Oops Too Hot!

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.