You spin me right round baby, right round.
April 10, 2006 7:50 PM   Subscribe

Resolution of a physics paradox; I figure that someone here knows far more than I do.

From a discussion with warcode:

There's a well known physics paradox about a long pole and a short barn--essentially, if a man with a twenty-foot long pole runs towards a 10 foot long barn with an open door at very high speeds, then from the frame of reference of the barn, the man is short enough that he can fit into the barn, and the door can be shut behind him. From his frame of reference though, the opposite is true and so he can't fit in.

Anyhow, the resolution of this paradox isn't really that difficult, but the one I'm curious about is in a similar vein:

Say you had a disc spinning fast. Very fast. From the standpoint of someone watching it, it seems like the disc should contract inwards--since each part of it is moving laterally very fast, they should shrink along the direction that they move, and so the outer part should contract inwards.

Now, say that you were to start with a disc of radius ten meters, spin it up fast enough that it shrinks to one meter, and put it in a hollow cylinder with radius five meters. This seems like it should be possible, but I wonder what it looks like from the point of view of the disc. Naively I would assume that it would see the cylinder spin around it, and so it would see the cylinder contracting and hence not be able to fit inside.

Of course there's a fundamental difference here, in that with the barn door example, both frames of reference are inertial, where in this case the disc's frame is quite clearly not inertial. So what actually happens? Are we correct in our analysis that the disc, from outside, will appear to shrink? And if so, what exactly would the cylinder look like to someone spinning with the disc?
posted by vernondalhart to Science & Nature (41 answers total)
 
i have a vague recollection this might be discussed in misner thorn + wheeler (which i don't own; personally i don't have a clue what the correct answer is).
posted by andrew cooke at 8:05 PM on April 10, 2006


ah. so i'm not alone :o)
posted by andrew cooke at 8:06 PM on April 10, 2006


I believe you're deeply confused about Fitzgerald Contraction. There is no shrinkage. What there is is a perception of shrinkage, which is different from different points of view because they're all observing the situation using light and light doesn't move infinitely fast. What the Special Theory of Relativity tells us is how much shrinking we will (or will not) see, but it also tells us that what we're seeing isn't true.

Take two spacecraft moving past one another at constant velocity. I see you as shrunk, and you see me as shrunk. The reason there's no contradiction is that neither of us is right. No one is actually shrunk; it just looks that way to each of us.

Take the barn door problem you described. One person can't control both doors to make them close "simultaneously" because there's a time delay in propagating the signal from the switch to the two doors. But what's more important is that what the guy at the switch sees is not what's really happening, because he's using light to see it. When he sees the near end of the pole as being just inside of the near door, it looks to him as if the other end of the pole is still inside at the far end, because the light he's using to see that is out of date. To him it looks at that instant as if the entire pole fits within the barn, but that's because what he's seeing isn't correct.

By the same token, the spinning disk you're describing doesn't shrink, and doesn't appear to be shrunk to anyone. What actually seems to be happening depends on who is looking and where they're standing, but all of them are seeing a fictitious representation of reality, in part due to the fact that the speed of light is not infinite, and in part because some of the frames of reference are accelerating and thus invalid.

Take a person sitting at the center of the disk. The outer edge of the disk is unquestionably measured as having a high velocity, but its radial velocity relative to the observer is zero (because it's traveling in a circle around the observer) and as a result there's no change in the latency of his observation caused by the non-infinite speed of light. Therefore he will not measure any shrinkage. As the disk speeds up, it will not seem to change size to him.

The reason that's different from an object moving in a straight line past him is because the distance between him and that object is changing with time, and thus the latency of his observation using light is also changing. In particular, the distance and latency of his observation of the far end of that object and the near end is non-zero. (Also note that an observer on the edge of the disk is in an accelerating frame of reference, whereas an observer on that passing object is in an inertial frame of reference.)

It turns out that mass increase and time slowing are equally aspects of measurement errors as long as the two frames of reference that disagree are both inertial.

