Geometrical meaning of Weyl tensor

[Benjam:n]

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

 

[robphy]

On my to-read list...
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

 

[A.T.]

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.

 

[WannabeNewton]

There are many different ways to interpret it. One easy way is in terms of the shear of null geodesics. Let $k^a$ represent the tangent vector field to a congruence of null geodesics. We want to find a way to describe the behavior of a neighboring collection of null geodesics in the congruence relative to each other and for this we need a spatial deviation vector connecting a given null geodesic in the congruence to infinitesimally nearby ones in the congruence. Let $p\in M$ and let $T^\perp _p(M)$ denote the subspace of $T_p M$ containing vectors orthogonal to $k^a(p)$. Now since all vectors in $T^\perp _p(M)$ which differ by a constant multiple of $k^a(p)$ represent the same physical deviation vector, we take the equivalence class $\hat{T}^{\perp}_{p}$ of vectors in $T^\perp _p(M)$ where $v^a\sim w^a$ if $v^a - w^a = ck^a$. Tensor fields over $M$ naturally give rise to tensor fields which at each point are tensors over $\hat{T}^{\perp}_{p}$; we denote them with a hat.

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.

 

[Benjam:n]

Thanks for the help as always wannabeNewton. Thanks to you (and others) I've got over my obsession with coordinates at last. I understand congruence's. Null congruence's are just the path that moves through no proper time. And geodesic congruence's are just where at each point in the local coordinate system with the flat metric the path looks like a straight line. I get lost at the point where you start talking about the equivalence classes. If you take all the vectors at a point which are perpendicular to the tangent vector of one of the paths, then why will there be vectors in that which differ only by a scalar factor of the tangent vector. surely if you add a bit of the tangent vector on to a perpendicular vector then it will no longer be perpendicular to the tangent vector and thus no longer be in that set?
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.

 

[WannabeNewton]

It will be easier if I start with time-like congruences instead because there the equivalence classes are not necessary. The Ricci and Weyl tensor can again be related to hydrodynamical quantities associated with time-like congruences although not in a manner as elegant as with null geodesic congruences. I'll try to be more thorough this time!

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 geodesic congruence i.e. one for which $\xi^b \nabla_b \xi^a = 0$. Pick a reference geodesic $\gamma_0$ in the congruence and at some event $p$ on the geodesic, imagine that an observer described by $\gamma_0$ erects a vector $\eta^a$ such that $\eta_a \xi^a = 0$ at $p$. Then, at $p$, $\eta^a$ represents the separation (i.e. displacement) vector from this observer to an infinitesimally nearby observer in the family. $\eta^a$ gets lie transported along the chosen observer's worldline i.e. $\mathcal{L}_{\xi}\eta^{a} = \xi^b \nabla_b \eta^a - \eta^b \nabla_b \xi^a =0$; this is because $\eta^a$ will always remain pointing from the chosen observer to the infinitesimally nearby observer. Note that $\xi^b \nabla_b (\xi^a \eta_a) = \eta_a \xi^b \nabla_b \xi^a + \xi^a \xi^b \nabla_b \eta _a = \eta ^b\xi^a \nabla_b \xi_a = 0$ so $\eta^a$ always remains orthogonal to $\xi^a$.

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.

 

[Benjam:n]

What is the Lie derivative? Is it just like the covariant derivative, but treating a different metric as the flat metric? Is there any way of explaining it that avoids lie derivatives? Or should I try learning lie derivative first. Does the lie derivative require lie algebra to understand?

 

[WannabeNewton]

The lie derivative condition isn't really important for the physics itself; it's just needed for calculations (e.g. to show that $\xi^b \nabla_b (\xi^a \eta_a) = 0$) and to make sense of what a connecting vector is mathematically. For now you can just think of the lie derivative of two vector fields as the commutator of the two vector fields; as far as vector fields go the two are equivalent. You don't need to know anything about lie algebras (the two are related of course).