# Large rotating disk.

1. Aug 25, 2011

### soothsayer

Imagine we have a spinning disk. Points near the center of the disk rotate slowly while those increasingly far out are moving faster. Now suppose we have a sufficiently large disk such that the points on the edge of it are moving relativistically, and would classically be reaching or surpassing the speed of light. How does SR reconcile this?

2. Aug 25, 2011

### Fredrik

Staff Emeritus
You obviously can't build a disk that's already rotating that fast, so you'd have to build a stationary disk, or one that's rotating slowly, and then try to give it a spin. If you use SR to calculate the work it would take to give the edge speed v, you would find that the result goes to infinity as v goes to c.

3. Aug 25, 2011

### Staff: Mentor

Assuming the disk is rigid enough so it can withstand the internal stresses, it has a limiting maximum angular speed:

$$\omega_{max} = \frac {c}{R}$$

where R is the radius of the disk. It can never actually reach this angular speed. As you make it spin faster and faster, the work you need to do increases without limit as Fredrik noted, and the internal stresses increase without limit. This is the ideal case of course. In practice the internal stresses tear the disk apart at some point.

4. Aug 25, 2011

### soothsayer

Fredrik, I understand it's impossible to get the disk to rotate at speed c, what I'm asking though, is that after some non-infinite work, eventually, the disk would reach a relativistic speed, like 0.2 c? or even 0.9 c. The outer points move relativistically while the inner points are not. So how would a phenomena like length contraction work when the disk is supposed to be a rigid object? I realize this is all theoretical in the first place.

jtbell, thanks for that, that explains a lot!

5. Aug 25, 2011

### Ben Niehoff

The answer is that the idea of a "rigid object" is inconsistent with relativity. Or to put it another way, there are no such things as rigid objects in relativity, even in theory.

6. Aug 25, 2011

### Staff: Mentor

Consider a linear length-contraction scenario first, for simplicity. A rod with rest-length 10 m traveling at speed 0.6c, normally gets contracted to 8 m, right?

Suppose the rod is made of rubber, and is being carried by two runners, one holding each end. As they run, each runner pulls on his end of the rod so as to stretch it out to 12.5 m in their reference frame. The rod now measures 10 m in the "stationary" frame, the same as its original rest-length. According to a stationary observer the length of the rod has not changed, but if he can inspect it carefully as it flies past, he might be able to observe signs of the internal strains in the rod (little cracks in the surface, or whatever).

A ring-shaped section of a rotating disk similarly has its moving "length" (circumference) kept at its original "rest length" by internal strains inside the disk, until the rotational speed becomes large enough that this is no longer possible, for the particular material the disk is made of.

7. Aug 25, 2011

### soothsayer

Ok, thanks Ben! I was actually just perusing wikipedia and came across the article on the Ehrenfest Paradox, which I've been reading up on and it's shed some light, though some of the mathematical/physical concepts used are a bit out of my league.

EDIT: ...Though it seems like actual physicists had enough problem with this paradox as it is, so I don't feel too bad ;)

Also, thank you to jtbell too.

8. Aug 25, 2011

### phinds

OK, but what is the implication of that? Does it mean the disk would come apart before reaching any significant portion of c ?

9. Aug 25, 2011

### PAllen

Of course. Let's do a simple calculation. Suppose a disk of 10 cm. Suppose we hope to spin it at .1 c. The acceleration at the rim would be 1000 trillion times the surface gravity. Way before relativistic corrections come into play, you reach fundamental material limits.

10. Aug 25, 2011

### phinds

Good. I assumed that to be the case but wasn't sure what the math might show. Thanks.

11. Aug 25, 2011

### HallsofIvy

In fact, if a disk is rotating relativistically (strictly speaking " rotate at speed c" doesn't even makes sense. c is a linear speed, not angular rotation. It is impossible to have a disk rotating a sufficient angular speed that points on the circumference move at c.), then the circumference contracts while the radius (perpendicular to the direction of motion) does not. Result- the geometry of a rotating disk is NOT Euclidean.

12. Aug 25, 2011

### Fredrik

Staff Emeritus
This depends on what exactly you mean by "the (spatial) geometry of a rotating disk". I think it's far from obvious how that should be defined. The mathematical object that has a non-Euclidean geometry in this scenario is a quotient manifold. It's a 3-dimensional manifold where each point corresponds to a world line of a particle component of the disk. If we define "the geometry of the disk" to be that manifold, then sure, it's not Euclidean. But in all these threads about this subject, I haven't seen a reason to think that this manifold is interesting other than as a mathematical curiosity. So I don't know why we would want to define that term that way. (There might be a good reason. I'm just saying that I haven't seen one).

I we instead define "space" as a hypersurface of simultaneity of a coordinate system, we don't get a non-Euclidean space in this scenario. The coordinate system that's rotating with the disk has the same hypersurfaces of simultaneity as the inertial coordinate system where the center of the disk is at rest.

13. Aug 26, 2011

### PAllen

Maybe another way to put this would be that the spacetime here is assumed to effectively flat, but a flat spacetime cannot accommodate a rigid rapidly spinning disc: For each atom to maintain equal distance to its near comoving neighbors requires the circumference must be less than 2 pi r (from the point of view of an inertial observer). This is impossible in flat spacetime, therefore a rigid, relativistically spinning disc is impossible.

14. Aug 26, 2011

### harrylin

Perhaps it's less confusing to simply give the solution that Lorentz derived (Nature 1921) for a rotating disk:

with v= velocity at the rim, the contraction (of radius and circumference) is 1/4 of the standard Lorentz contraction factor.

