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

[maline]

one of the potential issues with this is how to define derivatives of the velocity field if the subset we select is not continuous.

We are free to work with any subset we like, for instance a cylinder that includes the helix. If the whole cylinder, as a velocity field, is rigid then so is anybody contained in and moving with it. That was my point in post #78 above.

 

[maline]

Great to hear from you again OP! As you can see you've come up with a pretty tough paradox here.

If the helix had a minimum width tangentially this width would look wider in the frame where it is translating (slower angular velocity but same radius). This would move the center of energy back towards the original axis. Just brainstorming.

I'm not picturing this; can you describe it in more detail?

 

[AVentura]

If the helix wasn't made of a wire with zero thickness then the two observers would see different dimensions of that wire. They see different tangential velocities out at the radius r.

 

[AVentura]

Let's say in S an observer has a rotating cylinder that is translating along its axis. An observer traveling alongside it in S' sees a faster angular velocity and therefore length contraction tangentially, and it's twisted. Does he actually see a helix, saving him from Ehrenfest's paradox?

 

[PeterDonis]

If the whole cylinder, as a velocity field, is rigid then so is anybody contained in and moving with it.

I'm not 100% sure this addresses the derivative issue I brought up, but since I have no way of addressing that issue if it actually is an issue I'm going to assume that your statement is correct in what follows.

I'm going to take another crack at the angular momentum tensor, but this time in Cartesian coordinates, $t, z, x, y$. I'm not entirely sure my previous computation in cylindrical coordinates was correct, because I'm not entirely sure I took proper account of the effects of curvilinear coordinates. Rather than try to fix that, I'm going to just use coordinates where we know there are no such issues.

The angular momentum tensor, as I posted before, is given by $M^{ab} = X^a P^b - X^b P^a$, where $X^a$ is the 4-position vector and $P^a$ is the 4-momentum vector. I'm going to start by assuming constant mass density, which means we can factor out the mass density and use the 4-velocity field $U^a$ instead of the 4-momentum $P^a$.

The general approach I am going to try is to first evaluate the 4-momentum tensor at a particular event, and then integrate over the appropriate range of spatial coordinates in the $t = 0$ plane to evaluate the total angular momentum of the object at the instant $t = 0$. I'll start by doing this in the rest frame of the center of mass (i.e., the frame in which the motion is a "pure" rotation, no translation).

First we will evaluate the full cylinder. Here we have $X^a = \left( t, z, R \cos ( \omega t + \phi ), R \sin ( \omega t + \phi ) \right)$, where $R$ is the (constant) radius of the cylinder, $\omega$ is the (constant) angular velocity, and $\phi$ labels each worldline by its "angular position" in the cylinder at time $t = 0$, and so ranges from $0$ to $2 \pi$ since we are considering the full cylinder. We then have $U^a = \left( \gamma, 0, - \gamma \omega R \sin ( \omega t + \phi ), \gamma \omega R \cos ( \omega t + \phi ) \right)$, where $\gamma = 1 / \sqrt{1 - \omega^2 R^2}$.

The six independent components of $M^{ab}$ are then:

$$
M^{01} = - z \gamma
$$
$$
M^{02} = - \gamma R \left[ \cos ( \omega t + \phi ) + \omega t \sin ( \omega t + \phi ) \right]
$$
$$
M^{03} = - \gamma R \left[ \sin ( \omega t + \phi ) - \omega t \cos ( \omega t + \phi ) \right]
$$
$$
M^{12} = - z \gamma \omega R \sin ( \omega t + \phi )
$$
$$
M^{13} = z \gamma \omega R \cos ( \omega t + \phi )
$$
$$
M^{23} = \gamma \omega R^2
$$

