Stress-energy tensor of a wire under stress

[gregegan]

Since there are two competing contributions to the centroid-frame energy density as the material becomes more stretched, it makes perfect sense that this quantity should reach a peak and then decline.

Sorry, this statement was a bit careless. The energy density in the centroid frame actually falls with increasing omega, reaches a minimum, then starts to rise. The initial fall is simply because the hoop is being stretched, spreading its rest mass more thinly, an effect which at first will completely dominate over anything related to tension or potential energy.

But the behaviour of the radius, which first increases and then decreases, means that the total energy in the centroid frame first rises, then falls. The radius must fall eventually, because relativistic length contraction eventually overwhelms the expansion of the hoop material in its rest frame.

The gist of it is that there are so many competing terms and factors pulling in different directions that it's not all that surprising that the total energy is not monotonic in omega.

 

[pervect]

In combination with taking the total angular momentum of the hoop, what the math is telling me that at some value of omega, a spinning circular hoop will have less energy AND less angular momentum when we increase the angular velocity omega. This appears to be true for both Greg Egan's hyperelastic hooops, and the Born rigid hoops I analyzed.

What these equations would be telling us, if we interpret them literally and assume stability, is that a circular hoop spun up at some velocity would spin up even faster if we applied a negative torque to it, a torque that would reduce it's angular momentum and reduce its energy. I.e., if we applied some brakes to the hoop, the hoop would actually spin faster, transferring energy from the hoop to the brake pads. Energy which we could use to say, run a steam engine (or whatever). Eventually, the hoop would transfer a good proportion of its rest energy to the brake pads.

This is a pretty bizarre result. I have a strong suspicion, though, that an actual hoop would not so nicely provide us with a source of energy while shrinking its radius and increasing omega in a symmetrical manner. I have a strong suspicion that the assusmption of circular symmetry will fail, that the hoop will behave in a rather unstable manner under these conditions.

Unfortunately, I'm not quite sure how to go about analyzing the stability analytically. I think this boils down to evaluating the stability of $$\nabla_b T^{ab} = 0$$ around the operating point of the [tex]T^{ab}[/itex] for some particular rotating hoop.

 

[gregegan]

A question: in its own frame, the hoop is elongating, not shrinking. And there is a limit to the elongation factor. Previously, we had been assuming that the hoop simply broke when the maximum elongation factor was exceeded. But now, you are suggesting that that can't happen. So what does happen when the hoop speeds up enough that n < n_min?

That's a really good question!

The first part of the answer is that in the absolutely pure hyperelastic model, the force that holds the material together is an infinite-range force. You really can't break the hoop, no matter what you do to it, because you can never get one part of it out of the range of attraction of the rest. And the way the physics works out, at least for the case of zero-width hoops, this never actually clashes with the weak energy condition. It gets asymptotically close to violating it, but it never quite does so. Even as omega goes to infinity we always have n>n_min, where n_min is set by the weak energy condition rather than any property of the material.

OK, but what about something a bit closer to reality? What about a force that mimics the hyperelastic force out to a certain threshold, and then dies away? So long as the threshold is high enough, that ought to be able to take our hoop into the strange zone where its centroid-frame energy is less than its rest mass, but it doesn't seem to offer any guarantee that the hoop won't come apart after we get there, leading to the apparent paradox of explaining how the fragments can separate when they don't have enough energy.

I think the answer is that you have to consider the potential energy diagram for this force, and its derivative, which gives the tension. Along the x-axis is 1/n, the expansion factor for the material. Starting out at 1/n=1, the PE is 0, and flat, so the tension is zero. Initially the PE rises quadratically, with the tension increasing linearly. So far, this is just like a spring. If it went on like this forever, we'd have the idealisation of hyperelasticity.

But we want our new force to lose its grip eventually, which means we want the PE to stop climbing, hit a maximum, and come down to zero. To do that continuously, though, will have consequences for the tension. As the PE stops increasing and reaches its maximum, the tension will come back to zero.

But it's only the tension being so high that allowed the centroid-frame energy to fall below the rest mass. As soon as the tension starts to drop, there will be a barrier rising up in the centroid-frame energy that needs to be overcome in order for the material to expand any further. The material can break eventually, it can get free of the force holding it together, but only if energy is supplied to get it over this barrier.

The material can't "just break", as if someone instantly pulled the plug on the force. Of course the potential energy can change as rapidly as we like, but that will only make the wall of the centroid-frame energy barrier steeper.

 

[pervect]

I've had a couple of more thoughts:

The first thought is to model a "breakable" hoop that has a limit on pressure. I eventually came up with

for 0<s<4
P = (1/4)(s-1)(s-4)

for s>4
P=0

where s = 1/n is the "stretch factor"

Peak pressure occurs at s=2.5. The speed of sound should be less than 1 at s=1. (k should be .75 if you take the slope). The weak energy condition should be imposed naturally by the hoop elongating indefinitely and not have to be added artificially.

I won't go into the details of the calculation, but if I'm doing it correctly it appears that this hoop still has a maximum in the E curve, at s=1.62.

The second thought I had is to model the dynamics of the hoop. As a first step, we can keep circular symmetry (which makes the calculations much simpler) but simply allow the radius of the hoop to be a function of time. I haven't looked at the details yet as to how to do this, but I think it should allow us to create hoops with over the statically allowed maximum energy E, and study their evolution.

While E_max appears to be a limit on the amount of energy a static hoop can hold, it should be possible to create a dynamic hoop with more energy than that.

 

[gregegan]

The first thought is to model a "breakable" hoop that has a limit on pressure. I eventually came up with

for 0<s<4
P = (1/4)(s-1)(s-4)

for s>4
P=0

where s = 1/n is the "stretch factor"

That's a nice model! I'd been doing something with an exponential drop-off, but it was analytically intractable. I didn't want to use a piecewise-defined function, because I thought that would be a pain in the neck, but on reflection it's almost irrelevant, because for s>4 in your model we simply know that there are no stable solutions. This model's so much easier to work with; I managed to get quite simple analytic expressions for r, omega and E in terms of s.

if I'm doing it correctly it appears that this hoop still has a maximum in the E curve, at s=1.62

I assume you're using rho_0 = 1, in which case I get the same result. What's more, after reaching the maximum at s=1.62, I think E crosses the rest mass at s=2.19, hits a minimum of about 94% of the rest mass at s=2.45, then climbs up above the rest mass, long before breaking at s=4. (There's a second peak and fall just before the end, but it all takes place at much higher energy.) So as I guessed, even when the material model allows breakage, it can't happen until you get back above the rest mass.

Interestingly, at the same value for s as the energy minimum, r also has a minimum, and omega has a maximum. So as you drive the hoop along the path of monotonically increasing s, its radius first increases, then falls, then increases again; whereas its angular velocity first increases, then decreases. For a given omega less than the maximum possible value, there will always be two solutions with different radii, one much more stretched and energetic than the other.

