Stress-energy tensor of a wire under stress

[Chris Hillman]

Summarizing (?) the discussion so far

Hi, pervect,

Do you agree with my understanding of your current approach?:

1. You are working in flat spacetime with the stress-energy tensor of a model of a rotating hoop.

2. You are trying to compute the total mass-energy of a rotating hoop (constant nonzero angular velocity omega) as described by the inertial observer comoving with the centroid of the hoop. The idea is to compare with the total mass-energy of an identical but non-rotating hoop.

3. The world lines of the matter in the "before" hoop should be straight lines in Minkowski spacetime (vertical coordinate lines in the cylindrical chart). The world lines of the matter in the "after" hoop should agree with world lines belonging to one of the unique family, parameterized by omega, of Langevin congruences in Minkowski spacetime, which are stationary cylindrically symmetric with vanishing expansion tensor, i.e. they describe the world lines of a cylindrically symmetric configuration of rigidly rotating observers.

4. Given the global issues with integrating in the comoving Born chart, you are trying to integrate the kinetic energy (plus stored energy from the tension) along the hoop in the cylindrical chart (your "lab frame"). Specifically, you hope to find the mass-energy in the (constant omega) hoop in the obvious inertial frame and integrate over the hoop in a constant time slice. This should give the total mass-energy as described by the inertial observer comoving with the centroid of the hoop, which is stationary as described by this observer.

5. As far as I can see, you are not (yet) actually using any stress-strain relationship or indeed anything from the theory of elasticity. Rather, you are trying to impose "local rigidity" (vanishing expansion tensor of the world lines of the matter in the hoop).

6. As far as I can see, you avoided trying to model spinup phase and tried to guess what rigidly rotating hoop is equivalent to a given nonrotating hoop. As I understand it, you simply assumed that it will be possible to maintain rigidity during spin up (presumably via a complicated but unique congruence), and then deduced, based on the Lorentz contraction, that the diameter of the hoop, as described by the inertial observer, must have decreased by a specific ammount. IOW, your rotating hoop is supposed to have the same circumference as measured by comoving observers as the nonrotating hoop (since local rigidity is assumed to have been maintained throughout). You deduce that the mass-energy and angular momentum of a rotating hoop (with constant nonzero omega), as described by the inertial observer comoving with the centroid, are both smaller than that of a nonrotating hoop (zero omega).

As far as energy goes, the relativistically spinning disk appears to me now to be very much like the linearly accelerated one.

My objections to your current approach (as I understand it) are these:

1. My intuition tells me that your result can't be right: the total mass-energy of a rotating hoop should not be smaller than that of an "equivalent" nonrotating hoop, in any physically reasonable sense of "equivalent"!

2. To make the desired comparison between the two hoops, you need to be able to set up a physically reasonable equivalence between "before" and "after", i.e. between a nonrotating hoop (no doubt what this means!) and an equivalent but rotating hoop (nonzero omega, nonzero tension along the hoop). I maintain that the only "safe" way I see of doing this is to set up a simple material model and to try to model the spin-up phase. You maintain that you can evade this by insisting that local rigidity (vanishing expansion tensor of congruence of world lines of the matter in the hoop) must be maintained during spin-up, but it's not clear to me that this is possible. I showed that the congruence you imagine would have to be rather complicated.

3. You assumed the radius of the "equivalent" rotating hoop (constant nonzero omega) must be smaller by an amount deduced from the Lorentz contraction factor, but measuring circumference is problematical for the comoving observers. I think you are thinking of integrating length of a spacelike curve in a "constant time slice" in the comoving Born chart to compute this circumference, but there is no such slice. If we try to compute C as measured by comoving observers in the cylindrical chart, we run into global inconsistencies (draw the picture of alleged "space at a time" for hoop matter). So I maintain that you need to worry about measuring "distance in the large", and to confront the fact that theory shows clearly that results will depend upon the method of measurement used by the comoving observers (possibly one, possibly many acting together) and the definition they use to compute a distance from these measurements.

The correction factor due to tension is of order v^4. We have a volume reduction of sqrt(1-v^2) which is of order (1-v^2/2), and we have a density increase of exactly (1+v^2), which is different from the linearlly accelerated result of 1/(1-v^2) but is the same to order v^2.

The net result is a kinetic energy that agrees with the Newtonian result for both cases, i.e. a kinetic energy of m v^2/2 to second order.

I suggest backing off from a rotating hoop and starting over for a short rod which is linearly accelerated along the axis of the rod. Can you find a reasonable expression for the mass-energy of such a rod as described by inertial observers?

As we know from Rindler versus Bell congruence, such a rod can remain rigid only if we assume very special accelerations.

