\(\oint{B}d\,r=2\pi rB_o=\mu _{ o }\varepsilon _{ o }\cfrac { \partial \, E_{ A } }{ \partial t }\)
\(2\pi rB_o=\mu _{ o }\varepsilon _{ o }2\pi E_{ o }cos(\phi /2)\cfrac { d\, x_{ \bot } }{ d\, t } \cfrac { x^{ 2 }_{ \bot } }{ a_{ e } } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi,\theta)} { d\, (cos(\theta )) }\)
If we retain \(r\) and label it \(r_p\), the orbital radius of the nucleus then,
\(r_pB_o=\mu _{ o }\varepsilon _{ o }E_{ o }cos(\phi /2)\cfrac { d\, x_{ \bot } }{ d\, t } \cfrac { x^{ 2 }_{ \bot } }{ a_{ e } } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi,\theta)} { d\, (cos(\theta )) }\)
\(r_pB_o=\mu _{ o }\varepsilon _{ o }(\cfrac{e}{4\pi\varepsilon_or_e^2})cos(\phi /2)\cfrac { d\, x_{ \bot } }{ d\, t } \cfrac { x^{ 2 }_{ \bot } }{ a_{ e } } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi,\theta)} { d\, (cos(\theta )) }\)
Since,
\(\cfrac { d\, x_{ \bot } }{ d\, t } x^{ 2 }_{ \bot } =r_{ e }cos(\phi )\cfrac { d\phi }{ dt }(2r_{ e }sin(\phi /2))^2\)
\(\cfrac { d\, x_{ \bot } }{ d\, t } x^{ 2 }_{ \bot } =4r^{ 3 }_{ e }cos(\phi )\cfrac { d\phi }{ dt } sin^{ 2 }(\phi /2)\)
And so,
\(r_{ p }B_{ o }=\\\mu _{ o }\varepsilon _{ o }(\cfrac { e }{ 4\pi \varepsilon _{ o }a_{ e }r_{ e }^{ 2 } } )cos(\phi /2)(4r^{ 3 }_{ e }cos(\phi )\cfrac { d\phi }{ dt } sin^{ 2 }(\phi /2))\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) }\)
\(B_{ o }=\cfrac { \mu _{ o }e }{ 4\pi a_{ e } } \cfrac { r_{ e } }{ r_{ p } } sin(2\phi )sin(\phi /2)\cfrac { d\phi }{ dt } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) }\)
\(B_o\) is generated by the electron in circular motion of radius \(r_e\), so \(B_o\) is proportional to \(r_e\). Such a field decreases with distance from the plane containing the orbit of the electron, so \(B_o\) is inversely proportional to \(r_p\). Consider the energy contained in such a field.
\(U_B=\cfrac{1}{2}\cfrac{B^2_o}{\mu_o}\)
When the nucleus has a positive charge or a net positive charge of one by reciprocity,
\(r_e=r_p\) and so, \(\left(\cfrac{r_e}{r_p}\right)^2=1\)
\(U_B=\cfrac{1}{2\mu_o}\left( \cfrac { \mu _{ o }e }{ 4\pi a_{ e } } \right) ^{ 2 }\left( sin(2\phi )sin(\phi /2)\cfrac { d\phi }{ dt } \right) ^{ 2 }\left( \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) } \right) ^{ 2 }\)
\(U_B=\cfrac{\mu_o}{32}\left( \cfrac { e }{ \pi a_{ e } } \right) ^{ 2 }\left( sin(2\phi )sin(\phi /2)\cfrac { d\phi }{ dt } \right) ^{ 2 }\left( \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) } \right) ^{ 2 }\)
When such a system is subjected to heat, both \(r_e\) and \(r_p\) will oscillate and so does the magnetic field. The value \(\theta_o\) will also varies as \(r_p\) changes. When \(B_o\) collapses from a higher value to a lower value, energy is radiated. This radiated energy varies with the value of \(\theta_o\), given by the expression above.
