Drag and A Sense of Lightness

As the electron accelerates towards a positive charge, a perpendicular velocity component develops as a result of instability.  Due to the need to satisfy both conservation of momentum and conservation of total energy, as the electron speed increases continuously, there are unstable speed points along the way that the electron experience a oscillatory force perpendicular to its direction of travel.  This force redistribute the total kinetic energy and split the velocity into two perpendicular components in order that both conservation laws are satisfied.

Because of this perpendicular velocity component, the electron move in a helical path centered at the positive nucleus too.

We know that the electron experiences a drag force that eventually hold its velocity at terminal velocity,

drag force, $F_D=Av^2$

where $A$ is a constant and $v$, the electron velocity.

If we factor this force into the electron orbit around a positive charge.

$\cfrac { m_{ e }v^{ 2 } }{ r_{ e } } =\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }r^{ 2 }_{ e } } +Av^{ 2 }$

$(m_{ e }-Ar_{ e })r_{ e }v^{ 2 }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o } }$

$r_{ e }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }(m_{ e }-Ar_{ e })v^{ 2 } }$

Compare this with,

$\cfrac { m_{ e }v^{ 2 } }{ r_{ ec } } =\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }r^{ 2 }_{ ec } }$

$r_{ ec }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }v^{ 2 } }$

Is seem that the electron has lost mass,

$m_e>m_{ e }-Ar_{ e }$

If we take the ratio of the two values,

$\cfrac{r_{ec}}{r_e}=\cfrac{m_{ e }-Ar_{ e }}{m_{ e }}$

$\cfrac{r_{ec}}{r_e}=1-\cfrac{A}{m_e}r_{e}$

where $r_e$ is the measured atomic radius of hydrogen and $r_{ec}$ is the calculated value.

$r_{ ec }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }v^{ 2 } }$

$r_{ ec }$ = (1.602176565e-19)^2/(4pi*8.854187817e-12*9.10938291e-31*2*(299792458^2))=1.4090e-15m

Notice that the new terminal velocity, $2c^2$ has been used.

The measured $r_{e}$ = 5.3e-11 m

$\cfrac{r_{ec}}{r_e}$ = 1.4090e-15/5.3e-11=2.658e-5 m

$A=(1-\cfrac{r_{ec}}{r_e})\cfrac{m_e}{r_{e}}$

$A$ = (1-2.658e-5)*9.10938291e-31/5.3e-11=1.719e-20 Nm^-2s²

This drag coefficient seems low but for the mass of an electron (9.10938291e-31 kg) it is significant.