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

[maline]

Note that this time derivative is by coordinate time, unlike 4-force which is a derivative by proper time, so its value in our case is −ω2R^r−ω2Rr^-\omega^2 R \hat r times the mass density, without the γγ\gamma factor.

No, this doesn't work either. The $t$ derivative of momentum density at a constant spatial position $x^i$ does not include the acceleration of the mass elements as they move, so in particular the centripetal acceleration will not show up there at all. The force required for that must be expressed as a proper time derivative $\frac {dT^{\alpha t}} {d\tau}={T^{\alpha t}}_{,\beta}u^\beta$, also known as the Langrangian derivative of the 4-momentum density.
But I'm afraid I now am confused about how to work with this "force". What are its transformation properties? What is the correct way to write it as a derivative of stress? Can someone help me out here?

 

[PeterDonis]

the centripetal acceleration will not show up there at all

The centripetal acceleration can be derived from the kinematics alone--just compute the proper acceleration from the 4-velocity field. That gives the same answer for the helix as it does for the cylinder, because the 4-velocity fields are the same (unless there is something to my previous thought that there might be some issue with derivatives in the helix case because it is a discontinuous subset of the cylinder--but as I said before, I have no idea how to deal with that if it's an issue).

I now am confused about how to work with this "force". What are its transformation properties? What is the correct way to write it as a derivative of stress?

I'm not sure about how to work with this either; in fact I'm not sure how trying to work with the full stress-energy tensor in general will tell us anything new. That's why I've been trying to stick to the simplified relativistic model I've been using. I'm hoping that boosting that model into a frame that kinematically unwinds the helix will shed some light.

 

[pervect]

I don't think a rotating helix is a force-free motion, considering it as a purely classical non-relativistic problem. But I could be making a mistake. Hopefully it's obvious that if the motion isn't force free, then a good analysis of angular (or linear) momentum must consider the external torques and forces on the system.

To set up the problem as a purely classical problem, I consider a rigid central shaft, with tension bearing strings keeping the rotating masses rotating in the helix. This satisfies the force balance equation of the masses, we now need to consider the force balance equation on the central shaft. To have the motion of the rigid central shaft be force-free, we need two conditions. The total force on the central shaft must be zero, and the total torque on the central shaft must be zero.

Either condition alone appears to be possible to satisfy, but not both simultaneously.

To set up the problem, If we let the axis of symmetry of the cylinder be the z-axis, we can write the forces on the central shaft due to the strings as

$$f_x = -\cos kz \quad f_y = -\sin kz \quad f_z = 0$$

For convenience, we can set the pitch factor k equal to 1.

To maintain symmetry, we'll let z vary from -a to a. Then the force free condition (with the condition that the pitch factor k=1, which we will assume from this point onwards) turns out to be 2*sin(a) = 0, which has the obvious solutions a=$\pi, 3\pi$ etc, as expected.

The torques should be given by $\vec{f} \times \vec{z}$, the components of which should be:

$$ \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -\cos z & -\sin z & 0 \\ 0 & 0 & z\end{vmatrix}$$

Carrying out a similar integral for the torques as we did for the forces, we get the torque free condition as:

$$2 \sin a - 2\,a\,\cos a = 0$$

The solutions of the total force being zero don't satisfy the conditions that the torque be zero. The conditions that the torque be zero does has solutions (interestingly enough), but these solutions don't make the total force on the central shaft zero.

 

[PeterDonis]

To set up the problem as a purely classical problem, I consider a rigid central shaft, with tension bearing strings keeping the rotating masses rotating in the helix.

This is different to how I was doing it, but I think I can do a similar analysis in the model I've been using. I'll see.

 

[PeterDonis]

I think I can do a similar analysis in the model I've been using. I'll see.

This is interesting enough to sidetrack me from boosting my relativistic model, so I'll tackle it first.

