- #51

WannabeNewton

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Thanks for the reply Peter! There is actually such a relation but before I go into that I thought the following would help even further clarify what was discussed above: http://math.stackexchange.com/questions/428839/irrotational-vorticesI wonder if there is a similarly intuitive way of writing the expansion and shear; this would be particularly helpful for the ZAMO congruence in Kerr spacetime, because of the nonzero shear.

Note the first animation corresponds to a congruence of observers for which the angular velocity goes like ##\omega \sim 1/r^2## which is essentially what happens with the ZAMO congruence at least in the equatorial plane. The second animation on the other hand corresponds to a (rigid) congruence of observers who all orbit with the same angular velocity such as, in our case, the Killing field which agrees with a single ZAMO orbit. Each paddle wheel represents the local swarm of observers surrounding the fiducial observer centered on the paddle wheel, in the chosen congruence.

Lifting these animations to Kerr space-time or even Minkowski space-time is of course not entirely accurate because the animations assume comoving gyroscopes do not precess relative to spatial infinity which is not true in our case but the basic intuition is still correct, we just have to lift it to a completely local description in terms of comoving gyroscopes in Kerr space-time instead of relative to spatial infinity.

Anyways, I realized I made a

**HUGE**mistake before; if I spotted it before I could have saved us a lot of posts. It is

*not*true that ##L = 0 \Leftrightarrow \eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0##; rather it is only true that ##L = 0 \Rightarrow \eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0##, the converse does not hold. If it did then in flat space-time the only irrotational congruence would be a rigid inertial congruence but this is clearly not true as the ##\omega \sim 1/r^2## congruence above is also irrotational.

This actually clarifies things for us greatly. Indeed from the expression for ##\omega^{\alpha}## from the previous post we see that there are two trivial cases of congruences for which ##\omega^{\alpha} = 0##: when ##L = 0## everywhere in space-time and when ##L = \eta_t##. The ##L = 0## case in flat space-time is just that of inertial observers whereas the ##L = \eta_t## case corresponds to ##\omega = \frac{L}{g_{\phi\phi}} \sim 1/r^2## which we know from before.

So ##L = 0 \Rightarrow \eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0## is simply the statement that if a ZAMO carried a paddle wheel along with the flow of the ZAMO congruence then the paddle wheel will not rotate relative to comoving gyroscopes

*because*the ZAMOs are all locally non-rotating, in the sense that they all have no local orbital rotation relative to the space-time geometry. But the converse as we have seen is

*not*true.

Indeed ##\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0## simply means that the paddle wheel carried along the flow locally doesn't rotate but that doesn't mean each flow line can't have an orbital rotation relative to the local space-time geometry, it just means the orbital rotation, and hence shear, varies along the congruence in such a way that the paddle wheel doesn't rotate. It's just that for the ZAMOs there is no local orbital rotation at all so the paddle wheel won't rotate locally as a result i.e. ##\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0##.

I'm sorry for that critical mistake I made. But does this clarify things at all? I'll talk about the shear in a separate post.