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

[AVentura]

I guess there has to be some explanation, and not allowing the perfect rotation is the only thing it could be.

 

[jartsa]

I'll start with link describing the phenomenon:

https://books.google.com/books?id=WTfnBwAAQBAJ&pg=PA43&lpg=PA43&dq=relativity+twist+rotating+cylinder&source=bl&ots=C2SDJNPF2K&sig=ImFDYGm_0qK7JoDU0ulsxLk3sMU&hl=en&sa=X#v=onepage&q=relativity twist rotating cylinder&f=false

Quote:

"A cylinder rotating uniformly about the x' axis in the frame S' will seem twisted when observed in the usual second frame S, in which it not only rotates but also travels forward."​

Now picture that instead of a cylinder we have a helix in S'. The helix in S' has an integer number of curls in S', so its center of gravity is on the x' axis. It rotates independently in S'. If its pitch in S' is such that the twisting seen in S perfectly straightens it out then an observer in S sees a rod that is not on the x-axis rotating around the x-axis all by itself.

How can that be? Or, where have I gone wrong?

What's the problem with a rod doing such rotation? Conservation of momentum?

Let's say a bunch of clocks, each with one clock hand, are stacked up on the ground on a gravitating planet. When one clock reads 9, the next one reads 3, that's how the stack of clocks is supposed to stay in balance.

From a suitable frame all clock hands point at 3.

In that frame momentum flux in the clock stack is such that all those clocks pointing at 3 are experiencing forces from the adjacent clocks that left those clocks when they were pointing at 9. (Because that's what happens in the rest frame of the stack)

So stored momentum seems to be the answer in this case, probably it's the answer in the rotating helix case too.

 

[PeterDonis]

Basically that amounts to computing the kinematic decomposition

I still haven't done this completely, but I suspect from looking at a similar calculation for an ordinary rotating ring what it is going to say: that whereas an ordinary rotating ring is a rigid congruence (vorticity, but no expansion or shear), the rotating helix we've been describing is not: it has nonzero expansion or shear (not sure which), meaning that it is not a possible stable state for a freely spinning body--external forces would have to be applied, as Dale suspected. A freely spinning state of a helix (no expansion or shear, just vorticity), if there is one, would be described by a different congruence.

 

[Dale]

I suspect from looking at a similar calculation for an ordinary rotating ring what it is going to say: that whereas an ordinary rotating ring is a rigid congruence (vorticity, but no expansion or shear), the rotating helix we've been describing is not: it has nonzero expansion or shear (not sure which),

This seems likely to me. I would even hazard a guess that the expansion and/or shear oscillates with a frequency which is equal to the torque free precession frequency.

 

[AVentura]

Peter, Dale, I want to go ahead and thank you both for your posts and time. Just reading about Herglotz-Noether it's pretty clear what must be going on.

 

[maline]

It is true that a helix with a whole number of turns and a uniform mass distribution cannot rotate freely around its axis. Its principal direction (of minimal moment of inertia) is slanted toward the direction a quarter-turn away from each endpoint of the helix, so if it is rotated around the axis, the angular momentum vector will be slanted away from that principal direction, and the rotation will precess around the AM vector.
One way to avoid this is if the helix is infinitely long. In that (obviously very hypothetical) case, the rotation is possible, and there will indeed be an inertial frame in which we have a straight rod revolving around an axis parallel to itself. Momentum will not be conserved, but that is normal for mass distributions that extend to infinity. In particular, the shear tension in the rod is a component of the stress tensor that remains nonzero at infinity, so we can describe a sourceless "flow of force" from infinity that accelerates the rod toward the axis.
However, I'm not sure the OP's paradox has been adequately resolved. What about situations with nonuniform mass distributions along a finite helix? Is it true that there is no possible mass distribution that will allow free rotation around the axis? That seems quite surprising, but it seems like AVentura has proved it using relativity! Does anyone have a direct proof, a counterexample, or another resolution to the paradox?

 

[Dale]

It is true that a helix with a whole number of turns and a uniform mass distribution cannot rotate freely around its axis. Its principal direction (of minimal moment of inertia) is slanted toward the direction a quarter-turn away from each endpoint of the helix,

That is what I thought. Do you by any chance have a reference?

