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Einstein's field equation:

$$G_{\mu\nu} = \frac{8\pi G}{c^4}T_{\mu\nu} - g_{\mu\nu}\Lambda$$

I'm trying to understand each of the terms in this equation intuitively, but I'm struggling.

Basically, I want to understand how these equations allow me to predict the path of a particle, given the mass and energy distribution of a system.

I have some idea that $G_{\mu\nu}$ represents the curvature of spacetime, and that $T_{\mu\nu}$ represents the distribution of energy in the system, but it's not clear how.

Correct me if I'm wrong about any of this; I'm just starting.

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Try this from John Baez, the great explainiator. (don't blame him for my spelling) –  Michael Brown Apr 15 '13 at 12:00
That looks like a good read for later, but I'm kinda running low on time at the moment. :/ Is there perhaps a function of a set of particles which would give the gravitational force on a particle at a position, as in Newtonian gravity? –  user912 Apr 15 '13 at 12:09
you are running low on time and you want an intuition into einsteins field equations good luck with that :P. –  Prathyush Apr 15 '13 at 12:25
The kind of picture you are looking for, i.e the force as a function of possible positions is precisely the kind that einsteins said was a limiting approximation to a more complete theory. In a nutshell, what the field equations say is matter tells space time how to curve and the curvature of space time tells matter how to move. May I ask have you learnt geometry of curved manifolds formally? –  Prathyush Apr 15 '13 at 12:33
@user912 You solve the Einstein equations in the presence of whatever matter you have and then you solve the equations of motion for the matter (geodesic equations if there are no other forces than gravity). You have to do both together, self-consistently. It cannot be done in closed form except for the simplest cases, like the two body problem or a homogeneous isotropic universe. –  Michael Brown Apr 15 '13 at 12:46
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2 Answers

$G_{\mu \nu}$ is the Einstein tensor, and is calculated by the following:

$$G_{\mu \nu} = R_{\mu \nu} -\frac{1}{2}g_{\mu \nu}R$$

where $R_{\mu \nu}$ is the Ricci curvature tensor, $g_{\mu \nu}$ the metric tensor and $R$ is the trace of the curvature tensor with respect to the metric, i.e. $g^{\mu \nu}R_{\mu\nu}$, where the metric tensor with indices raised is related to the other by taking the inverse. I have never used the Einstein field equations to calculate paths chosen by a particle - you use the geodesic equations for that. The E.F.E, as you said, equate physical quantities in a system to curvature. One can be given a metric tensor and compute those quantities (which is rather tedious, as you will see if you fully expand the expressions for the terms), or solve the equations, i.e. obtain the metric (which describes the resultant manifold) given a $T_{\mu\nu}$.

You can also think of the Einstein field equations as the equations of motion of the metric tensor which arise when applying the principle of least action. The Einstein-Hilbert action is:

$$S=\frac{c^4}{16\pi G}\int d^4x \, \ R \sqrt{-g}$$

If you have a matter field, you add a lagrangian into the expression, apply the principle of least action, i.e. $\delta S =0$ and you will obtain a term which is defined as the stress-energy tensor equated to $G_{\mu\nu}$. Also, note that the stress-energy tensor $T_{\mu\nu}$ can be obtained via Noether's theorem which states that a continuous symmetry that a Lagrangian exhibits implies a conservation law. See Peskin and Schroeder's book on quantum field theory for a discussion and demonstration of this.

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Going from a stress-energy distribution to the motion of test particles is a long and arduous process in general.

To see how it works, you need to understand that the left-hand side of the Einstein equation is just a very concise shorthand for a function of the metric. You can back out this dependence using the standard definitions \begin{align} G_{\mu\nu} & := R_{\mu\nu} - \frac{1}{2} R g_{\mu\nu}, \\ R & := g^{\mu\nu} R_{\mu\nu}, \\ R_{\mu\nu} & := R^\lambda{}_{\mu\lambda\nu}, \\ R^\rho{}_{\sigma\mu\nu} & := \partial_\mu \Gamma^\rho_{\nu\sigma} - \partial_\nu \Gamma^\rho_{\mu\sigma} + \Gamma^\rho_{\mu\lambda} \Gamma^\lambda_{\nu\sigma} - \Gamma^\rho_{\nu\lambda} \Gamma^\lambda_{\mu\sigma}, \\ \Gamma^\sigma_{\mu\nu} & := \frac{1}{2} g^{\sigma\rho} \left(\partial_\mu g_{\nu\rho} + \partial_\nu g_{\rho\mu} - \partial_\rho g_{\mu\nu}\right). \end{align} At each step you can pause and ask for an interpretation. For instance, $\Gamma^\sigma_{\mu\nu}$ is the set of components that describe the difference between covariant differentiation and partial differentiation in your coordinate system, and $R$ is related to a local "radius of curvature" for your manifold.

As you can see from searching for partial derivatives, $G_{\mu\nu}$ will be a (probably horrendously nonlinear) function of the metric and its first and second derivatives. The Einstein equation, then, is just a partial differential equation that gives you the metric if all you know is the stress-energy.

After you have the metric, you can plug in to the geodesic equation $$ \frac{\mathrm{d}^2x^\mu}{\mathrm{d}\lambda^2} + \Gamma^\mu_{\rho\sigma} \frac{\mathrm{d}x^\rho}{\mathrm{d}\lambda} \frac{\mathrm{d}x^\sigma}{\mathrm{d}\lambda} = 0. $$ Okay, so maybe you don't need to get all the way down to the metric - knowing the connection coefficients is sufficient. But we can always make things more complicated again by adding non-gravitational accelerations to the right-hand side of the geodesic equation. In any event, this is another differential equation you would solve to get, together with the initial conditions, the trajectory of a particle.

In practice, you would have a computer do all of this numerically (which is still much harder than it sounds), or you would be handed values of $T_{\mu\nu}$ that have enough symmetry (perhaps the tensor is diagonal, or constant in spacetime, or spherically symmetric) to make the differential equations tractable.

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