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.
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