If they have a relative velocity of zero, then it's possible for them to keep comparing their results, because there won't be any change in communication latency -- and they'll both measure the same thing. If they have a non-zero relative velocity, each will measure something different, but their ability to compare their measurements will be distorted by the fact that they communicate using light at non-infinite speeds. I know this seems nonsensical, but the math works out perfectly.

The only way that they can start and end with the same distance between them (and thus compare their measurements without changes in latency) while still having nonzero relative velocities for a while, is if one or both accelerates. If only one of them does, then his frame of reference is invalid and his measurements are wrong. If they both accelerate, they're both wrong. And the degree to which they are wrong is a function of how much each of them accelerated.

That's how Special Relativity resolves the apparent contradiction in that case, which is why the "Twin Paradox" isn't a paradox. The twin who experiences acceleration also experiences time slowing; the one who stands still does not.

Put the two twins in separate space ships and make them accelerate apart at the same acceleration for a while, and then coast. Whose time shrinks? Answer? The question is meaningless. Using telescopes, each will see the other age more slowly, because as they move further and further apart the time taken for the image to propagate from one to the other will increase with time. If they never return to the same point (or same distance apart as when they began), then it isn't possible to ever directly determine whose time actually shrunk.

That's because in the Special Theory, there is no simultaneity at separate points in space. If you ask, "When this happens here what exactly is happening there?", then under the Special Theory it's not just that there's no way to find out. In fact the question itself is meaningless. The answer depends on what you do to find out, and it turns out that you'll get a different answer depending on what you do in order to find out.

If you have two events which happen at two separate points in space, all the ways of figuring out which one happened first involve either sending light back and forth or using acceleration to move together. Both of those are covered by the Special Theory in ways which correctly define what observers will see.

That is the case even when the points are quite close together -- such as the opposite ends of the pole being carried through the barn.
posted by Steven C. Den Beste at 8:26 PM on April 10, 2006


A small correction. When I wrote: "The reason there's no contradiction is that neither of us is right. No one is actually shrunk; it just looks that way to each of us."

That isn't actually the case. It's not that no one is shrunk, it's that there's no way for us to determine what's going on without someone undergoing acceleration, or someone sending a message using light, and both of those induce measurement errors which are precisely described by the Special Theory.
posted by Steven C. Den Beste at 8:38 PM on April 10, 2006


actually, i think you're confused. you most certainly can close both barn doors at the same time (in the barn's frame). time is well defined throughout the frame and you can arrange for clockwork mechanisms to do the closing at the correct moment. the contracted rod will fit inside the barn (instantaneously).

(in the rod's frame, the doors do not close at the same time, of course)

special relativity is more than an optical illusion. there is a real problem here with the disk.
posted by andrew cooke at 8:40 PM on April 10, 2006


ah, i missed your correction.
posted by andrew cooke at 8:41 PM on April 10, 2006


Response by poster: I'm actually glad you corrected that last bit, because otherwise you've contradicted everything I learned about special relativity over my undergraduate degree in physics. Well, not quite everything.

As I intimated in the original question, I have no issues with the barndoor paradox (or the twin one, for that matter) - those are relatively easy to explain.
"The reason that's different from an object moving in a straight line past him is because the distance between him and that object is changing with time, and thus the latency of his observation using light is also changing. In particular, the distance and latency of his observation of the far end of that object and the near end is non-zero. (Also note that an observer on the edge of the disk is in an accelerating frame of reference, whereas an observer on that passing object is in an inertial frame of reference.)"
Unless I'm misreading what you're writing, you're saying that length contraction is simply due to a latency in the amount of time light takes to get from the emitter to the observer. I was however under the impression that this was not the case--that from a fixed frame of reference, lengths are physically contracted. It's not that an object simply appears shorter, but from that frame of reference the object is shorter. Where the latency issue comes in is a further distortion in how an object appears (which I recall makes the object appear to bend somewhat), but the shrinkage itself isn't due to that distortion.

The Ehrentfest paradox is interesting; now that I think of it, I had heard of that one originally, and while these two aren't exactly the same, they are fairly similar.

