Equations of motion in general relativity: Einstein field equations vs geodesic equation

Question

It is said that the equations of motion of a theory are those whose solutions give the coordinates/trajectory of the system.

I was wondering: which is the correct equation of motion in the theory of general relativity?

The Einstein field equations $$ R_{\mu\nu} - \frac{1}{2} Rg_{\mu\nu} + \Lambda g_{\mu\nu} = \frac{8\pi G}{c^4} T_{\mu\nu}, $$ or $$ \frac{d^2x^{\alpha}}{d\lambda^2} + \Gamma^{\alpha}_{\mu\nu} \frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda} = 0, $$ which is the geodesic equation?

Both give the behaviour of objects along curved regions of spacetime so which one is the correct equation of motion of the theory?

Answer 1

It depends on what you mean. The equations of motion for the field theory of general relativity are the Einstein equations. The equation of motion for a freely falling particle within general relativity is the geodesic equation.

Both answers are correct, depending on what exactly you mean. Are you considering the relevant degrees of freedom to be those of the gravitational field or of the particle? Depending on your answer, you get a different meaning.

Answer 2

While these equations are equations of motion, they're not describing the same physics. The Einstein Field Equations (EFE) are resultant from applying the principle of least-action to the Einstein-Hilbert action plus a matter action

$$S_1 [g^{ab}, \phi] = \int \Big\{\frac{1}{16 \pi} R + \mathcal{L}_{\text{M}} \Big\} \sqrt{-g} \text{d}^4 x.$$

Asking $\delta S_1 = 0$ gives you

$$0 = \int \Big\{\frac{1}{16 \pi} \Big(R_{ab} - \frac{1}{2} R g_{ab} \Big) - \frac{1}{2} T_{ab} \Big\} \delta g^{ab} \sqrt{-g} \text{d}^4 x + \int \frac{\delta (\mathcal{L}_{\text{M}} \sqrt{-g})}{\delta \phi} \delta \phi \text{d}^4 x,$$

which results in $R_{ab} - \frac{1}{2} R g_{ab} = 8 \pi T_{ab}$. Since $R_{ab}$ and $R$ depend solely on the metric, solving the EFE gives you the metric tensor associated with your spacetime for a given stress-energy tensor $T_{ab}$. In this case, the variation is made within the space of metric tensors such that the stationary points within this space give you the equations of motion.

In contrast, the geodesic equation comes from applying the principle of least action to the line element

$$S_2 [x^a] = \int \sqrt{-g_{ab} \frac{\text{d} x^a}{\text{d} \lambda} \frac{\text{d} x^a}{\text{d} \lambda}} \text{d} \lambda.$$

Asking $\delta S_2 = 0$ gives you (assuming $\lambda$ is an affine parameter)

\begin{align} 0 &= \int \Big\{g_{ab} \frac{\text{d}^2 x^b}{\text{d} \lambda^2} + \Gamma_{abc} \frac{\text{d} x^b}{\text{d} \lambda} \frac{\text{d} x^c}{\text{d} \lambda} \Big\} \delta x^a \text{d} \lambda, \end{align}

which results in $\frac{\text{d}^2 x^a}{\text{d} \lambda^2} + \Gamma^a_{\ bc} \frac{\text{d} x^b}{\text{d} \lambda} \frac{\text{d} x^c}{\text{d} \lambda} = 0$. Solving the geodesic equation gives you the path an inertial particle will follow in your spacetime. Free-fall for temporal curves, and the paths rays of light follow for null curves. In this case, the variation is made within the space of curves $x^a (\lambda)$ such that the stationary points within this space give you the equations of motion. As you can see, the equations of motion for each of these actions do not follow from the same space.

Note that, for the geodesic equation, you need to know the metric tensor in order to be able to solve it; whilst with the EFE, these equations give you the metric tensor for your spacetime given a stress-energy tensor.