Last time we saw that a set of three Koide eigenvalues

$\lambda_{i} = \sqrt{m_{i}} = 1 + 2 \sqrt{r} \textrm{cos} (\delta + \omega_{i})$

gave a Koide ratio of

$\frac{(\lambda_{1} + \lambda_{2} + \lambda_{3})^{2}}{m_{1} + m_{2} + m_{3}} = \frac{3}{1 + 2r}$

and that the parameter $2r$ could be interpreted as $\textrm{tan}^{2} \phi$. For example, there exists a Koide triplet for the three neutral meson generations associated to an antibottom quark, namely light bottomonium $\eta_{b}$, $B_{d}^{0}$ and $B_{s}^{0}$. This triplet consists of the masses $5279$, $5367$, $9389$ MeV$/c^{2}$ and can be fitted with parameters

$\delta = 0.0218$

$2r = 0.0194$

and a scale parameter of $6551.3$. The conjugate of $\delta$ results in the same set of masses, with the two lightest values interchanged. The Koide eigenvalues in this example are

$(\lambda_{1}, \lambda_{2}, \lambda_{3}) = (0.8977, 0.9052, 1.1972)$

and one can verify that their sum equals $3$. Using all three parameters provides endless fun in searching for new mass triplets.

8 years ago

So if we add the $2/9$ to the phases, the two eigenvalue triplets become:

ReplyDelete$+ \delta: (1.1914, 0.8631, 0.9456)$

$- \delta: (1.1932, 0.8694, 0.9374)$

which give similar, but not equal, masses.