\(c^{ 2 }=\left| \cfrac { a^{ 4 }_{ e } }{ r^{ 4 }_{ e } } \cfrac { cos^{ 2 }(\phi /2) }{ sin^{ 2 }(\phi /2) } e^{ -\cfrac { 4r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) } } \right| ^{ \theta _{ 0 } }_{ 0 }\)
\( x=r_{ e }\sqrt { 2(1-cos(\phi )) } =2r_{ e }sin(\phi /2)\)
\( c^{ 2 }=\left| \cfrac { 4a^{ 4 }_{ e } }{ x^{ 2 } } cos^{ 2 }(\phi /2).e^{ -\cfrac { 2x }{ a_{ e }cos(\theta ) } } \right| ^{ \theta _{ 0 } }_{ 0 }\)
As \(\phi\) goes from \(0\) to \(\pi\), \(x\) goes from \(0\) to \(2r_e\). The \(\cfrac{1}{x}\) term explodes at \(x=0\).
It is too hot!
UNLESS, the electron and the \(T^{+}\) particle keep their distance. The \(T^{+}\) particle has a second spin, and spins with the same speed (possibly \(v=c\)) as the electron. In this way \(x\ne0\).
Since the expression for energy is the expression \(c^2\) integrated over \(\phi\),
\(c^{ 2 }=\left| \cfrac { 4a^{ 4 }_{ e } }{ x^{ 2 } } cos^{ 2 }(\phi /2).e^{ -\cfrac { 2x }{ a_{ e }cos(\theta ) } } \right| ^{\theta _{ 0 } }_{ 0 }=0\)
gives a turning point in \(\overline{U_B}\). Which occurs when,
as,
\(1-sin^{ 2 }(\phi /2)=1-\cfrac { x^{ 2 } }{ 4r^{ 2 }_{ e } } =cos^{ 2 }(\phi /2)\)
we have,
\( c^{ 2 }=\left| \cfrac { 4a^{ 4 }_{ e } }{ x^{ 2 } } \left( 1-\cfrac { x^{ 2 } }{ 4r^{ 2 }_{ e } } \right) e^{ -\cfrac { 2x }{ a_{ e }cos(\theta ) } } \right| ^{ \theta _{ 0 } }_{ 0 }=0\)
\(x^2=4r^2_e\), \(x=2r_e\)
When the \(T^{+}\) particle and the electron are at the opposite ends of a diameter along the orbit, \(U_B\) is at an extrememum.
How does the \(E\) field change and induce a \(B\) field when their relative positions are fixed? They are both spinning.
If this is true them the expression for \(U_B\) need only be evaluated only at \(\phi=\pi\), without the average over time. When \(\phi=\pi\),
\(\require{cancel}\)
\(\overline { U_{ B } } =\cfrac { \mu _{ o } }{ 2 } \left\{ \cfrac { qv }{ 4\pi r^{ 2 } } .\cfrac { r^{ 3 }_{ e } }{ a^{ 3 }_{ e } } \right\} ^{ 2 }\cfrac { a^2_{ e } }{ r^2_{ e } } \\ \left( { \left| \cfrac { 1 }{ cos^{ 2 }(\theta ) } e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) } }2cos^{ 2 }(\phi /2)sin(\phi /2) \right| ^{\theta _{ 0 } }_{ 0 }\\+\left| \cfrac {2 a_{ e } }{ r_{ e }cos(\theta ) } e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) } }cos(\phi /2) \right| ^{ \theta _{ 0 } }_{ 0 }\\ -\left| \cfrac { a^{ 2 }_{ e } }{ r^{ 2 }_{ e } } cot(\phi /2)e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) } } \right| ^{ \theta _{ 0 } }_{ 0 } } \right) ^{ 2 }\)
\(U_B=0\), very funny!
A plot of 2*cos(x/2)^2*sin(x/2), cos(x/2) and cot(x/2) are given below,
\(U_B\) is indeed zero at \(\phi=\pi\). But \(B_o=0\) also as \(U_B=\frac{B^2_o}{2\mu_o}\). The \(T^{+}\) particle will not be held in orbit.
But at \(\phi=\frac{\pi}{2}\), there is a point of inflection. where,
\(x=\sqrt{2}r_e\), \(cos(\theta)=\cfrac{x}{\sqrt{r^2_e+x^2}}=\cfrac{\sqrt{2}}{\sqrt{3}}\)
\(\overline { U_{ B } } =\cfrac { \mu _{ o } }{ 2 } \left\{ \cfrac { qv }{ 4\pi r^{ 2 } } .\cfrac { r^{ 3 }_{ e } }{ a^{ 3 }_{ e } } \right\} ^{ 2 }\cfrac { a^{ 2 }_{ e } }{ r^{ 2 }_{ e } } \\ \left\{ { \left( \cfrac { \sqrt { 2 } }{ 2cos^{ 2 }(\theta _{ 0 }) } +\cfrac { \sqrt { 2 } a_{ e } }{ r_{ e }cos(\theta _{ 0 }) } -\cfrac { a^{ 2 }_{ e } }{ r^{ 2 }_{ e } } \right) e^{ -\cfrac { \sqrt { 2 } r_{ e } }{ a_{ e }cos(\theta _{ 0 }) } }+\left( -\cfrac { \sqrt { 2 } }{ 2 } -\sqrt { 2 } \cfrac { a_{ e } }{ r_{ e } } +\cfrac { a^{ 2 }_{ e } }{ r^{ 2 }_{ e } } \right) e^{ -\cfrac { \sqrt { 2 } r_{ e } }{ a_{ e } } } } \right\} ^{ 2 }\)
\(\overline { U_{ B } } =\cfrac { \mu _{ o } }{ 2 } \left\{ \cfrac { qv }{ 4\pi r^{ 2 } } .\cfrac { r^{ 3 }_{ e } }{ a^{ 3 }_{ e } } \right\} ^{ 2 }\cfrac { a^{ 2 }_{ e } }{ r^{ 2 }_{ e } } \\ \left\{ { \left( \cfrac { 3\sqrt { 2 } }{ 4 } +\cfrac { \sqrt { 3 } a_{ e } }{ r_{ e } } -\cfrac { a^{ 2 }_{ e } }{ r^{ 2 }_{ e } } \right) e^{ -\cfrac { \sqrt { 3 } r_{ e } }{ a_{ e } } }+\left( -\cfrac { \sqrt { 2 } }{ 2 } -\sqrt { 2 } \cfrac { a_{ e } }{ r_{ e } } +\cfrac { a^{ 2 }_{ e } }{ r^{ 2 }_{ e } } \right) e^{ -\cfrac { \sqrt { 2 } r_{ e } }{ a_{ e } } } } \right\} ^{ 2 }\)
Another solution exists at \(\phi=-\frac{\pi}{2}\). But this would cause an imbalance, a loss of symmetry. But in a paired orbit, two electrons sharing two \(T^{+}\) particles will be like,
All the particles have the same angular velocity around the major orbit, and they keep their relative distances from each other. The \(T^{+}\) particles have two spins, one around the major orbit on which two other electrons are spinning. The other spin is perpendicular to the plane of the orbit around a \(B\) orbit generated by the revolving \(e^{-}\) particles.
This is how a material gain positive temperature particles and increases temperature.
Note: In this model a \(p^{+}\) particle in circular motion generates a \(g\) field that does not interfere with a \(T\) field.
