\(F=-\psi\)
\(F_{\rho}=-\cfrac{\partial\,\psi}{\partial x}\)
but,
\(F=\int_V{F_{\rho}}dV\)
So,
\(F=\int_V{-\cfrac{\partial\,\psi}{\partial x}}dV\)
\(F=\int_A{-\psi}dA\)
that means,
\(\psi=\int_A{\psi}dA\)???
This is because,
\(F\ne\int_V{F_{\rho}}dV\)
but,
\(F=\cfrac{1}{4\pi x^2}\int_V{F_{\rho}}dV\)
where \(x\) is the radius of the volume integrated over with, \(F\) is the force along a radial line \(x\) and it is the force distributed over the surface of the volume, \(V\).
\(\int_V{F_{\rho}}dV\)
is the sum of all \(F\)s over the area of the sphere.
\(\int_V{F_{\rho}}dV=4\pi x^2.F\)
And so,
\(\because dV=4\pi x^2.dx\) at \(x\)
\(F=\cfrac{1}{4\pi x^2}\int_0^x{-\cfrac{\partial\,\psi}{\partial x}}4\pi x^2.dx\)
\(F=\cfrac{1}{4\pi x^2}\int_0^x{-4\pi x^2}d\psi\)
as \(x\) is the independent variable.
\(\require{cancel}\)
\(F=\cfrac{1}{\cancel{4\pi x^2}}\cancel{4\pi x^2}.\int_0^x{-1}.d\psi\)
\(F=-\psi\) at \(x\)
What happen then to
\(\cfrac{\partial\,L}{\partial x}=-\cfrac{\partial}{\partial x}\left\{\int_V\psi dV\right\}=\int_V{-\cfrac{\partial\,\psi}{\partial x}}dV=\int_V{F_{\rho}}dV=F\)
from the post "On Second Thought... Potential Density" dated 24 Jul 2015?
\(F\) here is the force on the particle not the force due to the particle's \(\psi\) distributed around it. This force is more accurately expression as,
\(\psi_n-\psi(x)\)
\(F=F_n-F(x)\)
where \(F(x)\) is the force due to the containing particle. The force \(F\) on the orbiting particle is the result of the interaction between its energy density, \(\psi_n\) and the surrounding \(\psi(x)\) due to the containing particle.
Have a nice day.
