# Twisting of a rotating cylinder

1. Dec 29, 2013

### WannabeNewton

Hi guys. The following question is related to exercise 3.7 from Rindler's SR/GR text: http://postimg.org/image/8gdx0j3gd/

Consider first the following scenario: a Born rigid rod is parallel to the $x'$ axis of a frame $S'$ and accelerating uniformly with proper acceleration $a$ along the $y'$ axis of $S'$. Assume $at << c$ so that the equation of motion of the rod becomes $y' = \frac{1}{2}at'^2 + O(\frac{\alpha^2 t^2}{c^2})$.

Consider now a frame $S$ moving with velocity $-v$ along the $x'$ axis of $S'$. Performing a Lorentz boost to $S$, the equation of motion becomes $y = \frac{1}{2}a \gamma^2 (t - vx/c^2)^2$. Now in $S'$ the lines of simultaneity are given by $t' = \text{const.}$ and for such a line of simultaneity the observer at rest in $S'$ sees the shape $y' = \text{const.}$ as expected i.e. the rod is just a straight rod parallel to the $x'$ axis. In frame $S$, the lines of simultaneity are given by $t = \text{const.}$ and for such a line of simultaneity the observer at rest in $S$ sees the shape $y' = \alpha(1 - \beta x)^2$ where $\alpha,\beta$ are constants. So in $S$ the observer sees the rod in the shape of a parabola (technically part of a parabola because of the finite size of the rod).

With that in mind, consider now a hollow circular cylinder of radius $R$ rotating uniformly with angular velocity $\omega$ about the $x'$ axis of frame $S'$. Furthermore consider the tip of the clock hand of any circular cross section of the cylinder described in the exercise above as the arrowhead of the radius vector of the circular cross section; the equations of motion of the clock hand are $x' = \alpha, y' = R\sin \omega t', z' = R\cos\omega t'$ where $\alpha$ is the constant $x'$ position of the point particle relative to $S'$.

Boosting to $S$ we get $x = \gamma^{-1}\alpha+ vt, y = R\sin(\gamma \omega t - \gamma \omega vx/c^2), z=R\cos(\gamma \omega t -\gamma \omega vx/c^2)$.

As usual the angular velocity (frequency) of rotation changes to $\gamma \omega$ due to time dilation however the more important effect is the introduction of a phase factor $\varphi = -\gamma \omega vx/c^2$; in particular, $\frac{\mathrm{d} \varphi}{\mathrm{d} x} = -\gamma \omega v/c^2$. Hence in $S$ the clock hands all rotate with the same angular velocity $\gamma \omega$ but are all out of phase with one another due to the above phase gradient whereas in $S'$ the clock hands all rotated with the same angular velocity $\omega$ and did so in phase with one another.

Note that the clock hands all together still sweep out a cylinder: $y^2 + z^2 = R^2$. The obvious difference now is that the clock hands all move forward uniformly relative to an observer at rest in $S$ as per $x = \gamma^{-1}\alpha+ vt$ i.e. the entire cylinder moves forward.

Now if an observer at rest in $S'$ marked each clock hand with dark ink so as to draw a straight black line across the face of the cylinder parallel to the $x'$ axis, the black line will be twisted into a helix when viewed in $S$. Here's where I'm having trouble:

In the case of the rod, there is a clear visual deformation of the rod when viewed in $S$ because it goes from looking like a straight rod in $S'$ to looking like (part of) a parabola in $S$. Hence I can easily visualize the change in the rod when viewed between $S'$ and $S$. However in the case of the cylinder, I can't seem to visualize things quite as explicitly. As noted above, if a straight black line parallel to the $x'$ axis is drawn in $S'$, the line gets twisted into a helix in $S$ and an observer at rest in $S$ would see this helix as opposed to the straight line that an observer at rest in $S'$ sees.

But say I didn't make any mark at all on the cylinder whatsoever. Could an observer in $S$ actually see any difference in the cylinder from one in $S'$? Above I used the idea of the cylinder as being composed of a stack of circles along its height to make sense of the twisting by noting that the circles will start rotating out of phase from one another but this is just a conceptual aid. What if we simply had a rigid cylinder in and of itself-no markings and no imaginary rotating clock hands? What happens to its shape in $S$ when compared with its shape its shape in $S'$? Is there any way to make sense of any kind of "twisting" of the cylinder when viewed instantaneously in $S$ without the use of any kind of markings across the cylinder?

