Proton Emits Photon
 A proton-core that establishes a frame-of-reference and the origin for a three-dimensional space containing a photon.

Recall that the quark model of a proton is a bundle of twelve quarks

$\sf{p^{+}} \leftrightarrow \mathrm{4}\bar{\sf{d}} + \mathrm{4}\sf{b} + \mathrm{4}\bar{ \sf{t} }$

Let us combine this proton with an electromagnetic field such as one might find in an atom

$\mathscr{F} \leftrightarrow \mathrm{4}\sf{d} + 2\sf{m} + 2\sf{a} + \sf{e} + \sf{g} + 2\bar{\sf{m}} + 2\bar{\sf{a}} + \bar{\sf{e}} + \bar{\sf{g}}$

Then the excited proton has enough quarks to establish a spatial orientation and three-dimensional position. These quarks can be parsed as a photon

$\gamma _{\tiny{\bigcirc}} \leftrightarrow \mathrm{4}\sf{d} + 2\sf{m} + 2\sf{a} + \sf{e} + \sf{g} + \mathrm{4}\bar{\sf{d}} + 2\bar{\sf{m}} + 2\bar{\sf{a}} + \bar{\sf{e}} + \bar{\sf{g}}$

together with a baryonic proton-core noted by

+$\leftrightarrow \mathrm{4}\sf{b} + \mathrm{4}\bar{ \sf{t} }$

We could say that the proton has had an interaction with an electromagnetic field, and emitted a photon

$\sf{p^{+}} + \mathscr{F} \longrightarrow$+$+ \ \gamma _{\tiny{\bigcirc}}$

The proton-core does not contain any dynamic quarks, so its wavevector and wavelength are both nil

$\overline{\kappa} \, \large{(}$+$\large{)}$$\; = (0, 0, 0)and\lambda \, \large{(}+\large{)}$$\; = 0$

It does not contain any up or down seeds, so it has no angular momentum

$𝘑 \, \large{(}$+\large{)}\begin{align} \; \equiv \frac{ \, \left| \, N^{\mathsf{U}} - N^{\mathsf{D}} \, \right| \, }{8} =0 \end{align} and therefore the core of the proton is not rotating. This means that its orbital radius is also zero R \, \large{(}+\large{)}\begin{align} \; \equiv \frac{hc}{2 \pi } \frac{ \sqrt{𝘑 }}{E} = 0 \end{align}

So the proton-core is a point shaped, charged particle. We use it as a frame-of-reference F that is perfectly inertial and non-rotating. This frame is then employed to describe other particles in a three-dimensional space centered on the proton-core

$\sf{F} =$+

The photon $\gamma _{\tiny{\bigcirc}}$ is somwhere in this space. But since it has a balanced number of leptonic seeds

$N^{\, \mathsf{M}} \left( \gamma _{\tiny{\bigcirc}} \right) = N^{\, \mathsf{A}} \left( \gamma _{\tiny{\bigcirc}} \right)$and$N^{\, \mathsf{E}}\left( \gamma _{\tiny{\bigcirc}} \right) = N^{\, \mathsf{G}} \left( \gamma _{\tiny{\bigcirc}} \right)$

it does not have a definite spatial orientation or position. We cannot say exactly where it is unless it is absorbed by some more precisely located particle. The photon $\gamma _{\tiny{\bigcirc}}$ is a gamma-ray. Click here for a spreadsheet with more detail about three-dimensional quark arrangements and calculations of particle characteristics.

page revision: 110, last edited: 04 Feb 2019 18:51