# A physical description of some GR concepts

1. Jul 20, 2013

### R136a1

Can someone explain what the expansion, rotation, and shear of a time-like congruence are physically?

Last edited: Jul 20, 2013
2. Jul 20, 2013

### WannabeNewton

Sure. Let $\xi^{a}$ be a time-like congruence and $\gamma$ an integral curve of $\xi^a$; let $p\in \gamma(I)$. Take an orthonormal basis $\{\xi^{a},\eta^{a}_{1},\eta^{a}_{2},\eta^{a}_{3}\}$ for $T_p M$ such that $\mathcal{L}_{\xi}\eta^{a}_{i} = 0,\forall i$ (i.e. the $\eta^{a}_{i}$ are infinitesimal connecting vectors). The metric tensor can be written in this basis as $g_{ab} = -\xi_{a}\xi_{b} + \eta^{1}_{a}\eta^{1}_{b} + \eta^{2}_{a}\eta^{2}_{b} + \eta^{3}_{a}\eta^{3}_{b}$ so the spatial metric $h_{ab}$ is given by $h_{ab} = g_{ab} + \xi_{a}\xi_{b} = \eta^{1}_{a}\eta^{1}_{b} + \eta^{2}_{a}\eta^{2}_{b} + \eta^{3}_{a}\eta^{3}_{b}$. I now claim that the expansion $\theta = \nabla_{a}\xi^{a}$ represents the expansion of the infinitesimal space-time volume $V = \epsilon_{abcd}\xi^{a}\eta^{b}_{1}\eta^{c}_{2}\eta^{d}_{3}$ carried along $\gamma$. We have that $\xi^{n}\nabla_{n}\eta^{b}_{i} = \eta_{i}^{n}\nabla_{n}\xi^{b}$ so $\xi^{a}\nabla_{a}V = \epsilon_{abcd}\{(\xi^{n}\nabla_{n}\xi^{a})\eta^{b}_{1}\eta^{c}_{2}\eta^{d}_{3} + ...+(\eta_{3}^{n}\nabla_{n}\xi^{d})\xi^{a}\eta^{b}_{1}\eta^{c}_{2}\}$. Note that $\epsilon_{abcd}\eta^{b}_{1}\eta^{c}_{2}\eta^{d}_{3}$ must be parallel to $\xi^{a}$ at $p$ hence $\epsilon_{abcd}\eta^{b}_{1}\eta^{c}_{2}\eta^{d}_{3} = \alpha \xi^{a}\Rightarrow \alpha = -V$ at $p$. As a result, $\epsilon_{abcd}(\xi^{n}\nabla_{n}\xi^{a})\eta^{b}_{1}\eta^{c}_{2}\eta^{d}_{3} = -V\xi^{n}\xi_{a}\nabla_{n}\xi^{a} = 0$ at $p$. Handling the other terms in a similar manner, we end up with $\xi^{n}\nabla_{n}V = V(g_{r}^{n} + \xi_{r}\xi^{n})\nabla_{n}\xi^{r} = V\nabla_{n}\xi^{n} = V\theta$ as desired. If we imagine a small ball centered at $p$, $\theta$ tells us how its volume changes instantaneously.

As for the shear $\sigma_{ab}$, note that because it is symmetric we can find an orthonormal basis for $T_p M$ that orthogonally diagonalizes $\sigma_{ab}$ on top of the spatial basis vectors being lie transported along the integral curve of the time-like basis vector $\xi^{a}$ and as a result write $\sigma_{ab} = k_{1}\eta^{1}_{a}\eta^{1}_{b} + k_{2}\eta^{2}_{a}\eta^{2}_{b} + k_{3}\eta^{3}_{a}\eta^{3}_{b}$. $\sigma_{ab}$ is also traceless so that places a constraint on the eigenvalues $k_{i}$. Now assume that at $p$, the expansion and rotation vanish so that at $p$ we have $\xi^{n}\nabla_{n}\eta_{i}^{a} = k_{i}\eta^{a}_{i}$ after using the various conditions above. Now if we imagine a reference observer $O$, an infinitesimally nearby observer $O'$, and think of $\eta^{a}_{i}$ as a connecting vector between the two then the above tells us that $O'$ has an instantaneous velocity at $p$ relative to $O$ that is directed along $\eta^{a}_{i}$ so each $\eta^{a}_{i}$ defines an axis of instantaneous expansion/contraction with value $k_{i}$. Because of the trace constraint, expansion along one such axis must be sufficiently compensated for by contractions along the other axes so if we imagine a small ball centered at $p$, this ball will get distorted into an ellipse.

Rotation is the obvious one in the kinematical decomposition. It simply measures the extent to which the congruence fails to be hypersurface orthogonal i.e. $\omega_{ab} = 0$ if and only if $\xi_{[a}\nabla_{b}\xi_{c]} = 0$. Geroch gives a nice pictorial interpretation of this as follows: picture a normal twisted rope and note that there is no way you can slice the rope using a plane such that all the fibers of the twisted rope intersecting the plane are orthogonal to the plane.

3. Jul 20, 2013

### pervect

Staff Emeritus
I'll try a more "physical" and less rigorous description than Wanabee Newton's.

Consider a ball of coffee grounds - ala Baez, http://math.ucr.edu/home/baez/einstein/ "The meaning of Einstein's equation".

The worldlines of the individual grounds will trace out a time-like congruence. (The worldlines must be time like, and we'll assume the rest of the properties needed).

Then the ball can change it's volume. That's expansion. It can get squashed into an ellipsoid. That's shear. And, of course, it can rotate.

4. Jul 21, 2013

### R136a1

Thanks a lot pervect for giving a great intuition. And thanks miss wbn for giving a very neat mathematical derivation for the intuition!

5. Jul 21, 2013

### R136a1

May I ask why why $\omega_{ab} = 0$ if and only if $\xi^{a}$ is hypersurface orthogonal?

6. Jul 21, 2013

### WannabeNewton

Well first consider the twist $\omega_{a} = \epsilon_{abcd}\xi^{b}\nabla^{c}\xi^{d}$ of the time-like congruence. The rotation will be given in terms of the twist by $\omega_{ab} = \epsilon_{abcd}\xi^{c}\omega^{d}$. Hence if $\omega_{ab} = 0$ then $\xi^{c}\omega^{d} - \xi^{d}\omega^{c} = 0$ so $\omega_{a} = 0$. This then implies that $\xi^{[b}\nabla^{c}\xi^{d]} = 0$ so $\xi^{a}$ is hypersurface orthogonal by Frobenius' theorem. As for the converse, if $\xi^{a}$ is hypersurface orthogonal then $\xi^{a} = f\nabla^{a}g$ for smooth scalar fields $f,g$
so $\omega_{ab} = h_{[a}{}{}^{m}h_{b]}{}{}^{n}\nabla_{m}\xi_{n} = fh_{a}{}{}^{m}h_{b}{}{}^{n}\nabla_{[m}\nabla_{n]}g + h_{a}{}{}^{m}h_{b}{}{}^{n}\nabla_{[m}f\nabla_{n]}g = 0$ because $\nabla_{a}$ is torsion free (i.e. it commutes on smooth scalar fields) and $h_{a}{}{}^{m}\nabla_{m}g = f^{-1}h_{a}{}{}^{m}\xi_{m} = 0$.

Last edited: Jul 21, 2013
7. Jul 21, 2013

### R136a1

Thank you, that was a very good explanation!