Forces and more forces

This is a short post on forces, Coulombs, Gravitational and Plancks. I will be doing more on this in later posts but this is just a little result that I stumbled upon today. I am certain others have been down this road, but it is a nice result so I thought I would share it with you.

I am not going to go into full detail on how I got this one, it is mostly irrelevant to this post and I will cover it elsewhere. So to start, Planck force is defined as

$F_p =$ $\frac {c^4} {G}$   ...(1)

c- speed of light
G - gravitational constant

Coulombs force

$F_c =$ $\frac {e^2} {4 \pi \epsilon_0 r^2}$    ...(2)

e - charge on the electron
$1 / 4\pi \epsilon_0$ - Coulombs constant
r - distance between the electrons

Gravitational force

$F_g =$ $\frac {m_e^2 G} {r^2}$    ... (3)

$m_e$  - mass of the electron
r and G are the same as those above

Dividing (3) by (2) gives

$\frac {F_g} {F_c} = \frac {m_e^2 G 4 \pi \epsilon_0} {e^2}$    ...(4)

the r term has cancelled here.

Now let's take a look at $r_e$, the classic electron radius, given by

$r_e =$ $\frac {e^2} {4 \pi \epsilon_0 m_e c^2}$   ...(5)

We are going to do two things here, the first is to multiple (2) by $r_e$, while also replacing $r$ with $r_e$ to give

$F_c r_e =$ $\frac{e^2 r_e} { 4 \pi \epsilon_0 r_e^2}$

we can now cancel the $r_e$ on the right to give

$F_c r_e =$ $\frac{e^2} { 4 \pi \epsilon_0 r_e}$  ...(6)

now we put (5) into (6) to give

$F_c r_e =$ $\frac {e^2 4 \pi \epsilon_0 m_e c^2} { 4 \pi \epsilon_0 e^2}$

most of this cancels to leave

$F_c r_e = m_e c^2 = E$    ...(7)

Secondly put $r_e$ into (2) in place of r and sub in its value from (5) giving

$F_c =$ $\frac {e^2 (4 \pi \epsilon_0)^2 m_e^2 c^4} { 4 \pi \epsilon_0 e^4}$

after some cancelling this becomes

$F_c =$ $\frac { 4 \pi \epsilon_0 m_e^2 c^4} {  e^2}$ ...(8)

divide this bu $F_p$ to give

$\frac {F_c} {F_p} =  \frac { 4 \pi \epsilon_0 m_e^2 c^4 G} {   c ^4 e^2}$

cancelling leaves

$\frac {F_c} {F_p} =  \frac { 4 \pi \epsilon_0 m_e^2 G} { e^2}$    ...(9)

but this is just (4), so

$\frac {F_c} {F_p} =  \frac {F_g} {F_c}$    ...(10)

so when r = $r_e$ we have

$F_c^2 = F_p F_g$    ...(11)

Coulomb force squared is equal to the Planck Force multiplied by the gravitational force. Cool eh?

Wrote this while listening to this.