Properties of Born rigid congruence

[Ibix]

Why not just express everything in the disc inertial rest frame?

 

[cianfa72]

This is correct. However, that still leaves an infinite number of such vectors.

Ok, now I think I grasped what you said earlier in this thread: a Langevin congruence picks out a "specific" 2D local spacelike plane at each point P along congruence's worldlines (it is the 2D plane as defined in my previous post).

Then, given a pair of congruence's worldlines say $a$ and $b$, there is only one geodesic of flat spacetime pointing in a (spacelike) direction that lies in those local planes at point P on $a$ that intersect the $b$ member. However, since Langevin congruence is not hypersurface orthogonal, those spacelike local planes at P are not orthogonal to the congruence member's tangent vector at P.

 

[Ibix]

Why not just express everything in the disc inertial rest frame?

I mean, then it's obvious what vector at $Q$ is "the same as" one at $P$ in the relevant sense. In fact, if you work in polars it's completely trivial.

Then re-express that in terms of the "natural" Fermi-Walker transported frame field of a Langevin observer if you really want to.

 

[PeterDonis]

Why not just express everything in the disc inertial rest frame?

If you mean the inertial frame in which the disc's center is at rest, that is the frame in which the analysis in the Insights article is conducted (in cylindrical coordinates).

 

[PeterDonis]

a Langevin congruence picks out a "specific" 2D local spacelike plane at each point P along congruence's worldlines

Yes, but it's still a 2D plane. That's an infinite number of possible spacelike directions.

Then, given a pair of congruence's worldlines say $a$ and $b$, there is only one geodesic of flat spacetime pointing in a (spacelike) direction that lies in those local planes at point P on $a$ that intersect the $b$ member.

But there are an infinite number of possible $b$ members for which this will work. Which one do you pick?

 

[cianfa72]

But there are an infinite number of possible $b$ members for which this will work. Which one do you pick?

My point was different: to me congruence's member $a$ and $b$ are given/fixed in "advance".

 

[Ibix]

If you mean the inertial frame in which the disc's center is at rest, that is the frame in which the analysis in the Insights article is conducted (in cylindrical coordinates).

I was aiming the comment more at @cianfa72. Surely what "the same vector at a different event on a Langevin worldline" means (in the relevant sense) is obvious in those coordinates.

 

[PeterDonis]

to me congruence's member $a$ and $b$ are given/fixed in "advance".

What does this mean? The congruence has an infinite number of worldlines in it. How are you fixing any pair of them in advance?

I think you are not thinking very carefully.

 

[PeterDonis]

Surely what "the same vector at a different event on a Langevin worldline" means (in the relevant sense) is obvious in those coordinates.

There is no unique concept of "the same vector at a different event" even in flat spacetime. The advantage flat spacetime gives you is that parallel transport is not path dependent. But parallel transport is still only one of a number of possible transport laws that can be used to define "the same vector at a different event", and not all of them give the same answers even in flat spacetime.

For what the OP is trying to do, we have already established that parallel transport is not the right transport law to use.

 

[cianfa72]

What does this mean? The congruence has an infinite number of worldlines in it. How are you fixing any pair of them in advance?

Maybe I wasn't clear. In Langevin congruence take two members say $a$ and $b$ (e.g. the worldlines of two clocks riding on a ring). From a point P on worldline $a$ take a spacelike direction such that the spacelike geodesic (in flat spacetime) in that direction intersects the member $b$. Call $L$ the spacetime lenght of such geodesic segment.

Now move on point Q along the member $a$. What is the spacelike direction at Q such that the spacelike geodesic segment in that direction from Q that intersects the member $b$ has spacetime lenght $L$ as well ?

I believe the answer is simply the parallel transport in flat spacetime of the spacelike direction picked at point P from P to Q.

 

[PeterDonis]

I believe the answer is simply the parallel transport in flat spacetime of the spacelike direction picked at point P from P to Q.

This is not correct. Work it out and see. In fact the parallel transported spacelike direction at Q will in general not intersect worldline $b$ at all.

