### 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 $F_3$ are not magic. For example, the action of $F_3 F_2$ (the neutrino mixing matrix) on $M$ results in a $1 \times 2$ block matrix in terms of the square roots of the masses, because $F_3$ diagonalises, and $F_2$ then acts on a pair of mass eigenvalues. The fact that this matrix is not magic is the same as the statement that $m_1 \neq m_2$. The fact that it is not square magic follows from the statement that $s < 0$ in

$\textrm{cos} \delta = \frac{s - 6v}{s}$,

where $\delta$ is the angle shared by all three masses. This property is shared by the hadron fits.

But MUB operators such as $F_3$ are not magic. For example, the action of $F_3 F_2$ (the neutrino mixing matrix) on $M$ results in a $1 \times 2$ block matrix in terms of the square roots of the masses, because $F_3$ diagonalises, and $F_2$ then acts on a pair of mass eigenvalues. The fact that this matrix is not magic is the same as the statement that $m_1 \neq m_2$. The fact that it is not square magic follows from the statement that $s < 0$ in

$\textrm{cos} \delta = \frac{s - 6v}{s}$,

where $\delta$ is the angle shared by all three masses. This property is shared by the hadron fits.

## 2 Comments:

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.

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.

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