this wikiversity original research on
how gravity and inertia emerge from electromagnetic formulae
and
how they might be large-number electromagnetic interactions
has just started and you are about to read sketch notes unripe for print until this very warning is altered because the argument became more elaborated:
In his horribly typeset but brilliantly
http://vixra.org/pdf/1609.0217v3.pdf
Calculation of the gravitational constant G using electromagnetic parameters
2016-09-14 ©©-by jesus.sanchez.bilbao@gmail.com
peer reviewed, printed, and republished in
http://file.scirp.org/pdf/JHEPGC_2016122915423655.pdf
Journal of High Energy Physics, Gravitation and Cosmology, 2017, 3, 87-95
independent researcher Jesús Sánchez discovered
α
g
(
2
π
α
)
2
=
G
m
e
2
2
π
α
2
c
h
=
G
m
e
2
ϵ
0
α
π
q
e
2
=
G
m
e
2
c
h
π
2
q
e
4
ϵ
0
−
2
{\displaystyle {\frac {\alpha _{g}}{(2\pi \alpha )^{2}}}={\frac {Gm_{e}^{2}}{2\pi \alpha ^{2}ch}}={\frac {Gm_{e}^{2}\epsilon _{0}}{\alpha \pi q_{e}^{2}}}={\frac {Gm_{e}^{2}ch}{{\frac {\pi }{2}}q_{e}^{4}\epsilon _{0}^{-2}}}}
=
G
h
8
π
3
c
3
r
e
2
=
G
h
(
2
π
c
)
3
r
e
2
=
r
s
r
c
(
4
π
r
e
)
2
=
ℓ
P
2
2
π
r
e
2
{\displaystyle ={\frac {Gh}{8\pi ^{3}c^{3}r_{e}^{2}}}={\frac {Gh}{(2\pi c)^{3}r_{e}^{2}}}={\frac {r_{s}r_{c}}{(4\pi r_{e})^{2}}}={\frac {\ell _{P}^{2}}{2\pi r_{e}^{2}}}}
=
exp
(
2
α
π
4
−
1
2
α
)
=
e
−
2
1
α
−
α
π
2
=
e
−
96.891
{\displaystyle =\exp({\frac {{\sqrt {2}}\alpha \pi }{4}}-{\frac {1}{{\sqrt {2}}\alpha }})={\sqrt[{-{\sqrt {2}}}]{e}}^{{\frac {1}{\alpha }}-{\frac {\alpha \pi }{2}}}=e^{-96.891}}
=
2
−
139.784
=
10
−
42.079
=
8.33E-43
{\displaystyle =2^{-139.784}=10^{-42.079}={\texttt {8.33E-43}}}
using
exp
(
ln
(
x
)
)
=
x
{\textstyle \exp(\ln(x))=x}
[ edit | edit source ]
on
exp
(
ln
(
r
s
r
c
(
4
π
r
e
)
2
)
)
=
exp
(
2
α
π
4
−
1
2
α
)
{\displaystyle \exp(\ln({\frac {r_{s}r_{c}}{(4\pi r_{e})^{2}}}))=\exp({\frac {{\sqrt {2}}\alpha \pi }{4}}-{\frac {1}{{\sqrt {2}}\alpha }})}
using
exp
(
x
)
=
exp
(
x
)
{\textstyle \exp(x)=\exp(x)}
[ edit | edit source ]
on
ln
(
r
s
r
c
(
4
π
r
e
)
2
)
=
2
α
π
4
−
1
2
α
{\displaystyle \ln({\frac {r_{s}r_{c}}{(4\pi r_{e})^{2}}})={\frac {{\sqrt {2}}\alpha \pi }{4}}-{\frac {1}{{\sqrt {2}}\alpha }}}
on
0
=
2
α
π
4
−
ln
(
r
s
r
c
(
4
π
r
e
)
2
)
−
1
2
α
{\displaystyle 0={\frac {{\sqrt {2}}\alpha \pi }{4}}-\ln({\frac {r_{s}r_{c}}{(4\pi r_{e})^{2}}})-{\frac {1}{{\sqrt {2}}\alpha }}}
on
0
=
2
r
e
π
r
c
4
−
ln
(
r
s
r
c
(
4
π
r
e
)
2
)
−
r
c
2
r
e
{\displaystyle 0={\frac {{\sqrt {2}}r_{e}\pi }{r_{c}4}}-\ln({\frac {r_{s}r_{c}}{(4\pi r_{e})^{2}}})-{\frac {r_{c}}{{\sqrt {2}}r_{e}}}}
using
t
c
=
∫
0
c
1
2
−
(
v
(
t
)
c
)
2
∂
t
=
π
4
t
e