At his website, Greg Egan models a rod being accelerated by being tugged at one endpoint. He focuses on deriving the displacement of an elastic rod under these conditions, but he does give expressions for the stress-energy tensor in a frame comoving with the matter in the rod. His \vec{u}=\vec{e}_1 is the timelike unit vector in this frame, and his \vec{w}=\vec{e}_2 is the spacelike unit vector pointing along the axis of the rod. He writes down expressions for the stored energy and the tension (from a material model), then writes down the stress-energy tensor in our frame, then takes a (flat spacetime) divergence to obtain an equation which can be rearranged (as I understand it) to give either (a) an equation for the displacement, or (b) equations for the frame written in an ordinary cylindrical or cartesian chart for Minkowski spacetime. He finds an exact solution for (a) by assuming boost invariance. This models a rod which is static in the Rindler chart, so with trailing points being accelerated harder. I guess the implicit claim is that for a suitable elastic type material model, the tensions in the rod distribute themselves to make this possible.<br /> <br /> (BTW, I am not sure I understand his claims correctly, but I am seeking clarification from him.)<br /> <br /> <blockquote data-attributes="" data-quote="pervect" data-source="post: 1322050" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> pervect said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> The effect of the tension terms on the angular momentum density are much more significant, though. This is an effect of order v^2, and exactly cancels out the increase in density, i.e. the angular momentum density is just rho v r.<br /> <br /> However, the volume decreases as noted before, resulting in an order v^2 <b>decrease</b> in angular momentum, quite different than one might expect. This can be attributed to the shrinking radius of the hoop. </div> </div> </blockquote><br /> At the moment, I doubt that you are comparing the right pair of hoops. Indeed, I doubt that your proposed notion of setting up an equivalence between nonrotating hoops of radius R and stationary rotating hoops (of some different radius) makes sense.<br /> <br /> If I am right, it seems to me that one cannot evade postulating a material model. If so, the obvious choice would be to try to create a suitable model of an elastic material. Then, for small omega Newtonian effects should dominate. The question becomes: how do relativistic effects alter Newtonian predictions as omega increases? For reasonable elastic constants I think this question should be answerable, both for linearly accelerated short rods and for spinning hoops.<br /> <br /> I think there would be considerable interest in simple but reasonable (&quot;elastic material&quot;) models in gtr of the <i>interior</i> of (a) a tugged rod (constant acceleration of leading endpoint) (b) a rotating hoop (constant omega, so stationary spacetime), particularly if the latter could be matched to an exact stationary axisymmetric vacuum solution.

 

[pervect]

I agree with your understandings expressed in 1-5.

[add]Ooops, I was hasty in agreeing. Some of the fine print doesn't quite work.

6 is correct in that I don't have a detailed model of spin-up currently. I thought that you agreed that the congruence in 3) was unique, however?

I agree that the derivation needs to be improved in this area.

I definitely do NOT think that the energy decreases when you spin it up. My result for the energy was expressed in terms of v, the radial velocity of the hoop.

E = m_0 \left(1 + v^2\right) \sqrt(1-v^2)

here m_0 is the initial mass of the hoop.

where r_0 is the intial radius of the hoop.

The energy increases with v, the term of order v^2 is .5 m v^2

My observations related to "lower energy" were observations that coefficients of v^n of order 4 and higher in the expression for energy written as a polynomial in v are negative. I've actually only verified that the coefficients are negative for 4<=n<=48, howver, I haven't computed all of them :-).

We can find v as a function of \omega by solving the relationship

<br /> v = \sqrt{1-v^2} r_0 \omega<br />

this gives

<br /> v := \frac{{\it r_0 \omega}}{\sqrt { \left( {{\it r_0}}^{2}{\omega}^{2}+1 \right) <br /> }}<br />

The series expansion in terms of \omega turns out to be
<br /> (m+1/2\,m{{\it r\_0}}^{2}{\omega}^{2}-{\frac {9}{8}}\,m{{\it r\_0}}^{4<br /> }{\omega}^{4}+{\frac {25}{16}}\,m{{\it r\_0}}^{6}{\omega}^{6}-{\frac {<br /> 245}{128}}\,m{{\it r\_0}}^{8}{\omega}^{8}+O \left( {\omega}^{10}<br /> \right) )<br />

so unlike the series expansion for v, not all of the higher order terms are negative.

 

[pervect]

OK, let's define the congruence more precisely.

We initially have a point at r=r0, and some angle theta. Through this point there must be a uniqie curve in the congruence. We parameterize this curve using lab-frame coordinates

(r,theta) -> (r(t), theta + phi(t))

We require that r initially be at r0, however.

A feature of this congruence is that the difference in angles, theta, in the lab frame between two worldlines is independent of lab time.

in order to have rigid motion, the separation of worldlines (orthogonal to the worldlines) must be constant.

We assume that we can pick phi(t) in such a manner that dr/dt is as small as desired, that in the limit dr/dt is zero, therefore we can neglect it. This assumption can be checked

In this case, the velocity of the worldline will just be

v = r(t) * d(phi)/dt

Doing a Lorentz boost, it is intuitively obvious that the separation between worldlines passing through points (t,theta) and (t,theta+dtheta) is

r(t) dtheta / sqrt(1-v^2), where v = r(t) dphi/dt.

To do better than this bold statement, I'd have to draw some space-time diagrams. (add), but basically, all that is happening is that the separation between worldlines is the proper distance, and r*dtheta is the Lorentz contracted distance in the lab frame.

Because dtheta is invariant as a function of time, in order for the separation between worldlines to remain constant r(t) must equal sqrt(1-v^2).

[add]
Let's check the assumptions

Let's let r_0 = 1 for simplicity. Let \omega = \frac{d \phi}{dt}. Then we have using some of the results from earlier posts

v = \frac{\omega}{\sqrt{1+\omega^2}}
r = \sqrt{1-v^2} = \frac{1}{\sqrt{1+\omega^2}}

so
\frac{dr}{dt} = \frac{dr}{d\omega} \frac{d \omega}{dt} = -\frac{\omega}{\left( 1 + \omega^2 \right)^\frac{3}{2}} \frac{d \omega}{dt}

so by making \frac{d\omega}{dt} very small (quasi-stationary), we can make dr/dt small as well.

I suppose at this point I ought to learn how to compute the expansion scalar of this congruence to check that it's zero, at least in some limiting sense as d\omega/dt -> 0

 

[Chris Hillman]

Trying to clarify my objections

I agree with your understandings expressed in 1-5.

6 is correct in that I don't have a detailed model of spin-up currently. I thought that you agreed that the congruence in 3) was unique, however?

The Langevin congruence is the unique rigid "helical" congruence in Minkowski vacuum. That is, if we write down the frame for a general congruence in which the world lines form helices of fixed pitch depending upon r (i.e. the orbits are circles and observers move uniformly in each circle, but we don't specify how quickly a given observer moves around his circle), and if we then demand that the expansion tensor vanish, we obtain a differential equation, and solving this, we arrive at the Langevin congruence with fixed omega.

