I Properties of Born rigid congruence

[cianfa72]

Consider a Born rigid timelike congruence. Starting from any point P on each congruence's worldline take a spacelike direction at P and build the spacelike geodesic from P pointing in that direction. It will eventually intersect other congruence's worldlines.

Does the Born rigid condition imply that the spacetime lenght of such spacelike geodesic "segments" won't change regardless the point P chosen along a given congruence's member ?

 

[PeterDonis]

Consider a Born rigid timelike congruence. Starting from any point P on each congruence's worldline take a spacelike direction at P and build the spacelike geodesic from P pointing in that direction. It will eventually intersect other congruence's worldlines.

Does the Born rigid condition imply that the spacetime lenght of such spacelike geodesic "segments" won't change regardless the point P chosen along a given congruence's member ?

Not as you state it, because you haven't specified how the spacelike direction is to be taken.

I would suggest looking at some specific examples, such as the Rindler congruence and the Langevin congruence in flat spacetime, and seeing how you might try to do what you describe. For extra credit, you could then try some congruences in curved spacetimes.

 

[cianfa72]

Not as you state it, because you haven't specified how the spacelike direction is to be taken.

Yes, the spacelike direction I was talking about belongs to the orthogonal complement at each point P along the timelike congruence.

 

[PeterDonis]

the spacelike direction I was talking about belongs to the orthogonal complement at each point along the timelike congruence.

Have you tried the specific examples I gave using this specification?

 

[cianfa72]

Have you tried the specific examples I gave using this specification?

For Rindler congruence in 2D flat spacetime, the spacelike direction orthogonal to the congruence's member tangent vector at each point P is unique. If we draw its integral geodesic line in Minkowski standard inertial coordinate chart, the lenght of such spacelike geodesic "segment" between two given congruence's worldlines doesn't change regardless the point P picked.

I believe the same argument applies to Langevin congruence as well.

 

[PeterDonis]

For Rindler congruence in 2D flat spacetime, the spacelike direction orthogonal to the congruence's member tangent vector at each point P is unique. If we draw its integral geodesic line in Minkowski standard inertial coordinate chart, the lenght of such spacelike geodesic "segment" between two given congruence's worldlines doesn't change regardless the point P picked.

Yes.

I believe the same argument applies to Langevin congruence as well.

The Langevin congruence, unlike the Rindler congruence, is not hypersurface orthogonal. So you need to be more careful in that case.

 

[DrGreg]

in 2D flat spacetime, the spacelike direction orthogonal to the congruence's member tangent vector at each point P is unique

But in 4D spacetime, there are an infinite number of such directions to choose from. How do you relate the choices at two different Ps?

 

[PeterDonis]

in 2D flat spacetime

Note that you can't describe the Langevin congruence in 2D flat spacetime. So @DrGreg's question is already relevant even for that (still relatively easy) case.

 

[cianfa72]

The Langevin congruence, unlike the Rindler congruence, is not hypersurface orthogonal. So you need to be more careful in that case.

Pick a point P along a Langevin congruence's worldline in 4D flat spacetime and consider the orthogonal complement to the worldline's tangent vector at that point. Pick a spacelike direction in it and build the integral geodesic emanating from P. It is a straight line in Minkowski standard inertial coordinates that will intersect other congruence's members. Calculate the spacetime lenght of a such spacelike geodesic segment starting from P between two given congruence's worldlines/members.

Now pick a different point Q on the chosen congruence's worldline and repeat the above construction starting from it.

How the fact that Langevin congruence is not hypersurface orthogonal related to the result that the above calculated lenghts are not the same (note that Langevin congruence is Born rigid) ?

 

[PeterDonis]

repeat the above construction

You can't because there are an infinite number of possible results of "pick a spacelike direction in the orthogonal complement to the worldline's tangent vector", and you have given no way to pick a unique one.

How the fact that Langevin congruence is not hypersurface orthogonal related to the result that the above calculated lenght are not the same

That's not what I said. I said you need to first come up with a unique specification for how to pick the spacelike geodesics before we can even evaluate the question of whether the calculated lengths change or not.

