Stress-energy tensor of a wire under stress

[pervect]

I haven't thought about it much yet, but it ought to also be possible to take an approach where the state of a hoop is represented by its angular momentum, energy, and radius.

I.e. we can look at the phase space.

In fact I was doing this earlier, before I found the Lagrangian (but the results I got earlier are only approximate and not valid for arbitrary radial velocities).

 

[pervect]

Got an interesting new lead on stability.
http://www.pma.caltech.edu/Courses/ph136/yr2004/0410.1.K.pdf
sec $10.8

and
http://www.pma.caltech.edu/Courses/ph136/yr2004/0411.1.K.pdf

sec $11.35

which I think was mentioned by Stingray earlier on in the thread seems to suggest that the onset of instability of the hoop should occur at the same time a bifurcation of static solutions do. <snip>

I don't have any results yet from this approach, but it looks interesting and more productive than picking random points to attempt to analyze stabiity at.

 

[gregegan]

Got an interesting new lead on stability.
http://www.pma.caltech.edu/Courses/ph136/yr2004/0410.1.K.pdf
sec $10.8

and
http://www.pma.caltech.edu/Courses/ph136/yr2004/0411.1.K.pdf

sec $11.3.5

Thanks for the links to those course notes (by Kip Thorne, no less!). I hope to get a chance sometime to study these carefully.

I want to say a couple of things about the singular behaviour that sometimes appears in the pulsating hoops. Firstly, I accept now that this is irretrievably unphysical. Not only does the magnitude of the proper acceleration of the hoop material become infinite (according to the equations of motion), but there is literally no solution to the conservation equations with r greater than the critical value. So not even a finite jump in the angular velocity can save things.

I expanded the divergence equation with Leibniz's law, and there are three terms that contribute to the tangential component. One is just $\rho a$, one is the "naive force" on a unit element, $\nabla_w(P w)$, and the third is P (div w) w, a term where one might expect the tension to lower the effective mass. However, it turns out that the acceleration coordinates appear in all three terms, including the "naive force"; this isn't really all that surprising, because w and u are locked together by the orthonormality requirement, so derivatives of w are going to be linked to derivatives of u, i.e. acceleration. But in any case, the net effect is that despite no violation of the weak energy condition, the coefficients of both acceleration coordinates $a_r$ and $a_\omega$ in the tangential equation become equal to zero at the critical point. The radial equation independently gives a sensible answer for $a_r$, but the tangential equation basically becomes insoluble.

What I think this means is that the hyperelastic model (and any other material model that gives the same results) is physically impossible in this domain; the material must deviate from these idealised models before this point is reached. In other words, there are more stringent requirements than just the weak energy condition that need to be imposed.

I think there might be another compelling reason to declare that the models fail in this domain: I conjecture that at the critical point, the speed of some modes of information-carrying waves in the hoop will reach the speed of light. Of course we've ruled out the speed of pressure waves in the relaxed material exceeding the speed of light, but that's no guarantee that the kind of vibrations the hoop can carry under tension will also respect causality.

 

[pervect]

I agree that there is something unphysical going on in this domain (where the relativistic Lagrangian becomes singular) and think that your proposals as to what it might be are interesting and promising.

The obvious next step is to study vibrations in more detail, both in the Newtonian case and in the relativistic case.

 

[pervect]

I've had an annoying number tries to get the Newtonian Lagrangian right, but I think I have it. The following is nonlinear, so should work for any amplitude. The way I'm handling linearization is to put in a linear solution, and ignore terms of higher than linear order in $\delta$.

$$\matcal{L} = \frac{\eta1 \, (\eta2_{,1})}{2 \sqrt{\eta1^2 \, (\eta2_{,1})^2 + (\eta1_{,1})^2}} \left( \rho0 \left( (\eta1_{,0})^2 + \eta1^2 \, (\eta2_{,0})^2 \right) - k \left( \sqrt{\eta1^2 \, (\eta2_{,1})^2 + (\eta1_{,1})^2} - 1\right)^2 \right) \, d \phi$$

This can be seen as

$$\frac{\mathrm{volume-element}}{2 \, s} \left( \rho0 \, v^2 - k(1-s)^2 \right)$$
where v is the velocity of the segment of the wire $d\phi$

$$s = \sqrt{\eta1^2 \, \eta2_{,1}^2 + \eta1_{,1}^2}$$ is the stretch factor

and the volume element is $$r \frac{d \theta}{d \phi} d \phi$$ = $\eta1 \eta2_{,1}$

It is convenient to renormalize the above with $\rho0 = 1$ and k = $v_c$^2, however.

