# Riemannian curvature tensor [closed]

In Einstein's field equations, it includes only energy momentum tensor of the matter alone. However, it doesn't include the energy momentum tensor of the field. In Professor Hamber lectures on General theory of Relativity, he said that the energy momentum tensor of the fields are present in the Ricci curvature tensor and the Ricci scalar. In the Ricci tensor, there are products of Christoffel symbols.

1. Does this product imply that there is a self interaction (i.e) interaction of field with itself?

2. Also why energy momentum tensors are expressed as square of the derivatives of the metric tensor?

3. What about its first derivative and cube or fourth power of derivatives of metric?

## closed as too broad by John Rennie, Brandon Enright, Danu, ACuriousMind♦, Ben CrowellOct 5 '14 at 16:07

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• You've asked two different questions. Can you split your question into two different questions: maybe keep Q1 in this question and ask Q2/3 as a separate question. – John Rennie Oct 5 '14 at 9:33
• I don't understand that how can a tensor associated with GEOMETRY have anything to do with what CAUSES it? INTUITIVELY, the Ricci tensor only describes the trajectories to be followed (The curvature defines the trajectories), not the quantity following that trajectory ( the energy momentum tensor does that)...So I don't see any 'fields' associated with Ricci tensor other than the curvature. – GRrocks Oct 5 '14 at 15:33
• Also, how can bending of an abstract quantity have any energy? In other words, As long as you don't consider quantum uncertainties and dark matter/energy, how can the fabric of spacetime have energy? The objects in spacetime are the only ones with energy; the FABRIC itself cannot have any energy, or else, even empty space is supposed to have a gravitational field ( dark matter is a possible reason, but you don't associate it with the Ricci tensor) – GRrocks Oct 5 '14 at 16:05

The Einstein-Hilbert action, from which the Einstein field equations are derived, is given by,

$$S = \frac{1}{16\pi G_N} \int d^4x \, \sqrt{-g} \, \mathcal{R}$$

where $\mathcal{R}$ is the Ricci scalar, dependent on the metric $g_{\mu\nu}$ and its derivatives. The self-interaction and non-linearity of general relativity arises from this factor. To see this, let us work in natural units wherein $\hbar = c = 1$, and define the Planck mass $M_{pl}$,

$$8\pi G_N = M_{pl}^{-2}$$

The relevant coupling in the quantum theory is $1/M_{pl}$, so let us consider perturbations of the form,

$$g_{\mu\nu} = \eta_{\mu\nu} + \frac{1}{M_{pl}}h_{\mu\nu}$$

around flat spacetime with metric $\eta_{\mu\nu}$. We can then see that the action may be written schematically, neglecting messy index contractions as,

$$S= \int d^4x \, (\partial h)^2 + \frac{1}{M_{pl}} h (\partial h)^2 + \frac{1}{M_{pl}^2} h^2 (\partial h)^2 + \dots$$

Hence, we see that an infinite series of interaction terms are encoded in $S$, which include self-interactions of the field with itself. Whilst there are many issues to quantizing gravity, this is certainly one of them. In renormalization group language, the perturbation theory is an expansion in $E^2 / M^2_{pl}$, for some energy scale $E$, and so gravity is strong at high energies, and includes what are known as irrelevant couplings.