I guess when all is said and done, I'd forgotten to take into account that while the perimeter is moving laterally, the radius is not, and so from the lab frame of reference, the disc would not in fact contract inwards (although somewhat oddly, the perimeter still does shrink, I guess...).
posted by vernondalhart at 8:57 PM on April 10, 2006


Response by poster: andrew cooke: "actually, i think you're confused. you most certainly can close both barn doors at the same time (in the barn's frame). time is well defined throughout the frame and you can arrange for clockwork mechanisms to do the closing at the correct moment. the contracted rod will fit inside the barn (instantaneously)."

Actually, you're right. I'd forgotten what the resolution of that paradox was for a second. Whoops.
posted by vernondalhart at 8:59 PM on April 10, 2006


now i'm confused. my "you're confused" was addressed to scdb. anyway, bedtime here.
posted by andrew cooke at 9:10 PM on April 10, 2006


The easy way out (but also, I'm afraid, the correct one) is to note that a rotating coordinate system is, by definition, and accelerating one, and that therefore this problem should be solved by General Relativity, not Special Relativity.

It's a little late to be pondering this right now, but I'll think about it some in the morning and se what I can come up with.
posted by Johnny Assay at 9:15 PM on April 10, 2006


It's not that an object simply appears shorter, but from that frame of reference the object is shorter.

I cannot determine what is; I can only measure what I see, and I see with light, which doesn't move infinitely fast. That's precisely the point. Your statement requires an underlying assumption of simultaneity, and in Special Relativity there is no such thing.

It is possible to do the math and demonstrate that all the differences in perceived results in two inertial frames of references moving with a nonzero velocity relative to one another are the result of latency due to the fact that the speed of light isn't infinite. So which is shrunk and which is not? It isn't meaningful to discuss that because it requires the assumption of simultaneity.

This is actually something like wave functions in quantum theory, it turns out. Certain properties of photons (and other particles) are not actually determined until you measure them, or until some other event takes place which causes the wave function to collapse. (Or at least that's how some interpretations of the theory describe it.)

Likewise, certain questions about "what happened" in Special Relativity can only be definitively answered if two frames of reference come together and have zero relative velocity. Until they do, it's not just that people in each frame can't actually tell what's happening in the other, it's that it doesn't even make sense to talk about that. "When I do this here, what happens over there?" is a meaningless question because there is no particular point in time there which can definitively be correlated with "now" here.

Special Relativity says that each observer in each of the frames of reference will make measurements which are internally consistent, and tells us that they'll disagree, and tells us exactly how much and in what way. Which is right? As long as they're physically separated and moving relative to one another, that requires the assumption of simultaneity, and you're not permitted that assumption.

I think it appropriate to mention at this point that simultaneity is a sticking point between Relativity and Quantum theory, because some interpretations of Quantum theory require simultaneity, which Special Relativity forbids. That hasn't been resolved yet, so far as I know. (And it's extremely intriguing that they contradict one another, given that both of them have been so damned good at making surprising predictions very precisely.)

If experimental data confirms violations of Bell's Inequality, then Relativity is in deep trouble. So far such data as has been collected seems to suggest that this is the case, but I don't believe that this is universally acknowledged in the Physics community. It's highly controversial, and I hope that someone eventually does a definitive experiment. The result could end up as being as important to science as the Michelson-Morley experiment was, and could rototill physics just as much.
posted by Steven C. Den Beste at 9:36 PM on April 10, 2006


Another correction: "...come together and have zero relative velocity..."

Zero relative velocity is not required.
posted by Steven C. Den Beste at 9:51 PM on April 10, 2006


Response by poster: Now I'm sure what you're saying is a little off.

