The classical theory of spacetime geometry that we call gravity consists of the Einstein equation, which relates the curvature of spacetime to the distribution of matter and energy in spacetime. $ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$ Mathematically, how do the Einstein's differential equation of the curvature of spacetime come out of the Einstein's field equations, $$G_{\mu\nu}=8{\pi}T_{\mu\nu}$$. ?
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3$\begingroup$ I'm almost a layman in General Relativity, but I don't understand the question. The Einstein's field equations are a set of equations for the metric tensor, and the metric tensor is kind of a machine that gives you the scalar product of two 4-vectors. As I see it, the Einstein's field equations define in some sense the metric tensor as a solution of themselves, and the metric tensor can be used to compute scalar products. It's not that one implies the other or something like that. However nice question, I'll be waiting for the experts ;) $\endgroup$– J LCommented Nov 28, 2012 at 13:42
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6$\begingroup$ What do you mean with "Einstein's differential equation of the curvature of spacetime"? The Einstein field equations are equations for the metric and the metric gives you all the local information including the information of the curvature ( Riemann curvature tensor, ricci tensor,weyl tensor ...) $\endgroup$– ungeradeCommented Nov 28, 2012 at 13:43
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1$\begingroup$ So maybe the answer would be: the Einstein's field equations are the equations for the curvature so they don't have to come out $\endgroup$– ungeradeCommented Nov 28, 2012 at 14:05
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1$\begingroup$ just came to mind: the geodesic equation(search up on wiki) might be a more appropriate differential equation to ask for such a relationship $\endgroup$– resghCommented Nov 28, 2012 at 16:09
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5$\begingroup$ @Neo Please consider what everyone is saying outside of the answer you accepted. The expression you wrote for $ds^2$ is a definition, nothing more. It no more comes out of $G = 8\pi T$ than $\pi$ comes out of that equation. $\endgroup$– user10851Commented Dec 4, 2012 at 6:13
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3 Answers
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$ds^{2} = g_{ab}dx^{a}dx^{b}$ isn't the Einstein equation. It's just the equation for what arc length is. It's the definition of the metric tensor, pretty much. You're implicitly using it if you've ever done calculus in three dimensions.
The Einstein equation and the Einstein field equations are the same thing and they are expressible as:
$$R_{ab} - \frac{1}{2}Rg_{ab} = 8\pi T_{ab}$$
Full stop.
answered Nov 28, 2012 at 14:20
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1$\begingroup$ This got downvoted? Seriously? $\endgroup$ Commented Feb 5, 2013 at 17:02
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$\begingroup$ Before I comment, I should mention that I didn't downvote your answer. However, I would like to point out that you have not said what a and b are (or anything else) so your full stop is premature. $\endgroup$– MagpieCommented Mar 1, 2013 at 5:03
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1$\begingroup$ @Magpie: $a$ and $b$ are obviously indices of the tensors. The downvoter clearly can't use that as an excuse. $\endgroup$ Commented Jul 17, 2013 at 19:59
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$\begingroup$ @Dimension10 I don't think it's deserving of a downvote either but a and b could just as easily be subscript and g R and T have not been declared so they could also be anything at all, really. It is better to write things out explicitly so that equations have meaning and the answer is more complete. $\endgroup$– MagpieCommented Jul 18, 2013 at 1:19
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$\begingroup$ @Magpie: But most people recognise the equation as the EFE immediately, and know what each term is, metric, ricci curvature, SEM tensor and so on. . $\endgroup$ Commented Jul 18, 2013 at 4:58
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From Wiki:
$G_{\alpha \beta}= (\delta^\gamma_\alpha \delta^\zeta_\beta - \frac{1}{2} g_{\alpha\beta}g^{\gamma\zeta})(\Gamma^\epsilon_{\gamma\zeta,\epsilon} - \Gamma^\epsilon_{\gamma\epsilon,\zeta} + \Gamma^\epsilon_{\epsilon\sigma} \Gamma^\sigma_{\gamma\zeta} - \Gamma^\epsilon_{\zeta\sigma} \Gamma^\sigma_{\epsilon\gamma})$
Where
$\Gamma^\alpha_{\beta\gamma} = \frac{1}{2} g^{\alpha\epsilon}(g_{\beta\epsilon,\gamma} + g_{\gamma\epsilon,\beta} - g_{\beta\gamma,\epsilon})$
So, you see, the Einstein tensor, G, is a function of the metric g and the 1st and 2nd derivatives of the metric.
The simple looking equation, relating the Einstein tensor to the stress-energy tensor, that you quote in your question is, in fact, a system of differential equations written in compact form.
answered Nov 28, 2012 at 14:23
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$ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$ is not a result from general relativity, it is the definition of the metric tensor in general relativity.
The Einstien field equation is "Einstein's differential equation", which lets (which encodes the geometry of spacetime) you calculate the metric from the 4-d stress distribution and other properties of spacetime (since the EFE does not make a statement about the Weyl tensor -- see e.g. Wikipedia | the Schwarzschild solution).
answered Jul 18, 2013 at 5:14