Clearly the rod isn't physically being bent into a parabola when boosting from $S'$ to $S$ and the cylinder isn't physically being twisted when boosting from $S'$ to $S$ because such actions would induce stresses on the objects and possibly cause them to break which we know can't happen simply as a result of a Lorentz boost.

Last edited: Dec 30, 2013
2. Dec 30, 2013

### jartsa

Let's say we have a spinning accelerating rocket equipped with small rocket motors on the side walls, the small rockets keep the spinning rate constant everywhere. That's a constant coordinate spinning rate.

Physical stresses emerge on the rocket walls.

Then the small rockets run out of fuel simultaneously. The rocket twists itself into a shape where the internal stresses disappear.

Passangers on the rocket have some explanation for the forces that appear and the disappear. Outside observer has some other explanation. What is the explanation? Something to do with forces and relativity, I guess.

Last edited: Dec 30, 2013
3. Dec 30, 2013

### WannabeNewton

You have a habit of replying with these insanely random scenarios that have nothing to do with the scenario described in the original post. I would appreciate it if you could reply with something that at least tries to address (and hopefully answer) the questions raised in the original post within the contextual framework of the original post. Otherwise I don't see the point in replying. Thanks.

4. Dec 30, 2013

### Staff: Mentor

Your ideal rod has zero thickness in the y direction, so can be uniformly accelerated in that direction without any internal stresses. Assign it some non-zero thickness and frame-invariant physical stresses will appear. That puts the rod and the cylinder back on an equal footing, and suggests that helicity or straightness of the mark (we need a mark on the new improved non-zero-thickness rod before we can speak of it being straight or parabolic as well) is a coordinate artifact.

5. Dec 30, 2013

### Jonathan Scott

I don't understand the problem here. There is no "physical deformation" of the cylinder in going from one frame to another. It just looks different from another frame, but the physical shape and the forces that keep it in that shape only make sense in its rest frame.

6. Dec 30, 2013

### jartsa

Ok, so we have a cylinder without any marks on it. How can we see if the cylinder is twisted. I guess we can't.

If we have a rod so long that we can't see its ends, and without any marks on it, how can we see if the rod is length contracted? We can't see that.

So my answer is: Twisting and length contraction are very much same kind of things.

7. Dec 30, 2013

### Jonathan Scott

I'd disagree with that. The twisting is primarily related to the different definitions of simultaneity in the different frames.

8. Dec 30, 2013

### Staff: Mentor

There's a problem here - we have to maintain Born rigidity, and the forces that keep it in shape have to "make sense" in any frame.

9. Dec 30, 2013

### WannabeNewton

Why does the introduction of thickness to the rod put it on equal footing with the rotating cylinder? I'm assuming that the cylinder is also of zero thickness (a cylindrical shell if you like) so that if you unwrap it you just get a completely flat section of a plane. Sorry if I misinterpreted what you meant by "equal footing" in your post.

I'm also assuming that both the acceleration of the rod and the rotation of the cylinder have already existed indefinitely into the past i.e. we aren't taking a cylinder at rest and applying some angular acceleration or taking a rod at rest and applying some linear acceleration. This way we can easily constrain the motion to be Born rigid motion in which case the expansion tensor of the associated time-like congruence vanishes.

Now I agree with you that the deformation of the straight black line mark across the cylinder when viewed in $S'$ into a helix when viewed in $S$ is a purely coordinate-dependent deformation, just like the straight rod being deformed into a parabola; in fact both of these deformations derive from the difference in simultaneity planes between $S$ and $S'$. But if you take a look again at the problem statement: http://postimg.org/image/8gdx0j3gd/ it explicitly states that the cylinder will seem twisted when observed instantaneously in $S$.

My main questions are then: if I didn't make any markings whatsoever on the rotating cylinder how could an observer in $S$ possibly tell that the cylinder is "twisted"? In fact what does it even mean for the cylinder to be "twisted" if no markings are made? As noted the cylinder cannot physically twist between circular cross-sections because that would induce internal stresses within the cylinder and potentially break the coordinate-invariant Born rigidity constraint. So what's actually being "twisted" in the absence of any kind of marking and in the absence of radius vectors, one for each circular cross section, that act as clock hands which are parallel in $S'$ but non-parallel in $S$ as per $\frac{d\varphi}{dx}$?

