Consider the \(\psi\) of a particle expanding from \(x=x_a\) to \(x=x_c\), \(\psi\) is at velocity \(c\) along \(ix\), the gain in energy is,
\(\Delta E=E_{x_c}-E_{x_a}=2\pi x_cmc-2\pi x_amc\)
But,
\(\Delta E=\bar F.\Delta x=2\pi mc(x_c-x_a)\)
so,
\(\bar F=2\pi mc\)
We know that,
\(F=-\psi\)
where \(F\) the Newtonian force due to \(\psi\). The average force on \(\psi\) is,
\(\bar F =-(-\bar \psi)=\bar\psi=2\pi mc\)
as the particle's \(\psi\) expanded from \(x_a\) to \(x_c\). Through out this process \(KE=\cfrac{1}{2}mc^2\) remains unchanged.
The average acceleration,
\(\bar a_c=2\pi c\)
Since the particle is in circular motion, and \(\bar F\) is along a radial line,
\(\bar F=2\pi c=\cfrac{c^2}{\bar x}\)
and
\(\bar x=\cfrac{c}{2\pi}\)
What is \(\bar x\)? \(\bar F\) is a hypothetical average force, that extends up to \(\bar x\) and is zero beyond. But be warned, all mathematical interpretations are fiction next to bullshit. Bullshit on the other hand, bulls will testify, is for real.
Have a nice day.
