# Thin rotating disc under constant acceleration.

1. Aug 25, 2013

### yuiop

In a different thread the Herglotz Noether theorem was brought up and it was mentioned that this theory implies it is impossible for a cylinder rotating about its vertical axis to remain Born rigid in a gravitational field even at constant altitude. This is an extension of the claim that a Born rigid rotating disc cannot have linear acceleration and remain Born rigid. Now I accept and understand that:

1) A non rotating disc cannot be spun up and remain born rigid during the angular acceleration phase.
2) A vertical rotating cylinder that descends in a gravitational field cannot remain Born rigid.

What I do not understand is why a vertical cylinder rotating around its vertical axis cannot remain Born rigid if its rotation is constant and it remains at constant altitude in a gravitational field.

To try and break the problem down to its simplest elements, I would like to initially discuss/analyse the dynamics of an infinitesimally thin disc in the y/z plane that is rotating about the x axis, that experiences constant proper acceleration in the x direction.

First, in the context of artificial acceleration in flat (Minkowski) spacetime, the disc would appear to have increasing velocity in the x direction in a given inertial reference frame. To make discussion easier I will identify my statements/assumptions/considerations with letters.

a) I think we can also assume that as the velocity increases in the x direction, the rotation of the disc will slow down in proportion to the time dilation factor.
b) To an accelerating observer that remains at the centre of the disc, the rotation rate will appear to remain constant.
c) In the inertial reference frame, the increasing linear velocity and corresponding angular velocity slow down of the disc, implies that the radius and perimeter of the disc has to increase if the perimeter length of the disc is to remain constant as measured by observers on the rotating disc.
d) If (d) is true, then by the equivalence principle. this implies that a thin disc with constant rotation at constant altitude will expand or rip itself apart (if Born rigidity is not maintained). As far as I aware this effect as not been observed in everyday life or in a lab.
e) It may be that the expansion in consideration (c) does not occur. Study of successive Lorentz transformations of a rod shows that if the first transformation (to S') is parallel to the rod (say in the x direction) and the second (to S'') is orthogonal to the rod (say in the y direction), then Thomas precession does not occur and the length of the rod as measured in S'' is independent of the velocity component in the y direction and equal to the length of the rod as measured in the S' reference frame. This implies that if the disc is already rotating in the y,z plane about the x axis, and the constant linear acceleration is along the x axis (or upwards in a gravitational field) then length contraction effects above and beyond those already present due the rotation of the disc, should not occur.

OK, that's enough musings for now. Can anyone correct/expand on my thoughts here and explain what exactly happens in the special case of a infinitesimally thin disc that is rotating with constant angular velocity and linearly accelerated with constant proper acceleration along its rotation axis?

2. Aug 25, 2013

### Staff: Mentor

Yes.

If by "descending" you mean "freely falling" (as in, the center of mass is freely falling in the gravitational field), this is not correct; there's nothing preventing a freely falling rotating cylinder from remaining Born rigid (at least as long as its size is small enough that tidal gravity doesn't affect its motion measurably). All of the results quoted in the other thread about Born rigid motion being impossible were for objects that were both rotating *and* linearly accelerated; none of them were for objects that were rotating but whose centers of mass were freely falling.

If by "descending" you mean "slowly descending but accelerated", then as far as the theorems quoted in the previous thread are concerned, there's no real difference between this case and the case of constant altitude. According to the theorems, as long as both rotation and linear acceleration are present, Born rigid motion is impossible. See further comments below.

The theorems referred to in the other thread say that for the motion of any object to be Born rigid, the worldlines of each of its points must all be integral curves of a single Killing vector field. So the basic answer to your question is that there is no KVF that corresponds to the state of motion you are describing. (Again, this is true whether or not the rotating cylinder remains at constant altitude; the key is the presence of both rotation and linear acceleration of the center of mass.) Unfortunately, I haven't come up with a good way of describing, in intuitive physical terms, why this is the case. (Mathematically, it's easy to see for the Minkowski spacetime case, because all of the KVFs for Minkowski spacetime are easily enumerated, and it's evident that the worldlines of an object that is both rotating and linearly accelerated don't match up with any of them.)