If we allow omega to vary with time, the expansion tensor acquires a nonzero component \Theta[\vec{e}_1]_{44} which is positive for \omega_t &gt; 0, i.e. our observers are moving apart along \vec{e}_4, so the "comoving circumference" is increasing in this sense when we are increasing the angular velocity. Naturally this means that the Langevin congruence is no longer region when omega is conconstant. And each world line remains on a fixed cylinder of constant radius in the cylindrical chart, but their pitch varies. We can call this the variable pitch Langevin congruence.

You were turning the expansion on its head and arguing that, without modeling the spinup, we can imagine a spinup phase which is infinitesimally rigid throughout, i.e. the expansion tensor is zero throughout. I said it's not clear (at least not to me) that this is possible and provided some evidence that it might not be. But under your assumption, you said the comoving circumference should be the same, so the diameter measured by an inertial observer comoving with the centroid must be smaller, i.e. infinitesmal rigidity implies global contraction. In this case, during the spinup phase, each quasihelical world line stays on a quasicylinrical surface which shrinks in radius during the spinup phase, but before spin up they still look like vertical rays and after spinup they look like helices of constant pitch.

But I said your notion of circumference measured by the observers comoving with the hoop is incompletely specified, so your assumptions are invalid on at least two counts.

I definitely do NOT think that the energy decreases when you spin it up.

I asked you in a PM if you meant that the expression you found means the energy decreases for sufficiently small V, and I thought you said you did. I must have misunderstood your answer.

OK, now that we agree on this point, I know what your answer is and I see how you are coming up with it, but I don't think your answer is physically reasonable. Or at least, I don't think you have provided convincing reasoning that your assumptions about spinup are plausible. Or if you like, your assumptions about the equivalence between rotating hoops of radius R(\omega) as described by inertial observers comoving with the centroid and nonrotating hoops of radius R_0.

I said that it is clear that you need some way to establish such an equivalence. If you wish to avoid postulating a material model, I understand that your idea is to try to argue that there's some criterion like "keep expansion tensor zero throughout" which establishes the necessary equivalence. But then you'd have to show that spinups satisfying this condition exist and you'd have to argue that the implied material properties are not physically absurd (e.g. speed of sound smaller than speed of light).

So what about the angular momentum? Do you still say that rotating hoops which are spun up rigidly (assuming that's even possible) have decreasing angular momentum?

 

[Chris Hillman]

Clarification

by making \frac{d\omega}{dt} very small (quasi-stationary), we can make dr/dt small as well.

I think this is besides the point; it seems to me that all my objections still stand. I never said anything which involves the size of r_t, only the (varying) size of r.

I suppose at this point I ought to learn how to compute the expansion scalar of this congruence to check that it's zero, at least in some limiting sense as d\omega/dt -> 0

It's easy. In fact computing the expansion tensor for a given vector field is built into GRTensorII. When you installed that you should have obtained a (well written) manual which explains how to do this. I included the "spacetime definitions" (actually, frame definitions, which is at a higher level of structure!) I used so that you could quickly run your own computations as well as checking by hand (since these are simple examples, this is fortunately very easy, but you want to have confidence GRTensorII is doing what you expect).

I myself don't use the built in definition, I use grdef to define the expansion tensor directly. (In fact, I read in a whole bunch of useful definitions from a file before using GRTensorII, and read in more definitions from other files as needed.) In index gymnastics, if we write the velocity vector as a covector \vec{u} = u_j \partial_{x^j}, the coordinate basis components of the expansion tensor are u_{(j;k)} (symmetrize the covariant derivative). Now you can find the frame components. Since u is a unit vector, the expansion tensor is in fact a three dimensional symmetric tensor living in the hyperplane element spanned by the spatial vectors in the frame. See the book by Poisson, A Relativist's Toolkit, which offers a fine discussion of the kinematic decomposition of a vector field into acceleration vector, expansion tensor, and vorticity vector.

(You can also compute the expansion tensor directly in the frame by hand-- GRTensorII also has a built in command to compute covariant derivatives of a tensor wrt a frame vector , but you probably want to stick close to what you are most comfortable with for now!)

 

[pervect]

In this case, though, I don't really have a vector field. I have a one-parameter subfamily of curves (the parameter is theta_0, the value of theta at t=0) that do not fill all of space-time as a congruence should, but which do trace out the worldline of any point initally on the hoop.

 

[Chris Hillman]

In this case, though, I don't really have a vector field. I have a one-parameter subfamily of curves (the parameter is theta_0, the value of theta at t=0) that do not fill all of space-time as a congruence should, but which do trace out the worldline of any point initally on the hoop.

I guess you are worried that if you let r_0, \, \theta_0 vary (to make a three parameter family of curves which on dimensional grounds we hope will fill up some neighborhood without intersections), your curves might intersect?

 

[pervect]

While I think it is probably possible to define the "circumference" of the hoop by performing a branch cut along a worldline of the congruence, I think that it's also possible to avoid talking about the circumference simply by insisting that the worldlines maintain a constant distance (aka a constant "orthogonal deviation vector").

You keep bringing up the topic of the circumference, and I keep attempting to avoid it. If you really want to define a circumference, go ahead and define the circumference via a branch cut. The important point is that the branch cut be one of the worldlines of the congruence. Then we can talk about 'circumference vs time' if we really want to. (And I think we can obsserve that the circumference, in this sense, is constant, though I'd have to double-check this.) But I don't particular want to, I'd just as soon talk about the separation between close world-lines being constant.

To avoid some of the acceleration issues, we can imagine stopping the spinup process to perform our measurement of the separation of the worldlines, spin it up a little more, stop it again and check that our wordlines are still a constant distance apart, etc, though the only cure for dealing with the centripetial accleration is to chose worldlines close enough to each other initially.

