The Real Escape Artiste, Ionization Energy

From the previous post "Oops Too Hot" dated 14 May 2016,

the leading coefficient before $T^{-}$ that converts it to energy is,

$V=\left( \cfrac { q }{ 4\pi \varepsilon _{ o } } .C \right) ^{ 1/2 }\left( A-B.T^{ - } \right)$

$C=\cfrac { \tau _{ { E }o }(T^{ + }v_{ { T } }\times \hat { r } ) }{ 4\pi m_{ T }^{ 2 }c^{ 4 } }{T^{+}}^2$

$\cfrac { q }{ 4\pi \varepsilon _{ o } } .C=\cfrac { \tau _{ { E }o }q(T^{ + }v_{ { T } }\times \hat { r } ) }{ \varepsilon _{ o }(4\pi )^{ 2 }m_{ T }^{ 2 }c^{ 4 } } {T^{+}}^2$

$B=\cfrac { 1 }{ 4\pi \tau _{ o }r^{ 2 } }$

$\left( \cfrac { q }{ 4\pi \varepsilon _{ o } } .C \right) ^{ 1/2 }B=\left\{ \cfrac { \tau _{ { E }o }q(T^{ + }v_{ { T } }\times \hat { r } ) }{ \varepsilon _{ o }(4\pi )^{ 2 }m_{ T }^{ 2 }c^{ 4 } } T^{+} \right\} ^{ (1/2) }\cfrac { T^{+}  }{ 4\pi \tau _{ o }r^{ 2 } }$

$=\left\{ q(T^{ + }v_{ { T } }\times \hat { r } ) \right\} ^{ (1/2) }\sqrt { \cfrac { \tau _{ { E }o } }{ \varepsilon _{ o } }  } \cfrac { T^{+} }{ (4\pi )^{ 2 }m_{ T }c^{ 2 }\tau _{ o }r^{ 2 } }$

Consider the expression,

$\left\{ q(T^{ + }v_{ { T } }\times \hat { r } ) \right\} ^{ (1/2) }\sqrt { \cfrac { \tau _{ { E }o } }{ \varepsilon _{ o } }  } \cfrac { T^{+} }{ (4\pi )^{ 2 }m_{ T }c^{ 2 }\tau _{ o }r^{ 2 } }.T^{-}$

$=\sqrt { \tau _{ { E }o }(T^{ + }v_{ { T } }\times \hat { r } ) } \sqrt { \cfrac { q }{ \varepsilon _{ o } }  } \cfrac { 1 }{ 4\pi m_{ T }c^{ 2 } } .\cfrac { T^{ - } T^{+}}{ 4\pi \tau _{ o }r^{ 2 } }$

where an arbitrary $r_a$ has been added,

$=\sqrt { \cfrac { \tau _{ { E }o }(T^{ + }v_{ { T } }\times \hat { r } ) }{ 4\pi r^{ 2 }_{ a } }  } \sqrt { \cfrac { q }{ 4\pi \varepsilon _{ o }r^{ 2 }_{ a } }  } \cfrac { r^{ 2 }_{ a } }{ m_{ T }c^{ 2 } } .\cfrac { T^{ - }T^{+} }{ 4\pi \tau _{ o }r^{ 2 } }$

$=\cfrac { 1 }{ m_{ T }c^{ 2 } } \left\{ \sqrt { \cfrac { \tau _{ { E }o }(T^{ + }v_{ { T } }\times \hat { r } ) }{ 4\pi r^{ 2 }_{ a } }  } .\sqrt { r_{ a } }  \right\} \left\{ \sqrt { \cfrac { q }{ 4\pi \varepsilon _{ o }r^{ 2 }_{ a } }  } .\sqrt { r_{ a } }  \right\} \left\{ \cfrac { T^{ - }T^{ + }  }{ 4\pi \tau _{ o }r^{ 2 } } .r_{ a } \right\}$

The term,

$\sqrt { \cfrac { \tau _{ { E }o }(T^{ + }v_{ { T } }\times \hat { r } ) }{ 4\pi r^{ 2 }_{ a } }  } .\sqrt { r_{ a } } =\sqrt{E_{\small{T^{+}}}.r_a}$

is the square root of work done per unit charge in an electric field.  The term,

$\sqrt { \cfrac { q }{ 4\pi \varepsilon _{ o }r^{ 2 }_{ a } }  } .\sqrt { r_{ a } } =\sqrt{E_q.r_a}$

is also the square root of work done per unit charge in an electric field.  The product of this two term is the work done per unit charge in an electric field.  The term,

$\cfrac { T^{ - }T^{ + }  }{ 4\pi \tau _{ o }r^{ 2 } } .r_{ a }=T.r_a$

is the work done in a temperature field.  This multiplied by the leading term,

$\cfrac { 1 }{ m_{ T }c^{ 2 } }$

is dimensionless (have no units).  So, this gives correctly the unit for $V$ as work done per unit charge.

