Properties of Born rigid congruence

[PeterDonis]

one must rotate tetrad's spacelike vector fields about the 4-velocity

No, this doesn't make sense. The 4-velocity is a timelike vector that is orthogonal to the spacelike vectors. It is not a spacelike vector orthogonal to the plane of rotation, which is what a rotation axis is.

I don't grasp it from a mathematical point: if the generalized Fermi normal coordinates do not work on a worldtube about a given congruence's worldline, on the same ground the construction of coordinate chart involving the timelike KVF shouldn't.

You keep repeating this wrong statement without apparently even noticing the reasons I have already given you for why it is false. I have already pointed out to you the key difference between the 4-velocity field and the KVF. Multiple times. Go back and read my posts again.

is it always true that for a timelike Killing congruence the KVF and the direction of proper acceleration commute at each point along the congruence ?

I don't know, and I don't have time to try either proving this or finding a counterexample.

 

[cianfa72]

No, this doesn't make sense. The 4-velocity is a timelike vector that is orthogonal to the spacelike vectors. It is not a spacelike vector orthogonal to the plane of rotation, which is what a rotation axis is.

Ah ok, you are right. However such rotations leave invariant the timelike coordinate basis vector along the worldline.

I have already pointed out to you the key difference between the 4-velocity field and the KVF. Multiple times. Go back and read my posts again.

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 don't know, and I don't have time to try either proving this or finding a counterexample.

Ok, so for sure we can claim that the timelike KVF of Langevin congruence and the direction of proper acceleration do commute (alike the case of Rindler congruence). However, up to now, we have no proof that this result extend to any timelike Killing congruence.

 

[PAllen]

As I have commented in response to the OP, in general this will not always work. We have seen a counterexample in this thread: the Langevin congruence frame field given in the Insights article I referenced, where $\hat{p}_0$ is the 4-velocity of the worldlines and $\hat{p}_2$ always points radially outward, is not a valid coordinate basis because $\hat{p}_0$ and $\hat{p}_2$ do not commute. Because of the nonzero vorticity of the congruence, $\hat{p}_2$ is not Fermi-Walker transported along the worldlines.

It is possible, however, to use the timelike Killing vector field $K$ whose integral curves are the Langevin congruence worldlines as the timelike basis vector for a coordinate chart in which $\hat{p}_2$ is also a coordinate basis vector, since $K$ does commute with $\hat{p}_2$. This is still "similar" to the Fermi-Walker plus rotation construction you describe since $K$ is still tangent to the worldlines; it's just not a unit tangent to the worldlines.

I disagree. MTW demonstrates this construction working in full GR with arbitrary base world line and arbitrary rotation function. What is generally true is that congruence of resulting constant position lines is not hypersurface orthogonal, and the coverage is only for a tube around the world line. Note, Fermi Normal coordinates , and the rotation generalization, are not based on a starting congruence. Instead they are based on one origin world and one chosen tetrad, which is Fermi-walker transported, then rotated, then used as a basis for the hypersurface. See pages 327 to 332 of MTW.

I can look later for a paper that demonstrates my other claim: that whenever the base world line motion plus rotation are a possible Born rigid motion, then the constant position congruence in generalized Fermi-Normal coordinates is a Born rigid congruence. This result is considered somewhat well known.

 

[cianfa72]

I disagree. MTW demonstrates this construction working in full GR with arbitrary base world line and arbitrary rotation function. What is generally true is that congruence of resulting constant position lines is not hypersurface orthogonal, and the coverage is only for a tube around the world line.

You mean in bold the congruence of worldlines each described by constant spatial coordinates and varying coordinate time in the generalized Fermi normal coordinate chart being built with a coverage only for a worldtube around the picked worldline.

 

[PAllen]

You mean in bold the congruence of worldlines each described by constant spatial coordinates and varying coordinate time in the generalized Fermi normal coordinate chart being built with a coverage only for a worldtube around the picked worldline.

Correct.

