Recent content by gregegan

I've analysed a pulse with a bounded and continuous second derivative propagating through a relativistic hoop; movie at: http://www.gregegan.net/SCIENCE/Rings/SmoothPulseInHoop.gif and more details at: http://www.gregegan.net/SCIENCE/Rings/Rings.html While I can't pinpoint exactly...

We seem to agree 100% on the speed of sound in a straight rod under tension. I'm not sure that I fully understand your other calculations, but we seem to get roughly the same result, in as much as I find the phase velocity of one mode of vibrations in the hoop to hit c when s is approximately...

Calculating the behaviour of a step function in the hoop seems unmanageable, but I just calculated something else that's far simpler. The linearised relativistic equation for a longitudinal perturbation in an infinite straight-line string is just the usual wave equation -- with a speed of sound...

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

I found a nice parameterisation of the curves of constant angular momentum and energy for the pulsating hoop: r->Sqrt[r0^2*s^2*(1+K^2*(1-s^2))^2 - LM^2] / (1+K^2*(1-s^2)) gamma->(EM*r*r0) / (r0^2*s+K^2*(s-1)*(2*r^2-r0^2*s*(1+s))) omega->Sqrt[r0^2*s^2-r^2] / (gamma*r*r0*s) Here LM is...

That TeX might not have been the easiest thing to paste into Maple, so here's another try: c^3*m^2*(-1 + s)*(1 + K^2 - 4*K^2*s + 3*K^2*s^2)* (-(1 + K^2)^2 + 2*K^2*(1 + K^2)*s - 6*K^4*s^3 + 5*K^4*s^4) + c^2*m^2*(-1 + s)*(-1 - K^2 + K^2*s^2)*(-(1 + K^2)^2 + 2*K^2*(1 + K^2)*s +...

(edit: Removed unwieldy cubic in TeX; see next post for friendlier version.) For m=1, this does have an exact factor of (c+1), so there is a double root of -1. BTW, I said before that I hadn't found the singular behaviour except at points far from equilibrium, but I hadn't really looked at the...

I've written up what I've found so far about the relativistic vibrations in a new section at the end of the web page: http://www.gregegan.net/SCIENCE/Rings/Rings.html I've got an explicit polynomial for c, and the results concerning the regions where c has an imaginary part seem pretty...

I have found complex roots for c, now, though they appear at substantially higher angular velocities than the c=0 points. For example, for v_c^2=1/2, for m=2, the root c=0 appears at \omega=0.300, while the onset of complex roots for c appears at \omega=0.7236 (and ends at \omega=1.069)

Having calculated everything via the vanishing divergence of the stress-energy tensor, I've found a polynomial in c (with coefficients that are functions of s, the stretch factor). This has roots at c=0 for values of s that agree with your omega and r_eq values. Given that we've both arrived...

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

If m=2, the shape is roughly elliptical, isn't it? I'll see if I'm able to confirm this by a different route, hopefully sometime in the next few days. What I hope to do is a relativistic equivalent of the infinitesimal-element force balance that I used for the Newtonian case, which is a...

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

That's great news! To see why the roots are always real, first note that for m=1 the cubic factors into a quadratic and a linear term, with manifestly real roots for the quadratic. For higher values of m, note that: (a) as c goes to -infinity, the cubic goes to -infinity; (b) at c=0...

Yeah, the thing about that set of solutions is that alpha (the size of the longitudinal wave compared to the transverse) is very large, to the point where for the usual small value of delta I use, the hoop actually forms little loops, becoming a self-intersecting curve. I put in some code to...