εo=12c2ln(cosh(π))
after applying the √3 factor to account for 3D space,
εo=12c2ln(cosh(π))∗√3
εo=12∗2997924582∗ln(cosh(π))∗√3=3.9325e(−18)
Why was it not necessary to divide by the Durian constant AD=43π(2c)3=9.029022e26?
Because the elementary charge qe has absorbed the constant,
qe→qAD
Given one particle, q=1
qe=19.029022e26=1.107540e−27
since by definition,
εo=1μoc2=14π×10−7c2
If we adjust qe by the factor,
2ln(cosh(π))4π×10−7∗√3
to return to the definition of εo. ie,
12c2ln(cosh(π))∗√3→1μoc2=14π×10−7c2
As both denominator and numerator of qεoare multiplied by the same factor, we have,
qadj=qe∗2ln(cosh(π))4π×10−7∗√3
qadj=19.029022e26∗2ln(cosh(π))4π×10−7∗√3=2.493676e−21
and we further account for the fact that 77 particles coalesce up to the limit tanh(G√2mc2aψπ)=1
π→3.135009
ln(cosh(π))→ln(cosh(3.135009))
that is,
εo=12c2ln(cosh(3.135009))
qadj=19.029022e26∗2ln(cosh(3.135009))4π×10−7∗√3∗77=1.914991e−19
we compare this with the quoted value of e=1.602176565e−19, we are about 1.195 times off.
If we do not adjust for the factor √3,
qadj=19.029022e26∗2ln(cosh(3.135009))4π×10−7∗77=3.316861e−19
we are about twice off the quoted value,
qadj2=3.316861e−192=1.658431e−19
What happened? If the experiment to obtain e is with point charges then we may not have to factor in √3 for charged bodies in 3D space, but this results in a calculated elementary charge of twice the quoted value.
When we do factor in √3, the discrepancy seems to reflect the ratio between the Durian constant and Avogadro constant,
ADAv=9.029022e266.02214e26=1.49930
there is however no reason to adjust the Durian constant here. Furthermore, swapping Durian for Avogadro increases the calculated charge, qe that is already too high.
Otherwise, another constant hangs on my wall!
Note: Dividing by the Durian constant was necessary to account for entanglement.
