Stress-energy tensor of a wire under stress

[gregegan]

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

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

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

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

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

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

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

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

 

[gregegan]

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

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

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

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

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

 

[pervect]

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

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

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

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

 

[gregegan]

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

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

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

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

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

I think there are two aspects to consider:

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

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

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

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

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

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

 

[pervect]

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

your webpage

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

<br /> r(t,\phi) = r_e + \delta_r(t,\phi)<br />

<br /> \theta(t,\phi) = \phi + \omega t + \delta_{\theta}(t, \phi)<br />

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

Calculating the 4-velocity is reasonably obvious I think.

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

<br /> T = \rho \vec{u} \times \vec{u} + P \vec{w} \times \vec{w}<br />

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

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

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

But we need n and \vec{w}.

 

[gregegan]

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

your webpage

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

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

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

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

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

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

For small perturbations, the two approaches should agree.

 

[gregegan]

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

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

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

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

 

[pervect]

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

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

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

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

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

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

Since we also have

\vec{w} \cdot \vec{u} = 0 we also have
<br /> (\vec{p} + \alpha \vec{u}) \cdot \vec{u} = 0<br />

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

[edit]Looks like I still need to think about n, though. Let's try this again
d\phi[/itex] can be considred to be a one-form, which we could also write as \nabla_a \phi. It&#039;s easy to confuse this one-form with the scalar version, unfortunately.<br /> <br /> Since w^a is a unit vector, w^a d\phi will be a scalar, the new length. Since \phi is preserved during spin-up, the original length was just r_0 d\phi. The stretch factor s, 1/n, should be the new length over the original length, which means (?)<br /> <br /> s = 1/n = \frac{w^a d\phi}{r_0 |d \phi|} <br /> <br /> Something&#039;s still a little funky with that factor of r_0, though. Well, I need to think about it more.<br /> <br /> Assuming this n issue gets sorted out, I think I want to leave the linearization step till last.<br /> <br /> Because we are doing pertubations, we could take a Hamiltonian or Lagrangian density approach in terms of the displacement functions, but from what I&#039;m reading in Goldstein, this will just give us \nabla_b T^{ab} = 0 anyway (I&#039;m reading about pg 565).

 

[gregegan]

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

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

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

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

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

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

<br /> n = r_0 w^a\partial_a \phi <br />

 

[gregegan]

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

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

If we write:

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

then we simply have:

<br /> n = r_0/|y|<br />

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

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

<br /> s = 1/n = |y|/r_0<br />

 

[pervect]

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

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

[t,r,\theta,z]

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

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

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

also |y| = \gamma r

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

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

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

 

[gregegan]

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

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

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

<br /> n = r_0 w^a \partial_a \phi<br />

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

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

<br /> n=r_0/|y|<br />

 

[gregegan]

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

<br /> n = r_0 w^a \partial_a \phi<br />

even though we have an easier alternative.

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

<br /> [t,r,\theta] = [t_0, r(r_0), \phi + \omega t_0]<br />

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

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

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

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

<br /> w = \omega \gamma r \partial_t + \frac{\gamma}{r} \partial_{\theta}<br />

we can write n as:

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

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

<br /> \partial_{t_0} t = 1<br />
<br /> \partial_{t_0} \theta = \omega<br />
<br /> \partial_{\phi} t = 0<br />
<br /> \partial_{\phi} \theta = 1<br />

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

<br /> [\frac{\partial (t_0, \phi)}{\partial (t,\theta)}]=[\frac{\partial (t,\theta)}{\partial (t_0, \phi)}] ^{-1}<br />

which gives us:

<br /> \partial_t t_0 = 1<br />
<br /> \partial_t \phi = -\omega<br />
<br /> \partial_{\theta} t_0 = 0<br />
<br /> \partial_{\theta} \phi = 1<br />

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

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

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

 

[gregegan]

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

<br /> t(t_0,r_0,\phi) = t_0<br />
<br /> r(t_0,r_0,\phi) = r_e(r_0) + \delta_r(t_0,\phi)<br />
<br /> \theta(t_0,r_0,\phi) = \phi + \omega t_0 + \delta_{\theta}(t_0, \phi)<br />

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

<br /> \partial_{r_0} r_e(r_0) = 1<br />

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

We define:

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

we then have:

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

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

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

<br /> g(\partial_{r_0},\partial_{t_0}) = \frac{\partial \delta_r}{\partial t_0}<br />
<br /> g(\partial_{r_0},\partial_{r_0}) = 1<br />
<br /> g(\partial_{r_0},\partial_{\phi}) = \frac{\partial \delta_r}{\partial \phi}<br />

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

<br /> \delta_r(t_0) = \epsilon_r \sin{f t_0}<br />
<br /> \delta_{\theta}(t_0) = \epsilon_{\theta} \cos{f t_0}<br />

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

 

[pervect]

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

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

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

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

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

 

[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:

<br /> n=r_0/|y|<br />

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:

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

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:

<br /> c = -\omega<br />
<br /> \alpha = \frac{\rho \omega^2 {r_0}^2 - k}{m r_0 k}<br />

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

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

[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}<br /> }^{2}+{\omega}^{2}{r}^{2}}}}<br />

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

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<br />

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).