1
$\begingroup$

Context

Consider a lightlike congruence with velocity vector $k^\mu$ in a $D$-dimensional lorentzian manifold, and the optical decomposition:

$$\nabla_\mu k_\nu = \omega_{\mu\nu}+\sigma_{\mu\nu}+\tfrac{1}{D-2}h_{\mu\nu}\theta,$$

where $\nabla$ is the Levi-Civita connection and $h_{\mu\nu}$ is the $(D-2)$-dimensional transversal metric. In particular, $\omega_{\mu\nu}$ represents the antisymmetric part (known as twist tensor):

$$\omega_{\mu\nu}:=\partial_{[\mu}k_{\nu]}.$$

It is also usual to introduce the optical scalars. One of them is the "twist" of the congruence given by:

$$\omega:=\sqrt{\tfrac{1}{D-2}\partial_{[\mu}k_{\nu]}\partial^{\mu}k^{\nu}} \equiv \sqrt{\tfrac{1}{D-2}\omega_{\mu\nu}\omega^{\mu\nu}}.$$

Question

In [1], for example, where the calculations are done in four dimensions, it is shown that the nullity of the twist scalar $\omega$ is equivalent to the existence of wave fronts. But they take advantage of the dimension to build the typical null tetrad $\{t,\bar{t},m,k\}$.

My question is: is it a general property for any $D$? I mean, the implication:

$$\omega=0 \qquad{} \Rightarrow \qquad{}\text{existence of wavefront}.$$

Some thoughts

I see clearly for example that $\omega_{\mu\nu}=0$ implies that $k_\mu$ is locally an exact form and then, $k^\mu$ is a gradient field, and we can use this to build the wave fronts. So

$$\omega_{\mu\nu}=0 \qquad{} \Rightarrow \qquad{}\text{existence of wavefront}.$$

If there were an equivalence between $\omega_{\mu\nu}=0$ and $\omega=0$ in general dimensions (for null congruences at least), everything works. However I think the implication $\omega=0\Rightarrow \omega_{\mu\nu}=0$ is not true in general.

[1] W. Kundt, Zeitschrift für Physics, 163 (1961), 77-86.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.