This Born-rigid motion allows us to use the simplest possible model - a constant density wire that does not elongate at all with stress. This sort of motion is not possible for anything but a hoop of zero thickness, however, as has been previously noted.

The approach can be adjusted as needed to use a more elaborate material model if desired.

 

[pervect]

I think I may have a better handle on what you're looking for - a velocity field .

To fill all of space-time, we need to consider multiple hoops. Let's consider how r(t) varies, given that the initial radius was r0

We need
<br /> r(t) = \frac{r0}{\sqrt{1+r0^2 w(t)^2}}

By definition the theta component of the coordinate velocity is given by
\frac{d \theta}{d t} = w(t)we need to know what \frac{d r}{d t} is.

This is
\frac{d r}{d t} = \frac{dr}{dw} \frac{dw}{dt}

It turns out, unless I'm making an error, that
\frac{dr}{dw} = -\frac{r0^3 w}{(1+r0^2 w^2)^{3/2}} = -r^3 w

so

<br /> \frac{dr}{dt} = -r^3 w \frac{dw}{dt}<br />

Exactly what to do with this, I'm not sure :-). I suppose we need to convert this to a 4-velocity from an ordinary velocity for starters.

[add]
If we don't normalize v first, and just set dt/dt = 1, v(bup, cbdn) has a lot of zero components in the Langevian basis (with w replaced by w(t)).

with t=1, r=2, theta=3, and z=4

v^(1)_(3), v^(3)_(3) and v^(4)_(3) are all zero
v^(2)_(3) is nonzero however.

If we normalize v first, v^(1)_(3) no longer vanishes. Still a lot of zero components, but not as many.[add^2]
Well, this is getting nowhere fast. Basically I expect members of the the congruence to maintain the same separation (measured orthogonal to the congruence) from each other if they start on the same radius.

I do not expect the expansion tensor to vanish.

I expect the separation vector to rotate, too.

 

[Chris Hillman]

Still trying to clarify the nature of our current disagreement

While I think it is probably possible to define the "circumference" of the hoop by performing a branch cut along a worldline of the congruence, I think that it's also possible to avoid talking about the circumference simply by insisting that the worldlines maintain a constant distance (aka a constant "orthogonal deviation vector").

When you say "constant distance" here, clearly you still mean vanishing expansion tensor.

You keep bringing up the topic of the circumference
[of a rotating hoop, as described in some specified fashion by comoving observers riding on the hoop], and I keep attempting to avoid it.

I'm not trying to be difficult --- I'm trying to get you (and lurkers) to see that there is a serious and challenging issue here. Just one of many, unfortunately.

I only act like a philosopher exposing hidden assumptions when its really neccessary, i.e. when people are in fact arriving at incorrect results, or as in this case, when everyone seems to arrive at a different result, and to be convinced, through a failure of imagination, that only his own result can be correct. When a whole bunch of smart people have said "it's perfectly simple, and the result is R_j, where R_1, \, R_2, \dots are all different, this can be an indication that they are all confused. I think you know that I don't often say "there are serious issues here"--- most of the stuff which comes up in public forums really is pretty trivial once you've mastered the elementary stuff. This problem isn't like that--- it bites!

If you really want to define a circumference,

My point is that depending upon what you mean (see "distance in the large" again) by "define a circumference", this may be impossible.

To avoid some of the acceleration issues, we can imagine stopping the spinup process to perform our measurement of the separation of the worldlines, spin it up a little more, stop it again and check that our wordlines are still a constant distance apart, etc, though the only cure for dealing with the centripetial accleration is to chose worldlines close enough to each other initially.

This answers none of my objections. You haven't clarified, much less justified, how your alleged "equivalence" should work. Again, one of my points is that you want to perform a thought experiment in which you compare the mass-energy of a hoop, as computed by an inertial observer, "before" and "after" being spun-up to a constant angular velocity. You are claiming that the "after" hoop has smaller radius. To justify that you need to explain how to put an equivalence relation on rotating disks characterized by R, \omega, which declares certain disks to be equivalent via a particular kind of spin-up. You can certainly stipulate that the spin-up be very slow, or that it maintains vanishing expansion at all times, as long as you show your stipulations (1) yield realizable congruences (2) establish the necessary equivalence relation.

This Born-rigid motion allows us to use the simplest possible model - a constant density wire that does not elongate at all with stress. This sort of motion is not possible for anything but a hoop of zero thickness, however, as has been previously noted.

Are you still claiming that rigid acceleration acceleration is possible for a "thin" hoop (infinitesimal cross section)? I have provided (weak) mathematical evidence that this might not be true.

The approach can be adjusted as needed to use a more elaborate material model if desired.

I think our core disagreement concerns how to establish the necessary one-parameter equivalence relation on rotating hoops (constant omega). I think we agree that once an equivalence is established, we know how to compute the energy and angular momentum, and then using our equivalence we can compare the energy/ang.mom. of a hoop rotating with constant rate \omega with "an identical nonrotating hoop".

You wish to find some way of doing this which avoids any neccessity of postulating a material model; I doubt this can be done but am prepared to learn otherwise if I am wrong about that. My intuition is that postulating a material model may be unavoidable, but that a fairly simple elastic material will suffice to establish an equivalence.

There is another reason for my interest in material models: to (1) and (2) above I'd like to add (3) the equivalence should be "physically reasonable". Let me explain what I mean by that.

I feel that it is important not only to concoct a mathematically well-defined equivalence (and it is clear to me that there are many ways of doing this), but to justify one as being preferred on the grounds that it is the simplest physically reasonable method.

How do we assess whether an equivalence is physically reasonable? Very simple: the slow rotation limit should be an excellent approximation to a reasonable Newtonian model. Any equivalence which claims that even a tiny increase in rotation rate from zero will decrease the radius strikes me as physically objectionable, because no physically reasonable hoop would behave that way.

