A Brand New Attractive Force

Consider two electrons in parallel orbits,

$F_{ L }sin(\theta )=F_{ a }$

The attractive force between the electrons is the Lorentz's force.  $F_a$ is the force in the plane of the orbit that keeps the orbital radius constant due to the attraction of the neighboring orbit.

$tan(\theta )=\cfrac {\Delta r }{ 2a_{ e } }$

$F_L=\cfrac { n_{ e }q^{ 2 } }{ 4\varepsilon _{ o }(2a_{ e })^{ 2 } } cos^2(\theta )=\cfrac { n_{ e }q^{ 2 } }{ 4\varepsilon _{ o }(2a_{ e })^{ 2 } } \cfrac { 4a^2_e }{  (2a_{ e })^{ 2 }+(\Delta r)^{ 2 } } =\cfrac { n_{ e }q^{ 2 } }{ 4\varepsilon _{ o } } \cfrac { 1 }{ (2a_{ e })^{ 2 }+(\Delta r)^{ 2 } }$

Therefore,

$F_{ a }=\cfrac { n_{ e }q^{ 2 } }{ 4\varepsilon _{ o }(2a_{ e })^{ 2 } } \cfrac { (2a_{ e })^{ 2 } }{ (2a_{ e })^{ 2 }+(\Delta r)^{ 2 } } \cfrac {(\Delta r)}{ \sqrt { (2a_{ e })^{ 2 }+(\Delta r)^{ 2 } }  }$

$F_{ a }=\cfrac { n_{ e }q^{ 2 } }{ 4\varepsilon _{ o } } \cfrac { (\Delta r)}{ \left\{ (2a_{ e })^{ 2 }+(\Delta r)^{ 2 } \right\} ^{ 3/2 } }$

$\Delta r= r-r_e$

$F_{ a }=\cfrac { n_{ e }q^{ 2 } }{ 4\varepsilon _{ o } } \cfrac { r-r_e }{ \left\{ (2a_{ e })^{ 2 }+(r-r_e)^{ 2 } \right\} ^{ 3/2 } }$

A brand new attractive force! And we have,

$\cfrac { m_{ e }c^{ 2 } }{ r_{ e } } =Ac^{ 2 }+G\cfrac { m_{ a }m_{ e } }{ r^{ 2 }_{ e } }+\cfrac { n_{ e }q^{ 2 } }{ 4\varepsilon _{ o } } \cfrac { r-r_e }{ \left\{ (2a_{ e })^{ 2 }+(r-r_e)^{ 2 } \right\} ^{ 3/2 } }  -\cfrac { Zq^{ 2 } }{ 4\varepsilon _{ o }(r_{ e }-r_{ p })^{ 2 } } -\cfrac { T_{ n }T_{ e } }{ 4\pi \tau_{ o }r^{ 2 }_{ e } }$

a new expression for the forces in action on an orbiting electron.  Under normal circumstances,

$r=r_e$    and    $F_a=0$

But this force will definitely play a part in ionization.

$2a_e$ is twice the radius of an electron, and $n_e$ the number of electrons in orbit minus one.  We are assuming that the orbiting electrons are stacked closest possible in parallel orbits.