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

[AVentura]

I can only think of one function describing a straight rod revolving around the x-axis that has a center of mass on the x-axis. An infinitely long helix whose twists per unit length approaches zero. It wraps around one time at infinity, but is always flat locally. It happens to also keep its center of mass on the x-axis for any observer. Is it a wave function (or component of one)? You can ignore this post.

 

[Dale]

OK, so in S' the worldsheet of the helix can be written:
$$a' = \left(t',\theta p,\cos (\theta +t' \omega ),\sin (\theta +t' \omega
)\right) $$
Where units of distance are chosen such that the radius of the helix is 1 and units of time are chosen such that c=1. $\theta$ and $t'$ parameterize the worldsheet, $p$ is the pitch, and $\omega$ is the angular speed (we are limited to $-\frac{1}{2\pi} < \omega < \frac{1}{2\pi}$ to avoid the rim traveling faster than light).

Boosting to the unprimed frame and reparameterizing from t' to t gives:
$$ a = \left(t,\theta p \sqrt{1-v^2}-t v,\cos \left(\theta +\theta p v \omega +t
\sqrt{1-v^2} \omega \right),\sin \left(\theta +\theta p v \omega +t \sqrt{1-v^2}
\omega \right)\right) $$
The y and z coordinates are not a function of $\theta$ for $p v \omega = -1$. $\omega$ is constrained as mentioned above and $-1<v<1$, but $p$ can be any real number. So the helix is straight in the unprimed frame for $p=-\frac{1}{v \omega}$.

I would appreciate any checking of my math.

 

[AVentura]

Dale, is your ω the angular speed in S or S'?

 

[Dale]

Dale, is your ω the angular speed in S or S'?

In S', the frame where the helix is rotating but not translating.

 

[AVentura]

Thanks for doing that. I'll try and check it. If the length of one complete turn of helix in S' is p, then in S it's p/γ. It seems to agree with the linked book in the OP.

 

[Dale]

If the length of one complete turn of helix in S' is p,

I think the length of one turn is $p 2\pi$

 

[AVentura]

Is the definition of pitch not the length of one complete helix turn, measured parallel to the axis of the helix?

Should your worldsheet a' use θp/2π?

 

[Dale]

Maybe, I am not sure of the exact definition of pitch. Regardless, I don't like to needlessly throw in unnecessary numerical factors to clutter up the equations.

 

[PeterDonis]

I would appreciate any checking of my math.

I'm going to redo this in cylindrical coordinates $t, z, r, \theta$, and I'll leave in a constant $R$ for the radius of the helix and use $k$ for the analogue of its pitch; $k$ is the "spatial frequency" in radians per meter of the helix in the $z$ direction, i.e., how many radians the helix turns around in one unit of $z$. For the case described in the OP, $k = 2 \pi$, since the helix completes one full turn in one unit of $z$.

Before doing the case of a helix in S', I'm first going to do the case of the tips of the clock hands on the cylinder forming a straight line in S'. The worldsheet in this case is given by

$$
a' = \left( t', Z, R, \omega t' \right)
$$

This worldsheet is parameterized by $t'$ and $Z$: $Z$ labels each worldline (each tip of a clock hand), and $t'$ labels each surface of constant time (note that $t'$ is not the same as proper time along each worldline--we could parameterize by that instead, but it would just complicate the math).

Now we boost to frame S, moving with velocity $- v$ relative to S' (I put in the minus sign because it keeps the signs neater in the formulas I'm about to write--basically we're doing an inverse Lorentz transformation, treating S' as the "primed" frame moving with velocity $v$ relative to the unprimed frame S), and reparameterize by $t$. This gives:

$$
a = \left( t, \frac{Z}{\gamma} + v t, R, \frac{\omega}{\gamma} t - \omega v Z \right)
$$

This describes a helix.

Having seen how this works, it is easy to see that the reverse case works too. We start with a helix in S', described by the worldsheet:

$$
a' = \left( t', Z, R, \omega t' + k Z \right)
$$

where $k$ is the "spatial wavenumber" of the helix, i.e., the number of radians it turns through per unit of $z$. Then we boost to S with velocity $- v$ as above and reparameterize by $t$ to get:

$$
a = \left( t, \frac{Z}{\gamma} + v t, R, \frac{\omega}{\gamma} t - \omega v Z + k Z \right)
$$

We can see that $\theta$ is not a function of $Z$ if $v = k / \omega$, so the helix looks straight in S if it is moving with that velocity in the $z$ direction in S. This appears to be consistent with what Dale posted (but I think it's easier to see how it goes in these coordinates).

