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The Pinwheel Paradox

  1. Jul 29, 2004 #1
    What is the solution to the paradox below? I have a half-formed idea.

    GR tells us that a crew member at the bottom of a thrusting rocket ages slower than one at the top.

    Imagine a rocket in the form of a circle or pinwheel. The rocket is subdivided into floors, each of which contains a crew member. You are on one floor. The pinwheel is so big that its curvature is barely perceptible to a crew member (like earth’s curvature is barely perceptible to us on its surface). Let the pinwheel accelerate in rotation such that all the crew feels a 1g acceleration tangential to the pinwheel. The pinwheel is so big that the centripetal force felt is negligible.

    GR tells us that you age slower than those above you. But...and this is the paradox...you are above you (keep going upstairs and you end up where you started).
     
    Last edited: Jul 29, 2004
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  3. Jul 29, 2004 #2
    Where does GR tell us that those lower in the rocket will age slower? The equivalence principle tells us that a gravitational field and accelerated motion are considered equal, and the stronger the gravitational field the slower time passes due to the graviatational redshift. In a thrusting rocket, all parts of the rocket are accelerating at the same speed (otherwise it would crush or stretch) therefore the effects of general relativity, namely the grav. red ship and time slowing, are equal throughout the rocket. All observers, top or bottom age at the same rate. Have I missed something here?
     
  4. Jul 29, 2004 #3
    That is a very well known result which is deduced from Einstein's equivalence principle.
    There are several statements of the equivalence principle. Einstein's original statement of the equivalence principle was made in 1907, and holds exactly true today. It is as follows
    Thus observers who are inside the rocket ship cannot tell if they are in an accelerating frame of reference or in a uniform gravitational field.
    That is incorrect. All parts of the ship accelerate at different rates according to how high (what floor/stage) the part is inside the rocket.
    To be precise, the rocket Lorentz contracts. If all parts had the same acceleration then the rocket ships length as measure in any particular inertial frame, would remain constant. But we know that the faster a object moves the shorter it gets.
    Since this is a uniform g-field objects which are lower in the field run at a rate which is lower than objects high in the field. Therefore people in the nose of the rocket age faster than people in the tail of the rocket.

    Pete
     
  5. Jul 29, 2004 #4
    The resolution to your paradox is that you're assuming that you can ignore the new gravitational field which is present due to the rockets also being in an accelerating frame, i.e. the rotating frame. In the example you gave the effects cancel and as observed by an observer in the center of the pinwheel all the clocks will be sychronized.

    Pete
     
  6. Jul 29, 2004 #5

    Janus

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    But you aren't above yourself. The curvature over a small area may be imperceptible, but over the whole ship it is perceptible. The astronaut "above" you isn't directly above you, but a little to the side and tilted a little (compared to your line of acceleration), and the astronaut above him is a little more to the side, and each successive astronaut shifts and tilts more to the side. Eventually you get to an astronaut which, to you, is standing sideways. Then the astronauts start to tilt "head down" and become less and less "above" you, until you reach a point where the astronaut is at your height but "upside down". From then on the astronauts are "below" you. They contiune to tilt and start to shift back towards you until you get to one that is almost exactly below you.

    The thing is that you are trying to consider the line of acceleration as wrapping around in a circle; it doesn't. Each individual astronaut's line of acceleration is a straight line at a tangent to the circle, and he judges all of the other astronauts positions as relative to that line.
     
  7. Jul 30, 2004 #6
    Can you please elaborate? I don’t see how I’ve ignored the gravitational field since that is the basis for the paradox. I know your conclusion must be right because symmetry rules, and symmetry tells us that an observer at the center must see all floor clocks as synchronized. But to each crew member, who is experiencing a gravitational field, it seems that a clock above them must run faster. Can you elaborate upon how a clock above them does not run faster?
     
