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I'm trying to understand the Einstein Field equation equipped only with training in Riemannian geometry. My question is very simple although I cant extract the answer from the wikipedia page:

Is the "stress-energy" something that can be derived from the pseudo-Riemannian metric (like the Ricci tensor, scalar curvature, and obviously the metric coefficients that appear in the equation) or is it some empirical physics thing like the "constants of nature" that appear in the equation? Or do you need some extra mathematical gadget to specify it? Thanks and apologies in advance if this is utterly nonsensical. Also, as a non-physicists I'm not sure how to tag this either so sorry for that as well.

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1 Answer 1

Good question! From a physical perspective, the stress-energy tensor is the source term for Einstein's equation, kind of like the electric charge and current is the source term for Maxwell's equations. It represents the amounts of energy, momentum, pressure, and stress in the space. Roughly:

$$T = \begin{pmatrix}u & p_x & p_y & p_z \\ p_x & P_{xx} & \sigma_{xy} & \sigma_{xz} \\ p_y & \sigma_{yx} & P_{yy} & \sigma_{yz} \\ p_z & \sigma_{zx} & \sigma_{zy} & P_{zz}\end{pmatrix}$$

Here $u$ is the energy density, the $p$'s are momentum densities, $P$'s are pressures, and $\sigma$'s are shear stresses.

In its most "natural" physical intepretation, Einstein's equation $G^{\mu\nu} = 8\pi T^{\mu\nu}$ (in appropriate units) represents the fact that the curvature of space is determined by the stuff in it. To put that into practice, you measure the amount of stuff in your space, which tells you the components of the stress-energy tensor. Then you try to find a solution for the metric $g_{\mu\nu}$ that gives the proper $G^{\mu\nu}$ such that the equation is satisfied. (The Einstein tensor $G$ is a function of the metric.) In other words, you're measuring $T$ and trying to solve the resulting equation for $G$.

But you can also in principle measure the curvature of space, which tells you $G$ (or you could pick some metric and get $G$ from that), and use that to determine $T$, which tells you how much stuff is in the space. This is what cosmologists do when they try to figure out how the density of the universe compares to the critical density, for example.

It's worth noting that $T$ is a dynamical variable (like electric charge), not a constant (like the speed of light).

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There's nothing to stop you starting with some desired metric and working out what stress-energy tensor you need to achieve it. A good example of this is the Alcubierre metric (en.wikipedia.org/wiki/Alcubierre_drive). However if you start from the curvature you usually find the required stress-energy tensor isn't physically reasonable, as indeed it isn't in the case of the Alcubierre drive. –  John Rennie May 24 '12 at 6:42
    
Yeah, I'd meant for that sort of thing to be included in my second case. I put in a clarification. –  David Z May 24 '12 at 7:40
    
@DavidZaslavsky :Thanks so much and sorry for the delayed response, David and John. Thats exactly what I was looking for. In case you feel like answering another naive question, here's one more (you've already been quite generous so feel free to tell me to go read a physics book): So when people speak of a "solution" to the equation do they most often mean a local solution (i.e. in some coordinate chart) or do they mean the exhibition of an entire manifold whose metric and stress-energy satisfy the equation globally? Does T range over the tangent bundle of the observable universe or some ... –  rorypulvino May 25 '12 at 1:14
    
subset of interest to the person solving the equation? –  rorypulvino May 25 '12 at 1:14
    
My question could be more succinctly stated as follows "do 'solutions' to the EFE exhibit a model of the universe in its entirety or rather some particular coordinate patch with its peculiar arrangement of energy and matter and other 'stuff'?" –  rorypulvino May 25 '12 at 1:25

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