Properties of Born rigid congruence

[PeterDonis]

the covariant derivative of the 4-velocity

In other words, the proper acceleration. Yes.

the same as above for Rindler congruence, namely the particular invariant is the congruence's worldlines proper acceleration

Yes.

 

[cianfa72]

Ok, last point of the overall picture is: we have finally defined how to pick the relevant spacelike direction at each point P along the Langevin/Rindler congruence's worldline $a$ that intersects the congruence's given member $b$.

Now the question is: why all the spacelike geodesic segments emanating from point P along $a$ in that relevant direction at P have the same spacetime lenght in the underlying flat spacetime ?

 

[PeterDonis]

Now the question is: why all the spacelike geodesic segments emanating from point P along $a$ in that relevant direction at P have the same spacetime lenght in the underlying flat spacetime ?

You need to add the condition that the length of the geodesic segments is from worldline $a$ to a specific other worldline $b$ in the congruence; this would be a worldline that is radially inward from worldline $a$.

Given that, the answer is simple: because that's how the spacetime geometry works. Once you've proven something geometrically, it's pointless to ask why it's true any further. It's true because the geometry is what it is. There is no further answer.

 

[cianfa72]

You need to add the condition that the length of the geodesic segments is from worldline $a$ to a specific other worldline $b$ in the congruence;

Yes, between two specific assigned congruence's worldlines $a$ and $b$.

this would be a worldline that is radially inward from worldline $a$.

Sorry, why with the further condition above the worldline $b$ must be radially inward w.r.t. the worldline $a$ ?

 

[PeterDonis]

why with the further condition above the worldline $b$ must be radially inward w.r.t. the worldline $a$ ?

Because that's the direction in which the proper acceleration of worldline $a$ points. You have to have some way of picking a unique direction, and that's the one we've found.

 

[cianfa72]

Because that's the direction in which the proper acceleration of worldline $a$ points. You have to have some way of picking a unique direction, and that's the one we've found.

Yes, of course. However given/assigned two Langevin congruence's worldlines $a$ and $b$ there exist spacelike directions in $\{ \hat{p}_2, \hat{p}_3 \}$ planes at points P along $a$ such that all the spacelike geodesic segments emanating from there that intersect $b$ all have the same spacetime lenght.

However I agree with you that for those spacelike directions there isn't a definite way to pick them (as indeed in the case of proper acceleration direction).

 

[PeterDonis]

given/assigned two Langevin congruence's worldlines $a$ and $b$ there exist spacelike directions in $\{ \hat{p}_2, \hat{p}_3 \}$ planes at points P along $a$ such that all the spacelike geodesic segments emanating from there that intersect $b$ all have the same spacetime length.

Yes.

I agree with you that for those spacelike directions there isn't a definite way to pick them (as indeed in the case of proper acceleration direction).

There is a "definite" way to pick them: just find the direction that points from $a$ to $b$. (In the notation @Ibix used in an earlier post, find the appropriate coefficients $A$ and $B$ that multiply $\hat{p}_2$ and $\hat{p}_3$.) But only for the specific case of $b$ being radially inward from $a$ (i.e., $A = - 1$, $B = 0$ in the notation @Ibix used) will that direction have a simple description in terms of an invariant of worldline $a$.

 

[cianfa72]

There is a "definite" way to pick them: just find the direction that points from $a$ to $b$. (In the notation @Ibix used in an earlier post, find the appropriate coefficients $A$ and $B$ that multiply $\hat{p}_2$ and $\hat{p}_3$.)

Yes.

But only for the specific case of $b$ being radially inward from $a$ (i.e., $A = - 1$, $B = 0$ in the notation @Ibix used) will that direction have a simple description in terms of an invariant of worldline $a$.

Yes, definitely.

 

[cianfa72]

However given/assigned two Langevin congruence's worldlines $a$ and $b$ there exist spacelike directions in $\{ \hat{p}_2, \hat{p}_3 \}$ planes at points P along $a$ such that all the spacelike geodesic segments emanating from there that intersect $b$ all have the same spacetime lenght.

I believe one can check the above statement in Minkowski standard global inertial chart. The Langevin congruence's worldlines $a$ and $b$ are represented by helices and the flat spacetime (relevant) spacelike geodesics by straight lines in $\{ \hat{p}_2, \hat{p}_3 \}$ planes at points P along the (representative) of worldline $a$.

 

[PeterDonis]

I believe one can check the above statement in Minkowski standard global inertial chart.

Since for each individual case the statement is invariant, it will of course hold true in any chart.

Note, however, that if you use the global inertial chart in which the center worldline of the congruence (the one at $R = 0$) is at rest, the spacelike ${\hat{p}_2, \hat{p}_3}$ plane at any point P on any other worldline of the congruence will not lie in a surface of constant coordinate time in the chart. The radially inward spacelike geodesic that points in the direction of proper acceleration, however, will lie in a surface of constant coordinate time in the chart.

 

[PeterDonis]

There is a "definite" way to pick them: just find the direction that points from $a$ to $b$. (In the notation @Ibix used in an earlier post, find the appropriate coefficients $A$ and $B$ that multiply $\hat{p}_2$ and $\hat{p}_3$.)