 

[cianfa72]

That's not what I said. I said you need to first come up with a unique specification for how to pick the spacelike geodesics before we can even evaluate the question of whether the calculated lengths change or not.

Ah ok, we can do that by parallel transporting the spacelike direction picked at point P from P to Q along the given congruence's worldline.

 

[PeterDonis]

we can do that by parallel transporting the spacelike direction picked at point P from P to Q along the given congruence's worldline.

That would be one way, but not the only way. Will doing it this way make the spacelike length to a given other member of the congruence constant, or not? What do you think?

 

[cianfa72]

Will doing it this way make the spacelike length to a given other member of the congruence constant, or not? What do you think?

No, I don't think so. Which other way could there be to uniquely specify how to pick a spacelike direction in the 3D orthogonal complement at each point P along a worldline?

 

[PeterDonis]

I don't think so.

You're correct.

Which other way could there be to uniquely specify how to pick a spacelike direction in the 3D orthogonal complement at each point P along a worldline?

What other transport laws are there? What other properties of the worldline could you use?

(You can get one hint from considering what property of the Rindler congruence picks out a 2-D surface in spacetime. Note that even in 4-D spacetime there is still a Rindler congruence with a property that picks out 2-D surfaces.)

 

[cianfa72]

What other transport laws are there? What other properties of the worldline could you use?

Other transport laws I'm aware of are Fermi-Walker transport and Lie dragging/transport.

The Rindler congruence's members are integral orbits of a KVF and have zero expansion scalar. Perhaps these are useful properties to uniquely pick out a spacelike direction at each point along a congruence's member/worldline.

(You can get one hint from considering what property of the Rindler congruence picks out a 2-D surface in spacetime. Note that even in 4-D spacetime there is still a Rindler congruence with a property that picks out 2-D surfaces.)

Sorry, in 2D flat spacetime how can the Rindler congruence picks out a 2D surface when the spacetime itself has dimension 2 ?

 

[PeterDonis]

Other transport laws I'm aware of are Fermi-Walker transport and Lie dragging/transport.

Have you tried those to see what they say?

The Rindler congruence's members are integral orbits of a KVF and have zero expansion scalar.

So are the members of the Langevin congruence.

Perhaps these are useful properties to uniquely pick out a spacelike direction at each point along a congruence's member/worldline.

Since these properties are shared by both the Rindler and Langevin congruences, as above, it would not seem so.

in 2D flat spacetime how can the Rindler congruence picks out a 2D surface when the spacetime itself has dimension 2 ?

I didn't say in 2D flat spacetime, I said in 4D flat spacetime.

 

[cianfa72]

Have you tried those to see what they say?

What about Fermi-Walker transport? As far as I know, if one picks a spacelike direction $e_1$ at a point P along a worldline (a spacelike direction $e_1$ in the orthogonal complement w.r.t. the timelike tangent vector $e_0$ at that point P) then, using Fermi-Walker transport along the worldline, one gets a "non-rotated" version of $e_1$ at point Q.

 

[PeterDonis]

What about Fermi-Walker transport?

Again, have you tried it? Do the math and see.

using Fermi-Walker transport along the worldline, one gets a "non-rotated" version of $e_1$ at point Q.

For a particular sense of "non-rotated", yes. But is that sense of "non-rotated" the right one to make the spacelike distance you have defined constant for the Langevin congruence?

 

[cianfa72]

For a particular sense of "non-rotated", yes. But is that sense of "non-rotated" the right one to make the spacelike distance you have defined constant for the Langevin congruence?

Sorry, I really don't know. Perhaps it isn't the right one since the Langevin congruence is not "irrotational" -- i.e. it isn't hypersurface orthogonal.

 

[PeterDonis]

I really don't know.

That's why I suggested doing the math to see.

the Langevin congruence is not "irrotational" -- i.e. it isn't hypersurface orthogonal.

That does make a difference, yes.

 

[cianfa72]

That's why I suggested doing the math to see.

Langevin congruence's worldlines in flat spacetime are represented by helices in Minkowski standard inertial coordinates.