{\textstyle t_{c}=\int _{0}^{c}{\sqrt {1^{2}-({\frac {v(t)}{c}})^{2}}}\partial t={\frac {\pi }{4}}t_{e}}
[ edit | edit source ]
on
0
=
2
t
c
r
e
r
c
t
e
−
ln
(
r
s
r
c
(
4
π
r
e
)
2
)
−
r
c
2
r
e
{\displaystyle 0={\frac {{\sqrt {2}}t_{c}r_{e}}{r_{c}t_{e}}}-\ln({\frac {r_{s}r_{c}}{(4\pi r_{e})^{2}}})-{\frac {r_{c}}{{\sqrt {2}}r_{e}}}}
using
∫
∂
r
r
e
q
′
(
r
)
q
(
r
)
=
∫
∂
q
q
(
r
e
)
1
q
=
ln
(
q
(
r
e
)
)
{\textstyle \int _{\partial r}^{r_{e}}{\frac {q'(r)}{q(r)}}=\int _{\partial q}^{q(r_{e})}{\frac {1}{q}}=\ln(q(r_{e}))}
[ edit | edit source ]
on
0
=
∫
r
e
2
t
c
r
c
t
e
∂
r
−
−
2
r
s
r
r
c
(
4
π
r
)
2
r
s
r
c
(
4
π
r
)
2
∂
r
+
r
c
2
r
2
∂
r
{\displaystyle 0=\int ^{r_{e}}{\frac {{\sqrt {2}}t_{c}}{r_{c}t_{e}}}\partial r-{\frac {\frac {-{\frac {2r_{s}}{r}}r_{c}}{(4\pi r)^{2}}}{\frac {r_{s}r_{c}}{(4\pi r)^{2}}}}\partial r+{\frac {r_{c}}{{\sqrt {2}}r^{2}}}\partial r}
using
∂
r
2
∂
s
2
=
∂
r
∂
s
2
∂
r
=
∫
4
π
r
c
2
4
π
r
2
∂
r
=
−
r
c
2
r
{\textstyle {\frac {\partial r^{2}}{\partial s^{2}}}={\frac {\partial r}{\partial s^{2}}}\partial r=\int {\frac {4\pi {\frac {r_{c}}{\sqrt {2}}}}{4\pi r^{2}}}\partial r={\frac {-r_{c}}{{\sqrt {2}}r}}}
[ edit | edit source ]
on
0
=
∫
r
e
−
τ
∂
s
2
r
∂
r
2
∂
r
−
−
2
r
s
r
r
c
(
4
π
r
)
2
r
s
r
c
(
4
π
r
)
2
∂
r
−
∂
r
2
∂
s
2
r
−
2
∂
r
{\displaystyle 0=\int ^{r_{e}}-{\frac {\tau \partial s^{2}}{r\partial r^{2}}}\partial r-{\frac {\frac {-{\frac {2r_{s}}{r}}r_{c}}{(4\pi r)^{2}}}{\frac {r_{s}r_{c}}{(4\pi r)^{2}}}}\partial r-{\frac {\partial r^{2}}{\partial s^{2}}}r^{-2}\partial r}
using
∂
τ
2
∂
t
2
=
1
−
r
s
r
=
−
∂
r
2
∂
s
2
{\textstyle {\frac {\partial \tau ^{2}}{\partial t^{2}}}=1-{\frac {r_{s}}{r}}=-{\frac {\partial r^{2}}{\partial s^{2}}}}
[ edit | edit source ]
on
0
=
∂
r
∫
r
e
τ
∂
t
2
r
∂
τ
2
∂
r
−
−
2
r
s
r
r
c
(
4
π
r
)
2
r
s
r
c
(
4
π
r
)
2
−
∂
r
2
∂
s
2
r
−
2
{\displaystyle 0=\partial r\int ^{r_{e}}{\frac {\tau \partial t^{2}}{r\partial \tau ^{2}}}\partial r-{\frac {\frac {-{\frac {2r_{s}}{r}}r_{c}}{(4\pi r)^{2}}}{\frac {r_{s}r_{c}}{(4\pi r)^{2}}}}-{\frac {\partial r^{2}}{\partial s^{2}}}r^{-2}}
on
0
=
∫
r
e
∂
t
2
−
∂
s
2
−
∂
r
2
∂
s
2
∂
r
{\displaystyle 0=\int ^{r_{e}}{\frac {\partial t^{2}-\partial s^{2}-\partial r^{2}}{\partial s^{2}}}\partial r}
using
(
ϕ
,
θ
)
:=
(
0
,
0
)
{\textstyle (\phi ,\theta ):=(0,0)}
[ edit | edit source ]
on
d
s
2
=
(
(
R
d
Ω
)
↦
(
r
3
+
r
s
3
3
d
θ
2
+
sin
2
θ
d
ϕ
2
)
)
(
d
t
2
R
R
−
r
s
−
d
R
2
R
−
r
s
R
−
R
2
d
Ω
)
{\displaystyle ds^{2}=({R \choose d\Omega }\mapsto {{\sqrt[{3}]{r^{3}+r_{s}^{3}}} \choose d\theta ^{2}+\sin ^{2}\theta d\phi ^{2}})({\frac {dt^{2}}{\frac {R}{R-r_{s}}}}-{\frac {dR^{2}}{\frac {R-r_{s}}{R}}}-R^{2}d\Omega )}