# Computing the components of the Ricci scalar

After contracting the Riemann tensor to the Ricci tensor, I have a 2 times covariant tensor. Thus, before contracting this to the Ricci scalar, I need to use the metric tensor to transform it into a $$(1,1)$$ tensor.

What is the correct formula for the components of the Ricci scalar? I need to multiply a "column of columns" (metric tensor) with a "row of rows" (Ricci tensor), right? After that I need to calculate the trace of result.

For 2 dimensions I find this: $$R_{11} g^{11} + R_{21} g^{21} + R_{12} g^{12} + R_{22}g^{22}$$, but this looks strange to me.

• Do you mean $R_{11}g^{11} + R_{21}g^{21} + R_{12}g^{12} + R_{22}g^{22}$ ? Please use MathJax for formatting equations. Jan 24 at 11:53
• You have found $R_{\mu\nu} g^{\mu\nu}$ where the summation is implicit. That's not strange. Jan 24 at 12:13
• @Thomas: thanks for hint - and yes, that's what I meant.
– user326230
Jan 24 at 12:24
• @Connor: thank you for the comment. I thought I misunderstood everything - but maybe there is hope ... Thank you!!!
– user326230
Jan 24 at 12:26

You seem to be thinking of the tensors in terms of matrices, when sometimes (especially when doing computation) it can be more useful to just think about them as collections of numbers, without thinking too much about how to arrange them in rows or columns. I'll use index notation with explicit sums so that you can see better how the computation goes.

You earlier found the Ricci tensor by doing the computation $$R_{\mu\nu} = R_{\mu\sigma\nu}{}^{\sigma} = \sum_{\sigma} R_{\mu\sigma\nu}{}^{\sigma}.$$ Some authors put the upper index in other positions, but that's the general idea. The Ricci scalar is given by $$R = R_{\mu\nu}g^{\mu\nu} = \sum_{\mu,\nu} R_{\mu\nu}g^{\mu\nu}.$$

To compute it, all you need to do is to evaluate the double sum. Since addition is commutative and associative, you can do it in whichever order you prefer (the result is the same regardless of computing the sum over $$\mu$$ first of the sum over $$\nu$$ first).

Thinking in terms of matrices makes these operations somewhat more complicated, since you have to keep track of what is a row and what is a column everytime. The beauty of working with indices is that you don't need to go through all that trouble, and the notation pretty much solves all those issues for you. While it is a bit strange at first, it is quite useful on the long run.

• Wonderful Nickolas, thanks for helping. Yep, indices are the cooler solution, though, I wanted to understand the basics of the computation. Did it here again: (ooops - hehehe) physics.stackexchange.com/questions/690708/… - I fully agree that this way of thinking is not scaling and I definitely won't stay with it. As I said: once for understanding. Contracting a tensor removes information, Same is true for the Riemann Tensor. Thus, I wanted to understand the mathematical background as well as the physical one.
– user326230
Jan 29 at 20:34
• if I get things right, the physical background why the Ricci Tensor is good enough is, that the "forces" that squeeze and bend the object along its geodesic way are somewhat "harmonic". Is that a valid approach?
– user326230
Jan 29 at 20:43
• @Jerry I'm not sure I understand your question (the one in these comments). The Ricci tensor doesn't completely determine the curvature of spacetime physically, one still needs the Riemann tensor to do that. Are you asking why the Einstein Equations only involve the Ricci tensor and not the other components? If so, that would be a new question (which I'd be happy to answer). If not, then I didn't really understand what you are asking =/ Jan 29 at 21:43
• yes, the question is: why is the contracted version of Riemann Tensor (Ricci Tensor) good enough for the Einstein Equations. It contains less information - but obviously enough for this case. As far as I understand: the Metric Tensor can be used to rerive all Christoffel Symbols and therefore also both of these Tensors, right? As the Metric Tensor is also in the Einstein Equation, that might be the answer - but I'm unsure.
– user326230
Jan 30 at 6:35
• Mathematically the contraction means, that I let one of the bases-vectors and one of the dual-basis vectors to be orthogonal and so their product is the Kronecker Delta, right? The Riemann Tensor describes how the parallel transport of a "random" vector along 2 other vectors (in curved manifolds) ends in a new delta-vector. Looking at this scenario and comparing it to the Ricci Tensor, it looks to me as if we do not move a "random" vector but one of the base ones here, right?
– user326230
Jan 30 at 8:06