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Re: Spoonfeeding Field Equations
Posted:
Feb 19, 2013 2:01 PM
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On Feb 19, 7:44 am, Giovano di Bacco <gdb...@gmail.com> wrote:
> Tom Roberts wrote:
> > G = T
>
> > To derive it, vary the Lagrangian density, R.
>
> thanks, it seams you forgot the cosmological term
The Cosmological constant thing can be cloned off [T] where is it none
other than a negative mass density in vacuum --- just like the
possibility of such a case within the Poisson equation. <shrug>
Note: [G], [T] are matrices. In this case, they are both 4-by-4.
> however, i do not intend to derive them by myself, since they already are
> derived, they had one hundred years to do that
Deriving the field equations is extremely easy once you have the
Lagrangian. However, the Lagrangian that derives the field equations
has never been qualified as why it is a Lagrangian in the first place
and why the action it represents must be extremized. Since everything
is so bloodily sensitive to the Lagrangian, it is very ludicrous to
say the Lagrangian that derives the field equations is thoroughly
valid. <shrug>
> strange one cannot find the worlds most famous field equations anywhere on
> internet, not even here
As shocking as it may sound, that is because there are very few
physicists out there who actually understand the field equations.
They can look up the textbook and write down ([G] = [T]), but they
never can understand what [G] and [T] represent mathematically that
allow static, spherically symmetric, and asymptotically flat solutions
(such as the Schwarzschild metric) to be solved. <shrug>
For all practical applications, [T] is null, and the field equations
have never been verified when [T] is not null. The only instance
where [T] comes into play is cosmology where these clowns think they
can decide the wellbeing of the universe by tweaking [T] with the
Cosmological constant as its clone. <shrug>
In spherically symmetric polar coordinate system with static diagonal
metric, [G] consists of only 3 unique and ordinary differential
equations. Given the following spacetime geometry,
** ds^2 = c^2 M dt^2 ? P dr^2 ? Q dO^2
Where
** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2
Two of the 3 differential equations of [G] are:
** - M @^2Q@r^2 / (P Q) + M (@Q/@r)^2 / (4 P Q^2) + M @P/r @Q/@r / (2
P^2 Q) + M / Q
** (@Q/@r)^2 / (4 Q^2) + @M/r @Q/@r / (2 M Q) - P / Q
The last one is much more complex. If you are not yet bored and
twisting Koobee Wublee?s arm hard enough, He will post it. Hope this
helps. <shrug>
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