Consider the ψ of a particle expanding from x=xa to x=xc, ψ is at velocity c along ix, the gain in energy is,
ΔE=Exc−Exa=2πxcmc−2πxamc
But,
ΔE=ˉF.Δx=2πmc(xc−xa)
so,
ˉF=2πmc
We know that,
F=−ψ
where F the Newtonian force due to ψ. The average force on ψ is,
ˉF=−(−ˉψ)=ˉψ=2πmc
as the particle's ψ expanded from xa to xc. Through out this process KE=12mc2 remains unchanged.
The average acceleration,
ˉac=2πc
Since the particle is in circular motion, and ˉF is along a radial line,
ˉF=2πc=c2ˉx
and
ˉx=c2π
What is ˉx? ˉF is a hypothetical average force, that extends up to ˉx 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.