How do I conceptualize the difference between the Weyl tensor and Riemann Curvature tensor?

Currently, I am studying General Relativity from Sean Carroll's An Introduction to General Relativity: Spacetime and Geometry. In chapter 3 of this book, he develops the Riemann Curvature Tensor. I understand this tensor to be defined in two ways:

1. The failure of a vector to be parallel transported is related to this tensor.

2. The failure of the second covariant derivative to commute can be identified with this tensor.

Then, Carroll introduces the Ricci tensor $$R_{\mu \nu} = R^{\lambda}_{\mu \lambda \nu}$$ and the Ricci scalar $$R = R^{\mu}_{\mu} = g^{\mu \nu}R_{\mu \nu}.$$ Now, I see that the Ricci tensor and scalar contain the information about the trace of the Riemann tensor. Carroll then identifies the trace free parts with the Weyl tensor. He also says that the Ricci scalar fully characterizes the curvature. My question is why can we say that the Ricci scalar, which only contains information about the trace of the Riemann curvature tensor, fully characterizes curvature? When looking into this, I came across the following excerpt in wikipedia:

The Weyl tensor differs from the Riemann curvature tensor in that it does not convey information on how the volume of the body changes, but rather only how the shape of the body is distorted by the tidal force.

However, I do not see why this should be true given the way we obtained these tensors, nor do I see what this means for the context of curvature. Ultimately, I am wondering how I should conceptualize the difference between the Weyl tensor and Riemann Curvature tensor given the context of the question described above.

• He also says that the Ricci scalar fully characterizes the curvature. Please edit your question to provide the exact quote. What you have written seems incorrect to me. Commented Nov 4, 2023 at 5:31
• "the Ricci scalar fully characterizes the curvature" doesn't seem correct. First, it should be the Riemann tensor that characterizes the curvature, you need to contract the index to get a Ricci tensor first, and then you further contract the index to get the Ricci scalar, therefore the curvature cannot be fully characterized by the Ricci scalar. Second, the curvature will be partly determined by matter through the Einstein field equation. The part that is not determined by matter distribution is encoded in the Weyl tensor. Commented Nov 6, 2023 at 10:29
• For example, there is no matter outside of the horizon of the Schwarzscild solution, but the Ricci tensor vanishes, if you calculate the Wyel tensor, you will find the Weyl tensor is non-vanishing outside of the horizon. Another example is the gravitational wave which also leads to a nonzero Weyl tensor. Therefore you see the Weyl tensor is part of the Riemann Curvature tensor that is not determined by matter source. Commented Nov 6, 2023 at 10:33

Actually, the Weyl Tensor is the conformally-invariant part of the Riemann curvature and determines the shape of the field of light cones. In effect, it gives you the speed of light, in so doing. The remainder of the Riemann curvature tensor can be expressed in terms of the Ricci tensor and curvature scalar as: $$R_{μνρσ} = C_{μνρσ} + \frac{R_{μρ}g_{νσ} - R_{μσ}g_{νρ} - R_{νρ}g_{μσ} + R_{νσ}g_{μρ}}{n - 2} + R\frac{g_{μσ}g_{νρ} - g_{μρ}g_{νσ}}{(n - 1)(n - 2)},$$ for geometries of dimension $$n > 2$$.
For $$n = 3$$, the Cotton-Work Tensor plays a significant role, and I think it's also needed to determine the shape of light cones in that case. I'm not sure.