Stress-energy tensor of a wire under stress

[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:

<br /> \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}{<br /> {\it eta2_{,1}}}^{2}}-1 \right) ^{2} \right) d \phi<br />

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:

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

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}<br /> +2\,{\it \eta1\_1}\,{\it \eta1\_0}\,{\it \eta2\_1}\,{{\it \eta1}}^{2}{\it <br /> \eta2\_0}+{{\it \eta1}}^{2}{{\it \eta2\_1}}^{2}-{{\it \eta1}}^{2}{{\it <br /> \eta2\_1}}^{2}{{\it \eta1\_0}}^{2}+{{\it \eta1\_1}}^{2}}}{\sqrt {1-{{\it <br /> \eta1\_0}}^{2}-{{\it \eta1}}^{2}{{\it \eta2\_0}}^{2}}}}<br />

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:

<br /> r = r_e + \delta cos(m \phi)<br />

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

<br /> \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 e^{t}and e^{-t}, and if we try solutions of the form x = cos(omega t), we find omega is complex.&lt;br /&gt; &lt;br /&gt; Thus the linear methods we are using would find this hypothetical system unstable.&lt;br /&gt; &lt;br /&gt; 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:

<br /> Q^4 n^4 - 2 K^2 Q^2 n^3 + 6 K^4 n - 5 K^4<br />

where

<br /> K^2 = k / (2 \rho)<br />
<br /> Q^2 = 1 + K^2<br />

 

[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.

 

[pervect]

OK, it's been awhile, but I think I'm getting the same answer.

If v_c is the speed of sound in the unstretched rod, the maximum value of the stretch factor s where the speed of sound becomes equal to c is

<br /> {\frac {\sqrt {6+3\,{{\it v_c}}^{2}}}{{3 \it v_c}}}<br />

For a Newtonian hoop, the speed of sound would increase simply by the stretch factor s - the Lagrangian is the same, but one has to multiply the wave propagation speed for phi by the stretch factor s to get the actual physical wave speed through the medium.

For the relativistic hoop, we'd have something like:

eta(t,phi) = s0*phi + f(phi - beta*t)

The Lagrangian density of the hoop is
<br /> \mathcal{L} = -\rho(s) \, s \, \sqrt{1 - \eta_0^2} d \phi<br />

where
<br /> s = \frac{\eta_1}{\sqrt{1-\eta_0^2}}<br />

which sensibly reduces to \eta_1 in the non-relativistic limit, and

<br /> \rho(s) = \frac{2 + v_c^2 \, (s-1)^2}{2 \, s}<br />

Approximating the Lagrangian as a quadratic gives:

Lapprox = (-1/2*v_c^2*eta_1^2+1+1/2*v_c^2) eta_0^2 - v_c^2 eta_1^2

[add]Question - Can I really justify this approximation, though - eta_0 may not be small.We can divide this by the coefficient of eta_0^2 to find the effective velocity v_y^2, and set v_y * eta_1 = 1 to solve for the maximum stretch factor eta_1.

For v_c^2 = 1/2, this puts the maximum stretch factor at around 1.225, well before the energy peak.

Plugging the critical value of s into the equation for dr/ds and dE/ds seems to indicate that both the radius and the energy are increasing when s reaches its maximum allowable value.

 

[gregegan]

OK, it's been awhile, but I think I'm getting the same answer.

We seem to agree 100% on the speed of sound in a straight rod under tension. I'm not sure that I fully understand your other calculations, but we seem to get roughly the same result, in as much as I find the phase velocity of one mode of vibrations in the hoop to hit c when s is approximately 1.29, at least for short wavelengths (which is just the same point as where it happens for a rod).

I've been agonising a lot about trying to compute the rate of information flow -- the front velocity rather than the phase velocity -- but when I look at this plot:

http://www.gregegan.net/SCIENCE/Rings/RelativisticVibrationSpeed.gif

(explained a bit more on the web page
http://www.gregegan.net/SCIENCE/Rings/Rings.html
towards the end, in the section on relativistic vibrations)

where the curves of different colour represent different values of m, i.e. different wavelengths, it looks as if the phase velocity for the relevant modes is becoming almost independent of m by the time you get to m=5 (the magenta curves), so there really shouldn't be a huge amount of dispersion. There are a lot of quibbles here; we can't really take the linearised PDE seriously when m goes to infinity, as it needs to when we analyse a sharp-edged pulse. But it does seem quite plausible that the rate of information flow demanded by the model becomes superluminal around s=1.29, not only well before the energy peak (s=5/3, or 1.6667), but also well before the crisis point where the rate of change of angular momentum wrt omega is zero, which intersects the equilibrium states at s=1.617.

If this is true, it's interesting that things don't go haywire with the equations of motion as soon as the velocity of sound becomes superluminal! It's only when you stretch the hoop quite a bit beyond that point that the dynamics become completely absurd.

 

[pervect]

What I think is interesting is that from a purely Newtonian POV, that the speed of sound in the rod should increase with the stretch factor. Thus if you stretch a wire so it is 10% longer, the speed of sound in that wire should also increase by 10%. I've been meaning to redo this analysis in terms of force/body diagrams rather than a Lagrangian, but I haven't gotten around to it.

There are relativistic effects as well.

As far as the pertuabations go, I think that ideally they should have bounded first and second derivatives, though perhaps a step function in velocity (a bounded and low first derivative of position, but no bound on acceleration) might be acceptable.

 

[gregegan]

I've analysed a pulse with a bounded and continuous second derivative propagating through a relativistic hoop; movie at:

http://www.gregegan.net/SCIENCE/Rings/SmoothPulseInHoop.gif

and more details at:

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

While I can't pinpoint exactly where the hyperelastic model begins to imply superluminal signalling, it's clear from the behaviour of a pulse like this that the transmission of information is not significantly slower than the short wavelength limit of the linearised PDE, which becomes superluminal at exactly the same stretch factor where a straight string's speed of sound becomes superluminal.