I am trying to solve this kind of problem:

Consider the Milne model, i.e., the empty $ \kappa = −1 $ Friedmann model. Verify by a direct calculation that the expansion $ \Theta $ of the unit normal to the $ \tau = \mathrm{const}$ hypersurfaces satisfies $$ \dot{\Theta} + \Theta^2/3 = 0. $$

I started from Raychauduri equation

$$ \dot{Θ} + Θ^2/3 - \dot{u}^a_{;a} + 2(σ_{ab}σ^{ab} − ω_{ab}ω^{ab}) + R_{ab}u^au^b = 0 $$

Where for empty space $ R_{ab}=0 $ and geodesic flow $\dot{u}=0$.

Then I found that if the congruences are hypersurface orthogonal, then $\omega_{ab}=0$. But I don't really understand what that means(or why is it true), can someone please explain it to me please?


It can be shown that $\omega_{ab} = 0 \Leftrightarrow \omega^a \equiv \epsilon^{abcd}u_b \nabla_c u_d = 0$. The latter quantity is known as the twist (or vorticity). In a local inertial frame it is easy to see that $\vec{\omega} \sim \vec{\nabla}\times \vec{v}$ where $\vec{v}$ is the 3-velocity field of the flow.

This lends to the following interpretation of $\omega^a$ (and hence of $\omega_{ab}$). If an observer $\mathcal{O}$ in the congruence carries with them a local Lorentz frame $e_{\hat{\alpha}}$ such that the spatial axes $e_i$ are Fermi-Walker transported, so that these spatial axes constitute a set of mutually orthogonal torque-free gyroscopes, then the average rotation, relative to the gyroscopes, of a sphere of observers in the congruence centered on $\mathcal{O}$ will vanish.

Now, a geodesic flow in flat space-time is just a congruence of inertial observers. Furthermore, and more importantly, the expansion, just as in the usual FRW metric, is isotropic and homogenous meaning relative to a fiducial observer $\mathcal{O}$, the neighboring observers only have a relative radial velocity in the Milne coordinates. Thus they clearly do not rotate relative to $\mathcal{O}$, which is why $\omega_{ab} = 0$, basically by construction.

As for why hypersurface orthogonality implies $\omega_{ab} = 0$ on an intuitive level, it is simply because if $\omega_{ab} \neq 0$ then the worldlines of the congruence would be twisting around one another in which case it is impossible to find a non-intersecting family of surfaces that is everywhere orthogonal to the worldlines.

  • $\begingroup$ Thanks a lot. And so because we don't have any "deformation" in our expansion, that is the same reason why $ \sigma^2 = 0 $? $\endgroup$ – user35780 Jul 2 '15 at 14:44
  • $\begingroup$ Indeed, that is the case. $\endgroup$ – FenderLesPaul Jul 2 '15 at 14:58

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