I Properties of Born rigid congruence

[cianfa72]

Because "orthogonal" implies "linearly independent", and the latter is a sufficient condition for defining a coordinate basis vector field.

Yes, at any point an orthogonal vector field is linearly independent from the coordinate basis vector field associated to the timelike KVF's "adapted" coordinate time $t$.

However the question is: is such linearly independent vector field a coordinate basis vector field?

 

[PeterDonis]

the question is: is such linearly independent vector field a coordinate basis vector field?

Go read the last sentence of my post #100, which you quoted, again. Also read the last part of my post #97, which you quoted in your post #99, again, carefully.

 

[PeterDonis]

So actually $\mathscr{L}_{\hat{p}_0}\hat{p}_2$ does not vanish

Just to confirm by direct computation [Edit--this is not correct, see post #150 for a corrected computation]:

$$
\mathscr{L}_{\hat{p}_0}\hat{p}_2 = \nabla_{\hat{p}_0} \hat{p}_2 - \nabla_{\hat{p}_2} \hat{p}_0
$$

The first term is given in the Insights article. The second term is just $\partial_R \hat{p}_0 = ( \partial_R \gamma ) ( \partial_T + \omega \partial_\Phi ) = \gamma^3 \omega^2 R ( \partial_T + \omega \partial_\Phi ) = - A \hat{p}_0$ because there are no nonzero connection coefficients for $\nabla_{\hat{p}_2}$. So we have

$$
\mathscr{L}_{\hat{p}_0}\hat{p}_2 = A \hat{p}_0 + \Omega \hat{p}_3 - ( - A \hat{p}_0 ) = 2 A \hat{p}_0 + \Omega \hat{p}_3
$$

This result actually puzzles me somewhat. I would have expected the $\hat{p}_0$ terms to cancel, leaving just $\Omega \hat{p}_3$. However, $\partial_R \hat{p}_0 = \gamma^3 \omega^2 R ( \partial_T + \omega \partial_\Phi )$ does make sense since as we go further out along a radial line, the ordinary 3-velocity of a Langevin worldline increases in the $\Phi$ direction, so we expect both the $T$ and the $\Phi$ components of the 4-velocity to increase, and that is what the formula is saying. [Edit: see post #150 for a correction.]

 

[cianfa72]

You should be able to find a coordinate transformation to a new chart in which the KVF is just $\partial_t$ (hint: Born coordinates). Since coordinate transformations are invertible, the inverse of such a transformation establishes the statement I made for this particular case, and generalizing that method should make it evident that the statement I made holds generally.

Ok, the above boils down to the Straightening theorem that holds for any non-zero smooth vector field $K$. Namely, in a neighborhood of each point P in which $K(P) \neq 0$, there is a local coordinate chart such that $K=\partial / \partial_t$, $\mathscr{L}_{K} (\cdot) = \partial_t(\cdot)$ where $t$ is the timelike coordinate in that chart.

Because "orthogonal" implies "linearly independent", and the latter is a sufficient condition for defining a coordinate basis vector field.

Ok, is it basically given from a 2nd application of the above theorem ?

 

[cianfa72]

Btw, I didn't grasp why the wiki entry Straightening theorem states the following

The Frobenius theorem in differential geometry can be considered as a higher-dimensional generalization of this theorem

 

[PeterDonis]

the above boils down to the Straightening theorem

Hm, I didn't state the condition strongly enough. What I stated is equivalent to the straightening theorem, yes, and that theorem applies to any vector field, not just a KVF.

For a KVF we can make the stronger statement that in the chart where the KVF is $\partial_t$, all of the metric coefficients are independent of $t$. We actually were implicitly making use of that property as well in this discussion. That property does not hold for a vector field that is not a KVF.

is it basically given from a 2nd application of the above theorem ?

No. The straightening theorem by itself would let us find a chart in which $A \hat{p}_2$, the proper acceleration, was a coordinate basis vector. But it would not guarantee that the KVF is also a coordinate basis vector in that chart.

