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

[maline]

The key thing is to compare the above with the tensor we get from setting $\phi = k z$ in the above formulas for $U$ and $A$. This means there are now some more nonzero components that weren't there before

No, the $z$ in $\phi = k z$ is a parameter labeling a particular worldline. $\phi$ is not really a function of $z$ as a spatial coordinate, it just has a value chosen differently for each worldline. The partial derivative of $U$ in the $z$ direction is zero.
Again, the helix is kinematically nothing but a subset of the cylinder. We can treat $U$ and $A$ as being well-defined even outside the helix.

 

[PeterDonis]

the $z$ in $\phi = k z$ is a parameter labeling a particular worldline

I don't think so. In the cylinder case, $\phi$ labels the worldlines independently of $z$ (which is the other parameter labeling worldlines) because there are $2 \pi$ worth of worldlines at each value of $z$. In the helix case, there is only one worldline for each value of $z$, and $\phi = k z$ tells how the angular position of that worldline in the $\theta$ direction depends on $z$. So $z$ in the helix case is the only worldline parameter.

The partial derivative of $U$ in the $z$ direction is zero.

For the cylinder, I agree, because each worldline has nearest neighbors in the $z$ direction that has the same components of $U$; the only difference between the two is their $z$ parameter.

For the helix, I disagree, because there is no nearest neighbor of any worldline on the helix that has the same components of $U$; as you go along the helix, the components of $U$ change as a function of $z$.

I might be modeling the above feature incorrectly; it might be that it is not supposed to show up as a $z$ dependence of $U$ and therefore as extra components in the tensor I'm computing. But it ought to show up somewhere. If not in $U$, where?

 

[maline]

So $z$ in the helix case is the only worldline parameter.

Yes, my point is that when we write $\phi = k z$, we are using the symbol 'z' as a label rather than as a spatial coordinate.

as you go along the helix, the components of $U$ change as a function of $z$.

But you can't go along the helix without also changing your $x,y$ coordinates. Taking a partial derivative requires holding those constant.

I might be modeling the above feature incorrectly; it might be that it is not supposed to show up as a $z$ dependence of $U$ and therefore as extra components in the tensor I'm computing. But it ought to show up somewhere. If not in $U$, where?

I don't see that it should show up in the kinematics at all. We can always simply replace our helix with a rotating cylinder whose mass density is zero except at the points $\theta=kz+\omega t$. The mass density does not affect the kinematics, so the cylinder and helix, and any other body rigidly rotating about an axis, are kinematically indistinguishable.

 

[PeterDonis]

you can't go along the helix without also changing your $x,y$ coordinates. Taking a partial derivative requires holding those constant.

Hm, I'll have to think about that.

 

[maline]

No, the zzz in ϕ=kzϕ=kz\phi = k z is a parameter labeling a particular worldline. ϕϕ\phi is not really a function of zzz as a spatial coordinate, it just has a value chosen differently for each worldline. The partial derivative of UUU in the zzz direction is zero.

I just realized that the same goes for t as well- in $\phi+\omega t$, t is a parameter to specify a point along the worldline. $U_{3,0}$ and $U_{3,0}$ should be zero.

 

[pervect]

For what it's worth, I think that the relativistic treatment involving the angualr momentum and the stress-energy tensor of a rotating object is best addressed in a simpler scenario, like the relativistic rotating hoop. I am not aware of any peer-reviewed papers on that topic, but Greg Egan's has some discussion on the issue that appears reasonable (though not peer-reviewed.) If there's any interest his webpage on the topic is at is http://www.gregegan.net/SCIENCE/Rings/Rings.html. A discussion of that webpage probably belings in a new thread, though.

As far as the original issue goes, which only involves the twisting of a helix when it's boosted, that follows from the relativity of simultaneity. I don't think there's anything paradoxical about that observation, the claim that there was a paradox seems to be based on the idea that a helix would rotate in a torque-free manner around it's central axis. But it won't, not even in the non-relativistic case, so I think that puts that "paradox" to rest.

 

[TSny]

Very interesting thread. I had thought about the helix problem some years ago and ran into difficulties due to the complexity of the stresses and the need to choose a nonuniform mass density (or add point masses). The discussion here has been enlightening even though I have not been able to follow all of the arguments. It has led me to consider a simpler system that has some similarity to the helix and for which the stresses can be calculated.

The figure shows three point masses (two of mass m and one of mass 2m) connected by strings of length r to a "massless" rod of cross-sectional area A. In the unprimed frame, the system rotates with constant angular speed $\omega$. In the primed frame the masses are “lined up” (unwrapped helix) and the system translates with speed v as the system rotates. In the primed frame, the motion appears paradoxical since the net linear momentum and net angular momentum of the three particles vary with time. Yet there are no external forces or torques acting on the system.

This system is not as nice as the helix example, but it is simple enough that I believe the stresses in the rod can be worked out. You can show explicitly that the stresses contribute to the linear and angular momentum in the primed frame in such a way that total linear and angular momenta are constant. For me, these calculations have been long and tedious, and I could have made some mistakes or overlooked something. But it does appear to work out (as it must, of course).

Please ignore this post if you feel that this simplified system is not sufficiently similar to the helix paradox.

 

[maline]

It has led me to consider a simpler system that has some similarity to the helix and for which the stresses can be calculated.

Yes, this is a great example of the same paradox! Not only that, but the stresses in the crossbar look very much like the example I worked on in post #140. Hopefully we can do this in full!

One point of difference, though, is that the stress tensor is nonzero in the crossbar, which allows the momentum in the boosted frame to be nonzero on both sides of the axis of rotation. In the pure helix case, $T^{\alpha \beta}=0$ everywhere outside of the mass distribution. So the possibility of revolution around an axis parallel to the body seems even more difficult to resolve. That is also why I didn't want to use Pervect's model- a contunuum of point masses supported by strings like in your case, but tracing out the full helix. Nevertheless, your case is so much simpler that It's definitely worth working on before attempting a full, necessarily three-dimensional, helix.

You can show explicitly that the stresses contribute to the linear and angular momentum in the primed frame in such a way that total linear and angular momenta are constant. For me, these calculations have been long and tedious, and I could have made some mistakes or overlooked something. But it does appear to work out (as it must, of course).

I don't suppose you have those calculations saved somewhere, do you? If not, some pointers on how it's done would still be appreciated. In particular, can the idealization of infinite rigidity be used, or do we need to allow for deformation?

