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

[maline]

I think that @PeterDonis proposed method may still work.

I must admit that if Peter suggested a solution, it went right over my head! What was the direction? Is the rotating helix impossible, or is the transformation incorrect, or is the revolving rod somehow legal?

 

[Dale]

I must admit that if Peter suggested a solution, it went right over my head! What was the direction? Is the rotating helix impossible, or is the transformation incorrect, or is the revolving rod somehow legal?

He suggested writing down the congruence and then calculating the shear and expansion tensor. If those are not 0 then the motion cannot be performed by an object in a free motion.

 

[PeterDonis]

He suggested writing down the congruence and then calculating the shear and expansion tensor.

More precisely, the expansion tensor, which includes the expansion scalar (the trace of the tensor) and the shear tensor (the symmetric traceless part of the tensor).

I have done some calculations along these lines, but the problem I am having is how to distinguish the helix congruence from the "cylinder" congruence (the one describing a cylinder rotating about its axis). So far every way I have found of writing down the helix congruence gives me the same (zero) expansion tensor as the one for the cylinder (since the worldlines in the helix congruence are a subset of those in the cylinder congruence). I'm not sure how to capture in the math the fact that the helix congruence "twists" around the cylinder.

 

[maline]

So far every way I have found of writing down the helix congruence gives me the same (zero) expansion tensor as the one for the cylinder (since the worldlines in the helix congruence are a subset of those in the cylinder congruence).

Right, that's why I didn't even get that you thought there might be a solution in this direction. How could the rotating helix be kinematically impossible, when a helix can be considered as just a cylinder with a particular (singular) mass distribution?

 

[PeterDonis]

How could the rotating helix be kinematically impossible, when a helix can be considered as just a cylinder with a particular (singular) mass distribution?

No, a helix is not a cylinder with a particular mass distribution. A cylinder is symmetrical about its axis; that's what makes its free rotation kinematically possible. A helix is not symmetrical about its axis; that should make a difference somewhere in the physical model we use to describe it. I just have not figured out where.

 

[maline]

No, a helix is not a cylinder with a particular mass distribution. A cylinder is symmetrical about its axis; that's what makes its free rotation kinematically possible. A helix is not symmetrical about its axis;

From a purely kinematic perspective, can't you just imagine that we have a full cylinder, but that its mass density is zero except at the points making up the helix?

 

[PeterDonis]

From a purely kinematic perspective, can't you just imagine that we have a full cylinder, but that its mass density is zero except at the points making up the helix?

That might help for something like the angular momentum tensor, but it wouldn't help for the kinematic decomposition (expansion, shear, and vorticity), because that uses the 4-velocity, not the 4-momentum.

 

[maline]

That might help for something like the angular momentum tensor, but it wouldn't help for the kinematic decomposition (expansion, shear, and vorticity), because that uses the 4-velocity, not the 4-momentum.

That's exactly what I'm trying to say. The 4-velocity of the helix, at every point, is identical to that of the corresponding point in a rotating cylinder. The differences are only a question of mass distribution. So if a rotating cylinder is possible, that already includes a rotating helix, along with any other subset of the cylinder's worldlines.

 

[PeterDonis]

if a rotating cylinder is possible, that already includes a rotating helix, along with any other subset of the cylinder's worldlines

No, it doesn't. The general case you are thinking of is a given 4-velocity field with no restrictions on the mass distribution. A rotating cylinder--the case for which we know free motion is possible--is a particular instance of this general case in which the mass distribution is symmetrical about the axis (the usual assumption is that it is constant). Any mass distribution that is not symmetrical about the axis (which a helix is not) is not a rotating cylinder; it is a different particular instance of the general case, and we cannot conclude that it is possible as a free motion just because a rotating cylinder is.

 

[maline]

The general case you are thinking of is a given 4-velocity field with no restrictions on the mass distribution. A rotating cylinder is a particular instance of this general case in which the mass distribution is symmetrical about the axis (the usual assumption is that it is constant). Any mass distribution that is not symmetrical about the axis (which a helix is not) is not a rotating cylinder; it is a different particular instance of the general case.

But doesn't the possibility of a rotating cylinder show that the general case- simple rotation around an axis- is always kinematically possible?

 

[PeterDonis]

doesn't the possibility of a rotating cylinder show that the general case- simple rotation around an axis- is always kinematically possible?

Kinematically possible? Sure, with appropriate external forces applied. But the question I'm trying to answer is what is possible as a free motion, with no external forces applied. The rotating cylinder--mass distribution symmetrical about the axis--is possible as a free motion. I don't think the helix is, because of the unsymmetrical mass distribution. But I haven't been able to figure out how to model that asymmetry mathematically in order to test my conjecture.

 

[Dale]

From a purely kinematic perspective, can't you just imagine that we have a full cylinder, but that its mass density is zero except at the points making up the helix?

You can but then it is no longer axisymmetric

 

[maline]

The rotating cylinder--mass distribution symmetrical about the axis--is possible as a free motion. I don't think the helix is, because of the unsymmetrical mass distribution.

Oh, ok. So you are working with mass distributions, i.e. dynamics and not just kinematics. I think Dale misunderstood this:

Since it is based on the kinematics rather than the dynamics it would hold for any mass distribution.