However, you are missing the fact that I did not claim that we could find a coordinate chart in which $A \hat{p}_2$ was a coordinate basis vector and which had the other necessary properties. The claim I made was weaker. Go read the posts I told you to read, again, carefully.

 

[PeterDonis]

Btw, I didn't grasp why the wiki entry Straightening theorem states the following

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

 

[PAllen]

Perhaps of interest here is a generalization of Fermi-Normal coordinates. Standard Fermi-Normal coordinates are defined based on an arbitrary world line in possibly curved spacetime by constructing an arbitrary orthonormal tetrad (one of whose vectors must be the tangent of the world line) at a chosen origin point on the world line. Then this is Fermi-Walker transported along the world line. The foliation of hypersurfaces orthogonal to the world line is then coordinatized with the transported tetrad. Generally, this coordinate patch is only valid for a small tube around the origin world line.

This construction is often generalized by adding an arbitrary rotation $\omega(\tau)$ applied to the tetrad 'following' Fermi-Walker transport. This tends to shrink the coverage of the coordinates even more and also results that the coordinate stationary lines are not typically orthogonal to the foliation away from the origin.

With this given, the following is known: If the motion described by the origin world line plus the rotation is consistent with a valid Born-Rigid motion, then the coordinate stationary lines form a Born-rigid congruence. In this case, it is also true that proper distance between coordinate stationary lines within the foliation remains constant.

 

[cianfa72]

For a KVF we can make the stronger statement that in the chart where the KVF is $\partial_t$, all of the metric coefficients are independent of $t$. We actually were implicitly making use of that property as well in this discussion. That property does not hold for a vector field that is not a KVF.

Yes, definitely.

No. The straightening theorem by itself would let us find a chart in which $A \hat{p}_2$, the proper acceleration, was a coordinate basis vector. But it would not guarantee that the KVF is also a coordinate basis vector in that chart.

Yes.

However, you are missing the fact that I did not claim that we could find a coordinate chart in which $A \hat{p}_2$ was a coordinate basis vector and which had the other necessary properties. The claim I made was weaker.

I believe I grasped it. You meant that the timelike KVF $K$ for Langevin congruence is given from $K = \partial_T + \omega \partial_\Phi$ in Minkowski cylindrical coordinates.

The direction of Langevin congruence's proper acceleration at each point in that chart is $\hat{p}_2 = \partial_R$. Then, since $K$ has constant components in Minkowski cylindrical coordinates and $\partial_T, \partial_R, \partial_{\phi}$ are coordinate basis vector fields, the Lie bracket $[K, \partial_R]$ (i.e. Lie derivative) vanishes.

 

[PeterDonis]

the timelike KVF $K$ for Langevin congruence is given from $K = \partial_T + \omega \partial_\Phi$ in Minkowski cylindrical coordinates.

Yes.

The direction of Langevin congruence's proper acceleration at each point in that chart is $\hat{p}_2 = \partial_R$. Then, since $K$ has constant components in Minkowski cylindrical coordinates and $\partial_T, \partial_R, \partial_{\phi}$ are coordinate basis vector fields, the Lie bracket $[K, \partial_R]$ (i.e. Lie derivative) vanishes.

Yes.

However, that still doesn't address the claim I made about finding a coordinate chart in which the KVF $K$ is $\partial_t$, in which $\partial_R$ is also a coordinate basis vector.

 

[cianfa72]

However, that still doesn't address the claim I made about finding a coordinate chart in which the KVF $K$ is $\partial_t$, in which $\partial_R$ is also a coordinate basis vector.

Sorry to be boring, can you kindly restate why one can always find a local coordinate chart having those properties ?

 

[PeterDonis]

can you kindly restate why one can always find a local coordinate chart having those properties ?

If we have two vector fields $K$ and $E$ that we know are linearly independent and that we know commute, i.e., their Lie bracket vanishes, what does that say about our ability to find a coordinate chart in which both of them are basis vectors?

 

[cianfa72]

If we have two vector fields $K$ and $E$ that we know are linearly independent and that we know commute, i.e., their Lie bracket vanishes, what does that say about our ability to find a coordinate chart in which both of them are basis vectors?

