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

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1. May 4, 2017

### AVentura

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?

2. May 4, 2017

### Staff: Mentor

That can't happen; the twisting in question does not change the extent of the cylinder (or helix) perpendicular to the direction of motion.

3. May 4, 2017

### AVentura

I'm sorry, what do you mean by extent?

4. May 4, 2017

### Staff: Mentor

Note the key word "seem". What is actually going on here is relativity of simultaneity: in the frame in which the cylinder is moving forward as well as rotating, surfaces of constant time no longer cut the cylinder exactly perpendicular to its axis; instead they cut it at an angle. That's what creates the appearance of a twist. There is no actual stress applied to the cylinder (since all we're doing is changing coordinates, and changing coordinates can't change any physical observables).

5. May 4, 2017

### Staff: Mentor

The size of the cylinder or helix in the $y$ and $z$ directions (assuming the relative motion is in the $x$ direction).

6. May 4, 2017

### AVentura

I don't think it needs to. The apparent twisting changes the apparent pitch by itself.

Agreed. But if we stick to a cylinder, and just paint a helix on it in S', that painted line should have a different apparent pitch in S, correct? If not, then what exactly is meant by "seems twisted"?

7. May 4, 2017

### Staff: Mentor

It can't change a helix into a straight line, which is what you were claiming, without changing its extent in the $y$ and $z$ dimensions.

Yes. But that doesn't imply that the painted helix will become a straight line if we pick S with a large enough relative velocity.

8. May 4, 2017

### Staff: Mentor

I think that it could happen. I would have to work it out, but I am ~80% confident. Is it the integer number of turns in the center frame?

9. May 4, 2017

### AVentura

The pitch of a painted line is arbitrary. The required pitch in S' for a straight line in S is given in the OP link.

10. May 4, 2017

### Staff: Mentor

I am at least that confident that it can't happen. The reason is simple: a Lorentz transformation in the $x$ direction does not affect distances in the $y$ and $z$ direction. Making a helix into a straight line would require changing distances in the $y$ and $z$ directions--points which have nonzero separation in those directions in the original frame would have to have zero separation in the transformed frame.

11. May 4, 2017

### AVentura

I can paint two different helices on the same cylinder

12. May 4, 2017

### Staff: Mentor

Where?

13. May 4, 2017

### Staff: Mentor

So what? We are talking about one helix, not two.

14. May 4, 2017

### Staff: Mentor

Yes, but it has to be finite if the painted line is a helix. And as far as I can tell, a Lorentz transformation can only change one finite pitch to another finite pitch; it can't change a finite pitch into an infinite pitch, which is what "straight line" means.

15. May 4, 2017

### Staff: Mentor

Another item to think about: in a frame in which the helix is moving, it length contracts; so it seems like the pitch of the helix should increase (turns getting squeezed together), not decrease (turns getting stretched out) in a frame in which it is moving.

16. May 4, 2017

### Staff: Mentor

Any two points on the helix have different x coordinates, so distances between them can change under the Lorentz transform.

17. May 4, 2017

### AVentura

gamma*omega*v/c^2

This phenomenon is in addition to lorentz contraction

18. May 4, 2017

### Staff: Mentor

I didn't say the total distances between them couldn't change; I said the distances in the $y$ and $z$ directions couldn't change. Or, to put it more simply, the $y$ and $z$ coordinates of a given point on the helix can't change. But turning a helix into a straight line requires changing the $y$ and $z$ coordinates of at least some of its points.

19. May 4, 2017

### Staff: Mentor

That is the twist per unit length in S of something that is a stack of circles in S'. It is not the pitch in S' of something that is a straight line in S.

20. May 4, 2017

### AVentura

I think "twisting" implies that they do, but stay on the surface of the cylinder. If not, then what is twisting? Also remember there is more going on than Lorentz transformation. The surface of the cylinder is not a rest frame.

21. May 4, 2017

### AVentura

if you twist a helix the exact magnitude of its pitch, but in the opposite direction, you get a line.

BTW, you don;t need to make it a straight line for the question to remain. If you change it enough that its center of mass is not on the x-axis the helix should wobble.

22. May 4, 2017

### Staff: Mentor

I got this backwards: squeezing the turns together means decreasing pitch, and stretching them out means increasing pitch.

23. May 4, 2017

### Staff: Mentor

Um, what? I have no idea what you're talking about here.

Part of the problem might be the term "pitch". What are you using it to mean? The standard definition is the height of one complete turn, measured parallel to the axis. So twisting the helix "the exact magnitude of its pitch, but in the opposite direction", if anything, means squeezing all the turns into the same plane. That doesn't give you a line; it gives you a circle.

24. May 4, 2017

### AVentura

if the cylinder has a helix on it with one complete clockwise turn, and you grab the ends and twist it counterclockwise one turn, you get a straight line.

more importantly, if you turn it any, it will no longer have the same integer number of turns on the surface

25. May 4, 2017

### Staff: Mentor

I see what you are describing now, but it is not what a Lorentz transformation does. I think we are at the point where we really need to look at the math. I don't have time to try it now, but I strongly advise you to do it before posting further.