Too Hot! Keep Your Distance.

Consider the term,

$c^{ 2 }=\left| \cfrac { a^{ 4 }_{ e } }{ r^{ 4 }_{ e } } \cfrac { cos^{ 2 }(\phi /2) }{ sin^{ 2 }(\phi /2) } e^{ -\cfrac { 4r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) }  } \right| ^{ \theta _{ 0 } }_{ 0 }$

$x=r_{ e }\sqrt { 2(1-cos(\phi )) } =2r_{ e }sin(\phi /2)$

$c^{ 2 }=\left| \cfrac { 4a^{ 4 }_{ e } }{ x^{ 2 } } cos^{ 2 }(\phi /2).e^{ -\cfrac { 2x }{ a_{ e }cos(\theta ) }  } \right| ^{ \theta _{ 0 } }_{ 0 }$

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.