(((2)^(1/2)/(2*cos(x)*cos(x))+(2)^(1/2)*a/(cos(x))-a^2)*e^(-(2^(1/2))/(a*cos(x)))+(-(2^(1/2)/2)-2^(1/2)*a+a^2)*e^(-(2^(1/2)/a)))^2
from,
\(\overline { U_{ B } } =\cfrac { \mu _{ o } }{ 2 } \left\{ \cfrac { qv }{ 4\pi r^{ 2 } } .\cfrac { r^{ 2 }_{ e } }{ a^{ 2 }_{ e } } \right\} ^{ 2 }\\ \left\{ \left( { \cfrac { \sqrt { 2 } }{ 2cos^{ 2 }(\theta _{ 0 }) } +\cfrac { \sqrt { 2 } a_{ e } }{ r_{ e }cos(\theta _{ 0 }) } -\cfrac { a^{ 2 }_{ e } }{ r^{ 2 }_{ e } } } \right) e^{ -\cfrac { r_{ e }\sqrt { 2 } }{ a_{ e }cos(\theta _{ 0 }) } }+\left( -\cfrac { \sqrt { 2 } }{ 2 } -\sqrt { 2 } \cfrac { a_{ e } }{ r } +\cfrac { a^{ 2 }_{ e } }{ r^{ 2 }_{ e } } \right) e^{ -\cfrac { r_{ e }\sqrt { 2 } }{ a_{ e } } } \right\} ^{ 2 }\)
There is two unique regions of oscillation along \(\theta_0\) when,
\(\theta=0\)
and when,
\(\theta_0=\theta_d=double\,\,\, root\)
around the double root of \(U_B\). Together they give the absorption profile of the body. When \(a\) is low, as shown in the next plot, there is only one region of oscillation, and the absorption profile of the body is flat.
This is a plot of \(U_B\) with varies value of \(a\),
Low ratios of the size of the electron and its orbit, has low values of \(T^{+}\), the \(T^{+}\) particles are not strongly held in orbit but leaves readily. If a material has high \(a\) at high temperature, then it will retain \(T^{+}\) strongly and not feel hot to the touch. It is an heat insulator. If a material has low \(a\) at high temperature, then it will not retain \(T^{+}\) as strongly and will feel hot to the touch as it loses \(T^{+}\) particle readily.
Which leads us to, the \(T\) field around \(T^{+}\) and the \(E\) field generated by it.
Is the \(T\) field responsible for heat flow between a hot and cold object or is the \(E\) field responsible? But we know for sure that an \(E\) field directed at the orbiting \(T^{+}\) particle will push or pull it (the particle is spinning) into oscillations or caused it to be ejected from orbit.
Next stop, ohmic heating...
Note: For double roots,
\(\left( \cfrac { \sqrt { 2 } }{ 2 } (2+x)+\sqrt { 2 } \cfrac { a_{ e } }{ r_{ e } } \right) xe^{ x }=(1-e^{ x })\left( \cfrac { \sqrt { 2 } }{ 2 } +\sqrt { 2 } \cfrac { a_{ e } }{ r } -\cfrac { a^{ 2 }_{ e } }{ r^{ 2 }_{ e } } \right) \)
where \( \frac { 1 }{ cos(\theta _{ 0 }) } =1+x\)...Obviously \(x=0\) is a solution.
Anyway,
\(y=x\left( Ax+B \right) \frac { e^{ x } }{ (1-e^{ x }) } -C\)
where \( x=\frac { 1 }{ cos(\theta _{ 0 }) }-1 \).
\( A=\cfrac { \sqrt { 2 } }{ 2 } \)
\( B=\sqrt { 2 } (1+\cfrac { a_{ e } }{ r_{ e } } )\)
\( C=\cfrac { \sqrt { 2 } }{ 2 } +\sqrt { 2 } \cfrac { a_{ e } }{ r } -\cfrac { a^{ 2 }_{ e } }{ r^{ 2 }_{ e } } \)
is a good fit for the equation. The only physical constraint is \(\cfrac{a_e}{r_e}\), the ratio of electron radius to electron orbital radius.