The notation is compatible with Goldstein's. $\eta1$ is r, and $\eta2$ is $\theta$. ,0 represents differentiation with respect to time, ,1 represents differentiation with respect to $\phi$.

<snip>

Sanity checks:

r_eq comes out to the 'right' value of $v_c^2 / (v_c^2 - \omega^2)$
It appears to give the correct results for a pulsating hoop

I thought I had some stability results, but I realized that I'd better include a phase angle other than 90 degrees, so I'm going to retract those until I do the analysis.

[add]
OK, I may not have all of the solution modes, but the stability analysis with the current Lagrangian does! have unstable solutions, for instance:

$$r = r_{eq} + \delta e^{\alpha t} \cos(m(\phi-c\omega*t))$$
$$\theta = \phi + \omega t + \delta \beta e^{\alpha t} \sin(m(\phi - c \omega t)) + \delta \epsilon e^{\alpha t} \cos(m*(\phi-c \omega t))$$

with m=1; v_c := 1; omega := .1; alpha := 1.021129923; beta := .1807244597e-1; epsilon := -.1922035383; c := .9601444132;

I wasn't quite happy with complex c indicating an exponential solution, but this exhibits one explicitly.

 

[gregegan]

and the volume element is $$r \frac{d \theta}{d \phi} d \phi$$ = $\eta1 \eta2_{,1}$

But shouldn't this come from the arc length in polar coordinates?

$$\sqrt{r^2 + (\frac{dr}{d\theta})^2} d\theta$$

Unless I'm confused about something here, I think the formula you've used will work when r is constant, but not for perturbations in r.

I used to get complex / unstable solutions for Newtonian vibrations when I was assuming constant tension, but since I realized that the tension wasn't constant to first order and re-did the analysis, I now find only stable vibrations to first order. Actually, there were two things that happened simultaneously that made me realize my mistake; one was your earlier post in which you said you didn't find complex solutions, and the other was this applet I wrote:

http://www.gregegan.net/SCIENCE/Rings/SimpleHoopApplet.html

which simulates a hoop numerically, and which kept converging on solutions that were different from the ones I'd been predicting. It seems to agree with my new analysis, though.

Certainly for high tension (i.e. in the limit $\omega=v_c$, although what you call $v_c$ I've called $\omega_c$), both my new theoretical analysis and the results I'm getting from that applet agree with the c values you quoted in your earlier post: c=-1, c=1+2/m, c=1-2/m.

 

[pervect]

OK, I think you're right about the volume element. Let's see what happens if I correct it.

 

[pmb_phy]

OK, I think you're right about the volume element. Let's see what happens if I correct it.

Hi Pervect

I've been interested in what's going on in this thread. Its a very long thread and I was wondering if I could be of some use. Can you give me the short history of what this thread is about and how far you've gotten and what your goal is? Thanks

Best wishes

Pete

 

[pervect]

A short overview of the thread is this:

Greg Egan has worked on the relativistic disk and hoop in SR (i.e. no self gravity, just a disk or hoop in flat space-time), and I have done the same, and we basically get the same answers for things like the energy and angular momentum of a relativistic hoop. Greg has also worked out the relativistic disk, I have not. To do so we have been assuming certain relativistic models for elasticity such as the hyperelastic model.

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

and the early parts of this thread.

The solutions we both get exhibit a number of strange features that have fostered further work, such as a absolute maximum amount of energy that a hoop can hold at some particular critical value of omega. At some point, increasing omega decreases the angular momentum and the energy (!). This may seem very, very odd, unless one realizes that the analysis indicates radius of the hoop starts to shrink as omega increases for a strong enough material - after you realize this, it only seems very odd.

It has been agreed by both of us recently that the hyperelastic model is suspect if you "push it too hard" - part of the reason underlying the discussion is to find out where the model is valid and where it is not. So far we don't have any evidence that the energy and angular momentum peak is associated with pushing the model too hard (but we don't have a clear handle on the limits of the model yet, either).

Currently the discussion has shifted back to "simple" Newtonian hoops. What could be hard about that :-) Well, since we are interested in stability, the answer is quite a lot can be hard about that.

I'm leaning in a much different direction than Greg Egan is as far as stability goes is at the moment. I think that what happens is that the only stable shape of a rotating Newtonian hoop is in fact not circular, that the situation is similar to that of the playing card described by Kip Thorne in

http://www.pma.caltech.edu/Courses/ph136/yr2004/0410.1.K.pdf section $10.8 with some minor differences.

The difficulty is that I'm doing (attempting to do) a Lagrangian analysis, and Greg has done a different analysis. I want my Lagrangian analysis to give exactly the same answer that Greg's analysis does, then I'll be happy.

