User talk:Nimo8

Alpha

10:19, 7 January 2009 (UTC)

test 12:19, 5 January 2009 (UTC)

\alpha \ =\ {\frac {e^{2}}{\hbar c\ 4\pi \epsilon _{0}}}\ =\ {\frac {e^{2}}{\ 2h\ c\epsilon _{0}}}\ =\ {\frac {e^{2}c\mu _{0}}{2h}}\ ={\frac {1}{\sim 137}}

.

\------------------------------------------
(1)-
in "natural" units
: $\ \alpha ={\frac {e^{2}}{\hbar c}}={\frac {e^{2}2\pi }{hc}}$
: $e={\sqrt {\frac {hc\alpha }{2\pi }}}={\sqrt {\hbar c\alpha }}$

\---------------------------
(2)-
to ease the logik~flow pointing to the two field-epsilons $\varepsilon$

(μ0;ε0) ,their relation to C & The Identity

{c_{0}}^{2}\,\epsilon _{0}\,\mu _{0}=1.

treating $\left({\frac {e^{2}}{2h}}\right)$ as-an "Logic-Entity" ,one can see that

$\alpha \ =\left({\frac {e^{2}}{2h}}\right)((c)\mu _{0})=\left({\frac {e^{2}}{2h}}\right)\left({\frac {1}{(c)\epsilon _{0}}}\right)$

↓↓

$\left({\frac {e^{2}}{2h\alpha }}\right)=\left({\frac {1}{c\mu _{0}}}\right)=\left(c\epsilon _{0}\right)$

$\left({\frac {2h\alpha }{e^{2}}}\right)=(c\mu _{0})=\left({\frac {1}{c\epsilon _{0}}}\right)$

&↓↓

$\left({\frac {2h\alpha }{(c)e^{2}}}\right)=(\mu _{0})=\left({\frac {1}{c^{2}\epsilon _{0}}}\right)$

$\left({\frac {e^{2}}{(c)2h\alpha }}\right)=(\epsilon _{0})=\left({\frac {1}{c^{2}\mu _{0}}}\right)$

\---------------

$(2h\alpha )=(e^{2}c\mu _{0})=\left({\frac {e^{2}}{c\epsilon _{0}}}\right)$

↓↓

e={\sqrt {\frac {2h\alpha }{c_{0}\mu _{0}}}}={\sqrt {2h\alpha c_{0}\epsilon _{0}}}

.. ~
\---------------
Bohr_radius
$a_{0}={\frac {4\pi \epsilon _{0}\hbar ^{2}}{m_{e}e^{2}}}={\frac {\hbar }{m_{e}\,c\,\alpha }}={\frac {h^{2}\,\epsilon _{0}}{\pi \,m_{e}\,e^{2}}}={\frac {h}{2\pi \,m_{e}\,c\,\alpha }}$

where:

\epsilon _{0}\

is the permittivity of free space

\hbar \

is the reduced Planck's constant

m_{e}\

is the electron rest mass

e\

is the elementary charge

c\

is the speed of light in vacuum

\alpha \

is the fine structure constant

test

tutu אחת שלוש

Caption1
Caption2

08:48, 8 November 2009 (UTC)

Beta-&-Gamma-Functions

& zeta
Gamma-Function
$\ \Gamma (n)=(n-1)!={\frac {(n)!}{(n)}}$
&
$\Gamma (nx)=(nx-1)!={\frac {(nx)!}{(nx)}}$
&
$\Gamma (n+x)=(n+x-1)!={\frac {(n+x)!}{(n+x)}}$

Beta-Function by Gammas
$\mathrm {B} (x,n)={\frac {\Gamma (x)\,\Gamma (n)}{\Gamma (x+n)}}={\frac {(x)!\,(n)!}{(x+n)!}}\left({\frac {(x+n)}{(xn)}}\right)={\frac {(x)!\,(n)!}{(x+n)!}}\left({\frac {(1)}{(n)}}+{\frac {(1)}{(x)}}\right)$
~
for some reasn WiKI render the background of "formula" as "transparent"
and not white as it use to be
--12:20, 14 April 2010 (UTC)
.. solved with http://userstyles.org/styles/23389
~ ~ ~
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