 

[TSny]

Yes, the stresses in the rod in the unprimed frame are the same as for the "loaded beam" in your post 140. https://en.wikipedia.org/wiki/Bending_moment

I agree that the helix is much more interesting. But I think it's too difficult for me. I have been searching for some notes I made on the helix years ago, but have not found them. I do know that I had to give up.

For the three-mass example, I worked it out over the last several days and I have about 16 pages of notes that are still in rough shape. I used Tolman's old text Relativity, Thermodynamics and Cosmology for formulas showing how stress contributes to momentum. It is almost 2 AM here now, so I will wait on posting any details.

 

[maline]

The key thing we want is the tensor $\theta_{ab} + \omega_{ab} = A_a U_b + U_{a, b}$. (Normally this tensor is split up into its trace--the expansion scalar, its symmetric traceless part--the shear tensor, and its antisymmetric part--the vorticity tensor. We'll look at that very briefly once we have computed it.) Note the lower indexes; lowering the index is easy in Cartesian coordinates, since it just flips the sign of the 0 component (I'm using the −+++ signature convention, which is easier to work with).

For the cylinder, we have

​

$U_a = \left( - \gamma, 0, - \gamma \omega R \sin ( \omega t + \phi ), \gamma \omega R \cos ( \omega t + \phi ) \right)$

I see now that from the start, this is not the way to calculate the kinematic decomposition. We need to take partial derivatives by all the spatial coordinates, so $U$ needs to be defined in an open set and not just on a cylindrical shell. This means, for instance, that we cannot treat $\gamma$ as a constant. Also, there is no reason to use worldline labels at all. We need $U$ as a spacetime field, so it should be given directly in terms of the coordinates. I think the correct form, in $(t,z,x,y)$, should be $$U_\alpha=(-(1-\omega^2 x^2-\omega^2 y^2)^{-1/2},0,-(1-\omega^2 x^2-\omega^2 y^2)^{-1/2}\omega y,(1-\omega^2 x^2-\omega^2 y^2)^{-1/2}\omega x)$$ If the decomposition is done correctly, the result should be that $U$ is a Killing field, i.e. that $A_\alpha U_\beta + U_{\alpha, \beta}$ is purely antisymmetric (besides being orthogonal to $U^\beta$, which is true for any velocity field).
But maybe we should hold off on this "loose end" so as not to distract from @TSny 's work, which I am very interested to see.

 

[PeterDonis]

We need to take partial derivatives by all the spatial coordinates, so $U$ needs to be defined in an open set and not just on a cylindrical shell.

This would resolve my concern about derivatives, yes. I had thought of this but have not tried to work it out in detail yet. After confirming the kinematics, we would also need to redo the dynamics, including the mass distribution, with everything being functions of the coordinates; that gets messy because there are delta functions involved. I might try it in parallel with the other subthread that is going on.

 

[TSny]

I have rewritten my notes for the analysis of the linear momentum of the three particle system. It only involves the shear stress in the rod and is not too hard to work out. Here is a link to the notes
https://www.dropbox.com/s/2lhrbipzvs2exyv/Analysis of Linear Momentum.docx?dl=0
There are likely to be "typo" errors, but hopefully there are no severe conceptual errors. Let me know of any.

If there is a better way to share notes on the forum let me know. I'm not tech savvy.

The angular momentum is messier and involves normal stress as well as shear stress. I can work on cleaning up my notes on this if anyone would like to see them.

 

[maline]

Thank you @TSny !
So the idea seems to be that as the shear stress in the crossbar transfers momentum along the direction of motion in the primed (translating) frame, it contains a density of "hidden momentum". Since in this frame the crossbar is twisted, the integral of the hidden momentum turns out to be perpendicular to the plane containing the masses, the strings, and the axis of rotation. It exactly cancels the momentum of the masses, which are moving perpendicularly to that plane in their revolution around the axis. Relativity is weird!

This basic idea probably works for the helix case as well, except that it will be harder to work out.

We are still left with a revolution around an axis not containing the center of mass. Even if there is no violation of momentum conservation, is it really legal for the CoM to just wander around? Intuitively, where is the lateral motion of the mass "coming from"? I would appreciate seeing your work on the angular momentum tensor. Perhaps that will shed some more light.

One technical point is that you modeled the shear stress as being uniform throughout the cross-section of the bar, which is not really possible- it must vanish at the edges in the $y$ direction. But apparently this detail turns out to be irrelevant, so more power to you for not getting tangled with it!.

 

[TSny]

So the idea seems to be that as the shear stress in the crossbar transfers momentum along the direction of motion in the primed (translating) frame, it contains a density of "hidden momentum". Since in this frame the crossbar is twisted, the integral of the hidden momentum turns out to be perpendicular to the plane containing the masses, the strings, and the axis of rotation. It exactly cancels the momentum of the masses, which are moving perpendicularly to that plane in their revolution around the axis. Relativity is weird!

Yes, if $\vec{g} \,’$ is the density of momentum due to the shear stress in the primed frame, then $\vec{g} \,’$ points perpendicular to the axis of the rod and twists in direction as you move along the rod with a sudden change in direction at the center. The red arrows in the picture below show $\vec{g} \,’$ for different cross sections of the rod at time $t’ = 0$. (I hope I got these directions right.) You can see that when you add all the red vectors you get a net momentum in the negative $z'$ direction, as you said. The twisting of directions is of course due to relativity of simultaneity and the sudden change in direction in the middle of the rod is due to the discontinuous change in sign of the shear stress there.

[Picture edited June 5, 2017, to make a correction in the direction of the arrows]

We are still left with a revolution around an axis not containing the center of mass. Even if there is no violation of momentum conservation, is it really legal for the CoM to just wander around? Intuitively, where is the lateral motion of the mass "coming from"? I would appreciate seeing your work on the angular momentum tensor. Perhaps that will shed some more light.

It appears that you can actually show with a calculation that the centroid of the entire system in the primed frame is located on the x-axis! So, it does not revolve around the axis. Somewhat amazingly, there is a contribution to the location of the centroid of the system from the normal stress component $t_{xx}$ in the rod that “balances out” the location of the centroid of the three particles.

One technical point is that you modeled the shear stress as being uniform throughout the cross-section of the bar, which is not really possible- it must vanish at the edges in the $y$ direction. But apparently this detail turns out to be irrelevant, so more power to you for not getting tangled with it!.

My knowledge of elastic stress is shallow and sketchy. So, I could very well be overlooking something here.