The above components are at a particular event, labeled by its $X^a$ coordinates as given above. To evaluate the total angular momentum of the cylinder as a whole, at time $t = 0$, we first set $t = 0$ in the above, and then integrate over the spatial coordinates occupied by the cylinder at $t = 0$. The latter is a double integral:

$$
M_{\text{total}}^{ab} = \int_{-Z/2}^{Z/2} dz \int_0^{2 \pi} d\phi M^{ab} ( z, \phi )
$$

where we view each component of $M$ as a function of two variables, $z$ and $\phi$. Note carefully the limits of integration for $z$: we assume that the spatial origin, at $z = 0$, is the geometric center of the cylinder, and that its total length is $Z$, so we integrate over $z$ from $- Z / 2$ to $Z / 2$. It should be evident that, since the trig functions integrate to zero over the range of $\phi$, and since the function $z$ itself integrates to zero from $- Z / 2$ to $Z / 2$, the only component which ends up nonzero after integration is

$$
M_{\text{total}}^{23} = \gamma \omega R^2 Z
$$

This means that the total angular momentum of the cylinder at $t = 0$ lies purely in the x-y plane, which is what we expect. And in fact, the same will be true at every value of $t$, since plugging in any constant value of $t$ in the above does not change the evaluation of any of the integrals (it adds some trig function terms but they still integrate to zero).

Now for the case of the helix. Here the only difference from the above is that $\phi$, instead of being an independent variable labeling worldlines, is equal to $k z$, where $k$ is the "spatial turn frequency" of the helix (this is related to the pitch, but I'm not sure if it's identical, and that detail doesn't affect what follows). Since we have that the helix does one full turn in length $Z$, we have that $k = 2 \pi / Z$, so $\phi = 2 \pi z / Z$. So now, to find the total angular momentum of the helix, we have only a single integral, over $z$, but the function of $z$ we are integrating over is more complicated since it includes trig function terms, and they won't all vanish this time.

Let's look at the functions of $z$ we end up with and how they integrate. For $M^{01}$, there is no change from before: we still have $z$ integrated over $- Z / 2$ to $Z / 2$, which gives zero. Note that the reason for this is that $z$ is an odd function and is being integrated over an even domain. So we can focus only on the components that are even functions of $z$; there are three, $M^{02}$, $M^{12}$, and $M^{23}$ (which is unchanged from before).

So we need to evaluate two new integrals (plus we have one unchanged from before):

$$
M_{\text{total}}^{02} = - \gamma R \int_{-Z/2}^{Z/2} \cos ( \frac{2 \pi z}{Z} ) = - \gamma R \frac{Z}{2 \pi} \left[ \sin ( \frac{2 \pi z}{Z} ) \right]_{-Z/2}^{Z/2} = 0
$$
$$
M_{\text{total}}^{12} = - \gamma \omega R \int z \sin ( \frac{2 \pi z}{Z} ) = - \gamma \omega R \left[ - \frac{Z}{2 \pi} z \cos ( \frac{2 \pi z}{Z} ) + \frac{Z^2}{4 \pi^2} \sin ( \frac{2 \pi z}{Z} ) \right]_{-Z/2}^{Z/2} = - \gamma \omega R \frac{Z^2}{\pi}
$$
$$
M_{\text{total}}^{23} = \gamma \omega R^2 Z
$$

This is telling us two things. First, since $M^{02}$ vanishes, we don't have a "mass moment" in this frame, which is good because in this frame the cylinder is not translating, only rotating, and a nonzero "mass moment" would imply translation. Second, since $M^{12}$ does not vanish, the angular momentum is not solely in the x-y plane; it is partly in the x-z plane as well, at least at time $t = 0$. In other words, the plane of the angular momentum is not perpendicular to the axis of rotation.

In other words, the above analysis appears to be saying that the helix motion, considered dynamically, should not be a free motion, since the plane of its angular momentum is not perpendicular to the axis of rotation. If a helix were started with the assumed 4-velocity field, but no further external forces were applied, what should happen is that the axis of rotation should precess (so the 4-velocity field we assumed would not be valid after $t = 0$). Alternatively, in order to maintain the assumed 4-velocity field over time, external forces would have to be applied to change the angular momentum vector in the right way with time in order to keep the axis of rotation constant. (If we evaluate the above integrals for arbitrary $t$, we will see that indeed the plane of the angular momentum changes--in general it has components in both the x-z and y-z planes, as well as the x-y plane.)

I'll save for a follow-up post the question of whether we can, as maline suggested, construct a helix motion, with an axis of rotation as assumed here, that is a valid free motion by altering the mass distribution.

 

[maline]

Couple of points that I hope will add clarity:

First, since M02M02M^{02} vanishes, we don't have a "mass moment" in this frame, which is good because in this frame the cylinder is not translating, only rotating, and a nonzero "mass moment" would imply translation.

