$PE$ Ain't Physical Education

If,

$KE_{min}=\cfrac{1}{2}mv^2_{min}$

$KE=\cfrac{1}{2}mv^2$

what is $PE$?  Obviously,

$PE=\psi_n$

And we form the Lagrangian,

$L=\cfrac{1}{2}mv^2-\psi_n$

From the post "Nature Has Orbits" dated 19 Jul 2015,

$v^2=-\cfrac { 1 }{ m }  \psi _{ c } \cfrac { \psi _{ c }+u   }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }$

where $u=ln(cosh(x- x_{ z }))$

$v$ being complex being perpendicular to the radial line along $x$.

Therefore,

$L=-\cfrac{1}{2}m.\cfrac { 1 }{ m }  \psi _{ c } \cfrac { \psi _{ c }+u   }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }-\psi_n$

We see that $m$ cancels off immediately,  which is a good thing as this equation is for all types of $\psi$, in $t_c$, $t_g$ and $t_T$.

$L=-\cfrac { 1 }{ 2 } \psi _{ c }\cfrac { \psi _{ c }+u }{ u } { e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }-\psi _{ n }$

Given,

$\psi_c=\psi _{ n }-\psi_{max}$

$\psi_n=\psi _{ photon }+\psi_{o}$

$\psi_n$ is the elevated energy level of the particle from its ground state $\psi_o$ after receiving the photon $\psi_{photon}$.  $\psi_{max}$ is the maximum $\psi$ of the confining particle cloud.

$L=-\cfrac { 1 }{ 2 } \psi _{ c }\cfrac { \psi _{ c }+u }{ u } { e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }-\psi _{ c }-\psi _{ max }$

$L=-\cfrac { 1 }{ 2 } \psi _{ c }\left\{ \cfrac { \psi _{ c }+u }{ u } { e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }+2 \right\} -\psi _{ max }$

This is the Lagrangian of a particle being in the $\psi$ cloud of another particle of the same type, either positive or negative.  It is possible for both of them to be positive or negative, and for them to have different sign.

The relation,

$\cfrac{d}{dt}\left\{\cfrac{\partial\,L}{\partial\dot{x}}\right\}=\cfrac{\partial\,L}{\partial x}$

will lead us back to the Newtonian force but without the involvement of mass, $m$.  We note that

$F_{\rho}=-\cfrac{d\,\psi}{dx}$

where $F_{\rho}$ is the force density.

From the post "Opps! Lucky Me" dated 25 May 2015, the Newtonian force $F$,

$F=-\psi$

where $\psi$ is energy density and $F$ is in $N$, the newton.  This relation give us the resultant of two objects with interacting $\psi$,

$F_{1\,2}=\psi_1-\psi_2$  and

$F_{2\,1}=\psi_2-\psi_1$

irrespective of the notation, we have,

$F_{1\,2}=-F_{2\,1}$

Newton's Third Law of motion, equal opposite action and reaction.

Furthermore, in the Lagrangian,

$L=-\cfrac { 1 }{ 2 } \psi _{ c }\left\{ \cfrac { \psi _{ c }+u }{ u } { e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }+2 \right\} -\psi _{ max }$

the negative sign before the $T$ or $KE$ term allows for a negative sign without making mass negative forcefully.  And the $+2$ constant multiplied to $\psi_c$ allows for the term,

$\cfrac { \psi _{ c }+u }{ u } { e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }$

to dip two below zero, without introducing negative energy.

In the latter point, a negative $\psi$ results in, from the post "We Still Have A Problem" dated 23 Nov 2014,

$m_{\rho\,particle} c^2=m_\rho c^2-\int^{x_a}_{0}{\psi}dV$

where it was proposed the particle's total $\psi$ and its mass, $m_{\rho\,particle}$ in space add to give the mass, $m_\rho$ along the time dimension in which particle exist.

If $\psi\to-\psi$,

$m_{\rho\,particle} c^2=m_\rho c^2-\int^{x_a}_{0}{(-\psi)}dV=m_\rho c^2+\int^{x_a}_{0}{\psi}dV$

The particle gained mass density, above it inertia density (from $m_{\rho}c^2$), along the respective time dimension that it exist.

(We would want $-\psi$ to see a gain in mass, for example. Now the negative sign can come from (*) below.)

This could be the mechanism by which photon gain mass.

$0\gt \cfrac { \psi _{ c }+u }{ u } { e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }\gt-2$  -- (*)

and still have the same Lagrangian (the same physic behavior like a photon).  From, the velocity equation,

$v^2=-\cfrac { 1 }{ m }  \psi _{ c } \cfrac { \psi _{ c }+u   }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }=-\cfrac { 1 }{ m }(-\psi _{ c })\left|  \cfrac { \psi _{ c }+u   }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } } \right|$

and so,

$L=-\cfrac { 1 }{ 2 } \psi _{ c }\left\{2-\left|  \cfrac { \psi _{ c }+u }{ u } { e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }\right| \right\} -\psi _{ max }$

with condition (*), the Lagrangian is still of the same form, which implies that the particle behave the same as far as the principle of least action is concerned.

Since,

$\psi_c=\psi _{ n }-\psi_{max}$

when $\psi_{max}\gt\psi_{n}$ as we superimpose more particles of the same as the containing particle or remove the previously gained photon of a orbiting particle at its $v_{min}$.  We can effect (*), where the particles in orbit gain mass.  And because of the upper bound in (*), this occurs when $\psi_c$ is small and the orbiting particle is close to $x=x_z$ at low speed.

This is how we make heavy particles, lest they are still orbiting particles.

Have a nice day.

Note:

$m_{\rho\,particle} c^2=m_\rho c^2-\int^{x_a}_{0}{\psi}dV$

should be more accurately written as,

$c^2\int_V{m_{\rho\,particle}} dV =c^2\int_V{m_\rho} dV-\int_V{\psi}dV$

as volume integrals over the extend of the particle.