10. Dec 30, 2013

### Jonathan Scott

Born rigidity is defined by what happens in its rest frame.

What happens in the cylinder case doesn't need to involve any forces, in that you could just assume a row of parallel clock faces and note that locally they could be connected together by a rigid axle.

This is no different to any other example of difference of simultaneity, for example if a long object has blinking lights on it or bouncing balls.

There are similar cases in which the forces are quite interesting. For example, it may not be obvious that if you apply torque to one end of a rod, changing the energy of rotation of a system, this involves transmitting energy along the rod, which requires a force along the line of the axle. You can see what actually happens by considering the surface of the rod to be made up of triangles of struts which only support pushing and pulling but no shear forces, in which case you can show that the changes in Lorentz transformations during changes of velocity cause the push and pull forces to be temporarily unbalanced, giving rise to forces along the rod.

11. Dec 30, 2013

### WannabeNewton

That's not how one defines Born rigidity. If we have a time-like congruence of test particles with unit tangent field $\xi^{\mu}$ then the spatial distances relative to $\xi^{\mu}$ between infinitesimally separated particles is given by $h_{\mu\nu} = g_{\mu\nu} + \xi_{\mu}\xi_{\nu}$ and Born rigidity is defined geometrically by $\mathcal{L}_{\xi}h_{\mu\nu} = 0 \Leftrightarrow h^{\alpha}{}{}_{\mu}h^{\beta}{}{}_{\nu}\nabla_{(\alpha}\xi_{\beta)} = 0$. This condition obviously holds in all frames. The physical interpretation is simply that given any one reference particle in the congruence, the spatial distances between this particle to infinitesimally separated particles in the local Lorentz frame of an observer comoving with the reference particle remain constant.

See the last paragraph of post #9.

12. Dec 30, 2013

### Jonathan Scott

If you consider the cylinder to be made up of some material, then the view from some frame in which the cylinder is moving will show that material to be apparently subject to a shear effect aligned in such a way that bonds that would be in equilibrium when aligned with a straight line along the material at rest will appear to be in equilibrium instead when they follow the helical distorted shape. Obviously if the microscopic forces are transformed to any frame, they will still result in the same overall effects.

(I am of course assuming that the rotation speed is small enough that Born rigidity can still be approximated).

13. Dec 30, 2013

### Staff: Mentor

by "equal footing" I mean that they both can (and depending on the specifics of how they are accelerated, will) develop internal stresses that can be locally measured and are frame-invariant, so we can analyze both problems in terms of these stresses - we don't have to use the simultaneity-cursed notion of "shape".

In the absence of markings, the only possible definition of "twisted" would be the appearance of local stresses reported by stress-strain gauges placed at various points on the cylinder. My conclusion from the last (endless and inconclusive) rotating accelerated cylinder thread is that it is in fact impossible to uniformly accelerate a rotating cylinder without giving rise to such stresses.

14. Dec 30, 2013

### WannabeNewton

Thanks guys! Since you are both describing similar things, I'll just respond in unison:

(1) But here the rotating cylinder isn't being uniformly accelerating in any way. We simply have a rotating cylinder with no translational motion in $S'$ that gains uniform translational motion in $S$ (on top of speeding up the rotation rate due to time dilation) simply because $S$ has uniform translational motion relative to $S'$. So as you noted any kind of "shape deformation" that results after a simple Lorentz boost has to be purely coordinate-dependent under the guise of relativity of simultaneity.

(2) Here I'm just stating what's counter-intuitive to me (I haven't worked out the math yet) but if we went with the idea that the "twisting" caused local shear stresses in the cylinder and measured it by, for example, attaching a rigid rod between two points on the cylinder infinitesimally separated along the height of the cylinder, then looking at the phase gradient $\frac{d\phi}{dx} = - \gamma \omega v /c^2$ I can make the relative velocity $v$ between $S$ and $S'$ arbitrarily large (bounded above by $c$ of course) and as a result make the shear stress on the rod arbitrarily large and break it when viewed in $S$ but nothing would happen to the rod in $S'$. This seems to suggest to me (at least intuitively) that this "twisting" effect can't be related to local shear stresses because of such a dichotomy.