A nonzero mass moment means that the center of energy is not at the origin at time $t=0$. Since the origin is the only fixed axis of the rotation, the center of energy would then be revolving around the origin, which requires external force.

Second, since M12M12M^{12} does not vanish, the angular momentum is not solely in the x-y plane; it is partly in the x-z plane as well,

Just making sure everyone is aware that the x-z component of the tensor translates to a y component when the angular momentum is thought of as a 3-vector.

 

[PeterDonis]

A nonzero mass moment means that the center of energy is not at the origin at time $t=0$.

Yes, I misstated this, since I was specifically considering the instant $t = 0$.

Just making sure everyone is aware that the x-z component of the tensor translates to a y component when the angular momentum is thought of as a 3-vector.

Yes, if we confine ourselves to the 3 space dimensions, there is a 1-to-1 correspondence between antisymmetric 3-tensors and 3-vectors (more precisely, pseudovectors). In that language, the cylinder's angular momentum vector points purely in the $z$ direction, but the helix's angular momentum vector, if we assume the 4-velocity field given, has a $z$ component that is constant and $x$ and $y$ components that change with time, indicating that external forces are being applied.

The reason I didn't use this language in my previous post is that, in general, angular momentum in relativity is a 4-tensor, not a 3-tensor, and the antisymmetric tensor-vector correspondence doesn't hold in 4 dimensions. For the particular case of an object's rest frame, we can always choose the origin so that only the 3-tensor part of the angular momentum is nonzero; but we can't do that in any other frame, and since one of the things I want to consider is how whatever answer we get in the rest frame transforms to a frame in which the helix is moving, I don't want to use language that only works in the rest frame.

 

[PeterDonis]

I'll save for a follow-up post the question of whether we can, as maline suggested, construct a helix motion, with an axis of rotation as assumed here, that is a valid free motion by altering the mass distribution.

I haven't worked this out in detail mathematically, but my initial guess is that it is not possible, at least not by adding point masses.

Note first that the case $n = 1$ in maline's proposal results in the formula for $\theta$ for the point masses switching signs; the factor $n - \frac{3}{2}$ is negative for $n = 1$. That means, if I'm understanding his notation right, that the point masses would not be on the helix; they would be 180 degrees around the cylinder, in the transverse plane, from the helix points at their $z$ location (i.e., location along the axis).

Looking at the angular momentum formulas, we can ask whether there is some value $Z_0$ such that we can place point masses on the helix at $z = \pm Z_0$ and have the angular momentum due to them cancel out the $M^{12}$ component due to the helix at time $t = 0$. This does not seem possible because of the signs involved. The component $M^{12}$ due to the point masses would be

$$
M^{12} = - \left( \pm z_0 \right) \gamma \omega R \sin \left( \pm \frac{2 \pi z_0}{Z} \right)
$$

The two $\pm$ signs cancel since the $\sin$ function has the same sign, for the range under consideration, as $z$ itself. That means the sign of this $M^{12}$ term will be the same as the sign of the corresponding term due to the helix, so the two can't possibly cancel.

Note, again, that if we add $\pi$ to the argument of each $\sin$ function, which moves each point mass 180 degrees around the cylinder from its corresponding helix point (i.e., at the same $z$), the $\sin$ factors flip sign, so we could indeed set things up so the point masses canceled the $M^{12}$ terms due to the helix. But this does not satisfy the requirement that all masses involved have the same 4-velocity field as the helix (since obviously the two point masses would have 4-velocities of opposite sign to the corresponding helix points--another way of seeing how their angular momentum $M^{12}$ can cancel).

I suspect that this argument can be made more general, to cover any possible non-uniform mass distribution that is restricted to the helix congruence of worldlines.

 

[AVentura]

I'm not picturing this; can you describe it in more detail?

Nevermind my last 3 posts. They are incorrect.

 

[maline]

Note first that the case n=1n=1n = 1 in maline's proposal results in the formula for θθ\theta for the point masses switching signs; the factor n−32n−32n - \frac{3}{2} is negative for n=1n=1n = 1. That means, if I'm understanding his notation right, that the point masses would not be on the helix; they would be 180 degrees around the cylinder, in the transverse plane, from the helix points at their zzz location (i.e., location along the axis).

