# I Relativistic angular velocity

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1. Aug 6, 2017

### Tahmeed

Let's assume that a disk is rotating with relativistic speed in a frame. We can find the velocity of a particle using v=rw formula. However, what is the r in this formula? is it the radius of the disk in rest frame or in the lab frame??
And Is the magnitude of velocity same for all points of the disk that are situated at the same distance from the center of the disk?

2. Aug 6, 2017

### Ibix

By "in a frame" I take it you mean you are considering the rest frame of the wheel's centre as the "lab" frame. In that lab frame, think about one revolution. How far has a point on the disc travelled? What time did it take?

Again in the lab frame, what would be the implications if the speed changed around the disc? You might like to think about angular momentum and density.

3. Aug 7, 2017

### A.T.

Why would those be different?

4. Aug 7, 2017

### Tahmeed

Lorentz contraction?

5. Aug 7, 2017

### Staff: Mentor

Using the lab frame, the radius is perpendicular to the direction of motion everywhere.... So no length contraction of the radius.

6. Aug 7, 2017

### Bartolomeo

Imagine that a spool rotates with great angular velocity. An observer is at some distance from the plane in which the spool rotates. A wire is approaching along the tangent to the point of the rim. Speed of the wire is equal to the linear velocity of the point on the rim. The beginning of the wire touches the rim and the spool winds the wire on itself. The length $L$ of the wire is numerically equal to the length of the rim $L = 2 \pi r$. The spool does a complete turn and winds the entire wire around so that the beginning of the wire can touch its end. Then the spool slows the rotation to a full stop. If the radius remains unchanged, I wonder what happens to the wire? Generally speaking, when the spool stops, the length of the wire becomes equal to its proper length. Is it going to fall off the spool?

Last edited: Aug 7, 2017
7. Aug 7, 2017

### Ibix

The contracted length of the wire is equal to $2\pi r$, yes. Thus when it slows down the wire will be longer than the rim of the disc and will fall off. This is true even assuming (unrealistically) that the disc does not undergo elastic expansion when spinning.

If you imagine making many radial cuts into the disc (so it looks like a jet turbine), you would find that these cuts open up as the disc accelerates, since each blade of the "turbine" would length contract in the tangential direction. If you don't make cuts, the whole rim of the disc must "want" to length contract - but is unable to do so due to the symmetry of the situation. So you will find elastic stresses in the tangential direction.

Presumably a spinning hoop would try to decrease its radius due to Lorentz contraction (massively overwhelmed by elastic expansion and outright disintegration in practice, I should imagine).

8. Aug 7, 2017

### MikeLizzi

Uh?
Consider a disk at rest with respect to an inertial observer at its center. Then transform to a rotating reference frame still at the center of the disk.
There is no relativistic geometry change.

Note: The common formula for relativistic length contraction was derived for the condition where all points for the object in question had the same relative velocity.
No two atoms on a rotating disk have the same relative velocity. Therefore the common formula for length contraction does not apply. Try transforming all the atoms on the disk independently. As Nugatory pointed out the position vector of every single atom is always perpendicular to its velocity vector. All transformations are unity. (t, x, y) transforms to (t', x', y') where t = t' and x = x' and y = y'. The disk maintains its Newtonian geometry.

Think about a moving (length contracted) wire approaching the rim of a spinning disk. Whatever portion of the wire that you try to spool onto the disk immediately looses it length contraction. In other words you can't get it to spool at all.

9. Aug 7, 2017

### Ibix

Imagine a disc the radius of the Earth, spinning so that its rim is doing 0.866c - i.e. $\gamma=2$. You seem to be proposing that a 1m rule on the rim of the disc (pinned on with one nail at its centre) would have a length of 1m because it has "lost all length contraction" while an identical ruler floating past freely and instantaneously at rest with respect to the ruler on the disc would be length contracted by a factor of 2. And the same if the disc is one light year in radius, with the same 0.866c linear velocity at the rim. Or ten light years.

The states of motion of the ruler on the disc and the free-floating one are always instantaneously identical, and become indistinguishable for all practical purposes for longer and longer as the radius is increased. Yet you seem to be claiming that they would be a constant factor of two difference in their lengths.

10. Aug 7, 2017

### Staff: Mentor

Which geometry do you mean?

You can't change the geometry of spacetime by changing coordinates. So if you mean spacetime geometry, then yes, of course it doesn't change.

But you can change the geometry of space by changing coordinates, because different coordinates can split spacetime into "space" and "time" in different ways. Or, more subtly, the physical meaning of "space" can change even if the space-time split remains the same.

