Hydrogen, And Everyone Has His Spins

For hydrogen,

Consider the forces on the hydrogen nulceus,

$\cfrac{q.q}{4\pi\varepsilon_o(2{r}_{pe}sin(\theta))^2}=2\cfrac{q.q}{4\pi\varepsilon_o{r}^2_{pe}}sin(\theta)$

$\cfrac { 1 }{ (2sin(\theta ))^{ 2 } } =2sin(\theta )$

$sin(\theta )=\cfrac { 1 }{ 2 } ,\quad \quad \quad cos(\theta )=\cfrac { \sqrt { 3 }  }{ 2 } , \theta=30^o$

${r}_{ee}=2{r}_{pe}cos(\theta)=\sqrt{3}{r}_{pe}$

Both electrons are in circular motion about the axis joining the two positive nucleus at radius $\cfrac{1}{2}{r}_{ee}$,

$\cfrac{2{m}_{e}c^2}{{r}_{ee}}=2\cfrac{q.q}{4\pi\varepsilon_o{r}^2_{pe}}.cos(\theta)-\cfrac{q.q}{4\pi\varepsilon_o{r}^{2}_{ee}}$

$\cfrac { { m }_{ e }c^{ 2 } }{ { r }_{ pe }cos(\theta ) } =2\cfrac { q.q }{ 4\pi \varepsilon _{ o }{ r }^{ 2 }_{ pe } } .cos(\theta )-\cfrac { q.q }{ 4\pi \varepsilon _{ o }(2{ r }_{ pe }cos(\theta ))^{ 2 } }$

${ r }_{ pe }=\cfrac { q^{ 2 }cos(\theta ) }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \{ 2cos(\theta )-\cfrac { 1 }{ (2cos(\theta ))^{ 2 } } \}$

${ r }_{ pe }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \cfrac { \sqrt { 3 }  }{ 2 } \{ 2\cfrac { \sqrt { 3 }  }{ 2 } -\cfrac { 1 }{ 3 } \}$

${ r }_{ pe }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \{ \cfrac { 3 }{ 2 } -\cfrac { \sqrt { 3 }  }{ 6 } \}$

${ r }_{ pe }$ = (1.60217657e-19)^2/(4*pi*8.8541878176e-12*9.10938291e-31*299792458^2)*(3/2-(3)^(0.5)/6)

${ r }_{ pe }$ = 3.41e-15 m,   ${ r }_{ ee }$ = 5.91e-15 m,  $\theta$ = 30^o


The problem with this calculation is, why should not the hydrogen be in circular motion also?  If the hydrogen nucleus is in circular motion about the axis joining the two electrons, then instead of the first equation,

$\cfrac{{m}_{pr}c^2}{{r}_{pe}sin(\theta)}=2\cfrac{q.q}{4\pi\varepsilon_o{r}^2_{pe}}sin(\theta)-\cfrac{q.q}{4\pi\varepsilon_o(2{r}_{pe}sin(\theta))^2}$

${ r }_{ pe }=\cfrac { q^{ 2 }sin(\theta ) }{ 4\pi \varepsilon _{ o }{ m }_{ pr }c^{ 2 } } \{ 2sin(\theta )-\cfrac { 1 }{ (2sin(\theta ))^{ 2 } } \}$

${ r }_{ pe }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }{ m }_{ pr }c^{ 2 } } \{ 2sin^{ 2 }(\theta )-\cfrac { 1 }{ 4sin(\theta ) } \}$    ----(*)

Together with,

${r}_{ee}=2{r}_{pe}cos(\theta)$  and,

 $\cfrac { 2{ m }_{ e }c^{ 2 } }{ { r }_{ ee } } =2\cfrac { q.q }{ 4\pi \varepsilon _{ o }{ r }^{ 2 }_{ pe } } .cos(\theta )-\cfrac { q.q }{ 4\pi \varepsilon _{ o }{ r }^{ 2 }_{ ee } }$

or,

$\cfrac { { m }_{ e }c^{ 2 } }{ { r }_{ pe }cos(\theta ) } =2\cfrac { q.q }{ 4\pi \varepsilon _{ o }{ r }^{ 2 }_{ pe } } .cos(\theta )-\cfrac { q.q }{ 4\pi \varepsilon _{ o }(2{ r }_{ pe }cos(\theta ))^{ 2 } }$

${ r }_{ pe }=\cfrac { q^{ 2 }cos(\theta ) }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \{ 2cos(\theta )-\cfrac { 1 }{ (2cos(\theta ))^{ 2 } } \}$

${ r }_{ pe }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \{ 2cos^2(\theta )-\cfrac { 1 }{ 4cos(\theta ) } \}$    ----(**) 

Dividing (*) b (**)

$1=\cfrac{{m}_{e}}{{m}_{pr}}\begin{matrix}{2sin^{ 2 }(\theta )-\cfrac { 1 }{ 4sin(\theta ) }}  \\ \overline {  {2cos^2(\theta )-\frac { 1 }{ 4cos(\theta ) }}} \end{matrix}$

$\cfrac{{m}_{pr}}{{m}_{e}}$ = 1.67262178e-27/(9.10938291e-31) = 1836.152

If we let $x=sin(\theta)$,

$1836.152=\begin{matrix}{2x^{ 2 }-\cfrac { 1 }{ 4x}}  \\ \overline {  {2(1-x^2)-\cfrac { 1 }{ 4\sqrt{1-x^2} }}} \end{matrix}$

$x$ has 3 solutions,

$x$ = -0.0000778025    $x$ = -0.865838,
$\theta$ = -0.00445775489^o and -60^o

and

$x$ = sin(\theta) = 0.865898,  $\theta$ = 59.985^o 

This is twice the angle when the positive charges were considered not spinning.

${r}_{pe}$ =  (1.60217657e-19)^2/(4*pi*8.8541878176e-12*1.67262178e-27*299792458^2)*(2*(0.865898)^2-1/(4*0.865898))

${r}_{pe}$ = 1.85e-18 m,   ${r}_{ee}$ = 2*1.85e-18*(1-(0.865898)^2)^(0.5) = 1.85e-18 m

This distances are both shorter than the electron orbit a hydrogen atom by itself.  The configuration is a two equilateral triangles with a common side.  Both electrons and proton are in circular motion at light speed in planes that are perpendicular to each other, about a common center.  Would they collide?

Distance between positive charges ${r}_{pp}$ is,

${r}_{pp}=2*sin(\theta)*{r}_{pe}$ = 2*0.865898*1.85e-18 = 3.20e-18 m

The charges will always keep their distances and spin about the common center that is also their C.G.