On the Relativistic Twisting of a rotating cylinder (Max von Laue)

[TSny]

This simple straight loaded beam will be twisted into a helix when one does the appropriate Lorentz boost, so I assume that some of the results for helices (which I haven't had the time or inclination to look at in detail) will still be applicable using the simpler "loaded beam" variant of the problem.

I'm not following you here. Maybe you're thinking of a different "loaded beam" scenario. In the three-particle model of post 147, the beam lies along the x-axis (boost axis). The beam cannot be twisted into a helix by a boost along the axis.

I believe the beam must have a finite width to support the loads with finite stresses.

Yes, I agree.

I believe that an analysis based on the stress-energy tensor will show that sections of the beam under tension (in the direction of the boost) will have a negative energy density in the boosted frame, and sections of the beam under compression (in the direction of the boost) will have a positive energy density.

Yes, I think that's true. As Peter has pointed out, this would violate a postulate that the energy density must be positive in all frames of reference. In the case of the three-particle model, we can avoid the problem by simply allowing the rod to have mass. I initially chose a massless rod in order to make the "paradox" more dramatic. But we can add mass to the rod and the motion of the system will still appear paradoxical in the boosted frame in which the three masses are all lined up on the same side of the rod.

I don't think the stresses in the directions transverse to the boost should matter.

If the boost is along the x direction, then the stresses T_yy and T_zz in the rest frame will not matter. But the shear stresses, such as T_zx are important in resolving the paradox.

Because the beam will no longer be massless, in the boosted frame, the beam's contribution to the angular momentum must be evaluated, and I expect it will explain the "paradox".

In the three-particle system, the rod is along the x-axis. So, at first, you would think that it cannot contribute to angular momentum of the system. However, calculations show that the shear stress in the rod actually does contribute to the angular momentum of the system in the frame in which the particles are lined up. These contributions combine with the angular momenta of the particles to produce a total angular momentum of the system that has only an x-component and remains constant in time. This is consistent with zero torque acting on the system.

 

[TSny]

... ordinary matter satisfies all of the energy conditions, one of which is that the energy density is positive in any frame. Another is that the stress is less than or equal to the energy density in any frame.

Yes, this is a key point that I'm just now appreciating. For the flexible cable (or string) model of post #173, it is easy to check that the tension in the helical string must violate the stress condition if the helix can be unwound by a boost. In the notes that I link to in that post, equations (12) and (14) combine to show that the tension $\tau_0$ is given by

$\tau_0 = \lambda_0 \frac{1+u^2 v^2}{v^2} > \lambda_0 \,\,\,\,$ (in units where $c = 1$).

Here, $\lambda_0$ is the linear mass density of the string in the rest frame, $u$ is the tangential speed of rotation of an element of the string, and $v$ is the boost speed needed to unwind the helix.

The tension stress $\sigma$ is given by the tension divided by the cross-sectional area of the string. So, we get $\sigma > \rho _0$ where $\rho_0$ is the volume mass density.

I'm very inclined to think that the stress condition is going to be violated in any freely rotating helix that can be unwound by a boost.

 

[PeterDonis]

I'm very inclined to think that the stress condition is going to be violated in any rotating helix that can be unwound by a boost.

I think this is a possible resolution of the apparent paradox, yes.

 

[maline]

I'm very inclined to think that the stress condition is going to be violated in any freely rotating helix that can be unwound by a boost.

I think this is a possible resolution of the apparent paradox, yes.

Bravo! @AVentura, it looks like the answer to your paradox has been found!

I want to point out that violation of the weak energy condition can be thought of as an answer in two different ways. From the physical perspective, if it's true that a helix that is rotating fast enough must violate the energy conditions, then the "paradoxical" situation is just impossible. What will happen if you try to get the helix to rotate is that as the tension of fighting the centrifugal force builds up, the material will start to deform, generating elastic energy, which in turn contributes to the rest mass and the centrifugal force, requiring even more tension. No material will be able to supply the necessary tension, and the body will be destroyed.

From a more mathematical perspective, if we consider solutions that violate the energy conditions as valid (say using exotic matter), we get the result that the energy density will be negative in some parts of the body in some frames. In particular, if we can show that it is negative in the outer edge of the unwound helix in the boosted frame, that can answer the paradox as follows: The problem with a straight rod revolving around an axis parallel to itself is that the center of rotation, which must be the center of mass, is outside the (convex hull of) the body. But the fact that the center of a mass distribution must lie within the distribution depends critically on mass being nonnegative! So in "exotic matter" situations, we can indeed have a center of mass situated well outside the body, and revolve around it with zero linear momentum.