What geometry are you referring to? There is no single answer to what "the geometry of the disk" is. I suggest some reading up on the Ehrenfest paradox and Born coordinates. We have had some previous discussions of this here at PF; see, for example, this post (and the associated discussion in the thread):

11. Aug 7, 2017

### MikeLizzi

I was referring to the geometry of the disk as defined in newtonian geometry books. If I specify a disk of 1m diameter and you draw a circle on a piece of paper and labeled the diameter 1m, I would say you have drawn the correct geometry. If I specify a 1m square moving with x-velocity of .867c and you draw a rectangle with x dimension labeled .5m and y dimension labeled 1m I would say you have drawn the correct geometry. Wouldn't you?

I am familiar with the arguments with regard to Born Coordinates. That's why I presented the case of transforming to a rotating reference frame rather than spinning up the disk and having to consider centripetal forces.

12. Aug 7, 2017

### Staff: Mentor

We're talking about relativity here, not Newtonian physics. Your examples don't involve the issues that come up in the case of a relativistic rotating disk.

Born coordinates are the "rotating reference frame" you are transforming to. But by themselves, they don't address the question of what the "geometry" of the disk is. So just saying "let's transform to a rotating frame" does not address the geometry issue.

13. Aug 7, 2017

### MikeLizzi

Yes, I am. That's the reason the Ehrenfest paradox isn't a paradox. I can only assume that Ehrenfest didn't know how to transform points very well because he and everyone who read his material assumed behavior that is not predicted by the Lorentz Transformation.

If a 1m rule is pinned to a rotating disk such that it rotates with the disk, then every atom in the rule transforms the same way (t, x, y) --> (t', x', y') and t = t' and x = x' and y = y' because the position vector to each atom is perpendicular to each atoms velocity vector.
Do you not see that?

P.S. I specifically described the condition of an inertial observer transforming to a rotating reference frame to that mechanical stresses and strains would not participate in any calculation.

14. Aug 7, 2017

### Ibix

No, I don't see that. I notice that you chose not to quote the part of my post where I pointed out the extreme oddity of your position, that two identical rulers in almost identical circumstances (circumstances which are asymptotic to identical as the radius grows, in fact) behave radically differently. Is that fact something that you accept as reasonable? Or am I misrepresenting your position?
Perhaps I'm confused, then. Is the disc spinning in your thought experiment or not? If it's spinning there are going to be stresses and strains. If it's not spinning (but only being observed by someone spinning) then obviously there are no relativistic effects.

15. Aug 7, 2017

### Staff: Mentor

First, this transformation you refer to is not the transformation to Born coordinates. It is a Lorentz transformation using a boost velocity $v$ equal to the velocity vector of some particular chosen atom, in an inertial frame in which the center of the disk is at rest.

The problem with your logic is that this transformation is different for every atom; so once you pick one particular transformation, which you have to (there's no such thing as a Lorentz transformation with a different relative velocity at every point), only the chosen atom whose velocity is the boost velocity of the transformation acts as you describe. The position vectors for all of the other atoms will not be perpendicular to the boost velocity, and so will not transform as you describe.

16. Aug 7, 2017

### Staff: Mentor

This is true, but it can be ignored in the context of this discussion. We can describe the rotating disk purely kinematically, in terms of a congruence of worldlines, without having to get into the details of how, physically, the individual atoms in the disk are made to follow those worldlines. All of the key issues with the "geometry" of the rotating disk can be raised with the kinematics alone.

17. Aug 7, 2017

### MikeLizzi

What I'm saying is documented. Can't find a reference right now.

18. Aug 7, 2017

### Staff: Mentor

Then please find one ASAP. Otherwise further posts along the lines you are posting will receive a warning.

19. Aug 7, 2017

### Ibix

I agree mathematically. But I found it helpful when someone on here pointed out that the resolution to the paradox was that, while length contraction means that an elementary spinning ring might "want" to have a smaller circumference than when it's not spinning, stresses (in this case from the incompressibility of the parts of the disc inside the ring) can deform it to have some other circumference.

Similarly, a 5cm elastic band moving at $\gamma=2$ would have an unstressed length-contracted length of 2.5cm, but could perfectly well be stretched by a factor of 2.

20. Aug 7, 2017

### Staff: Mentor

That's true, but it doesn't change the fact that the "geometry" of a rotating disk involves complications that the geometry of a non-rotating disk does not. The presence of stresses changes the relationship between, say, the circumference of the disk before it was spun up with the circumference after; but it doesn't change the fact that the circumference is not just $2 \pi$ times the radius in the simple sense that it is for the non-rotating disk. The latter fact is an example of what I mean by kinematics.

