Stress-energy tensor of a wire under stress

[gregegan]

I'm afraid other obligations have limited my participation in this discussion, and will shortly end it completely. I've written up as much as I've been able to work out about pulsations and vibrations -- in both the Newtonian and relativistic cases -- at http://www.gregegan.net/SCIENCE/Rings/Rings.html#STABILITY

The idea I was working towards on the vibrations was to get a complete set of functions that could be used to synthesise arbitrary initial conditions for any small perturbation. I'm pretty confident that the solutions I found in both the Newtonian and relativistic cases that take the form of a traveling wave with a velocity exactly opposite the rotation of the hoop are correct; in other words, from a centroid-frame point of view, the hoop experiences an almost-stationary deformation of its shape, modified only by the longitudinal vibration which is 90 degrees out of phase with the transverse wave.

But curiously there are other traveling wave solutions in the Newtonian case, with different velocities, that don't seem to correspond to any simple relativistic equivalents. It's not hard to show that those "extra" Newtonian solutions do approximately satisfy the relativistic PDEs when r \omega is small, but what I can't see is the whole class of exact solutions of the relativistic PDEs whose low-velocity limit gives the Newtonian ones.

Anyway, I'd be curious to know if the PDEs pervect gets from his Lagrangian approach match the PDEs from my relativistic "F=ma" analysis when linearised.

 

[pervect]

I'm afraid that the equations I'm getting from the Lagrangian method for the general case are a real mess when expanded, which I haven't been able to do anything useful with, but I'll write them down here.

In a later post I'm also plan to do a comparison I promised Chris Hillman between the relativistic hoop and the Newtonian hoop, and also discuss a useful set of basis vectors for a "hoop riding" observer on the radially symmetrical expanding hoop similar to the Langevin basis vectors.

But onto the differential equations:

We start out with the wordlines of a point on the body as a function of the "body" coordinate phi, using a different notation that will be more compatible with Goldstein's (Classical mechanics) discussion of the continuous Lagrangian formulation.

r = \eta_{1}(t,\phi)
\theta = \eta_{2}(t,\phi)We will denote \frac{\partial}{\partial t} by appending ,0 : similarly ,1 will denote \frac{\partial}{\partial \phi}.

Thus \frac{\partial r}{\partial t} = \frac{\partial \eta_1}{\partial t} = \eta_{1,0}
and

\frac{\partial \theta}{\partial \phi} = \frac{\partial \eta_2}{\partial \phi} = \eta_{2,1}

Then we can write the stretch factor s as:

<br /> s = \sqrt {{\frac {{\eta_{1,1}}^{2}{{\eta_1}}^{2}{{\eta_{2,0}}}^{2}-<br /> 2\,{\eta_{1,1}}\,{\eta_{1,0}}\,{\eta_{2,1}}\,{{\eta_1}}^{2}{\eta_{2,0}}-<br /> {{\eta_1}}^{2}{{\eta_{2,1}}}^{2}+{{\eta1}}^{2}{{\eta_{2,1}}}^{2}<br /> {{\eta_{1,0}}}^{2}-{{\eta_{1,1}}}^{2}}{-1+{{\eta_{1,0}<br /> }}^{2}+{{\eta_1}}^{2}{{\eta_{2,0}}^{2}}}}<br />

and the Lagrangian density as

\mathcal{L} = -\rho(s) \eta_1 \eta_{2,1} \sqrt{1 - \eta_{1,0}^2} d \phi dt

here, for the hyperelastic hoop, the scalar function \rho(s) is

\rho(s) = \frac{\rho_0}{s} + \frac{k}{2s} + \frac{k s}{2} - k

and we require \eta_{2}(t, \phi=2 \pi) - \eta_{2} (t,\phi= 0) = 2 \pi for all t.

Given that s has been previously defined in terms of the various derivatives of \eta_1 and \eta_2, Lagrange's equations are then just

\frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \eta_{1,0}} \right) +\frac{d}{d \phi} \left( \frac{\partial \mathcal{L}}{\partial \eta_{1,1}} \right) - \frac{\partial \mathcal{L}}{\partial \eta_{1}} = 0<br />\frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \eta_{2,0}} \right) +\frac{d}{d \phi} \left( \frac{\partial \mathcal{L}}{\partial \eta_{2,1}} \right) - \frac{\partial \mathcal{L}}{\partial \eta_{2}} = 0<br />

 

[pervect]

OK, now for the promised comparison between the relativistic radially symmetrical hoop, and the Newtonian radially symmetrical hoop. Note that geometric units with c=1 will be used throughout.

We will allow the hoop radius r(t) and the hoop angular velocity \omega(t) to both be functions of time, however we will require the hoop to be radially symmetrical and the the angular velocity of any point on the hoop will depend only on time and not on position.

The associated Lagrangian for the hoop will be a function

L(r, v_r, \omega), where v_r = dr/dt

(We have omitted some variables such as time and angle present in a general Lagrangian but which are not present in our hoop).

