I Properties of Born rigid congruence

[cianfa72]

Think about it. There's no need to do any math. Here's a hint: what coordinate does $A$ depend on? And what is the Lie derivative of that coordinate along any worldline in the Langevin congruence?

Ok, $A$ depends only on $R$, i.e. proper acceleration $\vec P$ is $A(R)\partial_R$. In Minkowski cylindrical global coordinates we get $\hat{p}_0 = \gamma(R)\partial_T + \gamma (R)\omega \partial_{\phi}$.

To evaluate the relevant Lie bracket, I employed the following that holds in any coordinate system $$[V,U]^{\mu} = V^{\nu}\partial_{\nu}U^{\mu} - U^{\nu}\partial_{\nu}V^{\mu}$$ However for example the $T$ component of $\left [ \hat{p}_0, A(R) \partial_R \right]^T$ doesn't vanish identically $$- A(R) \partial_R \gamma (R)$$

 

[PeterDonis]

for example the $T$ component of $\left [ \hat{p}_0, A(R) \partial_R \right]^T$ doesn't vanish identically $$- A(R) \partial_R \gamma (R)$$

Hm, yes, that seems to be the case when I check the computation. So it might be that only the direction of the proper acceleration is Lie transported, not the overall proper acceleration vector.

That seems weird, though, because the magnitude of the proper acceleration is Lie transported along the worldlines: it is easily shown that the Lie derivative of the scalar function $A = - \gamma^2 \omega^2 R$ along Langevin worldlines is zero (since it's just the directional derivative along those worldlines, and since $R$ is constant along each worldline, $A$ is too). So if both the magnitude and the direction are Lie transported, why isn't the overall vector Lie transported?

Either we've missed something or my intuitions about Lie transport are not very good.

 

[cianfa72]

That seems weird, though, because the magnitude of the proper acceleration is Lie transported along the worldlines: it is easily shown that the Lie derivative of the scalar function $A = - \gamma^2 \omega^2 R$ along Langevin worldlines is zero (since it's just the directional derivative along those worldlines, and since $R$ is constant along each worldline, $A$ is too).

Indeed it reduces to the directional derivative of the function $A(R)$ along the $\hat{p}_0 = \gamma(R)\partial_T + \gamma (R)\omega \partial_{\phi}$ direction. $A(R)$ however doesn't depend on $T$ and $\phi$ coordinates hence it vanishes identically.

 

[PeterDonis]

Indeed it reduces to the directional derivative of the function $A(R)$ along the $\hat{p}_0 = \gamma(R)\partial_T + \gamma (R)\omega \partial_{\phi}$ direction. $A(R)$ however doesn't depend on $T$ and $\phi$ coordinates hence it vanishes identically.

Yes, and I had thought that the Lie derivative obeyed the product rule, so that we would have

$$
\mathscr{L}_{\hat{p}_0} \left( A \hat{p}_2 \right) = \left( \mathscr{L}_{\hat{p}_0} A \right) \hat{p}_2 + A \left( \mathscr{L}_{\hat{p}_0}\hat{p}_2 \right)
$$

So if both terms on the RHS vanish, I would have expected the LHS to vanish as well.

 

[PeterDonis]

if both terms on the RHS vanish

But on rechecking, the second term doesn't, because:

$\hat{p}_0$ is linear combination (with constant coefficients) of coordinate basis vectors

No, it isn't; $\gamma$ appears in the linear combination and $\gamma$ is not a constant, it depends on $R$. That's what we were missing. So actually $\mathscr{L}_{\hat{p}_0}\hat{p}_2$ does not vanish, and that is where the nonzero result for $\mathscr{L}_{\hat{p}_0} (A \hat{p}_2)$ comes from.

 

[cianfa72]

So actually $\mathscr{L}_{\hat{p}_0}\hat{p}_2$ does not vanish, and that is where the nonzero result for $\mathscr{L}_{\hat{p}_0} (A \hat{p}_2)$ comes from.

You're right indeed, as pointed out from you earlier, I was confused from the fact that the components of $\hat{p}_0$ in Minkowski cylindrical basis vectors are actually not constant (they depend on $R$# coordinate).

 

[PeterDonis]

I don't know if the following result is general: Lie bracket of 4-velocity and proper acceleration at the same point of a worldline vanishes.

As we have found, the result stated this way does not hold. However, I think a weaker result might: the Lie bracket of the directions of the 4-velocity and proper acceleration vanishes.

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.

 

[PeterDonis]

I think a weaker result might: the Lie bracket of the directions of the 4-velocity and proper acceleration vanishes.

For the case of a Killing congruence, in fact, I think the Lie bracket of the KVF and the proper acceleration itself (not just the coordinate basis vector in its direction) will vanish, because, although the magnitude of the proper acceleration is not constant, it cannot vary along the integral curves of the KVF, so its Lie derivative along the KVF must vanish, and hence by the product rule the Lie bracket of the KVF and the proper acceleration vector field vanishes.

The fact that this argument makes explicit use of the properties of the KVF makes me think that it will not generalize to other cases.

 

[cianfa72]

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.

Sorry @PeterDonis why the above is always true ?

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.

As above, why there is a coordinate basis vector field with that property ?

 

[PeterDonis]

why the above is always true ?

It's a generalization of the theorem, which you can find in many GR textbooks, that for a timelike KVF it is always possible to choose a coordinate chart with a time coordinate $t$ such that the KVF is $\partial_t$. We could do that for this case but it's easier to just use the cylindrical chart I used in the Insights article, in which the KVF is $\partial_T + \omega \partial_\Phi$.

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.

why there is a coordinate basis vector field with that property ?

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

 

[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.