I need to think over the above some more before responding to the rest of your statements about the thin disk in Minkowski spacetime.

3. Aug 25, 2013

### PAllen

Just to clarify, do you accept that rotating disk cannot be uniformly accelerated in an inertial frame, with gravity ignorable, while maintaining Born rigidity? This is the essence of Herglotz-Noether. So if you don't accept this, then the place to start is a derivation of this theorem. A modern (but not elementary) derivation is given in:

The Rich Structure of Minkowski Space

If, instead, you accept this, then it really seems obvious why a spinning disk cannot remain at rest in a gravitational field: it is undergoing proper acceleration in the locally Minkowski space. It is the free falling, spinning disk that could, at first glance, maintain born rigidity. In fact, the only factor I see preventing a free falling disk from remaining born rigidity is tidal forces.

[edit: I see Peter Donis beat me to all these points.]

Last edited: Aug 25, 2013
4. Aug 25, 2013

### yuiop

Yes, I meant "slowly descending but accelerated" and should of excluded the free falling case. Thanks for the correction ;)
Thanks. I look forward to your analysis. I am trying to find out if the none-infinitesimal thickness of a cylinder is an essential part of the failure of Born rigid rotation combined with linear acceleration. Is the infinitesimally thin disc a special exception?

5. Aug 25, 2013

### Staff: Mentor

Not according to the theorems I mentioned--at least, not if I'm understanding them correctly. Even the congruence of worldlines describing an infinitesimally thin disc is not a Killing congruence (i.e., the worldlines are not integral curves of a KVF).

6. Aug 25, 2013

### WannabeNewton

Each point on the disk is described by a time-like curve so the disk itself is described a time-like congruence i.e. a time-like vector field $V^{\mu}$ whose integral curves represent the worldlines of the points on the disk. If we have a disk uniformly accelerating along the $x$ axis while rotating in the $y$-$z$ plane, then one can show that the expansion tensor $\theta_{\mu\nu} = h^{\alpha}{}{}_{\mu}h^{\beta}{}{}_{\nu}\nabla_{(\alpha}V_{\beta)}$ does not vanish identically for this congruence (here $h_{\mu\nu}$ is the spatial metric). What this means is that if we take a given point $O$ on the disk and connect it to an infinitesimally nearby point $O'$ using a connecting vector $X^{\mu}$ with $X^{\mu}V_{\mu} = 0$ initially and $\mathcal{L}_{V}X^{\mu} = 0$ then the squared length $X^2 = (-X^{\mu} X_{\mu})$ of $X^{\mu}$, which initially is the squared spatial distance of $O'$ from $O$, changes at the next instant i.e. $\frac{d(X^2)}{d\tau} = V^{\mu} \nabla_{\mu} (X^2) = -2\theta_{\mu\nu}X^{\mu}X^{\nu} \neq 0$. Basically there are shear stresses that pull apart (or at least try to) neighboring points on the disk, on account of $\theta_{\mu\nu} \neq 0$. Born rigid motion is defined as $\mathcal{L}_{V}h_{\mu\nu} = 0$ and not surprisingly it can be shown easily that Born rigid motion occurs if and only if $\theta_{\mu\nu} = 0$.

I still can't fully get the intuition for why that happens though. Gron and others have detailed papers on the analysis of rotating disks that have a uniform tangential acceleration, wherein the expansion tensor is yet again non-vanishing, but I can't find anything on the analysis of rotating disks that have a uniform linear acceleration.

Last edited: Aug 25, 2013
7. Aug 25, 2013

### Staff: Mentor

No, it implies that a thin disk with constant rotation and constant altitude will experience shear stresses that will not be present in an accelerated disk that is not rotating, or a rotating disk that is not accelerated. The shear stresses would not necessarily have to exceed the breaking stress of the material.