And of course, for a thin rim the contraction would be nearly equal to the standard factor; and all without accounting for inertia.

Harald

15. Aug 26, 2011

### PAllen

I would like to clarify and weaken this comment. It is perfectly possible to posit a steady state of 'atom world lines' representing a disc whose rim atoms are moving anything less than c. Forgetting physical plausibility, a la Born rigidity, we assume each atom is self propelled with whatever arbitrarily large proper acceleration is needed to follow a chosen path. Then the only issue is you can choose either of the following:

1) If the world lines have isotropic separation measured by an inertial observer stationary with respect to the disc center, the distribution looks increasing non-isotropic to atoms approaching the rim.

2) If the world lines are chosen so every atom sees its nearest neighbors having isotropic distances, then, to the inertial observer, the distribution is increasingly non-isotropic toward the rim.

but you can't have isotropy from both points of view. Other than the asymptotically infinite force required to hold each atom on the chosen path, either steady state is mathematically consistent with relativity.

16. May 15, 2012

### kapildverma

I was also think of this since last few days during my vacation time. Lets assume that we have two discs close to each other rotating in opposite direction such that the edges of each of those are at 0.5c.

Will the two observers standing on the edges of the two discs see each other moving at c?

17. May 15, 2012

### A.T.

The spatial geometry of a rotating disk is measured by rulers at rest relative to the disc.

See my comments in this post:
I don't know why we would want to define spatial geometry that way in this case. Simultaneity is irrelevant here because you don't need clocks to measure constant distances, just rulers.

Last edited: May 15, 2012
18. May 15, 2012

### Staff: Mentor

As they pass each other, each one will see the other going past at a speed < c which can be calculated from the usual relativistic "velocity addition" equation. It's no different from having two spaceships traveling in opposite directions in a straight line.

19. May 15, 2012

### phinds

No. velocities do not add that way at relativistic speeds.

20. May 15, 2012

### yuiop

No they will each see the other observer moving at (0.5+0.5)/(1-0.5*0.5) = 0.8c.

21. May 15, 2012

### yuiop

Let's say we have metal hoop that is somehow spun up to a rim speed proportional to a gamma factor of 10. We have two observers, One is in a non rotating inertial reference frame in which the centre of the hoop is at rest. Call this frame S. The other is on the hoop and is in a rotating non inertial reference frame. Call this frame S'. Irrespective of whether the hoop expands or contracts as measured in either frame, an observer in frame S will always measure the circumference to be equal to 2*pi*r and the observer in frame S' will always measure the circumference to be equal to 20*pi*r. No observer sees the circumference contract relative to the radius.

In practice, for almost all known materials, a disc, hoop, cylinder or sphere will self destruct under rapid rotation long before relativistic effects become significant. Large massive bodies bound together by gravity may fare a little better.

22. May 15, 2012

### Staff: Mentor

Which coordinate system are you referring to as "the coordinate system that's rotating with the disk", and how are you determining its hypersurfaces of simultaneity? I take it you are *not* using Einstein clock synchronization; if you do that between two observers rotating with the disk, the hypersurfaces of simultaneity you obtain will not be the same ones as you would get by doing Einstein clock synchronization with two observers at the center of the disk and not rotating with it.

23. May 15, 2012

### Fredrik

Staff Emeritus
I just find it very strange to use the term "spatial geometry" for the geometry of a quotient manifold, instead of for the geometry of "space". To me, the term "space" can only refer to a subset of spacetime. The quotient manifold that has this "spatial geometry" isn't a subset of spacetime.

Maybe that manifold is useful somehow, and in that case I guess we should be talking about it, but why not call its metric something like "the ruler metric" instead of "the spatial metric"? It's not the metric of anything I would be comfortable calling "space".

I'm talking about a rotating coordinate system, e.g. the one defined by the following transformation from an inertial coordinate system in which the entire disk is at rest before it's given a spin.
$$\begin{pmatrix}t'\\ x'\\ y'\\ z' \end{pmatrix}=\begin{pmatrix}1 & 0 & 0 & 0\\ 0 & \cos\omega t & -\sin\omega t & 0\\ 0 & \sin\omega t & \cos\omega t & 0\\ 0 & 0 & 0 & 1\end{pmatrix}\begin{pmatrix}t\\ x\\ y\\ z \end{pmatrix}$$

By noting that by definition, t'=t.

A "hypersurface of simultaneity" is defined by a coordinate system. The mathematical object that describes the local experiences of each point of the disk is a frame field, not a coordinate system. I don't think this particular frame field determines anything that can be called a hypersurface of simultaneity.

Instead, the congruence of world lines of component parts of the disk determines a frame field, and the world lines are the points of a quotient manifold with a funny geometry. Since this manifold isn't a subset of spacetime, I find it very odd that people want to call it "space".

Last edited: May 16, 2012
24. May 15, 2012

### Staff: Mentor

Are these basically the Cartesian version of Born coordinates? See the Wiki page here:

http://en.wikipedia.org/wiki/Born_coordinates

Ah, got it. My terminology was confused.

Not a single one that is common to different points in the frame field, no. At each point of the frame field, the spatial vectors define a spacelike "hypersurface of simultaneity" (by means of the Einstein simultaneity convention) for that point, but it's a different hypersurface for each point.

25. May 16, 2012

### A.T.

That's the same. Rulers laid out at rest are measuring the spatial metric.
Well, then the concept of hypersurfaces of simultaneity is not usefull here, and appraently not generally applicable. Maybe it is better to use a more general one. From the other thread:

Last edited: May 16, 2012