The 4-position of the helix is given by

$$
X^a = \left( t, z, R \cos ( \omega t + k z ), R \sin (\omega t + k z ) \right)
$$

The 4-velocity field of the helix is given by

$$
U^a = \left( \gamma, 0, - \gamma \omega R \sin ( \omega t + k z ), \gamma \omega R \cos (\omega t + k z ) \right)
$$

where $k = 2 \pi / Z$ and $Z$ is the axial length (along $z$) of a single turn of the helix. The proper acceleration for this 4-velocity field is

$$
A^a = \left( 0, 0, - \gamma^2 \omega^2 R \cos ( \omega t + k z ), - \gamma^2 \omega^2 R \sin ( \omega t + k z ) \right)
$$

What we need to do now is to sum the force and torque over the entire helix and see whether they both vanish, which must be the case if the motion is a valid free motion. The two integrals in question are

$$
F^a_{\text{total}} = \int_{- NZ / 2}^{NZ / 2} \mu(z) A^a(z) dz
$$
$$
T^{ab}_{\text{total}} = \int_{- NZ / 2}^{NZ / 2} \mu(z) \left( X^a (z) A^b (z) - X^b (z) A^a (z) \right) dz
$$

Note that I have modeled the torque as an antisymmetric tensor rather than as a vector, for the same reason I did that for the angular momentum. Also, we evaluate the integrals at $t = 0$.

The force integral only has two nonzero components, $x$ and $y$; and the $y$ integral is an odd function over an even domain, which is zero. So we have, using the same simplifications as I used for the angular momentum in my earlier post:

$$
F^x_{\text{total}} = - 2 \gamma^2 \omega^2 R \int_{0}^{NZ / 2} \mu(z) \cos \left( \frac{2 \pi z}{Z} \right) = - 2 \gamma^2 \omega^2 R \sum_{k = 1}^N \sin \left( \pi k \right) W_k
$$

Since the $\sin$ will be zero for all terms, the weights don't matter and we have that the total force on the helix does vanish.

The torque integral turns out to also have only one component that is not an odd function over an even domain (note that $T^{23}$, the torque in the x-y plane, is an integral of the product $\cos k z \sin k z$, which is odd and therefore vanishes):

$$
T^{13} = - 2 \gamma^2 \omega^2 R \int_0^{NZ/2} \mu(z) z \sin \left( \frac{2 \pi z}{Z} \right)
$$

This turns out to be the same integral (except for some slightly different constant factors) as $M^{12}$, and therefore can be made to vanish by the same method as I showed in my previous post, by choosing appropriate weights for each piece of the helix.

So it looks to me like the total force and total torque on the helix can be made to vanish, so we can't rule out that it is a possible free motion on those grounds.

 

[PeterDonis]

I don't think a rotating helix is a force-free motion, considering it as a purely classical non-relativistic problem.

Looking at the relativistic analysis I just posted, the only force on an individual element of the helix (in my model, where there are no massless strings or massless central shaft) is the centripetal force that keeps the piece rotating about the $z$ axis with angular velocity $\omega$. This force is exerted by neighboring pieces of the helix: it is true that those pieces also exert sideways forces, but those should cancel leaving only the centripetal net force. That, at any rate, is what the kinematics, the proper acceleration vector, says. In this respect the helix appears to work exactly like a rotating ring, for which the centripetal force is also exerted by neighboring pieces of the ring.

 

[pervect]

Looking at the relativistic analysis I just posted, the only force on an individual element of the helix (in my model, where there are no massless strings or massless central shaft) is the centripetal force that keeps the piece rotating about the $z$ axis with angular velocity $\omega$. This force is exerted by neighboring pieces of the helix: it is true that those pieces also exert sideways forces, but those should cancel leaving only the centripetal net force. That, at any rate, is what the kinematics, the proper acceleration vector, says. In this respect the helix appears to work exactly like a rotating ring, for which the centripetal force is also exerted by neighboring pieces of the ring.

I'd agree that the force on an element is the centripetal force that keeps the piece rotating about the z axis.