 

[maline]

Do you by any chance have a reference?

No, I just worked it out directly. I did Physics 1 recently enough that I still remember what Euclidean 3-space looks like.
The calculation is pretty simple; does anyone have an interest in my writing it out?

 

[Dale]

The calculation is pretty simple; does anyone have an interest in my writing it out?

I would appreciate it!

 

[maline]

I would appreciate it!

For a given rotation vector $\vec \omega$, the contribution of a volume element $dV$, located at $\vec r$, to the angular momentum $\vec J$ around the origin is given by:$$d \vec J = \vec r \times d \vec p = \vec r \times [\rho (\vec r) dV (\vec \omega \times \vec r)] \\ = [\vec \omega r^2 - \vec r (\vec \omega \cdot \vec r)]\rho (\vec r) dV$$
This gives an explicit formula for the moment-of-inertia tensor:$$ J_i =I_{ij} \omega_j \Rightarrow \\I_{ij} = \int \,dV ~\rho (\vec r)(\delta_{ij} r^2 -r_i r_j)$$
For our purposes, we don't need to find the full tensor. We only need to know whether the axis of our helix is parallel to a principal direction (i.e. eigenvector) of the tensor- we set $\vec \omega$ along the axis and check whether $\vec J$ is parallel to it. The $\vec \omega r^2$ term is explicitly parallel to $\vec \omega$, so we only need to check the $\vec r (\vec \omega \cdot \vec r)$ term.
Let the helix be given by $\vec r(\theta)=(\cos \theta, \sin \theta, p \theta)$, where $-\frac L 2 \leq \theta \leq \frac L 2$, and let the mass distribution be given by $\lambda (\theta)$ (in units of mass per radian). This must be nonnegative and integrable, but it need not be continuous and may even include delta distributions (point masses). Let $\vec \omega=\omega \hat z$. Then $(\vec \omega \cdot \vec r(\theta))= \omega p \theta$, so our requirement is that the integrals of the $x$ and $y$ components of $\vec r (\theta)\omega p \theta\,\lambda(\theta)$ should vanish. Of course, we must also require that the $x$ and $y$ components of the integral of $\vec r(\theta)\lambda(\theta)$ should vanish, so that the center of mass will lie on the $z$ axis. Writing the $x$ and $y$ constraints separately, we have four requirements:$$\int_{-\frac L 2}^{\frac L 2} \,d \theta \,\cos \theta \,\lambda(\theta) =0~~(1)\\ \int_{-\frac L 2}^{\frac L 2} \,d \theta \,\sin \theta \,\lambda(\theta) =0~~(2)\\ \int_{-\frac L 2}^{\frac L 2} \,d \theta\, \omega p \theta \cos \theta \,\lambda(\theta) =0~~(3)\\ \int_{-\frac L 2}^{\frac L 2} \,d \theta\,\omega p \theta \cos \theta \,\lambda(\theta) =0~~(4)$$
From the symmetry of the problem, it if clear that if the requirements are satisfied by some $\lambda(\theta)$, they will also be satisfied by $\lambda(-\theta)$, and because the equations are linear in $\lambda$, the even sum $\lambda(\theta)+\lambda(-\theta)$ will be a solution as well. Thus without loss of generality we may assume that $\lambda(\theta)$ is an even function. Then the integrands in $(2)$ and $(3)$ are odd, so those two requirements are automatically satisfied. The integrands in $(1)$ and $(4)$ are even, so we may conclude that free rotation of a helix around its axis is possible iff there is a mass distribution $\lambda(\theta)$ such that $$\int_0^{\frac L 2}\, d\theta \cos \theta \,\lambda(\theta)= \int_0^{\frac L 2}\, d\theta\, \theta \sin \theta \,\lambda(\theta)= 0$$
For the case of uniform $\lambda$ we have $\int_0^{\frac L 2}\, d\theta \cos \theta=\sin \frac L 2=0\Rightarrow\frac L 2=n\pi, n\in \mathbb{N}$. But then integration by parts gives $\int_0^{n\pi}\,d\theta\,\theta\sin\theta= \left. -\theta \cos\theta\right|_0^{n\pi}+\int_0^{n\pi}\,d\theta\,\cos \theta=-n\pi\cos(n\pi)\neq 0$. Thus the angular momentum has a nonzero $y$ component.