Now the point is: at a point P along a given congruence's worldline/member (say $a$) pick a spacelike direction orthogonal to the (timelike) tangent vector at P. The spacelike geodesic built in that direction will intersect another member of the Langevin congruence (say $b$). Now take a different point Q along the same member $a$. In the orthogonal complement at Q I think there will be a unique (spacelike) direction such that the spacelike geodesic built in that direction will intersect the member $b$. Therefore the goal is to single out such a spacelike direction at Q starting from the spacelike direction picked at point P.

As you pointed out, Langevin congruence is not hypersurface orthogonal, therefore neighboring congruence's worldlines of member $a$ "twist/rotate" w.r.t. it.

If the above is correct then, from an intuitive point of view, Fermi-Walker transport might not be the right answer to get that unique spacelike direction at Q.

Edit: I know the definition of Fermi-Walker transport, namely $$ \frac {D_F X} {ds} = 0$$ where $X=e_1(\tau)$, however I'm in trouble to work out the math for this specific case.

 

[PeterDonis]

from an intuitive point of view, Fermi-Walker transport might not be the right answer to get that unique spacelike direction at Q.

Your intuition is correct; it isn't. But you should still take the time to learn how to work out the math. You might find this Insights article helpful:

https://www.physicsforums.com/insights/how-to-study-fermi-walker-transport-in-minkowski-spacetime/

 

[cianfa72]

Your intuition is correct; it isn't. But you should still take the time to learn how to work out the math. You might find this Insights article helpful:

https://www.physicsforums.com/insights/how-to-study-fermi-walker-transport-in-minkowski-spacetime/

Thanks, I read the insight and I have some questions on it. Is it fine to discuss it here?

For example how the vorticity vector $$
\vec{\Omega} = \frac{\gamma^2 v}{R} \hat{p}_1 = \gamma^2 \omega \hat{p}_1$$ is related to the frame base vectors spinning about Z axis relative to gyro-stabilized vectors.

 

[PeterDonis]

Is it fine to discuss it here?

It is now that I've moved this discussion to a new thread.

how the vorticity vector $$
\vec{\Omega} = \frac{\gamma^2 v}{R} \hat{p}_1 = \gamma^2 \omega \hat{p}_1$$ is related to the frame base vectors spinning about Z axis relative to gyro-stabilized vectors.

Have you actually read the Insights article? It addresses this exact point towards the end.

 

[cianfa72]

Have you actually read the Insights article? It addresses this exact point towards the end.

Yes, I read it. There you claim that
$$D_F \hat{p}_2 = \frac{\gamma^2 v}{R} \hat{p}_3$$ and $$D_F \hat{p}_3 = – \frac{\gamma^2 v}{R} \hat{p}_2$$ can be written using vorticity vector $\vec {\Omega}$ $$
\vec{\Omega} = \frac{\gamma^2 v}{R} \hat{p}_1 = \gamma^2 \omega \hat{p}_1$$ Is there a typo ? The latter includes only the vector $\hat{p}_1$.

In other words, we have a combination of the hyperbolic rotation in spacetime of $\hat{p}_0$ and $\hat{p}_2$ due to Fermi-Walker transport, and the extra rotation of $\hat{p}_2$ and $\hat{p}_3$ due to the nonzero vorticity.

So, the terms that include the "vorticity" scalar $\Omega$ represent the extra rotation due non-zero vorticity. So far so good.

Coming back to my original question how can I use the above to uniquely define a spacelike vector with the properties as in post #21 ?

 

[PeterDonis]

Is there a typo ?

No. $\hat{p}_1$ is the direction of the rotation axis, or, to put it another way that might help, it is the spatial direction that is orthogonal to the plane of rotation (which in this case is the $\hat{p}_2$, $\hat{p}_3$ plane). When you describe rotation using a vector, that is what the vector means.

 

[PeterDonis]

how can I use the above to uniquely define a spacelike vector with the properties as in post #21 ?

