 p.o.t.  C.L. (3  discrimination possible in a small portion of S-K 90% C.L region

Summary It is possible to exclude sin 2 2  at >90% C.L. in the following circumstances: –The value of  m 2 is greater than eV 2 –The value of sin 2 2  is at, or close to, the current S-K 90% lower bound –Total integrated p.o.t. > 7.4  There is a (very) small part of the S-K 90% allowed region where we can exclude sin 2 2  at >99% (assuming 25  p.o.t.) Life would be much better if the mean value of L/E for MINOS were lower by a factor of 2 –Should be an easier measurement off-axis