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.