  8. Jul 30, 2004 #7
    Of course you are right. I’m using that expression loosely to say that, if you find that you age slower than the crew member on the floor above you (well, almost perfectly above you) then there is a paradox because--since if you keep climbing upstairs then you’d return to your own floor--you must conclude that you age slower than yourself.

    Are you thinking that this is enough to explain how your clock runs at the same rate as the clock on the floor almost perfectly above you? The direction of the acceleration on the floor above you is slightly different than the direction of your acceleration. Is this enough to ensure that all floor clocks run at the same rate?
     
  9. Jul 30, 2004 #8
    I thought about this some more. That the direction of the acceleration on the floor above you is slightly different than the direction of your acceleration is not enough to ensure that all floor clocks run at the same rate.

    To know this, all that needs be asked is, “does light bend on my floor?” If yes, then lower clocks run slower (even within a single floor), which leads to the paradox. And the answer is yes, light does bend on my floor. So some other explanation resolves the paradox.
     
  10. Aug 6, 2004 #9

    sal

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    This is not a simple question. Let me try to provide a simple explanation, and see if it helps.

    First, here is a question whose answer includes the resolution of the "ring" part of the paradox: How fast are all the other astronauts on the ring aging?

    Since you've assumed "flat" space (no gravity) this question can be properly asked and answered: What is dtau/dt for every other astronaut in the ring in your MCRF as you sit in a particular room in the ring?

    Consider it this way: If you sit still, and you watch someone else climb up to the next room, and then he just keeps climbing, how rapidly will he age at each point on the ring? Initially, he ages faster than you, but at some point he'll be aging slower than you. When he gets back to you, it will all have evened out. So "climbing around the ring" doesn't really result in a paradox -- when you get "home" again you'll have aged the same amount as someone who just stayed put.

    But you may still object that the guy in the room above is getting progressively older as you watch him, but someone on the ground doesn't see that happening. Why is that? Pete says the effect of the rotation cancels the effect of the linear acceleration. Well, the ring isn't adding anything, so let's dispense with it. Let's just have two rocket ships, one in front of the other. From the POV of a stationary observer, they start accelerating simultaneously. Stationary-frame observations show the astronauts all age at the same rate, but the astronauts see each other aging at different rates. Who's wrong?

    Nobody. There's one more question you need to ask: They can't keep accelerating forever (they'll run out of fuel). So, when do they stop accelerating? If they stop simultaneously as seen by a stationary observer then they will not stop simultaneously as seen by the riders on the rockets! And the additional acceleration undergone by the astronauts who appear to shut off their engines last will exactly make up the difference in age, and will also make up the difference in velocity. (Note that the astronauts on ship A don't see ship B as traveling at the same velocity they're going at during acceleration, if they maintain a fixed following distance as seen by a stationary observer. This applies to the ring residents as well, if the radius of the ring is fixed as seen from the ground.)

    I hope this handwaving solution helps. I don't have the math on tap to actually prove this works out, unfortunately. Perhaps somebody else does.
     
  11. Aug 6, 2004 #10
    He ages slower than me at some point, but he also ages faster than each person immediately below him, who in turn ages faster than the person below them, and so on, until I am the person below them. So he also ages faster than me at that point.

    Let the ring contain 10 floors total, labeled A through J. I am on floor A. The guy on floor B ages faster than me. The guy on floor C ages faster than the guy on floor B, and so on. Comparing aging rates:

    A<B<C<D<E<F<G<H<I<J<A...

    The paradox is that A<A.

    You say that at some point he'll age slower than me. Say that point is when he is on floor H. We see above that H<A, so indeed he ages slower than me. But also H>A, so he also ages faster than me. I don’t see how the paradox is resolved. Let me understand that before I comment on the rest of your post.
     
    Last edited: Aug 6, 2004
  12. Aug 6, 2004 #11

    sal

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    The problem is that "<" isn't well defined in that expression. From your point of view, the rate at which each astronaut is aging is dtau/dt, where dtau is the change in their proper time and dt is the change in time in your MCRF -- the inertial frame which is momentarily moving with the same velocity you're moving with.