You're right that I assumed $N>1$, but if $n=1$, the masses are still on the helix, at points $z=\pm Z/4$ in your notation. The only problem is that their masses will be negative, $-2\lambda$, where $\lambda$ is the mass per radian of the uniform part of the helix.

 

[PeterDonis]

The only problem is that their masses will be negative

This is not physically possible, so I don't think it counts as a valid solution. I was assuming positive masses, which requires that in the $n = 1$ case, the masses are located as I described.

 

[PeterDonis]

I suspect that this argument can be made more general, to cover any possible non-uniform mass distribution that is restricted to the helix congruence of worldlines.

Here is a quick generalization of the argument: consider any mass distribution $\mu(z)$ along the helix. The angular momentum components at each event would then be multiplied by this function (I was basically assuming a constant function before and just not putting it in), and the total angular momentum components for the helix would include this function in the integrals over $z$. In order for the center of mass to remain in the same location, this function must be an even function of $z$. But putting an even function of $z$ into the integrals does not change which integrals give nonzero values, since the integrals are all over an even domain. So no possible mass distribution can make the helix congruence into a possible free motion.

 

[maline]

Here is a quick generalization of the argument: consider any mass distribution μ(z)μ(z)\mu(z) along the helix. The angular momentum components at each event would then be multiplied by this function (I was basically assuming a constant function before and just not putting it in), and the total angular momentum components for the helix would include this function in the integrals over zzz. In order for the center of mass to remain in the same location, this function must be an even function of zzz. But putting an even function of zzz into the integrals does not change which integrals give nonzero values, since the integrals are all over an even domain. So no possible mass distribution can make the helix congruence into a possible free motion.

Why are you ignoring the explicit counterexample I gave? The issue of the masses being negative is only for $n=1$. For any larger whole number of turns they will be positive, namely $\frac n {n-\frac 3 2}\lambda$.

 

[PeterDonis]

Why are you ignoring the explicit counterexample I gave?

I'm not ignoring it. I'm trying to see if I can arrive at a formulation of it in the relativistic case that gives the same answer you got. So far I can't.

The issue of the masses being negative is only for $n=1$.

The case $n = 1$ is a valid case, so if I can't get a relativistic formulation to work for that case, even if it worked for $n > 1$, there is an issue somewhere that needs to be understood.

In any event, the case of $n > 1$ in my relativistic formulation just corresponds to extending the range of $z$ to some integral multiple of the range I used. That doesn't change any of what I said about which integrals in the relativistic case are nonzero. So I don't see any difference between $n = 1$ and $n > 1$ in my formulation.

 

[AVentura]

Is the problem with n=1 just that there is nowhere on the helix to place the needed counterweights (using positive mass)?

Visual, for z<0 you can only put the weight on one side (say top). For z>0 you can only put on the bottom. With more curls you can put one on any side you want, then find the needed weight.

 

[PeterDonis]

Their contribution to the integral is equal and opposite to the above value

Looking back at this, I see how it is true for $n = 1$, but that's because the masses turn out to be negative. I don't see how it is true for $n > 1$, where the masses are positive; it seems to me that that makes their contributions to $J_y$ of the same sign as those of the uniform mass distribution in the original helix.

 

[maline]

Looking back at this, I see how it is true for $n = 1$, but that's because the masses turn out to be negative. I don't see how it is true for $n > 1$, where the masses are positive; it seems to me that that makes their contributions to $J_y$ of the same sign as those of the uniform mass distribution in the original helix.

AVentura got it right:

Is the problem with n=1 just that there is nowhere on the helix to place the needed counterweights (using positive mass)?

Visual, for z<0 you can only put the weight on one side (say top). For z>0 you can only put on the bottom. With more curls you can put one on any side you want, then find the needed weight.

In other words, you need to put positive masses at points where the sine function makes the sign of the contribution as desired. It doesn't work for $n=1$ because $x\sin x$ is positive for $-\pi\leq x\leq \pi$.

 

[maline]

I also want to point out that up to this point, the classical and relativistic treatments are essentially identical. We relabeled the "center of mass at time 0" and the angular momentum 3-vector as components of a rank-2 4-tensor, but that makes no difference until we actually carry out a Lorentz boost. The integral expressions for the components are the same as in the Newtonian case except that the mass density, in each expression, is multiplied by $\gamma(r)=(1-r^2\omega^2)^{-1/2}$. But since we are dealing with a cylindrical shell, $r=R$ is just a constant and so is $\gamma$, so it just gets factored out of the integrals.