 

[AVentura]

What will happen if you try to get the helix to rotate is that as the tension of fighting the centrifugal force builds up, the material will start to deform, generating elastic energy, which in turn contributes to the rest mass and the centrifugal force, requiring even more tension. No material will be able to supply the necessary tension, and the body will be destroyed.

Being a closed system, the increase in invariant mass could only come from the kinetic energy, right? That's possible if the radius increases, like a spinning figure skater opening her arms. A spinning ring could prevent this (right?), but why not a helix?

Since we are verbalizing theories here's mine: after setting any helix in motion elastic energy will move, putting some energy in a new place along the helix. This unbalances it. Any attempt to add mass to balance it back exacerbates the situation.

Stress doesn't contribute to the energy density. They are different components of the stress-energy tensor.

So elastic energy is different than stress? Sorry for the elementary question. I just know a compressed spring has more invariant mass than an otherwise relaxed one.

 

[PeterDonis]

the increase in invariant mass could only come from the kinetic energy, right?

Not necessarily, no. The bonds between neighboring atoms basically act like springs--putting them under tension increases the stored energy, even if the atoms aren't moving relative to each other (after some small adjustment due to the tension).

So elastic energy is different than stress?

Yes. The stress is a space-space component of the stress-energy tensor. But a body under stress also has a greater energy density, i.e., "elastic energy", so the time-time component of the stress-energy tensor also increases.

 

[TSny]

Here’s an attempt to show, or at least suggest, that the stress at certain points in the helix must be of order $\rho c^2$.

Consider the last half turn of the rotating helix and look at the angular momentum of this section. We choose a coordinate system such that this section is described at the present moment by $(x, y, z) = \left( a \theta, R \cos \theta, R \sin \theta \right)$, $0 < \theta < \pi$. So, the origin of this system is not the same as the origin that was used for the entire helix. Our new $y$ axis passes through one end of this last half turn of the helix. Assume the helix is rotating in this coordinate system at non-relativistic speed with angular speed $\omega$. This section of the helix has a linear mass density $\lambda$ which we take to be constant for this section.

The $z$ component of angular momentum of the section relative to the point (x, y, z) = (0, 0, 0) at this moment is found to be

$L_z = -\lambda \pi \sqrt{1+\frac{R^2}{a^2}} a^2 u \,\,\,\,$ where $u = \omega R$

The rate of change of this component of angular momentum is due to the rotation of the helix and has a magnitude given by

$|\frac{dL_z}{dt}| = \omega |L_z| =\lambda \pi \sqrt{1+\frac{R^2} {a^2}} a^2 u^2 / R$

The direction of $\frac{dL_z}{dt}$ is the positive $y$ direction. So, there must be a positive $y$ component of torque acting on this section of the helix which has a magnitude

$\tau_y =\lambda \pi \sqrt{1+\frac{R^2}{a^2}} a^2 u^2 /R$

For it to be possible to unwind the helix with a boost, we must have $a = \frac{c^2 R}{uv}$, where $v$ is the boost speed. So,

$\tau_y = \left(\lambda c^2 \right) \pi \sqrt{1+\frac{R^2}{a^2}} \frac{c^2}{v^2} R$

The forces acting on this section of the helix are due to the stresses at the end of the section that is connected to the rest of the helix. It is not hard to see that the only form of stress that can produce torque in the y-direction relative to the origin $(0, 0, 0)$ is a “bending moment”. This is because the end of the section where the stresses are acting is located on the $y$ axis. (I can try to add more explanation of this point, if necessary.)

The bending moment is given by $M_B = \int{T_{\tilde{x} \tilde{x}}\tilde {z} d\tilde{y} d\tilde{z}}$. The tildes refer to a coordinate system with origin at the center of the cross section of the rod with the $\tilde{x}$ axis perpendicular to the cross section of the rod and the $\tilde{y}$ axis parallel to the $y$ axis. The integration is over the cross section of the rod at $\theta = 0$. In order of magnitude, we can write this as $M_B \approx \left<|T_{\tilde{x} \tilde{x}}| \right> r A$ where $r$ is the radius of the rod, $A$ is the cross sectional area of the rod, and $\left< |T_{\tilde{x} \tilde{x}}| \right>$ represents an average of the absolute value of $T_{\tilde{x} \tilde{x}}$ over the cross section.

Thus, we must have $\left<|T_{\tilde{x} \tilde{x}}| \right> r A \approx \left(\lambda c^2 \right) \pi \sqrt{1+\frac{R^2}{a^2}} \frac{c^2}{v^2} R$.