But I don't think this matters to the analysis. The situation is one of dynamic balance, as seen for instance in tires. A tire can be perfectly balanced statically, but it must also be dynamically balanced to rotate without wobbling. So the question is - would a helix shaped "tire" be dynamically balanced, or not? The static balance of the helix is insufficient to answer this question.

According to wiki, the physical condition for torque free motion is that the axis of rotation be a principal axis of the moment of inertia tensor. This sounds right to me, I haven't tried to look it up in Goldstein (or another text) to confirm.

Whenever a rotor is forced to rotate about an axis that is not a principal axis, an external torque is needed.

https://en.wikipedia.org/w/index.php?title=Tire_balance&oldid=779604153

So, in this approach, we need to ask what the moment of inertia tensor of a helix is - specifically, we want to know if the z-axis of the cylinder is one of the principle axes of the moment of inertia tensor. If it is one of the principle axes, then the moment of inertia tensor should be diagonal.

We can write the moment of inertia tensor as

$$I^{ij} = \sum x^i x^j dm $$

where dm is a mass element.

If the off-diagonal elements of this tensor are zero, then the principle axes are aligned with the coordinate axes. Parameterizing the helix as we did before we have:

$$x = \cos s \quad y = \sin s \quad z = s$$

To maintain static balance, we'll let s range from $-\pi$ to $\pi$.

We can write $I^{yz}$ as being proportional to

$$\int_{-\pi}^\pi s \, \sin s \, ds$$

which is nonzero, demonstrating that the z-axis is not a principle axis of the moment of inertia tensor, and thus demonstrating that we need to apply an external torque to make a helix rotate around the z-axis.

 

[PeterDonis]

To maintain static balance, we'll let s range from $-\pi$ to $\pi$.

This is the case of one turn, which we already know is not a possible free motion. Try it for $s$ ranging from $- 2 \pi$ to $2 \pi$, which describes a helix of 2 turns; this gives a zero integral with appropriate weights for the segments (it's the same integral that comes up in the $M^{12}$ tensor component for the angular momentum).

We can write $I^{yz}$ as being proportional to
$$
\int_{-\pi}^\pi s \, \sin s \, ds
$$

You left out the mass distribution, which means you're assuming a uniform mass distribution. We already know this case is not a possible free motion. If you put in a non-uniform mass distribution with appropriate weights, as I did in my last few posts, you can get the components that are not in the x-y plane to vanish, since, as above, the integral that arises is the same one that has come up in my previous posts.

 

[PeterDonis]

the physical condition for torque free motion is that the axis of rotation be a principal axis of the moment of inertia tensor

I think what we have been showing in our last few exchanges is that there are at least three equivalent ways of stating this condition:

(1) The axis of rotation must be a principal axis of the moment of inertia tensor;

(2) The axis of rotation must be perpendicular to the plane of the angular momentum tensor;

(3) The total torque on the body, summed over all elements of the body, must be zero.

It looks, from my calculations, as if all three of these conditions give rise to the same integral (in this case the integral of $s \sin s$), so they will all three be satisfied, if they can be satisfied, by the same mass distribution and kinematics (where the latter means the same 4-position, 4-velocity, and 4-acceleration vectors).

 

[PeterDonis]

Here is a first stab at boosting my relativistic solution into a frame that unwinds the helix. We boost by a velocity $- v$, so that the helix in the new frame is moving at $v$ in the positive $z$ direction (note that this basically means we are treating the original frame, in which the helix center of mass is at rest, as the "primed" frame, even though the notation I am using labels it as unprimed and the new, boosted frame in this post as primed). Thus we have $t' = g(t + vz_0)$, $z' = g(z_0 + vt)$, where $g = 1 / \sqrt{1 - v^2}$, and I have used the notation $z_0$ to emphasize the fact that, in the original frame, each worldline in the helix congruence has a constant $z$ coordinate. This will help when we start looking at integrals, since we can still integrate over $z_0$ as the worldline label.