 

[Dale]

classical and relativistic treatments are essentially identical

I wouldn't say that. I think that we haven't done a fully relativistic treatment which I think would involve the stress energy tensor on the helix. We have instead done a classical treatment and some mixed treatments. This problem seems to be finding a flaw in the mixed analysis.

 

[PeterDonis]

you need to put positive masses at points where the sine function makes the sign of the contribution as desired. It doesn't work for $n=1$ because $x\sin x$ is positive for $-\pi\leq x\leq \pi$.

Yes, I see that. I'm working now on how to capture that, or something equivalent, in the relativistic formulation.

 

[PeterDonis]

I think that we haven't done a fully relativistic treatment which I think would involve the stress energy tensor on the helix

I'm not sure that's necessary; I think we can get away with just a variable mass distribution $\mu(z)$ along the helix, without having to do the full stress-energy tensor treatment. But I won't know for sure until I've tried it.

 

[AVentura]

Is the stress in the parts of the helix that are connected to any point masses infinite?

 

[maline]

Is the stress in the parts of the helix that are connected to any point masses infinite?

Good question. I think the answer is no: the 4-force density is the derivative (4-divergence) of the stress-energy tensor, $f^\alpha={T^{\alpha \beta}}_{,\beta}$. So the point force, being a "delta function" in the force density, can be caused by a jump discontinuity in the stress, whose derivative is a delta of the appropriate magnitude.

 

[pervect]

I'm afraid I haven't had time to keep up with this thread, but a couple of comments. First on the issue of rigidity.

The expansion scalar is a rank-0 tensor. The shear tensor is a rank 2 tensor. These tensors are both zero in some coordinate system, the Born coordinate system of a rigidly rotating disk or cylinder. <<wiki link for Born coordinates>>

A tensor that is zero in one coordinate system will be zero in all coordinate systems.

Therefore an object that is Born rigid in one coordinate system will be Born rigid in all coordinate systems, because one of several equivalent definitions of rigidity is that an object have a vanishing expansion and shear. And we know that a tensor that is zero in one coordinate system is zero in all.

One application of this is noting that changing the state of motion of the rotating cylinder by boost its motion along it's axis of symmetry can be regarded just as a coordinate transformation - the same cylinder viewed in two different coordinate systems moving relative to one another. Then we conclude that this boosting operation doesn't and can't change the rigidity of system.

I don't see what there is to argue about. Hopefully the argument is really over at this point, except possibly for a few holdouts who don't seem to be actually carrying out the math. Hopefully the argument will be sufficient to convince the moderators and science advisors - and anyone actually doing the math, which is my primary concern at this point.

The second point is that it might be useful to construct a fully relativistic model of the situation out of point masses, and massless strings, which exert the forces that hold the masses in place, in some convenient set of coordinates.

The point to realize is that the assumption that the strings are massless IS coordinate dependent. The tools we need to demonstrate and understand this are just an understanding of the stress energy tensor, and it's transformation properties. For a textbook discussion of some of the issues, see a discussion of the weak energy condition in your favorite textbook, specifically why a massless string under tension is considered to be "exotic matter" which has a negative energy density in some coordinates.

So, one convenient approach is to first create a system consisting of point masses and massless strings in a convenient coordinate system.

Evaluating the mass of the string after one changes coordinates will give one an idea of whether the contribution of the stresses in the string to the angular momentum of the system can be ignored in the new coordinates. In the coordinates in which the mass of the string is zero, the string will not contribute to the linear or angular momentum, but the same is not true (hopefully, this is obviously not true) in coordinates in which the mass of the string is not zero.

 

[PeterDonis]

I don't see what there is to argue about.

I don't think there's any dispute about the rotating cylinder--there appears to be general agreement that this is a Born rigid motion and that it is a possible "free" motion, i.e., its angular momentum and angular velocity vectors are parallel, so that no external forces are required to maintain the motion.

