# Spinning stick

1. Aug 10, 2009

### PaulRacer

If there was a stick the length of the diameter of earth spinning in space for the same amount of time at the same rate as earth, would it appear as a spiral?

2. Aug 10, 2009

### diazona

It would be a spiral of sorts in spacetime (actually more like a piece of rotini with a much looser spiral). But you couldn't actually see it like that without somehow getting "outside of space." Physically, it would always just look like a stick.

3. Aug 11, 2009

### Fredrik

Staff Emeritus
Great, now you made me hungry.

4. Aug 11, 2009

### PaulRacer

If the stick suddenly stopped spinning, would it take the same amount of time for it to "straighten out" as it took to "spiral" it or would it take a counter rotation to "unwind" it? None of the above?

5. Aug 12, 2009

### Fredrik

Staff Emeritus
I assume that we're starting and stopping the rotation by applying torque to points near the center of the stick. Hard to say if it would take longer to go from straight to the final weird shape, or from that weird shape back to straight, since the stick would oscillate around its equilibrium shape for much longer than it takes the effect of what we do at the center to propagate to the edges. The propagation speed of that effect should however be roughly the same in both cases, but not exactly the same, since the speed of sound in a material changes when the material is forcefully stretched. (I have never read anything about that last part, but it seems like a reasonable assumption).

6. Aug 12, 2009

### PaulRacer

Is it possible that there is any stored energy in this "twist" from the accumulated inertia?

7. Aug 12, 2009

### Staff: Mentor

From a top-down view of our solar system, the path traced-out by the stick would look like a spring laid on its side and crushed.
There is no such thing as "accumulated inertia", there is only inertia. Perhaps you are thinking of momentum, which is inertia times speed (in rotation, moment of inertia times rate of rotation). Or for energy, rotational kinetic energy.

...This doesn't appear to me to have anything to do with Relativity.

Last edited: Aug 12, 2009
8. Aug 13, 2009

### Fredrik

Staff Emeritus
I just realized that there's something about this that I don't quite understand. Suppose we take a very rigid rod (a rod made of diamond or something (yes, I know that there are no truly rigid objects in relativity)) and start spinning it very slowly. We do this by applying a small torque to a segment around the center. Let's say that we do it so slowly that it takes a million years to reach the angular velocity (at the center) of one radian per million years, and if there are still vibrations going on in the rod as a result of this, we'll wait another million years, or a billion years if we have to, for the vibrations to die down. It's clear that the rod can't be perfectly straight after this, but I don't see what exactly has caused the deformation.

I'll probably figure it out soon enough, but I wouldn't mind having someone else just tell me the answer.

9. Aug 13, 2009

### Staff: Mentor

After there is no more torque applied and the vibrations have been dissipated by damping, why can't the rod be perfectly straight?

10. Aug 13, 2009

### PaulRacer

Is this a rhetorical question? I know very little about physics but I am trying to understand relativity. All I have is speculation if it's not a rhetorical question.

11. Aug 13, 2009

### PaulRacer

Sorry, I was trying to build to something without using only my assumptions. I appologize for the terminology. I meant the possible accumulation of energy through the spinning of an object for a long period of time and how this relates to angular velocities and related spacetime implications. It just seemed possible to me that if the spacetime in the center of a rotating object was different than the ends then there would be a lag between the matter at the ends and the matter at the center resulting in a spacetime "tug". I have no idea so I thought I would ask physicists like you.

12. Aug 13, 2009

### Staff: Mentor

That question was for Frederick - I don't see why he thinks there is a twist.

13. Aug 13, 2009

### Staff: Mentor

I'm an engineer, but in any case, Newton's first law applies. There is no accumulation of or dissipation of energy in a constantly moving object.

14. Aug 14, 2009

### Fredrik

Staff Emeritus
I could be wrong about the twist. I'm imagining that we apply this procedure over and over again to increase the angular velocity more and more, as if we're naively trying to get the edges to eventually move faster than c. There seem to be two possibilities.

a) One is that the rod stays straight. In this case, the length of the rod sets an upper bound for the angular velocity. (The speed of the edges must be <c). The problem I have with this option is that when the rod is close to its upper bound angular velocity, we should still be able to apply more torque to the center to get the center parts to move faster, and now the rod has to bend and stay bent (or broken) after all the vibrations have died down. Nothing in relativity behaves this way. Light speed isn't a "barrier" that we suddenly "hit". All the effects that are large near the speed of light are present at small velocities too.

b) The other possibility is that the edges lag behind the parts near the center even at low angular velocities and, as the internal forces try to restore the length of each atomic-scale segment, the edges get pulled closer to the center, ensuring that they won't move faster than c, no matter how fast the parts at the center are moving. In this case, the distance between the edges and the center goes to zero as the speed of the parts near the center goes to c, so the rod wraps itself around the center point. A real rod would of course break long before that happens.