 

[maline]

It appears that you can actually show with a calculation that the centroid of the entire system in the primed frame is located on the x-axis! So, it does not revolve around the axis. Somewhat amazingly, there is a contribution to the location of the centroid of the system from the normal stress component txxtxxt_{xx} in the rod that “balances out” the location of the centroid of the three particles.

Indeed, it definitely does not make sense for the center of energy to "wander". Just as in the classical case, the motion of the center of energy in anyone Lorentz frame must indeed equal the total momentum in that frame, divided by the total energy. Here is a proof: the $i$ component of the center of mass at time $t_0$is given by $$\int_{t=t_0}\,dxdydz\, T^{tt} X^i$$
Take the partial derivative by $t$, and use local conservation of energy: $$ \frac \partial {\partial t} \left \{ \int_{t=t_0}\,dxdydz\, T^{tt} X^i \right \}=\int_{t=t_0}\,dxdydz\, {T^{tt}}_{,t} X^i=-\int_{t=t_0}\,dxdydz\, {T^{tj}}_{,j} X^i$$.

Now note that for any field $A^j$, we have $(A^jX^i)_{,j}={A^j}_{,j}X^i+A^j\delta^i_j$, so $$-\int_{t=t_0}\,dxdydz\, {T^{tj}}_{,j} X^i=-\int_{t=t_0}\,dxdydz\, {(T^{tj}X^i)}_{,j} +\int_{t=t_0}\,dxdydz\, T^{ti}$$
The first term is a 3-divergence, so using Gauss's theorem we can convert it to a boundary integral over a 2-surface surrounding our 3-volume at $t=t_0$. At long as our distribution in bounded in space, the boundary integral can be made to give zero. By symmetry of $T$, the second term is$$\int_{t=t_0}\,dxdydz\, T^{ti}=\int_{t=t_0}\,dxdydz\, T^{it}=P^i$$ which is what we wanted to show.

So if the only energy density in the system is at the point masses, they cannot move while the total momentum is zero. You are saying that there is an additional energy density somewhere in our massless, perfectly rigid crossbar? How so?

 

[TSny]

You are saying that there is an additional energy density somewhere in our massless, perfectly rigid crossbar? How so?

Yes. Consider the Lorentz transformation from the unprimed to primed frame: $T^{t't'} = \Lambda^{t'}_\mu \Lambda^{t'}_\nu T^{\mu \nu}$.

Using $T^{tx} = 0$ in the rest frame, you find $T^{t't'} = \gamma^2 (T^{tt} + T^{xx}v^2/c^4)$. For a "massless" rod, I think we can take $T^{tt} = 0$ in the rest frame of the rod. So, $T^{t't'} = \gamma^2 v^2/c^4 \, T^{xx}$.

So, the i ^th coordinate of the centroid in the primed frame will contain a contribution from the normal stress component $T^{x'x'}$ in the rod given by $\gamma^2 v^2/c^4 \int dx'dy'dz' X^{i'} T^{x'x'}$ divided by the total energy.

At time $t' = 0$, the masses are vertically above the rod. So, the integral of interest is $\gamma^2 v^2/c^4 \int dx'dy'dz' y' T^{x'x'}$. I find this to evaluate to a negative value that cancels the contribution of the three particles.

 

[TSny]

From post #61:

The helix has $n$ complete turns, parametrized as above by $\vec r (\theta) = ( \cos \theta , \sin \theta, p \theta)$ where $-n\pi \leq \theta \leq n\pi$. Start with a uniform mass distribution of $\lambda$ units per radian. As calculated there, if this rotates with $\vec \omega = \omega \hat z$, the angular momentum has a non-parallel component $J_y =\lambda \int_{-n\pi}^{n\pi} \, d\theta\, y(\theta) \omega z(\theta) =(-1)^n 2\pi n \omega p \lambda$. To fix this, we can simply add two point masses to the helix

Shouldn't the $(-1)^n$ be $(-1)^{(n+1)}$? If so, then unfortunately the required point masses will not lie on the helix, but rather on the opposite side of the axis.

 

[maline]

I see. The part of the bar with maximal compression stress is directly opposite the middle mass, so in the primed frame, it is on the side opposite all the masses. The side facing the masses will be under tension, so that will give a negative energy density with a positive $y$ value that also helps our balance.

Of course, negative energy is unphysical, but that comes from the idealizations of masslessness and perfect rigidity, as Pervect mentioned back in post #114. In real materials, the various elastic moduli must be bounded above by the mass density (up to constants) so that the speed of sound should be less than 1. Then any stress produces deformation inversely proportional to the relevant modulus, and the deformation stores (positive) energy quadratically, so the lower the mass density, the higher the minimal deformation energy. The total of (mass + deformation energy + energy from boosting the stress) will be positive in all frames and at all points.

But now, just as I feared, we have a resolution that cannot work for the helix case! There, it seems quite unavoidable that the center of mass will be within the body, away from the axis of rotation. We do not need to assume infinite rigidity, so the energy distribution should be purely positive.The revolution should be impossible!

From post #61:
Shouldn't the $(-1)^n$ be $(-1)^{(n+1)}$? If so, then unfortunately the required point masses will not lie on the helix, but rather on the opposite side of the axis.

No, the sign is correct. I assume you are asking because in post #60, the final result was $-\cos n\pi$ which equals $(-1)^{(n+1)}$. However, that calculation was for the integral of the second term in $d \vec J = [\vec \omega r^2 - \vec r (\vec \omega \cdot \vec r)]\rho (\vec r) dV$, which contributes negatively to the angular momentum. In that post I only needed to check whether the result was zero, so I didn't clarify the sign of the actual contribution. Good eye though!
Anyway, if you want to actually try working with a helix, Peter's model in post #118 is better than mine from post #61. It doesn't require point masses, so you can give the helix a finite thickness and have all the stresses at least be continuous.

 

[PeterDonis]

I had thought of this but have not tried to work it out in detail yet

Here is the first installment of my working it out in detail. I'm not going to give all of the calculations behind all this; I'm mainly just going to state the results. (I might write up an Insights post on the general technique, since the kinematic decomposition is a valuable tool and I can't be the only one that could use more practice at it.)