For the relativistic hoop, using http://www.citebase.org/fulltext?format=application%2Fpdf&identifier=oai%3AarXiv.org%3Agr-qc%2F0403073
as a guide, we can write the Lagrangian of the hoop as a function of a scalar field \rho which is defined as the total energy density (base + stored) in a frame field comoving with the hoop.

This can be written as just:

L_r(r,v_r,\omega) = \int_V -\rho(s) dV = -2 \pi \, r \, A \sqrt{1-v_r^2} \, \rho(s)

dV is a volume element, and A is a constant, equal to the cross-sectional area of the hoop in its rest frame. With a Poisson's ratio of zero, A will be constant in a comoving frame. But we are interested in the value of A in the lab frame, where we a are performing the integral. We note that A in the lab frame is not a function of \omega but is a function of v_r - in fact, A gets Lorentz-contracted by a factor of \sqrt{1-v_r^2}

In this expression, s(r,v_r,\omega) is the "stretch factor" representing how much the hoop has been elongated. The density of the hoop in its rest frame depends on how much it has been stretched.

The hyperelastic hoop gives one specific model for the value of \rho(s) which is

\rho(s) = \frac{\rho_0}{s} + \frac{k s}{2} + \frac{k}{2s} - k

The stretch factor s can be determined from the geometry of the hoop . If the initial radius of the hoop is unity, the stretch factor s is just

s = r \, \sqrt{ \frac{1 - v_r^2}{1 - v_r^2 - r^2 \omega^2} }

Some explanation for the origin of this factor will be given later in another post, but note that when \omega=0, the stretch factor is r, while when v_r=0, the stretch factor is r/\sqrt{1 - r^2 \omega^2}, which is sensible.

Putting this all together, for the hyperelastic hoop we get a relativistic Lagrangian of

L_r =A {\frac { \left( 2\,kr\sqrt {1-{{\it v_r}}^{2}}\sqrt {1-{{\it v_r}}^{2}-{r}^{2}{\omega}^{2}}-{r}^{2}k+{r}^{2}k{{\it v_r}}^{2}-k+k{{\it v_r}}^{2}+<br /> k{r}^{2}{\omega}^{2}-2\,{\it \rho0}+2\,{\it \rho0}\,{{\it v_r}}^{2}+2\,{<br /> \it \rho0}\,{r}^{2}{\omega}^{2} \right) \pi }{\sqrt {1-{{\it v_r}}^{2}-{<br /> r}^{2}{\omega}^{2}}}}<br />

The Newtonian Lagrangian, in comparison is just the difference between kinetic and potential energies:

L_n = \frac{M}{2} \left( v_r^2 + r^2 \omega^2 \right) - \frac{K}{2} \left(r - 1\right)^2<br />

In both expressions, we are assuming that the equilibrium radius of the hoop at \omega=0 is unity.

Expanding the relativistic Lagrangian around v_r=0 and \omega=0 we find that at v_r=0 and \omega=0 we have:

<br /> \frac{\partial^2 L_r}{\partial v_r^2} = -2\,\pi \, A \,kr+\pi \, A \,{r}^{2}k+\pi \, A \,k+2\,\pi \, A \,{\it \rho0}<br />

<br /> \frac{\partial^2 L_r}{\partial \omega^2} = \pi A \,{r}^{2}k+2\,\pi \, A \,{\it \rho0}\,{r}^{2}-\pi \, A \,{r}^{4}k<br />

<br /> \frac{\partial L_r}{\partial v_r} = \frac{\partial L_r}{\partial \omega} = \frac{\partial^2 L_r}{\partial v_r \,\partial \omega} = 0<br />

This, along with the value of L_r at v_r=\omega=0 allows us to taylor series expand L_r around the origin as follows:

<br /> Lr \approx -M - \frac{K}{2} \left(r-1\right)^2 + \frac{1}{2} \left(M + \frac{K}{2}\left(r-1\right)^2\right) v_r^2 + \frac{1}{2} \left(M + \frac{K}{2} \left( 1 - r^2\right) \right) r^2 \omega^2 + O(4)<br />

where

M = 2 \, \pi \, \rho0 A
K = 2 \, \pi \, k \, A

i.e. 2 \, \pi \, r_0 \, A \, \rho0 is the total mass M of the hoop where r_0=1, and similarly K provides the force constant in terms of k, A, and the initial circumference 2 \pi r_0 = 2 \pi of the hoop.

We can see that the relativistic Lagrangian approaches the Newtonian value, except for a constant factor of -M that does not affect the equations of motion, in the limit where v_r &lt;&lt;1 and \omega \, r &lt;&lt; 1 and when the stored energy is also less than the rest mass, i.e. in geometric units we must also have

\left| \frac{K}{2} (r-1)^2 \right| &lt;&lt; M \hspace{.5in} \left| \frac{K}{2}(1-r^2) \right| &lt;&lt; M

We note that for radial velocity, the coefficient of v_r^2 is half the total energy and approaches M/2 for small K. The relativistic correction makes this coefficient slightly greater than M/2 due to the energy stored in the hoop.

