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Topic:
Spoonfeeding Field Equations
Replies:
6
Last Post:
Feb 20, 2013 2:03 AM




Re: Spoonfeeding Field Equations
Posted:
Feb 19, 2013 2:01 PM


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 4by4.
> 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>



