I am just starting to learn general relativity. I don't understand why we use the Ricci curvature tensor. I thought the Riemann curvature tensor contains "more information" about the curvature. Why is that extra information so to speak irrelevant?
Answer
I think this question is more trivial than you think.
You should ask yourself why should the full Riemann tensor appear. I'll sketch a heuristic derivation of the field equations.
We know that with small velocities and a static field, the Poisson equation $$\Delta\phi=4\pi G\rho$$ is approximately satisfied. From special relativity we know that the mass/energy density $\rho$ must change with two Lorentz factors under a Lorentz transformation. Thus it is the time-time component of a rank two tensor $T_{\mu\nu}$. Using the equivalence principle, we promote this to a curved spacetime tensor. When we look for field equations, we demand that they be tensor equations. For one thing, this means we must have the same number of indices on both sides. We posit $$D_{\mu\nu}=\kappa T_{\mu\nu}$$ with $\kappa$ some constant. We don't know what $D_{\mu\nu}$ is, but the principle of covariant conservation fixes it to be the Einstein tensor. Note that the general form is the natural generalization of the Poisson equation.
You might propose an equation with more indices, such as $$R_{\mu\nu\rho\sigma}=\kappa'T_{\mu\nu}T_{\rho\sigma}$$ with some appropriate antisymmetrization scheme. What are the vacuum equations? They would be $$R_{\mu\nu\rho\sigma}=0$$ But this just says spacetime is flat! We know this is incorrect. Black holes are certainly vacuum solutions but are also certainly not flat spacetime solutions.
In summary, the Ricci tensor has the ability to vanish without the full Riemann tensor vanishing. The general form of the equations is determined by the Poisson equation to be a rank two equation. In my mind, these two facts are the most effective argument.
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