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:
Sanchez Exponential
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In his horribly typeset but brilliantly
helpful paper
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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
the equation
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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}
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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)}
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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 }}} using x = 0 + x {\textstyle x=0+x}
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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 }}} using α = r e / r c {\textstyle \alpha =r_{e}/r_{c}}
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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}}
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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}))}
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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}}}
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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}}}}
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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}} using . . . {\textstyle ...}
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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)}
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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 )} Milky Way Translation
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Thermomagnetic Maxwell Quaternion Potential
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Circular Conductor Cubic Cell
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References
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