However, for tangential velocity, the coefficient of (r \omega)^2 is actually lower than M/2 when r>1 and (1-r^2) is negative, though it approaches M/2 in the limit for small K. In the Newtonian limit we expect r>1. The explanation for this lies in the tension terms. Tension in the wire subtracts from the linear momentum in the direction of motion as discussed by Rindler in "Introduction to SR" in the section on continuum mechanics. This means that the tension terms subtract from the angular momentum of the wire. Since we know that {\partial L_r}/{\partial \omega} is equal to the angular momentum, we expect that there must be some relativistic modification of the coefficient of \omega^2 to reflect this reduction in angular momentum due to tension.

Similarly, except for higher order terms and the noted differences when the stored energy approaches the rest energy, the various quantities that can be calculated from the Lagrangian such as angular momentum, energy, and the equilibrium radius, also approach the Newtonian limit.

The Lagrangian for the wire allows us to calculate the time evolution of the hoop via Lagrange's equations. By choosing the relativistic or Newtonian Lagrangian, we can solve for respective equations of motion.For the radial velocity the equations of motion are:

\frac{d}{dt} \left( \frac {\partial L}{\partial v_r} \right) - \frac{\partial L}{\partial r} = 0

Thus the equilibrium radius of the hoop will occur when v_r=0 and \frac{\partial L}{\partial r} = 0.

For the angular velocity, the equations of motion simply say angular momentum is conserved:

\frac{d}{dt} \left( \frac{\partial L}{\partial \omega} \right) = 0

 

[pervect]

Finally, I want to present a useful set of basis vectors for the expanding hoop, to be compared with the Langevin set of basis vectors in for instance http://en.wikipedia.org/w/index.php?title=Born_coordinates&oldid=124648352

Consider a 1 parameter group of worldlines, expressed in a polar coordinate chart (t,r,theta) that form the worldsheet of our expanding hoop. We have two auxillary functions r(t) and \omega(t) that define the radius of the expanding hoop and the angular velocity of the hoop as a function of time. Then in terms of these functions, for every value of \phi we have an associated worldline in our cylindrical chart of

r = r(t)
\theta = \int \omega(t) dt + \phi

The first useful vector is the normalized 4-velocity of the worldline as a function of coordinate time, t. This is just

\vec{e0} = [\, {\frac {1}{\sqrt {1-{{\it v_r}}^{2}-{r}^{2}{\omega}^{2}}}} \frac{\partial}{\partial t} + {\frac {{\it v_r}}{\sqrt {1-{{\it v_r}}^{2}-{r}^{2}{\omega}^{2}}}} \frac{\partial}{\partial r} + {\frac {\omega}{\sqrt {1-{{\it v_r}}^{2} - r^{2}{\omega}^{2}}}}\frac{\partial}{\partial \theta}\,]<br />

The next useful vector is a spacelike vector which lies within the worldsheet of the hoop, but is perpendicular to the 4-velocity. This can be formed by taking a linear combination of a tangent vector \frac{\partial}{\partial \phi} which lies on the worldsheet but is not perpendicular to the 4-velocity, and the 4-velocity \vec{e0}. Normalized, this vector is just

\vec{e1} =[ \, {\frac {\omega\,r}{\sqrt { \left( 1-{{\it v_r}}^{2} \right) \left( 1-<br /> {{\it v_r}}^{2}-{r}^{2}{\omega}^{2} \right) }}}\frac{\partial}{\partial t} + {\frac {{\it v_r}\,<br /> \omega\,r}{\sqrt { \left( 1-{{\it v_r}}^{2} \right) \left( 1-{{\it v_r}<br /> }^{2}-{r}^{2}{\omega}^{2} \right) }}}\frac{\partial}{\partial r} + {\frac {1-{{\it v_r}}^{2}}{r<br /> \sqrt { \left( 1-{{\it v_r}}^{2} \right) \left( 1-{{\it v_r}}^{2}-{r}^{<br /> 2}{\omega}^{2} \right) }}} \frac{\partial}{\partial \theta} \, ]<br />

As mentioned by Greg Egan, if we compute the unnormalized vector y = \frac{\partial}{\partial \phi} + \alpha \vec{e0} that is perpendicular to \vec{e0}, the magnitude of y gives us the value of the stretch factor s. The value of s computed via this means was given in the previous post.The final useful vector is a normalized vector perpendicular to the worldsheet, i.e. perpendicular to both of the above vectors. This is just:

\vec{e2} = [ \,\frac{v_r}{\sqrt{1-v_r^2}} \frac{\partial}{\partial t} + \frac{1}{\sqrt{1-v_r^2}} \frac{\partial}{\partial r} \, ]