 

[Ibix]

There is no unique concept of "the same vector at a different event" even in flat spacetime.

Sure. So I was proposing a methodology that defines "the same vector at a different event" in a way that matches what I suspect @cianfa72 intends. I think the intention is that if a Langevin observer, $a$, points at another Langevin observer, $b$, and keeps pointing at them then a vector orthogonal to $a$'s four velocity and in the worldsheet of their arm should be "the same" by this definition. I think that if you work in the obvious polar coordinate system in the disc COM's inertial rest frame and its coordinate basis, this condition is trivial to express.

This is not Fermi-Walker transport, nor is it parallel transport. I don't know if it has a name.

 

[PeterDonis]

the intention is that if a Langevin observer, $a$, points at another Langevin observer, $b$, and keeps pointing at them

That is a requirement for what the OP wants to do, yes. The question is, if we have a vector at event P on $a$'s worldline that does this, how do we transport that vector along $a$'s worldline to another event Q so it keeps doing this? So far we have shown that neither parallel transport nor Fermi-Walker transport meet this requirement.

a vector orthogonal to $a$'s four velocity

This is a requirement, yes.

and in the worldsheet of their arm should be "the same" by this definition.

I'm not sure what this means.

I think that if you work in the obvious polar coordinate system in the disc COM's inertial rest frame and its coordinate basis, this condition is trivial to express.

Then can you express it?

 

[Ibix]

Then can you express it?

An observer sits in a chair that is bolted to a disc of radius $R$ that rotates at $\omega$. He is thus constrained to face the rotation axis. His four velocity expressed in non-rotating cylindrical polars centered on the rotation axis is $$\begin{eqnarray*}v^a&=&(v^t,v^r,v^\phi,v^z)\\&=&(\gamma_\omega,0,\gamma_\omega\omega ,0)\end{eqnarray*}$$where $\gamma_\omega$ is the Lorentz gamma factor associated with speed $\omega R$ and the metric is $\mathrm{diag}(1,-1,-r^2,-1)$. At any event he can define three unit spacelike vectors, #m$$\begin{eqnarray*}r'^a&=&(0,1,0,0)\\
\phi'^a&=&(\gamma_\omega \omega R,0,\gamma_\omega/R,0)\\
z'^a&=&(0,0,0,1)\end{eqnarray*}$$using the same coordinate system. These are mutually orthogonal and also orthogonal to his four velocity, so these four four vectors form a tetrad at the event.

If there is a mark painted somewhere on the disc the observer can point to it (if very, very strong), or even point in the direction where he sees it to be (that's a different direction, but I don't care here). If he keeps pointing at it, I think this action matches the intuitive notion of "pointing in the same direction" in this case. The observer could always express the direction he's pointing as $Ar'^a+B\phi'^a$, and we can define a one-parameter family of geodesics with these tangent vectors extending from the observer at all events he passes through. These geodesics will all pass through the same point (i.e. Langevin worldline) on the rim of the disc. The segments of geodesic from observer to point will all have the same length, and any pair of geodesics separated at the observer by Langevin proper time $\Delta \tau$ will also be separated by Langevin proper time $\Delta\tau$ at their other ends.

Note that the direction is not only expressible as a sum of those two tetrad vectors, the coefficients $A$ and $B$ are constants. Thus the tetrad field $v^a, r'^a,\phi'^a,z'^a$ is what we want to transport along the Langevin worldlines. This is not parallel transport (which would keep the tetrad fixed in an inertial frame) nor Fermi-Walker transport (which would precess with respect to these). As I said, I don't know if it has a name.

A couple of observations. First, many tetrads are possible. But the observer's four acceleration is parallel to $r'^a$ and he will feel a torque about $z'^a$ since he is not Thomas precessing, so this is a physically motivated choice. Second, given one tetrad, the rest of the field can be generated by the rotational and time-translational symmetries of this scenario.