A relativity paradox remains if there is any distribution satisfying the above conditions. Can anyone show directly that there is none?

 

[maline]

Well, sure enough! Here is an example of a helix that can rotate freely around its axis:
The helix has $n$ complete turns, parametrized as above by $\vec r (\theta) = ( \cos \theta , \sin \theta, p \theta)$ where $-n\pi \leq \theta \leq n\pi$. Start with a uniform mass distribution of $\lambda$ units per radian. As calculated there, if this rotates with $\vec \omega = \omega \hat z$, the angular momentum has a non-parallel component $J_y =\lambda \int_{-n\pi}^{n\pi} \, d\theta\, y(\theta) \omega z(\theta) =(-1)^n 2\pi n \omega p \lambda$. To fix this, we can simply add two point masses to the helix, at points $\theta = \pm \left( n- \frac 3 2 \right) \pi,$ with masses of $\frac n {n-\frac 3 2} \lambda.$ Their contribution to the integral is equal and opposite to the above value, so $J_y =0.$ They lie in the $yz$ plane, so they have no effect on $J_x=0$, nor on the $x$ component of the center of mass. Their $y$ coordinates are equal and opposite $(\pm 1)$, so the center of mass remains on the $z$ axis, and free rotation is possible.

So the paradox remains! In the frame boosted by $\vec v = - \frac 1 {\omega p} \hat z$, (which exists provided $\omega > \frac 1 p$, in turn requiring $p>1$), we have a straight rod, including the point masses, revolving around an axis parallel to itself, seemingly a violation of linear momentum conservation!

 

[Dale]

Very interesting work @maline!

 

[jartsa]

It is true that a helix with a whole number of turns and a uniform mass distribution cannot rotate freely around its axis. Its principal direction (of minimal moment of inertia) is slanted toward the direction a quarter-turn away from each endpoint of the helix, so if it is rotated around the axis, the angular momentum vector will be slanted away from that principal direction, and the rotation will precess around the AM vector.
One way to avoid this is if the helix is infinitely long. In that (obviously very hypothetical) case, the rotation is possible, and there will indeed be an inertial frame in which we have a straight rod revolving around an axis parallel to itself. Momentum will not be conserved, but that is normal for mass distributions that extend to infinity. In particular, the shear tension in the rod is a component of the stress tensor that remains nonzero at infinity, so we can describe a sourceless "flow of force" from infinity that accelerates the rod toward the axis.
However, I'm not sure the OP's paradox has been adequately resolved. What about situations with nonuniform mass distributions along a finite helix? Is it true that there is no possible mass distribution that will allow free rotation around the axis? That seems quite surprising, but it seems like AVentura has proved it using relativity! Does anyone have a direct proof, a counterexample, or another resolution to the paradox?

Let's consider what happens when we make finite helix rotate non-freely using rocket motors. At the ends of the helix we attach two precession preventing rockets which apply the same forces that an infinite helix would apply at the end of the short helix. And along the helix we attach million rotation initiating rockets (the helix is very long).

Now we observe the rotation initiation process. In the helix frame the rockets apply a torque and the helix gains angular momentum. In another frame rockets start non-simultaneously, which straightens the helix. Somehow the straight rod is able to gain angular momentum, and possesses angular momentum.

What happens if the rod/helix breaks to small segments? In the helix frame the parts fly apart. In the rod frame different parts of the rod start the linear motion at different times, and because of that the parts fly to different directions.

So I think the important problem here is: How does the rod have all the different momentums?

 

[AVentura]

I'm trying to follow you jartsa but I'm having trouble. Is the question how could such a helix come to have such a rotation? I'm not sure we should bother thinking about the rod. We know it cannot do what it appears to here.

If maline's analysis is correct there must be something else preventing the rotation. I don't really have the math skills to apply the Herglotz-Noether theorem, but I'd like to have a layman's understanding of it. I only understand born rigidity in the sense of linear acceleration (and nothing can naturally achieve it, correct?)

