\(m_ea_e=\cfrac{q^2}{4\pi\varepsilon_o r^2}\)
\(a_e=\cfrac{q^2}{4\pi\varepsilon_o m_e r^2}\)
\(\cfrac{d^2r}{dt^2}r^2=\cfrac{q^2}{4\pi\varepsilon_o m_e}\)
\(\iint _{ 0 }^{ r }{ r^{ 2 } } d^{ 2 }r=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e } } \iint _{ 0 }^{ t }{ 1. } (dt)^{ 2 }\)
\( \int _{ 0 }^{ r }{ \cfrac { r^{ 3 } }{ 3 } } dr=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e } } \int _{ 0 }^{ t }{ t+C } dt\)
\( \cfrac { { r }^{ 4 } }{ 12 } =\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e } } \left\{ \cfrac { { t }^{ 2 } }{ 2 } +Ct+B \right\} \) --- (1)
When \(t=0\), \(r=r_o\)
\( B=\cfrac { { r }_{ o }^{ 4 }\pi \varepsilon _{ o }m_{ e } }{ 3q^{ 2 } } \)
Differentiating (1) wrt time \(t\),
\( \cfrac { { r }^{ 3 } }{ 3 } \cfrac { dr }{ dt } =\cfrac { { r }^{ 3 } }{ 3 } v=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e } } \left\{ t+C \right\} \)
Suppose \(t=0\), \(v=0\) therefore \( C=0\) and we have
\( \cfrac { { r }^{ 3 } }{ 3 } v=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e } }t \)
\( \cfrac { { r }^{ 4 } }{ 12 } =\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e } } \left\{ \cfrac { { t }^{ 2 } }{ 2 } \right\} +\cfrac { { r }_{ o }^{ 4 } }{ 12 } \)
And if the electron is accelerated to velocity \(c\),
\( t=t_{ f }\) \(v=c\) \( r=r_{ f }\)
\( \cfrac { r^3_{ f } }{ 3 } c=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e } } t_{ f }\)
\({ t }_{ f }=\cfrac { 4\pi \varepsilon _{ o }m_{ e } }{ 3{ q }^{ 2 } }r^3_{ f } c\) --- (*)
and
\( \cfrac { r^4_{ f } }{ 12 } =\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e } } \left\{ \cfrac { { t_{ f } }^{ 2 } }{ 2 } \right\} +\cfrac { { r }_{ o }^{ 4 } }{ 12 } \)
\(\cfrac { t_{ f }^2 }{ 2 } =\cfrac { \pi \varepsilon _{ o }m_{ e } }{ { 3q }^{ 2 } } \{ { r_{ f } }^{ 4 }-{ r_{ o } }^{ 4 }\} \)
\(\cfrac { { t_{ f } }^{ 2 } }{ 2 } =\cfrac { \pi \varepsilon _{ o }m_{ e } }{ 3{ q }^{ 2 } } ({ r_{ f } }-{ r_{ o } })({ r }_{ f }+{ r }_{ o })({ r_{ f } }^{ 2 }+{ r_{ o } }^{ 2 })\)
\(r_f<(r_f+r_o)\) and \(r_f^2<({ r_{ f } }^{ 2 }+{ r_{ o } }^{ 2 })\)
\( { (r_{ f } }-{ r_{ o } }).r_f^3=\Delta r.r_f^3<\cfrac { 3{ q }^{ 2 } }{ 2\pi \varepsilon _{ o }m_{ e } } { t_{ f } }^{ 2 }\)
Substitute (*) into the above,
\( \Delta r<\cfrac { 8\pi \varepsilon _{ o }m_{ e } }{ 3{ q }^{ 2 } } { r_{ f } }^{ 3 }c^{ 2 }\)
\(\cfrac { 8\pi \varepsilon _{ o }m_{ e } }{ 3{ q }^{ 2 } }c^{ 2 }\)=8pi*8.854187817e-12*9.10938291e-31*(299792458)^2/(3*(1.602176565e-19)^2)=2.3658e14 m-5
If \(r_f\) is a thousand times the atomic radius of hydrogen then
\(r_f\) = 1000*5.29e-11
\( \Delta r\)<(5.29e-8)^3*2.3658e14
\( \Delta r\)<3.502233061962e-8 m
If \(r_f\) is a hundred times the atomic radius of hydrogen then
\(r_f\) = 100*5.29e-11
\( \Delta r\)<(5.29e-9)^3*2.3658e14
\( \Delta r\)<5.185e-11 m which means, an electron has to accelerated over a distance equal to the atomic radius of hydrogen in order to gain light speed at a distance of 100 times hydrogen radius from the hydrogen.
If \(r_f\) is a ten times the atomic radius of hydrogen then
\(r_f\) = 10*5.29e-11
\( \Delta r\)<(5.29e-10)^3*2.3658e14
\( \Delta r\)<3.502e-14 m which is a thousand times less that the atomic radius of hydrogen.
If \(r_f\) is a 2 times the atomic radius of hydrogen then
\(r_f\) = 2*5.29e-11
\( \Delta r\)<(1.058e-10)^3*2.3658e14
This shows that an electron takes very little distance to accelerate to light speed, \(c\). It is reasonable to assume that an electron in motion under the attraction of a positive charge is already at light speed.If \(r_f\) is a 2 times the atomic radius of hydrogen then
\(r_f\) = 2*5.29e-11
\( \Delta r\)<(1.058e-10)^3*2.3658e14
\( \Delta r\)<2.802e-16 m which is a hundred thousand times less that the atomic radius of hydrogen.