I didn't say you could. I just said that the Insights article might help you to work out the math for Fermi-Walker transport, which would in turn help you to see why your intuition is correct that Fermi-Walker transport is not the right transport law to define the spacelike vector you are trying to define.

 

[cianfa72]

No. $\hat{p}_1$ is the direction of the rotation axis, or, to put it another way that might help, it is the spatial direction that is orthogonal to the plane of rotation (which in this case is the $\hat{p}_2$, $\hat{p}_3$ plane). When you describe rotation using a vector, that is what the vector means.

Sorry, coming back to your Insight, I'm confused about the following: what can be written as "vorticity vector" $\vec{\Omega}$ ? $$
\vec{\Omega} = \frac{\gamma^2 v}{R} \hat{p}_1 = \gamma^2 \omega \hat{p}_1$$

 

[PeterDonis]

what can be written as "vorticity vector" $\vec{\Omega}$ ? $$
\vec{\Omega} = \frac{\gamma^2 v}{R} \hat{p}_1 = \gamma^2 \omega \hat{p}_1$$

I'm not sure what you're asking. The vorticity vector is right there in what you posted.

 

[cianfa72]

In the Insight, where the vorticity vector $\vec {\Omega}$ is actually used ?

In particular what is the meaning of the claim

This can be written as a “vorticity vector”: $$\vec{\Omega} = \frac{\gamma^2 v}{R} \hat{p}_1 = \gamma^2 \omega \hat{p}_1$$ which captures the failure of the basis vectors to be Fermi-Walker transported; in other words, relative to gyro-stabilized vectors, these basis vectors (I believe $\hat{p}_2$ and $\hat{p}_3$) are spinning about the Z axis

 

[PeterDonis]

where the vorticity vector $\vec {\Omega}$ is actually used ?

I'm not sure what you mean by "used". The point of the article is, among other things, to show how the vorticity vector is derived.

In particular what is the meaning of the claim

What is unclear about it? You do understand that "Fermi-Walker transported vectors" are the same thing as "gyro-stabilized vectors", right?

 

[cianfa72]

What is unclear about it? You do understand that "Fermi-Walker transported vectors" are the same thing as "gyro-stabilized vectors", right?

Yes, I do. Since $D_F \hat{p}_2 \neq 0$ and $D_F \hat{p}_3 \neq 0$ that means $\hat{p}_2$ and $\hat{p}_3$ vector in the frame field are not Fermi-Walker (FW) transported along the congruence.

I don't understand why you claim that, relative to gyro-stabilized vectors, these two basis vectors are spinning about Z axis.

 

[PeterDonis]

Since $D_F \hat{p}_2 \neq 0$ and $D_F \hat{p}_3 \neq 0$ that means $\hat{p}_2$ and $\hat{p}_3$ vector in the frame field are not Fermi-Walker (FW) transported along the congruence.

That's correct.

I don't understand why you claim that, relative to gyro-stabilized vectors, these two basis vectors are spinning about Z axis.

The vorticity vector points along the Z axis; that is the axis about which the $\hat{p}_2$ and $\hat{p}_3$ vectors are spinning relative to Fermi-Walker transported, i.e., gyro-stabilized, vectors.

 

[cianfa72]

Coming back to your post #27, the FW transport of spacelike basis vectors $\hat{p}_2$ and $\hat{p}_3$ along a Langevin congruence's member (say $a$) from point P to point Q doesn't give those frame field's basis vectors at Q.

I believe the right spacelike vector I'm interested in actually lies in the 2D plane spanned by $\hat{p}_2$ and $\hat{p}_3$ of the frame field at any point along the congruence's member $a$.

 

[PeterDonis]

the FW transport of spacelike basis vectors $\hat{p}_2$ and $\hat{p}_3$ along a Langevin congruence's member (say $a$) from point P to point Q doesn't give those frame field's basis vectors at Q.

Yes. You are restating what you said in the first paragraph of your post #32.

I believe the right spacelike vector I'm interested in actually lies in the 2D plane spanned by $\hat{p}_2$ and $\hat{p}_3$ of the frame field at any point along the congruence's member $a$.

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

 

[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 ?