You'd think I'd have the advantage as the Lagrangian analysis should be a lot easier than getting the free-body diagram right, but so far my track record has been not-so-great.

Greg has also done some sort of applet to study the problem of the Newtonian hoop - unfortunately, I don't quite understand it :-(.

As far as the direction I'm trying to go in, the following quote from Thorne about bending a playing card by pressing on it with two fingers describes the general situation:

A full discussion requires elastodynamics,
but basically what happens is that if we consider the behavior of small perturbations about
equilibrium, then for forces F < Fcrit, the perturbations decay exponentially. When F =
Fcrit, the card is neutrally stable. However, when F > Fcrit, straight cards constitute unstable equilibria. There is a however a unique stable equilibrium with a central displacement n that increases rapidly with increasing F. However, to find it requires including non-linear terms in the equation of elastostatic equilibrium.

My suspicion based on my Lagrangian analysis is that this is what is happening with the Newtonian hoop, but that there is no "critical frequency" at which this process starts, that for every frequency there is some optimum and non-circular "stable" shape of the hoop.

Amusingly, this behavior matched Greg's first results, but not his corrected results.

 

[gregegan]

Greg has also done some sort of applet to study the problem of the Newtonian hoop - unfortunately, I don't quite understand it :-(.

All the applet does is approximate a hoop as a number of point masses. The force on each point mass is computed by assuming that a piece of elastic material joins each mass to its two neighbours, and we approximate the behaviour of that small segment of elastic material by assuming that it will always lie in a perfect straight line, and that it will obey Hooke's law. So given the coordinates of all the masses it's extremely simple to compute the net force on each of them, and hence their acceleration vectors. (In order to improve the accuracy when integrating the motion over a finite time step, I also compute several higher-order rates of change, i.e. time derivatives of the acceleration as well as the acceleration itself.)

The applet let's you start the simulation with various small perturbations from the equilibrium as its initial conditions, namely those that I've computed theoretically for the continuum hoop. The applet seems to validate these, in that the time evolution of these small waves matches what the theory predicts. I only see any instability if I bring the tension down very low, i.e. if I lower omega to the point where the hoop is almost in its relaxed state -- and even then I think the reason for the instability is that these numerical perturbations are, of course, finite. (I believe that if I made them small enough, they'd be stable however low omega was -- but these are finite-precision numerical calculations, so there are limits to what they can say about very small perturbations.)

There are two reasons why I trust this applet. One is that it computes the total energy and angular momentum of the hoop as it goes, and these values stay very close to constant. (The only real exception is when I lower omega to the point where the hoop crumples up wildly, and I think the whole numerical approach becomes inaccurate there.) The other is that it actually exhibited "empirically" the c=1+2/m and c=1-2/m solutions for high tension, even when I was feeding it different starting conditions! I originally set it up with positions and velocities for the point masses that were compatible with different values for c -- in accordance with my old calculations -- but those starting conditions rapidly evolved into waves traveling at c=1+2/m or c=1-2/m. In other words, based solely on F=ma and Hooke's law, it exhibited precisely the theoretical results that you reached previously, and that I reached when I finally added variable tension to my calculations. I don't think that's any kind of coincidence or error.

 

[pervect]

Good news - I may have found the error in my analysis, but I want to do a lot more checking before I say more.

Meanwhile, I've noticed that in the hoop applet if I set 200 vertices, K=.5 and omega=.5 , m=3, cubic solution=1, it goes crazy.

 

[gregegan]

Meanwhile, I've noticed that in the hoop applet if I set 200 vertices, K=.5 and omega=.5 , m=3, cubic solution=1, it goes crazy.

Yeah, the thing about that set of solutions is that alpha (the size of the longitudinal wave compared to the transverse) is very large, to the point where for the usual small value of delta I use, the hoop actually forms little loops, becoming a self-intersecting curve. I put in some code to shrink delta specially for those solutions -- but depending on the value for omega you choose, it's still probably too big. It's hard to figure out a good algorithm for setting delta for the full range of parameters without making the perturbation invisible in some cases, and too big to count as "small" in others.

Maybe I should just pick a default but hand the choice over to the user ...

 

[pervect]

I'm actually getting the exact same cubic equation for c that you are now (yeah!), though I'm not quite sure how one makes sure the roots of this are real for all m.

 

[gregegan]

I'm actually getting the exact same cubic equation for c that you are now (yeah!), though I'm not quite sure how one makes sure the roots of this are real for all m.

That's great news!

To see why the roots are always real, first note that for m=1 the cubic factors into a quadratic and a linear term, with manifestly real roots for the quadratic.