The inverse transform is then $t = g(t' - v z')$, $z_0 = g(z' - vt')$. We use the second equation to find $z'$ as a function of $t'$, and the first to express the arguments of the trig functions as functions of $t'$. This gives us a 4-position vector in the new frame:

$$
X^a = \left( t', v t' + \frac{z_0}{g}, R \cos ( \frac{\omega}{g} t' - \frac{v}{g} z_0 + k z_0 ), R \sin ( \frac{\omega}{g} t' - \frac{v}{g} z_0 + k z_0 ) \right)
$$

If we set $k = v / g = v \sqrt{1 - v^2}$, then the trig functions no longer depend on $z_0$ and we have unwound the helix. The only constraint is that, for this to be possible, we must have $k \le 1/2$, since we have $0 < v < 1$ and the function $v \sqrt{1 - v^2}$ has a maximum value of $1/2$ in that interval, at $v = 1 / \sqrt{2}$.

With the helix unwound, the 4-position, 4-velocity, and 4-acceleration vectors are

$$
X^a = \left( t', v t' + \frac{z_0}{g}, R \cos ( \frac{\omega}{g} t' ), R \sin ( \frac{\omega}{g} t' ) \right)
$$
$$
U^a = \left( g \gamma, g \gamma v, - \gamma \omega R \sin ( \frac{\omega}{g} t' ), \gamma \omega R \cos ( \frac{\omega}{g} t' ) \right)
$$
$$
A^a = \left( 0, 0, - \gamma^2 \omega^2 R \cos ( \frac{\omega}{g} t' ), - \gamma^2 \omega^2 R \sin ( \frac{\omega}{g} t' ) \right)
$$

This does indeed look like a straight rod that is off axis and is rotating as it translates, which intuitively does not seem like a possible free motion. However, I'll save further comment until I've done the computations of the tensor components.

 

[PeterDonis]

Following on from my last post, here are the nonzero tensor components for an individual element of the helix at $t' = 0$. First, the angular momentum tensor $M^{ab} = \mu (z_0) \left( X^a U^b - X^b U^a \right)$:

$$
M^{01} = - \mu (z_0) \gamma z_0
$$
$$
M^{02} = - \mu (z_0) g \gamma R
$$
$$
M^{12} = - \mu (z_0) g \gamma v R
$$
$$
M^{13} = \mu (z_0) \frac{\gamma \omega R}{g} z_0
$$
$$
M^{23} = \mu (z_0) \gamma \omega R^2
$$

Already we can see that something is going on, since $M^{02}$ and $M^{12}$ are just constants times the even mass distribution function, with no $z_0$ in them at all, so when we integrate over the whole helix (or rod in this frame), they won't vanish, no matter what mass distribution $\mu$ we try. $M^{02}$ and $M^{13}$ will vanish because they are odd functions of $z_0$, but that's no help.

Next, the force vector $F^a = \mu(z_0) A^a$:

$$
F^1 = - \mu (z_0) \gamma^2 \omega^2 R
$$

This also won't vanish when integrated over the whole helix/rod, regardless of the mass distribution.

Finally, the torque tensor $T^{ab} = \mu (z_0) \left( X^a A^b - X^b A^a \right)$:

$$
T^{12} = - \mu (z_0) \frac{\gamma^2 \omega^2 R}{g}
$$

This also won't vanish.

So now we have a puzzle: in the original frame, we were able to choose a mass distribution that made all the necessary things vanish; but in this new boosted frame, we apparently aren't. So something is wrong somewhere. The question is: where?

 

[AVentura]

Is this possibly a manifestation of the Trouton–Noble paradox?

https://arxiv.org/abs/1206.4487
https://arxiv.org/abs/physics/0603110

 

[PeterDonis]

If we set $k = v / g = v \sqrt{1 - v^2}$, then the trig functions no longer depend on $z_0$ and we have unwound the helix

On going back over this, I realized that this part is wrong. It should be $k = \omega v$ to unwind the helix. This means we have to boost by $v = k / \omega = 2 \pi / \omega Z$. Everything else in my post should still be ok.