The second thought I had is to model the dynamics of the hoop. As a first step, we can keep circular symmetry (which makes the calculations much simpler) but simply allow the radius of the hoop to be a function of time.

Good luck! I might try something similar if I have time. One thing this ought to help reveal is whether even these symmetrical solutions are stable to small perturbations in r. In planetary dynamics, the nice thing to do is to fix the angular momentum L, turn omega into a function of L and r, and then use that to plot E as a function of r. Circular orbits then show up as the flat bottom of an energy trough. But in our context, we don't know yet if all our symmetrical solutions are in troughs, or if some of them sit on the top of ridges.

 

[pervect]

Yes, rho(0) was 1, and after the maximum at 1.62, E has a minimum below the rest mass at s=2.44, after which it starts rising. A rather strange-looking curve, especially considering how simple the defining function was.

I expect that at least some of these solutions represent unstable equilibrium solutions though, rather than a state that would actually represent a stable configuration of the hoop.

Certainly a Newtonian hoop would not be stable with a material that weakened as it stretched (say from s=2.5 to s=4).

Find out for sure which ones are stable in the relativistic case is the next big task (even the simpler subtask of considering only the stability of radially symmetrical hoops looks pretty involved.) While I think the equations needed should be generated from the divergence relations T^ab;b=0, T^ab won't be nearly as simple as it was before.

 

[gregegan]

Find out for sure which ones are stable in the relativistic case is the next big task (even the simpler subtask of considering only the stability of radially symmetrical hoops looks pretty involved.) While I think the equations needed should be generated from the divergence relations T^ab;b=0, T^ab won't be nearly as simple as it was before.

I think the stability question for axially symmetric hoops can be answered without computing the full dynamic equations that allow r to be a function of t.

I computed angular momentum L as a function of v^2 and r, solved the cubic in v^2 for some fixed L, and fed that v^2 into E, to get E as a function of r, for some fixed L. Note that this is all done without imposing div T=0, because the point is to look at adjacent states which are not stationary solutions (also, this is back in the hyperelastic model, not the breakage model).

The plots I get show the stationary solutions lying at the bottom of troughs for E, even when E is falling with increasing omega, and even when E < rest mass. In other words, perturbing r in either direction always means adding energy to the stationary solutions, so they ought to be stable, at least under perturbations that respect the axial symmetry.

If there's any instability it must involve the hoop losing its circular shape. It's conceivable to me that the stretched hoop might be vulnerable to crinkling at some point where expanding its length (and hence increasing the tension) lowers the overall energy, but I'm still trying to think of a reliable way to check this without a week's worth of algebra and/or numeric computations.

 

[Chris Hillman]

Reminder: Things to Bear in Mind

I am glad to see Greg Egan has taken up my discussion with pervect, and I hope that some of the many interesting and physically/philosophically/mathematically interesting issues related to rotating matter will be fruitfully discussed by them in this thread.

I look forward to following your further discussion. Silently, of course, lest Chris Hillmann wishes to exclude the public.

Ich, FYI, I feel more comfortable limiting my conversations here to individuals who have divulged their identity to me (perhaps by PM).

I don't know anything about your background or your motivations for commenting in this thread, but FYI the reason I feel that it would be best for posters other than Greg and pervect to keep silent (unlike middle Egyptian, English lacks a verb for keeping quiet--- is this why American tourists are so loud?) is that there are many subtle issues here which experience shows are difficult to explain to persons lacking a strong background in math, physics and even a philosophical bent. I don't wish to see either of them distracted by naive questions--- or even worse, foolish statements based on neglecting known technical, physical, or philosophical issues, particularly if these have already been mentioned earlier in the thread. I also feel that those who have made no attempt to get some sense of the vast literature on rotating relativistic matter are unlikely to play a helpful role here.

In recent days, I have been working on some related issues but am unhappy with the fruit of my labors (things less thoughtful investigators would sloppily label "exact solutions", but which I currently suspect are physically misleading), so I'll bow out of this thread now, although I hope everyone else will let pervect and Greg continue their discussion. But I'd like to leave this thread with one last attempt to stress that there are many subtle issues here, and failure to bear them all in mind will certainly result in conceptual errors, uneccessary confusion, physical absurdities, and nonsense generally.
So here are a few final hints for getting started on thinking about this stuff, mostly addressed to hypothetical intelligent lurkers who are intellectually capable of appreciating subtleties and of bearing multiple issues in mind:

Some important distinctions:

* density and other variables in strained versus unstrained material,

* frame fields (AKA anholonomic bases) versus coordinate bases in a given
chart,

* frame field versus corresponding congruence,

* Langevin frame proper (constant omega, observers move in circular orbits
with constant radius) versus the variable omega generalization (observers
move in constant radius circular orbits but their speed varies),

* congruence (fills up region of spacetime) versus the world sheet of a hoop (doesn't fill up a region of spacetime),

* Born chart ("rotating cyl. chart") versus cyl chart (used in this thread),

* Axel, the inertial observer stationary wrt centroid of disk/hoop, versus Barbarella, a hoop/disk riding observer (if constant omega, she is one of the observers whose world lines are given by Langevin congruence for that omega),

* radius, mass-energy, angular momentum of hoop measured by Axel (makes sense), versus the same as measured/computed by Barbarella (won't make sense, at least not without very careful qualification),

* clock synchronization by Axel and friends (makes sense) versus by Barbarella and friends (impossible even for a hoop--- c.f. Sagnac effect),

* Born rigid congruence (vanishing expansion tensor) versus other notions of "rigidity",

* pervect's position (no problem) versus my position (nothing shown either way) on pervect's claim that it is possible to define a notion of spinup of an elastic hoop (with radius expanding or contracting as described by Axel) which remains rigid throughout the evolution (in the sense that the world lines in the world sheet of the hoop can be enlarged to a Born rigid congruence),

* Alleged orthogonal spatial hypersurfaces for Langevin observers (doesn't exist, since Langevin congruence has nonzero vorticity) versus the quotient manifold (quotient of Minkowski spacetime by the Langevin congruence) (which does exist; indeed the Langevin-Landau-Lifschitz metric applies to this Riemannian three-manifold),

* multiple operationally distinct notions of "distance in the large" for accelerating observers even in flat spacetime--- c.f. problems with speaking carelessly about "the circumference of the hoop measured by Barbarella and friends",

* constant omega versus nonconstant omega (I discussed a generalization of Langevin congruence to variable omega, but these observers maintain constant radius as measured by Axel, so aren't suitable for discussing pervect's alleged "Born rigid" spinup of a hoop,

* crude conditions on "physical acceptability" like energy conditions, speed of sound, versus "physically realistic" models,

* making a computation versus interpreting it; a good physicist never omits the latter and in fact may spend most of his effort on this task,

* conclusions which depend upon choice of a physical model and those which do not; I feel that some important points require studying specific physical models and considering limits in order to have confidence that "any
reasonable model" would have such and such qualitative behavior.