Ah ok, as stated in J. Lee's book "Introduction to Smooth Manifold" if and only if two linearly independent vector fields commute in an open neighborhood, then there exists a local coordinate chart such that those vector fields are coordinate basis vectors.

 

[PeterDonis]

Ah ok, as stated in J. Lee's book "Introduction to Smooth Manifold" if and only if two linearly independent vector fields commute in an open neighborhood, then there exists a local coordinate chart such that those vector fields are coordinate basis vectors.

Yes. (Note that most GR textbooks state the same result, although most of them do not give a detailed proof.)

 

[cianfa72]

In at least one special case, which applies to the Langevin congruence, this result is easily shown: if the congruence of worldlines is a Killing congruence, i.e., if the worldlines are integral curves of a Killing vector field. The Langevin congruence worldlines are integral curves of $\partial_T + \omega \partial_\Phi$ in the coordinates used in the Insights article; this is a KVF because $\omega$ is constant. (Note, though, that it is not a 4-velocity field because it does not have unit magnitude; the normalization factor $\gamma$ is missing.)

If we call this vector field $K$, and we define $E = \hat{p}_2$, then the Lie bracket $[K, E]$ obviously vanishes by the argument you gave earlier (it's the Lie bracket of sums of coordinate basis vector fields with constant coefficients). And for a timelike Killing vector field $K$, we can always define a coordinate chart such that $K$ is the sum of coordinate basis vector fields with constant coefficients.

And since the proper acceleration of a worldline is always orthogonal to the worldline's tangent vector, we can always use its direction to define another coordinate basis vector field, which we can call $E$. So we can always set things up so the Lie bracket $[K, E]$ vanishes.

Sorry, I think I missed the point: in the general case in which the timelike congruence is a Killing congruence (therefore not in the specific case of Langevin congruence) even though the proper acceleration of a worldline $E$ is always orthogonal to the worldline's tangent vector (i.e. it is orthogonal at each point to the timelike KVF) hence it is linearly independent, that doesn't imply that the timelike KVF and $E$ actually commute.

 

[PeterDonis]

Sorry, I think I missed the point: in the general case in which the timelike congruence is a Killing congruence (therefore not in the specific case of Langevin congruence) even though the proper acceleration of a worldline $E$ is always orthogonal to the worldline's tangent vector (i.e. it is orthogonal at each point to the timelike KVF) hence it is linearly independent, that doesn't imply that the timelike KVF and $E$ actually commute.

Yes, you have missed the point: once again, I did not say the KVF and the proper acceleration of the worldline commute. I made a weaker claim. Go back and read my posts again, carefully. (Hint: did I say that $E$ itself is the proper acceleration vector?)

 

[cianfa72]

Yes, you have missed the point: once again, I did not say the KVF and the proper acceleration of the worldline commute. I made a weaker claim. Go back and read my posts again, carefully. (Hint: did I say that $E$ itself is the proper acceleration vector?)

No, you said that we can always use the direction of the proper acceleration at each point along the congruence to define another coordinate basis vector field $E$. My doubt is whether the above is always feasible.

 

[PeterDonis]

No, you said that we can always use the direction of the proper acceleration at each point along the congruence to define another coordinate basis vector field $E$. My doubt is whether the above is always feasible.

It is. Hint: Fermi normal coordinates.

 

[cianfa72]

Fermi normal coordinates.

Fermi normal coordinates are defined in such way that $e_0(\tau)$ points everywhere along the timelike congruence's 4-velocity and they require that the orthonormal spacelike vector fields $\{ e_a(\tau) \}$ picked must be Fermi-Walker transported along the congruence.

However we found that the above isn't true in general even if one considers just the direction of proper acceleration at each point (see also the case of Langevin congruence examinated earlier).

 

[PeterDonis]

Fermi normal coordinates are defined in such way that $e_0(\tau)$ points everywhere along the timelike congruence's 4-velocity and they require that the orthonormal spacelike vector fields $\{ e_a(\tau) \}$ picked must be Fermi-Walker transported along the congruence.

Yes.