Now Consider,

$A=\cfrac { G_{ { B }o } }{ 4\pi  } \cfrac { g^{ + }.v_{ { g } }\times \hat { r_{ g } }  }{ r_{ g }^{ 2 } }$

So,

$\left( \cfrac { q }{ 4\pi \varepsilon _{ o } } .C \right) ^{ 1/2 }A=\left\{ \cfrac { \tau _{ { E }o }q(T^{ + }v_{ { T } }\times \hat { r } ) }{ \varepsilon _{ o }(4\pi )^{ 2 }m_{ T }^{ 2 }c^{ 4 } }  \right\} ^{ (1/2) } \cfrac { G_{ { B }o } }{ 4\pi  } \cfrac { g^{ + }.v_{ { g } }\times \hat { r_{ g } }  }{ r_{ g }^{ 2 } }T^{+}$

$=\sqrt { \tau _{ { E }o }(T^{ + }v_{ { T } }\times \hat { r } ) } \sqrt { \cfrac { q }{ \varepsilon _{ o } }  } \cfrac { 1 }{ 4\pi m_{ T }c^{ 2 } } .\cfrac { G_{ { B }o } }{ 4\pi  } \cfrac { g^{ + }.v_{ { g } }\times \hat { r_{ g } }  }{ r_{ g }^{ 2 } }T^{+}$

$=\cfrac { 1 }{ m_{ T }c^{ 2 } } \left\{ \sqrt { \cfrac { \tau _{ { E }o }(T^{ + }v_{ { T } }\times \hat { r } ) }{ 4\pi r^{ 2 }_{ a } }  } .\sqrt { r_{ a } }  \right\} \left\{ \sqrt { \cfrac { q }{ 4\pi \varepsilon _{ o }r^{ 2 }_{ a } }  } .\sqrt { r_{ a } }  \right\} \cfrac { G_{ { B }o } }{ 4\pi  } \cfrac { g^{ + }.v_{ { g } }\times \hat { r_{ g } }  }{ r_{ g }^{ 2 } } T^{ + }$

The second and third terms are the same as the above, the last term,

$\cfrac { G_{ { B }o } }{ 4\pi  } \cfrac { g^{ + }.v_{ { g } }\times \hat { r_{ g } }  }{ r_{ g }^{ 2 } } T^{ + }$

is the work done in a $T$ field established by a spinning $g^{+}$ particle, which when multiplied by the leading term,

$\cfrac { 1 }{ m_{ T }c^{ 2 } }$

is dimensionless (without units).  The constant term in,

$V=\left(\cfrac{q}{4\pi\varepsilon_o}.C\right)^{1/2}\left( A-B.T^{ - } \right)$ where the sign of $T^{-}$ is positive.

$\left( \cfrac { q }{ 4\pi \varepsilon _{ o } } .C \right) ^{ 1/2 }A$

also gives the correct unit of work done per unit charge in an electric field.  The dependence of $V$ on $T^{-}$ is very important and have wide practical applications.

Since, $T^{-}$ are particles, $T^{-}\ge0$  that means

$V\le\left( \cfrac { q }{ 4\pi \varepsilon _{ o } } .C \right) ^{ 1/2 }A$

If this amount of energy is given to the electron it will escape the nucleus.  So, $V$ is the ionization energy of the electron held by the weak field of a spinning $g^{+}$ particle.  Increasing temperature by removing $T^{-}$ increases $V$, increases ionization energy.  At $T^{-}=0$ when no more negative temperature particles are left, $V$ is at its maximum, ionization energy is also at its maximum.

Strangely there is nothing much on the internet about ionization energy dependence on temperature.

Good night.

Note:  $V$ is negative as the electron is being held by the weak field at a potential below the reference $0V$ defined at infinity.  At $V$ maximum in the above discussion, the greatest amount of energy is required to remove the electron, so ionization energy is at its maximum.  What happens after there is no negative temperature particles?

Next stop change in phase...transistors can wait.