First of all, there is such thing as simultaneity in SR. If I pick a particular inertial frame of reference, then within that frame there is a perfectly well defined notion of simultaneity. Of course it doesn't extend in the least to any other frame, but there's no issue if you specify a particular frame. And while this seems somewhat of a trivialization, it isn't. It does allow us to speak of a very precise, exact timeline for spatially separated objects... including a spaceship flying through, or a rod rotating within that particular frame. So your assertion that all these bizarre effects are simply an artifact of this latency is pattently false - an observer within a frame can say quite a bit about distant objects, and in particular (in principle) exactly how far they are away from him at a particular instant, again, within that inertial frame of reference. This observer can adjust for any latencies, and so these contractions are not due to that, but due to the fact that said object is moving through his frame of reference at a non-zero velocity.

As such all your comments about the inability to tell "what happened" without two frames coming to meet at zero relative velocity are moot - I can say perfectly well what has happened within my frame, and so can you - sure, we might have disagreements, but both of us will be right - and all of these issues have nothing in the least to do with the fact that light travels at a finite velocity.
posted by vernondalhart at 9:55 PM on April 10, 2006


Quantum mechanics and special relativity go together just fine; in particular, the absence of a universal preferred notion of simultaneity is not a problem for quantum mechanics. Indeed, the many experimental triumphs of the Standard Model of particle physics are applications of quantum mechanics together with special relativity.

The difficulties in reconciling quantum mechanics and relativity appear only when one considers _general_ relativity.
posted by em at 9:55 PM on April 10, 2006


So your assertion that all these bizarre effects are simply an artifact of this latency is pattently false - an observer within a frame can say quite a bit about distant objects, and in particular (in principle) exactly how far they are away from him at a particular instant, again, within that inertial frame of reference.

No. What he can say is what he will observe about those objects at a distance. He cannot say how far away they are; he can only say how far away they appear to be. And if, for instance, they move, he won't know it until he sees it happen, using light that isn't traveling infinitely fast.

Saying how far away they are requires the assumption of a correlation of some point in time there with "now" here and you can't do that.

This observer can adjust for any latencies... but only by making the assumption of simultaneity. He can only do that by assuming that some particular instant of time there matches "now" here, and you can't actually do that.

Or rather, you can, and can get a consistent measurement and calculation by doing so. But it will disagree with what someone else measures in a different frame of reference because he will correlate a different instant of time there with "now" here.

The only way to resolve such contradictory calculations is by matching the frames of reference with each other in space, so that here and there are the same point.
posted by Steven C. Den Beste at 10:19 PM on April 10, 2006


If anybody has ever resolved a misunderstanding of Relativity by posting on any sort of internet discussion, that'd be news to me.

If you are really dedicated and interested, I find Reflections on Relativity to be an awesome book, available online for free. But it takes dedication. I still can't say I follow all of it, but it's certainly a better and deeper explanation than anything else I know of.

(And now I don't consider "spooky action at a distance" either "spooky" or "action at a distance", having read this book. It's worth it for that perspective alone, although I imagine many won't be convinced.)
posted by Jeremy Bowers at 10:22 PM on April 10, 2006


Yep. I was also going to say that Relativity and Quantum Mechanics have already been unified by Paul Dirac and others who have contributed to relativistic quantum field theory.

With all due respect, SDB, there's quite a few flaws in your whole write-up. I mean there's nothing wrong with thinking out loud, but perhaps Ask.Metafilter is not the best place for it? I mean, when you say things like:

Likewise, certain questions about "what happened" in Special Relativity can only be definitively answered if two frames of reference come together and have zero relative velocity


I dont think you understand that a frame of reference is, well, just that a (valid) frame of reference. There are working cosmological frames of reference - the CMB provides one. Also causality holds in all frames of reference. Of course "what happened" can be definitely answered - you just have to choose a frame of reference first.
posted by vacapinta at 10:23 PM on April 10, 2006


Response by poster: You're missing the point though. There is a notion of simultaneity in his frame of reference. So he can speak of those things locally.

