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.
Best Answer
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. .