2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.