Exploring Relativistic Spinning Wheels: Radius & Simultaneity Paradoxes

[michelcolman]

Suppose you make a big horizontal wheel, put markings at equal distances on the edge and also on the floor just outside the wheel. Now let the wheel spin at relativistic speeds.

If you are standing next to the wheel, the markings on the wheel will be closer together (because they are moving), so you would calculate the circumference to be less. After all, you can count the number of markings, and that is still the same. This means that, assuming the radius is unchanged, the circumference is no longer equal to 2 pi times the radius. How can that be explained?

Meanwhile on the disc, you would measure the circumference of the circle on the floor to be less than that of the disc, since the markers on the floor are closer together. How can that be? The enclosure would appear to be smaller than the disc?

And what about simultaneity? If you do some experiment sending laser beams around the circle and back, and basically checking to see if your own clock is simultaneous with itself seen from a distance, wouldn't you get into contradications with what the outside observer will say? Many straight line thought experiments are solved by considering there is no objective simultaneity, but if you go once around the disc, you have to agree on simultaneity. How do the paradoxes solve themselves then?

For example, what about the twin paradox? Both the observer on the disc, and the observer next to it, will see the other as moving, but there is no turnaround point to shift the reference frames to solve the paradox. Of course there's a different kind of assymmetry in the centripetal force, but how exactly does that solve the problem?

And do both observers agree on the radius of the disc, or does one see it as bigger than the other?

If you can either point me to a good article on this, or answer the questions yourself, I would be very grateful.

Thanks,

Michel Colman

 

[Dale]

How can that be explained?

By the simple fact that you cannot rigidly accelerate a disk in this manner. It must undergo some deformation in the process of changing its angular velocity.

 

[michelcolman]

I just stumbled upon the answer, I think.

http://math.ucr.edu/home/baez/physics/Relativity/SR/rigid_disk.html#star

Indeed, like DaleSpam said, a rigid disc cannot be brought to spin, so it must deform, explaing the mismatch with the markings on the side. The observer on the disc sees the circumference stretch, which is allowed because he is no longer in a euclidean geometry. To the outside observer, the markings match.

Furthermore, it appears that it is perfectly possible for your clock not to be synchronized with itself, at least, with the version of itself that you might observe using rays of light that are bounced via mirrors on the edge of the disc. The plane of simultaneity is sort of a "screw" of simultaneity. So while you are currently passing some particular mark, the version of you that you would meet if you went once around the disc, is next to a different mark "now".

I have to figure out a couple of more details, but I think I'm getting there.

Michel

 

[Q-reeus]

Michel (no 'a' in there somewhere?); I'm no relativity expert but the following may help:

If you are standing next to the wheel, the markings on the wheel will be closer together (because they are moving), so you would calculate the circumference to be less. After all, you can count the number of markings, and that is still the same. This means that, assuming the radius is unchanged, the circumference is no longer equal to 2 pi times the radius. How can that be explained?

Short answer; Unless it is allowed to tear apart, stresses necessarily act to maintain the disk as a disk in the 'rest frame'. Longer answer follows. First, there is an unspoken assumption that the disk is a solid circular disk of uniform thickness. It will be easier to first treat the case of a circular hoop - all the mass concentrated at the rim. A second unspoken assumption is that the disk circumference is restrained in some manner so as to be stress free, despite centripetal acceleration which would normally induce an outward expansion of circumference and radius vastly larger than any relativistic changes possible for any realistic solid (barring perhaps spinning neutron stars with a solid crust). This requires a frictionless circumferentially uniform inward radial pressure be exerted on the disk. The picture is then, from the point of view of the stationary frame, of a rim that has contracted by the usual SR inverse 'gamma factor' (1-(v/c)²)^1/2. This must be so because as seen from the stationary frame the constituent atoms in the rim are all Lorentz contracted in the direction of motion AND stress free (no circumferential stress to stretch things in either frame). But it is not possible then for the radius to remain unchanged - relativity is not magic. The two constraints - Lorentz contraction of the constituent matter, plus being stress free, automatically means the radius undergoes the same Lorentz contraction. So, FOR THE CONSTRAINTS IMPOSED TO MEET THE ASSUMPTIONS MADE, the disk/hoop simply shrinks by the inverse gamma factor, in the stationary frame. A uniformly thick disk is more complicated. Necessarily to avoid 'tears' the radius MUST contract by the same inverse gamma factor, but clearly Lorentz contraction is now a function of radius. The detailed force balance is then for the mathematically inclined and if you really want to delve in here's a link: http://arxiv.org/abs/physics/0211004

Meanwhile on the disc, you would measure the circumference of the circle on the floor to be less than that of the disc, since the markers on the floor are closer together. How can that be? The enclosure would appear to be smaller than the disc?

