\(\cfrac { 1 }{ 2 } m_{ e }c^{ 2 }+\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }r_a}=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }r_o }\) --- (1)
where the second term on the LHS is the potential energy of the electron at a distance \(r_a\) right above the nucleus.
The centripetal force, \(F_{c}\) is given by,
\(F_c=\cfrac{q^2}{4\pi\varepsilon_o r^2}sin(\theta)=\cfrac{m_e v^2}{r^{'}}\)
where \(v\) is the circular velocity of the electron.
\(\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o } } sin(\theta )\frac { { r }^{ ' } }{ r } =m_{ e }v^{ 2 }r\)
\( r=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }v^{ 2 } } sin^{ 2 }(\theta )\) --- (*)
At \(\theta=\frac{\pi}{2}\), ie. right above the nucleus,
\( r_a=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }v^{ 2 } }\)
Substitute into (1)
\(\cfrac { 1 }{ 2 } m_{ e }c^{ 2 }+m_{ e }v^{ 2 } =\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }r_o }\)
\(r_o=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_e(\frac{c^2}{2}+v^2)}\)
At \(r = r_o\),
\(\cfrac { v^{ 2 } }{ sin^{ 2 }(\theta ) } =\frac { 1 }{ 2 } c^{ 2 }+v^{ 2 }\)
\(v^{ 2 }=\frac { 1 }{ 2 } c^{ 2 }\cfrac { sin^{ 2 }(\theta ) }{ 1-sin^{ 2 }(\theta ) } \)
\(v^{ 2 }=\frac { 1 }{ 2 } c^{ 2 }tan^{ 2 }(\theta )\)
\(v=\cfrac { c }{ \sqrt{2} }tan(\theta)\)
Since \(v≤c\),
\(tan(\theta)≤\sqrt{2}\)
When \(v=c\), ie the electron is in circular motion with circular velocity \(c\),
\( tan(\theta_o)=\sqrt{2}\)
\(\theta_o\) = 0.9553 r
Since the electron takes very little distance to accelerate up to \(c\) (from the post "To Light Speed and Beyond"), the case of \(v<c\) does not happen.
Consider expression (*) again. A polar plot of r=0.28(sin(theta))^2 for the 0.9553^<theta<2.187
The electron will resonate about the positive charge in a spiral with the above curve as envelope. The circular speed of the electron remains at \(c\), the velocity component that is always perpendicular to the circular path oscillate between zero,\(c\), zero and \(-c\).
\(r_{small}=r_osin(\theta)=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_e(\frac{c^2}{2}+c^2)}sin(\theta_o)\)
\(r_{small}=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_ec^2}.\frac{2}{3}sin(0.9553)\)
\(r_{small}\) = (1.602176565e-19)^2/(4pi*8.854187817e-12*9.10938291e-31*(299792458^2))*(2/3)*sin(0.9553)
\(r_{small}\) = 1.5339e-15 m
And the largest at \(r_a\),
\( r_a=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }c^{ 2 } }\)
\(r_a\) = (1.602176565e-19)^2/(4pi*8.854187817e-12*9.10938291e-31*(299792458^2))
\(r_a\) = 2.818e-15 m
The ratio of these two radii is about 0.5443.
Under such oscillations, the hydrogen would seems to be wobbling. We have considered the drag force on the electron when we set the speed limit to be \(c\). Until next time...
