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Topic: The Math is still Not Ready
Replies: 8   Last Post: Mar 19, 2013 10:35 AM

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Koobee Wublee

Posts: 1,417
Registered: 2/21/06
Re: The Math is still Not Ready
Posted: Jan 7, 2013 1:42 AM
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On Jan 6, 10:52 am, Jimmy Kesler <jim...@yahoo.com> wrote:
> Koobee Wublee wrote:

> > For the language of convention in this case, [A] means a matrix with
> > elements [A]_ijk... or [A]^ijk...

>
> three indices matrix?


Yes, Jimmy. Three indices indicate rank-three matrices. Riemann
curvature tensors are rank-four matrices while both the Ricci and the
metric tensors are rank-two matrices. <shrug>

> where are the ten field equation then
>
> R_{\mu \nu} - {1 \over 2}g_{\mu \nu}\,R + g_{\mu \nu} \Lambda = {8 \pi G
> \over c^4} T_{\mu \nu}


The field equations actually involves with 4x4 matrices. Yes, they
are all the familiar rank-two matrices. These 4x4 matrices should
give you 16 elements, and each element is a partial differential
equations. When space and time are not intertwined like the
scientific communities have decided time and space should intertwine,
all correlations of elements between time and space become null. That
is what you are thinking of only 10 equations. However, for almost
all practical applications, only diagonal metrics are considered. In
doing so, there are only 4 differential equations you have to worry
about. <shrug>

> and how come the expansion is metric tensor dependent, ie the universe
> expand accordingly more at the regions with intense curvature (presence
> of matter); why not the universe expands flat? what tells it must expand
> like this?


Not sure what you are talking about, but the expansion of the universe
can be found with the de Sitter and the Friedman-LeMaitre-Robertson-
Walker metrics where both do not satisfy Newtonian law of gravity at
short distances. You can apply Koobee Wublee?s theorem of parallel
variations to find a modified de Sitter metric that satisfies the
Newtonian law of gravity. You can also modify the Schwarzschild
metric to allow an expanding universe at cosmic scales. <shrug>

> this was the first question
>
> and also, i dont understand a half of a metric tensor, lol, what on earth
> is a half of a metric tensor!!


Half a metric tensor is simply ([g] / 2) where [g] is a 4x4 matrix
describing spacetime. Not sure what is not to understand? <shrug>





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