From the point of view of an atom on the rim, the stationary frame is UNIFORMLY Lorentz contracted in the instantaneous direction of that atom. However a 'snap-shot' taken by a camera on the rim would be different to this, owing to complicated mix of return paths for various light rays (see eg. http://www.anu.edu.au/Physics/Searle/".

And what about simultaneity? If you do some experiment sending laser beams around the circle and back, and basically checking to see if your own clock is simultaneous with itself seen from a distance, wouldn't you get into contradications with what the outside observer will say? Many straight line thought experiments are solved by considering there is no objective simultaneity, but if you go once around the disc, you have to agree on simultaneity. How do the paradoxes solve themselves then?

This needs rephrasing a bit more carefully I think. Maybe you could provide an illustration to clarify what you mean - is this something not covered by Sagnac effect?

For example, what about the twin paradox? Both the observer on the disc, and the observer next to it, will see the other as moving, but there is no turnaround point to shift the reference frames to solve the paradox. Of course there's a different kind of assymmetry in the centripetal force, but how exactly does that solve the problem?

The purely formal resolution involves 'world-lines' and 'invariant intervals' and there's volumes of standard SR references on that point - whether it satisfies 'physical intuition' is another matter. It is an experimental fact that acceleration per se has NO EFFECT on time dilation. One particle on the rim of a flywheel experiencing a trillion 'g' will age precisely the same as another particle experiencing one trillionth of a 'g' provided only they have identical rim speeds as seen from some common 'rest' frame.

And do both observers agree on the radius of the disc, or does one see it as bigger than the other?

Since Lorentz contraction is always normal to the direction of motion in either frame, there is agreement on the radius - it is seen the same in both frames. This is a separate matter to the stresses that may induce 'absolute' changes as discussed above.

 

[michelcolman]

Michel (no 'a' in there somewhere?)

No, it's like "Michelangelo" only I'm not an angel

This needs rephrasing a bit more carefully I think. Maybe you could provide an illustration to clarify what you mean - is this something not covered by Sagnac effect?

OK, I'll give a specific example. First the straight line version:
Alice is moving away from Bob at 0.6c. They agree that, one minute after Alice's departure, Bob will send a signal to Alice. Alice will immediately send a reply saying when she received it.
From Bob's point of view, he sends the signal after 60 seconds, the signal takes 90 seconds to reach Alice, so she receives it after 150 seconds but her ("slow") watch is only indicating 120 seconds. He receives her return message "message received after 120 seconds" another 90 seconds later, so when his watch is indicating 240 seconds. This confirms his point of view that Alice has the slower watch.
Now from Alice's point of view, since Bob's watch is slower, he sent his signal late, after 75 seconds. The message takes 45 seconds to reach her, which is indeed confirmed when she reads 120 seconds on her watch. She sends the message back, but since Bob is moving away, it takes 180 seconds for the signal to get back. This means the total time is 300 seconds, which will of course be 240 seconds on Bob's "slow" watch.
In this example, the problem works out because Alice and Bob aren't getting together again unless one of them changes reference frames. Relativity of simultaneity saves the day. But what if Alice is not moving away in a straight line, but is on the rim of a spinning disc? With the signal being sent around the disc via a circular mirror attached to the edge? If she happens to just pass Bob when she receives his message, there can be no ambiguity about the time of reception.

Actually, I needn't even make it that hard. They needn't even send any signals, just compare watches on each turn and see which watch is slowest. If both have some reason to think the other watch is slower, they should soon find out which it is.

The purely formal resolution involves 'world-lines' and 'invariant intervals' and there's volumes of standard SR references on that point - whether it satisfies 'physical intuition' is another matter. It is an experimental fact that acceleration per se has NO EFFECT on time dilation. One particle on the rim of a flywheel experiencing a trillion 'g' will age precisely the same as another particle experiencing one trillionth of a 'g' provided only they have identical rim speeds as seen from some common 'rest' frame.

