I Properties of Born rigid congruence

[cianfa72]

Putting the above together, we see that the $\gamma \omega / R$ terms cancel and we are left with
$$
\mathscr{L}_U E = - \gamma^3 \omega^2 R \left( \partial_T + \omega \partial_\Phi \right)
$$ which equates to $A \hat{p}_0$.

Ok, in your work you used Ricci calculus with Latin indices (although Latin indices are typically used in abstract index notation).

Note that this is actually not the same result I had given in an earlier post; we can also see that $\nabla_{\hat{p}_2} \hat{p}_0 = \Omega \hat{p}_3$, which makes more sense than what I had incorrectly computed in that earlier post.)

From Insights article $\Omega = \gamma^2 \omega$ and $\hat{p}_3 = \gamma \omega R \partial_T + \frac{\gamma}{R} \partial_{\Phi}$, so the above calculation of $\nabla_{\hat{p}_2} \hat{p}_0$ is not equal to $\Omega \hat{p}_3$.

 

[cianfa72]

Look up the Herglotz-Noether theorem. It establishes what combinations of motion of one world line optionally plus rotation are compatible with Born rigidity. For example, inertial motion plus uniform rotation is compatible with Born rigidity. However, changing rotation is not compatible.

Above by motion of one worldline (in bold) you mean actually its path through spacetime.

I saw it on Wikipedia, however I didn't find many details. As far as I can understand, given a timelike worldline with non-zero vorticity, it is allowed to be member of a Born rigid timelike congruence if it is the integral curve of a timelike KVF.

 

[cianfa72]

Yes that's true, however we don't have a mathematical argument/reason that shows why it doesn't work for generalized Fermi normal coordinates (i.e. Fermi normal coordinates followed from rotations). We know that it works for timelike KVF in special cases like Langevin or similar congruences.

I think the problem with generalized Fermi normal coordinates applied to Langevin congruence is the following: it is true that the Fermi-Walker transported rotated tetrad's spacelike vectors are orthogonal to the 4-velocity at each point along the base worldline (and therefore the spacelike hyperplane spanned by them at each point along the base worldline contains the proper acceleration vector at that point). However since the Langevin congruence is not hypersurface orthogonal, that isn't true for the congruence's worldlines in the worldtube around the base worldline. This could be the reason why a such coordinate chart built around the base worldline doesn't work.

Possibly it would work for irrotational timelike congruences (i.e. with zero vorticity or hypersurface orthogonal).

 

[PeterDonis]

Ok, in your work you used Ricci calculus with Latin indices (although Latin indices are typically used in abstract index notation).

Whatever. I used the same notation I used in the Insights article.

From Insights article $\Omega = \gamma^2 \omega$ and $\hat{p}_3 = \gamma \omega R \partial_T + \frac{\gamma}{R} \partial_{\Phi}$, so the above calculation of $\nabla_{\hat{p}_2} \hat{p}_0$ is not equal to $\Omega \hat{p}_3$.

Yes, it is. The $\partial_T$ term is the same by inspection. The $\partial_\Phi$ term I calculated in post #150 just needs some algebra:

$$
\gamma^3 \omega^3 R + \frac{\gamma \omega}{R}
$$
$$
= \gamma^3 \omega^3 R + \frac{\gamma^3 \left( 1 - \omega^2 R^2 \right) \omega}{R}
$$
$$
= \gamma^3 \omega^3 R + \frac{\gamma^3 \omega}{R} - \gamma^3 \omega^3 R
$$
$$
= \frac{\gamma^3 \omega}{R}
$$

which is $\gamma^2 \omega$ times the $\partial_\Phi$ term of $\hat{p}_3$.

 

[cianfa72]

which is $\gamma^2 \omega$ times the $\partial_\Phi$ term of $\hat{p}_3$.

Ah yes, you are right

 

[PAllen]

I think the problem with generalized Fermi normal coordinates applied to Langevin congruence is the following: it is true that the Fermi-Walker transported rotated tetrad's spacelike vectors are orthogonal to the 4-velocity at each point along the base worldline (and therefore the spacelike hyperplane spanned by them at each point along the base worldline contains the proper acceleration vector at that point). However since the Langevin congruence is not hypersurface orthogonal, that isn't true for the congruence's worldlines in the worldtube around the base worldline. This could be the reason why a such coordinate chart built around the base worldline doesn't work.

Possibly it would work for irrotational timelike congruences (i.e. with zero vorticity or hypersurface orthogonal).

Since when do coordinates have to be hypersurface orthogonal? That is a rare special case. In the very first post I made here about generalized Fermin-Normal coordinates I pointed out that in most cases the coordinates were NOT hypersurface orthogonal.

 

[cianfa72]

That is a rare special case. In the very first post I made here about generalized Fermin-Normal coordinates I pointed out that in most cases the coordinates were NOT hypersurface orthogonal.

The tetrad's spacelike vectors are orthogonal to the 4-velocity along the base worldline by construction. However they are not in general off that base worldline w.r.t. the 4-velocity of congruence's worldlines inside the worldtube around it where they are defined.

In other words generalized Fermi normal coordinates for Langevin congruence do not give coordinates basis vector fields equal to vectors $\hat{p}_0, \hat{p}_2$ when evaluated off the base worldline inside the worldtube. I believe zero Lie derivative actually requires it in an open neighborhood.

 

[PAllen]

The tetrad's spacelike vectors are orthogonal to the 4-velocity along the base worldline by construction. However they are not in general off that base worldline w.r.t. the 4-velocity of congruence's worldlines inside the worldtube around it where they are defined.

True, and so what?

In other words generalized Fermi normal coordinates for Langevin congruence do not give coordinates basis vector fields equal to vectors $\hat{p}_0, \hat{p}_2$ when evaluated off the base worldline inside the worldtube. I believe zero Lie derivative actually requires it in an open neighborhood.

True, and so what? The Lie derivative condition is IF you want coordinate basis at every point to match some vector fields. But that says nothing about existence of coordinates that don’t have this property.

 

[cianfa72]

The Lie derivative condition is IF you want coordinate basis at every point to match some vector fields. But that says nothing about existence of coordinates that don’t have this property.

Yes, of course. That's the reason why generalized Fermi normal coordinates do not give a coordinate basis vector field that match up $\{ \hat{p}_0, \hat{p}_2 \}$ vector fields in a neighborhood of Langevin congruence's base worldline.