* things which have been well-defined (e.g. Born rigid, radar distance) versus things which so far have not been well-defined (pervect's alleged Born rigid spinup procedure, which may be related to an alleged notion of "rigid spin-up" suggested by Grunbaum and Janis, which I also currently consider unconvincing).

* exact solutions of ODEs mentioned by Greg, pervect and myself (typically hard to obtain) versus approximations via perturbation theory (which can also yield valuable physical insight),

* attempting str treatments (pervect and Greg) versus exploring gtr treatments (me only),

* Newtonian limit (str or gtr) versus weak-field limit (gtr); I advocated latter as a stepping stone to exact solutions in gtr. I expect to expend more work laying the foundation to interpret such solutions than in actually finding them.

Further general issues:

* what can be neglected? e.g a nonspinning inertial frame for Langevin observers will appear to spin wrt Axel as per Thomas precession.

* which idealizations are "physically acceptable"? "Physically reasonable?"

* what are the criteria for "physical acceptability", anyway?

* perturbation analysis is usually very helpful when things get confusing and formulas get messy, but choice of variables is critical, i.e. this is a delicate art.

And a general reminder:

The literature on rotating disks and hoops is large and spread over many decades, journals, and several languages. None of these authors have taken all relevant considerations into account, so none of them have provided fully correct treatments. Some have come much closer than others, however, in fact much of the literature consists of independently recommiting old errors.

All parties should bear in mind the advice of George Santayana, which I'll paraphrase as the warning that "those who [fail to study past errors] are condemned to repeat [them]." Study the literature, or else forfeit the honorable title of scholar! Grrr!

A good place to begin is the review paper by Gron and papers cited therein:
http://freeweb.supereva.com/solciclos/gron_d.pdf

Last but not least, this list is incomplete.

 

[gregegan]

It's conceivable to me that the stretched hoop might be vulnerable to crinkling at some point where expanding its length (and hence increasing the tension) lowers the overall energy, but I'm still trying to think of a reliable way to check this without a week's worth of algebra and/or numeric computations.

For what it's worth, I've now done some calculations which make me suspect that the hoop probably won't be vulnerable to crinkling, at least from small perturbations.

I computed L and E for a state where some small, arbitrary function delta*f(theta) is added to the radius. I found series expansions for both L and E to second order in delta; the coefficients for both included integrals over theta of f, f^2, and (f')^2. Numerically, I found that requiring L to be constant to second order always resulted in E being constant to 1st order, and with a +ve coefficient for the delta^2 term. In other words, just as with symmetry-preserving perturbations to r around the stationary solutions, these arbitrary small perturbations to the shape seem always to be of higher energy.

 

[pervect]

For what it's worth, I've now done some calculations which make me suspect that the hoop probably won't be vulnerable to crinkling, at least from small perturbations.

Are you at a point yet where you can make some predictions for the stability of the hoop, in either sense, near (after) the point where E reaches a maximum?

I think I'm getting the rather interesting result that for a sufficiently strong hoop, there is a limit to how much energy it can hold (the first peak of E on the curve) but this limit is governed by an implosion failure rather than an explosion failure.

As far as the dynamics go:

If I drop terms of order vr^2, where r is the radial velocity, I find that we only add T^{tr} to the stress energy tensor. This comes from the equation

$$T = \rho \vec{u} \times \vec{u} + P \vec{w} \times \vec{w}$$

and adding a radial component to v while letting w remain unchanged.

There is a second order term T^{rr}, also second order corrections to other terms, which I ignore.

Keeping only this linear terms the continuity equation says simply

$$\frac{\partial T^{01}}{\partial t} = r T^{22}$$

so all we need to do is look at the value of T^22 (in the lab frame) to determine whether or not the radial velocity v, which is proportional to T^{01}, accelerates or deaccelerates.

This gives us a rather physical interpretation of T^22 in the lab frame by the way - we previously calculated that this was zero, now we see again that the continuity equation requires this for an equilibrium hoop. Furthermore, its sign controls expansion or contraction - a positive sign means expansion (more precisely, a positive acceleration of rate of expansion dvr/dt, assuming vr is small).

T^22 ignoring second order corrections is just
$$\frac{\rho \omega^2 r^2+P}{r^2\left( 1-\omega^2 r^2 \right)}$$

so basically it's just the sign of $$\rho \omega^2 r^2+P$$ which should determine the direction of acceleration or deacceleration of expansion.

I am not terribly confident that radial symmetry would be maintained during the implosion process - I just don't see any stable state that a hoop with "too much energy" could reach. For instance, in the "breakable hoop" with P = .25(s-1)(s-4) we see that E eventually climbs up above the first peak, but IMO this portion of the curve is radially unstable.

I'm not too sure about what happens near the first valley of the E curve yet.

 

[gregegan]

The plots I get show the stationary solutions lying at the bottom of troughs for E, even when E is falling with increasing omega, and even when E < rest mass.

I just found something delightfully weird! It was premature of me to jump to the conclusion (on the basis of some numerically derived plots) that all the axially symmetric solutions were stable. There turns out to be a narrow range of parameters where they aren't.

To make the calculations more tractable, instead of trying to express E(r) for constant L explicitly, I used L=constant to implicitly define a relationship between v^2 and r, and then took derivatives of that equation to evaluate the first and second derivatives of E(r). (Conceptually this is all the same as using omega and r, rather than v^2 and r, but the algebra is much simpler using v^2.)

At the solution points, E'(r) was zero, which was reassuring, but the second derivative E''(r) was an expression not guaranteed to be positive. Close to the point where r reaches its maximum, there is a range of values where E''(r) at the solution point is negative, i.e. the solution lies on an energy ridge.

Here's the weird part, though: if you fall off the ridge in the direction of increasing r, you approach an edge to the energy curve, beyond which there are no states which have this value of L. Also, there's an energy trough just inside, so if a hoop started out on the ridge and fell inwards, it would get caught in that trough and presumably oscillate radially between two (quite close) r values. I don't think there's any run-away behaviour, leading to hoops either exploding or imploding, but at this point I wouldn't bet my life on it. Maybe it will take the complete dynamic equations for axially symmetric states to fully understand this, after all.

If anyone feels like checking this out, an example is k=0.32, rho_0=1 (this corresponds to K=0.4 for the K used on my web page), and this phenomenon occurs between n=0.510 and n=0.523, where n is the compression factor, which pervect showed how to use to parameterise the (r,omega) solutions.

 

[gregegan]

Are you at a point yet where you can make some predictions for the stability of the hoop, in either sense, near (after) the point where E reaches a maximum?

I think my last post partly answers that; in the region where r, L and E reach maxima (and of course they don't all reach maxima at exactly the same point), I see some definite instability even for axially symmetric states, but it looks to me as if it's probably contained on both sides.

I'm not studying the breakable hoop at this point, though; it sounds like you are?

I think I need to look more closely at the dynamics.

