From the previous post "Still Not",
∂ψ∂t=∂ψ∂x∂x∂t+∂ψ∂x′∂x′∂t=−2mμoqx(iv){∂2V∂x2}
Since
∂ψ∂x′∂x′∂t=∂ψ∂x′ii∂x′∂t=∂ψ∂x∂x∂t
as x=ix′, and
∂x∂t=iv=i∂x′∂t,
we have the gradient of the energy density around the particle modeled as a dipole as,
∂ψ∂x=−mμoqx{∂2V∂x2}
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,
∂V∂x=0
∂2V∂x2>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,
∫∂ψ∂x∂x=ψ=−∫mμoqx{∂2V∂x2}∂x
ψ=−mμoqx∂V∂x−∫2mμoqx2∂V∂x∂x
ψ=−mμoqx∂V∂x−2mμoqx2V−∫2mμoqx3V∂x
Since,
E=∂V∂x=q4πεox2
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=−q4πεox
We have,
ψ=−mμoqxq4πεox2+2mμoqx2q4πεox+∫2mμoqx3q4πεox∂x
ψ=m4πμoεox3+∫m2πμoεox4∂x=mc24π1x3−mc26π1x3
where 1μoεo=c2
ψ=mc212π1x3
and
∫b→∞aψ∂x=mc224π1a2=+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", ψ exist as waves. Since such energy density waves (ψ 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.