We would then expect the stress-energy tensor of our hoop to be given by the formula
T^{ab} = \rho \vec{e0} \times \vec{e0} + P \vec{e1} \times \vec{e1}

where \rho is the density of the hoop in its rest frame, and the pressure P must be in the plane of the worldsheet, i.e. in the direction of \vec{e1}[/tex]. The pressure normal to the worldsheet of the hoop must be zero. This gives the following stress energy tensor in a cylindrical coordinate chart.<br /> <br /> T^{ab} =&lt;br /&gt; \left[ \begin {array}{ccc} {\frac {\rho-\rho\,{{\it vr}}^{2}+P{\omega}^{2}{r}^{2}}{ \left( -1+{{\it vr}}^{2} \right) \left( -1+{{\it vr}}^&lt;br /&gt; {2}+{r}^{2}{\omega}^{2} \right) }}&amp;amp;{\frac {{\it vr}\, \left( \rho-\rho&lt;br /&gt; \,{{\it vr}}^{2}+P{\omega}^{2}{r}^{2} \right) }{ \left( -1+{{\it vr}}^&lt;br /&gt; {2} \right) \left( -1+{{\it vr}}^{2}+{r}^{2}{\omega}^{2} \right) }}&amp;amp;-&lt;br /&gt; {\frac {\omega\, \left( \rho+P \right) }{-1+{{\it vr}}^{2}+{r}^{2}{&lt;br /&gt; \omega}^{2}}}\\\noalign{\medskip}{\frac {{\it vr}\, \left( \rho-\rho\,&lt;br /&gt; {{\it vr}}^{2}+P{\omega}^{2}{r}^{2} \right) }{ \left( -1+{{\it vr}}^{2&lt;br /&gt; } \right) \left( -1+{{\it vr}}^{2}+{r}^{2}{\omega}^{2} \right) }}&amp;amp;{&lt;br /&gt; \frac {{{\it vr}}^{2} \left( \rho-\rho\,{{\it vr}}^{2}+P{\omega}^{2}{r&lt;br /&gt; }^{2} \right) }{ \left( -1+{{\it vr}}^{2} \right) \left( -1+{{\it vr}&lt;br /&gt; }^{2}+{r}^{2}{\omega}^{2} \right) }}&amp;amp;-{\frac {\omega\,{\it vr}\,&lt;br /&gt; \left( \rho+P \right) }{-1+{{\it vr}}^{2}+{r}^{2}{\omega}^{2}}}&lt;br /&gt; \\\noalign{\medskip}-{\frac {\omega\, \left( \rho+P \right) }{-1+{{&lt;br /&gt; \it vr}}^{2}+{r}^{2}{\omega}^{2}}}&amp;amp;-{\frac {\omega\,{\it vr}\, \left( &lt;br /&gt; \rho+P \right) }{-1+{{\it vr}}^{2}+{r}^{2}{\omega}^{2}}}&amp;amp;-{\frac {\rho&lt;br /&gt; \,{\omega}^{2}{r}^{2}+P-P{{\it vr}}^{2}}{{r}^{2} \left( -1+{{\it vr}}^&lt;br /&gt; {2}+{r}^{2}{\omega}^{2} \right) }}\end {array} \right] &lt;br /&gt;<br /> <br /> We can substitute for the density and pressure terms \rho and P for any desired model, such as the hyperelastic model - both \rho and P will be functions of the stretching factor s for any given model as has been previously discussed.<br /> <br /> By integrating the stress energy tensor by the volume of our expanding worldsheet at any time t, we should and do come up with the same expressions for total energy and also for angular momentum that we did using the previous Lagrangian approach. <br /> <br /> i.e in terms of the above stress-energy tensor<br /> <br /> energy = \sqrt{|T^{00} T_{00}|} * volume<br /> angular momentum = r \, \sqrt{|T^{02} T_{02}|} * volume<br /> <br /> Of course, since our metric is diagonal, T_{00} = g_{00}^2 T^{00} and T_{02} = g_{00} g_{22} T^{02}.<br /> <br /> In terms of the relativistic Lagrangian previously mentioned, the energy and angular momentum are just<br /> <br /> energy = \omega \frac{\partial L}{\partial \omega} + v_r \frac{\partial L}{\partial v_r} - L<br /> <br /> and<br /> <br /> angular momentum = \frac{\partial L}{\partial \omega}<br /> <br /> Note however that the volume of a hoop of radius r and proper cross sectional area A is<br /> &lt;br /&gt; V = 2 \pi r A \sqrt{1 - v_r^2}&lt;br /&gt;<br /> as previously noted, due to the Lorentz contraction in the radial direction in the lab frame.

 

[gregegan]

Unfortunately I can't seem to lay my hands on the quote right now, but one of the authors whose textbooks I have been studying remarks that something as simple as a vibrating and rotating hoop is quite tricky (in nonrelativistic elasticity) for technical reasons. I'm sure that results are published, but even a Newtonian discussion needs to be carefully examined, since its very easy to go astray by misinterpreting boundary conditions, etc.

I'm about to leave the net for a fortnight, but I just wanted to point out the somewhat surprising result I found in my Newtonian analysis of a vibrating, rotating hoop: there are unstable modes! I am assuming constant tension, which of course will not hold up for long if the mode is growing exponentially, so this behaviour might ultimately be constrained, but it's still interesting to find that an arbitrarily small perturbation of the right kind can, apparently, grow. The analysis is up on my web page, so feel free to examine it and let me know if you find anything wrong with it (please don't ask me to add Poisson's ratio, though; we're having enough trouble with our maximally idealised models).

Anyway, thanks to you both for a fun discussion.

 

[Chris Hillman]

Thanks for everything, Greg!