 

[cianfa72]

I can look later for a paper that demonstrates my other claim: that whenever the base world line motion plus rotation are a possible Born rigid motion, then the constant position congruence in generalized Fermi-Normal coordinates is a Born rigid congruence. This result is considered somewhat well known.

Sorry, what does it mean that the base worldline motion plus rotation are a possible Born rigid motion ?

 

[PAllen]

Sorry, what does it mean that the base worldline motion plus rotation are a possible Born rigid motion ?

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.

 

[PeterDonis]

MTW demonstrates this construction working in full GR with arbitrary base world line and arbitrary rotation function.

If that were true, it should be possible to have a valid coordinate chart with the vectors $\hat{p}_0$ and $\hat{p}_2$, as defined in my Insights article, as coordinate basis vectors. But that can't be possible, because those vector fields don't commute, as we have already shown in this thread. (Unless our computation of that was wrong; it wouldn't hurt to have an independent check.)

Of course one can always define a frame field (tetrad field) using an arbitrary rotation in addition to Fermi-Walker transport along an arbitrary worldline. But AFAIK that does not mean such a frame field will always be a valid coordinate basis. One can always find a coordinate basis whose vectors all point in the same direction as the frame field vectors; IIRC that is all MTW actually show. But I haven't reviewed that part of MTW in a while.

 

[PAllen]

If that were true, it should be possible to have a valid coordinate chart with the vectors $\hat{p}_0$ and $\hat{p}_2$, as defined in my Insights article, as coordinate basis vectors. But that can't be possible, because those vector fields don't commute, as we have already shown in this thread. (Unless our computation of that was wrong; it wouldn't hurt to have an independent check.)

Fermi-Normal coordinates don’t use either a vector field or frame filed in their construction. You Fermi-Walker transport one tetrad along a world line, rotate it, then use its spatial vectors as a basis to construct coordinates in the hypersurace orthogonal to the origin world line.

Of course one can always define a frame field (tetrad field) using an arbitrary rotation in addition to Fermi-Walker transport along an arbitrary worldline. But AFAIK that does not mean such a frame field will always be a valid coordinate basis. One can always find a coordinate basis whose vectors all point in the same direction as the frame field vectors; IIRC that is all MTW actually show. But I haven't reviewed that part of MTW in a while.

Again, you are focused on statements about fields. This is not how FN coordinates or their generalizations are constructed. This is described in detail in the reference I provided

 

[PeterDonis]

(Unless our computation of that was wrong; it wouldn't hurt to have an independent check.)

It's easy enough to rewrite the computation here, so I'll do it. In the cylindrical chart used in the Insights article, we have

$$
\hat{p}_0 = U = \gamma \left( \partial_T + \omega \partial_ \Phi \right)
$$

$$
\hat{p}_2 = E = \partial_R
$$

So $\mathscr{L}_U E = \nabla_U E - \nabla_E U$. We'll compute the two terms separately:

$$
U^a \nabla_a E^b = U^a \left( \partial_a E^b + \Gamma^b_{a c} E^c \right) = U^\Phi \Gamma^\Phi_{R \Phi} E^R = \frac{\gamma \omega}{R} \partial_\Phi
$$

$$
E^a \nabla_a U^b = E^a \left( \partial_a U^b + \Gamma^b_{ac} U^c \right) = E^R \left[ \partial_R U^T + \left( \partial_R + \Gamma^{\Phi}_{R \Phi} \right) U^\Phi \right] = \gamma^3 \omega^2 R \partial_T + \left( \gamma^3 \omega^3 R + \frac{\gamma \omega}{R} \right) \partial_\Phi
$$

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 $A \hat{p}_0$.

(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.)

 

[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.

 

[cianfa72]

This is a question that should be asked in a separate thread, probably in one of the math forums.

Anyway I believe the point I made in post #105 is the following: given a non-zero smooth vector field the Straightening theorem claims there exists a coordinate chart adapted to it in a neighborhood. In other words a smooth vector field is trivially always integrable.

On the other hand if you have two or more smooth vector fields, which are the conditions to be integrable (i.e. such that there exist integral immersed submanifolds) ? The answer is given by the Frobenius theorem.