    If you use that definition, then you are aging at rate 1, the guy upstairs is aging a little faster, and so forth, but at some point in the ring the rates decrease again. So you actually have A<B<C< ... >J>K>L> ... <Y<Z<A.

    What you did, instead, is tacitly assume the "<" operation is defined based on the rate of each astronaut's clock in the MCRF of the astronaut adjacent to him. And when you do it that way, you're really using a different "<" operator between each pair of astronauts, and the relationship is no longer transitive. Because, again, it's not one single relational operator -- it's a different operator for each pair of astronauts.

    Now, this probably sounds all wrong, because, you will say, "They can compare their clocks directly -- it's a fact that the guy upstairs is aging faster!" But there's a problem here: Two adjacent astronauts cannot compare their clocks "directly". At any particular time in any MCRF they're separated from each other by a spacelike interval, and they can only compare their clocks indirectly, by sending signals. Maybe the signal is as simple as one astronaut looking across a gap of 20 feet at the next one's clock, but that's sufficient: They are not sharing any events. And it's only at a shared event that we can talk about absolute clock synchronization or absolute differences in clock readings, and it's only between a pair of shared events that we can talk about absolute differences in clock rates.

    It's because of the fact that each pair of "adjacent" astronauts is actually separated by a spacelike interval that I keep talking about the clock rates in an MCRF. It's only in an inertial frame that we can talk sensibly about relative clock rates at distant locations, and none of the astronauts is actually in an inertial frame -- they're all accelerating! So to answer any relative-rate question, we really need to pick an inertial frame which is comoving with one of them, and pose the question in that frame.

    I'm afraid the breeze from my waving arms is getting a little stiff here. Is this helping at all?

    This problem is related to the problem of two spaceships moving on a circular path, going in opposite directions (the "counter-revolving twins" problem). I do happen to have a reasonably complete solution to that on tap, with graphs :smile:, at http://www.physicsinsights.org/revolving_twins.html. But I haven't worked out the case where everybody's going around the ring in the same direction, nor have I worked through the case of tandem acceleration; this paradox is the conflation of those two.
     
    Last edited: Aug 6, 2004
  13. Aug 6, 2004 #12
    This is a big help. I appreciate it!

    Again let me focus on the part of your post that I don’t understand. If I understand that part, I may be able to accept the rest of it.

    But indirectly seems good enough. If I look at the clock of the guy above me and see that his clock shows a later time than mine, then I know that it runs faster than mine (even though it took some time for the image of his clock to get to me). That’s enough to establish that A<B, which in turn is enough to establish that:

    A<B<C<D<E<F<G<H<I<J<A...

    I’m not sure what you mean by an absolute difference. But I can certainly determine whether the clock above me runs faster than mine. It does if it shows a later time than mine, because all clocks were synchronized before the ring began accelerating.

    Here’s another way to put that: If the guy on the floor above me is my twin, and I see that he is old while I’m still young, then I know for a fact that his clock is running faster than mine.
     
  14. Aug 6, 2004 #13

    sal

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    :smile:

    No, you don't know that -- not yet.

    You know it when you shake his hand, and it feels old. If you just see a picture of him, you still can't be sure.

    Part of the problem is relativity of simultaneity -- events that you view as simultaneous aren't necessarily simultaneous for someone in another FoR, so "now" is not well defined when discussing spacelike-separated events.

    The rest of the problem is that as long as you're separated by a spacelike interval, you need to get together again somehow before you can actually compare your clocks face-to-face. And the details of how you arrange that meeting really matter.

    In the standard "linear twins" paradox, when your twin arrives at Alpha Centaurus he's "younger than you" -- or so it seems. But somehow, you need to get together again before that statement becomes absolute. For consider: If he comes home, you'll find out he really is younger than you. But if instead, you fly out to meet him on Alpha Centaurus, you'll find that you're the same age! This is an important issue.