I'm afraid other obligations have limited my participation in this discussion, and will shortly end it completely.

Darn. My expository energies flagged in the other thread (not being able to insert new material in the old posts is such a drag!), but I am increasingly convinced that the long hard slog by first very carefully analyzing analogous Newtonian rotating disks/hoops or stretched bars, plus discrete models, is the only way. The literature provides ample evidence that even this is challenging, in fact some pretty famous mathematicians have gotten things wrong!

I've written up as much as I've been able to work out about pulsations and vibrations -- in both the Newtonian and relativistic cases -- at http://www.gregegan.net/SCIENCE/Rings/Rings.html#STABILITY

Excellent! Thanks much for making this effort. My intent has always been to catch up with you via the Newtonian route. Hopefully in the other thread I'll eventually explain your analysis of Newtonian vibrating hoops in my own way.

The idea I was working towards on the vibrations was to get a complete set of functions that could be used to synthesise arbitrary initial conditions for any small perturbation. I'm pretty confident that the solutions I found in both the Newtonian and relativistic cases that take the form of a traveling wave with a velocity exactly opposite the rotation of the hoop are correct; in other words, from a centroid-frame point of view, the hoop experiences an almost-stationary deformation of its shape, modified only by the longitudinal vibration which is 90 degrees out of phase with the transverse wave.

That's an interesting requirement. A Lie symmetry analysis of a system of differential equations, BTW, can often reveal the presence of interesting choices of parameters in a given problem. (See Stephani's textbook.)

But curiously there are other traveling wave solutions in the Newtonian case, with different velocities, that don't seem to correspond to any simple relativistic equivalents. It's not hard to show that those "extra" Newtonian solutions do approximately satisfy the relativistic PDEs when r \omega is small, but what I can't see is the whole class of exact solutions of the relativistic PDEs whose low-velocity limit gives the Newtonian ones.

Anyway, I'd be curious to know if the PDEs pervect gets from his Lagrangian approach match the PDEs from my relativistic "F=ma" analysis when linearised.

Although I seem to have stalled in my long hard slog approach, behind the scenes I think I have been continuing to make progress on understanding such difficulties, so perhaps at some point in the future we can lure you back!

If nothing else, I think we have succeeded in persuading some lurkers that the problem of the relativistic rotating hoop is nothing to sneeze at! In fact, even the Newtonian theory is rather intricate, subtle, and technically challenging (facts which elasticity books for engineers tend to cover up, which might partially explain the odd structural failure).

 

[pervect]

I've found some interesting difficulties by attempting to numerically solve my Lagrangian for the equations of motion with particular initial conditions.

I find that the simulation reaches points where the equations for the time evolution of the radially symmetric "pulsing" hoop become singular. This has happened for both the breakable hoop and the hyperelastic hoop.

The basic difficulty is this:

The angular momentum AM(r,vr,\omega) must be constant. However, there exist points for which

\frac{\partial AM}{\partial r} \neq 0
\frac{\partial AM}{\partial \omega} = 0
\frac{\partial AM}{\partial vr} = 0

and for which vr is not zero.

Because vr is not zero, r must change. But there is no way that \omega and vr can change to keep AM constant, because the partial derivatives of AM with respect to \omega and vr are both zero, while the partial derivative of AM with respect to r is nonzero.

An example of this occurs with the breakable hoop:

rho := (-1/12*s^3+5/8*s^2-s+35/24)/s

at

omega = .299
r =1.75564
vr = .1348649775

For the hyperelastic hoop
rho = (-2s+s^2+5)/4s

an example occurs at

omega = .236
r = 1.754
vr = .8034284089e-1

Interestingly enough, it seems that \frac{\partial AM}{\partial vr} = 0 \implies \frac{\partial AM}{\partial \omega} = 0

This is rather unphysical - my conclusion is that it is probably the assumption of radially symmetry that causes the problem.

i.e. in these particular examples, AM is dropping with increasing r (\partial AM / \partial r &lt; 0, vr>0). If we assume that the hoop moves in a radially symmetric manner, there is no way that we can maintain the value of L as r increases, it must drop. But there should be a way to keep L from dropping by allowing the hoop to assume a non-circular shape.

The motivation for the simulations that lead to the above observations might also be interesting. There appears to be a maximum amount of angular momentum, AM, that an equilibrium circular hoop can hold. However, by considering non-equilibrium circular hoops, we can form a hoop that has greater than the maximum allowed amount of angular momentum. Analysis of the simulation results indicate that the Lagrangian itself becomes singular as we attempt trace the time evolution of such a hoop (\omega=.8, r=.5, vr=0 will do for starting conditions for both models mentioned above).

 

[gregegan]

Have you considered working with s, the stretch factor, rather than r, as the "radial" degree of freedom? My own analysis of stability of pulsations was much simpler when I switched to s, because it has a monotonic relationship with omega.

BTW, I've written a little applet that does numerical simulations of a Newtonian hoop, and this has clarified the question of stability of vibrations. Basically, if the tension is high enough perturbations to the shape will be constrained (although there are some modes that grow exponentially at first from infinitesimal perturbations), but if the hoop is too close to its relaxed state, it will be vulnerable to crumpling.