And who cares if someone somewhere in a different frame of reference measures something different? If I see you flying by at speed v, then I measure the length of your ship to be shorter, and you measure the same for mine. Does it matter that we measure different lengths than each other? No. There's no contradiction there, and I've never claimed such. But within my particular frame of reference your ship is shorter. This in no way affects it's length as measure in any other frame of reference though, which I've never claimed it does.
posted by vernondalhart at 10:23 PM on April 10, 2006


With all due respect, SDB, there's quite a few flaws in your whole write-up. I mean there's nothing wrong with thinking out loud, but perhaps Ask.Metafilter is not the best place for it?

"With all due respect," I see no indications of respect in what you just wrote. I'd be more explicit but I'm trying to be a good citizen these days. But be informed that I'm seething right now.

I dont think you understand that a frame of reference is, well, just that a (valid) frame of reference. There are working cosmological frames of reference - the CMB provides one. Also causality holds in all frames of reference.

Sorry, not correct. The Cosmic Background is not a universal frame of reference, it is just one among many and is not privileged. There is no one privileged universal frame of reference. And causality has nothing to do with this discussion; it refers to sequences of events, not to simultaneity. Nothing I've written here contradicts causality.

Of course "what happened" can be definitely answered - you just have to choose a frame of reference first.

And if you choose a different frame of reference you'll calculate something different. Yes, you can "definitely answer" the question. I never said you couldn't. But you can't come up with the-one-and-only answer which is valid for everyone, because there is no such thing.

Relativity and Quantum Mechanics both run against our intuitions, and that's why a lot of this stuff is hard to grasp. When it comes to QM, the biggest intuitive block is the assumption of exactness, that reality is not fuzzy. When it comes to Relativity, the biggest intuitive block is the assumption of simultaneity.
posted by Steven C. Den Beste at 10:44 PM on April 10, 2006


It seems that you guys are neglecting the important counterpart to Lorentz contraction: time dilation.

Lorentz contraction is perfectly "real," in any meaningful sense, just as time dilation is "real." The effects have nothing to do with the "latency" of light, but with the fact that light moves at the same speed in any reference frame.

I care little about how long it took light to get to me from the upper atmosphere where a cosmic ray created a relativistic muon. I can still tell the thing lasted a lot longer than its rest lifetime. Likewise, it cares little that I even looked at it. The distance to the surface of the Earth was much shorter than I claimed. Both phenomena are completely real in their respective frames.

Equally, I can create a larger mass out of a relativistic electron-positron pair than I can out of a non-relativistic pair. The physical effects of relativity are real, period.

Neutrino oscillations provide one of my favorite examples--the existence of oscillations implies that neutrinos have mass not only because of (even non-relativistic) quantum theoretical models, but also because without mass, they could never have a well defined oscillation period.

As for the original question, I have no answer. I just wanted to protest the previous posts! I'll see what I can find out, though.
posted by dsword at 10:44 PM on April 10, 2006


Response by poster: In fairness, SCDB, Relativity and QM do run counter to intuition (although I'm honestly not sure why you keep bringing up QM, since it's irrelevant to the topic at hand). But it honestly seems like in this case that you're the one whose intuition for the subject is faulty.

No one else has claimed that observations in one frame are absolute to every frame. Any calculation and measurement done in a particular frame will of course only be valid in that frame. But by specifying a particular frame of reference, in effect you are referring to absolute data. So in that sense, it is a valid one-true-answer for everybody.

dsword: Although I hadn't explicitely mentioned time dilation, I figured that it was tacitly assumed that it came with what I was saying. Since the particular question, as near as I can tell, only involves relative lengths, it didn't seem necessary to mention otherwise.
posted by vernondalhart at 10:54 PM on April 10, 2006


Just to liven up the discussion, though... Consider what the disc looks like from a point on its edge. The points right next to it aren't moving at all and don't seem contracted, but the points across from it should be moving even closer to the speed of light than it sees the outer cylinder, and so should be even more contracted.

Blegh! Spinning things are so difficult to deal with... My intuition just jumps ship. Blah blah blah, Thomas precession, blah.
posted by dsword at 10:54 PM on April 10, 2006


"With all due respect," I see no indications of respect in what you just wrote.