We have three vector fields on Minkowski spacetime:

$$
X^a = \left( t, z, x, y \right)
$$
$$
U^a = \left( \gamma, 0, - \gamma \omega y, \gamma \omega x \right)
$$
$$
A^a = \left( 0, 0, - \gamma^2 \omega^2 x, - \gamma^2 \omega^2 y \right)
$$

where $\gamma = 1 / \sqrt{1 - \omega^2 x^2 - \omega^2 y^2}$. For help in computations to follow, we note that $\gamma_{, x} = \gamma^3 \omega^2 x$ and $\gamma_{, y} = \gamma^3 \omega^2 y$.

We now compute the tensor $D_{ab} = \theta_{ab} + \omega_{ab} = A_a U_b + U_{a, b}$. (Note, as before, the lower indexes; since we are working in Minkowski coordinates, that just means the signs of the $0$ components of all vectors are reversed.) This gives:

$$
D_{ab} = \begin{bmatrix}
0 & 0 & - \gamma^3 \omega^2 x & - \gamma^3 \omega^2 y \\
0 & 0 & 0 & 0 \\
\gamma^3 \omega^2 x & 0 & 0 & - \gamma^3 \omega \\
\gamma^3 \omega^2 y & 0 & \gamma^3 \omega & 0
\end{bmatrix}
$$

This is obviously antisymmetric, so we have zero expansion and shear and nonzero vorticity. So far so good.

Next we add a mass distribution, which we will first write simply as a function $\rho (t, z, x, y)$ which gives the mass density as a function of the coordinates. (We will leave out the functional dependence for simplicity.) We then have the 4-momentum density $P^a = \rho U^a$ and the 4-force $F^a = \rho A^a$. Note that these two expressions assume that $d\rho / d\tau = 0$, i.e., that along each worldline in the congruence, the mass density is constant; it only varies from one worldline to another. This assumption is valid for the scenario we are considering, but it's important to recognize that it is there.

We next compute the angular momentum density tensor $M^{ab} = \rho \left( X^a P^b - P^a X^b \right)$. This gives:

$$
M^{ab} = \rho \begin{bmatrix}
0 & - \gamma z & - \gamma \left( x + \omega t y \right) & - \gamma \left( y - \omega t x \right) \\
\gamma z & 0 & - \gamma \omega z y & \gamma \omega z x \\
\gamma \left( x + \omega t y \right) & \gamma \omega z y & 0 & \gamma \omega \left( x^2 + y^2 \right) \\
\gamma \left( y - \omega t x \right) & - \gamma \omega z x & - \gamma \omega \left( x^2 + y^2 \right) & 0
\end{bmatrix}
$$

Finally, we compute the torque density tensor $T^{ab} = \rho \left( X^a A^b - X^b A^a \right)$. This gives:

$$
T^{ab} = \rho \begin{bmatrix}
0 & 0 & \gamma^2 \omega^2 t x & \gamma^2 \omega^2 t x \\
0 & 0 & - \gamma^2 \omega^2 z x & - \gamma^2 \omega^2 z y \\
- \gamma^2 \omega^2 t x & - \gamma^2 \omega^2 z x & 0 & 0 \\
- \gamma^2 \omega^2 t y & - \gamma^2 \omega^2 z y & 0 & 0
\end{bmatrix}
$$

More to come in follow up posts.

 

[PeterDonis]

For this second installment, I'm going to do the same computations as before, but in a frame in which the center of mass is moving in the positive $z$ direction with velocity $v$. We then have the vector fields as follows:

$$
X'^a = \left( t', z', x', y' \right)
$$
$$
U'^a = \left( g \gamma, g \gamma v, - \gamma \omega y', \gamma \omega x' \right)
$$
$$
A'^a = \left( 0, 0, - \gamma^2 \omega^2 x', - \gamma^2 \omega^2 y' \right)
$$

where $g = 1 / \sqrt{1 - v^2}$. Note that $g$ is constant everywhere.

The tensors are as follows:

$$
D'_{ab} = \begin{bmatrix} \\
0 & 0 & - g \gamma^3 \omega^2 x' & - g \gamma^3 \omega^2 y' \\
0 & 0 & g \gamma^3 \omega^2 v x' & g \gamma^3 \omega^2 v y' \\
g \gamma^3 \omega^2 x' & - g \gamma^3 \omega^2 v x' & 0 & - \gamma^3 \omega \\
g \gamma^3 \omega^2 y' & - g \gamma^3 \omega^2 v y' & \gamma^3 \omega & 0
\end{bmatrix}
$$

This is still obviously antisymmetric, so we have zero expansion and shear and nonzero vorticity, as expected since a Lorentz transformation should not affect the symmetry properties of a tensor.

For the other two tensors, we write the mass distribution as $\rho' (t', z', x', y')$, indicating that we have to transform it into the new coordinates. We'll save details of that for future posts.

$$
M'^{ab} = \rho' \begin{bmatrix}
0 & - g \gamma \left( z' - v t' \right) & - g \gamma \left( x' + \omega t' y' \right) & - \gamma \left( y' - \omega t' x' \right) \\
g \gamma \left( z' - v t' \right) & 0 & - g \gamma \left( v x' + \omega z' y' \right) & g \gamma \left( - v y' + \omega z' x' \right) \\
g \gamma \left( x' + \omega t' y' \right) & g \gamma \left( v x' + \omega z' y' \right) & 0 & \gamma \omega \left( x'^2 + y'^2 \right) \\
g \gamma \left( y' - \omega t' x' \right) & - g \gamma \left( - v y' + \omega z' x' \right) & - \gamma \omega \left( x'^2 + y'^2 \right) & 0
\end{bmatrix}
$$

$$
T'^{ab} = \rho' \begin{bmatrix}
0 & 0 & g \gamma^2 \omega^2 t' x' & g \gamma^2 \omega^2 t' x' \\
0 & 0 & - g \gamma^2 \omega^2 z' x' & - g \gamma^2 \omega^2 z' y' \\
- g \gamma^2 \omega^2 t' x' & - g \gamma^2 \omega^2 z' x' & 0 & 0 \\
- g \gamma^2 \omega^2 t' y' & - g \gamma^2 \omega^2 z' y' & 0 & 0
\end{bmatrix}
$$

More to come.

 

[TSny]

The helix has $n$ complete turns, parametrized as above by $\vec r (\theta) = ( \cos \theta , \sin \theta, p \theta)$

$J_y =\lambda \int_{-n\pi}^{n\pi} \, d\theta\, y(\theta) \omega z(\theta) =(-1)^n 2\pi n \omega p \lambda$.