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

 

[pervect]

Unfortunately, replacing s with r in the relativistic Lagrangian doesn't seen very straightforwards, because s is a function of r and dr/dt in the relativistic case:

s = r \, \sqrt{ \frac{1 - v_r^2}{1 - v_r^2 - r^2 \omega^2} }

Perhaps there's some way to write L(s, ds/dt, \omega) but it isn't obvious to me.

It may be possible to solve the angular momentum (at least for the hyperelastic hoop) for \omega, which would allow for a standard effective potential treatment. The problems with this approach are that there are multiple solutions, warnings about "solutions being lost" from Maple, and the fact that the solutions offered by Maple are very long, complex, and difficult to work with.

There appears to be some minor differences between my results for the pulsating hoop and Greg Egan's. I attribute these differences to small but perhaps important differences in our expressions for angular momentum and energy. Greg has basically assumed that the volume of the hoop is 2 pi r A, and assumed that A is constant by ignoring Lorentz contraction of A due to the radial velocity. I made the same assumptions earlier - when I did, I got the same results Greg did. Our results for energy and momentum density agree, the discrepancy lies in the volume that they are multiplied by to get the total energy and angular momentum. However, I found that I have to use my Lorentz contracted volume element to get a self-consistent Lagrangian formulation.

I still believe that the numerical simulations and an analysis by hand of the points where they fail demonstrates that the problem with the Lagrangian approach is that the Lagrangian does actually become singular.

 

[gregegan]

There appears to be some minor differences between my results for the pulsating hoop and Greg Egan's. I attribute these differences to small but perhaps important differences in our expressions for angular momentum and energy. Greg has basically assumed that the volume of the hoop is 2 pi r A, and assumed that A is constant by ignoring Lorentz contraction of A due to the radial velocity. I made the same assumptions earlier - when I did, I got the same results Greg did. Our results for energy and momentum density agree, the discrepancy lies in the volume that they are multiplied by to get the total energy and angular momentum. However, I found that I have to use my Lorentz contracted volume element to get a self-consistent Lagrangian formulation.

Have you found anything qualitatively different on the stability question (for axially symmetric states)? Although I agree that it's absolutely necessary to include the Lorentz contraction when attempting to compute the detailed time evolution of a pulsation, my aim was just to establish the stability or otherwise of various equilibrium states. To that end, what I computed was the total energy and angular momentum of non-equilibrium, instantaneously stationary hoops. Given that the stability is a question of the shape of this curve as seen in the limit of an infinitesimal perturbation, the Lorentz contraction shouldn't matter. And even for the question of the "finite containment" of the unstable equilibria, it's hard to see how the "energy ridge inside a valley" can be transformed into something else by relativistic effects.

I'll have to think harder about the singularity you're finding, but my own hunch is still that this is down to the choice of variables, and not that there's anything physically impossible about a perfectly symmetrical hoop undergoing pulsations that encompass the "exotic points" here: the energy maximum, or the radial maximum. My Newtonian analysis suggests that a perfectly circular shape is not stable -- even at high tension there will be a contained instability -- but that shouldn't really matter; implausibly confining a system to a state of high symmetry shouldn't lead to anything physically impossible!

 

[pervect]

For pulsations near the equilibrium points, I don't think there's any problem with your analysis.

However, it appears to me that the difference is important in regards to the singular points I'm finding, which your analysis doesn't appear to find (though it wasn't intended to be that general).

Your derivative of angular momentum with respect to omega doesn't vanish unless the weak energy condition is violated - mine, however does. And that's the problem.

For instance for the hyperelastic hoop (k=1/2, rho0=1) at omega = .236, r=1.754, vr=.08034284089 I find a singular point with s=1.928. Your derivative of angular momentum with respect to omega doesn't vanish there - mine does.

 

[pervect]

The trial solutions Greg suggests do work if I ignore terms of \delta^2 and higher, howver I get real values for c, so our PDE's probably aren't the same, and one or both of us is probably making some silly error :-(.

To normalize the problem, I set \omega=1 which implies that \rho = k.
With this normalization, I'm currently getting two sets of solutions: c = -1

or

\alpha = \frac{1}{r_{eq}} \hspace{.5 in} c = 1 + \frac{2}{m}
\alpha = -\frac{1}{r_{eq}} \hspace{.5 in} c = 1 - \frac{2}{m}

for Greg's trial solutions (with my slightly different notation)

<br /> r = \eta_1(t,\phi) = r_{eq} + \delta \cos(m(\phi - c \, \omega \, t)) <br />
<br /> \theta = \eta_2(t,\phi) = \phi + \omega \, t + \delta \, \alpha \, \sin(m(\phi - c \, \omega \, t)) <br />

c=-1 is an exact solution. The second solution satisfies the first partial differential equation to order \delta^3, the second to order \delta^2.

 

[gregegan]

The trial solutions Greg suggests do work if I ignore terms of \delta^2 and higher, howver I get real values for c, so our PDE's probably aren't the same, and one or both of us is probably making some silly error :-(.