 

[pervect]

First off, let me say that I appreciate your time and help, and if I sounded a little frustrated, it was probably because I was getting a little frustrated. But the problem is not your help, which I do appreciate, the problem is the problem itself.

I got some more sleep and sent a PM, and also came up with some new ideas. For the benefit of any hardy souls still with us (if any), I'm saying that the expansion tensor won't vanish on the entire 4-d manifold, it will only vanish on the submanifold generated by the congruence of paths passing through a circle. The disk must be of zero thickness. The expansion tensor of an actual disk of nonzero thickness does not vanish, this is why the expansion tensor of the manifold does not vanish, it only vanishes on the submanifold.

At first this didn't seem all that helpful, but then I realized we could always find the induced metric on the submanifold and calculate the tensor there.

Hopefully this will be a little clearer. Anyway, time for lunch.

 

[Chris Hillman]

Aha, a time zone clue!

First off, let me say that I appreciate your time and help, and if I sounded a little frustrated, it was probably because I was getting a little frustrated. But the problem is not your help, which I do appreciate, the problem is the problem itself.

You didn't sound frustrated. But discussing this problem is exhausting, because there are so many things to bear in mind.

I got some more sleep and sent a PM, and also came up with some new ideas. For the benefit of any hardy souls still with us (if any), I'm saying that the expansion tensor won't vanish on the entire 4-d manifold, it will only vanish on the submanifold generated by the congruence of paths passing through a circle. The disk must be of zero thickness. The expansion tensor of an actual disk of nonzero thickness does not vanish, this is why the expansion tensor of the manifold does not vanish, it only vanishes on the submanifold.

I agree that a "thin hoop" should be easier to study than a "thin disk", and I agree that in the former case, you have the option of defining a congruence in which the expansion tensor vanishes except for world lines which actually correspond to matter in the hoop. I still don't see how to obtain such a congruence however, subject to (1), (2).

If you (or I) can find one or otherwise show they exist, subject to establishing a unique equivalence, I'd want to move onto (3). That's why I've been studying Greg Egan's treatment of a linear accelerated elastic rod in flat spacetime.

At first this didn't seem all that helpful, but then I realized we could always find the induced metric on the submanifold and calculate the tensor there.

Whew!--- for a moment I thought you meant "the" (nonexistent) hyperslice orthogonal to the world lines in the variable omege type Langevin congruence.

(One of the elementary points in the full Ehrenfest paradox is that because we have a stationary timelike congruence we can form a quotient manifold, obtaining the Langevin-Landau-Lifschitz metric, but this is certainly not a hyperslice (submanifold)! And the LLL metric of course gives rise (as does any Riemannian metric) to too many notions of "distance in the large" (integrate length along multiple paths), none of which are particularly interesting physically. But in his PM, pervect says this submanifold is the "world sheet" of the hoop, which he says should gradually shrink as we spin it up. So, in the standard cylindrical chart, this world sheet would look like a cylinder in t,r,phi space--- we suppress the z coordinate--- whose radius is decreasing as we move upwards, i.e. increase time coordinate.)

OK, I agree that to realize your program, you need to set the expansion tensor to zero on this submanifold. And then show this gives a unique equivalence. Then given a spinning hoop, you'd know which nonspinning hoop (which radius) to compare it with. Then we could study slow rotation limit and compare with Newtonian analyses to argue over (3).

 

[pervect]

Here's what I'm getting.

the cylindrical metric is -dt^2 + dr^2 + r^2 dtheta^2 + dz^2

We define some function w(t) to represent spin-up, and we set

r(t) = 1/sqrt(1+w(t)^2)

We want to consider a submanifold having only (t,theta,z) and eliminating r. To calculate the induced metric on this submanifold, we just do some algebra.

The result we get is (for convenience I've removed the explicit dependence of w with t)

dt^2 = (-1 + (w dw/dt)^2/(1+w^2)^3) dt^2 + dtheta^2/(1+w^2) + dz^2

Now, we need to argue that we can make dw/dt as small as we like, and that in the limit where dw/dt is negligible, representing a very slow spin up, that our induced metric becomes

ds^2 = -dt^2 + dtheta^2/(1+w^2) + dz^2

I.e we get a series of different line elements, which converge to this unique induced metric when we take the limit dw/dt -> 0.

We need to do this _before_ we calculate the expansion scalar.

I'm not sure how mathematically rigorous this step is, but it seems at least plausible.

With this assumption, I get zero for the expansion scalar assuming I'm doing the calculation correctly.

The velocity field (unnormalized) I use is just (t,theta,z) = (1,w(t),0), which represents the curves in the congruence parameterized by time.

I pass this velocity vector to grnomralize first, then compute the expansion scalar.

 

[pervect]

Finally, if we make

r(t) = f(w(t))

and we repeat the above analysis, upon setting the expansion scalar to zero we get

f(w)^3*w + df/dw = 0, which has the solutions

f = 1/sqrt(w^2+ C)

Setting f=1 at w=0 yields the original expression. So if the original analysis holds up, it generates a unique answer for the radius of the disk vs w given the initial radius at w=0.

 

[Chris Hillman]

Time to study some previous work!

Hi, pervect,

I think it's time to visit the library, since we are approaching the point of reinventing the wheel, or more precisely, varieties of spinup procedures for attempted treatments of relativistic hoops. To mention just three relevant papers:

G. L. Clark, "The Problem of a Rotating Incompressible Disk", Proc. Camb. Phil. Soc. 45 (1949): 405.

For a small strain limit, compares a Newtonian with relativistic analysis of the spin-up of an elastic disk. Clark and other authors find that the radius increases, as described by inertial observer comoving with centroid, hereafter "Axel", for "axle observer". This is obviously relevant to our hoop discussion re (3) physical reasonableness of a proposed "equivalence" between rotating and unrotating disks.