It seems there should be a general theorem concerning how a relativistic center of mass (?) transforms between frames. Similar to how angular momentum in the direction of motion is unchanged in boosts, but perpendicular AM is.

 

[maline]

We know it cannot do what it appears to here.

Is that completely clear? I'm a bit confused about this. The off-diagonal spatial components of the stress tensor are nonzero, so doesn't that turn into a "hidden momentum" in the boosted frame? Perhaps our rod always has zero momentum in the $x,y$ directions, despite its revolution? I'm probably talking nonsense, so somebody please help clear this up.

I tried a bit to work out what the stress tensor looks like, but I got stuck because in a one dimensional body, there are (IIUC) only three degrees of freedom for the spatial components of the tensor, so we get a system of three linear ODE's in three variables (the force density at a point is the derivative of the stress along the length of the body), but then we can only satisfy three boundary conditions, and we have six because all components of the stress must vanish at both ends of the body. It seems our helix needs to be three dimensional, and that makes it pretty scary to work with.

(and nothing can naturally achieve it, correct?)

I think you are referring to the fact that any angular acceleration must distort a body & violate Born rigidity, because different parts of the body length-contract differently.

It seems there should be a general theorem concerning how a relativistic center of mass (?) transforms between frames.

Yes, I would also love a clear reference on this topic!

 

[PeterDonis]

It seems there should be a general theorem concerning how a relativistic center of mass (?) transforms between frames. Similar to how angular momentum in the direction of motion is unchanged in boosts, but perpendicular AM is.

The general theorem, to the extent that there is one, says that you have to use the relativistic angular momentum tensor. I posted some computations of that for the helix earlier in this thread.

 

[AVentura]

I think you are referring to the fact that any angular acceleration must distort a body & violate Born rigidity, because different parts of the body length-contract differently.

Yes. But we don't have angular acceleration in this case. I can see how that could affect how the helix came to be in this state though.

The general theorem, to the extent that there is one, says that you have to use the relativistic angular momentum tensor. I posted some computations of that for the helix earlier in this thread.

I see, thanks.

 

[jartsa]

I'm trying to follow you jartsa but I'm having trouble. Is the question how could such a helix come to have such a rotation? I'm not sure we should bother thinking about the rod. We know it cannot do what it appears to here.

If maline's analysis is correct there must be something else preventing the rotation. I don't really have the math skills to apply the Herglotz-Noether theorem, but I'd like to have a layman's understanding of it. I only understand born rigidity in the sense of linear acceleration (and nothing can naturally achieve it, correct?)

It seems there should be a general theorem concerning how a relativistic center of mass (?) transforms between frames. Similar to how angular momentum in the direction of motion is unchanged in boosts, but perpendicular AM is.

Well, I think a helix can have such a rotation - with just a little help from two rockets at the ends of the helix. (such rotation is possible for an infinitely long helix, so we take a finite clip of that helix and emulate the two removed parts by two rockets)

But, a real rod can not have such sideways motion (acceleration) as the 'rod', because we know that a rod inside a helix can not get out of the helix by moving sideways.

I guess the above is really hard to follow. I was saying that if we accelerate an axially moving real rod sideways, the rod will either experience axial stress or the rod will turn. But that does not apply to a 'rod' - an object that is a helix in its rest frame. (it does not apply to that odd motion that the 'rod' does according to me, and according to you can not do)

So (according to me) a 'rod' and a real rod don't follow the same laws, so maybe it's not so surprising if a 'rod' moves around a little bit oddly?

 

[maline]

The post by maline in #61 seems to increase the feeling of confusion in this thread.

Well yes! I showed in that thread that the suggested resolution of the paradox fails. I am hoping someone will come up with something, otherwise we have a serious contradiction on our hands!

 

[Dale]

I think that @PeterDonis proposed method may still work. Since it is based on the kinematics rather than the dynamics it would hold for any mass distribution. So I wouldn't go to "serious contradiction" yet. I would go to "requires a fully relativistic treatment".

 

[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.