The average energy over one period, (power) is given by,
\(\bar{U_B}=\cfrac{1}{T}\int^{T}_{0}{U_B}dt=\cfrac{1}{T}.\int^{T}_{0}{\cfrac{\mu_o}{32}\left( \cfrac { e }{ \pi a_{ e } } \right) ^{ 2 }\left( sin(2\phi )sin(\phi /2)\cfrac { d\phi }{ dt } \right) ^{ 2 }\left( \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) } \right) ^{ 2 }}dt\)
\(\bar{U_B}=\cfrac{1}{T}\cfrac{\mu_o}{32}\left( \cfrac { e }{ \pi a_{ e } } \right) ^{ 2 }.\int^{\phi_T}_{0}{\left( sin(2\phi )sin(\phi /2) \right) ^{ 2 }\left( \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) } \right) ^{ 2 }}\cfrac { d\phi }{ dt } d\phi\)
\(\bar{U_B}=\cfrac{\omega}{T}\cfrac{\mu_o}{32}\left( \cfrac { e }{ \pi a_{ e } } \right) ^{ 2 }\int^{\phi_T}_{0}{\left( sin(2\phi )sin(\phi /2)\right) ^{ 2 }\left( \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) } \right) ^{ 2 }}d\phi\)
This is the integral of two functions involving \(\phi\) since,
\(h(\phi,\theta)=e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) } }\)
Consider,
\(\Psi=\int { \left( sin(2\phi )sin(\phi /2) \right) ^{ 2 }\left( \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) } \right) ^{ 2 } } d\phi \)
\( \Psi=\int{ \left( 4sin(\phi /2)cos(\phi /2)cos(\phi )sin(\phi /2) \right) ^{ 2 }\left( \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) } \right) ^{ 2 } } d\phi \)
\(\Psi=-\int { \left( 4sin^2(\phi /2)cos(\phi ) \right) ^{ 2 }\left(\cfrac { a_{ e } }{ r_{ e } }\right)^2.\\ \left( \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 3 }(\theta ) } \left( -h(\phi ,\theta )\cfrac { r_{ e } cos(\phi /2)}{ a_{ e }cos(\theta ) } \right) } { d\, (cos(\theta )) } \right) ^{ 2 } } d\phi \)
Becasue,
\( h(\phi ,\theta )=e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) } }\)
\( \cfrac { dh(\phi ,\theta ) }{ d\phi } =e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) } }\left(-\cfrac { 2r_{ e }}{ a_{ e }cos(\theta ) }\right) cos(\phi /2)\cfrac { 1 }{ 2 } \)
\( \cfrac { dh(\phi ,\theta ) }{ d\phi }=-h(\phi ,\theta )\cfrac { r_{ e }cos(\phi /2) }{ a_{ e }cos(\theta ) } \)
We have,
\( \Psi=-\int { 16sin^4(\phi /2)cos^2(\phi )\left(\cfrac { a_{ e } }{ r_{ e } }\right)^2\left( \cfrac { d }{ d\phi } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 3}(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) } \right) ^{ 2 } } d\phi \)
From,
\( \cfrac { r_{ p } }{ x_{ \bot } } =tan(\theta )\)
\( x_{ \bot }=2r_{ e }sin(\phi /2)cos(\phi /2)=r_{ e }sin(\phi )\)
\( r_{ p }=r_{ e }sin(\phi )tan(\theta ),\quad but\quad r_{ p }=r_{ e }\)
\( cot(\theta )=sin(\phi )\)
\( -sin(\theta )d\theta =cos(\phi )sin^{ 3 }(\theta )d\phi =dcos(\theta )\) --- (*)
Despite of this relationship, \(\phi\) and \(\theta\) are still independent variable. The integration over \(\theta\) is taken for each value of \(\phi\) as \(\phi\) revolves around the orbit defined by \(r_e\).
Substitute (*) into \(\Psi\),
\(\Psi=-16\left(\cfrac { a_{ e } }{ r_{ e } }\right)^2 \int { sin^4(\phi /2)cos^2(\phi )\left( \cfrac { d }{ d\phi } \int _{ 0 }^{ \theta_o}{ \cfrac { 1 }{ cos^{ 3 }(\theta ) } } h(\phi ,\theta ) { cos(\phi )sin^{ 3 }(\theta )d\phi } \right) ^{ 2 } } d\phi \)
\( \Psi=-16\left(\cfrac { a_{ e } }{ r_{ e } }\right)^2\int { sin^4(\phi /2)cos^2(\phi )\left( \cfrac { 1 }{ cos^{ 2 }(\theta ) } h(\phi ,\theta ){ sin^{ 3 }(\theta ) } \right) ^{ 2 } } d\phi \)
\(\Psi=-16\left(\cfrac { a_{ e } }{ r_{ e } }\right)^2\cfrac { sin^{ 6 }(\theta ) }{ cos^{ 4 }(\theta ) } \int {sin^4(\phi /2)cos^2(\phi )\left( h(\phi ,\theta ) \right) ^{ 2 } } d\phi \)