Finally, I've been careful to not call my observer a Langevin observer. He is following a Langevin worldline but he is not Fermi-Walker transporting his tetrad, and I am not sure if Langevin observers are formally points or tetrads.

 

[cianfa72]

An observer sits in a chair that is bolted to a disc of radius $R$ that rotates at $\omega$. He is thus constrained to face the rotation axis. His four velocity expressed in non-rotating cylindrical polars centered on the rotation axis is $$\begin{eqnarray*}v^a&=&(v^t,v^r,v^\phi,v^z)\\&=&(\gamma_\omega,0,\gamma_\omega\omega ,0)\end{eqnarray*}$$where $\gamma_\omega$ is the Lorentz gamma factor associated with speed $\omega R$ and the metric is $\mathrm{diag}(1,-1,-r^2,-1)$. At any event he can define three unit spacelike vectors, $$\begin{eqnarray*}r'^a&=&(0,1,0,0)\\
\phi'^a&=&(\gamma_\omega \omega R,0,\gamma_\omega/R,0)\\
z'^a&=&(0,0,0,1)\end{eqnarray*}$$using the same coordinate system. These are mutually orthogonal and also orthogonal to his four velocity, so these four four vectors form a tetrad at the event.

From the Insight $\gamma_\omega$ should be $1 / \sqrt{1 – \omega^2 R^2}$ even though the signature chosen there is $\mathrm{diag}(-1,1,r^2,1)$.

If there is a mark painted somewhere on the disc the observer can point to it (if very, very strong), or even point in the direction where he sees it to be (that's a different direction, but I don't care here). If he keeps pointing at it, I think this action matches the intuitive notion of "pointing in the same direction" in this case. The observer could always express the direction he's pointing as $Ar'^a+B\phi'^a$, and we can define a one-parameter family of geodesics with these tangent vectors extending from the observer at all events he passes through. These geodesics will all pass through the same point (i.e. Langevin worldline) on the rim of the disc. The segments of geodesic from observer to point will all have the same length, and any pair of geodesics separated at the observer by Langevin proper time $\Delta \tau$ will also be separated by Langevin proper time $\Delta\tau$ at their other ends.

By "point" on the rim of the disc above (bold is mine) you mean the associated Langevin timelike congruence's worldline/member, right ?

Note that the direction is not only expressible as a sum of those two tetrad vectors, the coefficients $A$ and $B$ are constants. Thus the tetrad field $v^a, r'^a,\phi'^a,z'^a$ is what we want to transport along the Langevin worldlines. This is not parallel transport (which would keep the tetrad fixed in an inertial frame) nor Fermi-Walker transport (which would precess with respect to these).

Finally, I've been careful to not call my observer a Langevin observer. He is following a Langevin worldline but he is not Fermi-Walker transporting his tetrad, and I am not sure if Langevin observers are formally points or tetrads.

We said that the tetrad field that includes the Langevin worldline timelike 4-velocity and the spacelike vectors $r'^a$ and $\phi'^a$ at a point P is not Fermi-Walker transported since $D_Fr'^a \neq 0$ and $D_F\phi'^a \neq 0$. Why do you think that a Langevin congruence's worldline must F-W its tetrad ?

 

[Ibix]

By "point" on the rim of the disc above (bold is mine) you mean the associated Langevin timelike congruence's worldline/member, right ?

Yes, exactly as I said in the exact bit you highlighted.

Why do you think that a Langevin congruence's worldline must F-W its tetrad ?

I don't think it must. I don't know if observer, in this context, is typically defined to include a "natural" tetrad. It might be because there is interesting behaviour of the tetrad. But it might not. I don't know. That's why the sentence you quoted says "I am not sure".

 

[cianfa72]

I don't think it must. I don't know if observer, in this context, is typically defined to include a "natural" tetrad. It might be because there is interesting behaviour of the tetrad. But it might not. I don't know. That's why the sentence you quoted says "I am not sure".

Ah ok. To me an observer was just a synonym for "timelike worldline" in the congruence (Langevin congruence in this specific case) without reference to any tetrad/frame it carries with itself.

