# Index free decomposition of derivative timelike congruence

1. Mar 7, 2015

### center o bass

Suppose we have a general timelike congruence of curves with tangent vector field $V$, then the standard decomposition of the covariant derivative in index form (see e.g. Hawking and Ellis' "Large scale structure of space and time" equation 4.17) is given by
$$V_{a;b} = \omega_{ab} + \sigma_{ab} + \frac{1}{3} \theta h_{ab} - \dot U_a U_b$$
where $\omega_{ab}, \ \sigma_{ab}, \ \theta, \ \dot U_a$ respectively represent the rotation, shear, expansion and acceleration of the congruence.

QUESTION: Is there a decomposition analogous to this but in an index-free language? If so I would very much appreciate a reference.

2. Mar 7, 2015

### Ben Niehoff

Surely you can just erase the indices...I'm not sure what the issue is.

It's probably easier to work with the previous equations that express the twist, shear, and expansion in terms of $\nabla V$. All of the terms are clearly defined beginning in Chapter 4, it shouldn't be hard to "translate" them into index-free language.

3. Mar 7, 2015

### center o bass

How would you define $\omega_{ab}, \sigma_{ab}$ and $\theta$ by $\nabla V$ in an index-free language?

4. Mar 7, 2015

### martinbn

Trace, symmetric part, antisymmetric part.

5. Mar 7, 2015

### center o bass

Yes, but the trace is taken relative to $h_{ab}$. How would one express this on an index free from?

6. Mar 7, 2015

### Ben Niehoff

Well, if you turn back a page or two, $h$ is defined:

$$h = g - V \otimes V$$
Or you can just think of $h$ as the induced metric on the orthogonal hypersurfaces to the path.

I think you might actually have better luck looking back at the description of Fermi propagation, since the twist and shear are defined relative to that. Also, it's already given in index-free language. The Fermi-propagated orthonormal frame gives you a basis for $h$.

It also turns out, however, that $\theta$ is just the divergence of $V$. Since we've already constrained $V$ to be a unit vector, this operation kind of automatically restricts the trace to be over $h$.