    In the case of the ring, what happens when you climb up into the compartment above? When you meet the astronaut there face to face, is he the same age as you, or is he younger? The only simple way to find the answer is ask the person on the ground: what does the person in the middle of the ring see as you climb around the ring?

    I said to start with that your age isn't affected by climbing, but I realize now that is wrong -- I was mistaken :redface:. When you get into the compartment above you, you will find, unless I'm (again) mistaken, that the astronaut there is indeed older than you. But the person who watches from the ground will tell you it's because you accelerated in order to move into the upper compartment, so your average velocity has been larger than his, and so you have aged less! In other words, the climb itself affects you.

    If you climb all the way around the ring, your average speed during the climb will be higher than the average speed of the rest of the astronauts, and you will find that you are younger than someone who remained in your compartment the whole time. If you climb down around the ring, your average speed will be lower, and consequently you'll age faster than the rest of the astronauts on the trip around the ring.

    If, on the other hand, the whole ring stops accelerating simultaneously in the stationary frame, and you then look at the astronaut in the next compartment, you'll find he's the same age as you -- as long as you didn't do any climbing during the acceleration phase! From your point of view, the compartments on either side of you will stop accelerating at different times, which will account for the fact that your ages end up the same. Quite a different picture from what the person on the ground saw!

    Ain't nuthin' simple 'bout time dilation.

    (I'll be signing off for the evening in a few minutes, BTW, so, 'till tomorrow...)
     
  15. Aug 6, 2004 #14
    I can be sure. If my twin truthfully tells me that he is 40 years old and I am only 20 at the moment I receive his message, then he is definitely older than me and I can be sure that his clock ran faster than mine. There’s no ambiguity.

    Let’s forget about anyone moving around the ring. That only complicates the experiment. Just let each person report their time elapsed to the person below them. Let me get a report that shows that the guy above me has aged more than I have by the moment that I am reading his report. Then we have the paradox.
     
  16. Aug 6, 2004 #15

    sal

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    Hmmm .... yes, you're certainly right. If you receive a time reading from someone else which is ahead of your current time, then their clock is unambiguously ahead of yours. I haven't been looking far enough into this.

    This is beginning to look to me like it's going to require a full solution in order to really understand it, but perhaps I'm still not seeing it right.


    Bear with me; I'm going to repeat what you just said, spelling it out in block letters, for my own understanding.

    I send a picture of my clock down to the cabin below me. My clock reads 12:00 in the picture. Suppose the next astronaut's clock reads 11:59 when he receives it. He sends a picture of his clock down one level; by symmetry, it must be received by an astronaut whose clock reads 11:58. He also includes the picture I sent. When the pictures make it all the way around the ring, I receive a picture of my own clock showing 12:00 -- but at that point, again by symmetry, my actual clock must read 11:50.

    So, it's a contradiction.

    Therefore, I'm wrong, and PMB was right to start with: you cannot see a "future" reading on the clock of the astronaut in the compartment above yours. Since you can see a future reading on the ship ahead of you in the case of tandem acceleration (or at least I think you can -- haven't actually worked it out) the effect of the rotation must be the key.

    Needless to say, this is not a solution :smile: it's just a statement of what the solution must say!

    If I think of an arm-waving way of proving you can't see a future time on a clock "above" you, I'll post it. Otherwise, unless someone else posts an analytic solution (or a link to one) there will be a long delay (some number of days, most likely) while I try to come up with something solid.

    A mathematical solution shouldn't be too hard, since all we really need is the "porthole" view: What does each astronaut see, looking at all the others? That can be done entirely in the stationary frame (no MCRF need be involved). It will still take some time, though.

    Fun stuff...
     
  17. Aug 6, 2004 #16

    sal

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    OK, I thought about it a little longer and realized that, from the point of view of the stationary frame, the solution is indeed obvious :redface:.

    Viewed by a stationary observer, by symmetry all the clocks in the ring must read identically at all times, since they all have identical acceleration histories.

