I want to find the spherically symmetric, static solutions to Einstein's equations

$$ R_{\mu \nu} - \frac{1}{2}Rg_{\mu \nu} = 0 $$

in four dimensions using the metric

$$ g_{\mu \nu}dx^{\mu}dx^{\nu} = -A(r)dt^2 + B(r)\left[ dr^2 + r^2 \left( d\theta^2 + \sin^2\theta d\phi^2\right)\right] $$

My question is: What is the fastest way of doing that? I am eliminating the terms by using the obvious simplifications such as "only $r$ and $\theta$ derivatives may survive" or "off-diagonal metric elements give zero" but it is still so lengthy and complicated. It took more than an hour for me to find the $tt$ equation. So, I want to know if there is a faster way to handle this kind of equations. I will be grateful if you can help.


closed as off-topic by ACuriousMind, Kyle Kanos, Brandon Enright, JamalS, Neuneck Feb 18 '15 at 7:26

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  • 1
    $\begingroup$ en.wikipedia.org/wiki/Cartan_formalism_%28physics%29 $\endgroup$ – Ryan Unger Feb 17 '15 at 19:44
  • $\begingroup$ I know this method although I didn't use it when calculating the tt-equation. So, is this the best way of doing that? $\endgroup$ – sahin Feb 17 '15 at 19:50
  • $\begingroup$ People will argue all day over which way is the best to calculate the Ricci tensor. I think Cartan is faster than the standard method. Some use the null tetrad method. It's really subjective. $\endgroup$ – Ryan Unger Feb 17 '15 at 19:51
  • $\begingroup$ Alright then, I will try it by using Cartan's equations. Thanks for the help. $\endgroup$ – sahin Feb 17 '15 at 19:54
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    $\begingroup$ If you are going to do this more than once you might like to consider using a computer algebra system like Maxima maxima.sourceforge.net or its windowed version wxMaxima andrejv.github.io/wxmaxima $\endgroup$ – m4r35n357 Feb 17 '15 at 20:19

I would recommend using Mathematica to calculate curvatures, unless there's a good reason to do it by hand (for example, perhaps you want to calculate the curvatures for a metric while keeping the dimension general). It's not hard to write your own code to do this, and I think it's a nice idea actually. I also have found this code to be very useful: http://www.inp.demokritos.gr/~sbonano/RGTC/. It's good enough to handle differential forms also.

The solution you'll find for your ansatz above is the Schwarzschild solution, but you've written it in non-standard coordinates known as isotropic coordinates. The second term in parentheses is just flat space in spherical coordinates.

If you are calculating the curvatures by hand for a simple warped product metric like this one, there's a slick trick you can use. If you preform a Weyl transformation,

$ds'^2 = \Omega^2 ds^2$, with $\Omega^2 = B^{-1},$

then the resulting metric is very simple:

$ds'^2 = -\frac{A}{B} dt^2 + \sum_{i=1}^3 dy_i^2, $

and the curvature of this new metric is very easily to calculate since it's a direct product (and one of the products is flat space!). Then, using the Weyl transformation formula for the curvature tensor allows you to find the curvature for the original metric $ds^2$. This formula can be found in any GR textbook.

  • $\begingroup$ Thank you. Actually, I am not good at computer software much, so I should spend a lot of time to learn them. Considering my lack of time I will choose to do the calculation by hand. Besides I should show most of my work written. But I will think of the Weyl transformation method you mentioned seriously, it is something new to me and seems simple and useful. Thanks a lot. $\endgroup$ – sahin Feb 18 '15 at 7:59

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