That's what threw me off.

But isn't it the case that for a rigid cylindrical shell with arbitrary mass distribution, rotating in its rest frame around the axis, the conditions for relativistic free rotation reduce to those of the nonrelativistic case, because all points on the shell have the same "gamma factor"? Is that's correct, then the distribution I mentioned in post #61, which can rotate freely in Newtonian physics, should be able to do so in SR as well.

 

[Dale]

@maline I don't know of any argument against your reasoning

 

[PeterDonis]

the distribution I mentioned in post #61, which can rotate freely in Newtonian physics

This claim would seem to be a "paradox" in Newtonian physics, since as you said in post #61, it appears to violate linear momentum conservation, which is a valid conservation law in Newtonian physics. So maybe we first need to figure out whether this claim is actually true in Newtonian physics, or whether there is in fact some flaw in the reasoning in post #61 (or some other factor involved that that post does not address).

 

[maline]

This claim would seem to be a "paradox" in Newtonian physics, since as you said in post #61, it appears to violate linear momentum conservation, which is a valid conservation law in Newtonian physics.

The apparent violation of momentum conservation is in the relativistic boosted case, because the helix becomes a rod. In Newtonian physics, and seemingly in the SR rest frame as well, the motion is unproblematic.

In Newtonian physics this is not usually considered a problem or an "unsurmountable paradox" because the "infinitely long" mathematical idealization is an accepted ordinary procedure

This is not relevant at all; the rigid body we are discussing is finite (although its mass distribution happens to be singular, which may or may not be important).

we can see how a general rigid body like a helix(without the particular symmetries of a disk or a cylinder that allow the rotation to remain a Killing motion) is not allowed to freely rotate with translation in SR

We only need to show that free rotation without translation, i.e. in the rest frame, is possible. The Lorentz transformation will tell us what happens in other frames. Only in our scenario, it seems to be giving an absurd answer.

 

[maline]

That's what we're trying to figure out.
The theorem says that any Born rigid motion with nonzero vorticity must be a Killing motion. But that doesn't help unless we can verify that the helix motion in question is or is not a Killing motion. That's what I asked if you have a mathematical proof of; you said of the helix that "its points don't describe a Killing motion when rotating", i.e., that the helix motion is not a Killing motion. Do you have a mathematical proof of that? The Herglotz-Noether theorem is not such a proof because it doesn't tell you whether or not a particular motion with nonzero vorticity is or is not a Killing motion; it just says that if it isn't a Killing motion, it can't be Born rigid.

Hold it. Why is this whole question about Born rigidity, Killing fields, and Herglotz-Noether still an issue? I thought we were in agreement, in Peter's post #85, that our rotating helix definitely is both Born rigid and Killing, being that its velocity field is simply a subset of the standard Killing field for rigid rotation, namely, in cylindrical coordinates $(t,r,\theta,z)$, the field $u^\alpha =(\gamma (r),0,\gamma(r) r \omega,0)$, where $\omega$ is a constant, $\gamma(r)=(1-r^2\omega^2)^{-1/2}$, and $0\leq r<1/\omega$.

Peter agreed there that the only doubt was about the dynamical issue of whether such a motion can be free, for any particular mass distribution. Herglotz-Noether has nothing to say about that question- it is a purely kinematic theorem giving conditions for rigidity, i.e whether the motion distorts the body. Issues of mass, momentum and force do not enter the theorem at all.

So can we please lay Born rigidity to rest for the remainder of this thread, and focus on the aspects of angular momentum and center of energy?

 

[pervect]

Hold it. Why is this whole question about Born rigidity, Killing fields, and Herglotz-Noether still an issue? I thought we were in agreement, in Peter's post #85, that our rotating helix definitely is both Born rigid and Killing, being that its velocity field is simply a subset of the standard Killing field for rigid rotation, namely, in cylindrical coordinates $(t,r,\theta,z)$, the field $u^\alpha =(\gamma (r),0,\gamma(r) r \omega,0)$, where $\omega$ is a constant, $\gamma(r)=(1-r^2\omega^2)^{-1/2}$, and $0\leq r<1/\omega$.

I'd agree with this. This also implies that expansion and shear for the congruence should be zero, because it's rigid. Which matches the calculations, I gather.

So can we please lay Born rigidity to rest for the remainder of this thread, and focus on the aspects of angular momentum and center of energy?

As I recall, center of energy is just frame dependent, and this is an example of said frame dependence. Angular momentum is still conserved, of course.

 

[PeterDonis]

I thought we were in agreement, in Peter's post #85, that our rotating helix definitely is both Born rigid and Killing

I originally thought it was, but in post #85 I only said I thought it was "kinematically possible...with appropriate external forces applied". Not all motions that meet that description are Born rigid Killing motions.

being that its velocity field is simply a subset of the standard Killing field for rigid rotation

Yes, but one of the potential issues with this is how to define derivatives of the velocity field if the subset we select is not continuous. And you have to have well-defined derivatives of the velocity field in order to evaluate Killing's equation.

 

[AVentura]

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.

 

[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.