\(v^2=-\cfrac { 1 }{ m } \psi _{ c } \cfrac { \psi _{ c }+u }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }\)
and,
\(0\gt \cfrac { \psi _{ c }+u }{ u } { e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }\gt-2\) --- (*)
this implies that the velocity is once again along \(x\) not perpendicular to \(x\).
Also,
\(L=-\cfrac { 1 }{ 2 } \psi _{ c }\left\{2-\left| \cfrac { \psi _{ c }+u }{ u } { e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }\right| \right\} -\psi _{ max }\)
is not indicative of the type of motion the particle is in. Furthermore even with (*) and \(\psi_c\lt0\), the orbiting particle is at most at ground state, \(\psi_o\), its energy density is not negative.
However, if we make \(\psi_n=\psi_{max}-\delta\), where \(\delta\gt0\) then from,
\(v^2=-\cfrac { 1 }{ m } \psi _{ c } \cfrac { \psi _{ c }+u }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }\)
we have,
\(v^2=-\cfrac { 1 }{ m } .(-\delta)\cfrac { -\delta+u }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }\)
\(v^2=+\cfrac { 1 }{ m } \delta \cfrac { -\delta+u }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }\)
\(v^2=-\cfrac { 1 }{ m } \delta \cfrac { -\delta+u }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }\)
and we let \(\delta=-\psi\),
\(v^2=-\cfrac { 1 }{ m } (-\psi) \cfrac { \psi+u }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }\) --- (**)
the particle is then still in orbit about \(x=x_z\), but with "\(-\psi\)". The factor,
\(\cfrac { \psi+u }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }\)
is dimensionless. The change in sign for \(v^2\) is adjusted by taking the negative root. Graphically,
the particle have "\(-\psi\)" and that results in a heavy particle. This happen when the transition to a lower energy state upon the release of a photon go past \(\psi_{max}\) of the containing particle. In effect the orbiting particle given its velocity \(v\) and location \(x\), is squeezed by the surrounding \(\psi\) of the containing particle.
Must the orbiting particle passes beyond its ground state \(\psi_o\), and be in the absolute \(-\psi\) region, for it to manifest heavy mass? As far as the expression (**) is concerned, no. The particle is in "\(-\psi\)" when it transits pass \(\psi_{max}\), it may actually be above \(\psi_o\). Once we obtained (**) we may write,
\(c^2\int_V{m_{\rho\,particle}} dV =c^2\int_V{m_\rho} dV-\int_V(-{\psi})dV=c^2\int_V{m_\rho} dV+\int_V{\psi}dV\)
Mathematically, \(-\psi\) is just as valid and the particle just manifest more mass. But is taking the negative root justifiable?
Fiction is my game. In my dreams, yes I can!
