Stress-energy tensor of a wire under stress

[gregegan]

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

I'm not sure if you're worried that the analysis we're doing is somehow presupposing that the forces on the hoop will always be radial? I don't believe that's the case. Given the perturbations in r and $\theta$, the approach we're taking will include a force, and an acceleration, in whatever direction in the z=0 plane that the mixture of the changing local direction of the tangent to the hoop and the changing magnitude of its tension implies.

Maybe what's worrying you here is the fact that we're neglecting shear? I think that in the limit of a very narrow ring undergoing small perturbations of shape, it's reasonable to continue to omit shear. We could always add it in as yet another degree of freedom -- with an associated potential energy due to the material's resistance to shearing -- but it seems like begging for more work when we've yet to fully analyse the simpler case, and I don't see any reason to believe that shear will make any qualitative change to the hoop's behaviour close to equilibrium.

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

I read the section you cited in MTW, but that's really primarily concerned with the general-relativistic consequences of a lower-dimensional (delta-function-valued) stress-energy tensor: which measures of spacetime curvature are discontinuous across the surface and which aren't. Here, of course we're not using T as a source of any field, but we want to make sure we're doing the right thing with the divergence.

My understanding of what we're doing is that, in effect, we're looking at T inside an infinitesimally thin layer of material, and imposing the condition that nothing outside that layer is exerting any force on it. That's the way I reached the hoop solution from the ring solution on my web page; the boundary of a ring has zero pressure orthogonal to it, and if you take the limit of bringing the inner and outer boundaries together, then there is zero orthogonal pressure everywhere.

In the case of a perturbed hoop, we're aligning the sole pressure (tension) with the vector w, which is a tangent to the world sheet. This is declaraing that nothing is pushing on the hoop except other bits of hoop. But that doesn't mean that there's any restriction on the directions in which it can be pushed; since it's free to wiggle around arbitrarily in the plane, there's no direction, in principle, in which it can't be pushed.

 

[pervect]

The physical experience of a hoop-riding observer would be described by the vectors $$\hat{u}, \hat{w}, \hat{z}$$, and a fourth vector which we haven't given a name yet which would be perpendicular to all of them.

In our previous analysis, this vector was $$\hat{r}$$, but $$\hat{r}$$ is no longer perpendicular to $$\hat{w}$$.

This bothers me. For instance, how can we be sure that we are modelling a hoop of constant thickness? Because r is not perpendicular to the hoop, delta-r won't be a constant, for instance (though it would be constant in our unperturbed hoop).l

The delta function that models the distribution of matter in the hoop vs r won't have a constant area in our perturbed hoop because dr isn't constant.

 

[gregegan]

The physical experience of a hoop-riding observer would be described by the vectors $$\hat{u}, \hat{w}, \hat{z}$$, and a fourth vector which we haven't given a name yet which would be perpendicular to all of them.

In our previous analysis, this vector was $$\hat{r}$$, but $$\hat{r}$$ is no longer perpendicular to $$\hat{w}$$.

This bothers me. For instance, how can we be sure that we are modelling a hoop of constant thickness? Because r is not perpendicular to the hoop, delta-r won't be a constant, for instance (though it would be constant in our unperturbed hoop).l

The delta function that models the distribution of matter in the hoop vs r won't have a constant area in our perturbed hoop because dr isn't constant.

I'm not using delta functions anywhere myself, but if you want to use them, even just conceptually, imagine them as having a constant integral when integrated normal to the world sheet.

But the only point where we actually need to integrate anything is when computing the total energy or angular momentum, which will obviously need to be done with care for configurations that aren't axially symmetric.

Maybe my comments about the choice of $r_0$ as a third coordinate were a bit confusing. We need some third coordinate to work with that will never be degenerate, and assuming we never distort the hoop so much that a stretch of it lies along a radius, the way I defined $r_0$ will meet that need. But other than convenience, the choice of third coordinate is completely arbitrary (so long as you compute the metric components for it correctly, of course, and use that metric for your covariant derivatives and divergence). It has no physical meaning. All the physics is concentrated in the non-zero contributions to the stress-energy tensor, which come from $rho$, u, P, and w. It's the choice of w alone that defines the orientation of each small segment of the hoop; that w is not generally orthogonal to $\partial_{r_0}$ is simply a fact of life that is recorded in the metric.

