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Can anyone give me a geometrical interpretation of the weyl curvature tensor?

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- Thread starter Benjam:n
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Can anyone give me a geometrical interpretation of the weyl curvature tensor?

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Pirani and Schild

"Geometrical and Physical interpretation of the Weyl Conformal Curvature Tensor"

http://oai.dtic.mil/oai/oai?verb=getRecord&metadataPrefix=html&identifier=AD0260083

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A.T.

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Can anyone give me a geometrical interpretation of the weyl curvature tensor?

http://en.wikipedia.org/wiki/Weyl_tensor

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. The Ricci curvature, or trace component of the Riemann tensor contains precisely the information about how volumes change in the presence of tidal forces, so the Weyl tensor is the traceless component of the Riemann tensor.

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WannabeNewton

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We then define the shear of the null geodesic congruence as ##\hat{\sigma}_{ab} = \widehat{\nabla_{(a}k_{b)}} - \frac{1}{2}\theta \hat{h}_{ab}## where ##\theta = \hat{h}^{ab}\widehat{\nabla_{a}k_{b}}## is the expansion of the congruence and ##\hat{h}_{ab}## is, at each point, the metric tensor on ##\hat{T}^{\perp}_{p}##. The physical interpretations of ##\hat{\sigma}_{ab}## and ##\theta## are the exact same as for timelike geodesic congruences. If we imagine a sphere being carried along any given null geodesic in the congruence, ##\theta## measures the volume expansion of the sphere and ##\hat{\sigma}_{ab}## measures the deformation of the sphere into an ellipse.

It is easy to show that ##k^{c}\nabla_{c}\hat{\sigma}_{ab} = -\theta \hat{\sigma}_{ab}+ \widehat{C_{cbad}k^{c}k^{d}}## where ##C_{cbad}## is the Weyl tensor. Note in particular that if ##\theta = 0## then we just have ##k^{c}\nabla_{c}\hat{\sigma}_{ab} = \widehat{C_{cbad}k^{c}k^{d}}##. So the Weyl tensor determines the rate of change of the shear of the null geodesic congruence along the congruence itself. See Wald chapter 9 for more details.

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Is T just a coordinate system or is it a whole new manifold?

Theta - this sounds very like the Ricci tensor, which when contracted twice with the vector for the direction in which the ball is moving and multiplied by Volume gives the second derivative of volume? - Are they the same or different. Because the first derivative of the volume is zero isn't it, because of the existence of Gaussian normal coordinates.

- #6

WannabeNewton

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Imagine we have a swam of (non-colliding) test particles in some region of space. The worldlines of the test particles in the swarm form a (normalized) vector field, ##\xi^a##, which fills some open subset ##U## of a space-time ##(M,g_{ab})##, such that through each event ##p## in ##U## there passes exactly one worldline (because the particles in the swarm are non-colliding). This is called a time-like congruence in ##U##.

Now define the tensor field ##h_{ab} = g_{ab} + \xi_{a}\xi_{b}##. Let ##p\in M##; ##T_p M## denotes the tangent space to ##M## at ##p##. If ##V^a \in T_p M##, then ##h^{a}{}{}_{b}V^b = V^{a} + \xi_b V^b \xi^a## is the projection of ##V^a## onto the subspace ##T^{\perp}_p M \subseteq T_p M## of vectors perpendicular to ##\xi^a## at ##p## because ##h^{a}{}{}_{b}V^b\xi_a = V^{a}\xi_a - \xi_b V^b = 0##. We can think of ##h_{ab}## as a spatial metric relative to a family of observers comoving with the swarm of test particles (i.e. a family of observers whose 4-velocity field is also ##\xi^a##). We can essentially use the swarm of test particles and the comoving family of observers interchangeably.

Consider now a time-like

Now we can define the (relevant) hydrodynamical quantities. The expansion ##\theta## is defined as ##\theta = h^{ab}\nabla_a \xi_b = (g^{ab} + \xi^a \xi^b)\nabla_a \xi_a = \nabla_a \xi^a## i.e. it is just the divergence of the 4-velocity field. One can show that if an observer carries an infinitesimal space-time volume ##V## along his worldline, then ##\theta = \frac{1}{V}\xi^a \nabla_a V## i.e. if you imagine ##V## as describing a small spherical volume of test particles from that swarm mentioned earlier (centered on the chosen observer) then ##\theta## measures how these particles expand away from or contract towards said observer. The shear ##\sigma_{ab}## is defined as ##\sigma_{ab} = \nabla_{(a}\xi_{b)} - \frac{1}{3}\theta h_{ab}## (the curved brackets in ##\nabla_{(a}\xi_{b)}## represent symmetrization, if you haven't seen that before). If you consider again a small spherical volume of test particles centered on the chosen observer, then ##\sigma_{ab}## measures the deformation of this spherical volume into an ellipsoidal volume. The third quantity ##\omega_{ab} = \nabla_{[a}\xi_{b]}## measures the (local) rotation of the congruence (the square brackets in ##\nabla_{[a}\xi_{b]}## represent antisymmetrization); ##\omega_{ab}## is essentially just the curved space version of the curl from vector calculus.

If we imagine a situation where ##\sigma_{ab} = \omega_{ab} = 0##, the evolution equation for ##\theta## comes out to ##\xi^{a}\nabla_{a}\theta = -\frac{1}{3}\theta^{2} - R_{ab}\xi^a \xi^b## so the Ricci tensor determines the rate of change of the expansion along the congruence when both the rotation and shear vanish. Now if we imagine a situation where ##\theta = \omega_{ab} = 0## and ##R_{ab} = 0## (i.e. vacuum) then the evolution equation for ##\sigma_{ab}## becomes ##\xi^{a}\nabla_{a}\sigma_{ab} = -\sigma_{ac}\sigma^{c}{}{}_{b}+ \frac{1}{3}h_{ab}\sigma_{cd}\sigma^{cd} + C_{cbad}\xi^{c}\xi^{d}##. As you can see, it isn't as nice as before but you can still make note of the fact that in vacuum, the Weyl tensor determines the rate of change of the shear along the congruence when both the rotation and expansion vanish. Clearly if we have non-vanishing expansion, rotation, and/or shear (and are not necessarily in vacuum) then the evolution equations are much more complicated and ##R_{ab}, C_{abcd}## both come into play for the evolution of all the aforementioned hydrodynamical quantities.

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WannabeNewton

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