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Of course, keeping the circumference of the hoop constant means its radius must shrink.

Informally, this would be an approximation of a 'rigid' hoop, though of course one can't change the state of rotation of a rigid body in special relativity. For a thin enough, hoop, though, we should be able to approximate a rigidity.

We'll start with a standard cylindrical metric in 2 spatial dimensions and geometric units:

$$-dt^2 + dr^2 + r^2\,d\phi^2$$

Following Egan's lead from <<link>>, we'll introduce Langevin vectors u and w. See also <<wiki link>>

$$u = \frac{1}{\sqrt{1-r^2\omega^2}} \left( \partial_t + \omega \, \partial_\phi \right) \quad w = \frac{1}{\sqrt{1-r^2\omega^2}} \left( \omega \, r\,\partial_t + \frac{1}{r} \partial_\phi \right)$$

Then we can write the stress energy tensor in a coordinate basis as:

$$T = \rho \, u \otimes u + P \, w \otimes w$$

where ##\rho## is the density of the hoop and P is a negative number equal to the magnitude of the tension in the hoop, as seen by the Langevin observer. Both ##\rho## and P are only functions of r, they do not depend on t or ##\phi##.

By setting ##\nabla_a T^{ab}=0##, we can solve for the pressure, ##P = -\rho \, r^2\,\omega^2##

Substituting for P as a function of ##\rho## yields the following stress energy tensor in a coordinate basis

$$T^{ab} = \begin{bmatrix} (1 + r^2\omega^2) \rho & 0 & \rho \omega \\ 0 & 0 & 0 \\ \rho \omega & 0 & 0 \end{bmatrix}$$

Introducing an orthonormal basis of one-forms for the cylindrical metric, i.e. dt, dr, and ##r\,d\phi##, we can write the stress-energy tensor in an orthonormal basis as:

$$T^{\hat{a}\hat{b}} = \begin{bmatrix} (1 + r^2\omega^2) \rho & 0 & -\rho \,r \, \omega \\ 0 & 0 & 0 \\ -\rho \,r \,\omega & 0 & 0 \end{bmatrix}$$

Next, we need to solve for how the radius of the hoop decreases as we spin it up while keeping it's circumference constant.

Keeping the circumference of the hoop constant demands that

$$r_0 = \frac{r_i}{\sqrt{1-r_i^2\omega^2}}$$

where ##r_0## is the radius of the hoop at ##\omega=0## and ##r_i## is the shrunk radius of the spun-up hoop.

Solving for ##r_i##, we find.

$$r_i (\omega) = \frac{r_0}{\sqrt{1+r_0^2\omega^2}}$$

By integrating ##T^{\hat{0}\hat{0}}## over the 'volume' of the hoop, ##2 \pi r_i## we get the total energy E

$$E = 2\,\pi\,r_i \,\rho \sqrt{1 + r_i^2\,\omega^2} = 2\,\pi\,r_0 \,\rho \frac{1 + 2\,r_0^2\,\omega^2}{(1+r_0^2\,\omega^2)^\frac{3}{2}}$$

E seems sensible, it gives the expected results for energy in the limit where ##\omega## is small. The value for E when ##\omega=0## is just ##2\,\pi\,r_0\,\rho##.

Integrating the angular momentum density r ##\times \rho \, r\,\omega## over the volume ##2 \pi r_i## appears to be more problematical, because the angular momentum reaches a peak, then starts to decrease.

$$L = \frac{2 \, \pi \, r_0^3 \, \omega \, \rho } { \left( 1 + r_0^2\,\omega^2 \right) ^ \frac{3}{2}}$$

It's unclear if I've made some error, or how to interpret this if no error was made.