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**

## No comments:

## Post a Comment

Watch the water. π¦