These stresses could in principle be measured in a lab, but I think (though I haven't run any actual numbers) that with a 1 g acceleration, in order to make the additional shear stress measurable you would need to have an angular velocity well beyond what we could achieve with known materials (i.e., the disk would tear itself apart purely due to the stresses induced by trying to spin it up, even if it was freely falling, so we would never get to the point where we had it in a state where we could test for the additional shear stress due to the combination of rotation and acceleration).

8. Aug 26, 2013

### WannabeNewton

I don't quite follow this part. Even in a general setting, why would a dichotomy between two different observers solely lead to a coordinate independent fact about a certain tensor equation e.g. $\theta_{\mu\nu} = h^{\alpha}{}{}_{\mu}h^{\beta}{}{}_{\nu}\nabla_{(\alpha}V_{\beta)}$? In this case, the expansion tensor $\theta_{\mu\nu}$ of the time-like congruence defined by the linearly accelerating, rotating disk is, in particular, not equal to 0 in the comoving coordinates attached to the center of the disk. So why physically does an observer comoving with the center of the disk measure shear stresses between neighboring points on the disk and, if the conditions are extreme enough, see the disk start to crack?

Last edited: Aug 26, 2013
9. Aug 27, 2013

### yuiop

This is basically what I am looking for here. A good intuitive physical picture of what actually happens to the disc. Does it expand and experience stresses similar to those caused by reaction to centripetal forces.

This is good news. Can we assume that for constant angular velocity of disc and constant proper linear acceleration, that the H/N stresses are constant and do not increase without bound over time? If so, then it might be possible to conceive of a tall rotating cylinder in a gravitational field, as discussed in the other thread, that remains intact as long as the radius of the cylinder is small enough, the rotation rate is slow enough and the gravitational and tidal forces are not too excessive.
I agree this is a major hurdle for carrying out an actual experiment in a lab. I was hoping we discuss what would happen in principle (hypothetically) if we ignore centripetal stresses as is usually done when discussing the Ehrenfest paradox. i.e. assume a very strong (but not infinitely rigid) material. It might be worth considering a rotating molecule/nucleus/neutron star, although measurements would be difficult in practice.

I have much to learn before I can understand everything you have posted here, but I can see that we can get a yes/no answer to the Born rigidity question. However, we do not seem to be able to quantify the extent of these forces as we can with centripetal force for example.
THis is essentially what I was hoping to find out in this thread.

In statement (c) I was musing that the disc should expand if its rotation rate was slowing down according to the inertial observer, but on further analysis that is probably incorrect. To the inertial observer, the angular velocity is decreasing, but the linear velocity is increasing and both velocities affect the time dilation of an accelerating clock on the disc rim and in effect cancel each other so there is no corresponding change in length contraction. An intuitive way of looking at this, is to consider the case of a disc rotating at w rpm, whose centre of mass is at rest in ref frame S. If we boost to the POV of frame S' that is moving at 0.6c relative to S, the rotation rate is 0.8w rpm. If we now boost to the point of view of frame S'' travelling at 0.8c relative to frame S, the disc slows down to 0.6w rpm, but in all these reference frames the radius and circumference of the disc remains unchanged. Of course there is no actual proper linear acceleration of the disc in this example and that may be the crucial difference. Does a rotating disc effectively act as an accelerometer?

To further illustrate, consider 2 identical discs that are linearly at rest in S and wrt to each other and rotating at the same angular velocity w about the x axis. Now we accelerate one disc to 0.8c relative to the other and then let it coast inertially at 0.8c. By retrospectively looking closely at these 2 discs after everything has settled down and stabilised, can we tell which disc actually accelerated by looking for physical differences in the discs?