The reason we know we're modelling a hoop of constant (infinitesimal) thickness is because the density of rest mass and potential energy are modified only by the single factor, n, and that factor is determined entirely by w. How thick the hoop "really is" is hidden in whatever recipe we use to compute $\rho_0$ and k.

But when, say, we take derivatives of T to compute the divergence, then although it makes a difference to the individual computations what coordinate system we choose -- including what we're using as the 3rd coordinate -- everything physical that follows from the calculations is independent of the coordinate system, and tied only to the physical directions of u and w. I'm not taking the derivative of any delta functions here; rather I'm assuming the existence of the tensor on an open set containing the world sheet, which is treated as lying in the interior of an arbitrarily thin hoop. We don't need to ask "how thin or thick", we just need the derivatives on the world sheet itself to be correct, and if we've defined n properly, then they will be.

Actually, it was probably a very bad choice for me to call the 3rd coordinate $r_0$, because when I write:

$$n=r_0/|y|$$

that's a fixed number, the unique relaxed radius of the hoop, not a spacetime coordinate. So that's potentially very confusing, and I should switch to a different name, $r_1$. Taking a derivative wrt $r_0$ of n would not be good ... that sloppy naming might actually be what's messing with my latest stability calculations. I'd better go and see if that's the case!

 

[gregegan]

Taking a derivative wrt $r_0$ of n would not be good ... that sloppy naming might actually be what's messing with my latest stability calculations. I'd better go and see if that's the case!

It turns out that wasn't a problem at all; because of where it appears in the stress-energy tensor, all derivatives of n vanish anyway.

The real problem, I've finally realized, is that if you have independent perturbations of size $\delta$ in r and $\theta$ away from the equilibrium values, the energy contains a term that is linear in $\delta$, whereas you really need the energy to have a zero first derivative in the perturbation for a linearised analysis to work.

What I know works is when you fix angular momentum and perturb a single variable; then the energy change is quadratic in the perturbation. I suppose if you go to high enough order you ought to be able to impose conservation of angular momentum as part of satisfying div T = 0, but that means solving non-linear PDEs.

There must be an easier way to construct two independent perturbations that lead to a quadratic energy change, but it's not obvious to me at this point.

 

[pervect]

Someone, somewhere must have done a thorough Newtonian analysis of vibrating rotating hoops. Probably we need to find and expand on those results rather than re-invent them.

Perhaps vibration of a rotating hoop will be the same as vibration of a string, but I have my doubts. I'm mainly concerned about coriolis forces which will be present in a rotating hoop but not in a classical vibrating string. I suspect that these forces should cause coupling between vibrational modes and longitudinal pressure waves. but that's really just a guess on my part at this point.

Off on a different tangent:

We know that for an equilbirum hoop, -P = (r * omega^2) rho. We can consider non-equilbirum hoops by modifying this relationship. For instance, we can add extra tension to the hoop by adding detachable dead weight. If we remove the dead weight, we'd have a hoop would contract.

If we take -(P+load) = (r omega)^2 rho then tension = (r * omega^2) rho+ load, so a positive load should represent extra tension, unless I'm making a sign error, that is.

Plotting the maximum energy with such a load, I find that negative values of load increase the maximum storable energy stored in the hoop, while positive values of load decrease the maximum energy, at least for the one case I tried (with the breakable hoop).

This suggests that if we create a hoop with too much energy by means of a load (which would have to be negative, i.e. something other than the dead weight I mentioned) that it would explode outwards, rather than implode (again,unless I'm making a sign error).

Note that I'm not including the energy associated with the loading mechanism in the "energy of the hoop", so for dead-weight loading, I'm only counting the energy of the hoop and not the energy in the dead weight. When we cut the strings tying the dead weight to a hoop, we'd have a contracting hoop with more tension than it needs to be in equilibirum and a total energy E.

 

[pervect]

Another so far unsuccessful attempt: I've been playing with a simpler example, but I'm finding it also intractable:

Consider
r = r(t)
theta = phi + alpha(t)

so that d (alpha)/dt = at t=0 is omega

This is a radially symmetric "pulsing" hoop.