I guess b) is a partial answer to the question I asked yesterday. When we apply torque to the part near the center, the rod is stretched, and whatever shape it finally settles down to will be the result of internal forces trying to restore the distance between nearby atoms to what it was before...plus a correction due to the centrifugal force. It looks like it would be hard to actually calculate the shape of the rod.

By the way, if there's a grammar nazi present, please explain to me if it's "there seem to be two" (131K Google hits) or "there seems to be two" (83K Google hits, and one of the first is an advanced linguistics document).

Last edited: Aug 14, 2009
15. Aug 14, 2009

### Fredrik

Staff Emeritus
The rod will be heated a bit when we apply the torque to the center, but we'll wait long enough for the rod to cool down again.

The problem is hard enough in Minkowski space. I'd rather not worry about spacetime curvature and that sort of thing.

16. Aug 14, 2009

### DrGreg

Fredrik, I think Russ was confused because earlier, I think, this thread was talking of applying a torque to set the rod in motion and then releasing the torque to let it continue spinning. Whereas you are talking about applying a continuous torque forever. Undoubtedly the rod would have to twist into some sort of spiral shape (provided it doesn't break into pieces) as that's really the only way the transverse accelerating force can be transmitted to the ends of the rod. By how much the "effective radius" shrinks would depend on how the rod deforms transversally and stretches longitudinally under this sort of tension.

Without doing any sort of calculation, applying constant torque would have to result in an ever decreasing angular acceleration, even near the centre, perhaps leading to some finite limit of angular velocity (as $t \rightarrow \infty$)? That is, it needn't necessarily have to collapse to the centre. It just converges (as $t \rightarrow \infty$) to some spiral shape with radius r angular velocity c/r and infinite angular momentum and rotational energy. The value of r (compared with the initial length of the rod) would depend on the structural properties of the rod, and I'm suggesting it needn't be zero. (In fact it couldn't be zero unless a black hole formed!)

17. Aug 14, 2009

### Fredrik

Staff Emeritus
Actually, what I had in mind when I wrote #8 is that we apply a torque for a while and then let it go. It seems to me that the rod will still end up being twisted, after all the vibrations have died down (for the reasons I tried to explain in #14, where I did talk about repeating the procedure forever, or at least long enough for some of the relevant speeds to get close to c).

18. Aug 14, 2009

### DrGreg

I see what you mean now. When you release the torque, it ought to (eventually) untwist back into a straight rod, but you're saying it can't because of the speed of light restriction?

But as it starts to untwist, there will be an outward shift of mass, and by conservation of angular momentum, a decrease in angular velocity, and so the radius can increase without the linear speed of the rod-ends increasing beyond c.

A straight rod whose ends were moving at c would have infinite angular momentum, so any shape with expanding radius but finite angular momentum will just reduce its angular velocity to maintain its finite angular momentum. (As any ice-skater or ballerina knows.)

So I see no reason why the rod shouldn't straighten out completely if you wait long enough. It will just rotate more slowly than you'd expect without relativity.

19. Aug 14, 2009

### Staff: Mentor

If you have the rod spinning very fast and you apply more torque to the center, it will accelerate more and twist while it is accelerating. Then when you release the torque it will stop accelerating and straighten out. This has nothing whatsoever to do with Relativity. It doesn't matter how fast or slow the rod is rotating, it will always be true.

Perhaps what is confusing you is the idea that maybe the center is rotating at a higher angular rate than the edges? Well the way that has to resolve is that when you release the torque, the edges continue to accelerate and the center decelerates until it straightens out.
I don't understand what you're saying here. When a rod bends, the end will be at a shorter distance from the center than when it is straight, which changes the angular momentum. Is that what you're getting at? Again, this has nothing to do with relativity: When the torque is released and the rod straightens, angular momentum will be conserved and the rod will be spinning at whatever rate conservation of momentum requires.

If you had a very flexible rod and accelerate it very fast, it might coil into a tight spiral at a very high rpm. But when you release it and it straightens out, it would be spinning at a much lower rpm due to conservation of momentum.
No, it isn't hard. If there is no torque, there is no force perpendicular to the rod, so the rod must be straight. Centrifugal force will stretch the rod a little, but that effect probably isn't that tough to calcluate.

So may be the simplest way around this is to ask this: if you think the rod is bent when spinning freely, what is the nature of the bending force? Where is it applied an how? Keep in mind that the only way to bend the rod is to apply a force perpendicular to the rod and that means you are applying a torque!

20. Aug 14, 2009

### Fredrik

Staff Emeritus
Hm, OK I think I get it. If we take a (straight) rod that's already spinning so fast that the edges are moving at 0.99c, and we apply more torque for a while to get the center spinning faster and then wait until the vibrations have died down, the edges will be moving at (say) 0.999c, the center will have slowed down to the same angular velocity as the (new) angular velocity of the edges, and the rod is straight again.