Indeed since the "twisting" effect derives entirely from the relativity of simultaneity I don't think there can be any induced stresses simply due to the "twisting"; the effect after all is mathematically codified by nothing more than phase shifts between the rotations of the imaginary clock heads represented by the imaginary radius vectors of consecutive circular cross sections of the cylinder when viewed instantaneously in $S$.

If there is no way to explicitly visualize this purported "twisting" of the rotating cylinder without using any kind of markings across the cylinder then that's ok with me, I was just wondering if there was a more direct way of characterizing this "twisting" akin to the way the straight rod in $S'$ explicitly becomes parabolic when viewed instantaneously in $S$. I just couldn't tell how to interpret the text's statement that the cylinder seems twisted when viewed instantaneously in $S$ in the absence of any kind of physical markings across the cylinder.

Thanks again.

Last edited: Dec 30, 2013
15. Dec 30, 2013

### yuiop

I agree, but just for completeness, it should be noted that it was generally concluded in this forum, that it was possible for a rotating cylinder that is being accelerated along its axis of rotation, to have Born rigid motion and not violate the Herglotz-Noether theorem. Such a cylinder would be stressed, but the stress can be finite and time independent.

16. Dec 30, 2013

### Jonathan Scott

I've been assuming that the rotating cylinder was moving with constant velocity along its axis, so I'll take it that the parts about acceleration and stresses only apply to Nugatory's response.

Yes, I agree with this.

There are obviously no additional stresses on bonds within the material due to the different frame of reference. If you want to describe the stable equilibrium structure at the microscopic level of the rigid object as seen in the moving frame you have to take into account the fact that the velocity of a point on the surface of the cylinder is not actually parallel to the motion, and neither is the Lorentz transformation relative to the rest frame of that point.

17. Dec 30, 2013

### WannabeNewton

Thanks Jon!

Just to make sure I have a definitive answer to this, let me just restate my main purpose in making this thread:

If you (Jon) or anyone else can clarify the above question for me once and for all then I'd be content

18. Dec 30, 2013

### Jonathan Scott

I'm guessing that the author sees a cylinder as being implicitly marked with labels indicating where the points originally came from.

(... and also assumes some form of illumination to see it, at an appropriate wavelength, and so on. And that's without getting into the debate about the differences between what we calculate in a given frame of reference and what we actually see!).

19. Dec 30, 2013

### pervect

Staff Emeritus
Well, I think the best mathematical representation of the 4-dimensional cylinder is a congruence of worldlines. That definition does allow us to place a mark on the cylinder, and trace out the worldline of that mark.

So I'd say "yes" to allowing marks. I think your comments about how defining rotation requires us to be able to make marks appears to be sensible as well, though I haven't pondered it at depth.

There are some other possible representations for a cylinder. Some people may think of a cylinder as being a quotient manifold, where you identify the 4-d wordlines of the marks as a point in some abstract space. (I call this process, informally, "smashing the worldline to a point"). It seems to me that this notion is only physically sensible when you have the appropriate timelike symmetry (a timelike killing vector), and furthermore that the "smashing" smashes the worldline of the orbit of the Killing vector to a point. In this case the "smashing" process is eliminating the notion of time from an object that has a time symmetry. Trying to eliminate time from an object without time symmetry doesn't seem to be well motivated.

As far as stresses go - the stresses in the cylinder are represented by the stress-energy tensor. If you transform the stresses between the two coordinate systems, you should see that pressures, for instance, transform as the appropriate component of the stress-energy tensor.

The stresses won't be zero - something is making the cylinder accelerate. But there shouldn't be any stress component that isn't just a transformed version of the stress components in the S frame.

Coming to grasp with this intuitively seems hard. The first step I'd suggest is finding and writing down the metric in both coordinate systems, then trying to divine the motion from the metric. In general, to even attempt to have an intuitive understanding of the dynamics requires that one know the metric.

The other point of finding the metric is to tell when the "smashing" process makes sense. (Recall, this is the informal name I gave for forming a quotient manifold where one eliminates time).

When your metric coefficeints are independent of time, you have a wise coordinate choice where the objects description in terms of coordinates isn't changing with time, and the smashing process is well motivated. You may have an object that posses the requisite time symmetries for a successful smashing, but if your coordinate choice doesn't respect the physical symmetry, it does not make sense to "smash" it into a quitoent manifold using the coordinate time.