The error's mine! I was assuming that tension was constant to first order in delta, but it turns out that's only true for the c=-1 solutions.

I've redone the calculation allowing for variable tension, and I now always get three real solutions from the cubic (details on my web page) plus c=-1. In the high-tension limit I agree with your solutions (with the cubic giving c=-1 a second time, in that limit), though more generally I think there are four distinct solutions for c. (For m=1, though, the cubic again gives c=-1 as one of its solutions).

 

[gregegan]

I've found some interesting difficulties by attempting to numerically solve my Lagrangian for the equations of motion with particular initial conditions.

I find that the simulation reaches points where the equations for the time evolution of the radially symmetric "pulsing" hoop become singular.

I derived equations of motion by setting the divergence of the stress-energy tensor to zero, and I get the same results as you describe, including the behaviour of the derivative of the total angular momentum wrt to v_r and \omega, with a singularity at the numerical value you quote (for the hyperelastic case).

But although this is a pain in the neck technically, I'm not sure that it signifies anything unphysical. What I find is that when the derivative of the angular momentum with respect to \omega (and simultaneously wrt v_r) is zero, the derivative of \omega wrt time is infinite. Hard as this is to deal with (especially numerically), is there anything unphysical about a plot of \omega vs time being instantaneously vertical, so long as the integral is bounded? The physical evolution of the state ought to be able to pass through this point, with the apparent problem of the zero derivative of the angular momentum with respect to \omega being overcome by the infinite derivative of \omega wrt time: in other words, the angular momentum gets to stay constant because the product of those two derivatives is really bounded but non-zero, and they work together to keep the angular momentum constant by cancelling out the contribution due to the non-zero derivative of the angular momentum wrt r.

I'd hope that with a bit of work the differential equations for the time evolution could be shown, either with a change of variables or some heavy-duty analysis, to be soluble across the singularity. This might turn out to be a naive hope, but I don't think a singularity in the time derivative of \omega per se can be taken to be a fatal, physical flaw in the model without further exploration.

 

[pervect]

I derived equations of motion by setting the divergence of the stress-energy tensor to zero, and I get the same results as you describe, including the behaviour of the derivative of the total angular momentum wrt to v_r and \omega, with a singularity at the numerical value you quote (for the hyperelastic case).

Glad to have the confirmation.

But although this is a pain in the neck technically, I'm not sure that it signifies anything unphysical. What I find is that when the derivative of the angular momentum with respect to \omega (and simultaneously wrt v_r) is zero, the derivative of \omega wrt time is infinite.

I'm not quite sure what to make of this behavior yet, but I know I don't like it.

Hard as this is to deal with (especially numerically), is there anything unphysical about a plot of \omega vs time being instantaneously vertical, so long as the integral is bounded?

I hadn't really considered the possibility that the jumps might be bounded - my intuition was (is still) pointing in a different direction.

Assuming that the jumps are bounded (I'll try and look at this more closely), I still find this a bit problematical. We have bits of matter changing velocity in zero time, this would require infinite accelerations.

I would think this would require either infinite forces, or zero mass. The system is close to the weak energy limit, but not quite at it.

The physical evolution of the state ought to be able to pass through this point, with the apparent problem of the zero derivative of the angular momentum with respect to \omega being overcome by the infinite derivative of \omega wrt time: in other words, the angular momentum gets to stay constant because the product of those two derivatives is really bounded but non-zero, and they work together to keep the angular momentum constant by cancelling out the contribution due to the non-zero derivative of the angular momentum wrt r.

I'd hope that with a bit of work the differential equations for the time evolution could be shown, either with a change of variables or some heavy-duty analysis, to be soluble across the singularity. This might turn out to be a naive hope, but I don't think a singularity in the time derivative of \omega per se can be taken to be a fatal, physical flaw in the model without further exploration.

 

[pervect]

OK, more on the singularity problem.

If you take the expression for angular momentum from the Lagrangian approach (i.e. including the correction terms in vr for the volume) for the hyperelastic hoop with k = (1/2) and rho0 = 1 and r_0 = 1 I get

<br /> AM = -{\frac {{r}^{2}\omega\,\pi \, \left( -5/2+5/2\,{{\it vr}}^{2}+5/2\,{r<br /> }^{2}{\omega}^{2}+1/2\,{r}^{2}-1/2\,{r}^{2}{{\it vr}}^{2} \right) }{<br /> \left( 1-{{\it vr}}^{2}-{r}^{2}{\omega}^{2} \right) ^{3/2}}}<br />

I'm using AM rather than L to avoid confusion with the Lagrangian.

The maximum angular momentum occurs when d AM/ d \omega = 0. This also implies that d AM / d vr = 0. This occurs when

<br /> \omega = {\frac {\sqrt { \left( 2\,{r}^{2}+5 \right) \left( 5-5\,{{\it vr}}^{2<br /> }-{r}^{2}+{r}^{2}{{\it vr}}^{2} \right) }}{ \left( 2\,{r}^{2}+5<br /> \right) r}}<br />

Substituting and simplifying (assuming vr^2 < 1 and r < 5) one gets the simple expression for AM_max

<br /> 1/9\,\sqrt {3}\pi \, \left( 5-{r}^{2} \right) ^{3/2}<br />

This is a monotonically decreasing function of r, and independent of vr. (It might be interesting to generalize this result).

