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Melody, would you check my work, please.

 

 

$$\Displaystyle \Text ${It's really me, Heather, but I'm from the future.
Twenty years ago Mr. Diaz gave some basic math equations to think about for the rest of our lives. He wanted us to explain in basic terms how this is derived. It took me a long time, but I now have a pretty good understanding of this. Some in put on this would be great. Starting with Einsteins basic field equation:}$$

$$.\ \vspace*{1em} \\
\Displaystyle R_{\mu \nu} - {1 \over 2}g_{\mu \nu}\,R + g_{\mu \nu} \Lambda = {8 \pi G \over c^4} T_{\mu \nu}\\

\Displaystyle \text {I know} \mathrm{ R_{\mu \nu}} \;$ is the Ricci curvature tensor. And \; \mathrm{g_{\mu \nu}}\; $ {is the metric tensor,} \mathrm {\Lambda} $\; {is the cosmological constant,} G\, is Newton's gravitational constant, c\, is the speed of light in vacuum, R\ {is the scalar curvature and \mathrm {T_{\mu \nu}} \; ${is the stress-energy tensor. Ect, etc . . . }$$

$$\Displaystyle \Text ${Now, if we dissect the equations that dfine Einstine's manifolds --in particular the vacuum field obtained when T is at zero, could we not also dfine this?}$$

$$\Displaystyle { G_{\mu \nu} \\

\vspace{1em}\\

\noindent \Text {as equal to this} \\

\ R_{\mu \nu} - {1 \over 2}R g_{\mu \nu}. \\

\noindent \Text {It seems reasonable, right?}\\


\Text {Now, if we take the symmetric second-rank tensor that is a function of the metric}\\

\ G_{\mu \nu} + g_{\mu \nu} \Lambda = {8 \pi G \over c^4} T_{\mu \nu}. \\

\Text {OK everything so far is consistent and easy.}\\

\Text {Now if we use geometrized units and set G = c = 1, then the above equation can be written as this:}$$

$$G_{\mu \nu} + g_{\mu \nu} \Lambda = 8 \pi T_{\mu \nu}\ \\

\Text {Now, taking the cosmological constant as an independent parameter, and the term in the field equation algebraically to the other side, This now is part of the stress-energy tensor:}\\

T_{\mu \nu}^{\mathrm{(vac)}} = - \frac{\Lambda c^4}{8 \pi G} g_{\mu \nu} \\

\Text {This results in making the vacuum energy a constant, given by}\\

\rho_{\mathrm{vac}} = \frac{\Lambda c^2}{8 \pi G} \\

\Text {Now, if true, then the terms "cosmological constant" and "vacuum energy" can be used interchangeably in general relativity.}\\$$

 

$$\Text ${One more thing: Though it looks like Einstein field equations are non-polynomial since they contain the inverse of the metric tensor, the equations can be arranged so that they contain only the metric tensor and not its inverse. Here is the determinant of the metric in 4 dimensions:}\\

\det(g) = \frac{1}{24} \varepsilon^{\alpha\beta\gamma\delta} \varepsilon^{\kappa\lambda\mu\nu} g_{\alpha\kappa} g_{\beta\lambda} g_{\gamma\mu} g_{\delta\nu} \\

\Text {See how this fits neatly in the Time-Space Continuum? }\\




\Text {Is this accurate, or did I make a mistake?}$$

 Apr 30, 2015
 #1
avatar+4709 
0

How have you adapted to LaTeX I hardly understand the basics....?!

 

It's hard

.

...

 Apr 30, 2015
 #2
avatar+906 
0

you're not me from the future.......

i fail to see how you are me

if you really are me, then you must send me all my darkest and most deep secrets that I have now (or 20 years ago in your case), and it must be a private message

let's see how this goes!

:p

 May 1, 2015
 #3
avatar+118587 
0

Well HeatherF, it looks like you are going to have a good education     

That is nice to know    LOL 

 May 1, 2015

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