For higher values of m, note that:

(a) as c goes to -infinity, the cubic goes to -infinity;

(b) at c=0, the cubic is always positive (for m>1);

(c) at $c=\omega_c/\omega$, the cubic is always negative;

(d) as c goes to +infinity, the cubic goes to +infinity.

So the cubic changes sign three times, and hence must have three real roots.

 

[pervect]

Let me document the correct Lagrangian density here:

$$\mathcal{L} =\frac{1}{2} \left( {{\it eta1_{,0}}}^{2}+{{\it eta1}}^{2}{{\it eta2_{,0}}}^{2}- {{\it v_c}}^{2} \left( \sqrt {{{\it eta1_{,1}}}^{2}+{{\it eta1}}^{2}{ {\it eta2_{,1}}}^{2}}-1 \right) ^{2} \right) d \phi$$

This is normalized - one can think of it as being for a hoop with a density of 1, and a velocity of sound of $v_c$.

eta1 can be understood as being the radial coordinate r, eta2 can be understood as being the angular coordinate $\theta$.The density is defined in terms of the body coordinate $\phi$ which ranges from 0 to 2 Pi

Thus every value of $\phi$ defines a worldline (t, eta1=r, eta2=$\theta$), the time evolution of that worldline is determined by its Lagrangian density. One can think of $\phi$ as being the angular coordinate $\theta$ of the hoop before it was "spun up".

Since eta1 and eta2 are functions of t and $\phi$, partial derivatives with respect to these variables exist. These are denoted by _,0 for $\partial / \partial t$, and _,1 for $\partial / \partial \phi$.

The textbook discussion of Lagrangian densities I used as a reference was Goldstein, "Classical Mechanics", chapter 12.

Is there any interest in me writing out, in more detail, the procedure for turning the above Lagrangian density into actual partial differential equations?

(I should add that a sign error in this conversion process was my final error).

 

[gregegan]

Is there any interest in me writing out, in more detail, the procedure for turning the above Lagrangian density into actual partial differential equations?

My own feeling on this is that I wouldn't be game to try to do this kind of Lagrangian analysis without reading Goldstein myself anyway, so it's probably not worth your effort to type out all the derivatives for this particular example.

(I should add that a sign error in this conversion process was my final error).

The trick with sign errors is always to make an even number of them.

 

[pervect]

OK, so onto the relativistic case. All we need to do is to replace the Newtonian Lagrangian with the relativistic Lagrangian.

The Newtonian Lagrangian density was basically:

$$\mathcal{L} = \frac{\rho0 v^2 - k (s-1)^2}{2 s} s \, d\phi$$

where s was the Newtonian "stretch factor", and v was the velocity. We can see this has the form of T-V, and that the volume element is s $d \phi$. This is logical, because with $r_0=1$, the unstretched length of the wire was $d \phi$, so if you stretch the wire by a factor of s, it's new length is just $s \dphi$ - and the area of the wire remains constant.

The relativistic Lagrangian is in some respects even simpler

$$\mathcal{L} = -\rho dV = \left( -\frac{rho0}{s} -\frac{k(s-1)^2}{2s}\right) dV$$

where dV is the relativistic volume element. The Newtonian case indicates the "easy way" to find the relativistic volume element. The original volume associated with the wire element was $A d \phi$, A being the area This gets multiplied by s, to give the proper volume $A \, s \, d \phi$. No matter how we orient the wire, it's volume in the lab frame gets multipled by a relativistic factor of $1 / \gamma$ from its proper volume - one might attribute this to reduction in A, or reduction in length, depending on the orientation of the wire, but the total volume change is given by the same factor.

Thus we can write:

$$\mathcal{L} = \left( -\frac{rho0}{\gamma} -\frac{k(s-1)^2}{2 \gamma} \right) d \phi$$

We can see that the kinetic energy terms enter in the series expansion of gamma multiplying the mass, in more or less the usual manner.

Next, we need to compute the relativistic stretch factor s. This involves finding the 4-velocity and the tangent vector $\partial / \partial \phi$ and adding a multiple of the 4-velocity to the tangent vector to make a vector perpendicular to the 4-velocity. The length of this new vector, y, is the stretch factor. This is a rather involved calculation, I get:

$$s = {\frac {\sqrt {-{{\it \eta1\_1}}^{2}{{\it \eta1}}^{2}{{\it \eta2\_0}}^{2} +2\,{\it \eta1\_1}\,{\it \eta1\_0}\,{\it \eta2\_1}\,{{\it \eta1}}^{2}{\it \eta2\_0}+{{\it \eta1}}^{2}{{\it \eta2\_1}}^{2}-{{\it \eta1}}^{2}{{\it \eta2\_1}}^{2}{{\it \eta1\_0}}^{2}+{{\it \eta1\_1}}^{2}}}{\sqrt {1-{{\it \eta1\_0}}^{2}-{{\it \eta1}}^{2}{{\it \eta2\_0}}^{2}}}}$$

This does reduce to the Newtonian stretch factor if the velocity terms (*_0) are set to zero.

