From the previous post "Still Not",
\(\cfrac{\partial\,\psi}{\partial\,t}=\cfrac{\partial\,\psi}{\partial\,x}\cfrac{\partial\,x}{\partial\,t}+\cfrac{\partial\,\psi}{\partial\,x^{'}}\cfrac{\partial\,x^{'}}{\partial\,t}=-\cfrac{ 2m } { \mu _{ o }q x}(iv)\left\{ \cfrac { \partial ^{ 2 }\, V }{ \partial \, x^{ 2 } } \right\} \)
Since
\(\cfrac{\partial\,\psi}{\partial\,x^{'}}\cfrac{\partial\,x^{'}}{\partial\,t}=\cfrac{\partial\,\psi}{\partial\,x^{'}}\cfrac{i}{i}\cfrac{\partial\,x^{'}}{\partial\,t}=\cfrac{\partial\,\psi}{\partial\,x}\cfrac{\partial\,x}{\partial\,t}\)
as \(x=ix^{'}\), and
\(\cfrac{\partial\,x}{\partial\,t}=iv=i\cfrac{\partial\,x^{'}}{\partial\,t}\),
we have the gradient of the energy density around the particle modeled as a dipole as,
\(\cfrac{\partial\,\psi}{\partial\,x}=-\cfrac{ m } { \mu _{ o }q x}\left\{ \cfrac { \partial ^{ 2 }\, V }{ \partial \, x^{ 2 } } \right\} \)
The particle has a energy density wave around it and emanating from it along the radial line, The change in energy density with radial distance is given by the above expression. If we start from a potential minimum,
\(\cfrac { \partial \, V }{ \partial \, x }=0 \)
\(\cfrac { \partial ^{ 2 }\, V }{ \partial \, x^{ 2 } }\gt 0 \)
then the energy density around the particle decreases with radial distance away from it. The total energy in such a scenario is not infinite as,
\(\int{\cfrac{\partial\,\psi}{\partial\,x}}\partial\,x=\psi=-\int{\cfrac{ m } { \mu _{ o }q x}\left\{ \cfrac { \partial ^{ 2 }\, V }{ \partial \, x^{ 2 } } \right\}}\partial\,x \)
\(\psi =-\cfrac { m }{ \mu _{ o }qx } \cfrac { \partial \, V }{ \partial \, x } -\int { \cfrac { 2m }{ \mu _{ o }qx^{ 2 } } \cfrac { \partial \, V }{ \partial \, x } } \partial x\)
\( \psi =-\cfrac { m }{ \mu _{ o }qx } \cfrac { \partial \, V }{ \partial \, x } -\cfrac { 2m }{ \mu _{ o }qx^{ 2 } } V-\int { \cfrac { 2m }{ \mu _{ o }qx^{ 3 } } V } \partial x\)
Since,
\( E=\cfrac { \partial \, V }{ \partial \, x }=\cfrac { q }{ 4\pi \varepsilon _{ o }x^{ 2 } } \quad \)
where \(E\) is in the positive \(x\) direction, (the negative end of the dipole is spinning behind the positive end; both travelling in the \(x\) direction).
\( V=-\cfrac { q }{ 4\pi \varepsilon _{ o }x } \quad \)
We have,
\( \psi =-\cfrac { m }{ \mu _{ o }qx } \cfrac { q }{ 4\pi \varepsilon _{ o }x^{ 2 } } \quad +\cfrac { 2m }{ \mu _{ o }qx^{ 2 } } \cfrac { q }{ 4\pi \varepsilon _{ o }x } +\int { \cfrac { 2m }{ \mu _{ o }qx^{ 3 } } \cfrac { q }{ 4\pi \varepsilon _{ o }x } } \partial x\)
\( \psi = \cfrac {m}{ 4\pi \mu _{ o } \varepsilon _{ o }x^3}+\int { \cfrac { m }{ 2\pi \mu _{ o }\varepsilon _{ o }x^{ 4 } } } \partial x=\cfrac { mc^{ 2 } }{ 4\pi } \cfrac { 1 }{ x^{ 3 } } -\cfrac { mc^{ 2 } }{ 6\pi } \cfrac { 1 }{ x^{ 3 } } \)
where \(\cfrac{1}{\mu _{ o }\varepsilon _{ o }}=c^2\)
\(\psi =\cfrac { mc^{ 2 } }{ 12\pi } \cfrac { 1 }{ x^{ 3 } } \)
and
\(\int _{ a }^{ b\rightarrow\infty }{ \psi }\, \partial x=\cfrac { mc^{ 2 } }{24 \pi } \cfrac { 1 }{ a^{2} }=+finite\)
where \(a\) is the radius of the particle, possibly the radius of the spinning part of the dipole.
A positive finite answer to the last integral suggests that such a energy density field is possible. From the post "Maybe Not", \(\psi\) exist as waves. Since such energy density waves (\(\psi\) waves) are possible, we can then postulate:
The two energy waves formulated in the post "Maybe Not"; one that envelops the dipole and the other that radiates from it, are the reasons why such a particle has wave property.