Calculating u, w, n, and even T without the material model isn't too bad (T = rho(t) u x u + P(t) w x w)

dealing with the result of setting the divergence equal to zero appears to be problematical, even before adding in the material model, i.e. rho(s(t)) and P(s(t)).

[add]
But I think maybe an approach based on keeping the angular momentum and total energy of the hoop constant could still give results.

 

[gregegan]

Someone, somewhere must have done a thorough Newtonian analysis of vibrating rotating hoops. Probably we need to find and expand on those results rather than re-invent them.

I didn't find any literature on this, but (after much banging my head against a brick wall) I found some small-amplitude vibrational modes of a rotating elastic Newtonian hoop myself. This is quite cool; I think I'll put the details up on a web page when I get a chance.

You can look for solutions of the form:

$$r = r_e + \delta \cos{( m(\phi-c t) )}$$
$$\theta = \phi + \omega t + \delta \alpha \sin{( m(\phi-c t) )}$$

where $\delta$ is small, and m is a non-zero integer in order to meet the continuity condition at $\phi = 2\pi$. I've assumed that the tension everywhere is constant, and equal to the equilibrium value, which should be reasonable if m is not large and hence the spatial derivative as well as the amplitude of the perturbation is small.

Plugging this into F=ma and requiring it to hold to first order in $\delta$ gives two polynomials in c and $\alpha$. There are some messy solutions involving the roots of a cubic, which I'll spare you, but also a nice simple solution:

$$c = -\omega$$
$$\alpha = \frac{\rho \omega^2 {r_0}^2 - k}{m r_0 k}$$

As pervect suggested, this involves both transverse and longitudinal vibrations. I couldn't find any purely transverse or purely longitudinal solutions.

 

[pervect]

OK, I think I'm finally getting results for the symmetric vibrating hyperelastic hoop, and they seem to match Greg Egan's.

I haven't checked my calculations over very thoroughly, though. If the calculations are correct, any instability is apparently not a radially symmetric instability, at least not at the one particular point I examined.

As mentioned, we assume

r = r(t)
theta = phi + alpha(t)

$$\omega = \frac{d \alpha}{dt}$$

The details follow. Most of the variables are self-explanatory, except possibly vr, which is just dr/dt. Some of the computer generated latex formatting is a bit funky.

We start of with a cylindrical coordinate chart, and do the previously discussed calculations for u, w, y, and s.

The hyperelastic model gives us rho(s) and P(s), and combined with u and w this gives us the stress-energy tensor by previously discussed formulas.

The total energy of the hoop can be expressed as an intergal of $T^{00}$ over the volume, i.e.

$$E = \int g_{00} T^{00} dV = 2\,{\frac {\pi \,r \left( \rho-\rho\,{{\it vr}}^{2}+P{\omega}^{2}{r}^{ 2} \right) }{ \left( -1+{{\it vr}}^{2}+{\omega}^{2}{r}^{2} \right) \left( -1+{{\it vr}}^{2} \right) }}$$

The total angular momentum of the hoop is a similar integral
$$L = r \int \sqrt{g_{22} g_{00}} T^{02} dV = -2\,{\frac {\pi \,{r}^{3}\omega\, \left( \rho+P \right) }{-1+{{\it vr} }^{2}+{\omega}^{2}{r}^{2}}}$$

[add]The reasoning behind this is that the i^th component of momentum is $$|p|^2 = p^i p_i$$

and $$p^i = T^{0i} a_0$$, a being a unit timelike vector.

thus we can write
$$|p|^2 = T^{0i} a_0 T_{0i} a^0 = g_{00} g_{ii} T^{0i} T^{0i} a^0 a_0$$

Since a is a unit vector, taking the square root gives us the expression above.

So one factor of r comes from conversion of linear momentum to angular momentum, another factor of r comes from $\sqrt{g_{00} g_{22}}$, and the last factor of r comes from the volume element, dV = r dr d theta.