 

[gregegan]

T^22 ignoring second order corrections is just
$$\frac{\rho \omega^2 r^2+P}{r^2\left( 1-\omega^2 r^2 \right)}$$

so basically it's just the sign of $$\rho \omega^2 r^2+P$$ which should determine the direction of acceleration or deacceleration of expansion.

Sorry to be thick, but I'm completely lost here! The numerator of the formula above will rarely, if ever, be zero. Are you claiming that whenever this expression is non-zero, the hoop is undergoing radial acceleration in one direction or another? That can't be true, or there'd be no constant-radius solutions at all; v_r would always be changing, even if it was initially zero.

What is it I'm missing here?

 

[pervect]

I think my last post partly answers that; in the region where r, L and E reach maxima (and of course they don't all reach maxima at exactly the same point), I see some definite instability even for axially symmetric states, but it looks to me as if it's probably contained on both sides.

I'm not studying the breakable hoop at this point, though; it sounds like you are?

I think I need to look more closely at the dynamics.

Typically r reaches a maximum first - I've always found L and E reach a maximum at the same time, however. I suspect that the later (L&E reaching a maximum at the same time) is required for consistency of the physics, that any torque that increases L must be in the same direction as omega and hence increase E, and that any torque that decreases L must decrease E, i.e. that the slopes of L and E must have the same sign for physical reasons. If they must have the same sign, they must switch signs at the same time.

 

[pervect]

I see at least one very fundamental point that probably needs to be addressed in the Gron paper, conveniently on-line in draft (?) form at

http://freeweb.supereva.com/solciclos/gron_d.pdf

This is a true, but I think potentially misleading remark made by Gron that I think fuelled some of the earlier rather long discussion of the spin-up.

Gron writes:

3. Due to the relativity of simultaneity Born rigid rotating motion of a ring with angular acceleration represents contradictory boundary conditions.

Going into more detail, one can see that it is indeed not possible to EXACTLY accelerate a ring in a Born rigid manner as Gron argues in the section on "Kinematical solution of Ehrenfest’s paradox".

Note that this also applies to the thin ring, not the disk.

What I believe *IS* possible, however, is that by taking the limit as w(t) increases very slowly (the limit in which one takes an infinite amount of time to spin-up the disk) one can also make the change in the proper proper distance ds between two nearby points during the spin-up process less than $r_ch \, ds$, where $r_ch$ is an arbitrarily small number. Note that any pun about r_ch being a small number might well be intentional.

So while the spin-up process is not (cannot) be perfectly Born rigid, it can approach the ideal as closely as desired.

 

[gregegan]

Typically r reaches a maximum first - I've always found L and E reach a maximum at the same time, however. I suspect that the later (L&E reaching a maximum at the same time) is required for consistency of the physics.

Yeah, you're right. I checked the derivatives of both E and L wrt n, and it's precisely the case.

I'm having a lot of trouble following what you've done in your dynamics calculations. Before taking the divergence of T, what quantities do you allow to become functions of t? In principle everything is now a function of t: r, omega, rho, P (though of course the material model will link those latter two to the behaviour of r). Are you holding some of these constant as an approximation, in order to look at perturbations around the equilibrium solutions, as opposed to computing whole dynamic trajectories?

 

[gregegan]

Are you at a point yet where you can make some predictions for the stability of the hoop, in either sense, near (after) the point where E reaches a maximum?

To give a more complete answer than I gave before:

As we decrease n from 1, i.e. increase the stretching of the hoop, if you look at the second derivative of E(r) for L held constant:

(1) Initially, E''(r) is +ve, i.e. the stationary solutions are stable.
(2) Before we hit E_max, E''(r) blows up to +ve infinity. This need not be anything pathological in the physics; it just means that at this particular stationary solution, there is literally no perturbation in r possible that will keep the angular momentum unchanged.
(3) E''(r) is then -ve (coming back towards zero from -ve infinity); in this region the stationary solution is unstable, sitting on an energy ridge. However it is contained on the lower-r side by an energy trough, and on the higher-r side by conservation of angular momentum, because there is a maximum r beyond which no more states exist with this L.
(4) Exactly at E_max, E''(r) is zero. The lower-r energy trough and the ridge merge into a "shelf", a point of inflection in the curve (just like the graph of y=x^3), which resists movement to lower r but is unstable under perturbations that increase r. However, once again conservation of angular momentum seems to contain the hoop from exploding, because the curve of constant L stops at a finite r.
(5) Immediately after this, E''(r) is +ve again, and the stationary solutions are stable again.

 

[pervect]

Sorry to be thick, but I'm completely lost here! The numerator of the formula above will rarely, if ever, be zero. Are you claiming that whenever this expression is non-zero, the hoop is undergoing radial acceleration in one direction or another? That can't be true, or there'd be no constant-radius solutions at all; v_r would always be changing, even if it was initially zero.

What is it I'm missing here?

OK, besides being terse, I was in somewhat of a hurry, so let's take a closer look at this. I'm definitely not infallible.

What I did was first to approximate T^ab in a cylindrical coordinate chart (t,r,theta,z) by assuming that only the following terms were nonzero.

T^{rr}, T^{rt}, T^{r theta}, T^{theta theta}

This is all being done in a coordinate basis, so the basis vectors are not normalized.

The second term was added by the assumption that I had a radial velocity, and I argued that T^{rr} could be excluded as it was of second order in radial velocity.

All of these terms were assumed to be functions of r and t, and independent of theta and z.

I did not substitute for the actual functions yet, I kept things symbolic.

[add]You may have already noticed that if we don't refrain from substituting, we get an awful mess that's hard to interpret.

The continuity equation

$$T^{ab}{}_{;b}$$ or $$\nabla_b T^{ab}$$

(using Grtensor to calculate the covariant derivatives) then gave me:

$$-\frac{\partial T01(r,t)}{\partial t} + r T22(r,t) = 0$$

So the only thing being differentiated with t is T01. I didn't substitute in the terms for the variation of T^01 with w, etc. as I probably should have - but if the outgoing radial momentum T01 is positive and accelerating, or negative and deaccelerating, we know the system is unstable. We are looking for feedback that drives T01 to zero, and the equilibrium case has T01 = 0 and not changing with time.

I was rather pleased that there were no obvious "delta-function" type terms resulting from the differentiation of a step function in this particular component of the continuity equation - if you include T^{22}, this is not the case, for instance, you'll see $\frac{\partial}{\partial r}$ of a non-zero function. I haven't been keeping really close tract of the delta functions though, I've been assuming that if the continuous part works out correctly, the delta functions will too.

There are some less obvious delta-function type terms in the first term, though. As the hoop expands, T01 evaluated at a specific (r,t) will jump suddenly from 0 to a non-zero value, implying that its time derivative has a delta function.

 

[gregegan]

Thanks very much for the explanation, I was confused about your coordinate system but it's all clear now. (I also had a weird mental block and failed to recognise why T^22 will be zero in equilibrium solutions ... )

On the stability front ... at least with the axially symmetric constraint, I find even the breakable hoop to be stable away from its energy extrema. I expect its first energy peak to be, er, quasistable in the same odd way that the hyperelastic hoop is, but I can't determine anything yet about the energy valley or the second, higher peak; the algebra is so horrendous that I'm running out of memory in Mathematica.