Please bear with me. If $y(\theta) = \sin \theta$ and $z(\theta) = p \theta$, then for the case $n = 1$ we have the integral $\int_{-\pi}^{\pi} \, \theta \sin \theta \, d\theta$ which is positive. But $(-1)^n = -1$, which is negative.

 

[PeterDonis]

Following on from my previous posts, I now want to try out some specific assumptions for the mass distribution $\rho$. I'll start with the cylinder. We assume that the cylinder is of radius $R$ and axial length $Z$, and has total mass $m$. The mass distribution will be:

$$
\rho = \frac{m}{2 \pi R Z} \delta \left( x^2 + y^2 - R^2 \right) \theta \left( \frac{Z}{2} + z \right) \theta \left( \frac{Z}{2} - z \right)
$$

where $\delta$ is the Dirac delta function and $\theta$ is the Heaviside step function. (We could also model the $z$ dependence by just limiting the integration over $z$ to the interval $- Z / 2 \le z \le Z / 2$ by hand.)

Let's look at how this behaves when we integrate the angular momentum and torque over the entire cylinder for the instant $t = 0$. We first observe that we can integrate separately over $z$ and over $x, y$, and that the $z$ dependence of $\rho$ is even; so any component that is odd in $z$ will integrate to zero. Next, we observe that when integrating over $x$ and $y$, the distribution given by $\rho$ is even in both variables, so any term that is odd in both variables (for example, $x$ alone with no $y$ factor, or vice versa) will integrate to zero. That gets rid of every term except the 2-3 term in the angular momentum, which integrates to

$$
M^{23}_{\text{total}} = \frac{m}{2 \pi R Z} \int dz \, \theta \left( \frac{Z}{2} + z \right) \theta \left( \frac{Z}{2} - z \right) \int dx dy \, \delta \left( x^2 + y^2 - R^2 \right) \gamma \omega \left( x^2 + y^2 \right) = m \gamma \omega R^2
$$

Now we look at how this goes in the primed frame. Here we have to transform the density $\rho$, which becomes

$$
\rho' = \frac{m g^2}{2 \pi R Z} \delta \left( x'^2 + y'^2 - R^2 \right) \theta \left( \frac{Z}{2g} + z' - v t' \right) \theta \left( \frac{Z}{2g} - z' + v t' \right)
$$

where $g = 1 / \sqrt{1 - v^2}$, as before. At the instant $t' = 0$, the only nonzero term is still the 2-3 term in the angular momentum, and it integrates, if I've gotten the factors of $g$ right, to $m g \gamma \omega R^2$.

The next installment will look at how this goes for the helix.

 

[PeterDonis]

For this installment, we'll look at a mass distribution for the helix and see how it goes. First we'll look at uniform mass distributions.

We will assume that a single turn of the helix has axial length $Z$ and that there are $N$ turns, and that the radius of the helix is $R$. The mass distribution I propose, in the center of mass frame, is:

$$
\rho = \frac{m}{N \sqrt{4 \pi^2 R^2 + Z^2}} \delta \left( x^2 + y^2 - R^2 \right) \delta \left( \tan^{-1} \frac{y}{x} - \omega t - \frac{2 \pi z}{Z} \right)
$$

where for simplicity I have left out the step functions in $z$ and just assumed that we limit the integration over $z$ appropriately. The easiest way to integrate this is to change variables from $x, y$ to $r, \theta$, which gives

$$
\rho = \frac{m}{N \sqrt{4 \pi^2 R^2 + Z^2}} \delta \left( r - R \right) \delta \left( \theta - \omega t - \frac{2 \pi z}{Z} \right)
$$

Now we can separate the integrals as before, and in fact we will end up with the same integrals that I did earlier (modulo some constant factors), with the same results. But let's try it now in the primed frame, picking just the right velocity $v$ to unwind the helix:

$$
\rho' = \frac{m g^2}{N \sqrt{4 \pi^2 g^2 R^2 + Z^2}} \delta \left( x'^2 + y'^2 - R^2 \right) \delta \left( \tan^{-1} \frac{y'}{x'} - \frac{\omega}{g} t' \right)
$$

Once again, it looks like we can change variables as above and separate the integrals so that things work out as before.

In the next installment I'll look at non-uniform mass distributions.

 

[maline]

Please bear with me. If $y(\theta) = \sin \theta$ and $z(\theta) = p \theta$, then for the case $n = 1$ we have the integral $\int_{-\pi}^{\pi} \, \theta \sin \theta \, d\theta$ which is positive. But $(−1)n=−1$, which is negative.

Again, $\lambda \omega p \theta \sin \theta$ is the $y$ component of the term $\rho (\vec \omega \cdot \vec r) \vec r$. But the angular momentum density is given by $d \vec J = \rho [r^2 \vec \omega - (\vec \omega \cdot \vec r)\vec r]dV$, so you get the opposite sign.

 

[maline]

Let's look at how this behaves when we integrate the angular momentum and torque over the entire cylinder for the instant t=0.

I don't think the local angular momentum tensor in the form you gave, $M^{\alpha \beta}=\rho (X^\alpha U^\beta-X^\beta U^\alpha)$, can be integrated over a hypersurface to give a tensor. To integrate over constant $t$, what you need is the $t$ components of the rank-3 angular momentum density tensor, $M^{\alpha \beta t}=X^\alpha T^{\beta t}-X^\beta T^{\alpha t}$. But the "kinetic" part of $T^{\alpha \beta}$ is given by $\rho U^\alpha U^\beta$, so the $t$ components $T^{\alpha t}=\rho U^\alpha U^t=\gamma \rho U^\alpha \ne p^\alpha$. The extra gamma factor gives additional weight to the parts that are moving faster in your frame (the frame where $t$ is constant on your hypersurface of integration).

By the way, i think it's important to spell out that $\rho$ here refers to the local rest mass density scalar, $\rho=T^{\alpha \beta}U_\alpha U_\beta$ rather than the coordinate dependent energy density $T^{tt}$, which I think is also sometimes denoted by $\rho$. The scalar is useful for defining local momentum and force densities as tensors, but the fact the relationship between $\rho$ and $T^{tt}$ varies with $U$ means that these tensors are not good for hypersurface integrals. I also am not sure how to write the conservation laws using such local tensors instead of the density tensors, where you take the 4-divergence using the additional index.

