\(\cfrac{d\,KE}{d\,t}=\cfrac{1}{2}m\cfrac{dv^2}{d\,t}=mv\cfrac{d\,v}{d\,t}=v.F\)
We consider again an electron in circular orbit around a hydrogen nucleus,
\(\cfrac{q^2}{4\pi \varepsilon_o r_e^2}=v.\cfrac{m_ev^2}{r_e}\) --- (1)
assuming that the electron is in light speed \(v=c\), and not \(v=2\pi c\) (not yet and not necessary),
\(\cfrac{q^2}{4\pi \varepsilon_o r_e}=m_ec^3\)
So,
\(r_e=\cfrac{q^2}{4\pi \varepsilon_o.m_ec^3}\)
where \(m_e\) is the electron mass, \(q\) is the electron charge.
We have a lot of a doubts upon the valid value for \(\varepsilon_o\), however, in calculations for ratios such as \(\cfrac{r_p}{r_e}\) or \(\cfrac{m_p}{m_e}\) where \(\varepsilon_o\) cancels, we can ignore the value for the constant.
With this mind, the ratio of \(\theta_\psi\) of a big particle (\(n=77\)), to a basic particle is,
\(r_{\small{Q}}=\cfrac{\theta_{\psi}}{\theta_{\psi\,c}}=\cfrac{a_{\psi}}{a_{\psi\,c}}=\cfrac{3.135009}{0.7369}=4.25432\)
But both particles have the same inertia \(\cfrac{1}{c}\), under the action of the force field, in the time dimensions. Should they have the same inertia in the space dimension?
When expression (1) is written down, the hydrogen nucleus is stationary and the electron with reference to the nucleus as a center, orbits at a radius \(r\). But the nucleus under the attraction of the electron should respond physically too.
To consider the motion of the nucleus together with the circular motion of the electron...we have a dance.
In both cases,
\(\cfrac{q^2}{4\pi \varepsilon_o r_p^2}=v.\cfrac{m_pv^2}{r_p}\)
\(r_p=\cfrac{q^2}{4\pi \varepsilon_o.m_pc^3}\)
And so,
\(\cfrac{r_p}{r_e}=\cfrac{m_e}{m_p}\)
This expression suggests that \(r_e\gt r_p\). In which case, we have instead,
Both of which are just guesses. One scenario is just as awkward as the other because they are equivalent, one electron is \(\pi\) rad out of phase compared to the other scenario.
Since, \(f=\cfrac{c}{2\pi r}\) as we have assumed that they are both at light speed,
\(\cfrac{r_p}{r_e}=\cfrac{m_e}{m_p}=\cfrac{f_e}{f_p}\)
If \(f_p\gt f_e\) then it is even more likely that the black colored proton crosses the orbit of the electron and be outside of the wider orbit, as the left part of the diagram shows. One the other hand, if \(f_e\gt f_p\) then the red colored electron can also cross the black orbit be outside of the proton orbit. When this happens the attraction force acting as the centripetal force must do work against the changing momentum of the particle in an orbit bent away from its line of action, and force the orbit inward. The particle slows, the orbital radius (inter-particle distance) decreases...and the electron eventually falls into the nucleus.
And the whole universe collapses.
If the inter-particle distance increases, at near light speed the particles will break attraction and not be in orbit around each other.
When \(f_p\ne f_e\), no matter where the starting positions of the particles are, one of the particle relative to the other will move closer to the orbit intersections at \(A\) or \(B\),
When either one of the two particles is outside the confine of the other's orbit, either the inter-particle distance decreases at light speed and the system collapses or the inter-particle distance increases at near light speed and the system breaks apart.
Could the particles move in perpendicular planes?
No, a second force will be needed for one particle to be in orbit around a perpendicular plane. Without such a force, the motions of both particles will be in the same plane.
\(f_p=f_e\) is not the only possible conclusion, Suppose,
\(v_p\ne v_e\)
that the particles are not of the same speed in spin. In that case,
\(\cfrac{r_p}{r_e}=\cfrac{m_e v_e^3}{m_p v_p^3}\)
\(\cfrac{f_e r_e}{f_p r_p}=\cfrac{v_e}{v_p}\)
and so,
\(\cfrac{m_e f_e^3}{m_p f_p^3}=\cfrac{r_p^4}{r_e^4}\) --- (*)
This expression still cannot guarantee that one particle does not move closer to the orbital intersections and crosses outside.
This is the conventional solution,
where the proton is considered much more massive that it is in a slow spin about an axis not in its interior, with the electron in wide orbit around it. Expression (*) still applies. The particles does not spin as if there is a rod holding them, but they interact with each other through their fields.
This does not indicate in any way, why the mass of a proton is more than the mass of an electron.
The wanted conclusion \(m_e=m_p\) takes a delay. And so they danced...