Sorry for the lenght - and I hope it's not too unclear.

20. Dec 31, 2013

### Staff: Mentor

Hi WBN,

Here is my take on this. To have Born-rigid rotation implies that the angular velocity is constant, so there is no deformation. All that you have is that a line which is spatially straight in one frame will be spatially helical in another frame. Conversely, under some constraints if you draw a helical line there should be a frame where it is straight. There is no sense of twisting or strain involved, just different shaped lines.

Now, if you have a non-rotating cylinder and draw a straight line on that cylinder then it will be straight in all frames. If you then begin rotating it, there absolutely must be some deformation. It is impossible to maintain Born-rigidity. That deformation will lead to a helical line in some frames but the material is deformed in all frames.

21. Dec 31, 2013

### WannabeNewton

I completely agree with you Dale. In my opinion the exercise would have been much better in demonstrating the dependence of spatial shapes of rigid objects on simultaneity slices of inertial observers if we just had a rigid rod in uniform circular motion about the $x'$ axis of the frame $S'$.

The motion of the rod can be given parametrically by $r(s',t') = \langle s', R\sin \omega t', R\cos \omega t'\rangle$ in $S'$, where $s' \in [0,L]$ with $L$ being the length of the rod in $S'$ which is of course constant in $S'$ since the motion is completely transverse to the length of the rod. The spatial shape of the rod in $S'$ is gotten by intersecting the worldtube of the rod with the simultaneity slice $t' = \text{const.}$ of $S'$ so in particular the spatial shape of the rod at $t' = 0$ is simply given by $\langle s', 0, R \rangle$ in $S'$ which is just a straight rod as we would expect.

Substituting in for the coordinates of frame $S$, which is moving backwards with velocity $-v$ relative to $S'$, we get $r(s,t) = \langle \gamma (s - vt), R\sin (\gamma \omega t - \gamma \omega vs /c^2 ), R\cos (\gamma \omega t - \gamma \omega vs /c^2 )\rangle$. Intersecting the worldtube of the rod with the simultaneity slice $t = 0$ of the frame $S$ we get the spatial shape $\langle \gamma s, -R\sin \gamma \omega vs /c^2 , R\cos \gamma \omega vs /c^2 )\rangle$ which is a helix.

So in $S'$ the rod has the spatial shape of a straight line at an instant in $S'$ whereas in $S$ the rod has the spatial shape of a helix at an instant in $S$.

I personally think this much better illustrates the effect of relativity of simultaneity on the "distortion" of spatial shapes because if you instead base the exercise on a cylinder rotating about the $x'$ axis (like the textbook referenced in this thread did), which we can think of as a collection of rods laid out around a circle in the $y'$-$z'$ plane, then while reach rod would look like a helix instantaneously in $S$ as shown above, the entire collection of helical rods would still form the exact same rotating cylinder as in $S'$ (except now its moving forward uniformly) so there's no way to even distinguish the two unless as you stated someone draws an explicit black line across the cylinder parallel to the $x'$ axis.

Would you agree? And thanks for the reply.

22. Dec 31, 2013

### WannabeNewton

No it was perfectly clear pervect, thanks for the very detailed reply. Your proposed method of analysis of this exercise reminded me a lot of the analysis of the uniformly rotating disk wherein we have the usual disk shape when "viewed" instantaneously in a background global inertial frame as per the simultaneity associated with Einstein synchrony but we have a geometrically curved spatial shape for the disk when "viewed" instantaneously relative to the time-like congruence of observers corotating with the disk using the local radar simultaneity. It's a very elegant mode of analysis of SR and GR problems of this nature in my opinion. I will definitely try to perform an analysis of the rotating cylinder in similar fashion to that of the rotating disk using time-like congruences.

23. Jan 2, 2014

### WannabeNewton

By the way just to clarify what I meant in the quoted paragraph above, I've attached two diagrams with the striation curves on the cylinder made explicit, one for the $S'$ frame and one for the $S$ frame. Clearly if the striation curves on the cylinder were not made explicit, the effect of simultaneity on spatial shapes of rigid objects would seem non-existent in the case of the cylinder when compared to the obvious nature of the effect in the case of a single rod.

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