From the post "Perhaps Temperature Is..." dated 13 May 2016,
\(T=\cfrac { T^{ - } }{ 4\pi \tau _{ o }r^{ 2 } } \)
\( T=\cfrac { G_{ { B }o } }{ 4\pi } \cfrac { g^{ + }.v_{ { g } }\times \hat { r_{ g } } }{ r_{ g }^{ 2 } } \)
\( E=\cfrac { \tau _{ { E }o } }{ 4\pi } \cfrac { T^{ + }v_{ { T } }\times \hat { r } }{ r_{ { T } }^{ 2 } } \)
The field due to \(T^{-}\) act against, the spinning \(g^{+}\) particle and pulls the captured \(T^{+}\) in the weak field, increasing \(r_T\),
\(\cfrac { m_{ T }c^{ 2 } }{ r_{ { T } } } =\cfrac { G_{ { B }o }T^{+} }{ 4\pi } \cfrac { g^{ + }.v_{ { g } }\times \hat { r_{ g } } }{ r_{ g }^{ 2 } } -\cfrac { T^{ - } T^{+}}{ 4\pi \tau _{ o }r^{ 2 } } \)
where the sign of \(T^{-}\) has already been accounted for.
Substituting for \(r_T\),
\(\cfrac { 1 }{ r_{ { T } } } =\cfrac { {T^{+}}^2 }{ m_{ T }c^{ 2 } } \left( \cfrac { G_{ { B }o } }{ 4\pi } \cfrac { g^{ + }.v_{ { g } }\times \hat { r_{ g } } }{ r_{ g }^{ 2 } } -\cfrac { T^{ - } }{ 4\pi \tau _{ o }r^{ 2 } } \right) \)
\( E=\cfrac { \tau _{ { E }o } }{ 4\pi } \cfrac { T^{ + }v_{ { T } }\times \hat { r } }{ r_{ { T } }^{ 2 } } =\cfrac { \tau _{ { E }o } }{ 4\pi } (T^{ + }v_{ { T } }\times \hat { r } )\cfrac {{T^{+}}^2 }{ m_{ T }^{ 2 }c^{ 4 } } \left( \cfrac { G_{ { B }o } }{ 4\pi } \cfrac { g^{ + }.v_{ { g } }\times \hat { r_{ g } } }{ r_{ g }^{ 2 } } -\cfrac { T^{ - } }{ 4\pi \tau _{ o }r^{ 2 } } \right) ^{ 2 }\)
\(E=C \left( A-B.T^{ - } \right) ^{ 2 }\)
where,
\( C=\cfrac { \tau _{ { E }o }(T^{ + }v_{ { T } }\times \hat { r }) }{ 4\pi m_{ T }^{ 2 }c^{ 4 } }{T^{+}}^2 \)
\( B=\cfrac { 1 }{ 4\pi \tau _{ o }r^{ 2 } } \)
\( A=\cfrac { G_{ { B }o } }{ 4\pi } \cfrac { g^{ + }.v_{ { g } }\times \hat { r_{ g } } }{ r_{ g }^{ 2 } } \)
\(T^{+}\) is fixed in the inner orbit of the nucleus. \(T^{-}\) is the negative temperature particle free to flow.
Since,
\(V=-\cfrac{q}{4\pi\varepsilon_o r_e}\)
\(V^2=\cfrac{q}{4\pi\varepsilon_o}.\cfrac{q}{4\pi\varepsilon_o r^2_e}=\cfrac{q}{4\pi\varepsilon_o}.E\)
\(V^2=\cfrac{q}{4\pi\varepsilon_o}.C \left( A-B.T^{ - } \right) ^{ 2 }\)
\(V=\left(\cfrac{q}{4\pi\varepsilon_o}.C\right)^{1/2}\left( A-B.T^{ - } \right) \)
So, energy, \(E_i\)
\(E_i\propto\left( A-B.T^{ - } \right)\)
where the sign of \(T^{-}\) has already been accounted for.
As \(T^{-}\) increases, temperature \(T\) decreases, \(E_i\) decreases. As \(T^{-}\) decreases, temperature \(T\) increase, \(E_i\) increases.
This corrects a previous post "Perhaps Temperature Is..." dated 13 May 2016.
