Thursday, November 11, 2021

This Is How We Do It - FREE - Full Rydberg Equation Extracted, part 1

The Rydberg constant for hydrogen, from:
$R_M=$ Rydberg constant for reduced Mass(?)
$R_\infty=$ Rydberg constant for heavy atoms
$R_H=$ Rydberg constant for Hydrogen
$m_e=$ mass of electron
$M=$ Mass of nucleus
$m_p=$mass of proton
$\alpha=$ Fine structure constant
$r_p=$ proton radius
$\epsilon_0$ = Permittivity of free space
$h=$ Planck's constant
$c=$ speed of light
$e=$ electron charge, fundamental charge
$\pi=$ 3.1415926535897932384626433...
$1=$  1.0000000000000000000000000...

From wikipedia page on the Rydberg constant:
$R_M={R_\infty\over {1+{m_e\over M}}}$
$R_H=R_\infty{1\over {1+{m_e\over m_p}}}$
$R_H=R_\infty{m_p\over {m_e+m_p}}$
$R_H=R_\infty{1\over {1+{m_e\over m_p}}}$
https://en.wikipedia.org/wiki/Hydrogen_spectral_series

This post is simply to show how to derive the Full Rydberg Equation (FRE), a polynomial that has been solved via the sign-flip numeric iteration algorithm for all terms in the equation.  This solution can only be performed with the Full Rydberg Equation as the equation for the Rydberg constant is incomplete and not solvable for the terms which compose the equation. The stabilty and uniqueness of the solution can be shown via computer numeric solution and will be available soon in The Excel VBA / macro spreadsheet.

This derivation makes use of, as a starting point,: $$R_M={R_\infty\over {1+{m_e\over M}}}$$ however, it can also begin with the wave equation, Schrรถdinger's wave equation, as done in previous posts and whitepapers linked throughout this blog. For simplicity and clarity, this is how we do it:
$R_M={R_\infty\over {1+{m_e\over M}}}$
$R_M(1+{m_e\over M})=R_\infty$
$(1+{m_e\over M})={R_\infty\over R_M}$
$1={R_\infty\over R_M}-{m_e\over M}$
For hydrogen, 1H @ 0°K:, $M=m_p$
$1={R_\infty\over R_H}-{m_e\over m_p}$
${m_p\over m_e}={\alpha^2\over\pi r_pR_H}=1836.15267$ - see derivation-of-proton-to-electron-mass-ratio
$1={R_\infty\over R_H}-{\pi r_pR_H\over \alpha^2}$
$R_\infty={m_ee^4\over 8{\epsilon_0}^2h^3c}$
$1={{m_ee^4\over 8{\epsilon_0}^2h^3cR_H}}-{\pi r_pR_H\over \alpha^2}$


4 comments:

  1. The equation has to be formulated carefully in preparation for solving for the roots using numerical methods - iteration.

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  2. Mr. Leslie Jackson of ELHS (East Liverpool High School) circa 1978 first made me aware of this problem...

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  3. The proton radius solution points towards advanced physics such as Dan Winter's and Nassim Haramein's work and many others.

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  4. Perhaps that is why mainstream refuses to provide a theoretical solution to the proton radius puzzle since it would imply they have lost the pebble... Grasshopper has snatched the science pebble from the false master

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