Properties of Born rigid congruence

[PeterDonis]

I would say it does

The question you are answering in your first paragraph here--do we have a well defined meaning for "same vector"--is not the question the OP is trying to answer, although it is necessary to establish the overall framework of the scenario the OP is using.

The question the OP is trying to answer is the one I posed (again) at the bottom of post #59. (More precisely, that question is the remaining missing piece of the overall picture the OP is trying to put together.)

 

[PeterDonis]

"does 'the combination of rotation and time translation that generates the Langevin congruence' answer your 'how is the vector transported' question"

Only if you can define how the Langevin congruence transports its basis vectors. What transport law does it use?

 

[cianfa72]

Only if you can define how the Langevin congruence transports its basis vectors. What transport law does it use?

As @Ibix pointed out, the vector field we're interested in along worldline $a$ is given as linear combination $\hat{P} =A \hat{p}_2 + B \hat{p}_3$ where $A,B$ are constant. Therefore we need to find a sort of derivative $D_Z$ along Langevin congruence's worldlines such that $D_Z\hat{P}=0$.

 

[PeterDonis]

As @Ibix pointed out, the vector field we're interested in along worldline $a$, is given as linear combination $\hat{P} =A \hat{p}_2 + B \hat{p}_3$ where $A,B$ are constant.

Yes, any such vector will work. One of them, though--the one with $A = -1$ and $B = 0$, has a particular useful property in terms of the Langevin worldlines. Can you see what it is? (Hint: the relevant spacelike vector for the Rindler congruence, the one that lies in the 2-plane you described earlier, has the same property.)

Therefore we need to find a sort of derivative $D_Z$ along Langevin congruence's worldlines such that $D_Z\hat{P}=0$.

Yes.

 

[Ibix]

Only if you can define how the Langevin congruence transports its basis vectors. What transport law does it use?

Parallel transport plus a rotation around the z axis?

 

[PeterDonis]

Parallel transport plus a rotation around the z axis?

Fermi-Walker transport plus a z axis rotation described by the vorticity vector is one way to describe it, yes. (Note that the z axis rotation only applies to the vectors you call $r'^a$ and $\phi'^a$.)

However, at least for the special case I mentioned in post #64, there is a simpler way of describing the transport.

 

[cianfa72]

Yes, any such vector will work. One of them, though--the one with $A = -1$ and $B = 0$, has a particular useful property in terms of the Langevin worldlines. Can you see what it is?

Its useful property is that it always points inward in the radial direction $- \partial_R$, so the transport law we are looking for is such that transports it along the Langevin worldlines.

(Hint: the relevant spacelike vector for the Rindler congruence, the one that lies in the 2-plane you described earlier, has the same property.)

Sorry, the 2-plane I described earlier (namely the 2-plane spanned from $\{ \hat{p}_2, \hat{p}_3 \}$) was for Langevin congruence not for Rindler one.

 

[cianfa72]

Of course it is. Consider: at how many points will the timelike curve $b$ intersect the spacelike $\{ \hat{p}_2, \hat{p}_3 \}$ plane?

The point here is that, for example, in Minkowski standard global inertial chart the Langevin congruence's worldline $b$ is represented by an helix and of course spacelike planes in flat spacetime by planes in that chart. Therefore in that chart (hence in spacetime) there is only one intersection event/point between them.

 

[PeterDonis]

Its useful property is that it always points inward in the radial direction $- \partial_R$

Ok, but that's still a coordinate-dependent specification. Is there some invariant that points in that direction? (Hint: the Insights article tells you the answer.)

the 2-plane I described earlier (namely the 2-plane spanned from $\{ \hat{p}_2, \hat{p}_3 \}$) was for Langevin congruence not for Rindler one.

I know that. I was pointing out a similarity between the Langevin congruence and the Rindler one, in terms of a particular invariant that picks out a spacelike direction that satisfies the property you want it to satisfy.

 

[cianfa72]

Ok, but that's still a coordinate-dependent specification. Is there some invariant that points in that direction? (Hint: the Insights article tells you the answer.)

Yes, it is the covariant derivative of the 4-velocity of Langevin congruence's worldlines (i.e. their proper acceleration), namely $$
\nabla_{\hat{p}_0} \hat{p}_0 = K \hat{p}_2$$ with $K = – \gamma^2 \omega^2 R$

I was pointing out a similarity between the Langevin congruence and the Rindler one, in terms of a particular invariant that picks out a spacelike direction that satisfies the property you want it to satisfy.

Ok, so the same as above for Rindler congruence, namely the particular invariant is the congruence's worldlines proper acceleration $$
\nabla_{\hat{e}_0} \hat{e}_0 = \frac{1}{X} \hat{e}_1$$ using the notation in the Insights article.

 

[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?