 

[Chris Hillman]

Clarification

Not everyone knows that the distinguished mathematician Fermat was also a judge and a sometime ambassador who handled important and sensitive negotiations, apparently with great success. Alas, it seems that they no longer make such mathematical diplomats! Certainly I seem to be formed from a different die

I don't wish to see either of them distracted by naive questions

Ich kindly provided me with information which removed my concern so far as he is concerned. Yes, I am very aware that I would clearly be much happier posting in a forum where the other posters are known and trusted individuals. I am working on it!

All I was really trying to say was: please study the WP articles in the versions I cited, Greg's web pages, Gron's review paper, and ideally at least some of the papers cited therein, plus this thread to date, before speaking up, because this is subtle stuff so we don't want to encourage people to reinvent hexagonal wheels.

OK, getting back on track: a quick comment:

I see at least one very fundamental point that probably needs to be addressed in the Gron paper, conveniently on-line in draft (?) form at
http://freeweb.supereva.com/solciclos/gron_d.pdf
This is a true, but I think potentially misleading remark made by Gron that I think fuelled some of the earlier rather long discussion of the spin-up.

[snip]

Going into more detail, one can see that it is indeed not possible to EXACTLY accelerate a ring in a Born rigid manner as Gron argues in the section on "Kinematical solution of Ehrenfest’s paradox".

Right. Studying Gron at the outset can really save a lot of needless fuss by clearing up a lot of misconceptions likely to be shared by most physicists when they first think about this stuff.

Just to be clear, I don't consider Gron's paper definitive at all, just the best review currently available. (To be truthful, it might be "the best out of a field of one", but he did a lot of work and it really is a pretty good review.) As I mentioned, Gron completely overlooks the crucial issue of multiple operationally significant notions of distance in the large for accelerating observers, and he appears to confuse quotient manifolds and submanifolds when he discusses the Langevin-Landau-Lifschitz metric (which describes "the geometry of a rigidly rotating constant omega disk" as a Riemannian manifold, but this is the quotient of Minkowski spacetime by the Langevin congruence, not any submanifold!).

What I believe *IS* possible, however, is that by taking the limit as w(t) increases very slowly (the limit in which one takes an infinite amount of time to spin-up the disk) one can also make the change in the proper proper distance ds between two nearby points during the spin-up process less than $r_{\em ch} \, ds$, where $r_{\rm ch}$ is an arbitrarily small number.

You said that before, but it is not clear to me that this can be controlled over the entire spinup. I.e. at the end of the spinup we need omega to be measurably positive.

Note that any pun about $r_{\rm ch}$ being a small number might well be intentional.

I don't get it. PM?

So while the spin-up process is not (cannot) be perfectly Born rigid, it can approach the ideal as closely as desired.

I don't see why.

 

[gregegan]

On the stability front ... at least with the axially symmetric constraint, I find even the breakable hoop to be stable away from its energy extrema. I expect its first energy peak to be, er, quasistable in the same odd way that the hyperelastic hoop is, but I can't determine anything yet about the energy valley or the second, higher peak; the algebra is so horrendous that I'm running out of memory in Mathematica.

I finally whipped the algebra for the breakable hoop into shape, reducing the question of stability to the sign of a ratio of two polynomials. I still need to use numeric methods to see when this ratio changes sign, because the polynomials are of order 5 and 6, but with high precision arithmetic I'm reasonably confident of the results, which are stable under changes of precision past a certain point.

As you'll recall, as you increase the strain factor s the energy of the breakable hoop rises to a peak, falls from there to a minimum (which lies below the rest mass), rises up to a second peak which is higher than the first one, falls a bit (but not below the rest mass this time), and then the hoop breaks. There are equilibrium solutions everywhere until it breaks at s=4, but the question is whether these are stable or unstable.

My calculations give the following picture of the stability of the breakable hoop (assuming it's constrained to remain axially symmetric):

(1) Except where mentioned below, the equilibrium solutions are stable.

(2) Around the first energy peak, the solution is unstable, but constrained in the same way as the hyperelastic hoop: an energy trough not far away in the direction of decreasing r keeps it from shrinking too much, and conservation of L keeps r from increasing too much, because the constant-L curve does not extend past a certain r. [Just to be 100% clear, when I talk about energy rising and falling around a solution like this, I'm no longer referring to the plot of E vs s for the equilibrium solutions; rather, I'm imagining that we've computed L for a particular equilibrium solution, then, holding L fixed at that value, we're computing E as r varies, without requiring the states we're considering to be in equilibrium. This is similar to the kind of perturbation that demonstrates why a circular planetary orbit is stable.]

(3) A similar thing happens close to the energy minimum, except the roles of increasing and decreasing r are reversed. That is, there is a small region where the equilibrium solutions are unstable, but an energy trough not far away in the direction of increasing r keeps the hoop from expanding too much, and conservation of L keeps it from shrinking too much, because the constant-L curve does not continue below a certain r.

(4) From the second, higher energy peak, the solutions become unstable, with nothing to constrain them from increasing r. So once the hoop is over that second peak, the tiniest perturbation will see it rapidly enlarge and then break.

 

[gregegan]

In case anyone else has been wondering exactly how to reconcile what I've been saying about stability with what pervect has calculated for dynamic hoops, I've just realized that they're in perfect agreement.

I've been looking at the total energy of hoops, and checking the sign of the second derivative of E wrt r (holding L constant) near an equilibrium solution; if it's positive, I claim the equilibrium is stable.

Pervect has explained how the sign of the T^22 term in the stress energy tensor (where the coordinate 2 is theta in lab-frame polar coordinates) determines the sign of the second rate of change of the hoop's radius; of course this is zero at an equilibrium. So for the equilibrium to be stable, we need the derivative of T^22 wrt r to be negative (again holding L constant), so that any small change in r is resisted.

It turns out that these two conditions are (as you'd hope) precisely the same. It's clear on physical grounds that they ought to be, unless there's something seriously wrong with the whole analysis, but I've done explicit calculations now, and the 2nd derivative of the energy is just a negative constant times the first derivative of T^22 (or to be precise, of the equivalent coordinate in an orthonormal basis, rather than the cylindrical coordinate basis).

Or putting it another way, the force associated with this tangential pressure, T^22, can be seen to come directly from the derivative wrt r, holding L constant, of the hoop's total energy.

 

[pervect]

What I'm not quite sure of is how to figure out T for the perturbed hoop.

If we pluck a string, the vibrations are perpendicular to the string - can we say the same for the hoop? Can we assume that (for small vibrations) the tension in the hoop is constant? What about the density, rho? (constant or non-constant), and a similar question for the vector w in the expression

$$T = \rho \vec{u} \times \vec{u} + P \vec{w} \times \vec{w}$$

is that indepedent of our radial pertubation, or dependent on it?