W. H. McCrea, "Rotating Relativistic Ring", Nature 234 (1971): 399.

McCrea studies a hoop made of a material in which speed of sound (or better say the speed of p-waves?) equals speed of light (an elastic material variant of something often called a "stiff fluid" in the gtr literature), and finds R=R_0/\sqrt{1-R_0^2 \, \omega^2} &gt; R_0 for radii measured by Axel "before" R_0 and "after" R the spinup. (The elastic deformation would be even larger for "softer" materials.)

A. Grunbaum and A. I. Janis, ''The Geometry of the Rotating Disk in the Special Theory of Relativity", Synthese 34 (1977): 281.

The authors introduce a spin-up condition ensuring that tangential stresses vanish and conclude a disk spun-up in this way has smaller radius (as described by an inertial observer comoving with the centroid). For our hoop problem, this would correspond to trying to spin up the hoop without introducing any tangential tensions, in fact I think it may correspond to the spinup procedure you are trying to formulate.

Another possible spin-up procedure we should both think about would be impulsive tangential blows delivered around the hoop, simultaneously and equal magnitude as described by Axel. As we know from twin paradox, impulsive blows can be more confusing than helpful, but we should consider this anyway.

Many of the other papers discussed in Gron's review are also relevant to our discussion, but I think these three might be particularly important for us. But bear in mind some generalizations about the literature on rotating disks and hoops:

1. None of the published papers are, in my view, fully correct,

2. The reason for this is that none of the authors have borne in mind all relevant considerations (for example, we haven't discussed the issue of whether Thomas precession mucks up our analysis--- I think not, but contrary views have been expressed!),

3. The worst papers come from authors who think "it's all perfectly simple" if you just think of it like they do.

So trust nothing, verify everything! Yep, whole lotta work for a simple seeming problem.

For example, while Gron's review is excellent, he fails to consistently distinguish between "congruence orthogonal hyperslice" and "quotient by congruence" manifolds (respectively impossible and possible for the constant omega Langevin congruence). He also fails to discuss the issue of multiple operationally significant notions of "distance in the large" for accelerating observers (and all rotating observers are accelerating, even if they have constant angular velocity).

In reading Gron, be careful to recall that most of his discussion doesn't involve spinup at all, but rather comparing a disk of radius R as described by Axel using the cylindrical chart with omega zero and nonzero, or the same as described by hoop riding observers using the Born chart. Note too the distinction between imagining small Born rigid measuring rods and the material of the disk itself; Einstein's analysis imagines rigid rods sliding on the disk, and would correspond roughly to a notion of circumference measured by the hoop riding observers which I called "pedometer" distance. But never forget that clocks can't be synchronized for these hoop riding observers (c.f. Sagnac effect).

In the paper by McGregor cited by Gron, M argues that the elastic potential energy should be fourth order in the rim velocity measured by Axel which we can try to verify and use in discussing physical reasonableness of proposed spinup procedures.

I should also say that I am still thinking about trying to adapt Greg Egan's analysis to spinning hoops.

 

[Chris Hillman]

Bulletin for lurkers: pervect and I (and Greg Egan!) are still working on this, but right now I at least am reviewing the literature, playing with some putative exact solutions I have found, with alternative congruences, charts, and so on.

I forgot to mention M"arzke-Wheeler coordinates for accelerated observers in Minkowski spacetime, which are an important addition to the roster. See gr-qc/0006095, which discusses MW coordinates for the Langevin observers.

Greg tells me he has found an exact solution (given in terms of an ODE) modeling a rotating hoop, using his stress-strain assumptions, which exhibits this behaviour under spinup: initially the hoop expands, as seen by Axel (the inertial observer comoving with the centroid) as per Newtonian theory, but eventually a relavistic correction counteracting the centrifugal force (roughly speaking) becomes significant. Thus, as described by Axel, the diameter of the hoop increases, but not as quickly as in Newtonian theory.

I rather easily found another exact solution (also given in terms of an ODE), using the stress-strain relation proposed by Clark for a "stiff elastic solid" (in which the propagation speed of p-waves is c, while the propagation speed of s-waves is less than c), which might model the interior of a linearly accelerated rod (I'm still thinking about whether or not I believe this). According to Clark, in such a stiff solid, we should have
\rho = \frac{\mu}{1+n}, \; p = n \; \rho
where n is "the dilitation", by which I assume he means the strain, which here is one-dimensional, and \mu is the rest state density of the material. The very first Rindler type Ansatz I tried quickly leads to a solution in which we use n as the master variable, so that everything is expressed in terms of n and n_x.

 

[Chris Hillman]

I have to interrupt my work on this for a few hours, but I just wanted to say that Greg Egan has just put up his analysis of his own model of a relavistic hoop treated as an elastic solid at http://gregegan.customer.netspace.net.au/SCIENCE/Rings/Rings.html
His analysis confirms my intuition: the hoop expands as per Newtonian analysis, but as the angular velocity increases, the counteracting relativistic effect first noticed by Einstein means that the rate of increase is slower than in Newtonian physics.

Note that the unit vectors he writes
\vec{e}_t, \; \vec{e}_r, \; \vec{e}_\Phi
are the ones I wrote (in the cylindrical chart for Minkowski vacuum)
\vec{e}_1 = \partial_t, \; \vec{e}_3 = \partial_r, \; \vec{e}_4 = \frac{1}{r} \, \partial_\phi
He is treating an elastic solid in Minkowski vacuum, i.e. ignoring gravitational effects in order to clarify the dynamics of a rotating disk or annnulus or hoop. His models should relate to ones considered by Clark in his earlier Proc. Cambr. Phil. Soc. paper, which is cited in the later paper I cited.

