Electrons In Orbit Again

Consider an electron in orbit, together with thermal gravity, with velocity  \(v\),

\(\cfrac{m_ev^2}{r_e}=\cfrac{q^2}{4\pi\varepsilon_or^2_e}+Av^2-\cfrac{T_eT_n}
{4\pi\tau_or^2_e}\)

\(m_{ e }v^{ 2 }r_{ e }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o } } +Av^{ 2 }r^{ 2 }_{ e }-\cfrac { T_{ e }T_{ n } }{ 4\pi \tau _{ o } } \)

This is a quadratic in  \(r_e\),

\( Av^{ 2 }r^{ 2 }_{ e }-m_{ e }v^{ 2 }r_{ e }+\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o } } -\cfrac { T_{ e }T_{ n } }{ 4\pi \tau _{ o } } =0\) ---(1)

Since  \(Av^2\gt0\),  this equation has a minimum point and at most 2 roots.

\(r_e=\cfrac { m_{ e }v^{ 2 }\pm \sqrt { (m_{ e }v^{ 2 })^{ 2 }-4Av^{ 2 }(\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o } }-\cfrac { T_{ e }T_{ n } }{ 4\pi \tau _{ o } } ) }  }{ 2Av^{ 2 } } \)

\(r_e=\cfrac { m_{ e }\pm \sqrt { m^{ 2 }_{ e }-\cfrac { A }{ \pi v^{ 2 } } (\cfrac { q^{ 2 } }{ \varepsilon _{ o } } -\cfrac { T_{ e }T_{ n } }{ \tau _{ o } } ) }  }{ 2A } \)

When we have a stable electron orbit, there would be only one admissible root, this suggest a double root or a positive root where the negative root is inadmissible.  For the latter case, this implies the y-intercept is negative,

\(\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o } } -\cfrac { T_{ e }T_{ n } }{ 4\pi \tau _{ o } }\le0 \)

\(\cfrac { q^{ 2 } }{\varepsilon _{ o } } \le\cfrac { T_{ e }T_{ n } }{ \tau _{ o } } \)

This suggests, under this circumstance the electrostatic force is less then the thermal repulsion force.  It is drag that keeps the electrons in orbit.  There is one valid orbital value but it is not at its minimum,  ie.  not at the lowest potential,  \(PE_{re}\).  It would seem that increasing  \(T\)  increases  \(r_e\)  and eventually causes the atom to ionize, then plasma as discussed previously will not be formed.

In the case of a positive double root,

\(r_e=\cfrac { m_{ e } }{ 2A } \)

This marks an orbit where all forces balance and it is also the lowest orbital energy possible.

\(\cfrac{d\,PE_{re}}{d\,r_e}\require{cancel}\cancelto{0}{\cfrac{d\,r_e}{d\,x}}=\cfrac{d\,PE_{re}}{d\,x}=0\)

In this case, the discriminant of expression (1) is zero,

\( (m_{ e }v^{ 2 })^{ 2 }-4Av^{ 2 }(\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o } }-\cfrac { T_{ e }T_{ n } }{ 4\pi \tau _{ o } } )=0\)

\(v^{ 2 }\left\{ m^{ 2 }_{ e }v^{ 2 }-4A(\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o } } -\cfrac { T_{ e }T_{ n } }{ 4\pi \tau _{ o } } ) \right\} =0\)

Since,  \(v\ne0\),

\(m^{ 2 }_{ e }v^{ 2 }=4A(\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o } } -\cfrac { T_{ e }T_{ n } }{ 4\pi \tau _{ o } } )\)

For the above expression to be valid,

\(\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o } }\gt \cfrac { T_{ e }T_{ n } }{ 4\pi \tau _{ o } } \)

This is true, for the minimum double root to rest on the x-axis, the y-intercept has to be positive.  This situation is achieved by lowering  \(T\).  We can simply further,

Since,

\(r_e=\cfrac { m_{ e } }{ 2A } \),    \( 2A r_e={ m_{ e } } \)

\(v^{ 2 }A=\cfrac { 1 }{ 4\pi r_{ e }^{ 2 } } (\cfrac { q^{ 2 } }{ \varepsilon _{ o } } -\cfrac { T_{ e }T_{ n } }{ \tau _{ o } } )\quad \)

This suggests the drag force balances out the electrostatic and thermal repulsive force.  The drag force has switched side without switching sign.  It is now resisting the attractive force between the electron and the nucleus.

The case of a double root, the electron in orbit has the lowest potential.   \(T\)  is low enough for the electrostatic force to be greater than the thermal repulsive force.  More importantly the inclusion of the term  \(T_eT_n≤ T^2\)  allows atomic properties with changing temperature to be investigated.