Putting this all together, I took a look for solutions with c=0, which should indicate the onset of bifrucation. We didn't see any bifrucation in the Newtonian case in spite of my earlier remaks, but I expect it in the relativistic case.

And it appears that there is a critical rotation rate at which bifrucation occurs (i.e. soultions with c=0). I didn't find any for m=1. For m=2, and v_c^2 = 1/2, this was omega = 0.3002283099 and r_eq = 1.122536155.

m=3 had a higher critical omega, .5270577924 for the same v_c.

So if I did everything right (and I'm not making any guarantees), I expect that an unsupported hoop with a stifness such that the speed of sound is .707c will at about omega=.3 find a new stable equilbrium in a four-lobed shape rather than a circular shape.

 

[gregegan]

I expect that an unsupported hoop with a stifness such that the speed of sound is .707c will at about omega=.3 find a new stable equilbrium in a four-lobed shape rather than a circular shape.

If m=2, the shape is roughly elliptical, isn't it?

I'll see if I'm able to confirm this by a different route, hopefully sometime in the next few days. What I hope to do is a relativistic equivalent of the infinitesimal-element force balance that I used for the Newtonian case, which is a matter of setting the divergence of the stress-energy tensor to zero.

BTW, are you able to get a relativistic polynomial in c from your Lagrangian analysis that converges on the Newtonian one?

 

[pervect]

From 0 to 2pi, cos(theta) has 2 maxima (at 0 and Pi) so m=1 should be elliptical, and m=2 should be four-leafed (four maxima, four minima)

Maybe we need to think about why 1/2 integer values of m are ruled out, or if they are ruled out? (Would the center of mass be at the centroid?)

I don't have a closed form solution for r_eq for the relativistic case, so it's not quite that straightforwards to get a polynomial for c.

Looking at the Lagrangian convinced me that the results should be the same for the relativisitc vs Newtonian case as long as the speed of sound << 1 and the velocities are also << 1. But I'll try and work out an example for a low v_c and report back.

 

[pervect]

OK, I took v_c = 2e-5 (slightly higher than the velocity of sound in steel). Setting r_eq to 2, I found omega := .1414213563e-4;

For m=1 relativistic I found
3.449489737, -.9999128315, -1.000087187, -1.449489723

for m=1 Newtonian for comparison:
-1., -1., 3.449489742, -1.449489742

for m=2 relativistic, I found
0.07530941840, 2.342079597, -1.000000004, -1.417389017

for m=2, Newtonian for comparison
-1., .07530941822, 2.342079601, -1.417389019

 

[gregegan]

From 0 to 2pi, cos(theta) has 2 maxima (at 0 and Pi) so m=1 should be elliptical, and m=2 should be four-leafed (four maxima, four minima).

At t=0, the radius is:

$$r = r_e + \delta cos(m \phi)$$

If m=1 the radius is increased by $\delta$ at $\phi=0$, and decreased by $\delta$ at $\phi=\pi$. Depending on $\alpha$ there can be some further distortion, but in the limit of small $\alpha$ this is roughly just a circle displaced in the +ve x direction.

If m=2, the radius is increased twice, at $\phi=0$ and $\phi=\pi$, and decreased at $\phi=\pi/2$ and $\phi=3\pi/2$. In other words, the hoop is stretched along the x-axis and compressed along the y-axis, into (roughly) an ellipse.

It's easy to see this with the applet, most clearly with the c=-1 solutions.

 

[pervect]

You're right - m=2 will be an ellipse, not 4-lobed.

 

[pmb_phy]

Alas, this thread is far too along for me to contrubute in even a miniscule way. I wish I could join in though. Sounds fascinating. Perhaps I'll find time to read this carefully from the beginning.

Pete

 

[gregegan]

And it appears that there is a critical rotation rate at which bifrucation occurs (i.e. soultions with c=0). I didn't find any for m=1. For m=2, and v_c^2 = 1/2, this was omega = 0.3002283099 and r_eq = 1.122536155.

m=3 had a higher critical omega, .5270577924 for the same v_c.