 

[PeterDonis]

Is this possibly a manifestation of the Trouton–Noble paradox?

That might be a piece of the puzzle, yes. At the very least you've reminded me that my equations for the force are not correct. I need to rework those with the correct relativistic relationships.

 

[PeterDonis]

The weights $W_k$ are all positive. The contributions to the integrals for different pieces of the helix have different signs because the trig functions involved change sign from one piece to another. But that only happens for $N > 1$.

It might be helpful for me to briefly expand on how this works. Here are the key steps. We start with the following integral (where I have removed irrelevant constant factors to simplify things for this post):

$$
I = \int_{- N \pi}^{N \pi} \mu(z) z \sin z dz
$$

We first rewrite the integral by observing that the integrand is even and the integral is over an even domain; that let's us do this:

$$
I = 2 \int_{0}^{N \pi} \mu(z) z \sin z dz
$$

Next we separate the integral into $k$ pieces, as follows:

$$
I = 2 \sum_{k=1}^{N} \int_{\left( k - 1 \right) \pi}^{k \pi} \mu(z) z \sin z dz
$$

Now we adopt a particular form for $\mu(z)$; we assume that it is a constant $W_k > 0$ for each piece, but can vary from piece to piece. That gives us:

$$
I = 2 \sum_{k=1}^{N} W_k \int_{\left( k - 1\right) \pi}^{k \pi} z \sin z dz
$$

Finally, we evaluate the integral in each piece, which gives

$$
I = 2 \sum_{k=1}^{N} W_k \left[ z \cos z + \sin z \right]_{\left( k-1\right) \pi}^{k \pi}
$$

The $\sin z$ term vanishes, and the $z \cos z$ term gives a factor $\left( 2 k - 1 \right) \left( -1 \right)^k$, for a final result:

$$
I = 2 \sum_{k=1}^{N} \left( -1 \right)^k \left( 2 k - 1 \right) W_k
$$

Obviously the condition $I = 0$ has no solution for $N = 1$, but it can be solved for any $N > 1$.

 

[PeterDonis]

At the very least you've reminded me that my equations for the force are not correct. I need to rework those with the correct relativistic relationships

Actually, on looking back over these, I think they are correct. There is no force along the direction of the boost (the $z$ direction), so issues with how forces transform don't come into play.

 

[AVentura]

Is there a possible false assumption that force and acceleration are the same direction? I know nothing about this, I ask because this has been presented as a resolution to Trouton–Noble.

Also, in the first paper I linked the abstract claims a "suitable treatment of angular momentum and simultaneity" as a resolution. Sounds awfully related.

 

[PeterDonis]

Is there a possible false assumption that force and acceleration are the same direction?

If a component of the force is parallel to the boost direction (which is the $z$ direction in this case), then yes, the force and acceleration will not be parallel except in one particular frame. In the Trouton-Noble case (also called the right angle lever paradox), one of the forces is parallel to the boost direction, so the resolution of the paradox does require properly taking into account that the force and the acceleration are not parallel in all frames.

But that isn't the case in this problem--the force is entirely in the $x$ and $y$ directions, so it remains parallel to the proper acceleration in any frame which is boosted only in the $z$ direction. So unfortunately I don't think the resolution of Trouton-Noble will help here. It was a good thought though.

 

[PeterDonis]

I realized on reading back over the thread that I never actually computed the kinematic decomposition for either the cylinder or the helix. So I'm going to do that now. As we'll see, this only adds to the confusion.