The open question is about the case of the rotating helix, considered as a subset of the cylinder. There appears to be agreement that, if the mass distribution is uniform along the helix, this motion, while it is Born rigid, is not a possible "free" motion, because the angular momentum and angular velocity are not parallel; external forces would be required to maintain this motion. However, we have not resolved the question of whether there is a non-uniform mass distribution along the helix that would make the motion "free", by removing the non-parallel component of the angular momentum. @maline gave a Newtonian scenario that appears to have this property; I have been working on a relativistic model but have not yet come up with one that has the property.

If there is a relativistic model that has the above property--a helix rotation about the axis of the helix that is a "free" motion--then we have a possible "paradox" in the relativistic case: there will be some frame in which the helix transforms into a straight rod that is off of the axis of rotation (basically, imagine painting the helix on the rotating cylinder and then transforming into a frame in which the painted-on helix is "unwound" so that it is just a straight line down the cylinder that rotates with the cylinder). The latter motion does not seem like it could possibly be a "free" motion, but the property of being a "free" motion, requiring no external forces to maintain, should be invariant under Lorentz transformations.

it might be useful to construct a fully relativistic model of the situation out of point masses, and massless strings

That's not the approach I am taking. The approach I am taking is to model the helix as a congruence of worldlines (a subset of the cylinder congruence), paramterized by $z$, the axial coordinate along the cylinder in the center of mass frame (the frame in which the cylinder/helix only rotates, with no translation). The 4-position and 4-velocity vectors of this congruence as a function of $z$ are known and have been posted earlier in this thread. I am then modeling the mass distribution as a function $\mu(z)$, whose units are mass per unit length, along the helix. If I am able to find such a function that makes the relativistic angular momentum in this frame have only one nonzero component, $M^{23}$ (in the x-y plane, perpendicular to the axis of rotation), I am then going to boost the corresponding solution into the moving frame that, kinematically, "unwinds" the helix, to see what things look like in that frame.

 

[PeterDonis]

Post #106 has the relativistic analysis I have done so far, for the case of a constant mass distribution:

https://www.physicsforums.com/threa...inder-max-von-laue.913600/page-6#post-5762960

 

[Dale]

The connection with the constraints is that the time reversibility and foliability of the geometry that they enforce guarantee the vanishing of the momentum transfer(killing motion:the rotating cylinder is an example of this).

Got it. Those conditions guarantee that there exists some coordinate system such that the stress energy tensor is diagonal. Of course, you can choose poorly adapted coordinates but the nice coordinates exist.

 

[PeterDonis]

As a first step at analyzing the case with a non-constant mass distribution, consider the following modifications to the model I gave in post #106.

First, we extend the model to cover more than one turn of the helix. A simple way to do that is to extend the limits of integration to $\pm NZ / 2$, where $N$ is the number of turns.

Second, we observe that the nonzero integrals are even functions over an even domain, which means we can rewrite them as integrals from $0$ to $NZ / 2$ and multiply by $2$. This leaves the final answers the same but helps with the next step.

Next, we allow for a non-constant mass distribution. Instead of trying to work with a generic function $\mu(z)$, I'm first going to try a simpler model. View the helix as composed of "segments", which appear in the integrals (rewritten as in the second step above) as pieces each occupying an interval of $Z / 2$, so that piece $k$ goes from $(k - 1) Z / 2$ to $k Z / 2$, and $k$ runs from $1$ to $N$. (Note that, except for $k = 1$, each "piece" actually represents two disconnected segments of the helix, one in the negative $z$ range and one in the positive $z$ range, which happen to make identical contributions to the integrals as written according to the first step above, i.e., before changing the limits as in the second step above. "Piece" $k = 1$ represents the one-turn helix that was modeled in post #106.) We then give each piece $k$ a "weight" $W_k > 0$ that represents its contribution to the mass of the helix. We won't try at this point to normalize the weights or worry about the exact units; their relative magnitudes are all that will matter in what follows. This is equivalent to assuming that, within each piece, the mass distribution is uniform, i.e., the value of $\mu$ is constant within each piece, but its value can vary from one piece to another.

(Note: This is not the same as the point mass method that @maline used in an earlier post, placing additional mass in points lying on the $x$ axis to avoid making $J_x$ nonzero. I think it still works because the mass distribution of each piece is symmetrical about the $x$ axis, so it should not affect the $M^{13}$ total angular momentum component--note that I am using the $t, z, x, y$ coordinate ordering, so $x$ is coordinate 2, perpendicular to the 13 plane.)