I am still mulling my putative exact solution modeling a linearly accelerated elastic solid body (using Clark's "stiff solid" material model in which the speed of p-waves equals the speed of light, while the speed of s-waves is smaller).

 

[Chris Hillman]

Referring the my post #46: MW coordinates for a Langevin observer are a pain to work with, but the eprint cited gives a nice figure. To obtain the MW chart determined by a proper time (T) parameterized timelike curve in Minkowski vacuum, given two events on the curve with T=T_1 and T=T_2, where T_1 < T_2, draw the forward light cone at T=T_1 and the past light cone at T=T_2, and observe that their intersection is a boosted two-sphere. All the events on this two-sphere have MW coordinates
\tau = \frac{T_2+T_1}{2}, \; \sigma = \frac{T_2-T_1}{2}
and the other two MW coordinates are determined by putting some chart on the sphere. Similarly for all other events. Exercise: the MW constant time surfaces are always orthogonal to the world line used to define them. Why doesn't this contradict the Frobenius theorem if it belongs to a congruence with nonvanishing vorticity.

 

[pervect]

I'll have to compare Greg Egan's results with mine in more detail. A couple of interesting points stand out to me, though from an initial reading.

Egan predicts that hoops can shrink when they are spun up. There is nothing to constrain the angular velocity for the relativistic case in his model (or in my rigid model), therfeore one can increase omega as much as one wishes. If r did not shrink, a point on the circumference of the hoop would exceed the speed of light at some value of omega. One must either have r shrinking or some natural limit on omega (a limit which does not appear to exist).

I would argue with Greg Egan's wording when he concludes the fact that r eventually shrinks means the ring is under compression, however - the ring always be in tension in its own frame field.

 

[gregegan]

I would argue with Greg Egan's wording when he concludes the fact that r eventually shrinks means the ring is under compression, however - the ring always be in tension in its own frame field.

You're quite right, of course! The ring will always be under tension. I've amended the web page, and added a curve to the final plot showing the tension, which continues to increase but appears to approach an asymptote as omega goes to infinity.

 

[gregegan]

You're quite right, of course! The ring will always be under tension. I've amended the web page, and added a curve to the final plot showing the tension, which continues to increase but appears to approach an asymptote as omega goes to infinity.

The tension does approach a horizontal asymptote, which turns out to be connected to the weak energy condition. So in principle omega can increase indefinitely, but any real material would have a breaking strain that fell short of the absolute ceiling imposed by the weak energy condition, so there would be some finite omega at which it would have to be torn apart.

Still, it's nice to imagine spinning a sufficiently strong ring so rapidly that it managed to fit inside a smaller radius than it started out with!

 

[pervect]

Hi, and welcome to PF, gregegan!

I believe that your equations can be solved exactly for the rotating hoop if one takes n as the primary variable.

1/n is (for lurkers) the amount by which the hoop stretches, i.e. n=.5 represents the material elongating to twice its length, see Greg's original webpage at

http://gregegan.customer.netspace.net.au/SCIENCE/Rings/Rings.html

I'm getting the following. Let \rho_0 be the initial density of the hoop, and let k be the Young's modulus as per the above webpage

Then we can write the stress-energy tensor in your u,r,w frame which we've been calling the Langevian frame here using geometric units as \rho,P, where \rho is the density and P is the pressure

<br /> \rho = n \rho_0 + \frac{k}{2} n \left(1-\frac{1}{n}\right)^2<br />

<br /> P = k \left(1-\frac{1}{n}\right)<br />

this follows most directly from your other webpage

http://gregegan.customer.netspace.net.au/SCIENCE/Rindler/SimpleElasticity.html

We observe that there is a minimum n factor for which the weak energy condition is satisfied, this is

<br /> n_{min} = \sqrt{\frac{k}{2 \rho_0 + k}}<br />

While n is a minimum, 1/n is the amount by which the hoop expands, so this represents a condition of maximum expansion.

Given the above, can then solve for the tangential velocity of the hoop via the relationship:

v^2 = -P/\rho

This relationship between presssure(tension), density, and the radial velocity v of the hoop was derived by setting the divergence of the stress-energy tensor to zero.

This simplifies (using Maple) to
<br /> v^2 = {\frac {2 \,k \left( 1-n \right) }{k \left( 1-n \right) ^{2}+2\,{n}^{2}<br /> {\it \rho_0}}}<br />

We notice that when n=nmin, v=1, and that for n>nmin, a series expansion puts v^2<1, as it should be. So for any velocity <1, the positivity of the stress energy tensor is (just barely) satisfied.

Now we can find the radius of the hoop explicitly via the relationship

<br /> r = \frac{r_0}{n} \sqrt{1-v^2}<br />

How this was derived takes a little explaining. Basically, the difference in angles between any two points on the hoop is assumed to be constant as the hoop is spun up. So two points 1 degree apart initially in the lab frame will always be 1 degree apart. This is part of the spinup process which was discussed at some length in previous posts.

We can then say that the intial proper length of a small section of the hoop before spinup was r_0 d\theta. After spin up, the proper length is \frac{r d \theta}{\sqrt{1-v^2}}

But we know that the ratio must equal 1/n, hence the above equation.

We can find r explicitly by substituting the value we just calculated for v^2 into the formula above. We can also find \omega = v/r as a function of n the same way.

I get
<br /> r = \frac {r_0}{n}\,\sqrt {{\frac {2\,n^{2}{\it \rho_0}+k n^{2}-k}{2\,n^{2}{<br /> \it \rho_0}+k n^{2}-2\,kn+k}}<br />

<br /> \omega = \frac{n}{r_0}\sqrt {\frac {2 \, k \left( 1-n \right) }{2\,{n}^{2}{\it \rho_0}-k \left( 1-{n}^{2} \right) }}<br />

we note that \omega goes to infinity just as n=nmin when the weak energy condition is violated.