Having calculated everything via the vanishing divergence of the stress-energy tensor, I've found a polynomial in c (with coefficients that are functions of s, the stretch factor). This has roots at c=0 for values of s that agree with your omega and r_eq values. Given that we've both arrived at exactly the same numbers via such different routes, I'm fairly confident that we've come up with correct equations here.

So far I can't find any complex roots for c, but I haven't rigorously shown that there are none. (I do find axially symmetric exponential solutions of my PDE in the region of instability we previously identified for pulsations; what I haven't found are any exponentially growing solutions that involve a change of shape.)

So it appears that these c=0 solutions don't indicate any change to a new, more stable shape. When omega lies precisely on the critical value, small deformations in the shape with the appropriate m will neither grow exponentially, nor oscillate back towards the circular shape; however, as soon as omega exceeds the critical value, all the roots for c will become non-zero again, and there will be a restorative force back towards circularity.

 

[gregegan]

So far I can't find any complex roots for c, but I haven't rigorously shown that there are none.

I have found complex roots for c, now, though they appear at substantially higher angular velocities than the c=0 points.

For example, for $v_c^2=1/2$, for m=2, the root c=0 appears at $\omega$=0.300, while the onset of complex roots for c appears at $\omega$=0.7236 (and ends at $\omega$=1.069)

 

[pervect]

I can confirm that for m=2, the roots for c appear to be real for both omega=.28 and omega=.32.

I'm having some difficulties (possibly numeric) with the roots of c for m=1 and omega = .28. (I've got some roots near -1 that are evaluating as complex, but the magnitude of the complex part is suspiciously small).

I also have one value of c for omega=.28 which is 4.1797 which appears to make c*omega*r_eq > 1, I'm not sure if this is a problem, but I think this implies a FTL phase velocity.

One other thing I want to look at in more detail is what effect the presence of the pertubation with c=0 has on energy and angular momentum.

 

[gregegan]

I can confirm that for m=2, the roots for c appear to be real for both omega=.28 and omega=.32.

I'm having some difficulties (possibly numeric) with the roots of c for m=1 and omega = .28. (I've got some roots near -1 that are evaluating as complex, but the magnitude of the complex part is suspiciously small).

I also have one value of c for omega=.28 which is 4.1797 which appears to make c*omega*r_eq > 1, I'm not sure if this is a problem, but I think this implies a FTL phase velocity.

One other thing I want to look at in more detail is what effect the presence of the perturbation with c=0 has on energy and angular momentum.

I've written up what I've found so far about the relativistic vibrations in a new section at the end of the web page:

http://www.gregegan.net/SCIENCE/Rings/Rings.html

I've got an explicit polynomial for c, and the results concerning the regions where c has an imaginary part seem pretty robust. Basically, after the section where there are unstable pulsations (which comes just before the energy maximum), there's a gap where everything is stable, and then a series of overlapping regions where c has complex roots for successive values of m and so the circular shape will be unstable.

The FTL phase velocity doesn't imply FTL information transfer, of course, because these oscillations come from globally-prepared initial conditions. An initially localised perturbation will tell us how fast information is travelling, but that's going to be quite complicated to analyse. And in terms of understanding what's going on with the singularities you found in certain large pulsations, there's the added complication that we might only see anything pathological if we look at perturbations imposed on top of the pulsating hoop, rather than on top of any equilibrium state. When I looked at finite but small-ish pulsations -- including ones that encompassed the maximum energy, and ones that encompassed the maximum radius -- I didn't see the singular behaviour, so it might be that this pathology is only going to show up when the hoop is quite far from the equilibrium state.

 

[pervect]

Upping the number of digits in the calculations I made it pretty clear that there isn't any actual issue with c becoming complex.

I can post the set of simultaneous equations I use to solve for c and the multiplicative factor if there is any point to doing this, though I would guess that your closed form equation for c as a function of s is superior and that there wouldn't be much point.

Actually, it might be somewhat helpful if you could paste an expression for your cubic for the relativistic hoop, that I could cut and paste it into maple to examine. The webpage shows the equation as an image, so I'd have to type it in, and it's a bit long.

I'm pretty sure from the results that there isn't any exponentially growing instability - could there be any linear growing instability? For instance, could we have a double root at c=-1, and would this cause some linear growing solution like $\delta \, t \, cos(...)$ to become a solution of the differential equations? (I can't quite confirm from the numerical results I have that there is a double root at c=-1, but it's at least close to being a double in some of the regions I was evaluating numerically).

As somewhat of an interesting aside, considering a hypothetical double-humped effective potential of V(x) = x^4/4 - x^2/2 with an associated differential equation of

$$\frac{d^2 x}{d t^2} + x^3 - x=0$$

which was a translation of figure 1.4 from one of Kip Thorne's examples into math, made it clear that this was NOT happening for the hoop.

