What besides the metric do you need to set up the EFEs and the geodesic equation?

Question

One of my professors wrote on the board

(1) Mass tells spacetime how to curve $\to$ Metric/Einstein Field Equations

(2) Spacetime tells mass how to move $\to$ Geodesic equation

Suppose I am given the following metric:

$$ds^2 = -c^2dt^2 + dl^2 + (k^2 +l^2)(d\theta^2+\sin^2\theta d\phi^2)$$

Given this metric, do I have enough information to get useful information from

(1) the Einstein Field Equations

(2) the geodesic equation

In other words, besides the metric, how many other pieces of information do I need before I can set up the Einstein Field Equations and the geodesic equation and start extracting information about how test particles will behave in this space?

Remark:

Apologies for any abuse of terminology. I am just learning this stuff, so please correct me if I didn't state the question well.

Answer 1

From Wald (1984)

The entire content of general relativity may be summarized as follows: Spacetime is a manifold $M$ on which there is defined a Lorentz metric $g_{ab}$. The curvature of $g_{ab}$ is related to the matter distribution in spacetime by Einstein's equation $$R_{ab}-\tfrac{1}{2}Rg_{ab}=8\pi T_{ab}$$

Note that the geodesic equation follows from the field equations. To prove this, let the energy-momentum tensor be that of incoherent dust: $$T^{ab}=\rho u^a u^b$$ The field equations imply $\nabla_a T^{ab}=0$ and thus $$\nabla_a(\rho u^a)u^b+\rho\nabla_u u^b=0$$ Contract this with $u_b$. Note that $u_b\nabla_u u^b=0$ because $\nabla u_b u^b=\nabla(-1)=0.$ Thus $\nabla_a(\rho u^a)=0$ and $$u^b\nabla_b u^a=0$$ which is the geodesic equation. To actually solve the geodesic equation, you need to know the metric. To know the metric, you need to solve the Einstein field equations. There are some cases where the energy-momentum tensor itself depends on the metric (a simple example is that of a perfect fluid). The Einstein field equations are horribly nonlinear partial differential equations for the metric. You are not given the metric. Once you have the metric, you can solve the geodesic equation for the test particles.

So how do we solve the field equations? This is a question that Einstein asked himself the moment he wrote them down. There are several techniques, but nothing foolproof. A method of approximation exists in the Post-Newtonian formalism. (See, e.g., Weinberg (1972) Ch. 9 or Straumann (2013) Ch. 6.) In Ch. 7 of Wald, he outlines a few programs for obtaining solutions with various symmetries that reduce the number of equations you have to solve. Given initial data, there is a method for extending the solution to all future times. This is called the Cauchy problem and is detailed in Wald Ch. 10 and Hawking & Ellis (1973) Ch. 7. Of course, the literature listed here is just a sliver of what's out there and I'm certainly no expert in solving partial differential equations.

Answer 2

If all you want is to examine the motion of test particles (or test fluids, or test quantum fields), then you have everything you need, and the process is simple. You know the covariant and contravariant metric components $g_{\mu\nu}$ and $g^{\mu\nu}$, so you can calculate the connection coefficients $$ \Gamma^\sigma_{\mu\nu} = \frac{1}{2} g^{\sigma\rho} \left(\partial_\mu g_{\nu\rho} + \partial_\nu g_{\rho\mu} - \partial_\rho g_{\mu\nu}\right). $$ Then the differential equation governing the spacetime position $x^\mu$ of test particles is the geodesic equation: $$ \frac{\mathrm{d}^2x^\mu}{\mathrm{d}\lambda^2} + \Gamma^\mu_{\rho\sigma} \frac{\mathrm{d}x^\rho}{\mathrm{d}\lambda} \frac{\mathrm{d}x^\sigma}{\mathrm{d}\lambda} = 0. $$

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For going the other direction (matter telling spacetime how to curve), you would start with other information. Once you have the metric, your job is done. You typically start with the stress-energy tensor $T_{\mu\nu}$. Then as I explained in an answer to a related question, you set $8\pi T_{\mu\nu}$ equal to $G_{\mu\nu}$, where we have \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}. \end{align} This allows you to solve for the connection coefficients (or the metric components if you want), at least in theory.

If you start with the metric, one thing you could do is run the above equations in reverse and find out what $T_{\mu\nu}$ must be; that is, you can see what energy/momentum/pressure/shear is consistent with your spacetime.

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If you specify the metric on a 3D spacelike slice of spacetime, you can ask how the metric evolves forward in time, both due to pure spacetime effects and due to the changing matter distribution. This is described by the ADM equations. In practice, this is complicated enough that it is always done numerically (and figuring out the numerics for even simple setups took decades after the equations were first written down!)

Answer 3

From the metric alone you have enough information to solve the Einstein Field equations for the Stress Energy tensor (assuming no cosmological constant), as well as enough information to write the geodesic equation and have a differential equation for test particle motion (assuming no outside force). This can be seen by the form of the EFE (in units where $G=c=1$);

$$G_{\mu \nu} = 8\pi T_{\mu\nu}.$$

The $G_{\mu\nu}$ here is the Einstein Curvature tensor, which is dependent only on the metric. Usually $G_{\mu\nu}$ is defined in terms of the Riemann Tensor, which is a function of the Christoffel Symbols, which are determined completely by the metric.

The exact equations for this progression are below;

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

where $R$ is the trace of $R_{\mu\nu}$,

$$R_{\alpha\beta} = {R^\rho}_{\alpha\rho\beta} = \partial_{\rho}{\Gamma^\rho_{\beta\alpha}} - \partial_{\beta}\Gamma^\rho_{\rho\alpha} + \Gamma^\rho_{\rho\lambda} \Gamma^\lambda_{\beta\alpha} - \Gamma^\rho_{\beta\lambda}\Gamma^\lambda_{\rho\alpha} ,$$

and finally

$$\Gamma^i{}_{k\ell}=\frac{1}{2}g^{im} \left(\frac{\partial g_{mk}}{\partial x^\ell} + \frac{\partial g_{m\ell}}{\partial x^k} - \frac{\partial g_{k\ell}}{\partial x^m} \right).$$

Similarly the geodesic equation states

$$\frac{d^2\gamma^\lambda }{dt^2} + \Gamma^{\lambda}_{\mu \nu }\frac{d\gamma^\mu }{dt}\frac{d\gamma^\nu }{dt} = 0.$$

So if you have $\Gamma$ you have a second order differential equation for the particle motion, $\gamma$. Assuming you add on appropriate initial conditions, the metric will uniquely determine particule path.