Finally, I want to present a useful set of basis vectors for the expanding hoop, to be compared with the Langevin set of basis vectors in for instance
http://en.wikipedia.org/w/index.php?title=Born_coordinates&oldid=124648352
Consider a 1 parameter group of worldlines, expressed in a polar coordinate chart (t,r,theta) that form the worldsheet of our expanding hoop. We have two auxillary functions r(t) and \omega(t) that define the radius of the expanding hoop and the angular velocity of the hoop as a function of time. Then in terms of these functions, for every value of \phi we have an associated worldline in our cylindrical chart of
r = r(t)
\theta = \int \omega(t) dt + \phi
The first useful vector is the normalized 4-velocity of the worldline as a function of coordinate time, t. This is just
\vec{e0} = [\, {\frac {1}{\sqrt {1-{{\it v_r}}^{2}-{r}^{2}{\omega}^{2}}}} \frac{\partial}{\partial t} + {\frac {{\it v_r}}{\sqrt {1-{{\it v_r}}^{2}-{r}^{2}{\omega}^{2}}}} \frac{\partial}{\partial r} + {\frac {\omega}{\sqrt {1-{{\it v_r}}^{2} - r^{2}{\omega}^{2}}}}\frac{\partial}{\partial \theta}\,]<br />
The next useful vector is a spacelike vector which lies within the worldsheet of the hoop, but is perpendicular to the 4-velocity. This can be formed by taking a linear combination of a tangent vector \frac{\partial}{\partial \phi} which lies on the worldsheet but is not perpendicular to the 4-velocity, and the 4-velocity \vec{e0}. Normalized, this vector is just
\vec{e1} =[ \, {\frac {\omega\,r}{\sqrt { \left( 1-{{\it v_r}}^{2} \right) \left( 1-<br />
{{\it v_r}}^{2}-{r}^{2}{\omega}^{2} \right) }}}\frac{\partial}{\partial t} + {\frac {{\it v_r}\,<br />
\omega\,r}{\sqrt { \left( 1-{{\it v_r}}^{2} \right) \left( 1-{{\it v_r}<br />
}^{2}-{r}^{2}{\omega}^{2} \right) }}}\frac{\partial}{\partial r} + {\frac {1-{{\it v_r}}^{2}}{r<br />
\sqrt { \left( 1-{{\it v_r}}^{2} \right) \left( 1-{{\it v_r}}^{2}-{r}^{<br />
2}{\omega}^{2} \right) }}} \frac{\partial}{\partial \theta} \, ]<br />
As mentioned by Greg Egan, if we compute the unnormalized vector y = \frac{\partial}{\partial \phi} + \alpha \vec{e0} that is perpendicular to \vec{e0}, the magnitude of y gives us the value of the stretch factor s. The value of s computed via this means was given in the previous post.The final useful vector is a normalized vector perpendicular to the worldsheet, i.e. perpendicular to both of the above vectors. This is just:
\vec{e2} = [ \,\frac{v_r}{\sqrt{1-v_r^2}} \frac{\partial}{\partial t} + \frac{1}{\sqrt{1-v_r^2}} \frac{\partial}{\partial r} \, ]
We would then expect the stress-energy tensor of our hoop to be given by the formula
T^{ab} = \rho \vec{e0} \times \vec{e0} + P \vec{e1} \times \vec{e1}
where \rho is the density of the hoop in its rest frame, and the pressure P must be in the plane of the worldsheet, i.e. in the direction of \vec{e1}[/tex]. The pressure normal to the worldsheet of the hoop must be zero. This gives the following stress energy tensor in a cylindrical coordinate chart.<br />
<br />
T^{ab} =<br />
\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}}^<br />
{2}+{r}^{2}{\omega}^{2} \right) }}&amp;{\frac {{\it vr}\, \left( \rho-\rho<br />
\,{{\it vr}}^{2}+P{\omega}^{2}{r}^{2} \right) }{ \left( -1+{{\it vr}}^<br />
{2} \right) \left( -1+{{\it vr}}^{2}+{r}^{2}{\omega}^{2} \right) }}&amp;-<br />
{\frac {\omega\, \left( \rho+P \right) }{-1+{{\it vr}}^{2}+{r}^{2}{<br />
\omega}^{2}}}\\\noalign{\medskip}{\frac {{\it vr}\, \left( \rho-\rho\,<br />
{{\it vr}}^{2}+P{\omega}^{2}{r}^{2} \right) }{ \left( -1+{{\it vr}}^{2<br />
} \right) \left( -1+{{\it vr}}^{2}+{r}^{2}{\omega}^{2} \right) }}&amp;{<br />
\frac {{{\it vr}}^{2} \left( \rho-\rho\,{{\it vr}}^{2}+P{\omega}^{2}{r<br />
}^{2} \right) }{ \left( -1+{{\it vr}}^{2} \right) \left( -1+{{\it vr}<br />
}^{2}+{r}^{2}{\omega}^{2} \right) }}&amp;-{\frac {\omega\,{\it vr}\,<br />
\left( \rho+P \right) }{-1+{{\it vr}}^{2}+{r}^{2}{\omega}^{2}}}<br />
\\\noalign{\medskip}-{\frac {\omega\, \left( \rho+P \right) }{-1+{{<br />
\it vr}}^{2}+{r}^{2}{\omega}^{2}}}&amp;-{\frac {\omega\,{\it vr}\, \left( <br />
\rho+P \right) }{-1+{{\it vr}}^{2}+{r}^{2}{\omega}^{2}}}&amp;-{\frac {\rho<br />
\,{\omega}^{2}{r}^{2}+P-P{{\it vr}}^{2}}{{r}^{2} \left( -1+{{\it vr}}^<br />
{2}+{r}^{2}{\omega}^{2} \right) }}\end {array} \right] <br /><br />
<br />
We can substitute for the density and pressure terms \rho and P for any desired model, such as the hyperelastic model - both \rho and P will be functions of the stretching factor s for any given model as has been previously discussed.<br />
<br />
By integrating the stress energy tensor by the volume of our expanding worldsheet at any time t, we should and do come up with the same expressions for total energy and also for angular momentum that we did using the previous Lagrangian approach. <br />
<br />
i.e in terms of the above stress-energy tensor<br />
<br />
energy = \sqrt{|T^{00} T_{00}|} * volume<br />
angular momentum = r \, \sqrt{|T^{02} T_{02}|} * volume<br />
<br />
Of course, since our metric is diagonal, T_{00} = g_{00}^2 T^{00} and T_{02} = g_{00} g_{22} T^{02}.<br />
<br />
In terms of the relativistic Lagrangian previously mentioned, the energy and angular momentum are just<br />
<br />
energy = \omega \frac{\partial L}{\partial \omega} + v_r \frac{\partial L}{\partial v_r} - L<br />
<br />
and<br />
<br />
angular momentum = \frac{\partial L}{\partial \omega}<br />
<br />
Note however that the volume of a hoop of radius r and proper cross sectional area A is<br />
<br />
V = 2 \pi r A \sqrt{1 - v_r^2}<br /><br />
as previously noted, due to the Lorentz contraction in the radial direction in the lab frame.
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.