Look...all I mean is that the poster asked a question that involves General Relativity and you dived into answering questions nobody has asked and then throwing QM into the mix. I'm just not sure what you're trying to get at here other than, as I said, thinking out loud. I'm not sure how any of this is helping anyone "find an answer" - the stated purpose of the Green.

But you can't come up with the-one-and-only answer which is valid for everyone, because there is no such thing

We all understand this. You're preaching to the choir. We're just saying - so what? This is no revelation.

Relativity and Quantum Mechanics both run against our intuitions, and that's why a lot of this stuff is hard to grasp.


Yes and no. The equations are quite exact. If you model matter as probability waves instead of discrete points and if you model SR interactions using 4-vectors, its all quite easy to grasp. Now whether that corresponds to how we usually intuit reality is of course a different question.
posted by vacapinta at 11:25 PM on April 10, 2006


Is it just me or is Steven C. Den Beste totally out of his depth? It seems like he's saying Relativity is just an optical illusion caused by the speed of light. If that were the case, you could experiment with relativity by using sound waves and objects moving at the speed of sound.

I'm actually glad you corrected that last bit, because otherwise you've contradicted everything I learned about special relativity over my undergraduate degree in physics. Well, not quite everything.

Um, yeah. Steven, what qualifications do you have to be making these claims?

And then there's this:

This is actually something like wave functions in quantum theory, it turns out. Certain properties of photons (and other particles) are not actually determined until you measure them, or until some other event takes place which causes the wave function to collapse. (Or at least that's how some interpretations of the theory describe it.)

Mixing relativity with quantum physics? I thought they were incompatible.


----
As far as the question itself, obviously no disk could ever spin that fast, so the question is moot. :P
posted by delmoi at 11:40 PM on April 10, 2006


I asked a prof once about what happens when a disc starts spinning near light-speed. The first thing that he pointed out was that the simple modelling equations called special relativity didn't apply, because angular velocity implies constant acceleration.

He then started writing things on a chalkboard. What things? Things that were very incomprehensible to my feeble little brain, which had only just squeaked through differential equations a semester prior.

He explained to me for the better part of an hour, so presumably he had some understanding of what such a disc could do.

You can start to get a handle on this difficulty by analysing the sentence 'a disc starts spinning near light speed.' If you're standing at the hub of the disc, a point A on the edge of the disc has an instantaneous linear speed tangentially at 0.9c relative to the hub, and a point B diametrically opposed to A is moving in the opposite direction at, again, a speed of 0.9c.

Now consider the disc as if you were standing on point A. Not only are you experiencing inward acceleration, invalidating special relativity; but your linear speed relative to the hub, to point B, and in fact to every other point on the disc is subject to relativistic effects, which are different depending on where on the disc those points lie.

It makes a big hairy ugly awful mess; what it does not do is neatly 'shrink from a radius of ten meters to a radius of one meter.' In fact, what it does do depends on where you're looking at it from.

Some neutron stars apparently spin fast enough to produce extremely bizarre relativistic effects, but I regret extremely that I don't understand the equations that purport to describe these behaviors.
posted by ikkyu2 at 11:57 PM on April 10, 2006


Best answer: Huh, how do you like that: a google search for "spinning relativistic disc" brings up this.
posted by ikkyu2 at 12:02 AM on April 11, 2006


Say you had a disc spinning fast. Very fast. From the standpoint of someone watching it, it seems like the disc should contract inwards--since each part of it is moving laterally very fast, they should shrink along the direction that they move, and so the outer part should contract inwards.

Now, say that you were to start with a disc of radius ten meters, spin it up fast enough that it shrinks to one meter, and put it in a hollow cylinder with radius five meters. This seems like it should be possible, but I wonder what it looks like from the point of view of the disc. Naively I would assume that it would see the cylinder spin around it, and so it would see the cylinder contracting and hence not be able to fit inside.


I think you're wrong. The "Shrinkage" happens in the direction of travel, and the direction of travel is perpendicular from any line drawn from the center of the disk, so there will be no shrinking at all of the disk inwards.

this wikipedia article has the formulas you need to figure out exactly where a point will appear to be in relation to your reference point. Solve first for a single orbiting point, figure out what that would do and then imagine a whole bunch of those.

