### Mass Update

Carl Brannen has been surreptitiously posting Koide mass formulas for pi meson and (the lightest) rho meson triplets at PF. For $n = 1,2,3$ and that damned number $\delta \simeq \frac{2}{9}$, the square root mass eigenvalues (for the same choice of units) all take the form

$\lambda_{n} = v + 2s \cdot \textrm{cos} (\delta + \frac{2n \pi}{3})$

where the parameters $v$ and $s$ must be set to

lepton: $v = \frac{1}{\sqrt{2}} , s = 1$

pi: $v = \frac{6}{5} , s = \frac{-3}{4}$

rho: $v = \frac{10}{7} , s = \frac{-1}{3}$

Now I must find time to check these against the PDG data...

$\lambda_{n} = v + 2s \cdot \textrm{cos} (\delta + \frac{2n \pi}{3})$

where the parameters $v$ and $s$ must be set to

lepton: $v = \frac{1}{\sqrt{2}} , s = 1$

pi: $v = \frac{6}{5} , s = \frac{-3}{4}$

rho: $v = \frac{10}{7} , s = \frac{-1}{3}$

Now I must find time to check these against the PDG data...

## 2 Comments:

I also downloaded the LaTeX template for Foundations of Physics and am busily typing.

Right now I'm still trying to figure out how much I should put in a paper. Clearly I have to put the Koide formula in there, but I'm not sure if I should put my parallel solution to the weak hypercharge and weak isospin problem (i.e. the set of quadratic equations whose solutions are these quantum numbers).

Along that line, I finally figured out why it is that the up and down quark have the same mass. A better way of describing the situation is that they have the same mass vector. (Recall Michael Rios' comment? Did I remember his name right?)

The mass vector has two components, the v(valence) and the s(sea). To get the up and down the same, these two components must add to the same total mass vector. Then masses will be the same. But the electic charge depends on the v (or valence) part of the vector, so the up and down quark are distinguished by electric charge. This sort of reminds me of J = L+S, I wonder if there are some similar calculational techniques.

Yes, a degenerate eigenvalue set can arise from delta=0, for which one obtains

m1 = v + 2s

m2 = m3 = v - s

(if I got my cosines right). I was thinking about the proton and neutron last night, whilst lying awake like a child the night before Xmas (Neutrino08 starts this afternoon). Maybe I will write another post after I plug some numbers in. Anyway, note that if v = s then there is only one non-zero mass equal to 3s, and there is a dual case where delta = pi where there is only one zero mass and two equal masses. Lots of nice symmetry to play with.

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