What this means is that as the hoop expands, the maximum value of angular momentum decreases. (When r^2=5 it can't hold any angular momentum - I believe this is where the weak energy condition is being violated).

So if we start out with a small hoop with enough angular momentum, [edit]the equations of motion from the Lagrangian tell us that its expansion accelerates, but the above equation sets a limit on how much the hoop can expand while remaining radially symmetrical.

I assume that this implies that the hoop must expand in a non-symmetric manner, but I don't understand this at the level of cause and effect - it still seems like a very strange result.

 

[pervect]

Another comment. I don't have a lot of confidence in the results yet, but it seems that the relativistic hoop may have complex solutions for 'c' even for low values of omega, i.e. omega=.1, according to the Lagrangian approach. The Lagrangian hasn't gotten any less messy, but I substitute in the trial solution, and find the coefficient of order \delta in the result by taking the partial derivative with respect to \delta evaluated at \delta=0. This gives an expression that is just long and involved, rather than unmanageable.

 

[gregegan]

So if we start out with a small hoop with enough angular momentum, [edit]the equations of motion from the Lagrangian tell us that its expansion accelerates, but the above equation sets a limit on how much the hoop can expand while remaining radially symmetrical.

I assume that this implies that the hoop must expand in a non-symmetric manner, but I don't understand this at the level of cause and effect - it still seems like a very strange result.

I agree with all the specific formulas for angular momentum etc. in this post. What I'm not clear about is precisely why (and with how much confidence) you believe the hoop ever reaches the singular value. The simplest resolution to this problem would be if the hoop never actually expands that far, and if the only reason you think it does get that far is a numerical solution to the equations of motion, it could be that the numerical solution is behaving badly prior to the singularity and over-estimating how large r becomes before the hoop begins to contract.

Do you find that the hoop is at least decelerating radially before it reaches the singular value, even if it's unclear exactly where it stops expanding and starts contracting?

(I'm trying some numerical simulations myself, by solving for \omega in terms of angular momentum and substituting that into the energy to get an equation involving only r and v_r. Mathematica is grinding away, but on my 300Mhz computer this might take days ...)

 

[gregegan]

The simplest resolution to this problem would be if the hoop never actually expands that far...

If I solve for \omega in terms of angular momentum (using the initial starting conditions you suggested in an earlier post, r=1/2, \omega=4/5, v_r=0), substitute \omega into the total energy, equate the total energy to the constant value set by the starting conditions, and then set v_r=0 to look for points where the hoop reverses its expansion ... this definitely doesn't happen before the hoop reaches the singularity. So regardless of the accuracy of any numerical simulation of the time evolution, I can't see how the singularity can be avoided.

(We know the r value the hoop must have when it reaches the singularity, by equating AM_max to our constant AM and solving for r. By substituting the critical expression for \omega into the total energy we can equate that to our constant energy and solve for v_r. And then we can feed those two values back into the expression for \omega. Doing this, I get r=1.7542296, v_r=0.0670504, \omega=0.236135155, i.e. close to the original values you gave for the singularity, though v_r is a little less.)

 

[pervect]

I am relying on the numerical simulations to say that the hoop actually reaches the critical points.

I don't have any reason to question the numerical simulations at this point, but they are numerical simulations.

I'd suggest rather than attempting to solve for omega, that you solve simultaneously and numerically

[d AM / dt = 0], [d pr / dt = fr]

where AM is the expression angular momentum, i.e. AM = \partial L / \partial \omega, (add) pr is the radial momentum and fr is the generalized force, i.e. pr = \partial L / \partial v_r, fr = \partial L / \partial r - that's what I did. (Though in some respects, it might be better if you did something different than what I did).

In more detail, I took the equation for L(r,vr,omega), evaluated the above quantities as partial derivatives, and substituted the variables (r,vr,omega) with functions of time (r(t), vr(t), omega(t)), which I then put into Maple's built-in numerical simulator.

As far as checking the simulation, one can check the angular momentum AM and the energy h (h = omega*AM + vr*pr - L) to see if it's conserved - it seemed to be.

I did check the initial conditions I mentioned, and it turns out that fr < 0 a t the critical point I specified,

omega = .236, r=1.754, vr=.0834284089

which essentially means that the hoop is de-accelerating radially at that point - near that point (actually near vr=0) we can write pr as

<br /> pr \approx ( 7.573105388\,{\it vr}+ 3.892083308\,{{\it vr}}^{3}+ 2.919429833\,{{<br /> \it vr}}^{5}+O \left( {{\it vr}}^{7} \right) )<br />

so you're right in that it's "trying" to stop expanding, but it has too much radial momentum.

 

[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.

<br /> \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<br />

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:

<br /> r = r_{eq} + \delta e^{\alpha t} \cos(m(\phi-c\omega*t))<br />
<br /> \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))<br />

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?

<br /> \sqrt{r^2 + (\frac{dr}{d\theta})^2} d\theta<br />

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.