For instance, if we linearize this, the cubic term disappears, and we have the unstable

x'' - x = 0

with solutions [itex]e^{t}[itex] and $e^{-t}$, and if we try solutions of the form x = cos(omega t), we find omega is complex.

Thus the linear methods we are using would find this hypothetical system unstable.

It does serve as an interesting example of how a non-linear term added to an unstable linear system can make it into a stable system at a different operating point.

 

[pervect]

I just remembered something. In another thread, a user found instabilities in the three-body problem via simulation and also in a textbook that I didn't find with a linear analysis.

See https://www.physicsforums.com/showpost.php?p=1353020&postcount=4
the textbook was "The Three-Body Problem" by Christian Marchal

Unfortunately I never got to the root of the problem (is that a pun?). I have a suspicion, though, that it may be related to the issue of repeated roots.

 

[gregegan]

Actually, it might be somewhat helpful if you could paste an expression for your cubic for the relativistic hoop, that I could cut and paste it into maple to examine. The webpage shows the equation as an image, so I'd have to type it in, and it's a bit long.

I'm pretty sure from the results that there isn't any exponentially growing instability - could there be any linear growing instability? For instance, could we have a double root at c=-1, and would this cause some linear growing solution like $\delta \, t \, cos(...)$ to become a solution of the differential equations? (I can't quite confirm from the numerical results I have that there is a double root at c=-1, but it's at least close to being a double in some of the regions I was evaluating numerically).

(edit: Removed unwieldy cubic in TeX; see next post for friendlier version.)

For m=1, this does have an exact factor of (c+1), so there is a double root of -1.

BTW, I said before that I hadn't found the singular behaviour except at points far from equilibrium, but I hadn't really looked at the algebra closely enough. If you take either the denominator of the angular acceleration, or the derivative of the angular momentum with respect to omega, set v_r to 0, substitute the versions of r and omega parameterised by n, and factor the polynomial in n ... you get a quartic, one of whose roots is real, positive, and in the physically valid range. For k=1/2, rho=1, r_0=1, this value of n yields:

omega = 0.56583358
r = 1.19312053
s = 1/n = 1.617309826

This s value is after the point where the radius has its maximum, but before the point where the energy has its maximum; it lies within the region where pulsations are unstable.

While this equilibrium state of the hoop is itself perfectly OK, if you perturb it by adding a tiny non-zero v_r, the hoop can't actually change radius while preserving angular momentum. I still can't see the deep reason why this is true -- whether the hyperelastic model is somehow implying a violation of causality, or whether there's something else entirely that we're missing. But it might be helpful to have this equilibrium point to study.

If you want the exact quartic for the general case, it's:

$$Q^4 n^4 - 2 K^2 Q^2 n^3 + 6 K^4 n - 5 K^4$$

where

$$K^2 = k / (2 \rho)$$
$$Q^2 = 1 + K^2$$

 

[gregegan]

That TeX might not have been the easiest thing to paste into Maple, so here's another try:

c^3*m^2*(-1 + s)*(1 + K^2 - 4*K^2*s + 3*K^2*s^2)*
(-(1 + K^2)^2 + 2*K^2*(1 + K^2)*s - 6*K^4*s^3 + 5*K^4*s^4) +
c^2*m^2*(-1 + s)*(-1 - K^2 + K^2*s^2)*(-(1 + K^2)^2 + 2*K^2*(1 + K^2)*s +
4*K^2*(1 + K^2)*s^2 - 22*K^4*s^3 + 17*K^4*s^4) +
(-1 - K^2 + K^2*s^2)*((1 + K^2)^2 + (1 + K^2)*(-4 - 6*K^2 + m^2 + K^2*m^2)*
s + 14*K^2*(1 + K^2)*s^2 - 2*K^2*(3 + 8*K^2 + m^2 + K^2*m^2)*s^3 +
9*K^4*s^4 + K^4*(-2 + m^2)*s^5) +
c*(-3*(1 + K^2)^3 + (1 + K^2)^2*(4 + 14*K^2 + m^2 + K^2*m^2)*s -
K^2*(1 + K^2)*(11 + 19*K^2 + 8*m^2 + 8*K^2*m^2)*s^2 +
K^2*(1 + K^2)*(-2 - 10*K^2 + 5*m^2 + 5*K^2*m^2)*s^3 +
K^4*(31 + 55*K^2 + 16*m^2 + 16*K^2*m^2)*s^4 -
K^4*(12 + 62*K^2 + 13*m^2 + 13*K^2*m^2)*s^5 - K^6*(-31 + 8*m^2)*s^6 +
K^6*(-6 + 7*m^2)*s^7)

I checked whether multiplying the c=-1, m=1 solutions by t works in the PDE, and it does! Both f and g need the factor of t. But what I think this represents is just (to first order in delta) inertial motion of the hoop's centre of mass. It's not a linearly growing perturbation in the shape, it's just an invariance of the solution under a transformation to a moving reference frame. In the Newtonian case, it ought to hold exactly, but in the relativistic case it's the linearised version of boost-invariance.