However we found that the above isn't true in general even if one considers just the direction of proper acceleration at each point (see also the case of Langevin congruence examinated earlier).

That's true. But think about it: if you can construct Fermi normal coordinates, you can also add to those coordinates a time-dependent rotation of the axes so as to keep one of them always pointed in the direction of the proper acceleration.

And in fact for the Langevin congruence we already know such a coordinate chart exists, one in which the congruence worldlines are "at rest" (i.e., the Langevin KVF is $\partial_t$) and one of the spatial axes always points in the direction of proper acceleration (technically it's the "radial" axis that always points outward instead of inward, but that's just a minus sign). I told you the name of this chart in one of the hints I gave you in a previous post.

 

[cianfa72]

That's true. But think about it: if you can construct Fermi normal coordinates, you can also add to those coordinates a time-dependent rotation of the axes so as to keep one of them always pointed in the direction of the proper acceleration.

Ok, we construct Fermi normal coordinates starting from an orthonormal basis picked at a point and Fermi-Walker transport it along the timelike congruence. Does the above added time-dependent rotation define a coordinate chart as well ?

I told you the name of this chart in one of the hints I gave you in a previous post.

Ah yes, the chart you were talking about is the Born chart. I missed the point that in such a chart (in which the Langevin congruence's worldlines are "at rest") the Langevin timelike KVF is $\partial_t$ and the direction in which the proper acceleration always points is $\partial_R$, therefore they commute (i.e. their Lie derivative vanishes).

 

[PeterDonis]

we construct Fermi normal coordinates starting from an orthonormal basis picked at a point and Fermi-Walker transport it along the timelike congruence. Does the above added time-dependent rotation define a coordinate chart as well ?

Is a rotation of axes a valid coordinate transformation from one chart to another?

 

[PeterDonis]

the chart you were talking about is the Born chart.

Yes, exactly.

 

[cianfa72]

Is a rotation of axes a valid coordinate transformation from one chart to another?

You mean a rotation of just the direction of proper acceleration along the timelike congruence's worldlines. This is time-dependent (since the congruence's members can be parametrized by the coordinate time) rotation and yes it is a valid transformation

 

[cianfa72]

I think the argument above is much more general: for every timelike congruence the 4-velocity and the direction of proper acceleration at any point along them does commute (Lie derivative vanishes).

 

[PeterDonis]

for every timelike congruence the 4-velocity and the direction of proper acceleration at any point does commute (Lie derivative vanishes).

The directions of those two things indeed always commute. But the vector fields themselves in general do not. For example, the KVF $K$ for the Langevin congruence and the direction of proper acceleration $E = \partial_R$ commute; but the 4-velocity field $\hat{p}_0$ of the Langevin congruence does not commute with $E$ (we showed that earlier in this thread).

 

[cianfa72]

The directions of those two things indeed always commute. But the vector fields themselves in general do not.

Ah ok, so the fact that 4-velocity field has always norm $-1$ doesn't qualify it as a "direction".

 

[PeterDonis]

the fact that 4-velocity field has always norm $-1$ doesn't qualify it as a "direction".

I'm not sure what you mean. The KVF $K$ points in the same direction as the 4-velocity $\hat{p}_0$. The difference between them is in their magnitudes, not that one defines a "direction" and the other doesn't.

 

[cianfa72]

The KVF $K$ points in the same direction as the 4-velocity $\hat{p}_0$. The difference between them is in their magnitudes, not that one defines a "direction" and the other doesn't.

The point I was trying to make is that if we take the 4-velocity (that is normalized in length) along the timelike congruence and the direction of proper acceleration (that's normalized as well) then, starting from an orthonormal set of vectors at a point that includes the former as unit timelike vector and the latter as one of the orthonormal spacelike vectors, we can build Fermi normal coordinates adapted to them (in which the above two are coordinate basis vector fields). Therefore they commute. So far so good.

Now if we apply your argument about time-dependent axis rotation on proper acceleration direction along the congruence (leaving the timelike direction invariant) then we should get a coordinate chart in which both the 4-velocity and the proper acceleration direction are coordinate basis vectors, hence they must commute.

