From the post "One Plus One Plus One",
\(m_\rho=\sqrt{m_g}.\sqrt{2}cG\)
If the particle \(m_g\) is an 2D entity then instead,
\(m_g=Am_\rho\)
\(m_{ \rho }=\cfrac { 1 }{ 2Ac^{ 2 }G^{ 2 } }\)
\(m_{ \rho }=\cfrac { 1 }{ 2\pi c^{ 2 }a^{ 2}G^{ 2 } }\)
where \(a\) is the radius of the disc that defines \(m_g\). And \(m_{ \rho }=0\) is also valid.
The particle is two dimensional because \(\psi\) is oscillating between two space dimensions only.