One aspect that has not really been touched on in this thread is Thomas precession. If we accelerate the disc linearly in the x direction and THEN rotate it around the x axis, then when looking at a short section on the edge of the disc, Thomas precession tends to twist that section out of the rotation plane. This does not rotate the axis of the disc as a whole because for each section on the edge of the disc there is a corresponding section on the opposite side of the disc that is trying to twist in the opposite direction. These opposing twists may cause stresses in the disc. However, if we rotate the disc first and THEN accelerate linearly along the rotation axis, there is no Thomas precession effect. In a gravitational field this could be achieved approximately, by initially rotating the disc very far away from the gravitational source (so that the linear acceleration is negligible) and then moving the disc lower into the gravitational well.

10. Aug 27, 2013

### yuiop

Just curious if anyone has seen this paper? http://arxiv.org/abs/0810.0072 It claims to simplify the analysis of rotating objects in GR by implementing the concept of Rigid Quasilocal Frames. Essentially they consider the surface of rotating solid rather than the interior and specify an alternative form of rigid motion that makes analysis easier, rather than give up when the Herglotz-Noether theorem suggests that Born Rigid motion is impossible.

11. Aug 27, 2013

### Staff: Mentor

I hadn't seen it but it looks interesting, thanks for the link!

One thing that occurs to me just on skimming the abstract is that, if we only consider the boundary of an object, not its interior, we have no way of taking into account internal stresses within the object, and since it's those internal stresses that signal the presence of "non-rigid" motion (e.g., shear stress), this method basically "solves" the problem by ignoring it. However, there might be more details in the body of the paper that address this, so I'll postpone further comment until I've read it in full.

12. Aug 27, 2013

### Staff: Mentor

That's what I think, based on the math (the shear tensor being nonzero), but I'm not entirely sure whether the stresses vary with time. See below.

That was my initial intuition, but on reflection I'm not sure. It's clear from the math that there is an effect due to the combination of rotation and linear acceleration, that is not present when either linear acceleration or rotation are present in isolation. The problem is distinguishing between stresses that are due to the individual pieces of the disc *responding* to this effect, and stresses that are due to the individual pieces of the disc *resisting* this effect.

Put another way, the problem is figuring out to what extent the congruence of timelike worldlines that I described, and that WannabeNewton wrote down equations for, describes the *actual* motion of the pieces of the disc, versus only the "idealized" motion of the pieces of the disc, which is then altered by internal forces within the disc. If the congruence that I described represents the actual motion of the pieces of the disc, then I think the internal stresses are *not* constant for constant angular velocity; they will increase over time until the disc is torn apart. (Basically, the situation is like that in the Bell Spaceship Paradox; the only difference is that here it is shear rather than expansion that causes the increasing internal stress.)

But an actual disc might have internal forces that resist this, and cause the pieces of the disc to follow different worldlines than those expressed by the congruence I described; however, I'm not clear about just how much leeway there is for the pieces of the disc to do that while still following timelike worldlines. It seems quite possible to me that there is *no* set of timelike worldlines that the pieces of the disc could follow which would lead to a constant shear stress, rather than one which grows with time. But I haven't yet come up with a way to resolve the question one way or the other.

This isn't the way I understand Thomas precession. [Edit: Actually I missed something here--see follow-up post.] First, Thomas precession doesn't cause stress within a rotating disc; it's just a way of expressing what "non-rotating" actually means, in terms of Fermi-Walker transport. Second, Thomas precession acts wholly within the rotation plane of the disc.

Here's a way to visualize what Thomas precession means. Suppose I put a gyroscope on the edge of a rotating disc, supported by a frictionless bearing so that the gyro's orientation is not affected by the motion of the disc itself (the bearing applies a force at the gyro's center of mass to keep it riding around with the disc). At some time t = 0, the gyro is pointing directly at some distant object (such as a star), which is at rest relative to infinity.

If Newtonian physics were exactly correct, the gyro would remain pointing at the distant star forever, as it rode around on the edge of the disc. However, when we take relativity into account, the gyro will slowly precess in the retrograde direction: that is, the direction it points will slowly rotate, relative to the distant star, in the same plane as the disc itself, but in the opposite direction.