Note that, even though such a direction only has a simple description in terms of invariants of worldline $a$ for one particular choice (radially inward), the transport law that preserves such a direction along worldline $a$ is the same for all of them. Can you see what transport law that is? We've eliminated parallel transport and Fermi-Walker transport. But there's one other transport law we've discussed in other threads.

 

[cianfa72]

Note, however, that if you use the global inertial chart in which the center worldline of the congruence (the one at $R = 0$) is at rest, the spacelike ${\hat{p}_2, \hat{p}_3}$ plane at any point P on any other worldline of the congruence will not lie in a surface of constant coordinate time in the chart.

Here the point is that $\partial_T$ ($T$ is the coordinate time of the Minkowski global inertial chart in which the congruence's center worldline is at rest) is orthogonal to $\hat{p}_1=\partial_Z$, hence if local spacelike planes $\{ \hat{p}_2, \hat{p}_3 \}$ were all on spacelike hypersurfaces of constant coordinate time $T$, then there would be integral submanifolds orthogonal to the Langevin congruence (note that vector fields $\hat{p}_1, \hat{p}_2, \hat{p}_3$ are orthogonal each other at each point, hence linearly independent). However we know Langevin congruence is not hypersurface orthogonal so the above is ruled out.

Note that, even though such a direction only has a simple description in terms of invariants of worldline $a$ for one particular choice (radially inward), the transport law that preserves such a direction along worldline $a$ is the same for all of them. Can you see what transport law that is? We've eliminated parallel transport and Fermi-Walker transport.

The only transport law left is Lie dragging/transport

 

[cianfa72]

Note, however, that if you use the global inertial chart in which the center worldline of the congruence (the one at $R = 0$) is at rest, the spacelike ${\hat{p}_2, \hat{p}_3}$ plane at any point P on any other worldline of the congruence will not lie in a surface of constant coordinate time in the chart. The radially inward spacelike geodesic that points in the direction of proper acceleration, however, will lie in a surface of constant coordinate time in the chart.

Ok, the above holds true only for any radially inward spacelike geodesic that points in the direction of proper acceleration at each point P along Langevin congruence's members. Of course, for a given congruence's member $a$, each of such inward spacelike geodesic at different point along it, will lie in different spacelike hypersurfaces of different coordinate time $T$.

 

[PeterDonis]

we know Langevin congruence is not hypersurface orthogonal so the above is ruled out.

Yes.

The only transport law left is Lie dragging/transport

Yes, indeed! So can you, for example, show that each worldline of the Langevin congruence Lie transports its proper acceleration vector?

for a given congruence's member $a$, each of such inward spacelike geodesic at different point along it, will lie in different spacelike hypersurfaces of different coordinate time $T$.

Yes.

 

[cianfa72]

Yes, indeed! So can you, for example, show that each worldline of the Langevin congruence Lie transports its proper acceleration vector?

Yes, the relevant transport condition is $$\mathcal L_{\hat{p}_0} \left (\nabla_{\hat{p}_0}{ \hat{p}_0}\right ) = - K \mathcal L_{\hat{p}_0} \hat{p}_2 = - K \left [ \hat{p}_0,\hat{p}_2 \right ] = 0$$ since $\hat{p}_0$ is linear combination (with constant coefficients) of coordinate basis vectors and Lie bracket is linear (Lie bracket of any coordinate basis vector field vanishes).

 

[PeterDonis]

Yes, the relevant transport condition is $$\mathcal L_{\hat{p}_0} \left (\nabla_{\hat{p}_0}{ \hat{p}_0}\right ) = - K \mathcal L_{\hat{p}_0} \hat{p}_2 = - K \left [ \hat{p}_0,\hat{p}_2 \right ] = 0$$ since $\hat{p}_0$ is linear combination (with constant coefficients) of coordinate basis vectors and Lie bracket is linear (Lie bracket of any coordinate basis vector field vanishes).

For Lie transport of $\hat{p}_2$, yes, this works fine. But what about Lie transport of the vector $A \hat{p}_2$, where $A$ is the magnitude of the proper acceleration, as given in the Insights article? $A$ depends on $R$, so $A \hat{p}_2$ is not a coordinate basis vector times a constant and your argument as given here does not work. (You called it $K$ in your argument quoted above, but you failed to note that it is not a constant.) But you can still compute the Lie bracket directly. Or, more easily, you could find a similar argument that does work for the particular vector $A \hat{p}_2$.

 

[cianfa72]

For Lie transport of $\hat{p}_2$, yes, this works fine. But what about Lie transport of the vector $A \hat{p}_2$, where $A$ is the magnitude of the proper acceleration, as given in the Insights article? $A$ depends on $R$, so $A \hat{p}_2$ is not a coordinate basis vector times a constant and your argument as given here does not work. (You called it $K$ in your argument quoted above, but you failed to note that it is not a constant.)

Ah yes, you are right. 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.

 

[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.

I don't think such a result would always apply. However, that doesn't mean it doesn't apply to this particular case.

 

[cianfa72]

I don't think such a result would always apply. However, that doesn't mean it doesn't apply to this particular case.

Ah ok, without doing the explicit math, which is the argument that works for the particular vector field $A\hat{p}_2$ ?

 

[PeterDonis]

Ah ok, without doing the explicit math, which is the argument that works for the particular vector field $A\hat{p}_2$ ?

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?

 

[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:

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

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.

 

[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