 

[gregegan]

What I'm not quite sure of is how to figure out T for the perturbed hoop.

If we pluck a string, the vibrations are perpendicular to the string - can we say the same for the hoop? Can we assume that (for small vibrations) the tension in the hoop is constant? What about the density, rho? (constant or non-constant), and a similar question for the vector w in the expression

$$T = \rho \vec{u} \times \vec{u} + P \vec{w} \times \vec{w}$$

is that indepedent of our radial pertubation, or dependent on it?

I'm not very happy with what I've done with this myself so far, but I'll mention some ideas that I've been mulling over for Version 2, if you don't mind that they haven't been road-tested and might turn out to be misguided. I'm unlikely to have time to do any more serious work on this until the weekend.

I think there are two aspects to consider:

(i) pressure waves that travel tangentially around the hoop, displacing material purely by changing its angular coordinate; and

(ii) "plucked string" type waves that involve purely radial displacements of the material.

Now, large enough mode (i) waves will induce a radial acceleration, because the change in tension will mess up the balance between tension and centrifugal force, but I think it's still meaningful to consider pressure waves that are so small that this can be ignored, and this mode can be analysed separately.

With case (ii), the usual treatments of small waves in a string under tension assume constant tension, and I don't see any reason why that's any less reasonable here. So I'd leave the tangential tension unchanged, and the vector w, and consistent with this, stick to the approximation that nothing has been stretched or compressed in the tangential direction.

However, the trick to getting a sensible tensor T is going to be to treat this plucking as a new degree of elastic freedom, with its own contribution to the potential energy density and a consistent pressure associated with that. (If you follow those rules, whole components of the divergence vanish identically without any further effort; if you don't, you find it's impossible to make div T vanish.) In other words, we need to add another simple "spring equation" term to rho, and we need to add another pressure term, associated with the appropriate spatial direction, to T.

So my guess for case (ii) would be to add some function f(theta,t) to r, leave all the tangential stuff unchanged for now, and then figure out a nice simple model that turns f into a contribution to rho, and a new pressure, in much the same manner as you would when working out the force and energy associated with an orthogonally plucked string.

 

[pervect]

I'm still hung up on computing the scalar n and the vector $\vec{w}$ as per

your webpage

It's easy enough to come up with a displacement model

$$r(t,\phi) = r_e + \delta_r(t,\phi)$$

$$\theta(t,\phi) = \phi + \omega t + \delta_{\theta}(t, \phi)$$

This gives r and $\theta$ as a function of time a point that started out at the equilibrium value of r $r_e$ and $\theta = \phi$, the $\delta_r$ and $\delta_{\theta}$ being the pertubation. We can even try and simplify things by making $\delta_{\theta}$ 0 if it helps significantly as you suggest, but it's probably worth a try to keep them both and see if we can formally separate them.

Calculating the 4-velocity is reasonably obvious I think.

The scalar n should give us $\rho$ and P. And we also need $\vec{w}$. If we have the 4-velocity u, $\rho$, and P, and also $\vec{w}$, we can compute the stress-energy tensor via

$$T = \rho \vec{u} \times \vec{u} + P \vec{w} \times \vec{w}$$

And setting $$T^{ab}{}_{;b}=0$$ should give us the differential equations.

One other point - we probably want to linearize everything, to get rid of anything quadratic in a $\delta$ or a $\partial / \partial \delta$.

Then we should have a linearized differential equation and the linearized stability analysis should be easy enough.

But we need n and $\vec{w}$.

 

[gregegan]

I'm still hung up on computing the scalar n and the vector $\vec{w}$ as per

your webpage

Start with the world lines of the material in the hoop. u is tangent to the world lines, and w is orthogonal to u but tangent to the world sheet of the hoop. Then 1/n is found by taking the w component of the vector separating the world lines of two hoop particles whose relaxed separation is 1, i.e. it tells you how far apart they are in a frame co-moving with the hoop.

What I was contemplating doing was something closer to the 2+1 dimensional approach I use in my page on rotating rings. Even though this hoop is so skinny that we can think of it as essentially 1-dimensional in some contexts, I think the fact that it is being deformed radially means that it can usefully be analysed by means of a second pressure/tension term, i.e.

T = rho u (X) u + p_1 w (X) w + p_2 r (X) r

where rho must now include potential energy for two degrees of freedom. Where the problem doesn't quite follow the "fully 2-d" nature of an elastic ring or disk is that we're not considering the hoop to have any radial thickness, or at least none that can actually vary, so we don't have compression factors n_1 and n_2, we just have a single factor n that modifies the densities, i.e. converts them from relaxed to physical.

So I guess you've actually got a choice here. If you want to make w weave back and forth to follow the curve of the perturbed hoop exactly, I think you'll get everything you need from the same rho and p as functions of n that we've used for hoops so far, because w, and hence n, will register the full perturbation. So you could try it that way if you prefer, and if the computation turns out to be tractable.

The other way, which is what I was thinking might be simpler, would be to lock w to the shape of the unperturbed hoop, and then introduce a second pressure term, p_2, with its own contribution to rho, which would be computed as functions of the radial perturbation.

For small perturbations, the two approaches should agree.

 

[gregegan]

Start with the world lines of the material in the hoop. u is tangent to the world lines, and w is orthogonal to u but tangent to the world sheet of the hoop. Then 1/n is found by taking the w component of the vector separating the world lines of two hoop particles whose relaxed separation is 1, i.e. it tells you how far apart they are in a frame co-moving with the hoop.

If you'd like a less abstract and slightly more operational version of this:

$$u = normalised (\partial_t)$$
$$w = normalised (\partial_{\phi} + g(u, \partial_{\phi}) u)$$
$$n = r_0 \partial_w \phi = r_0 w^a \partial_a \phi$$

So u is just the normalised tangent to the world line, w is a unit vector which lies in the plane spanned by u and $\partial_{\phi}$, i.e. the tangent plane to the world sheet of the hoop, and is orthogonal to u. And since $r_0 \phi$ is just the relaxed, original distance between points on the hoop, $r_0 \partial_w \phi$ is the derivative of that original distance along a tangent to the current hoop, as measured in a co-moving frame, so it tells us how much the material has been compressed.

 

[pervect]

If you'd like a less abstract and slightly more operational version of this:

$$u = normalised (\partial_t)$$
$$w = normalised (\partial_{\phi} + g(u, \partial_{\phi}) u)$$
$$n = r_0 \partial_w \phi = r_0 w^a \partial_a \phi$$

So u is just the normalised tangent to the world line, w is a unit vector which lies in the plane spanned by u and $\partial_{\phi}$, i.e. the tangent plane to the world sheet of the hoop, and is orthogonal to u. And since $r_0 \phi$ is just the relaxed, original distance between points on the hoop, $r_0 \partial_w \phi$ is the derivative of that original distance along a tangent to the current hoop, as measured in a co-moving frame, so it tells us how much the material has been compressed.

