Second Guessing Again

I was wrong.  From the post "Pound To Rescue Permittivity" dated 30 May 2016,

$\cfrac{q}{\varepsilon_o}=4\pi a^2_{\psi}m.2c^2ln(cosh(\pi))$

where we generalized to

$\cfrac{q}{\varepsilon_o}=4\pi a^2_{\psi}m.2c^2ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }a_{\psi}))$

but for the case of a basic particle, from the post "New Discrepancies And Hollow Earth" dated 23 Jun 2016,

$ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }a_{\psi\,c}))=\cfrac{1}{4}$

and a big particle,

$ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }a_{\psi\,\pi}))=1$

The charge of a basic particle and a big particle differ by a factor of $\cfrac{1}{4}$.

So, in the post "And They Danced" dated 18 Jul 2016,

$\cfrac{1}{4}\cfrac{q^2}{4\pi \varepsilon_o r_e^2}=v.\cfrac{m_ev^2}{r_e}$

the new factor $\cfrac{1}{4}$ still cancels in the ratios of mass and orbital radii derived here,

$\cfrac{r_p}{r_e}=\cfrac{m_ev_e^3}{m_pv_p^3}$

But as we rearrange,

$\cfrac{q^2}{4\pi \varepsilon_o r_e^2}=v.\cfrac{4m_ev^2}{r_e}$

$m_e\rightarrow 4m_e$

because we have assumed that the electric charge on a electron in orbit is the same as the electric charge of a free electron.  If the experimenter has also made the same assumption, this would introduce a factor of four to the quoted experiment value of the mass of electron.  The quoted mass of an electron is,

$m_e=\cfrac{1}{4}m_e^{'}$

where $m_e^{'}$ is the value for electron mass used here.

The published result of the ratio of electron mass to proton mass is,

$\cfrac{m_e}{m_p}=\cfrac{1}{1836.152 673 89}$

If we replace $c$ with $2\pi c$ as the electron is in circular motion,  the expression

$\cfrac{q^2}{4\pi \varepsilon_o r_e}=m_ec^3$

from the post "And They Danced" dated 18 Jul 2016, becomes

$\cfrac{q^2}{4\pi \varepsilon_o r_e}=m_e(2\pi c)^3$

When we associated the extra factor with $m_e$,

$\cfrac{q^2}{4\pi \varepsilon_o r_e}=(2\pi)^3m_e.c^3$

$m_e\rightarrow (2\pi)^3m_e$

The missing factor of $2\pi$ accounts for circular motion.  This factor cancels as we take the ratio, but if the experiment to obtain the mass of an electron is based on linear speed, the factor of $2\pi$ for circular motion will reduce mass by a factor of $(2\pi)^3$ here.

$m_e=(2\pi)^3m_e^{'}$

Since the proton as the nucleus is in spin, and we consider its rotational kinetic energy with the use of moment of inertia, $I$

$I=\cfrac{1}{2}\left(\cfrac{2}{5}m\right)r^2\omega^2=\cfrac{1}{2}\left(\cfrac{2}{5}m\right)v^2$

$m\rightarrow \cfrac{2}{5}m$

we introduced the factor $\cfrac{2}{5}$ to the mass of the proton, $m_p$.  The actual mass should be

$m_p=\cfrac{5}{2}m_p^{'}$

where $m_p^{'}$ is the value for proton mass used here.

And if we accept the results from the post "And They Danced" dated 18 Jul 2016,

$\cfrac{m_e f_e^3}{m_p f_p^3}=\cfrac{r_p^4}{r_e^4}$

and,

$\cfrac{m_e }{m_p }=\cfrac{1}{77}$

with the following adjustments,

$m_e\rightarrow \cfrac{1}{4}m_e$

due to the assumption that the particles have the same charge magnitude.  And

$m_p\rightarrow \cfrac{5}{2}m_p$

due to the use of mass instead of moment of inertia as the proton is in spin.

And also, for linear speed as opposite to circular motion,

$m_e\rightarrow (2\pi)^3m_e$

we have,

$\cfrac{m_e}{m_p}\rightarrow \cfrac{\cfrac{1}{4}m_e (2\pi)^3}{\cfrac{5}{2}m_p }=\cfrac{1}{77}$

$\cfrac{m_e }{m_p }=\cfrac{1}{77*(2\pi)^3*\cfrac{2}{5}*\cfrac{1}{4}}$

or

$\cfrac{m_p }{m_e }={77*(2\pi)^3*\cfrac{2}{5}*\cfrac{1}{4}}=1909.98$

and if we have,

$\cfrac{m_e }{m_p }=\cfrac{1}{74}$

$\cfrac{m_p }{m_e }={74*(2\pi)^3*\cfrac{2}{5}*\cfrac{1}{4}}=1835.57$

and if,

$\cfrac{m_e }{m_p }=\cfrac{1}{75}$

$\cfrac{m_p }{m_e }={75*(2\pi)^3*\cfrac{2}{5}*\cfrac{1}{4}}=1860.37$

and if,

$\cfrac{m_e }{m_p }=\cfrac{1}{76}$

$\cfrac{m_p }{m_e }={76*(2\pi)^3*\cfrac{2}{5}*\cfrac{1}{4}}=1885.18$

Which seems to adjust for the experimental value of the ratio of proton to electron mass, but the introduction of moment of inertia adds complications to the original expression,

$\cfrac{m_e f_e^3}{m_p f_p^3}=\cfrac{r_p^4}{r_e^4}\rightarrow\cfrac{I_e f_e^3}{I_p f_p^3}=\cfrac{r_p^4}{r_e^4}$

where $I_e$ and $I_p$ are moment of inertia of the electron and proton respectively.

Lunch time!