21. Aug 7, 2017

### MikeLizzi

Perhaps I did not explain my position accurately.
Below are 2 videos of a simulated car, one where the observer is at rest with respect to the road and one where the observer is at rest with respect to the car.
I have seen both versions on various web sites.
My versions were produced by my own relativistic physics engine. The only calculations that engine makes are Lorentz Transformations and Velocity Composition. (I don't know anything else)
Does anybody have an issue with either of these?

http://www.relativitysimulation.com/Documents/Car_from_Driver.webm

P.S. My understanding is this thread was talking about the second situation where the observer is at rest with respect to the car/wheel axle. In that case the wheel is circular.

22. Aug 7, 2017

### Staff: Mentor

These don't appear to be relevant for the discussion in this thread.

Nobody is disputing that in this thread. But you are talking about an inertial frame in which the center of the wheel is at rest. The discussion in this thread has also been about what things look like for an observer rotating with the rim of the wheel. Your simulation does not deal with that at all.

23. Aug 8, 2017

### Ibix

It's easy to show that clocks on the rim of a rotating disc tick slow by a factor of $\gamma$ compared to clocks at rest in the axle frame. You just observe that the line element in cylindrical polars is $c^2d\tau^2=c^2dt^2-(dr^2+r^2d\theta^2+dz^2)$ and that for a clock on the disc at radius r=R, dr=dz=0 and $d\theta=\omega dt$, immediately recovering $d\tau=\sqrt{1-(R\omega/c)^2}dt$. $R\omega$ is, of course, the linear speed of the clock.

Mount a laser on the rim of the disc and a mirror some angle $\theta_1$in the direction of rotation away. The laser points (nearly) tangentially, and we add mirrors to the rim of the disc so that the beam is reflected along (a polygonal approximation to) the rim until it meets the mirror at $\theta_1$, which returns it to its source by the same rim-tracing path. Add more mirrors to make the polygon a curve in the limit of infinite mirrors.

When the disc is at rest, a laser pulse returns in time $T=2R\theta_1/c$.

When the disc is rotating with angular velocity $\omega$ in the lab frame and assuming it is constrained not to have changed radius in that frame, let the laser pulse be emitted at time zero. It is easy to calculate when the pulse catches up with the mirror - it's when the distance travelled by the laser pulse ($ct$) is equal to the distance travelled by the mirror ($R\omega t$) plus its head start ($R\theta_1$). Let that be time $t_1$, satisfying $R\omega t_1+R\theta_1=ct_1$, which canbe solved to get $t_1=R\theta_1/(c-R\omega)$. The pulse returns to the laser at time $t_2$ which, by a similar argument, satisfies $(R\omega+c)(t_2-t_1)=R\theta_1$. We can eliminate $t_1$ to get $t_2=2R\theta_1 /c(1-(R\omega/c)^2)$.

That was in the lab frame. But we already established that a clock on the rim ticks slowly by a factor of $\gamma$, so the flight time of the pulse as measured by a clock attached to the laser is $t'_2=2R\theta_1/c\sqrt{1-(R\omega/c)^2}=\gamma T$. Assuming that an observer on the rim wishes to maintain that the round trip speed of light is c, he concludes that the distance to the target mirror has grown by a factor of $\gamma$. That is to say, if he lays a series of short rulers along the rim from laser to mirror, he can fit more of them ($\gamma$ times more) in when the disc is rotating than when it is not.

Conclusion: In the lab frame, rulers on the rim of the disc are shorter when it is rotating, by exactly the same factor as would be naively expected from the linear speed of the rim.

Last edited: Aug 8, 2017
24. Aug 8, 2017

### Laurie K

That is verified by the axle velocity between events in these 2 images from this thread. https://www.physicsforums.com/threads/rolling-rings-in-sr.920107/

The thread shows a solution for Grøn's Fig. 9 part C relativistic rolling ring problem. The points on the rim are considered to be in the same locations as the tips of length contracted wheel spokes. This solution removes Born rigidity concerns by making the wheel have zero thickness (z=0) while using a wheel frame an axle frame and a road frame (to plot the emission point locations).

A good reference if you want to dig further is Øyvind Grøn's paper "Space geometry in rotating reference frames: A historical appraisal".
http://areeweb.polito.it/ricerca/relgrav/solciclos/gron_d.pdf

25. Aug 8, 2017

### MikeLizzi

I have not compared the data you listed to that generated by my program, but the diagram looks consistent with my video of the Car with Observer on the road. Kudos.