 

[PeterDonis]

To integrate over constant $t$, what you need is the $t$ components of the rank-3 angular momentum density tensor, $M^{\alpha \beta t}=X^\alpha T^{\beta t}-X^\beta T^{\alpha t}$. But the "kinetic" part of $T^{\alpha \beta}$s given by $\rho U^\alpha U^\beta$ so the $t$ components $T^{\alpha t}=\rho U^\alpha U^t=\gamma \rho U^\alpha \ne p^\alpha$.

You're assuming that the stress-energy tensor is that of a perfect fluid, since that's the one for which the "kinetic" part is $\rho U^a U^b$. I'm not sure that approach works if we use delta functions; we would have to model the cylinder and the helix as continuous substances occupying an open region of spacetime. I'm trying to avoid doing that if at all possible since it greatly complicates the math.

Heuristically, though, I think you're correct that there should be an extra factor of $\gamma$ in the 4-momentum $P^a$, since the individual pieces of matter whose worldlines are given by $U^a$ are not at rest in the frame we are using. That doesn't change which components are nonzero, since there are no derivatives in the angular momentum or torque tensors; it just adds an extra factor of $\gamma$ to those tensors.

i think it's important to spell out that $\rho$ here refers to the local rest mass density scalar, $\rho=T^{\alpha \beta}U_\alpha U_\beta$ rather than the coordinate dependent energy density $T^{tt}$

Yes, I think that's correct--and that's another way of seeing why the formulas should have a factor of $\gamma$ added, as above. If $\rho$ is the mass/energy density measured by an observer moving with the matter (in this case the rotating cylinder/helix), then the mass/energy density measured by an observer at rest in the frame we are using will be $\gamma \rho$.

the fact the relationship between $\rho$ and $T^{tt}$ varies with $U$ means that these tensors are not good for hypersurface integrals

You can still do the integrals, you just have to be careful to specify the integrands taking into account the effects discussed above.

I also am not sure how to write the conservation laws using such local tensors instead of the density tensors, where you take the 4-divergence using the additional index

I'm not sure what you mean. The fundamental conservation law is $T^{ab}{}_{; b} = 0$, which is local (valid at every event, in any frame), and involves a "density tensor" (the SET). Any other conservation law, using a tensor derived from $T^{ab}$, can be derived from this one. But conservation laws that are not local, i.e., that hold on some foliation of the spacetime into spacelike hypersurfaces, will involve integrals over those hypersurfaces, so you have to be careful to specify the integrands, as above.

The case we're discussing is actually pretty simple, since we are in flat spacetime and are using inertial coordinate charts, which means there are no connection coefficients and the local conservation law is just $T^{ab}{}_{, b} = 0$. We also have obvious foliations into spacelike hypersurfaces, namely those of constant coordinate time in whatever frame we choose. So the issues that arise in curved spacetimes or non-inertial charts don't arise here.

 

[PeterDonis]

I think you're correct that there should be an extra factor of γγ\gamma

Yes, I think that's correct

Actually, on looking back at the formulas I gave for $\rho$, I'm not sure the extra factor of $\gamma$ should be there, because I defined $\rho$ using the rest mass $m$ of the object as a whole. That rest mass already includes the internal kinetic energy due to the motion of parts of the object, as long as those motions average out to zero, i.e., as long as we are in the center of mass frame. So actually I was defining $\rho$ as $T^{tt}$ (actually, the mixed component $T^t{}_t$, strictly speaking, since that's the one that doesn't include any effects of the metric or the metric signature). The mass/energy density measured by an observer rotating with the internal parts, by my definition, would be $\rho / gamma$.

As I said before, none of this affects which components of the various tensors are nonzero, so it doesn't affect the key question of what is necessary, dynamically, to make the motion in question a free motion.

 

[TSny]

Again, $\lambda \omega p \theta \sin \theta$ is the $y$ component of the term $\rho (\vec \omega \cdot \vec r) \vec r$. But the angular momentum density is given by $d \vec J = \rho [r^2 \vec \omega - (\vec \omega \cdot \vec r)\vec r]dV$, so you get the opposite sign.

Ok, I've got the signs settled now and your final answer for $J_y$ in #61 looks good.
I will now move on to Peter's more general modeling.

 

[maline]

So actually I was defining $ρ$ as $T^{tt}$

Okay, glad we got that clarified. But note that with that definition, multiplying $\rho$ by a tensor such as $U$ does not give a tensor quantity.

The mass/energy density measured by an observer rotating with the internal parts, by my definition, would be $ρ/\gamma$.

Actually I think the ratio should be $\gamma^2$, one gamma factor for the length contraction and another for the kinetic energy. This is consistent with $T^{\alpha \beta}=\rho U^\alpha U^\beta$ for the case of dust (using the scalar $\rho$).

As I said before, none of this affects which components of the various tensors are nonzero, so it doesn't affect the key question of what is necessary, dynamically, to make the motion in question a free motion.

This is true so long as those integrals that give zero do so for each cylindrical shell separately. Until now our distributions have been confined to a single cylindrical shell, so these details were not important, but sooner or later we will have to move on to a 3D model with a stress tensor, and we probably will not be able to deal only with particular shells.

Ok, I've got the signs settled now and your final answer for $J_y$ in #61 looks good.
I will now move on to Peter's more general modeling.

Don't forget the elephant in the room: these analyses, both for the center of mass and for the angular momentum, are dealing only with the movement of the mass. But the helix, even in the CoM frame, is a moving body under stresses that are not perpendicular to the motion. That gives a whole other sector, of both energy and momentum, where has of yet we have no idea what happens. It may well be that after taking these into account, there is no possible mass and stress distribution that will allow a free rotation!

 

[PeterDonis]

note that with that definition, multiplying $\rho$ by a tensor such as $U$ does not give a tensor quantity

Yes, but we can fix that by getting the factors of $\gamma$ right. To put it another way, we can view $\rho$ as I wrote it as $T^{ab} (e_0)_a (e_0)_b$, where $(e_0)$ is the timelike basis vector of our coordinate chart; physically this would be the mass/energy density measured by an observer at rest in the chart (i.e., relative to the center of mass of the object). To correct it to $T^{ab} U_a U_b$, which is what we want if we are going to use $\rho U^a$ as the momentum density $P^a$, just means multiplying or dividing by appropriate factors of $\gamma$.

Actually I think the ratio should be γ2γ2\gamma^2, one gamma factor for the length contraction and another for the kinetic energy. This is consistent with $T^{\alpha \beta}=\rho U^\alpha U^\beta$ for the case of dust (using the scalar $\rho$).