Right, thanks. If we let
$$\vec{p} = \partial_{\phi}$$

then
$$\vec{w} = normalize(\vec{p} + \alpha \vec{u})$$

with $\alpha$ being some constant (I think this is your g? We're running out of constants, but I don't want to confuse it with the metric tensor).

Since we also have

$$\vec{w} \cdot \vec{u} = 0$$ we also have
$$(\vec{p} + \alpha \vec{u}) \cdot \vec{u} = 0$$

Using the -+++ sign convention implies
$$\vec{u} \cdot \vec{u} = -1$$
so
$$\alpha = \vec{p} \cdot \vec{u}$$

[edit]Looks like I still need to think about n, though. Let's try this again
[tex]d\phi[/itex] can be considred to be a one-form, which we could also write as $$\nabla_a \phi$$. It's easy to confuse this one-form with the scalar version, unfortunately.

Since $w^a$ is a unit vector, $w^a d\phi$ will be a scalar, the new length. Since $\phi$ is preserved during spin-up, the original length was just $r_0 d\phi$. The stretch factor s, 1/n, should be the new length over the original length, which means (?)

$$s = 1/n = \frac{w^a d\phi}{r_0 |d \phi|}$$

Something's still a little funky with that factor of r_0, though. Well, I need to think about it more.

Assuming this n issue gets sorted out, I think I want to leave the linearization step till last.

Because we are doing pertubations, we could take a Hamiltonian or Lagrangian density approach in terms of the displacement functions, but from what I'm reading in Goldstein, this will just give us $$\nabla_b T^{ab} = 0$$ anyway (I'm reading about pg 565).

 

[gregegan]

If we let
$$\vec{p} = \partial_{\phi}$$

then
$$\vec{w} = normalize(\vec{p} + \alpha \vec{u})$$

with $\alpha$ being some constant (I think this is your g? We're running out of constants, but I don't want to confuse it with the metric tensor).

g is the metric tensor! g(u,p) is just another way of writing the dot product between vectors u and p, sometimes used to make it clearer that the dot product is not just the usual Euclidean one, but depends on the particular metric, which of course in this case is just the flat spacetime one. So what I've written for w in order to make it orthogonal to u is the same as the answer you get.

Since $w^a$ is a unit vector, $w^a d\phi$ will be a scalar, the new length.

No, the scalar product of w and $d\phi$, which you're writing here as $w^a d\phi$ (and which is just the derivative of $\phi$ in the direction of w, or $w^a\partial_a \phi$) is actually $n/r_0$. Why? Because original distance is $r_0 \phi$, and the "faster" original distance changes as you move in the w direction, the more compressed the material. If n=1, then as you move in the w direction, original distance changes at the same rate as current distance, so you get a derivative of 1. If n>1, i.e. the material is compressed, then you get a change of 1 in the original distance after traveling just (1/n) in the w direction, so the derivative is the reciprocal of that, n. So you have:

$$n = r_0 w^a\partial_a \phi$$

 

[gregegan]

$$u = normalised (\partial_t)$$
$$w = normalised (\partial_{\phi} + g(u, \partial_{\phi}) u)$$
$$n = r_0 \partial_w \phi = r_0 w^a \partial_a \phi$$

Ah, I've just realized that there's an even simpler way to get n. (It's hard to keep all the definitions straight sometimes, because we're working in a 1+1-dimensional submanifold within 2+1 spacetime, and certain things that work in other contexts either don't work here, or need extra care; for example, $d\phi$ isn't actually defined on the whole spacetime, because $\phi$ is only defined on the submanifold, i.e. the hoop's world sheet.)

If we write:

$$y = \partial_{\phi} + g(u, \partial_{\phi}) u$$
$$w = y/|y|$$

then we simply have:

$$n = r_0/|y|$$

This saves computing all the tedious partial derivatives of $\phi$ with respect to the spacetime coordinates.

Why is this formula true? Because $\partial_{\phi}$ is a vector that takes us from one world line to another that is an (infinitesimal) unit of $\phi$ away, and y is just the projection of $\partial_{\phi}$ into the co-moving frame's definition of space. If we divide by $r_0$ to turn the separation from a unit of $\phi$ to a unit of original distance, we simply get:

$$s = 1/n = |y|/r_0$$

 

[pervect]

Let's use an unperturbed hoop so that
$$\theta = \phi + \omega t$$

and calculate the components of all of these vectors in a cylindrical coordinate basis.

[t,r,$\theta$,z]

Letting $$\gamma = \frac{1}{\sqrt{1 - r^2 \omega^2}}$$

$$u^a = \left[ \gamma, 0, \gamma \omega, 0, \right]$$
$$y^a = \left[\omega \gamma^2 r^2,0,\gamma^2,0 \right]$$
$$w^a = \left[\omega \gamma r,0,\frac{\gamma}{r},0 \right]$$
$$\partial_a \phi = d \phi = \left[0,0,1,0 \right]$$

the last for $d\phi$ being a one-form and not a vector.

also |y| = $\gamma r$

The values for u and w match Chris Hillman's Langevian chart
http://en.wikipedia.org/w/index.php?oldid=53957524$$n = r_0 / |y| = \frac{r_0}{\gamma r}$$
seems OK , so I'll use that. If for instance r=r_0, then n=$1/\gamma$ or $s = \gamma$.

But I can't see where I'm going wrong with the earlier expression
$$n = r_0 w^a d\phi = \gamma \frac{r_0}{r}$$

I suspect I'm not taking the "special care" you mentioned.

 

[gregegan]

But I can't see where I'm going wrong with the earlier expression
$$n = r_0 w^a d\phi = \gamma \frac{r_0}{r}$$

I suspect I'm not taking the "special care" you mentioned.

Yeah, the problem is that $\phi$ and $d\phi$ aren't actually defined except on the 2-dimensional world sheet of the hoop, so I was sloppy when I wrote:

$$n = r_0 w^a \partial_a \phi$$

because I should have made it clear that we ought to be summing over components of w and coordinate derivatives in the 2-dimensional world sheet, not in the wider spacetime coordinates. Because w is a tangent to the world sheet, this directional derivative is perfectly well defined, but it's easy to trip up in actually computing it.

But rather than rambling on about that at length, I'll just say it's safer, less confusing, and probably easier to justify intuitively to use:

$$n=r_0/|y|$$

 

[gregegan]

On reflection, I think it probably is worth saying a little about how we can use the formula:

$$n = r_0 w^a \partial_a \phi$$

even though we have an easier alternative.

Strictly speaking, $\phi$ has only been defined on the world sheet of the hoop. We have world lines of the form:

$$[t,r,\theta] = [t_0, r(r_0), \phi + \omega t_0]$$

where we're using cylindrical polar coordinates in flat spacetime, and I'm pretending there's no z coordinate (a) to save typing, and (b) because I'm afraid pervect will want to add a perturbation to it.

