# Rotating rigid disk

Discussion in 'Physics & Math' started by IggDawg, Jan 21, 2003.

1. ### IggDawgRegistered Senior Member

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Someone help. This is one of those precious few things I can't seem to get over. I was reading The Elegant Universe a while back, and this issue is still bugging me. he kinda left it hanging.

Here's the issue: Let's say you (Bob) and a friend (Mary) are on a merry-go-round. A third friend (Dave) watches from off the merry-go-round. assume the merry-go-round is perfectly rigid.

Initially, everyone takes measurements. Bob measures the cicumfrence from on the merry-go-round by laying a ruler down head to tail several times. He measures the total length as X. Mary measures the radius. she measures it as she measures is as X/(2*pi*r). Dave measures the cicumfrence from off the merry-go-round ("disk" hereafter) using the same head-to-tail method Bob used. he gets X as the cicumfrence.

Spin the disk at a speed such that special relativity comes into play... let's say .3c (assume they don't die for the sake of the problem

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). Everyone takes measurements again. here's where it gets funky.

a) Mary measures the radius. Because she is not measuring along the direction of travel, her ruler is not contracted. she measures X/(2*pi*r) as she did when the disk was at rest.

b) Dave measures the disk from a stationary position. he goes head-to-tail with the ruler very close to the spinning disk. he measures the same cicumfrence, X when he comes back to his starting point.

c) Bob measures the disk from onboard. what happens here? the circumfrence of the disk should be shorter because of length contraction, as should Bob's ruler. so one would assume Bob would measure X as the circumfrence since both his ruler and the disk would be contracted by the same amount due to SR. But Dave measured the circumfrence as X with his ruler which is NOT contracted. and the circumfrence is not effected. the disk would still appear from the air to have a circumfrence of X and a radius proportional to X. so what's up?

-IggDawg

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3. ### CrispGone 4everRegistered Senior Member

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1,339
Hi IggDawg,

Some points:
- There is no perfectly rigid object.
- SR does not apply here, rotating frames are not inertial.

Bye!

Crisp

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5. ### chrootCrackpot killerRegistered Senior Member

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2,350
This is why I advise people to stay away from Hawking's books.

Hopefully she'll be measuring r = X / (2 pi), not X / (2 pi r). In any event, where is she? At the hub? Along the circumference? The bottom line is that the entire disk is corotating. There are no relative velocities on the surface of the disk. The circumference is turning faster than the center, but those are all lateral velocities, at right angles to any measurement, and thus don't affect those measurements. For Bob and Mary, the disc looks the same no matter what the speed (except for, of course, the fictitious inertial foces -- centrifugal force -- that they feel when it's spinning).
Well, if Dave is looking along the plane of the disc, he'll see it foreshortened in the direction he's looking. It's hard to imagine how he'd measure the local circumference of a disc when he is by definition a distant observer. In any event, if he does something like measuring the circumference by looking at the disc's angular velocity per unit time, given its radius, he will always arrive at the same circumference measurement. The truth is that the disc is foreshortened into an ellipse.
Perhaps this is just a rehash of a). On the disc, no observer will notice anything changing. Outside the disc, the observer will measure different quantities depending upon the inclination of the disc to his line of sight.

- Warren

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7. ### letheRegistered Senior Member

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Re: Re: Rotating rigid disk

and greene

8. ### IggDawgRegistered Senior Member

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I read popsci books to get the general theory on areas I want to get into, so I have an idea how to visualize what I'm going to be studying. I always come out of it more confused than anything else

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. is there any good way to break into string theory without going the popsci route? I'm reading supersymmetry by Kane. is that any good? I know my stuff for the most part, but I have issues opening new areas if I can't visualize them first.

-IggDawg

PS - I'm working on a response to your replies. thanks for answering so quickly.

9. ### chrootCrackpot killerRegistered Senior Member

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2,350
No, there's no way to "break into" string theory. This is equivalent to asking:

Is there any way to break into playing violin in the London Orchestra?

The answer is, that, no -- it takes years of hard work. Start with the basics and work your way up.

- Warren

10. ### James RJust this guy, you know?Staff Member

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37,772
The rotating disk thing is difficult.

As the disk rotates, outside observers see the outer edges of the disk moving faster than the parts closer to the centre. Relativistically, the outer edges must undergo length contraction to a greater extent than the inner regions. The radius remains unchanged.

If the edges are to contract whilst the radius stays constant, then it is not longer true that the circumference is given by 2 pi r. In fact, c < 2 pi r. How can that be? Remembr we're measuring on the material of the disk. If we measure from just above it (ignoring the spinning disk below) we get the usual result, so something physically happens to the disk as it rotates.

What happens is that the edges of the disk try to "separate". The usual forces which normally hold the disk together fight this. As the disk rotates, tension develops in the material. In practice, the material of the disk will deform to "fill" the gaps which would be created. i.e. the disk "thins out" at the edges.

What of an infinitely-rigid disk? No thinning out is possible; therefore we conclude that an infinitely rigid disk cannot rotate. Alternatively, we assert that infinitely rigid disks can't exist.

11. ### (Q)Encephaloid MartiniValued Senior Member

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20,855
In response to Born's rigid disc:

SR

we see that infinitesimally close particles must keep the same proper distance. So in the original rest frame, they suffer Lorentz contraction in the transverse direction but none in the radial direction. The circumference contracts but the radius doesn't.

Ehrenfest

GR

When "spun up" wouldn't the centripetal force cause the particles to contract in the radial direction and not in the transverse - acceleration is directed towards the center of the disc ?

12. ### (Q)Encephaloid MartiniValued Senior Member

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20,855
James, Chroot, Lethe

Perhaps I'm missing something here - what is is ?

13. ### James RJust this guy, you know?Staff Member

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(Q):

Centripetal forces are provided by electromagnetic forces which hold the disk together. They do not compress the disk when it is spun, but simply keep the disk together. If the disk is spun too fast, it will break apart.

This has nothing to do with relativistic length contraction.

14. ### (Q)Encephaloid MartiniValued Senior Member

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20,855
James

Thanks. Sorry, I wasn't very clear. I didn't mean relativistic length contraction. I was reading this paper and began thinking about what happens to the particles in the disc under constant acceleration towards the center of the disc.

What you're saying is that the disc will fly apart before any accelerated effects take place. Correct ?

http://www34.homepage.villanova.edu/robert.jantzen/research/articles/bjdfproc.pdf

15. ### James RJust this guy, you know?Staff Member

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What you're saying is that the disc will fly apart before any accelerated effects take place. Correct ?

No. I'm not saying that.

What accelerated effects do you expect?

(I had a quick look at the paper you linked, but I didn't really understand it.)

16. ### (Q)Encephaloid MartiniValued Senior Member

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20,855
In the spinning disc there is a constant acceleration towards the center of the disc. Particles in the disc would appear to be elongated in the direction of acceleration (radial) according to GR.

In the explanation I quoted above, Ehrenfest claims that the particles appear contracted in the transverse direction (SR) and NOT the radial. The circumference contracts but the radial does not.

You also said, "Relativistically, the outer edges must undergo length contraction to a greater extent than the inner regions. The radius remains unchanged."

I keep thinking that the radial WILL be affected, it will elongate somewhat under this constant acceleration.

As well, if we assume the radial a straight line when the disc is at rest, it will appear curved AND elongated when 'spun up.'

As you stated, "As the disk rotates, outside observers see the outer edges of the disk moving faster than the parts closer to the centre." Therefore, the particles nearer the center appear to be accelerating more than the particles on the outside edge. So the radial should appear curved and elongated.

Most likely I am missing something here - I just can't put my finger on it.