One Plus One Plus One

From the addition of $\psi$ and $F_\rho$,

Given that,

$F_{\rho}=i\sqrt { 2{ mc^{ 2 } } }\,G.tanh\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x-x_z) \right)$

It seems that, for charge,

$q_\rho=\sqrt { 2{ mc^{ 2 } } }\,G=\sqrt{m_e}.\sqrt{2}c.G$

without further scaling, charge per unit volume, where $m=m_e$ is the mass of one charge particle.

And for gravity particle,

$m_\rho=\sqrt { 2{ mc^{ 2 } } }\,G=\sqrt{m_g}.\sqrt{2}cG$

mass per unit volume, where $m=m_g$ is the mass of one gravity particle.

Simplifying further,  (Note:  It is possible that $m_g$ is 2D not a 3D entity)

$m_g=Vm_\rho$,

where $V$ is the volume of the particle.

$m_{ \rho  }=\sqrt { Vm_{ \rho  } } .\sqrt { 2 } cG\\ m^{ 2 }_{ \rho  }=m_{ \rho  }2Vc^{ 2 }G^{ 2 }$

$m_{ \rho  }\left( 1-m_{ \rho  }2Vc^{ 2 }G^{ 2 } \right) =0$

$m_{\rho}=0$

the solution allows for particles that are mass-less.

$m_{ \rho  }=\cfrac { 1 }{ 2Vc^{ 2 }G^{ 2 } }=\cfrac { 3 }{ 8\pi c^{ 2 }a^{ 3 }G^{ 2 } }$  this is wrong if the particle is a 2D entity.

where $a$ is the radius of the particle.

For the case of photon,

$p_\rho=\sqrt { 2{ mc^{ 2 } } }\,G=\sqrt{m_p}.\sqrt{2}c.G$

$p_\rho$ has yet to be defined, per unit volume; and $m=m_p$ is a hypothetical photon mass.  Unless photon has a hypothetical mass, the derivation for $F_\rho$ would not apply to photons.  $\psi$ would not be valid for photons either.

From the post "Bouncy Balls, Sticky Balls, Transfer Of Momentum", photons can transfer momentum, in the process expending $\psi$.  It is reasonable to define a hypothetical photon mass, with an understanding that all collisions involving photons are inelastic.  In the case where $\psi$ is totally lost, photon collisions are totally inelastic collisions.