Now, suppose we paint a radial stripe on the disc; this stripe defines a spatial vector that always points "radially outward" relative to the disc. Suppose the outer end of this stripe passes directly underneath the gyro, and everything is transparent so we can see the relative orientation of the gyro and the stripe from above. If we look down from above, as the disc rotates and the gyro rides around on its edge, we will see the gyro and the stripe rotating relative to each other. However, there will be no stress in the tangential direction due to this relative rotation; the only stress, on either the disc itself or the gyro, is radially inward (because that's the direction of the proper acceleration).

Last edited: Aug 27, 2013
13. Aug 27, 2013

### WannabeNewton

and here: http://www.nature.com/nature-physci/journal/v235/n61/pdf/physci235175a0.pdf

Also, Thomas precession is present even when the disk is not linearly accelerating but just rotating, relative to a global inertial frame (recall that the Thomas precession is due to the non-commutativity of successive non-parallel Lorentz boosts). Of course that isn't to say that the effect is exactly the same in magnitude as when the disk is indeed linearly accelerating.

EDIT: Thanks for the article, yuiop! I just saw that :)

Last edited: Aug 27, 2013
14. Aug 27, 2013

### Staff: Mentor

Thinking this over, I think it can't be, and the difference might be what yuiop was referring to when talking about "twisting" the disk out of its rotation plane.

If the disk's CoM is moving inertially, then a point on the edge of the disk is subjected to Lorentz boosts all in the same plane, but with changing direction within that plane. So the resultant of the successive boosts will be a rotation about an axis perpendicular to the plane.

However, if the disk's CoM is being accelerated in a direction perpendicular to the plane of rotation, then a point on the edge of the disk is subjected to Lorentz boosts with two components, the one due to rotation, within the plane, and the one due to linear acceleration, perpendicular to the plane. Also, the 3-velocity of a point on the disk's edge is has a component within the plane of rotation and a component perpendicular to that plane (at least, it does once the disk has accelerated for more than an instant relative to a given inertial frame). So the cross product between the velocity and the acceleration vectors, and the magnitude of the velocity vector, will both change, and since those are what determine the Thomas precession, the latter will also change. (This means it will also change with time, as the disk accelerates.)

15. Aug 27, 2013

### Staff: Mentor

I don't think this can be right; if the disk is rotating while it is being linearly accelerated, then the argument in my previous post applies and Thomas precession must be (a) present, (b) not the same as for the case of a rotating disk whose CoM moves inertially, and (c) changing with respect to time.

I'm not sure, however, how any of this relates to stresses in the disk; as I said a couple of posts ago, Thomas precession by itself doesn't cause any stresses, it's just a definition of what a "non-rotating frame" means (more precisely, it defines how a local definition of a "non-rotating frame", using gyroscopes, *differs* from the Newtonian expectation that such a frame should always point to exactly the same location at infinity).

16. Aug 27, 2013

### WannabeNewton

Yeah that's what I was thinking as well. The precession is given by $\dot{S} = \frac{\gamma - 1}{\beta^{2}}S\times (\boldsymbol{\beta}\times \boldsymbol{\dot{\beta}})$. In 3-vector notation, the 3-velocity $\boldsymbol{\beta}$ would be like $(0,\beta_{\phi},\beta_{z})$ and $\boldsymbol{\dot{\beta}}$ would be like $(\dot{\beta}_{r},0,\dot{\beta}_{z})$ so the cross product would be like $(\beta_{\phi}\dot{\beta}_{z}, -\beta_{z}\dot{\beta}_{r},-\beta_{\phi}\dot{\beta}_{r})$. So at $t = 0$ if we attach a gyroscope to a differential element of the disk such that it (and hence the associated spin vector $S$) points radially outward, then initially (assuming that at $t = 0$ the disk is already in linear motion relative to the global inertial frame that this is all written in) the precession $\dot{S}$ is proportional to $(0,-S_r \beta_{\phi}\dot{\beta}_{r}, -S_r \beta_{z}\dot{\beta}_{r})$ so there would be a component of the precession pointing out of the plane of rotation.

