\(F_{ h }=Fsin(\theta )=\cfrac { q_{ p }q }{ 4\pi \varepsilon _{ o }r^{ 2 } } sin(\theta )\)
from the post "Not This Way", is that \(q_p\) don't exist.
Even if the photon is a dipole, its net charge from afar is zero.
There is one other possibility, that is for the matter-antimatter annihilation process to be slow. That popular literature on explosive matter/anti-matter reaction may not be true. Take the case of a hydrogen atom, when the electron collide into the proton nucleus,
the rate of annihilation along the charge-time line is,
\(E_a=\cfrac { d\, q }{ dt } =\cfrac { \, d(q_{ pr }+q_{ e }) }{ dt } =\cfrac { dm }{ dt } =\cfrac { \, d(m_{ pr }+m_{ e }) }{ dt } \)
And \(E_a\) is slow. The reside charge on each of the particles forms a dipole and the energy released from \(E=mc^2\) make this both a very hot particle and a electric dipole. This is a \(H\) plasma particle, \(p_p\), that is experiencing a observable decay. What would its decay half life, \(p_pT\) be? This hot dipole is a likely candidate for photon, both mechanisms for acceleration to light speed/terminal speed (as a dipole or hot particle) can apply to this particle.
Moreover since,
\(\cfrac { \, d(q_{ pr }+q_{ e }) }{ dt } =\cfrac { \, d(m_{ pr }+m_{ e }) }{ dt } \)
we have a charge mass equivalence,
\(\int^0_{q_{ pr }+q_{ e }}{1.}d\,q=\int^{m_{ pr }-m_{ e }}_{m_{ pr }+m_{ e }}{1.}d\,m\)
\(-(q_{pr}+q_{e})=-2m_{e}\) and \(q_{pr}=q_{e}\)
we should have,
\(q_{e}=m_{e}\)
the resulting neutral hot particle has mass \(m_{pr}-m_{e}\) , likely a neutron. In this instance, both mass and charge are treated as inertia whether they are on the positive or negative time line. We also have,
\(q_{pr}=m_{e}\)
that all the positive charge in a proton is from a mass of \(m_e\).
It is more likely,
\(q_{e}=M_cm_{e}\) where \(M_c\) is a scaling factor that also adjust for unit dimension, ie charge per unit mass, C kg-1.
