# Vanishing of Weyl Tensor Contraction

Within the context of Einstein space-times, we know that the contraction of the Weyl tensor across a set of indices always vanishes, like so :

$$C{^{\alpha }}_{\mu \alpha \nu }=0$$

From a purely mathematical standpoint this should be straightforward ( but perhaps tedious ) enough to prove from the definition of the conformal tensor in terms of the Riemann tensor and its contractions. However, I am wondering what the physical and/or geometric meaning and significance - if any - of this vanishing contraction really is ? I am a very visual person and learner, so an intuitive geometric understanding of this would be very helpful to me.

• Looking here, (and here) you will understand, that the Weyl tensor is not an indicator of change of volume, but an indicator of change of shape. Jan 21, 2014 at 20:39
• You can think of the Weyl tensor as the trace-free part of the Riemann tensor. Then its trace should vanish.
– MBN
Jan 22, 2014 at 11:00

If you choose a local inertial frame at a specific point of space-time, the metric tensor, around this point, is :

$g_{ij}= \delta_{ij} - \frac{1}{3} R_{ikjl} x^k x^l + O(x^3) \tag{1}$

And the space-time volume element (corresponding to the square root of the determinant of the metric) is :

$d\mu_g = (1 - \frac{1}{6} R_{jk} x^j x^k + O(x^3)) ~d\mu_{Euclidean}\tag{2}$

The fact that the contraction of the Weyl tensor is zero, that is $C_{jk}=0$, looking at equation $(2)$, means that the $C_{jk}$ part of $R_{jk}$ is zero, so the Weyl tensor does not contribute to modifications of (infinitesimal) space-time volume. However, equation $(1)$ indicates you, that the Weyl tensor is participating to the modification of the metrics, because the $C_{ikjl}x^kx^l$ part of $R_{ikjl} x^k x^l$ is not zero.

So, finally, the Weyl tensor participates to a modification of the metrics, but without participating to the modification of a (infinitesimal) space-time volume, so the Weyl tensor is associated to modifications of the shape of(infinitesimal) space-time volume, but without modification of volume.

Typically, this involves tidal forces, gravitational waves. For instance, for a (basic) gravitational wave (here we suppose that the Ricci tensor is zero, and the Riemann tensor equals to the Weyl tensor) propagating along the $z$ axis, considering an infinitesimal space-time volume, the physical $\delta x$ could increase, and the physical $\delta y$ could decrease, so the shape is changing, however one variation is compensating the other, so that the infinitesimal space-time does not change. .

Let $$\oplus$$ denote the Kulkarni–Nomizu product. Then, in four dimension, the Weyl tensor is defined as $$C = R - \frac{1}{2}\left(\mathrm{Ric} - \frac{\operatorname{tr}_gR}{n}g\right)\oplus g - \frac{s}{24}g \oplus g$$ Here, $$g$$ is the metric, $$R$$ is the Riemann tensor. $$Ric$$ is the Ricci curvature tensor. Thus, the Weyl tensor expresses the tidal force that a body feels when moving along a geodesic. When this vanishes, it means that in a geodesic $$\gamma$$ in spacetime, the tidal force that any body feels when moving along $$\gamma$$ is $$0$$. Also, the metric is locally conformally flat: there exists a local coordinate system in which the metric tensor is proportional to a constant tensor. Thus, is the Weyl tensor vanishes, then the spacetime is conformally flat. By the definition of a conformally flat pseudo-Riemannian manifold:

Definition: If for each point $$x$$ in spacetime $$M$$, there exists a neighborhood $$U$$ of $$x$$ and a smooth function $$f$$ defined on $$U$$ such that $$(U, \mbox{e}^{2f}g)$$ is flat (i.e. the curvature of $$\mbox{e}^{2f}g$$ vanishes on $$U$$).

One can thus say that the spacetime can conformally be mapped to flat space, the simplest being Minkowski spacetime.

A more mathematically oriented paper is http://www.pnas.org/content/69/9/2675.full.pdf . A physical paper is M. A. Singer, 1990: Flat twistor spaces, conformally flat manifolds and four-dimensional field theory. Comm. Math. Phys. Volume 133, Number 1 (1990), 1-215. Available online here.

• Thank you for your time on this. I understand this, but my question was more specifically about the contraction of the Weyl tensor, rather than the tensor itself - any ideas whether or not there is a geometric meaning to its vanishing ? Jan 22, 2014 at 8:30
• @MarkusHanke: Given that the Weyl tensor itself vanishes, it is trivial to show that the contraction $C^\alpha_{\mu\alpha\nu}$ vanishes. Thus, whatever results I posted in the answer holds even for the contracted Weyl tensor.
– user28355
Jan 23, 2014 at 0:32