Yes, I think you're right.

sooner or later we will have to move on to a 3D model with a stress tensor

I'm still hoping we can avoid that level of complication. But see below.

the helix, even in the CoM frame, is a moving body under stresses that are not perpendicular to the motion

This is the part I'm not sure how to capture. I have been hoping that we can somehow capture it using the distributions I gave, limited to the cylindrical shell. But it is not captured in any of the tensors I have computed so far; neither the kinematic decomposition, nor the angular momentum as I've computed it, nor the torque, depend on purely space-space components of the stress-energy tensor (pressure and shear stress), which is what you are talking about.

The only other possibility I can see is to move to the full 3-index angular momentum tensor, which uses the SET directly, and see if any of its components differ between the cylinder and the helix.

 

[maline]

Ok, I want to take a crack at the full stress tensor analysis. I think the cleanest way to go about this is to work in Born coordinates, that is, cylindrical polar coordinates that rotate with the body. In these coordinates the body is stationary. That helps us in several ways. The most important advantage I see is that the stress-related components of the SET take their "natural form" in this chart, which is that of the the purely spatial 3-tensor $-\sigma^{ij}$. In fact, the whole SET (assuming no distortion and no EM fields) is just $$T^{\alpha \beta} =\begin{pmatrix} \rho & 0 \\ 0 & -\sigma^{ij} \end{pmatrix}$$
So, here goes. The transformation from Born coordinates $(t,r,\phi,z)$ to rectangular $(t',x',y',z')$ is: $$ t'=t,~x'=r \cos(\phi+\omega t) ,~y'=r \sin(\phi+\omega t), ~z'=z$$
The partial derivatives are: $${J_\alpha}^\beta =\frac {\partial X'^{~~\beta}}{\partial X^\alpha}=\left( \begin{array}{cccc} 1 & -\omega r \sin(\phi+\omega t) & \omega r \cos(\phi+\omega t) & 0 \\ 0 & \cos(\phi+\omega t) & \sin(\phi+\omega t) & 0 \\ 0 & -r \sin(\phi+\omega t) & r \cos(\phi+\omega t) & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) $$ Therefore the metric tensor is: $$g_{\alpha \beta}={J_\alpha}^\mu \,\eta'_{\mu \nu} \,{J_\beta}^\nu= \begin{pmatrix} -1+\omega^2 r^2 & 0 & \omega r^2 & 0 \\ 0 & 1 & 0 & 0 \\ \omega r^2 & 0 & r^2 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$ I plan to work on this a bit at a time. The next step is to write the formula for the covariant 4-divergence of a rank 2 contravariant tensor in these coordinates. We can apply this to the SET and get 4 equations (PDE's) we will need to solve.

If anyone else wants to continue where I leave off, please go right ahead.

 

[maline]

All right, so it seems the covariant 4-divergence of a rank 2 tensor does not have a simple form (using $\sqrt {-g}$, for instance) except where the tensor is antisymmetric, which the SET obviously is not. We will have to use the Christoffel symbols explicitly.

The nonzero partial derivatives of the metric are: $$g_{tt,r}=2r\omega^2~~~g_{t\phi,r}=g_{\phi t,r}=2r\omega~~~g_{\phi \phi ,r}=2r$$
We wish to calculate the Christoffel symbols, using the explicit formula $\Gamma^\alpha_{\beta \gamma}=\frac 1 2 g^{\alpha \mu}(g_{\mu \beta,\gamma}+g_{\gamma \mu,\beta}-g_{\beta \gamma, \mu})$. For this we also need the contravariant metric tensor. Matrix inversion gives: $$g^{\alpha \beta}=\begin{pmatrix} -1 & 0 & \omega & 0 \\ 0 & 1 & 0 & 0 \\ \omega & 0 & r^{-2}-\omega^2 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$
Now for the Christoffel symbols: $$\Gamma^r_{tt}=-\frac 1 2 g^{rr}g_{tt,r}=-r\omega^2~~~\Gamma^r_{\phi\phi}=-\frac 1 2 g^{rr}g_{\phi\phi,r}=-r~~~\Gamma^r_{t \phi}=\Gamma^r_{\phi t}=-\frac 1 2 g^{rr}g_{t \phi,r}=-r\omega \\ \Gamma^t_{tr}=\Gamma^t_{rt}=\frac 1 2 g^{tt}g_{tt,r}+\frac 1 2 g^{t \phi}g_{\phi t,r}=-r\omega^2+r\omega^2=0 \\\Gamma^t_{\phi r}=\Gamma^t_{r \phi}=\frac 1 2 g^{tt}g_{t \phi,r}+\frac 1 2 g^{t \phi}g_{\phi \phi,r}=-r\omega+r\omega=0 \\ \Gamma^\phi_{tr}=\Gamma^\phi_{rt}=\frac 1 2 g^{\phi t}g_{tt,r}+\frac 1 2 g^{\phi \phi}g_{t \phi,r}=r\omega^3+\frac \omega r-r\omega^3=\frac \omega r\\ \Gamma^\phi_{\phi r}=\Gamma^\phi_{r \phi}=\frac 1 2 g^{\phi t}g_{t \phi,r}=\frac 1 2 g^{\phi \phi}g_{\phi \phi,r}=r\omega^2+\frac 1 r -r \omega^2=\frac 1 r$$
All the others are clearly zero because all the metric derivatives involved are zero.

Now we can work out the covariant divergence equations, representing local energy/momentum conservation. We assume all derivatives by $t$ are zero, and $T^{ti}=T^{it}=0$ everywhere for each spatial index $i$. For the "energy" equation, we have $${T^{t \alpha}}_{;\alpha}={T^{t \alpha}}_{,\alpha}+\Gamma^t_{\alpha \mu}T^{\mu \alpha}+\Gamma^\alpha_{\alpha \mu}T^{t \mu}=0$$All of these terms vanish independently, which is intuitive. Our body is "stationary" in these coordinates and no "energy" is "flowing" anywhere.

For the "$z$ momentum", we get $${T^{z \alpha}}_{;\alpha}={T^{z \alpha}}_{,\alpha}+\Gamma^z_{\alpha \mu}T^{\mu \alpha}+\Gamma^\alpha_{\alpha \mu}T^{z \mu}={T^{zi}}_{,i}+\Gamma^\phi_{\phi r}T^{zr}=\\={T^{zi}}_{,i}+\frac 1 r T^{zr}=0$$This is the same as with a truly stationary body in ordinary polar coordinates: the "flow" in the $r$ direction has a change in "density" as the $\hat r$ vectors spread apart.