I would imagine that it might be different depending on where you were in relation to the position of the disk, like are you off to the side, or looking at it from the center?
posted by delmoi at 12:04 AM on April 11, 2006


No, delmoi. That article is for points in inertial reference frames. The points on the disk are being accelerated; they are not in inertial frames.

If you apply the equations you linked to this situation, you end up with a disc with a decreasing circumference but a constant radius. Rigidity (or point-to-point non-deformability, if you will) of a body is not possible even in special relativity; but the "solution" of such a disc is actually a mathematical impossibility.

The rather surprising solution, according to GR, is that a rotating disc cannot exist in the flat Minkowski space that your article gives as a prerequisite; its rotation (and the - surprisingly - infinite stresses it engenders during its ideal, massless, rigid motion) actually warp space-time locally so that geometry becomes non-Euclidean. This monstrous corollary to the Einstein field equations is called the Lense-Therring effect.
posted by ikkyu2 at 1:03 AM on April 11, 2006


And, frankly, I'm appalled, and my head hurts.

I'm going to bed.
posted by ikkyu2 at 1:04 AM on April 11, 2006


its rotation (and the - surprisingly - infinite stresses it engenders during its ideal, massless, rigid motion) actually warp space-time locally so that geometry becomes non-Euclidean.

I'm not quite sure how a massless object can warp spacetime, but I'll take your word for it :)
posted by delmoi at 1:37 AM on April 11, 2006


In the rotating reference frame of the disk, the circumference experiences relativistic shrinkage but the diameter does not. The non-Euclidean spacetime geometry of the disk's accelerated reference frame is such that the disk's circumference no longer equals pi times its diameter.

This isn't really so weird. Related effects occur on the surface of the Earth, if you make all your measurements along that surface and if your circles are big enough. For example, the equator is (roughly) only twice as long as a "diameter" plotted through either pole.
posted by flabdablet at 3:31 AM on April 11, 2006


Re: the barn door issue (as opposed to the spinning disk), I'm with delmoi, the Fitzgerald contraction is in the line of movement in any case, and the pole would still appear to both observers to be the same length vertically, although becoming elliptical in section to the 'stationary' fellow at the barn door.

A good read on it is "The theory of relativity" paper by Clement Durell. Explains a lot in minimal math, but with abundant illustration and prose. Requires concentration and about 2 hours. Does not approach the disk problem, unfortunately.
posted by FauxScot at 4:32 AM on April 11, 2006


as i sad earlier, i'm out of my depth with this, but i'm going to say the following because i suspect it's one of the few moments in my life i'll be able to do so with some semblance truth: baez covers pretty well my initial thoughts on the question :o)

in particular, he concludes: To settle the question definitively, it seems one has to perform a full-blown, hairy GR calculation. Perhaps someone has done this; perhaps someone has turned the vague notion of "infinitely rigid" into a formula for a stress-energy tensor, plugged that into the Einstein field equations, and solved.

i'd suggest people read ikku2's link.
posted by andrew cooke at 4:57 AM on April 11, 2006


s/ku/kyu/
posted by andrew cooke at 4:57 AM on April 11, 2006


reading everything again, although not terribly carefully, scdb is correct to point out that the lorentz (i'm sorry; i'm old) contraction should act along the circumfrence rather than the radius; he was correcting errors in the question that i skimmed over, but they're not enough to remove the underlying problem.

the underlying problem is, in case it's not clear, that if you follow that line of reasoning you end up with a circumfrence that is less than 2 pi * radius. now that in itself is not the end of the world, because we have a rotating frame and so expect general relativity to be important (and the special relativity intuition/hand-waving to be only a rough guide to what actually happens), and in general relativity we may have a curved space. the obvious initial idea is that space is indeed curved - this is something like a circle drawn on the earth where the circumfrence of the equator is much smaller than you'd expect from the "radius" you get if you measure the distance along the curved earth from the pole to the equator.