 

[PeterDonis]

He is thus constrained to face the rotation axis.

In other words, he is facing in the direction that, in the notation of the Insights article, would be $- \hat{p}_2$. In your notation, it would be $- r'^a$. The tetrad you define is the same one that is defined in the Insights article. And as the Insights article shows, this tetrad is indeed not Fermi-Walker transported along the worldlines of the congruence.

 

[PeterDonis]

I don't know if observer, in this context, is typically defined to include a "natural" tetrad.

To me an observer was just a synonym for "timelike worldline" in the congruence (Langevin congruence in this specific case) without reference to any tetrad/frame it carries with itself.

Both meanings can be found in the literature. But regardless of how you define "observer", we have a tetrad (or more precisely a tetrad field--the Insights article calls it a frame field, which is the same thing). Does this help answer the question the OP is trying to address?

 

[cianfa72]

The tetrad you define is the same one that is defined in the Insights article. And as the Insights article shows, this tetrad is indeed not Fermi-Walker transported along the worldlines of the congruence.

Ok, by definition of F-W derivative, $D_F\hat{p}_0 = 0$ for the 4-velocity $\hat{p}_0$ of any worldline in any timelike congruence. Then the Insights article shows that $D_F\hat{p}_2 \neq 0, D_F\hat{p}_3 \neq 0$ hence that "overall" tetrad field defined there is not F-W transported along the Langevin congruence's worldlines.

 

[cianfa72]

Does this help answer the question the OP is trying to address?

As pointed out earlier in this thread, fixed two Langevin congruence's members $a$ and $b$ (e.g. the timelike worldlines of ring riding clocks), there are spacelike directions belonging to the 2D local planes spanned from frame basis spacelike vectors $\{ \hat{p}_2, \hat{p}_3 \}$ at each point P along the worldline $a$ such that the spacelike geodesics drawn in each of those directions intersect for sure the member $b$.

 

[PeterDonis]

there are spacelike directions belonging to the 2D local planes spanned from frame basis spacelike vectors $\{ \hat{p}_2, \hat{p}_3 \}$ at each point P along the worldline $a$ such that the spacelike geodesics drawn in those directions intersect for sure the member $b$.

Yes, there is such a spacelike direction for a given worldline $b$ at each point P along worldline $a$. How is the vector that defines this spacelike direction transported along worldline $a$?

 

[cianfa72]

Yes, there is such a spacelike direction for a given worldline $b$ at each point P along worldline $a$. How is the vector that defines this spacelike direction transported along worldline $a$?

First question: at each point P along worldline $a$, is the spacelike geodesic pointing along a direction in $\{ \hat{p}_2, \hat{p}_3 \}$ plane with the property that intersects the worldline $b$ unique ? I believe yes, it is.

Second point: such unique vector at each point P along worldline $a$ is not F-W transported or Parallel transported (according flat spacetime) along the worldline itself.

 

[PeterDonis]

at each point P along worldline $a$, is the spacelike geodesic pointing along a direction in $\{ \hat{p}_2, \hat{p}_3 \}$ plane with the property that intersects the worldline $b$ unique ? I believe yes, it is.

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

such unique vector at each point P along worldline $a$ is not F-W transported or Parallel transported (according flat spacetime) along the worldline itself.

Yes, we've established how it's not transported. I am asking you how it is transported.

 

[Ibix]

we have a tetrad (or more precisely a tetrad field--the Insights article calls it a frame field, which is the same thing). Does this help answer the question the OP is trying to address?

I would say it does: we have a rigorous definition of "the same vector" at different events on the worldline, and given that definition the answer to the question in the OP is yes. From the symmetry it's also obvious that using other definitions of "the same vector" the answer will be "no" in general.

So IMO the remaining questions are "does 'the combination of rotation and time translation that generates the Langevin congruence' answer your 'how is the vector transported' question" and "does the implied notion of 'the same vector' match @cianfa72's notion of the phrase".

 

[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.