 

[PeterDonis]

if we take the 4-velocity (that is normalized in length) along the timelike congruence

This would be the vector $\hat{p}_0$ in the notation of the Insights article.

and the direction of proper acceleration (that's normalized as well)

Not in general, no. It happens to be a unit vector in this particular case, the vector $E = \partial_R = \hat{p}_2$. But it won't always be.

starting from an orthonormal set of vectors at a point that includes the former as unit timelike vector and the latter as one of the orthonormal spacelike vectors, we can build Fermi normal coordinates adapted to them (in which the above two are coordinate basis vector fields). Therefore they commute.

No, they don't. You can check explicitly that, in the notation of the Insights article, $\hat{p}_0$ and $\hat{p}_2$ do not commute.

The point you are missing is that $\hat{p}_2$ is not Fermi_Walker transported along the Langevin worldlines. That is explicitly shown in the Insights article. If you insist on using $\hat{p}_0$ as your timelike basis vector in Fermi normal coordinates, they only work for spacelike basis vectors that are Fermi-Walker transported.

You are also missing the fact that the claim I made was weaker: I did not claim that you can have a coordinate chart with $\hat{p}_0$ and $\hat{p}_2$ as basis vectors. I only claimed that you can have a coordinate chart with the KVF $K$ and $\hat{p}_2$ (i.e., the vector field $E$) as basis vectors. In such a chart, $\hat{p}_0$ will always point in the "time" direction, but that's not the same as being the timelike coordinate basis vector.

Now if we apply your argument about time-dependent axis rotation on proper acceleration direction along the congruence (leaving the timelike direction invariant) then we should get a coordinate chart in which both the 4-velocity and the proper acceleration direction are coordinate basis vectors, hence they must commute.

My argument was for a weaker claim than the one you are making here (which is false). See above.

 

[cianfa72]

You are also missing the fact that the claim I made was weaker: I did not claim that you can have a coordinate chart with $\hat{p}_0$ and $\hat{p}_2$ as basis vectors. I only claimed that you can have a coordinate chart with the KVF $K$ and $\hat{p}_2$ (i.e., the vector field $E$) as basis vectors. In such a chart, $\hat{p}_0$ will always point in the "time" direction, but that's not the same as being the timelike coordinate basis vector.

Sorry, I don't understand which is the difference in taking the congruence 4-velocity vs timelike KVF in the construction of Fermi normal coordinates when the congruence is a Killing congruence.

Both 4-velocity $e_0$ and timelike KVF $K$ have the same direction. $e_0$ and the orthogonal proper acceleration unit vector can be used to define Fermi normal coordinates adapted to congruence by Fermi-Walker transport such tetrad along the congruence starting from an "initial" point. After such process is completed, we can add a possibly axis rotation to the spacelike vector field obtained by Fermi-Walker transporting the unit proper acceleration vector of the tetrad from the starting point (to align it to the proper acceleration direction at each point). In the coordinate chart we get this way both 4-velocity and proper acceleration direction result to be coordinate basis vectors.

Where am I wrong ? Thank you.

 

[PeterDonis]

I don't understand which is the difference in taking the congruence 4-velocity vs timelike KVF in the construction of Fermi normal coordinates when the congruence is a Killing congruence.

The difference is in whether the vector field commutes with $\partial_R$. The KVF does, the congruence 4-velocity doesn't.

Both 4-velocity $e_0$ and timelike KVF $K$ have the same direction.

Yes, but not the same magnitude. The magnitude of the 4-velocity is a function of $R$. The magnitude of the KVF is not.

$e_0$ and the orthogonal proper acceleration unit vector can be used to define Fermi normal coordinates adapted to congruence by Fermi-Walker transport such tetrad along the congruence starting from an "initial" point.

Yes, but only for Fermi-Walker transported basis vectors, as I said. The vector $\hat{p}_2 = \partial_R$ is not Fermi-Walker transported along the worldlines of the congruence. Again, this was shown in the Insights article.

