# What do the entries of the Einstein Tensor mean?

So if I understand correctly a tensor is something that transforms under certain laws and can be imagined as a combination of two vectors, e.g. the Stress tensor is a combination of surface normal vector and momentum-flux/force vector. For example the $\tau_{xy}$ is the force in the y-direction on a surface with a normal vector in the x-direction.

The Stress-Energy tensor is a generalization of the stress-tensor which incorporates the time-dimension, i.e. $T_{01}$ is the kinetic energy in the x-direction or the '4-momentum flow in the x-direction on a surface with a normal vector in the t-direction'.

Anyway, my question is about the left-hand-side of the Einstein Field Equations. If I understand correctly that is called the Einstein Tensor, which is composed of the Ricci tensor and the Metric tensor. I know it describes the curvature of spacetime, but I'm getting tired of that explanation and I would like to correctly understand what exactly the Einstein tensor is describing, i.e. which two vectors it combines and how it describes/quantifies spacetime-curvature.

The Stress-tensor is a simple example that is similar to the Stress-Energy tensor, is there something the same for the Einstein Tensor? For example a two-dimensional 'Einstein Tensor' that describes the curvature of a 2-sphere?

• This is not a good understanding of what a tensor is, which is a multilinear function of vectors and one-forms. As such your question is probably not answerable. – tfb Mar 2 '17 at 19:34
• My question is simply how to understand what the entries of the Einstein Tensor stand for. I agree that my understanding of Tensors is not very good, but I don't think my question is unanswerable; because that would mean that no one can understand that tensor, I guess. – B2q Mar 2 '17 at 21:40
• It's answerable, but the answer is 'learn enough differential geometry': that's not really something that's going to fit in an answer here. – tfb Mar 2 '17 at 22:04
• Yes you are right, but learning differential geometry is a daunting task that has not a lot of elementary and simple books on it (which is probably not possible). It amazed me how easy the Right-Hand-Side of the equation is to understand, thus I hoped and guessed the Left-hand-side would also be possible to grasp for college-level physics/math – B2q Mar 2 '17 at 22:26

Seeing that you don't want to go into too much differential geometry, I will be qualitative. Roughly speaking, the Riemann tensor measures how a tensor field would change upon parallel transport.

Roughly speaking, for vectors, parallel transport is a translation along some curve such that the vector remains unchanged with respect to the curve. In 3D flat space, this would just be trivial translation of an 'arrow' along a curve without change in magnitude or direction. However, in curved space, such a transport will in general result in a change in the direction of the vector (with respect to our coordinates). You may convince yourself by doing this action on the surface of a ball. The Riemann tensor for curved spaces will therefore be non-identically-zero, but is identically zero for flat space.

Einstein himself is often quoted as having said that the left hand side of his equations is made of marble while the right hand side is made of wood.

• Actually the OP asked the meaning of the Einstein tensor, not the meaning of the Riemann tensor. – magma Mar 5 '17 at 0:20
• Thanks for your comment magma. You are right, I suppose there is also the metric, which describes the geometry of spacetime. – John Mar 5 '17 at 0:41

In their original form, the Einstein field equations simply said the following,

$$T_{\mu\nu} - \frac{1}{2}g_{\mu\nu}T = \frac{1}{\kappa}R_{\mu\nu}$$

where $g_{\mu\nu}$ is the covariant metric tensor, and $T$ is the quantity $g_{\mu\nu}T^{\mu\nu}$, where Einstein summation is used. $\kappa$ is a constant.

This was the original form of the Einstein field equations, which express a relation between the information proferred by the stress energy tensor, or rather a complicated polynomial thereof on the left and the Ricci tensor, you on the right. By some algebra, one can recover from these equations, the form of the Einstein field equations you have mentioned in the question.

There probably is no meaning to the components of the Einstein tensor, which is simply obtained by subjecting the original form of the field equations to some algebra.

• Arguably the first 3-4 paragraphs are totally unnecessary and introducing very basic GR which the OP is familiar with to be able to pose this question. Your entire answer is basically $G_{ab}$ has no meaning as it results from an algebraic manipulation of the field equations, which is really a comment. – JamalS Mar 3 '17 at 16:19
• @JamalS You're right, I removed most of the answer. Sorry. – Cynthia's Light Mar 3 '17 at 17:11