Then, $\left< |T_{\tilde{x} \tilde{x}}| \right> \approx \left(\rho c^2 \right) \pi \sqrt{1+\frac{R^2}{a^2}} \frac{c^2}{v^2} \frac{R}{r}$, where $\rho = \lambda/A$ is the volume mass density of the section.

Since $r < R$ and $v < c$ we find

$\left<|T_{\tilde{x} \tilde{x}}| \right> \sim \left(\rho c^2 \right)$

 

[maline]

So elastic energy is different than stress? Sorry for the elementary question. I just know a compressed spring has more invariant mass than an otherwise relaxed one.

As you say, a deformed body has more invariant mass. From a microscopic perspective, this is because the atoms are moved away from their equilibrium positions, forcing the electrons into wavefunctions that do not do as well at minimizing the electromagnetic energy. So you can say that elastic energy is really EM field energy.

Stress is the force derived from the gradients of this energy. As long as the deformations are small, the energy is approximately quadratic in the deformation, so the stress is linear in the deformation (strain). This is "Hooke's law" familiar from discussions of springs. In more detail, typical materials have three elastic moduli, relating the stress linearly to the tensile, shear, and bulk strains respectively. For simplicity, let's think of the strain as one dimensional (stretching or compressing in the $x$ direction with no change in the perpendicular directions). Then $$\frac {\Delta V} V = \frac {A\Delta L_x} {AL_x}= (1/K) \sigma_{xx}$$ where $\Delta V$ and $\Delta L_x$ are the changes in volume and in length in the $x$ direction, respectively, $A$ is the cross sectional area in the $y,z$ directions, $K$ in the bulk modulus, and $\sigma_{xx}$ is the tension or stress resisting further deformation.

For idealized "perfectly rigid" materials, $K\rightarrow\infty$, so there can be unlimited stress with no (or infinitesimal) strain, and therefore no elastic energy. But when $K$ is finite, the elastic energy is given by $$\Delta E= \int_{L_x}^{L_x+\Delta L_x}\,dl\,A\sigma_{xx}=A\int_{L_x}^{L_x+\Delta L_x}\,dl\,K\frac{l-L_x} {L_x}=\frac{AK}{2L_x}(\Delta L_x)^2=\frac {KV} 2 {\left(\frac {\Delta L_x}{L_x}\right)}^2=\\=\frac {KV} 2 {\left(\frac {\sigma_{xx}}K\right)}^2=\frac V {2K} \sigma_{xx}^2$$ Thus the added energy density is proportional to the stress squared, and inversely proportional to the bulk modulus.

However, the speed of sound in a solid is given by $\sqrt {\frac K \rho}$, where $\rho$ is the rest mass density. In relativity this must be less than $c$, so we must have $K< \rho c^2=E/V$. Therefore $$2 \frac E V \frac{\Delta E} V =\frac{\Delta(E^2)}{V^2}> \sigma_{xx}^2$$ and in particular, the total energy density will always remain more than the stress. This is how the weak energy condition is "enforced".

 

[maline]

However, the speed of sound in a solid is given by $\sqrt {\frac K \rho}$, where $ρ$ is the rest mass density. In relativity this must be less than $c$, so we must have $K< \rho c^2=E/V$.

I should add that this bound is purely theoretical. Real materials come nowhere close to this value. For instance, diamond has a very high value of $K$, given as $443~ [GPa]$. But $\rho c^2$ would be something like $3.17 \times 10^{11} [GPa]$.

 

[AVentura]

I never dreamed the answer would be this complicated. Even though the gist of the answer seems simple, I think. Thank you for the explanation @maline. Thank you @TSny, @PeterDonis. Thank you everyone.
,

 

[PeterDonis]

I never dreamed the answer would be this complicated.

I think that's a sign that it was a good question.

 

[AVentura]

Thank you Peter.

Side note about this "relativistic twisting", I came across this while thinking about how it transforms wave functions. If you treat the real and imaginary components of the wave function as space, it ensures all observers see a wavelength that corresponds to the momentum that they would observe. At least for a non-localized free particle. The wave function is just a helix. Is this surprising? If not, is it already theorized that wave functions live on the Kaluza-Klein "cylinder"?

 

[PeterDonis]

The wave function is just a helix

A helix if you equate the axis of the helix with a single space dimension (note that this is based on an idealized "particle" that can only move in one space dimension), and the two dimensions transverse to the helix with the complex plane, yes. But that's very different from an actual helix in actual 3-dimensional space; the "space" in question is the abstract Hilbert space of complex functions of a single real variable.

is it already theorized that wave functions live on the Kaluza-Klein "cylinder"?

Not that I'm aware of; I don't know of any useful analogy between Kaluza-Klein type geometric spaces and the Hilbert spaces used in quantum mechanics.