Thus I also find that a hyperelastic material satisfying the weak energy condition can (just barely) be "spun up" to a point where it shrinks.

I think that the total energy E and angular momentum J of the spun-up hoop are also of some interest - in fact, this was what originally motivated me to try and perform these caclulations.

This can be found by converting the stress-energy tensor back into the lab frame, and integrating over the volume. I have not yet done these calculations for the hyperelastic hoop yet. I think that my simpler "Born rigid" model may even provide more insight here, being more tractable (though less physical).

 

[pervect]

While the formulas above should now be stable (it took me a while to get some of the latex to come out correctly), and I have checked them against my worksheet for typos, they still probably need to be checked to make sure they actually satisfy Greg Egan's original differential equations. I haven't done that yet, I'm not quite sure of the best way to proceed.

 

[gregegan]

While the formulas above should now be stable (it took me a while to get some of the latex to come out correctly), and I have checked them against my worksheet for typos, they still probably need to be checked to make sure they actually satisfy Greg Egan's original differential equations. I haven't done that yet, I'm not quite sure of the best way to proceed.

Hi pervect, there's no need to check your formulas against the differential equation; because you're looking at a zero-width "hoop", not a finite-width ring, you can just plug them into the scalar equation I derived for that special case.

I did this, and they do satisfy the equation. It is also possible to solve the equation analytically for r in terms of r0 (a pretty messy cubic for r^2) or for r0 in terms of r (which I give explicitly, it's just a quadratic), but your way of parameterising the solution space has a nice physical meaning.

 

[pervect]

Great! Glad to hear we're getting the same results.

As far as the energy and momentum goes for the rotating hoop go

The energy (if geometric units are used, this is also the mass) of the hoop when it was not spun up was just

2 \pi r_0 \rho_0 A

where we've introduced A, the cross-sectional area of the hoop.

Using the results I got way back in post #10 (also in #19)

the energy of the spun-up hoop should be

2 \pi r \rho A (1+v^2)

A shouldn't change as the hoop is spun up, and we can use the expressions for \rho[/itex], r and v^2 as a function of n already derived.<br /> <br /> If there were no tension in the hoop, we would see a factor of \gamma^2 term multiplying the density, rather than (1+v^2).<br /> <br /> Similarly, the angular momentum should be just be p x r, where p is the linear momentum (the integral of the linear momentum density T^{0i} ) or<br /> <br /> L = 2 \pi r^2 v \rho A = 2 \pi r^3 \omega \rho A<br /> <br /> Here the tension in the disk introduces terms which exactly cancel the usual factor of \gamma^2 so that it does not appear in the final expression.

 

[pmb_phy]

I'm not sure whether you addressed this or not but let me mention it just in case it didn't come up. Any physical loop must expand when it is spun up. You'd have to take a counter action to force the radius to maintain a radius of your desire.

Pete

 

[pervect]

If you spin up a hoop of constant radius, and mark two points on the surface of the hoop, you will find that the proper distance between these points increases at high enough relativistic velocities.

Thus a "Born rigid" hoop will have its radius contract when it is spun up, but of course such a hoop is not very physical.

Adding elasticity to the hoop requires a more detailed model, which Greg Egan has supplied, in the form of a hyperelastic hoop. Hyperelastic hoops are more physical than Born hoops, however the caclulations show they still shrink at high enough velocities, just as the less physical Born rigid hoops did.

Are hyperelastic hoops "physical"? It's hard to say, but in the region above, where the hoop is contracting, they still satisfy the weak energy condition, and they also still have a dynamic "speed of sound" in the material lower than 'c'.

We don't have any materials nearly strong enough to actually exhibit this sort of effect, however - one would need materials strong enough that their tension could approach their density.

I did some back of the envelope calculations for a carbon nanotube, for instance, and determined that elastically they'd stretch about 10% in length at their yield point, which would optimistically be 8-9 km/sec radial velocity. At this velocity there would be aroudn a part per billion relativistic contraction.

 

[pervect]

Here's one of many possible points where the radius just starts to shrink with increasing n or increasing omega, i.e. dr/dn = 0.

rho0=1
k = .5730994
n = .7

This represents a material where the velocity of sound is about 3/4 the speed of light

This gives rho = .737 and P = -.246, so it's within the weak energy condition by a fairly large margin.

I haven't fully explored the solution space (dr/dn=0). I've seen possibilities with a lower speed of sound in the material that are closer to violating the weak energy condition because of a higher stetch factor, and materials which have a lower ratio of P/rho that have a higher sound velocity.

 

[MeJennifer]

Thus a "Born rigid" hoop will have its radius contract when it is spun up, but of course such a hoop is not very physical.

There is no such thing as a Born rigid rotating hoop, not even in theory!

In an accelerating Born rigid rod all surfaces orthogonal to the direction of acceleration enjoy a coherent proper distance but an incoherent proper time. Alternatively an accelerating rod that maintains a coherent proper time between these surfaces would undergo stress and possibly break since the distances between the surfaces cannot be held coherent.

But in the case of a rotating hoop there are no surfaces of coherent distance or coherent time, so Born rigid rotation is not just practically but also theoretically impossible.

 

[pervect]

There is no such thing as a Born rigid rotating hoop, not even in theory!

In an accelerating Born rigid rod all surfaces orthogonal to the direction of acceleration enjoy a coherent proper distance but an incoherent proper time. Alternatively an accelerating rod that maintains a coherent proper time between these surfaces would undergo stress and possibly break since the distances between the surfaces cannot be held coherent.

But in the case of a rotating hoop there are no surfaces of coherent distance or coherent time, so Born rigid motion is not just practically but also theoretically impossible for a rotating hoop.

I think my answer to this (in case you didn't guess, I disagree) is pretty much on record in my earlier posts. Perhaps we can entice Greg Egan into giving his view on this matter. (And, perhaps not, we'll see.).