 

[gregegan]

I found a nice parameterisation of the curves of constant angular momentum and energy for the pulsating hoop:

r->Sqrt[r0^2*s^2*(1+K^2*(1-s^2))^2 - LM^2] / (1+K^2*(1-s^2))

gamma->(EM*r*r0) / (r0^2*s+K^2*(s-1)*(2*r^2-r0^2*s*(1+s)))

omega->Sqrt[r0^2*s^2-r^2] / (gamma*r*r0*s)

Here LM is the hoop's angular momentum divided by its rest mass, EM is the hoop's total energy divided by its rest mass, and gamma is 1/Sqrt[1-v_r^2]. I haven't substituted the values for r and gamma into the formulas for gamma and omega here, but doing that gives all three quantities solely in terms of s. The only messy thing is finding the domain(s) for s, such that these quantities are all real; that boils down to numeric solutions of a high-order polynomial in s. At the endpoints of each domain for s, gamma will be equal to 1, i.e. v_r=0.

Plotting these curves in (omega, r, v_r) space, they generally form loops topologically -- sometimes two loops for a given (LM,EM) pair. In the "nice" cases, what happens is that r is a monotonic function of s across the domain, so the state of the hoop hits the endpoint of the domain where v_r=0, switches sign for v_r, and retraces the same states in the opposite direction.

But in the pathological cases, r reaches a maximum as a function of s in the middle of a domain, where v_r is non-zero. So although the states that conserve angular momentum and energy still form a loop in (omega, r, v_r) space, v_r is not zero at the maximum r value, and it's physically impossible to traverse the whole loop.

 

[pervect]

At the moment, I'm out of ideas. I did stumble across a paper which at least talks about the existence of solutions, though I found it so hard to follow that I gave up:

http://www.citebase.org/fulltext?format=application%2Fpdf&identifier=oai%3AarXiv.org%3Agr-qc%2F0411145

if I take the abstract at face value, the existence of solutions is only been proven for 'small' omega.

 

[gregegan]

At the moment, I'm out of ideas. I did stumble across a paper which at least talks about the existence of solutions, though I found it so hard to follow that I gave up:

http://www.citebase.org/fulltext?format=application%2Fpdf&identifier=oai%3AarXiv.org%3Agr-qc%2F0411145

if I take the abstract at face value, the existence of solutions is only been proven for 'small' omega.

I'll see if I can make anything of this, though if even the existence of equilibrium solutions for low angular velocities is such heavy lifting it might not shed much light on the weird problems we get once our hoops are allowed to pulsate.

The work of Beig and Schmidt is the background for this nice PhD:

http://www.arxiv.org/abs/gr-qc/0605025

which I sort of came to terms with, at least to the point where I could "dumb-down" most of his fully-GR equations into simpler flat spacetime ones that agreed with my own treatment.

I suppose I ought to be computing the speed at which a step function perturbation moves around a hoop -- though given the difficulty of that computation I've been holding out in the hope of finding a simpler way to pin down the problem!

 

[gregegan]

I suppose I ought to be computing the speed at which a step function perturbation moves around a hoop -- though given the difficulty of that computation I've been holding out in the hope of finding a simpler way to pin down the problem!

Calculating the behaviour of a step function in the hoop seems unmanageable, but I just calculated something else that's far simpler. The linearised relativistic equation for a longitudinal perturbation in an infinite straight-line string is just the usual wave equation -- with a speed of sound that depends on the tension, and is equal to the speed of light when the tension is a factor of 1/Sqrt(3) times the ceiling on the tension set by the weak energy condition.

So it seems there could be a good reason to expect a much lower limit on the tension than we've been assuming so far. If that 1/Sqrt(3) factor is reliable, it does actually cut in before all the problems start (and sadly, before all the fun stuff with the energy peak).

I wish I'd done that calculation a few months ago ... !

edit: Just for the record, "when the tension is a factor of 1/Sqrt(3) times the ceiling on the tension set by the weak energy condition" isn't quite right; it's the stretch factor that is 1/Sqrt(3) times its weak-energy ceiling when the speed of sound hits the speed of light.