The stretch factor s (with an initial starting radius of r0=1) is:
$$\sqrt {{\frac {{r}^{2} \left( -1+{{\it vr}}^{2} \right) }{-1+{{\it vr} }^{2}+{\omega}^{2}{r}^{2}}}}$$

These can be massaged to give a system of equations as follows:
$$vr^2 = 1-{\omega}^{2}{r}^{2}-2\,{\frac {\pi \,{r}^{3}\omega\, \left( \rho+P \right) }{{\it L0}}}$$
$$vr^2 = 1+{\frac {{s}^{2}{\omega}^{2}{r}^{2}}{{r}^{2}-{s}^{2}}}$$
$$vr^2 = 1-\frac{P{\omega}^{2}{r}^{2}} {\left( -\rho+{\frac {{\it E0}\,{r}^{2}\omega\, \left( \rho+P \right) }{{\it L0}}} \right) }$$

We can regard $\omega$, r, P, and $\rho$ as functions of s.

We pick an operating point via the values of E0 and L0.

We can solve for r(s) and $\omega(s)$ by setting the three different expressions for vr^2 eqal to each other. This basically gives two equations for two unknowns. There isn't one solution, however, we do have to pick a solution where vr^2 is positive.

The results were that vr^2 was found to be positive only for a narrow region around the chosen operating point near equilibrium at s=1.8 in the hyperelastic hoop with k=.5 and rho0=1 (using numerical methods to find the solutions, and pick only solutions which were positive at s=1.8). So the hoop could not leave a very narrow region of s.

[add]
To give even more details:

k=.5, rho=1 at s=1

E0 := 7.275034568;
L0 := 3.920507627-.001;

this is the approximate equilibrium value for E0, and a value slightly lower than the equilibrium value for L0 at s=1.8

The region where vr^2 was positive was between
[r = 1.125734773, omega = .6910327647, s = 1.791533434],
[r = 1.091399912, omega = .7304883285, s = 1.808022749]

at the two listed points, vr was zero. Note that when omega is at its maximum, r is at its minimum.

At s=1.8, [r = 1.109158959, omega = .7100359127] and vr^2 = .1205520e-3

 

[pervect]

Next up: can we come up with a Lagrangian (or possibly a Hamiltonian) for the hoop, i.e. find some Lagrangian

$$L(r,vr,\alpha,\omega)$$ that gives the same equations of motion

For instance, we might try L = total energy - 2 * elastic stored energy

if we can figure out the elastic stored energy. Also, Goldstein makes some remarks that leads me to believe there may be a way to write the conjugate (cannonical) momentum p directly from the stress-energy tensor.

 

[pervect]

I have a feeling that we should be able to somehow integrate the Lagrangian density referenced in

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

to come up with a noncontinuous Lagrangian for the hoop, but I don't quite see how to make it work yet.

It looks like $\script{L}$ should be just (ks/2 + k/2s - k) for the hyperelastic hoop (?).

 

[pervect]

I ran across this while reading the above:

Lemma 6. A necessary and sufficient condition for the strain to define a
tensor on B is that the four-velocity uμ is Born-rigid, i.e. that the Lie derivative
of metric projection orthogonal to uμ in the uμ direction vanishes.

This may not matter for the hoop, because we can approach Born rigidity - but for the disk, it seems potentially problematical. I may or may not be confused by interpreting the four-velocity described in the paper as an actual vector field.

 

[pervect]

Not sure if Greg is still with us, but I may be getting somewhere with the Lagrangian approach.

[edit] or maybe not.

I was able to get the correct expression for angular momentum using L = - 2 Pi r rho, but it doesn't seem to give the right value for the energy function (i.e. the Hamiltonian)

$$h = \omega \frac{\partial L}{\partial \omega} + vr \frac{\partial L}{\partial vr} - L$$

which should be equal to the energy E I calculated earlier.

The papers seem to be suggesting using L = ne = e/s rather than rho - however, this didn't make the angular momentum come out the same as the previous calculation which I'm using as a reference.

 

[pervect]

I found another paper which suggests that $$\script{L}$$ should indeed be $-\rho = -n(\rho_0 + \epsilon)$, with $\epsilon$ being the stored energy, namely

http://www.citebase.org/fulltext?format=application%2Fpdf&identifier=oai%3AarXiv.org%3Agr-qc%2F0403073

eq 58

 

[pervect]

Aha! I think I've finally got it. There was a rather subtle error in my previous calculation. The volume of the expanding hoop is:

V = 2*Pi*r*sqrt(1-vr^2)

My previous calculations omitted this factor of sqrt(1-vr^2). This happens because of the Lorentz contraction of the thickness of the hoop in the lab frame due to its radial velocity vr, making the volume of the hoop slightly lower.