    Suppose a signal is emitted from one astronaut, and it is received by another astronaut. It's emitted at a particular time on the ring, and that corresponds to a particular time in the stationary frame. The moment when it's received corresponds to a later time in the stationary frame, and therefore also must correspond to a later time in the ring ... and since all clocks in the ring agree from the PoV of the stationary frame, that includes the clock of the astronaut who receives the signal.

    Therefore the receiving astronaut's clock will read later than the clock of the sending astronaut at the moment when the signal was sent.

    If you look at the clock of the astronaut in the compartment above yours, then you are receiving a signal -- you received photons which traveled from his clock to you. When you receive that "signal", your clock will read later than his did when the signal was sent. In other words, if you look at the astronauts in the adjoining compartments, you will always see their clocks reading earlier than your clock does. You can only look into the past, not the future.

    And you will never see the astronaut in the compartment above you aging faster than you ... unless, of course, you climb up into his compartment, in which case your acceleration history no longer matches his, and you'll be younger as a result of the difference.

    Deriving this result from the point of view of someone on the ring would be harder. However, the resolution, as I stated it here, is entirely in terms of specific events and the elapsed proper time between them in various frames. The geometry of general relativity guarantees that the intervals between specific events on particular worldlines will be the same no matter what coordinates we use (because we carefully transform the metric tensor when we change coordinate systems to assure this result), so we can be sure in advance of doing the calculations that we'll get the same result no matter how we compute it. It's still interesting to do it from the other points of view; I'm just pointing out why we don't have to "hang fire" until it's been done that way to find out if there's really a contradiction!

    Sorry for all the confusion on my part.

    An interesting question which still remains is whether the same argument applies to two spaceships accelerating in tandem, identically as viewed by a stationary observer. I think it does but until I actually write some things down on paper I'm not going to be certain, and right now I need to shut down for the night.
     
  18. Aug 6, 2004 #17
    I agree, they must, not only by symmetry but also because otherwise we have a paradox. But, since in a rocket it seems that we could indeed have a situation where at age 20 I receive a message from my twin above me telling me that he’s 40, then the solution must elaborate and explain how that cannot happen in the pinwheel.

    Here is what I wrote before your last post. It still seems applicable. In any case, like you, I have to shut down for the night.

    You got it! Next time I put forth this paradox I’ll explain it that way.

    I see it a bit differently. Yes, all the crews’ clocks must elapse time at the same rate (or else we have a paradox). But I don’t think rotation is the key. I think rotation is irrelevant.

    The paradox arises if crew members on 2 adjacent floors elapse time differently. Thus we can disregard the rest of the ring as superfluous and focus just on a segment of the ring that contains 2 adjacent floors. Now, for our purposes, the only significant difference between such segment and a normal rocket is the curvature of the segment which makes the floors accelerate in slightly different directions.

    Consider that the curvature of the segment can be as slight as desired whereas the acceleration that the crew members feel can in principle be arbitrarily high. If the acceleration is a million g’s and the curvature of the segment cannot be detected with the finest available instruments, then you're going to have a hard time explaining how there is no difference in elapsed time as you would expect between floors of a rocket accelerating at a million g’s. In this case the segment and the rocket differ only negligibly. You'd have a hard time explaining a significantly different result.

    That is why I think that there is nothing about the rotation that solves the paradox. The above tells me that in the case of tandem acceleration, a normal rocket, there is no difference in elapsed time between floors when crew members on all floors feel the same acceleration; that is, when the acceleration measured in a MCRF on every floor is the same.

    The pinwheel paradox boils down to Bell’s spaceship paradox. Here is this paradox stated:

    I put a detailed solution to this paradox in laymen’s terms here.