\(\Psi= 4\cfrac { a^3_{ e } }{ r^3_{ e } } \cfrac { sin^{ 3 }(\theta ) }{ cos^{ 2 }(\theta ) }. \int { sin^{ 3 }(\phi /2)cos(\phi )\left( h(\phi ,\theta ) \right) ^{ 2 } } \left( -\cfrac { 4r_{ e }sin(\phi /2) }{ a_{ e } } \cfrac { 1 }{ cos^{ 2 }(\theta ) } \right) cos(\phi )sin^{ 3 }(\theta )d\phi \)
Substitute \(*\) into \(\Psi\) again,
\(\Psi= 4\cfrac { a^3_{ e } }{ r^3_{ e } } \cfrac { sin^{ 3 }(\theta ) }{ cos^{ 2 }(\theta ) } \int { sin^{ 3 }(\phi /2)cos(\phi )\left( h(\phi ,\theta ) \right) ^{ 2 }\left( -\cfrac { 4r_{ e }sin(\phi /2) }{ a_{ e } } \cfrac { 1 }{ cos^{ 2 }(\theta ) } \right) } dcos(\theta )\)
\(\Psi= -4\cfrac { a^3_{ e } }{ r^3_{ e } } \cfrac { sin^{ 3 }(\theta ) }{ cos^{ 2 }(\theta ) } \int{ sin^{ 3 }(\phi /2)cos(\phi )\left( h(\phi ,\theta ) \right) ^{ 2 }\left( -\cfrac { 4r_{ e }sin(\phi /2) }{ a_{ e } } \cfrac { sin(\theta ) }{ cos^{ 2 }(\theta ) } \right) } d\theta \)
Because,
\( h(\phi ,\theta )^2=e^{ -\cfrac { 4r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) } }\)
\(\cfrac { d\left( h(\phi ,\theta ) \right) ^{ 2 } }{ d\theta } =h(\phi ,\theta )^2\left(-\cfrac { 4r_{ e }sin(\phi /2) }{ a_{ e }}\cfrac{sin(\theta) }{cos^2(\theta ) }\right)\)
And so,
\(\Psi=- 4\cfrac { a^3_{ e } }{ r^3_{ e } } \cfrac { sin^{ 3 }(\theta ) }{ cos^{ 2 }(\theta ) } \int { sin^{ 3 }(\phi /2)cos(\phi )\cfrac { d\left( h(\phi ,\theta ) \right) ^{ 2 } }{ d\theta } } d\theta \)
\(\Psi=-4 \cfrac { a^3_{ e } }{ r^3_{ e } } \cfrac { sin^{ 3 }(\theta ) }{ cos^{ 2 }(\theta ) } \cfrac { d }{ d\theta } \int { sin^{ 3 }(\phi /2)cos(\phi ) } \left( h(\phi ,\theta ) \right) ^{ 2 }d\theta \)
\(\Psi= -4\cfrac { a^3_{ e } }{ r^3_{ e } } \cfrac { sin^{ 3 }(\theta ) }{ cos^{ 2 }(\theta ) } sin^{ 3 }(\phi /2)cos(\phi )\left( h(\phi ,\theta ) \right) ^{ 2 }\)
And the average energy over one period (power) in the \(B\) field is,
\(\bar{U_B}=-\cfrac{\omega}{T}\cfrac{\mu_o}{8}\left( \cfrac { e }{ \pi a_{ e } } \right) ^{ 2 }\cfrac { a^3_{ e } }{ r^3_{ e } }\left\{\cfrac { sin^{ 3 }(\theta ) }{ cos^{ 2 }(\theta ) }\right\}^{\theta_o}_{0}2 \left\{sin^{ 3 }(\phi /2)cos(\phi )\left( h(\phi ,\theta_o ) \right) ^{ 2 }\right\}^{\pi}_{0}\)
\(\bar { U_{ B } } =-\cfrac { \omega ^{ 2 } }{ 2\pi } \cfrac { \mu _{ o } }{ 8 } \left( \cfrac { e }{ \pi } \right) ^{ 2 }\cfrac { a_{ e } }{ r^{ 3}_{ e } } \cfrac { sin^{ 3 }(\theta _{ o }) }{ cos^{ 2 }(\theta _{ o }) } 2\left\{ -1.\left( h(\pi ,\theta _{ o }) \right) ^{ 2 } \right\} \)
\(\bar { U_{ B } } =\cfrac { \mu _{ o }\omega ^{ 2 }e^{ 2 }a_{ e } }{ 8\pi ^{ 3 }r^{ 3 }_{ e } } \cfrac { sin^{ 3 }(\theta _{ o }) }{ cos^{ 2 }(\theta _{ o }) } \left( h(\pi ,\theta _{ o }) \right) ^{ 2 }\)
\(\bar { U_{ B } } =\cfrac { \mu _{ o }\omega ^{ 2 }e^{ 2 } a_{ e }}{ 8\pi ^{ 3 }r^{ 3 }_{ e } } \cfrac { sin^{ 3 }(\theta _{ o }) }{ cos^{ 2 }(\theta _{ o }) } e^{ -\cfrac { 4r_{ e } }{ a_{ e }cos(\theta _{ 0 }) } }\)
A plot of (sin(x))^3*e^{-1.1/cos(x)}*1/(cos(x))^2 for 0<x<π/2 is given below,
There is no catastrophe. When subjected to high perturbations,( high temperature etc), both orbits oscillate. High \(\theta_o\) corresponds to high orbit. When the charge return from high values of \(\theta_o\), higher amount of energy are released as the potential energy difference between those at high orbits and that at the centers of the oscillation are higher. So, high \(\theta_o\) corresponds to higher radiated energy.
If this interpretation is correct then there is no need for discrete amount of energy either.
