Since this is simply an extension to basic pre-existing quantum mechanics, there is no need to reformulate or revisit all of quantum mechanics - that exercise is left up to the reviewer.
This equation can be solved to get R_E or we can use and earlier post get get the expression for R_E I developed in 1987-1991: The Answers
R_E={\alpha^2\over \pi R_H}
Since we are dealing presently with mono-protonic situations (single proton hydrogen in a low energy, non-ionized, unperturbed state) R_H is used rather than R_{\infty}.
What is this electron radius and why is it so huge? From the equation
M_PR_P=4LM=M_ER_E
one can see that the torque / spin of the proton is matched by the electron - for every action there is an equal and opposite reaction, and the length is such that the product of the electron radius and the electrom mass EQUALS the same 4LM (Haramein's M_PR_P=4LM equation where
M_P=Mass\;of\;Proton
R_P=Radius\;of\;Proton
L=Planck\;Length
M=Planck\;Mass
M_E=Mass\;of\;Electron
R_E=Radius\;of\;Electron
.
.
.
and this balances both the electron and proton to a mass ratio of
\mu={\alpha^2\over {\pi r_p R_H}}=1836.15267
R_E={\alpha^2\over \pi R_H}=1.54463707 10^{-12} m = 1.54463707 pm
Radius\;of\;Proton={4LM \over M_P}=0.841235641fm
The mass of the electron is
M_E={M_PR_P\pi R_H\over \alpha^2}
Arc length, when dealing with radians, is simply the product of the radius and the angle subtended. Since we are doing the electron for 180 degrees, or \pi radians: S=\pi Radius
S_{E1}={\alpha^2 \over R_H}=4.85262048pm
For comparison, if we made an arc length using the proton radius:
S_{proton radius - arc}=R_P \pi= {4Lsqrt(hbar*c/G)*\pi \over M_P}=2.64281971fm
theses values are way shorter than the latest plasmon wavelength observed in single-wall nanotubes.
More later on interpreting this.
The Surfer, OM-IV
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