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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.

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  • $\begingroup$ 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. $\endgroup$
    – Trimok
    Jan 21, 2014 at 20:39
  • $\begingroup$ You can think of the Weyl tensor as the trace-free part of the Riemann tensor. Then its trace should vanish. $\endgroup$
    – MBN
    Jan 22, 2014 at 11:00

2 Answers 2

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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. .

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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.

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  • $\begingroup$ 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 ? $\endgroup$ Jan 22, 2014 at 8:30
  • $\begingroup$ @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. $\endgroup$
    – user28355
    Jan 23, 2014 at 0:32

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