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

$$
U_a = \left( - \gamma, 0, - \gamma \omega R \sin ( \omega t + \phi ), \gamma \omega R \cos ( \omega t + \phi ) \right)
$$
$$
A_a = \left( 0, 0, - \gamma^2 \omega^2 R \cos ( \omega t + \phi ), - \gamma^2 \omega^2 R \sin ( \omega t + \phi ) \right)
$$

The nonzero components of the tensor are

$$
\theta_{20} + \omega_{20} = A_2 U_0 + U_{2, 0} = \left( \gamma^3 - \gamma \right) \omega^2 R \cos ( \omega t + \phi )
$$
$$
\theta_{30} + \omega_{30} = A_3 U_0 + U_{3, 0} = \left( \gamma^3 - \gamma \right) \omega^2 R \sin ( \omega t + \phi )
$$
$$
\theta_{22} + \omega_{22} = A_2 U_2 = \gamma^3 \omega^3 R^2 \cos ( \omega t + \phi ) \sin ( \omega t + \phi )
$$
$$
\theta_{33} + \omega_{33} = A_3 U_3 = - \gamma^3 \omega^3 R^2 \cos ( \omega t + \phi ) \sin ( \omega t + \phi )
$$
$$
\theta_{23} + \omega_{23} = A_2 U_3 = - \gamma^3 \omega^3 R^2 \cos^2 ( \omega t + \phi )
$$
$$
\theta_{32} + \omega_{32} = A_3 U_2 = \gamma^3 \omega^3 R^2 \sin^2 ( \omega t + \phi )
$$

I'm not going to try to interpret this in detail; I think it ends up showing zero expansion and shear (the zero expansion is evident from the fact that the $22$ and $33$ components cancel each other) and nonzero vorticity, as expected, but I'm not familiar enough with how to interpret this tensor to be able to read that off from the above. The only comment I have is that the idea is that this tensor at a given event should be orthogonal to the 4-velocity at that event, and the 4-velocity in this case has a $\theta$ component as well as a $t$ component, so the spacelike surface orthogonal to it is not just a surface of constant $t$.

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, while all of the components above are still there and still look the same:

$$
\theta_{21} + \omega_{21} = U_{2, 1} = - \gamma \omega k R \cos ( \omega t + k z )
$$
$$
\theta_{31} + \omega_{31} = U_{3, 1} = - \gamma \omega k R \sin ( \omega t + k z )
$$

What I'm wondering is if those extra components mean that the congruence describing the helix, unlike the one describing the cylinder, has nonzero shear, because of the "twist" of the helix (basically, that a given point on the helix does not have neighboring points in all directions, but only in two directions, and those directions are at an angle to the direction of motion due to rotation). I don't have an answer to this question, but I wanted to post the computations above (which also could stand some checking) to at least make it clear what questions I think we're dealing with.

 

[maline]

I'd like to elaborate on a point that I made a few days ago, without sufficient explanation:

we can easily extend this example to a fully three-dimensional one: just multiply this distribution by an arbitrary function of $r$, defined on an interval $R_1≤r≤R_2$, and by another arbitrary function of $\theta$ defined on $\theta_1\leq\theta \leq\theta_2$. We may as well choose the angular function to be constant, but choosing different radial functions may be useful.

The reason I am interested in making the body three-dimensional is that I still expect that working out the stress-energy tensor will turn out to be necessary for a resolution of our paradox. If so, our one-dimensional helix model will not suffice, because our helix must support shear stress (i.e. resist bending), and this is impossible for a truly one-dimensional body.

The reason shear stress is required (unlike, for instance, the case of a rotating ring) is that our body has free endpoints. The net force on the endpoints, like that on the rest of the body, must be centripetal, in the $-r$ direction, which is perpendicular to the helix curve. Along the interior of the helix this can be provided by tensile stress, because the change in the tangent direction along the curve is indeed in the $-r$ direction. Therefore, as with a rotating ring, taking the derivative of a tensile stress of constant magnitude gives a net centripetal force. At the endpoints, however, the tensile stress must go to zero- otherwise we have a jump discontinuity in the stress which implies a delta-function of force, while the mass distribution remains finite. Even if we added point masses to absorb that force, it would cause them to accelerate in the direction tangent to the helix, ruining our rigidity. It seems clear that we should indeed let the tension go to zero at the ends, and that the centripetal force will be provided, at least near the ends, by a shear stress. In other words, a rotating ring can be replaced by a flexible wire and still retain its shape, but a rotating helix cannot- it will "unwind" under the centrifugal force.

