Arcadian Functor

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Marni D. Sheppeard

Sunday, February 22, 2009

M Theory Lesson 264

Recall that circulants are always magic as well as square magic in the sense that the sum of squares along a row or column is a fixed constant. In particular, any Koide mass matrix M has this property.

But MUB operators such as F3 are not magic. For example, the action of F3F2 (the neutrino mixing matrix) on M results in a 1×2 block matrix in terms of the square roots of the masses, because F3 diagonalises, and F2 then acts on a pair of mass eigenvalues. The fact that this matrix is not magic is the same as the statement that m1m2. The fact that it is not square magic follows from the statement that s<0 in

cosδ=s-6vs,

where δ is the angle shared by all three masses. This property is shared by the hadron fits.

2 Comments:

Blogger CarlBrannen said...

Okay, that was too mysterious. What is s and v. Oh wait. That's my notation for
m = v + s cos (delta + 2n pi/3).

The hadrons have s<0 but the leptons took s>0 and still had m1 not equal to m2.

There are two ways to get m1=m2 corresponding to sqrt(m1) = + or - sqrt(m2). The usual way would be to have delta = 0. The sqrt(m1) = -sqrt(m2) way depends on s/v.

February 23, 2009 2:02 AM  
Anonymous Anonymous said...

Yeah, sorry, Carl. It was too much trouble to write out the matrices today. Anyway, I'm not sure these kind of products will help. I started with the magic conditions on a certain matrix, and derived the conditions on s and v from there.

Remote Kea

February 23, 2009 8:49 AM  

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