With these adjustments, the two integrals under consideration are modified as follows:

$$
M^{12} = \gamma \omega R \frac{Z^2}{\pi} \sum_{k=1}^N \left( -1 \right)^k \left( 2 k - 1 \right) W_k
$$

$$
M^{23} = \gamma \omega R^2 Z \sum_{k=1}^N W_k
$$

Notice that the only changes from before are the sums over the weights.

We can see, first, that $M^{23}$ remains positive and is basically unaffected by all this, since the sum of the weights still has to give the overall mass of the helix (again, without worrying about normalization or exact units). We can also see that, for $N = 1$, $M^{12}$ is unchanged from before (as expected, since our assumption about the mass distribution leaves the case of one turn the same--but even allowing the mass distribution to vary within a piece would not allow any cancellation in $M^{12}$ for the case $N = 1$, since the sign of all contributions within a given piece is the same). But for the case $N > 1$, we can now see the possibility of arranging the weights so as to make $M^{12}$ vanish. For example, for the case $N = 2$, if we have $- W_1 + 3 W_2 = 0$, or $W_1 = 3 W_2$, then $M^{12}$ vanishes. This matches up with what @maline posted earlier in the thread in the Newtonian case.

What I think I was missing before is that, even though the mass distribution $\mu(z)$ is even, which means it can't make any of the integrals that were zero before nonzero, it can still, for the case $N > 1$, make a nonzero integral zero, if there is the possibility of different segments of the helix making contributions of different signs. The above analysis shows that that is indeed possible.

The next step will be to transform the $X^a$ and $P^a$ vectors into a frame in which the helix is "unwound" (kinematically), and see what the angular momentum tensor looks like in that frame. I'll address that in a follow-up post.

 

[maline]

Excellent! Now we know that the paradox is not due to using point masses. Not only that, but 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 $R_1\leq r\leq R_2$, 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.

I think that we haven't done a fully relativistic treatment which I think would involve the stress energy tensor on the helix.

I think we can get away with just a variable mass distribution mu(z) along the helix, without having to do the full stress-energy tensor treatment.

If it's the case that all we need to do is check that the center of energy and the angular momentum are both on the z-axis and therefore are maintained by the rotation, in other words that $M^{tx}=M^{ty}=M^{xz}=M^{yz}=0$, then I agree with Peter. The full expression for $M$ is: $$M^{\alpha \beta}=\int_V\, (X^\alpha T^{\beta \gamma}-X^\beta T^{\alpha \gamma})\,d\Sigma_\gamma$$ where $V$ is the spacelike hypersurface representing the "present" and $d\Sigma_\gamma$ is the volume element (one-form) normal to it. Since we are considering surfaces of constant $t$, $d\Sigma_\gamma$ is purely in the $t$ direction, so we can set the index $\gamma$ as $t$. Then only the $T^{\alpha t}$ components of the tensor contribute, and those are the mass and momentum densities we have been considering.

So we now have clear examples where the rotation does conserve $M^{\alpha \beta}$, regardless of the stress. Is that sufficient for free motion to be possible, or are there other conditions that may involve the stress?

BTW, I need to correct what I wrote before to @AVentura:

the 4-force density is the derivative (4-divergence) of the stress-energy tensor

That is clearly wrong, ${T^{\alpha \beta}}_{,\beta}=0$ always; that is conservation of energy & momentum. What I meant to say is $\frac \partial {\partial t} T^{i t}=-{T^{i j}}_{,j}$ (the momentum part of the conservation equation). In words: the rate of change of the momentum density is minus the 3-divergence of the stress. 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 $-\omega^2 R \hat r$ times the mass density, without the $\gamma$ factor.

 

[PeterDonis]

I'm not seeing how $M^{12}$ can be made to vanish

It can't be made to vanish unless the helix has more than one full turn. What having multiple turns does is add segments of the helix that make contributions to $M^{12}$ that are of opposite signs. With a uniform mass distribution, this can't make $M^{12}$ vanish because the contributions from the segments are of different magnitudes; but allowing a different mass for each segment let's you adjust the masses so the contributions of opposite sign cancel. I gave a specific illustration for the case $N = 2$, a helix of two full turns.