The reason you can't have shear stress in a 1D body is that the stress tensor is symmetrical. If, for instance, the $T^{yx}$ term is nonzero then so is the $T^{xy}$ term. But in a one-dimensional body, lying along the $x$ axis, we can't have $x$-momentum flowing in the $y$ direction- there's nowhere for it to go! Trying to calculate the total $x$-force on any point, we would have to take a derivative in the $y$ direction. We would be stuck with the unphysical "derivative of the delta function".
The following is a bit of an aside, but here is an example of how shear stress works. Let the body lie along the $x$- axis at $-1\leq x\leq1$, and suppose two forces, each $+1\hat y$, are applied at $x=\pm 1$ while a force $-2\hat y$ is applied at the origin. It is clear that the total force and total torque are both zero, so assuming the body remains rigid, it will not move.

But momentum conservation is primarily a local law (in fact in GR, according to most formulations it's only a local law), and we can think of each force as a constant local "input" of momentum. The rod is "absorbing" $y$-momentum at the ends and must transport this momentum to the origin, where it is canceled by the negative force. This "flow of momentum" is (one way of looking at) stress, and the material must have strength to support it. Since the momentum being carried is not in the direction of the flow, it is shear stress rather than tensile stress. It shows up in the stress-energy tensor as a nonzero $T^{yx}$ term- a flow of $y$-momentum in the $x$ direction. In this example it will be $+1$ for negative $x$ and $-1$ for positive $x$. But it can't work in one dimension!

To make the shear stress possible, the body must have thickness in the $y$ direction, say $0\leq y\leq 0.1$. At the top and bottom surfaces, we will indeed have $T^{yx}=T^{xy}=0$ to avoid jump discontinuities in the $y$ direction causing infinite $x$-force. Within the thickness, T^{yx}=T^{xy} will be positive for negative $x$ and negative for positive $x$.

Now let's consider the local force balance. At every point except where the forces are applied, we must have zero net force, meaning $\frac \partial {\partial x} T^{xx} +\frac \partial {\partial y} T^{xy} =0$ and $\frac \partial {\partial x} T^{yx} +\frac \partial {\partial y} T^{yy} =0$. We do not need $T^{yx}$ to change in the $x$ direction except at $x=0$, so we can set $\frac \partial {\partial x} T^{yx} =\frac \partial {\partial y} T^{yy} =0$. But $T^{yx}$, and therefore also $T^{xy}$, does change in the $y$ direction. For negative $x$, $\frac \partial {\partial y} T^{xy}$ is positive in the lower part of the body and negative in the upper part, and the opposite for positive $x$. Therefore $\frac \partial {\partial x} T^{xx}$ must have the opposite sign at each point. $T^{xx}$ must vanish at the ends of the body unless outside forces are applied. So along the bottom edge, $T^{xx}$- the tensile stress in the $x$ direction, will become more and more negative going from the ends toward the center. This stress can reach values much higher than the $T^{yx}$ term that is actually handling the force!

Sure enough, our body- let's say a wooden shelf supporting a weight that is concentrated in the center- is most likely do deform by stretching in the horizontal direction, at the point on the bottom surface directly beneath the weight. There is also an equally large positive term along the top surface, but wood is more likely to fail under tension than under compression.
Note that the maximum tension is greater the longer the shelf is, and less the thicker the shelf is, because $x$, $\frac \partial {\partial y} T^{xy}$ is smaller. This corresponds to experience.

I also should point out that stress is usually expressed as a 3-tensor $\sigma^{ij}$, which is basically the same as the stress-related part of the spatial terms in $T^{\alpha \beta}$, but with opposite sign: positive for tension and negative for compression.

 

[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.