Pions
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Here are some quark models of pions. All pions are built-up around the heart of familial seeds shown on the left. Particles with a different spin or charge are modeled by attaching various quarks around this common kernel. Then excited states are obtained by adding even more quarks. The pions may share other quarks in common, beyond these familial seeds. But this nugget is the minimum necessary to distinguish the pions from other nuclear particle families. Particles are classified on this basis. WikiMechanics analyzes the physics of pions using chains of events noted by $\Psi = \left( \sf{\Omega}_{1}, \sf{\Omega}_{2}, \sf{\Omega}_{3} \ \ldots \ \right)$ where each repeated cycle $\, \sf{\Omega} \,$ is composed of the following quarks.

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All the foregoing quark models perfectly replicate the quantum numbers of pions. They also accurately represent observed values for lifetimes, widths and the mass. With very few exceptions, they are within experimental uncertainty. But with so many quarks it is difficult to see how the models work. So to view the underlying pattern, we remove most of the quark/anti-quark pairs. These $\begin{align} \sf { q \overline{q} } \end{align}$ pairs are needed for equilibrium. Without them, many excited states are so unstable that they are not measurable. But the field of $\begin{align} \sf { q \overline{q} } \end{align}$ pairs obscures the minimum number of quarks required to identify a particle and account for its mass. These minima are called core coefficients. They show more clearly how excited pions are built-up over blocks of the same baryonic quarks.

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For more detail about calculations please see these spreadsheets. Experimentally observed values are taken from this referenceXlink.png.

The pions are great little particles because they are small enough to see how the excited states are structured as the total number of quarks is increased. And we can also notice some other details. For example, pions present an uncluttered relationship between baryonic and electronic quarks. There are no muonic quarks in the ground states. So they are a good case for considering the baryonic quarks to be bound together by a simple electric force with no magnetic component. Also, uniquely among all nuclear particles, $H-U =0$ for the negative pion. This makes makes $\pi ^{-}$ an especially pointy charge.

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