Relativistic Hoop w/ Constant Circumference: Solving Egan's Results

[pervect]

I wanted to re-do Egan's results for relativistic hoops with a different model, one with a relativistic hoop with a constant circumference. The idea is that by having the circumference not change as we spin up the hoop, we shouldn't be storing energy by "stretching" the material of the hoop. Putting the material of the hoop under tension without changing it's length shouldn't do any work.

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.

 

[pervect]

Just a quick note - the expression for E is also reaching a maximum as a function of $\omega$, then dropping. I could understand this happening if the weak energy condition were being violated, but I don't think it is, $\rho$ is always greater than P. There could be an error in the calculation somewhere, though I also recall similar issues with Egan's hoops that were "more stretchy".

Basically, in the lab frame, the result says the volume of the hoop shrinks due to the radius of the hoop contracting, and this dominates at high values of $\omega$. My calculation of the $r_i$, the shrunken radius, is a bit casual, but I don't see an error, yet.

 

[PeterDonis]

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

No, this is not correct. You seem to be assuming flat spacetime. In flat spacetime the radius of a rotating hoop (under an appropriate interpretation of "radius", namely the radius in the quotient space geometry) is larger than the Euclidean radius for its circumference, not smaller.

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

The radius of the hoop is not the same as the coordinate $r$. The coordinate $r$ of the hoop stays the same as it is spun up; in Born coordinates $r$ is the radius that an inertial observer at the center of the hoop would assign to the hoop, and by construction in your scenario that remains constant as the hoop is spun up. The physical radius of the hoop (again, under the appropriate quotient space interpretation) has to be calculated from the quotient space metric.

 

[pervect]

No, this is not correct. You seem to be assuming flat spacetime.

Yes.

In flat spacetime the radius of a rotating hoop (under an appropriate interpretation of "radius", namely the radius in the quotient space geometry) is larger than the Euclidean radius for its circumference, not smaller.

If we take the spatial metric in the quotient space from Ruggiero's "Relative space: space measurments on a rotating disk", [[link]] , in the rotating coordinates r', $\varphi$, the spatial line element $d\sigma$ is:

$$d\sigma^2 = dr'^2 + \gamma^2\,r'^2\,d\varphi^2$$

here $v = r'\, \omega$, so $\gamma = 1/\sqrt{1-v^2} = 1/\sqrt{1-r'^2\omega^2}$ Ths $\gamma$ > 1, and we see that the spatial distance $d\sigma$ between two points along the circumference is scaled by a factor $\gamma$ > 1, as the metric coefficient changed from $r^2 d\phi^2$ to $\gamma^2 \,r^2 \, d\varphi^2$. Radial distances are not affected.

We can also do this with projection operators, where we have the projected spatial metric
$$h_{ab} = g_{ab} + u_a u_b$$

We already have $u^a$, this is the vector

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

Lowering the index, and carrying out the calculation (using the cylindrical metric previously given) yields:

$$h_{ab} = \begin{bmatrix} 1 & 0 \\ 0 & \frac{r^2}{1-r^2\,\omega^2} \end{bmatrix}$$

Again, we see that radial distances are not affected, while the cicumferential distances increase.

For a third approach, if we place unit length rods end-end around the circumference, in the rotating frame we can fit more rods. In the lab frame, the rods appear length contracted.

Rods placed radially do not length contract, as the velocity is perpendicular to the radius. So we need the same number of rods in the radial direction, but more around the circumference.

So the circumference , as counted by the number of unit length rods we can fit around the outside of the hoop, is larger for a given radius. The radial measurement in the same in the lab frame and the rotating frame. This means that to keep the circumference in the rotating frame constant, so that we have the same number of rods (and hence the same amount of matter), we need a smaller radius.

Note that Egan's hyperelastic hoops also start to decrease in radius when they are spun up fast enough [[link]]

The radius for the Newtonian case goes to infinity as ω approaches K √2 / r0. In the relativistic case, r reaches a maximum and then declines

 

[PeterDonis]

we need a smaller radius

Ah, you're right, I was thinking of it backwards.

 

[pervect]

What motivated this thread for me was an off-hand remark in another thread about how to approach the relativistic hoop. Looking over what I wrote, I thing that needs the most attention in my an attempt at an analysis is a much more thorough discussion of the process of spinning up a hoop that was intially at rest. This is especially important as part of my motivation is to avoid discussion of the limitations (basically, failures) of the idea of rigid bodies with regards to rotating frames.

However, if we ignore the issues associated with "spinning up" the hoop, I think the conceptual approach for finding the stress energy tensor of the hoop is conceptually sound, though that doesn't rule out errors in the detailed calculations.

Basically, there are three "frames", defined by a set of basis vectors, that are of interest. The two most important are the frame of the Langevin observer [[link]], which can be regarded as the local frame moving along with a small section of the rotating hoop. The other frame of interest is the lab frame, an orthonormal set of basis vectors $\hat{r}, \hat{\phi}, \hat{t}$ in the lab frame.

Two of the Langevin bais vectors, which I denoted as $\vec{u}$ and $\vec{w}$, can be regarded as the Lorentz booss of $\hat{t}$ and $\hat{\phi}$. The other basis vector in the Langevin basis, $\hat{r}$, is unaffected by the Lorentz boost.

The third frame is of lesser importance, it's the lab frame in a coordinate basis rather than in an orthonormal basis. The third frame is convenient because it allows a relatively easy calculation of the divergence of the stress-energy tensor, $\nabla_a T^{ab}$ in that frame.

Having done the analysis, I can now see why I don't think the "naive" approach to the hoop works correctly. Basically, the approach that I am calling "naive" will give a stress-energy tensor in the lab frame that looks something like

$$T^{ab} = \rho \vec{u} \otimes \vec{u}$$

The problem with this approach is that divergence of the stress-energy tensor doesn't vanish, as it must. Adding the tension in the hoop solves this issue.

The correct approach is to write

$$T^{ab} = \rho \vec{u} \otimes \vec{u} + P \vec{w} \otimes \vec{w}$$

The proper choice of P makes the divergence of the stress-energy tensor vanish.

We can see definite differences in the physical predictions when we include the tension terms.

With what I am calling the "naive" approach, the density of energy in the lab fame in an orthonormal basis would be
$$T^{\hat{0}\hat{0}} = \frac{\rho}{1-r^2\omega^2} = \gamma^2 \rho$$

An even more naive (and still wrong approach) might give the energy densty as $\gamma \rho$ rather than $\gamma^2 \rho$.

When one takes into account the effects of the tension in the hoop, one finds that the energy density in lab frame is lower. Namely, from my detailed calculation.

$$T^{\hat{0}\hat{0}} = (1 + r^2 \omega^2)\,\rho$$

There are similar effects that lower the momentum density, which I won't belabor further.