Suppose that we now apply more torque to get the center (say the first meter) to rotate at a 1% greater angular velocity, and then force the center to rotate with that angular velocity forever. (Note that the acceleration period is rather short, but then the center is held at a fixed angular velocity forever...and yes, this is a different scenario than the one I've been considering until now). In this case the rod can't straighten out. If it could, the edges would be moving faster than c. It also can't settle down to one specific shape. If it did, then it would have a constant angular momentum. But it can't have a constant angular momentum, because we're constantly applying a torque. To see why we must be applying a torque, just think about the fact that if we let the rod go, the center would slow down as the rod straightens out. So by keeping the center at a fixed angular velocity, we're doing work against the internal forces that are trying to straighten out the rod.

Since the rod can never settle for one specific shape, it must continue to get more and more deformed until it breaks.

21. Aug 14, 2009

### Staff: Mentor

Yes.
Yes, that sounds reasonable. The key there is while you aren't accelerating the midpoint of the rod anymore, you are accelerating the edge and so you are applying a torque and bending the rod. From the point of view of a person near the center of the rod, the outer edge wouldn't seem to be accelerating much, but in terms of the increase in momentum, it could still be accelerating quite a bit.

22. Aug 14, 2009

### PaulRacer

I was wondering more about a rod spinning in space like the earth and the warped spacetime within. What would happen after all of this spinning if the stick rapidly shortened itself to singularity like a black hole does?

23. Aug 14, 2009

### DrGreg

Now, instead of applying a constant torque, you're applying whatever variable torque is necessary for the centre to maintain a constant angular velocity $\omega$. Well, at first, yes, the rod will continue to deform (assuming it doesn't break). But as it deforms, the maximum radius of the whole deformed structure will shrink. If the radius ever shrunk below $c/\omega$, there would no longer be any relativistic objection to the tips moving with angular velocity $\omega$, and then zero torque would be required to maintain the motion. But without a torque the rod would straighten out again. So I guess what happens is the radius tends to $c/\omega$ as $t \rightarrow \infty$. Does that sound plausible?

24. Aug 15, 2009

### PaulRacer

The only torque I was talking about was to get the rod spinning in space. The speed of light is obviously a brake of sorts but only because time slows at such a rate at high velocities that the speed of light at the edges relative to the center is multiplied. This is due to the fact that (if I understand it correctly) time would theoretically stop at the speed of light. Therefore the speed of light would become infinite if you could travel at c because speed is measured in units per hr./min./sec. If this is the case, the rod wouldn't need torque to deform, only to be spinning for a long period of time because of warped spacetime due to angular velocities. Once the spacetime is warped and the rod is twisted like a spiral, how could it possibly straighten if time was equalized throughout the rod by stopping the spin? If you stopped the spinning, more time would have still passed relatively in the center due to slowed time at the higher velocities near the ends. Seems to me that the only way to straighten it is a counter rotation for the identical amount of time at the identical velocity. If this sounds somewhat reasonable I have more questions. If not, I guess I have many more questions.

25. Aug 15, 2009

### Fredrik

Staff Emeritus
What do you mean? It sounds like you answered the question in the question. ("If I go to New York, what city will I be in?" ) Only a black hole would remain. Besides, if the rod is massive enough to form a black hole, I think that would happen long before we can get it spinning that fast.

I don't see a way to derive the results we're talking about from time dilation alone, and the last part of that sentence is wrong (if I have understood what you're saying).

You shouldn't say things like that. You can talk about what happens to other variables in the limit v→c, but there's no such thing as "at the speed of light". Massive particle's can't even move that fast. Massless particles can (they can't move at any other speed), but there's no natural way to associate an inertial frame with their motion. See my posts in this thread for more on that.

c doesn't go to infinity in the limit v→c.

I don't really follow your reasoning, but you seem to be confusing GR stuff with SR stuff. There is no gravity in SR, and therefore no curvature, but the rod still gets twisted due to the finite propagation speed of the deformation we're doing to the center. This doesn't really have anything to do with relativity. To see a truly relativistic effect, we must do something special. We can e.g. keep applying torque to the center to keep it rotating at an angular velocity that's too fast for the edges to keep up. (See my last post before this one). There's nothing in pre-relativistic physics that would prevent this rod from straightening itself out, but SR says that the rod will continue to get more and more deformed.

A general relativistic treatment of this problem would be extremely complicated. The rod's mass would curve spacetime, making it bend under its own weight (if it's large enough), and the rotation of this massive object with a weird shape would give a very complicated contribution to spacetime curvature. I don't think that matters much though, because either the rod is massive enough to form a black hole before we can even get it spinning fast, or it doesn't have enough mass for its contribution to spacetime curvature to change the results much. If we apply more and more torque to the rod, it will break under the centrifugal forces long before its mass and motion gives us a large contribution to spacetime curvature.