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- Thread starter ghc
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The well-known solutions ( well known to me, that is ) were not derived by starting with a EMT and solving the Einstein field equations for the metric. The Schwarzschild metric has zero Ricci tensor and so represents a vacuum near a spherically symmetric, non-rotating source. The spacetime derives its properties from spherical symmetry alone. I highly recommend the English translation of Scharzschilds paper.

The basic cosmological solution is derived from the principles of homogeneity ( same to all observers) and isotropy ( no preferred direction). It is a happy accident that the Einstein tensor of this line element ( Robertson-Walker) turns out to describe a space filled with non-interacting, pressureless ( no collisions) dust. This leads to the Friedmann equations which link the expansion rate to the energy density, ( and look where that got us - Dark Matter ).

There are probably numerical methods available to solve the EFE, but it's doubtful whether closed algebraic solutions even exist for most general EMTs.

The basic cosmological solution is derived from the principles of homogeneity ( same to all observers) and isotropy ( no preferred direction). It is a happy accident that the Einstein tensor of this line element ( Robertson-Walker) turns out to describe a space filled with non-interacting, pressureless ( no collisions) dust. This leads to the Friedmann equations which link the expansion rate to the energy density, ( and look where that got us - Dark Matter ).

There are probably numerical methods available to solve the EFE, but it's doubtful whether closed algebraic solutions even exist for most general EMTs.

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I have another question: is there a formula in which the Riemann tensor is related only to the metric and its derivatives? if yes, can this equation give numerical solutions to EFE?

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[tex]R^{r}_{mqs} = \Gamma ^{r}_{mq,s}-\Gamma ^{r}_{ms,q}+\Gamma ^{r}_{ns}\Gamma ^{n}_{mq}-\Gamma ^{r}_{nq}\Gamma ^{n}_{ms}[/tex]

where

[tex]\Gamma^m_{ab}=\frac{1}{2}g^{mk}(g_{ak,b}+g_{bk,a}-g_{ab,k})[/tex]

The comma before an index indicates partial differentiation. The Ricci tensor is the contraction of the Riemann tensor on the first and third indices, and appears in the EFE.

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I don't know enough about numerical relativity to answer your question. There ought to be information available on the internet, so a search would probably get something useful.

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haushofer

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ghc,

I forgot to mention that there's a very good free math program called Maxima, which runs on most OSs. It has a package called ctensor that can calculate most of the GR tensors, when given a metric. It also does frame calculations ( something you should investigate if you're not familiar with the concept ).

You can get Maxima and it's GUI interfaces xMaxima and wxMaxima here.

http://maxima.sourceforge.net/

Coming back to solving the EFE. I'm told by a correspondent who doesn't post here that the solution of the EFE is greatly simplified if you use the method of frame fields, aka "local Lorentz frames", aka "using Cartan's method of moving frames") . He gives a 'recipe' for this -

M

I forgot to mention that there's a very good free math program called Maxima, which runs on most OSs. It has a package called ctensor that can calculate most of the GR tensors, when given a metric. It also does frame calculations ( something you should investigate if you're not familiar with the concept ).

You can get Maxima and it's GUI interfaces xMaxima and wxMaxima here.

http://maxima.sourceforge.net/

Coming back to solving the EFE. I'm told by a correspondent who doesn't post here that the solution of the EFE is greatly simplified if you use the method of frame fields, aka "local Lorentz frames", aka "using Cartan's method of moving frames") . He gives a 'recipe' for this -

I haven't studied this myself yet, and I may have made mistakes in transcription.First, write down an frame Ansatz based upon assuming a combination of spacetime symmetry and what kind of motion you expect. For example, assume spherical symmetry and spherically symmetric collapse of spherical shells of test particles which our observers are riding on, a very simple frame Ansatz is:

[tex]

\begin{array}{rcl}

\vec{e}_1 & = & \partial_t - f \, \partial_r \\

\vec{e}_2 & = & \partial_r \\

\vec{e}_3 & = & \frac{1}{r} \, \partial_\theta \\

\vec{e}_4 & = & \frac{1}{r \, \sin(\theta)} \, \partial_\phi

\end{array}

[/tex]

where f is an undetermined function of r only, saying the rate of falling toward the origin depends only on r. Note that this Ansatz also says that differences in the t coordinate correspond to elapsed proper time measured by any of our infalling observers.

Second, demand that the Einstein tensor as measured by these observers have a given form, say that of an nonnull electrovacuum appropriate for a radial electrostatic field

[tex]

G^{ab} = \frac{q^2}{r^2} \, diag (1,-1,1,1)

[/tex]

where q is the charge. The great thing about using frames is that tensors like the EM stress-energy tensor have exactly the same form and behave the same way under boosts and rotations at an event as in flat spacetime.

In this example, we find an ODE for f which is easily solved giving

[tex]

f = \sqrt{ \frac{2m}{r} - \frac{q^2}{r^2} }

[/tex]

If we had demanded instead that the stress-energy tensor be the sum of the previous EM term plus a Lambda term

[tex]

G^{ab} = \frac{q^2}{r^2} \, diag (1,-1,1,1) + \frac{3}{R^2} \, diag \, (1,-1,-1,-1)

[/tex]

we would have found

[tex]

f = \sqrt{ \frac{2m}{r} - \frac{q^2}{r^2} + \frac{r^2}{R^2} }

[/tex]

M

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