I've parameterised the world lines along their length by a parameter I've called $t_0$. This happens to be exactly equal to t everywhere it's defined, but it's worth picking a different name for it, for a reason: the vector $\partial_{t_0}$ is completely different from the vector $\partial_{t}$, so if we use the same name for them (as I did in an earlier post), there's a potential for confusion. If you have two coordinate systems, and one coordinate in system A happens to be identical to a coordinate in system B, then that does not mean that the corresponding coordinate derivatives are the same; the meaning of $\partial_{t}$ is "take the derivative while varying t and holding the values of all other coordinates constant", so it will depend on the entire coordinate system, not just the values of t at different points.

OK, now what we have is a 2-dimensional submanifold, the world sheet of the hoop, with two different coordinate systems on it: there's t and $\theta$, which are just inherited from the spacetime coordinates, and there's $t_0$ and $\phi$, which are related to the world lines. This is actually a much nicer situation than if the hoop was perturbed, in which case there would be no simple, standard coordinate system on the whole of spacetime such that two coordinates neatly covered the world sheet while the third one was constant on it. In that case, the simplest thing would probably be to artificially extend $\phi$ away from the world sheet by treating $r_0$ as a coordinate, much as if we were looking at a finite-width ring.

Anyway, given that we do have two simple coordinate systems for the world sheet, and that the vector w, being a tangent to the world sheet, can be written easily as:

$$w = \omega \gamma r \partial_t + \frac{\gamma}{r} \partial_{\theta}$$

we can write n as:

$$n = r_0 w^a \partial_a \phi = r_0 (\omega \gamma r \partial_t \phi + \frac{\gamma}{r} \partial_{\theta} \phi)$$

So what we need in order to evaluate n by this method are the partial derivatives $\partial_t \phi$ and $\partial_{\theta} \phi$. These do not come to us immediately from the spacetime coordinates for a point on the worldsheet; what they give us are the partial derivatives:

$$\partial_{t_0} t = 1$$
$$\partial_{t_0} \theta = \omega$$
$$\partial_{\phi} t = 0$$
$$\partial_{\phi} \theta = 1$$

To obtain the partial derivatives we actually want, which go the other way, we invert the matrix of partial derivatives, i.e. we use:

$$[\frac{\partial (t_0, \phi)}{\partial (t,\theta)}]=[\frac{\partial (t,\theta)}{\partial (t_0, \phi)}] ^{-1}$$

which gives us:

$$\partial_t t_0 = 1$$
$$\partial_t \phi = -\omega$$
$$\partial_{\theta} t_0 = 0$$
$$\partial_{\theta} \phi = 1$$

When we plug the values for the partial derivatives of $\phi$ into our formula for n, we get:

$$n = r_0 w^a \partial_a \phi = r_0 (-{\omega}^2 \gamma r + \frac{\gamma}{r}) = \frac{r_0}{\gamma r}$$

So this agrees with our other formula for n, but it took a lot more work ... and with a perturbed hoop, it would be even more work. So it's nice to understand both ways, but use the easier one.

 

[gregegan]

I thought I'd report a tentative first result from a stability analysis I'm doing based on pervect's displacement model:

$$t(t_0,r_0,\phi) = t_0$$
$$r(t_0,r_0,\phi) = r_e(r_0) + \delta_r(t_0,\phi)$$
$$\theta(t_0,r_0,\phi) = \phi + \omega t_0 + \delta_{\theta}(t_0, \phi)$$

This maps what I'll call "hoop coordinates", $t_0$, $r_0$ and $\phi$ into cylindrical spacetime coordinates. Although this is set up to work with three material coordinates as if we were dealing with a ring, we assume that:

$$\partial_{r_0} r_e(r_0) = 1$$

so that we really just have a hoop, since that condition tells us that the material is undistorted in the radial direction.

We define:

$$\alpha = g(\partial_{t_0},\partial_{t_0}) = -1+(\frac{\partial \delta_r}{\partial t_0})^2 + r^2 (\omega + \frac{\partial \delta_{\theta}}{\partial t_0})^2$$
$$\beta = g(\partial_{\phi},\partial_{\phi}) = (\frac{\partial \delta_r}{\partial \phi})^2 + r^2 (1 + \frac{\partial \delta_{\theta}}{\partial \phi})^2$$
$$\chi = g(\partial_{t_0},\partial_{\phi}) = \frac{\partial \delta_r}{\partial t_0} \frac{\partial \delta_r}{\partial \phi} + r^2 (1 + \frac{\partial \delta_{\theta}}{\partial \phi})(\omega + \frac{\partial \delta_{\theta}}{\partial t_0})$$

we then have:

$$n = r_0 \sqrt{\frac{\alpha}{\alpha \beta - \chi^2}}$$
$$u = \frac{1}{\sqrt{-\alpha}} \partial_{t_0}$$
$$w = \frac{n}{r_0} (-\frac{\chi}{\alpha} \partial_{t_0} + \partial_{\phi})$$

So far I'm only looking at the hyperelastic model, not the breakable model.

I computed the divergence in a hoop coordinate basis; we already have most of the components of the metric, but we also need:

$$g(\partial_{r_0},\partial_{t_0}) = \frac{\partial \delta_r}{\partial t_0}$$
$$g(\partial_{r_0},\partial_{r_0}) = 1$$
$$g(\partial_{r_0},\partial_{\phi}) = \frac{\partial \delta_r}{\partial \phi}$$

I then took a first-order series expansion of the three divergence components. These yield two independent PDEs, and if we restrict the perturbations to be axially symmetric, i.e. having no dependence on $\phi$, then there appear to be stable solutions of the form:

$$\delta_r(t_0) = \epsilon_r \sin{f t_0}$$
$$\delta_{\theta}(t_0) = \epsilon_{\theta} \cos{f t_0}$$

everywhere, including at the energy peak. I can't quite see how to reconcile this with my previous analysis, because although that suggested there'd be no runaway solutions near the energy peak, it did imply a localised instability, suggesting that the hoop's radius would accelerate away from the equilibrium position there if it was displaced from it (albeit to be slowed down shortly afterwards). There must be something subtle going on here ... either that, or there's an error in my calculations.

 

[pervect]

If we take an elliptical hoop, we expect non-radial forces from a Newtonian analysis, which should imply non-zero $\phi$ components in the forces.

For instance, if we consider a planet orbiting the sun, we know that the forces are central, we also know that the orbits are ellipses with the sun at one focus.

Thus we know that for an ellipse, forces point towards the focus, and thus can't point towards the center (centroid) of the ellipse. So if we have an elliptical hoop rotating around its centroid (rather than its focus), we know that the forces won't all be radial (in a Newtonian analysis). And I think this should imply non-zero $\phi$ components of motion in the SR case (as well as the Newtonian case).

So I suspect a problem here. I've been thinking along roughly similar lines, but not getting anywhere useful.

I've been hoping that MTW's discussion of "junction condition" and surface stress-energy tensors would enlighten me (pg 551-556 for anyone who has the book), but so far it hasn't :-(.