17. Aug 27, 2013

### yuiop

I set up a little spreadsheet to calculate successive Lorentz boosts to study the non commutative nature. It was here that I discovered that if we boost a rod lying on the x axis in the x direction and then boost it in the y direction, that Thomas precession does not occur (but it does if we do the boosts in the opposite order). I am not sure if that is widely known. I am also not always sure how we know which boost occurs first in non trivial situations. As for the difference between boosting to different reference frames and actually accelerating the disc linearly, that is the big question! This is what Peter is addressing below.

I tentatively agree, although I not sure that it changes over time.

That leaves me wondering what does cause the stresses in the disc. I guess one important difference between successive Lorentz boosts and continual proper linear acceleration is how spatial measurements are made by disc observers. If radar measurements are made between two neighbouring points on the disc, the linear and angular velocity of the disc at the time the radar signal is emitted is different from the velocities when the radar signal return. This may be the crucial difference from constant inertial velocity case.

P.S. I would love an opinion from either of you on my gravitational differential gear thingy in the tall cylinder thread ;)

18. Aug 27, 2013

### Staff: Mentor

See WN's latest post, he writes out the math, which makes it easier to see: the components he writes out are obviously changing with time (remember he's writing them in a single inertial frame, in which the velocity and acceleration of any point on the disk both change with time).

Which stresses? As I said before, Thomas precession by itself doesn't cause any stresses; it's just a way of making precise what a "non-rotating" frame is, and how the definition differs from the Newtonian expectation when relativistic effects are included.

Yes, I think this is an important difference.

I haven't posted on that because I don't think it really raises any new issues that weren't raised by the original cylinder scenario, and I'm not sure I'm ready to tackle the curved spacetime case again until we've beaten the flat spacetime case, if not to death, at least to within an inch of its life.

19. Aug 27, 2013

### jartsa

Let there be a small thin plate, rotating around a long pole, the radius of the orbit is large.
Picture:
O .
(On the left plate from above, on the right pole from above.)

We accelerate the plate linearly along the pole, so that it stays as Born rigid as possible, by giving every part of the plate a push, simultaneously in the plate frame.

This will cause the plate to tilt, according to the pole. The trailing edge tilts in the direction of the push. (because that side is pushed first, in the pole frame)

Now we weld many plates together, so that they form a ring. We make the ring to rotate. Then we apply same kind of force to each plate as I described above. Those forces would make the plates to tilt, to various directions, but the internal stresses keep the plates alingned.

Edit: Oy yes, that kind of tilting may happen when we are increasing the linear acceleration, but the question was about constant acceleration.

Last edited: Aug 28, 2013
20. Aug 27, 2013

### pervect

Staff Emeritus
I'd agree
I agree (though I don't have a textbook reference for this point, which would be nice to mae sure I'm not overlooking something).

I disagree, and your mixing up of frames (the inertial frame and whatever it is you mean by "measured by observers on the rotating disk") is going to confuse the issue massively.

The rest of the argument hinges on this point.

I'd say that d most likely isn't true. Assuming I'm understanding you correctly.

21. Aug 28, 2013

### yuiop

I think I paraphrased too much. Let us say there is an inertial frame S, that the centre of the rotating disc is initially at rest in. The inertial observer O remains at rest wrt to S and the disc rotating around the x axis of S, is accelerated along the x axis. Observer O' is comoving with the centre of the disc and remains at rest wrt reference frame S' which is moving linearly wrt frame S along the x/x' axes. The last observer O'' is at rest in the rotating reference frame S'' of the disc and is positioned on the edge of the disc. Observer O'' measures distances between reference points on the disc. If these remain constant then he confirms that Born rigidity is maintained.