This is true if, as a non-accellerating observer, you are observing different accellerating observers. However, if you are accellerating yourself, all clocks at different distances around you will be doing strange things. I suppose that solves the spinning disc problem too, with the spinning clock definitely ticking slower than the stationary one. Seen from inside, the stationary clock is faster because it is not experiencing the same gravity. Or something like that. Right?

Since Lorentz contraction is always normal to the direction of motion in either frame, there is agreement on the radius - it is seen the same in both frames. This is a separate matter to the stresses that may induce 'absolute' changes as discussed above.

That makes sense.

 

[michelcolman]

Seen from inside, the stationary clock is faster because it is not experiencing the same gravity. Or something like that. Right?

Actually, the stationary clock is probably ticking faster because it is moving along the screw of simultaneity. The time difference of one turn around the screw is added to its actual time (which should be slower), so it becomes faster.

I thik I got it correctly now. Right? ;-)

 

[Q-reeus]

Made a boo boo earlier, writing:
"Since Lorentz contraction is always normal to the direction of motion in either frame..." Wrong. That should have read "Since Lorentz contraction normal to the direction of motion is absent in either frame...". Sorry about any confusion.

Actually, I needn't even make it that hard. They needn't even send any signals, just compare watches on each turn and see which watch is slowest. If both have some reason to think the other watch is slower, they should soon find out which it is...Actually, the stationary clock is probably ticking faster because it is moving along the screw of simultaneity. The time difference of one turn around the screw is added to its actual time (which should be slower), so it becomes faster.
I thik I got it correctly now. Right? ;-)

Agreed, an easier task; clock on the disk will tick slower, by the inverse gamma factor. And as I think you are alluding to, it boils down to just compute and compare the integral of the proper time interval. Three cheers for Wikipedia: http://en.wikipedia.org/wiki/Proper_time"

To be honest rotating disk and Twin Paradox 'paradoxes' are still being fought over, and I'm not one to adjudicate. For instance, I wrote earlier that acceleration does not affect time dilation. This is backed up eg. http://www.phys.ncku.edu.tw/mirrors/physicsfaq/Relativity/SR/clock.html On the other hand, Einstein wrote a paper in 1918 claiming a proper resolution of the Twin Paradox required recourse to GR and accelerations, but only by artificially matching the accelerations to a 'pseudo gravitational field'. See eg. http://www.lifenotes.org/Twin%20Paradox%20Explanations.pdf (see The Equivalence Principle Analysis). Another reference that explains and scathingly criticizes the logic is http://www.ias.ac.in/currsci/dec252005/2009.pdf (beware this author is a believer in a universal reference frame).

Some contrary views: http://worknotes.com/Physics/SpecialRelativity/TwinParadox/page2.aspx"
http://worknotes.com/Physics/SpecialRelativity/TwinParadox/ap3.aspx?print=true"
Sorry if this is confusing but while the philosophical issues are still debated, predicted results are agreed upon.

 

[bcrowell]

Here is a discussion I wrote: http://www.lightandmatter.com/html_books/genrel/ch03/ch03.html#Section3.4 (subsection 3.4.4) It gives references to more detailed papers by Grøn and Dieks, which are available online.

 

[michelcolman]

Thanks for all the replies and links, I'm beginning to see a bit more clearly now, it's just about starting to make sense. The fact that it's impossible to synchronize all clocks from the spinning point of view (without a kind of branch cut discontinuity somewhere) was a big eye-opener.

I might even be starting to understand gravitational time dilation now...

Thanks!

Michel

 

[yuiop]

Thanks for all the replies and links, I'm beginning to see a bit more clearly now, it's just about starting to make sense. The fact that it's impossible to synchronize all clocks from the spinning point of view (without a kind of branch cut discontinuity somewhere) was a big eye-opener.

While it is true that is impossible to synchronise all the clocks on the disc using the Einstein synchronisation convention, it is nevertheless possible to synchronise all the clocks on the disc simply by putting a clock at the centre and using a signal from the central clock, to start all the peripheral clocks.