No Charge But Thermal Gravity To The Rescue

The problem with an expression like,

$F_{ h }=Fsin(\theta )=\cfrac { q_{ p }q }{ 4\pi \varepsilon _{ o }r^{ 2 } } sin(\theta )$

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.