So the Lagrangian density is just $$\mathcal{L} = -\rho$$, where $\rho$ is to be interpreted for this purpose as a scalar function. Physically, one can think of $\rho$ as the density of the hoop in its rest frame (said density including stored energy) - as it is a simple scalar, it does not transform in the same manner as the stress-energy tensor. (This does not match with some previous usages of the symbol on my part.)

That's all there is to it! Just as described at http://www.citebase.org/fulltext?format=application%2Fpdf&identifier=oai%3AarXiv.org%3Agr-qc%2F0403073

Note, in passing, that this Lagrangian gives the correct Lagrangian for a mass m moving at a uniform velocity v:

L = -m \sqrt(1-v^2)

because if the volume of the mass was V in its rest frame, it shrinks in a frame moving at velocity v by a factor of 1/gamma, i.e. sqrt(1-v^2).

Thinking along these lines is what finally alerted me to the fact that I was missing this factor of sqrt(1-vr^2) - the hoop is expanding, and this volume adjustment factor needs to be included to get the correct Lagrangian from an expanding hoop.

Using this approach, the Lagrangian of the expanding hoop:

r= r(t)
theta = alpha(t)

omega = d alpha/dt
vr = dr/dt

is just

L (r,vr,alpha,omega) = -2 Pi r sqrt(1-vr^2) rho

i.e. -rho*volume

For the hyperelastic hoop

rho =(-2*k*s+k*s^2+k+2*rho_0)/2s

and s is determined by the geometry, for the uniformly radially expanding hoop s is uniform and equal to

s= (r/r_0) * sqrt ( (1-vr^2)/(1-vr^2-r^2*$\omega$^2) )

The above expressions gives a rather complex expression for L(r,vr,$\omega$) which can be solved in the usual manner to give the time evolution of the hoop using Lagrange's equations.

Two invariants of motion, the angular momentum and energy of the hoop, can be calculated from the Lagrangian as follows.

angular momentum = $$\frac{\partial L}{\partial \omega}$$

Because the Lagrangian is not a function of the angle $\alpha$, dL/dt = 0.

energy = $$\omega \frac{\partial L}{\partial \omega} + vr \frac{\partial L}{\partial vr} - L$$

this is the usual "energy function", which is also constant vs time since the Lagrangian is not a function of itme. It's numerically equal to the Hamiltonian, except that it's written in terms of coordinates and their time derivatives rather than the momentum.

With the correct expression for the volume, the results from this Lagrangian match the (corrected) results from the stress-energy tensor.

Equilibrium solutions can be found from $$\frac{\partial L}{\partial r} = 0$$ and vr=0.

Time evolution of the radius can be explicitly solved for directly from Lagrange's equation

$$\frac{d}{d t} \left( \frac{\partial L}{\partial vr} \right) = \frac{\partial L}{\partial r}$$

The other equation of motion, as previously mentioned, is that the angular momentum $$\frac{\partial L}{\partial \omega}$$ is constant.

Having a Lagrangian formulation, it should be now be possible to much more simply analyze the general case for "crinkling" sorts of instabiity, at least if one likes nonlinear partial differential equations :-).

In some future post I may describe a convenient set of basis vectors for the expanding hoop, but I think this post is already long enough.

 

[Chris Hillman]

Quick remark on Newtonian background

Hi, pervect,

Someone, somewhere must have done a thorough Newtonian analysis of vibrating rotating hoops. Probably we need to find and expand on those results rather than re-invent them.

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 hope to pick up the other thread with deformations of elastic beams and then the issue of stability wrt small perturbations, which is more sophisticated than the "upper bound on strain" approach I have been using so far to assess when Something Bad is About to Happen.

I wouldn't trust anything anyone says about relativistic elasticity unless I have carefully studied the Newtonian limit, discrete models with springs, and so on, BTW, so I hope you will take the time to compare the results of Greg and yourself with a Newtonian analysis. One thing I can say is that Greg's neglect of Poisson's ratio typically introduces errors involving a factor of two or so (assuming the material of interest has $\nu \sim 1/5 --- 1/3$).

