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OK, it's been awhile, but I think I'm getting the same answer.
If v_c is the speed of sound in the unstretched rod, the maximum value of the stretch factor s where the speed of sound becomes equal to c is
[tex]
{\frac {\sqrt {6+3\,{{\it v_c}}^{2}}}{{3 \it v_c}}}
[/tex]
For a Newtonian hoop, the speed of sound would increase simply by the stretch factor s - the Lagrangian is the same, but one has to multiply the wave propagation speed for phi by the stretch factor s to get the actual physical wave speed through the medium.
For the relativistic hoop, we'd have something like:
eta(t,phi) = s0*phi + f(phi - beta*t)
The Lagrangian density of the hoop is
[tex]
\mathcal{L} = -\rho(s) \, s \, \sqrt{1 - \eta_0^2} d \phi
[/tex]
where
[tex]
s = \frac{\eta_1}{\sqrt{1-\eta_0^2}}
[/tex]
which sensibly reduces to [itex]\eta_1[/itex] in the non-relativistic limit, and
[tex]
\rho(s) = \frac{2 + v_c^2 \, (s-1)^2}{2 \, s}
[/tex]
Approximating the Lagrangian as a quadratic gives:
Lapprox = (-1/2*v_c^2*eta_1^2+1+1/2*v_c^2) eta_0^2 - v_c^2 eta_1^2
[add]Question - Can I really justify this approximation, though - eta_0 may not be small.We can divide this by the coefficient of eta_0^2 to find the effective velocity v_y^2, and set v_y * eta_1 = 1 to solve for the maximum stretch factor eta_1.
For v_c^2 = 1/2, this puts the maximum stretch factor at around 1.225, well before the energy peak.
Plugging the critical value of s into the equation for dr/ds and dE/ds seems to indicate that both the radius and the energy are increasing when s reaches its maximum allowable value.
If v_c is the speed of sound in the unstretched rod, the maximum value of the stretch factor s where the speed of sound becomes equal to c is
[tex]
{\frac {\sqrt {6+3\,{{\it v_c}}^{2}}}{{3 \it v_c}}}
[/tex]
For a Newtonian hoop, the speed of sound would increase simply by the stretch factor s - the Lagrangian is the same, but one has to multiply the wave propagation speed for phi by the stretch factor s to get the actual physical wave speed through the medium.
For the relativistic hoop, we'd have something like:
eta(t,phi) = s0*phi + f(phi - beta*t)
The Lagrangian density of the hoop is
[tex]
\mathcal{L} = -\rho(s) \, s \, \sqrt{1 - \eta_0^2} d \phi
[/tex]
where
[tex]
s = \frac{\eta_1}{\sqrt{1-\eta_0^2}}
[/tex]
which sensibly reduces to [itex]\eta_1[/itex] in the non-relativistic limit, and
[tex]
\rho(s) = \frac{2 + v_c^2 \, (s-1)^2}{2 \, s}
[/tex]
Approximating the Lagrangian as a quadratic gives:
Lapprox = (-1/2*v_c^2*eta_1^2+1+1/2*v_c^2) eta_0^2 - v_c^2 eta_1^2
[add]Question - Can I really justify this approximation, though - eta_0 may not be small.We can divide this by the coefficient of eta_0^2 to find the effective velocity v_y^2, and set v_y * eta_1 = 1 to solve for the maximum stretch factor eta_1.
For v_c^2 = 1/2, this puts the maximum stretch factor at around 1.225, well before the energy peak.
Plugging the critical value of s into the equation for dr/ds and dE/ds seems to indicate that both the radius and the energy are increasing when s reaches its maximum allowable value.
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