To Light Speed and Beyond

However, an electron under the attraction of a single positive charge experiences acceleration given by,

$m_ea_e=\cfrac{q^2}{4\pi\varepsilon_o r^2}$

$a_e=\cfrac{q^2}{4\pi\varepsilon_o m_e r^2}$

$\cfrac{d^2r}{dt^2}r^2=\cfrac{q^2}{4\pi\varepsilon_o m_e}$

$\iint _{ 0 }^{ r }{ r^{ 2 } } d^{ 2 }r=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e } } \iint _{ 0 }^{ t }{ 1. } (dt)^{ 2 }$

$\int _{ 0 }^{ r }{ \cfrac { r^{ 3 } }{ 3 }  } dr=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e } } \int _{ 0 }^{ t }{ t+C } dt$

$\cfrac { { r }^{ 4 } }{ 12 } =\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e } } \left\{ \cfrac { { t }^{ 2 } }{ 2 } +Ct+B \right\}$ --- (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

$\Delta r$<2.802e-16 m   which is a hundred thousand times less that the atomic radius of hydrogen.

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.