 

[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:

$$s = \sqrt {{\frac {{\eta_{1,1}}^{2}{{\eta_1}}^{2}{{\eta_{2,0}}}^{2}- 2\,{\eta_{1,1}}\,{\eta_{1,0}}\,{\eta_{2,1}}\,{{\eta_1}}^{2}{\eta_{2,0}}- {{\eta_1}}^{2}{{\eta_{2,1}}}^{2}+{{\eta1}}^{2}{{\eta_{2,1}}}^{2} {{\eta_{1,0}}}^{2}-{{\eta_{1,1}}}^{2}}{-1+{{\eta_{1,0} }}^{2}+{{\eta_1}}^{2}{{\eta_{2,0}}^{2}}}}$$

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

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

 

[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

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

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

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

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:

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

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

$$\frac{\partial L_r}{\partial v_r} = \frac{\partial L_r}{\partial \omega} = \frac{\partial^2 L_r}{\partial v_r \,\partial \omega} = 0$$

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:

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

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 $v_r <<1$ and $\omega \, r << 1$ and when the stored energy is also less than the rest mass, i.e. in geometric units we must also have

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

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

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

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

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

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 [itex]\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.

$$T^{ab} = \left[ \begin {array}{ccc} {\frac {\rho-\rho\,{{\it vr}}^{2}+P{\omega}^{2}{r}^{2}}{ \left( -1+{{\it vr}}^{2} \right) \left( -1+{{\it vr}}^ {2}+{r}^{2}{\omega}^{2} \right) }}&{\frac {{\it vr}\, \left( \rho-\rho \,{{\it vr}}^{2}+P{\omega}^{2}{r}^{2} \right) }{ \left( -1+{{\it vr}}^ {2} \right) \left( -1+{{\it vr}}^{2}+{r}^{2}{\omega}^{2} \right) }}&- {\frac {\omega\, \left( \rho+P \right) }{-1+{{\it vr}}^{2}+{r}^{2}{ \omega}^{2}}}\\\noalign{\medskip}{\frac {{\it vr}\, \left( \rho-\rho\, {{\it vr}}^{2}+P{\omega}^{2}{r}^{2} \right) }{ \left( -1+{{\it vr}}^{2 } \right) \left( -1+{{\it vr}}^{2}+{r}^{2}{\omega}^{2} \right) }}&{ \frac {{{\it vr}}^{2} \left( \rho-\rho\,{{\it vr}}^{2}+P{\omega}^{2}{r }^{2} \right) }{ \left( -1+{{\it vr}}^{2} \right) \left( -1+{{\it vr} }^{2}+{r}^{2}{\omega}^{2} \right) }}&-{\frac {\omega\,{\it vr}\, \left( \rho+P \right) }{-1+{{\it vr}}^{2}+{r}^{2}{\omega}^{2}}} \\\noalign{\medskip}-{\frac {\omega\, \left( \rho+P \right) }{-1+{{ \it vr}}^{2}+{r}^{2}{\omega}^{2}}}&-{\frac {\omega\,{\it vr}\, \left( \rho+P \right) }{-1+{{\it vr}}^{2}+{r}^{2}{\omega}^{2}}}&-{\frac {\rho \,{\omega}^{2}{r}^{2}+P-P{{\it vr}}^{2}}{{r}^{2} \left( -1+{{\it vr}}^ {2}+{r}^{2}{\omega}^{2} \right) }}\end {array} \right]$$

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.

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.

i.e in terms of the above stress-energy tensor

energy = $$\sqrt{|T^{00} T_{00}|}$$ * volume
angular momentum = $$r \, \sqrt{|T^{02} T_{02}|}$$ * volume

Of course, since our metric is diagonal, $T_{00} = g_{00}^2 T^{00}$ and $T_{02} = g_{00} g_{22} T^{02}$.

In terms of the relativistic Lagrangian previously mentioned, the energy and angular momentum are just

energy = $$\omega \frac{\partial L}{\partial \omega} + v_r \frac{\partial L}{\partial v_r} - L$$

and

angular momentum = $$\frac{\partial L}{\partial \omega}$$

Note however that the volume of a hoop of radius r and proper cross sectional area A is
$$V = 2 \pi r A \sqrt{1 - v_r^2}$$
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 ($\partial AM / \partial r < 0$, 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)

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

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

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

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

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

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

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

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

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

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