For the "$\phi$ momentum": $$ {T^{\phi \alpha}}_{;\alpha}={T^{\phi \alpha}}_{,\alpha}+\Gamma^\phi_{\alpha \mu}T^{\mu \alpha}+\Gamma^\alpha_{\alpha \mu}T^{\phi \mu}=\\={T^{\phi i}}_{,i}+\Gamma^\phi_{\phi r}T^{r \phi}+\Gamma^\phi_{r \phi}T^{\phi r}+\Gamma^\phi_{\phi r}T^{\phi r}=\\= {T^{\phi i}}_{,i}+\frac 3 r T^{\phi r}=0$$The additional two $T^{\phi r}=T^{r \phi}$ terms are because "$r$ momentum" needs to shift its direction as it "flows" along the curve in the $\phi$ direction, and because the magnitude of a unit of "$\phi$ momentum" changes with $r$.

For the "$r$ momentum": $$ {T^{r \alpha}}_{;\alpha}={T^{r \alpha}}_{,\alpha}+\Gamma^r_{\alpha \mu}T^{\mu \alpha}+\Gamma^\alpha_{\alpha \mu}T^{r \mu}=\\={T^{r i}}_{,i}+\Gamma^r_{tt}T^{tt}+\Gamma^r_{\phi \phi}T^{\phi \phi}+\Gamma^\phi_{\phi r}T^{r r}=\\={T^{r i}}_{,i}-r\omega^2 \rho-r T^{\phi \phi}+\frac 1 r T^{rr}=0$$The $r\omega^2 \rho$ term is, of course, the centrifugal force- the source of all our stress. It is the only place that the mass distribution enters the equations. The T^{\phi \phi} term is because of the "$\phi$ momentum" changing direction as it flows, and it expresses the fact that the body, in resisting centrifugal expansion, will be under tension in the $\phi$ direction. In the case of a rotating ring, for instance, this tension is the only stress needed to maintain rigidity. That's why I claimed a few posts ago that the rotating ring can just as well consist of a flexible rope and still be a "rigid body".

We have arrived at a very surprising result: the form of these equations is exactly the same whether the speed $\omega r$ is tiny, or highly relativistic! No $\gamma$ factors appear. In fact $\omega$ itself does not appear except in the centrifugal force term. This implies that any solution for the stresses in a slow rotation- a purely classical problem- can simply be multiplied by a constant $(\omega_2^2/\omega_1^2)$ to be valid for any value of $\omega$!

In particular, Peter's solution for a freely rotating helix, in post #118, should certainly work for slow rotations (if it is modified to make the helix 3D, which is simple). Whatever the stresses are in that case, they satisfy local energy-momentum conservation. So we can always transfer that solution to Born coordinates, multiply by the appropriate factor, and get a fully relativistic solution. If it satisfies the local conservation laws then it will automatically be a free motion as well. So freely rotating relativistic helices definitely can be described! We are still stuck with our paradox!

 

[TSny]

I’m still working on this. I agree that there are probably many possible choices for variable mass densities that would produce freely rotating helices. But the nature of the stresses in the helix are causing me problems. There will be compression, tension, shear, and bending moments that vary in a complicated way along the helix.

I did run across some notes that I made years ago on a helical system with simple stress, but with the disadvantage of requiring external forces at the ends of the helix.

https://www.dropbox.com/s/20kfa57au...ting Helix Paradox edited 5_30_2017.docx?dl=0

The helix consists of a flexible cable of uniform mass. The only stress is tension in the cable. As the helix rotates, it maintains its shape with just the tension and the external forces at the ends. The model assumes the rotation speed of the helix is nonrelativistic.

In the frame of reference where the helix is rotating but not translating along its axis, the forces on the ends have magnitudes equal to the tension and tangent to the helix. In the frame in which the helix is unwound into a straight line, the forces on the end are equal in magnitude, opposite in direction, but are not parallel to the cable. The result is that the net force on the cable is zero but there is a nonzero torque (couple). The paradox is double: (1) How can the cable rotate around the coordinate axis if there is zero net external force? (2) How does the cable maintain its orientation parallel to the coordinate axis despite the external torque which appears to be in a direction that would change the orientation?

The notes show, I hope, how the contribution of the tension to the momentum and angular momentum in the unwound coordinate system resolves the paradox. Even though the cable revolves around the coordinate axis, there is no linear momentum associated with this rotatory motion. The system only has a constant linear momentum parallel to the cable. So, no external force is required. However, in general there is a changing angular momentum which is shown to equal the applied torque. So, it appears to all work out.

The obvious shortcoming of this model is the need for the external forces at the ends of the helix. Also, in the notes, I assume that the rotational speed of the helix is nonrelativistic to simplify the calculations.

I don’t know if anyone will want to wade through the details of the notes. I tried to keep it very elementary. But the calculations are inelegant and tedious. I used several different coordinate systems, including some that are comoving with an element of the helix. Sorry for not including any diagrams in the notes.

 

[pervect]

I’m still working on this. I agree that there are probably many possible choices for variable mass densities that would produce freely rotating helices. But the nature of the stresses in the helix are causing me problems. There will be compression, tension, shear, and bending moments that vary in a complicated way along the helix.

This thread has gotten too long for me to follow, but I thought your simplified version of the helix which was a loaded beam was a promising approach and simpler than the other approaches I'd seen. Unfortunately, I'm not familiar with the stresses in a loaded beam in any detail, so I wasn't able to proceed further.

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 believe the beam must have a finite width to support the loads with finite stresses. 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. I don't think the stresses in the directions transverse to the boost should matter. 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".

 

[PeterDonis]

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

This is not correct if the beams are made of ordinary matter, since 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.

Assuming a "massless" beam should not change the above, because "massless" does not mean "zero energy density in the rest frame", and objects made of massless particles still obey the energy conditions. "Massless", in terms of the stress-energy tensor, means that the SET is something like that of the electromagnetic field, as given, for example, here:

https://en.wikipedia.org/wiki/Electromagnetic_stress–energy_tensor

Note that the key property due to masslessness is that the SET is traceless, not that the energy density is zero.

In order to violate the energy conditions and have negative energy density in some frame, the beam would have to be made of exotic matter, but I don't think that was the intent of the model, and I would not recommend making such an assumption anyway.