You agreed that the angular velocity of the disc slows down as measured by O as the linear velocity of the disc increases relative to frame S. Ordinarily when the rotation rate of a disc is changed (no linear motion) the ratio of the circumference to the proper radius has to change. This is the well known concept that it is impossible to spin up a disc and maintain Born rigidity of the disc. However, if we have a disc with constant angular velocity but linearly at rest in a given irf and then do a Lorentz boost along the spin axis of the disc, the outcome is different. The angular velocity appears to change, but there is no corresponding changes in the proper dimensions of the disc and Born rigidity is maintained in this case. The question is which case best represents the disc that experiences proper linear acceleration? As mentioned in an earlier post, performing a Lorentz boost with velocity -v does not necessarily produce the same results as actually physically accelerating an object to velocity v relative to the initial unboosted irf.

I think Peter is tending to the opposite conclusion, but the jury is still out.

22. Aug 28, 2013

### pervect

Staff Emeritus
If we look at the disk in what you call frame S', the frame that's co-moving with the disk, it doesn't "slow down".

I'm not sure how you are defining "the" circumference of the disk, the term is notorious for different definitions. But if we use a definition that is independent of the observer, then it seems obvious to me that the circumference must remain the same - because it's really easy to compute in frame S', and in frame S' it's constant.

Basically, we just have a disk, and [strike]we are[/strike] I am demanding that what we define as "the circumference" be the same before and after a Lorentz boost. The problem is formulated in terms of an "active" boost, which confuses things a bit, but if we start by demanding that our definition of "circumference" be unchanged under a passive boost I think we will start to see some resolution.

So ultimately I suspect the whole argument is about how to compute the circumference of a disk. Which has been the topic of a lot of other threads, and is notorious for causing a lot of confusion (due to the problem previously mention of defining precisely what it actually means in a manner that allows one to actually calculate it.)

I'm not sure how much I want to rehash all that, but I suppose if you go to the trouble of spelling out your defintion, I might get motivated to try and spell out mine.

On a related note I suppose I should review the expansion tensor approach and see what I get with the congruence that I mentioned earlier.

Last edited: Aug 28, 2013
23. Aug 28, 2013

### PAllen

I am wondering about different approach to understanding this problem (unfortunately, I only pose it - I don't have time now to pursue it), via Fermi-Normal coordinates. Fermi-Normal coordinates can be specified for a world line undergoing arbitrary motion and rotation (the more extreme, the smaller the region covered). The origin of such coordinates can be taken as the 'center' of a body. A (not necessarily born rigid) body can be defined as the congruence of constant coordinate position world lines, within a boundary. Questions:

1) For cases where Born rigid motion is possible (e.g. uniform acceleration; rotation without acceleration), is this FN-coordinate body a Born rigid body?

2) For other cases, is there a meaningful attribute (less abstract than, e.g., the expansion tensor) of this FN defined body that shows how it is not Born rigid?

I would like to think the answer to (1) is yes, and that thus this FN body is the 'closest you can come' to a rigid motion; and it is Born rigid when that is possible.

24. Aug 28, 2013

### PAllen

In reference to the above idea, I found the following paper:

http://arxiv.org/abs/1103.4475

It is of interest in general for this thread. Further, it suggests (1) above is true, but that final idea is not useful. It provides an alternative to Fermi congruences that can be achieved by perturbation of Fermi congruences that the author's argue is 'as rigid as possible' for cases more general than Born rigid motion.

25. Aug 28, 2013

### Staff: Mentor

I don't think this is quite right. A single worldline can't undergo "rotation"; that requires a congruence of worldlines. And F-N coordinates can't be constructed in the way you are describing for an arbitrary congruence, only for one in which the spatial basis vectors that point from one worldline to "neighboring" worldlines in the congruence are Fermi-Walker transported, i.e., in which the spatial basis vectors are "nonrotating". In other words, only a non-rotating congruence (one whose twist is zero) can be described by a set of worldlines that maintain constant spatial coordinates in the Fermi Normal chart that is centered on the "origin" worldline of the congruence.