No Cigar

This is one of those things that just seems so close that it must mean something. Take a look at this.

$ m_p k_e = 195.61406 = 10^8 / 511210 $     ...(1)

$ m_p = \sqrt \frac {\hbar c} {G}   $     ...(2)
$ k_e = \frac {1} {4 \pi \epsilon_0} $    ...(3)

G - Gravitational constant - 6.67384 .10^-11 m³ kg^-1 s^-2
$ \hbar $ - reduced Planck constant - 1.0545717.10^-34 kg m² s^-1 C^-2
c - speed of light - 299792458 m s^-1
$ \epsilon_0 $ - permittivity of free space - 8.9875517.10⁹ kg m³ s^-2 C^-2

now granted this is going to start looking like some numerology exercise, but bare with me. It's the 511210 that caught my eye and this is why

$ \frac {m_e c^2} {e} = V_e =  510998.9   eV $    ...(4)

$ 5110210 / 510998.0 = 1.000415 $, that is pretty close to unity. So close in fact that it would only involve a small change in the Gravitational constant for it to work. Try this, say we replace the 511210 with equation (4) to give

$ m_p k_e \approx  \frac {e 10^8} {m_e c^2} = \frac {10^8} {V_e} $    ...(5)

sub in (2) and (3)

$  \sqrt \frac {\hbar c} {G} \frac {1} {4 \pi \epsilon_0} =  \frac {e 10^8} {m_e c^2} $    ...(6)

using (1), (2) & (3) we can show that

$ G = \frac {m_e^2 \hbar c^5} {e^2 (4 \pi \epsilon_0)^2  10^{16}} $    ...(7)

giving a value

G = 6.66837 .10^-11 m³ kg^-1 s^-2

That is less than 1% difference. So I thought I would give it an hour to see if there was anything to this. After all if I understood where the 10⁸ comes from maybe we would have something. So lets do some more rearranging to get charge to mass ratio

$ \frac {e} {m_e} = \frac {m_p c^2} {4 \pi \epsilon_0  10^8} =   \frac {E_p} {4 \pi \epsilon_0  10^8}  $   ....(8)

$E_p$ - Planck energy. This can easily be rearranged using

$ E_e = m_e c^2 $    ...(9)

and

$ c \mu_0 = 1 / c \epsilon_0 $    ....(10)

$ \mu_0$ - permeability of free space - $4\pi  10^{-7}$

to give

$ E_p E_e = e  10^{15}  $   ...(11)

so what is the $10^{15}  $? Let's take a look at the dimensions. Energy is Joules, e - charge is in Coulombs, so if we use

$ d = 10^{15}  $

$ E_p E_e = e  d  $   ...(12)

d has the units $ J^2 / C = V^2 C $, so let us try

$ d = V_e V_p q_p $    ... (13)

$V_p$ - Planck voltage - $1.04295   10^{27} V$
$q_p$ - Planck Charge - $1.875545   10^{-18} C$
$V_e$ - this is electron energy in electronVolts

so (12) becomes

 $ E_p E_e = V_p q_p  e V_e $   ...(14)

but

$ E_p = V_p q_p  $  and  $ E_e =  e V_e $ by definition!

using (9) and dividing the above we get

$ \frac {q_p V_p} {m_p} = \frac {e V_e} {m_e} = c^2 $   ...(15)

using (15) we get

$ m_p = {q_p V_p} \frac{m_e}{e V_e}$    ...(16)

multiplying by $k_e$ and rearranging gives

$ m_p k_e V_e = \frac {m_e} {e} q_p V_p k_e $    ....(17)

so substituting from (5) we should have

$ 10^8 \approx \frac {m_e} {e} q_p V_p k_e $    ....(18)

doing the calculation

$ \frac {m_e} {e} q_p V_p k_e = 99858416 \approx 10^8$    ....(19)

So that is what our $10^8$ from (5) actually is! Had I been a little smarter and looked at (15) first then I could have saved myself a couple of hours. Maybe next time.

We can go a little further with this, but I think that this is probably a good place to finish.