however, in general relativity, mass/energy tells space how to curve (and space tells mass/energy how to move) and there seems to be nothing here stopping you from saying "assume the disk is massless". which is annoying, bceause then you don't have any mass/energy to do the curving. so space cannot be curved after all!

the problem seems to be that physically you cannot have something that is both rigid and massless. which is very odd, because if you're a physicist you're used to "the maths still working" even when you do things like pretend that there is no mass.
posted by andrew cooke at 5:15 AM on April 11, 2006


Best answer: there seems to be nothing here stopping you from saying "assume the disk is massless". which is annoying, bceause then you don't have any mass/energy to do the curving.

One of the problems here is that you're applying terms from SR, including Lorentz contraction and energy, to a problem where they aren't adequate to model the physical phenomena. The disk is rotating, which means that it is accelerating, which means energy. Furthermore, the massless disc, undergoing rigid motion at relativistic speeds, is experiencing internal stresses that are terrific (and strain that is negligible on an infinitesimal scale.) That means more energy. What this means is a lot of nonzero terms in the stress-energy tensor, which is the 4x4 matrix that is used instead of a scalar E in the general field equations.

It all smells like a big kludge to me, and it leads to some surprising predictions. For instance, when mass exceeds a certain amount, we all know that gravity causes a singularity to form- a black hole, a point mass of radius 0. However, if that mass was rotating when the black hole was formed, that angular momentum does not go away. It does, however, become dimensionless, so you have a black hole which is rotating, but in no particular direction.

Fascinatingly, if it's rotating fast enough, objects that enter such a black hole's ellipsoidal ergosphere are forced to spin along with the black hole due to this local warping of space-time (Lense-Thirring effect). Such objects gain energy in the form of momentum (and slow the black hole's rotation) but because they have not yet entered the event horizon, they can leave the black hole's vicinity, taking their new energy with them.
posted by ikkyu2 at 9:09 AM on April 11, 2006


i was about to correct a mistake in ikkyu2's post and see i've made the same error myself. i said "because we have a rotating frame and so expect general relativity to be important". strictly speaking, that's not true. a rotating frame implies acceleration, but acceleration alone does not imply that general relativity is necessary - any physics undergraduate has calculated the relativistic motion of an accelerating rocket.

however, there is something strange about rotating frames - see mach's principle and godel's cosmology. the whole thing is a rat's nest of complications....
posted by andrew cooke at 10:09 AM on April 11, 2006


Right, andrew, but as you say, rotating frames don't work with SR. The problem is not that any inertial frame can't be worked out with judicious use of the SR equations; the problem is that applying the SR equations to a rotating frame leads to perfectly contradictory results.

This is what I was trying to hint at when I pointed out that the circumference of the disc is shown to decrease while the radius is unaffected; if you really work it out, you can't even tell how much the circumference decreases, because you can obtain different values, given the same start conditions, depending on how you apply the equations.

If the equations, properly applied to the same set of initial conditions, lead to two different results that can't be compatible, then they are not valid models. (And there's no reason to expect they would be, because rotating discs don't adhere to the conditions specified for those equations' use.)
posted by ikkyu2 at 10:14 AM on April 11, 2006


oh, ok - i missed your reference to lense-thirring above. i guess i'm just bothered by "don't adhere to the conditions specified for those equations' use", because it's not clear to me what those conditions are. rather, it seems more natural to say that rotation somehow entails mass, hence the need for gr. but i may be confused.
posted by andrew cooke at 10:27 AM on April 11, 2006


Response by poster: ikkyu2: Thanks for your answers. That pretty much completely resolves it.
posted by vernondalhart at 12:22 PM on April 11, 2006


I can't believe no one has yet posted:

There once was a fencer named Fisk
Whose thrusting was exceedingly brisk.
So fast was his action
The Fitzgerald Contraction
Soon shortened his epee to a disk.
posted by Rumple at 12:55 PM on April 16, 2006


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