After such process is completed, we can add a possibly axis rotation to the spacelike vector field

Yes, but if you insist on using the 4-velocity as your timelike basis vector, you might no longer have a valid coordinate chart. But if you use the KVF as your timelike basis vector, you will have a valid coordinate chart no matter what rotation you do.

In the coordinate chart we get this way both 4-velocity and proper acceleration direction result to be coordinate basis vectors.

Once again, this is false. I already told you why. Go read my last post again, carefully.

 

[cianfa72]

Yes, but only for Fermi-Walker transported basis vectors, as I said. The vector $\hat{p}_2 = \partial_R$ is not Fermi-Walker transported along the worldlines of the congruence.

Maybe I wasn't clear: my point is define Fermi normal coordinates for a worldtube about a given congruence's worldline (say $a$) starting from a point P along it picking there the timelike 4-velocity and the proper acceleration direction of the worldline $a$ passing there. Starting from this point P, using Fermi-Walker transport law, one actually transports such a tetrad along the worldline $a$ defining in such way an adapted Fermi normal coordinate chart. By definition the tetrad field one gets this way is Fermi-Walker transported along the worldline $a$.

Yes, but if you insist on using the 4-velocity as your timelike basis vector, you might no longer have a valid coordinate chart. But if you use the KVF as your timelike basis vector, you will have a valid coordinate chart no matter what rotation you do.

Ah, why this holds true using KVF as timelike basis vector and does not using 4-velocity as timelike basis vector ?

 

[PeterDonis]

Maybe I wasn't clear

You're being clear about what you're saying. You're just not thinking carefully enough about what is and is not implied by what you're saying.

By definition the tetrad field one gets this way is Fermi-Walker transported along $a$.

Sure. But this tetrad field will not have one of its basis vectors always pointing radially for the Langevin congruence. That's what the congruence having nonzero vorticity means.

why this holds true using KVF as timelike basis vector and does not using 4-velocity as timelike basis vector ?

I gave you the answer in my post. Go read it again, carefully.

 

[cianfa72]

I gave you the answer in my post. Go read it again, carefully.

Sorry Peter, which post in particular are you referring to ?

Surely I'm wrong, however a time-dependent axis rotation should be always a legitimate coordinate transformation in both cases.

 

[PAllen]

Yes, but not the same magnitude. The magnitude of the 4-velocity is a function of $R$.

As far as I know, 4-velocities are defined to always have a fixed magnitude, generally either c or 1.

 

[PeterDonis]

which post in particular are you referring to ?

Several of them. Read them all carefully. You have not been doing that; instead you keep repeating the same wrong statements and missing the same things I keep saying to correct you.

a time-dependent axis rotation should be always a legitimate coordinate transformation in both cases.

It does not appear that that is true. To see why in detail one would need to look at the specific transformations involved.

 

[PeterDonis]

4-velocities are defined to always have a fixed magnitude

I was speaking loosely in the context of the discussion. To be more precise: the magnitude of the coordinate basis vectors included in the 4-velocity is not constant, so the normalization factor that is required to make the magnitude of the 4-velocity fixed is a function of $R$. That normalization factor affects the Lie bracket.

 

[PeterDonis]

This construction is often generalized by adding an arbitrary rotation $\omega(\tau)$ applied to the tetrad 'following' Fermi-Walker transport.

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.

 

[cianfa72]

This construction is often generalized by adding an arbitrary rotation $\omega(\tau)$ applied to the tetrad 'following' Fermi-Walker transport. This tends to shrink the coverage of the coordinates even more and also results that the coordinate stationary lines are not typically orthogonal to the foliation away from the origin.

Ok, that implies that to bring the Fermi-Walker transported "initial spacelike proper acceleration vector at P on the picked congruence's worldline" always in the direction of proper acceleration at each point along the worldline, one must rotate tetrad's spacelike vector fields about the 4-velocity (the latter is left invariant from such rotations).

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.

Yes, we "learned by example" that @PAllen full construction does not always result in a coordinate basis for a worldtube about the picked congruence's worldline.

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.

Sorry to be boring, 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.

Btw, 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 ?

 

[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.)