However, we still haven't resolved the dilemma I ended with in my last post: what happens to the center of mass and the axis of rotation of the helix in S? To address that, we need to look at the angular momentum tensor of the helix. This tensor is $M^{\alpha \beta} = X^\alpha P^\beta - X^\beta P^\alpha$, where $X$ is the position 4-vector and $P$ is the momentum 4-vector. If we assume that the helix has a constant rest mass per unit length, we can leave that out and just use the 4-velocity vector $U$ (which means we're actually looking at the angular momentum per unit rest mass, which is fine).

The position 4-vector $X$ is just the expressions we saw above. The 4-velocity is the derivative of those expressions with respect to proper time. We don't have an explicit expression for proper time, but we can finesse that by using the chain rule and writing $g$ for $dt / d\tau$, i.e., for the "gamma factor" relating coordinate time to proper time. Since this factor is the same for all the worldlines in the helix, this won't cause a problem.

Let's first look at the helix in S', where it is rotating but not translating. Here we have

$$
X' = \left( t', Z, R, \omega t' + k Z \right)
$$
$$
U' = \left( g', 0, 0, g' \omega \right)
$$

The angular momentum tensor is then

$$
M' = \begin{bmatrix}
0 & - g' Z & - g R' & - g' k Z ) \\
g' Z & 0 & 0 & g' \omega Z \\
g' R & 0 & 0 & g' \omega R \\
g' k Z & - g' \omega Z & - g' \omega R & 0
\end{bmatrix}
$$

Now let's look at it in S, in the case where $v$ is just right to straighten the helix:

$$
X = \left( t, \frac{Z}{\gamma} + \frac{k}{\omega} t, R, \frac{\omega}{\gamma} t \right)
$$
$$
U = \left( g, g \frac{k}{\omega}, 0, g \frac{\omega}{\gamma} \right)
$$
$$
M = \begin{bmatrix}
0 & - g \frac{Z}{\gamma} & - g R & 0 \\
g \frac{Z}{\gamma} & 0 & 0 & g \frac{Z \omega}{\gamma^2} \\
g R & 0 & 0 & g \frac{R \omega}{\gamma} \\
0 & - g \frac{Z \omega}{\gamma^2} & - g \frac{R \omega}{\gamma} & 0
\end{bmatrix}
$$

I'll leave this to simmer for a bit and then follow up with some more thoughts.

 

[Dale]

if v=k/ω

So I think that we agree with our p and k related by $p=-1/k$.

Regarding the "paradox". I don't think that is a relativity issue, I suspect that it stems from a misunderstanding of non-relativistic rotation. Specifically, I don't think that a single helix will freely spin about its axis. I think that it requires a balanced pair of external forces to keep it from precessing.

 

[AVentura]

I don't think that a single helix will freely spin about its axis. I think that it requires a balanced pair of external forces to keep it from precessing.

That would have to be true for any helix then, with any fractional number of turns as well, because there exists an observer that sees it that way.

 

[Dale]

That would have to be true for any helix then, with any fractional number of turns as well, because there exists an observer that sees it that way.

I would have to work out the math to be sure, but I think that it is only balanced for an integer number of turns. For non integer number of terms I believe that the force is unbalanced.

 

[AVentura]

I thought in your previous post (#45) you were saying that even an integer number of turns would be unbalanced. I was asking, "how about fractional ones then?"

Personally I think an integer number can be spun on it's axis. In S' ω could be slow.

 

[Dale]

Personally I think an integer number can be spun on it's axis.

I don't believe that this is true without some external forces. I think that this mistaken belief is the source of the "paradox"

 

[PeterDonis]

I don't think that a single helix will freely spin about its axis. I think that it requires a balanced pair of external forces to keep it from precessing.

This could be it. I've been trying to come up with a way of using the angular momentum tensor to check this, but I'm not familiar enough with the math involved. Another possibility would be to see if the congruence of worldlines we've been describing meets the conditions of the Herglotz-Noether theorem, which can be viewed as describing the possible stationary states of a rotating object, i.e., the states that don't require external applied forces to maintain them. Basically that amounts to computing the kinematic decomposition, which I'm more familiar with but don't have time to do at the moment.

 

[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".