    A summary of the solution to Bell’s spaceship paradox: In the lab-frame the ships individually length-contract, with their noses staying a constant distance apart. A thread that initially connects the ships would eventually break. The breakage occurs because the ships accelerate at the same constant rate and this increases the distance between the nose of the pursuing ship and the tail of the pursued ship (from any frame). Likewise, if a single ship accelerates in such a way that all parts of it accelerate at the same constant rate, then the ship is being physically stretched and will eventually break apart.

    This tells me that any given segment of the pinwheel is being physically stretched, and moreover, that this stretching cancels out the otherwise expected difference in elapsed time between floors.

    If that is true, though, it seems to conflict with general relativity, which tells us that an observer who measures the circumference of a rotating pinwheel will find that it has an excess circumference; that is, a circumference greater than 2 * pi * radius. With this solution to the pinwheel paradox, no excess circumference would be found.
     
    Last edited: Aug 6, 2004
  19. Aug 7, 2004 #18

    sal

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    I (finally!) agree. Tandem spaceships, with constant separation in the stationary frame, behave identically. An identical argument to the one I used comparing emission and reception times of a photon viewed from the stationary frame shows that the trailing ship cannot ever see a "future" time on the leading ship's clock.


    Yes, exactly!

    Consider the case of two ships. If the distance between them is allowed to contract then the trailing ship must accelerate at a greater rate than the leading ship, in order to "catch up" with it. Consequently, the trailing ship's average velocity is larger, and that that ship's elapsed proper time will be smaller than the leading ship's elapsed proper time as a result.

    So, the tricky bit here is that it's a ring, not that the path is curved (subtle distinction!).

    Here's an interesting twist on it: Instead of forcing the ring to stretch (as I've been assuming we're doing all along) let the ring shrink, and the radius decrease, as it spins up. Now it looks like the case of a single contracting spaceship, right? But in the contracting spaceship, the astronaut in the nose ages faster than the one in the tail -- why doesn't that happen here?

    Because in the spaceship, the back accelerated more than the front. In the case of the ring, all segments follow the same (spiral) path, so all have the same acceleration histories, and there is no difference in their elapsed proper times.

    And again, the centripetal acceleration appears to play no role.


    Beware all statements about GR which are couched in English rather than equations.

    In flat space (no gravity) GR and SR are the same theory. The only additional items GR brings to the table in flat space are the mathematical tools to deal with accelerated frames -- the predictions it makes are necessarily identical to SR's, because the Riemannian geometry it uses assures that the metric will, in all cases, be the Lorentz metric, merely transformed to whatever coordinates are in use. In other words, in flat space, SR is GR, with the additional restriction that we'll only ever switch to coordinate systems in which the metric still looks like diag(-1,1,1,1) -- and those are exactly the coordinate systems one can get to by using the Lorentz transforms. A non-Lorentz transform results in a different matrix for the metric, and forces one to explicitly pay attention to it.

    But regardless of which math you use (SR's flat math or GR's more general geometry), if you spin up a disk, the edge will contract, and either the edge stretches (as in our ring), the middle gets squished (and the edge moves toward the center), or the disk cracks. Note also that if the edge is allowed to stretch, then it will indeed be "too long" compared to the radius -- it will be longer than 2*pi*r. That's certainly true in the ring example we've been discussing.

    * * *

    Here's another aspect to the ring problem.

    Consider a beam of light traveling from the compartment ahead of you, into your compartment. You're accelerating while the beam moves, and in consequence, it's going to be blue-shifted. But that's an identical statement to saying you see the clock in the compartment ahead running faster than your clock!

    So, you do, indeed, see the clock "ahead" of you ticking at a higher rate than your clock. But when you first looked at it, you saw a time in the past relative to your clock due to the signal travel time between compartments, and even though it's ticking faster than your clock, it will never quite catch up to your clock.

    You can see this, once again, by looking at it from the PoV of the stationary frame. The chord of the circle traversed by a light ray going from one compartment to the compartment "behind" it decreases in length as time goes by, so the number of "wavecrests" stored along the chord must also decrease; that change shows up as a blue-shift. In addition, the trailing compartment's clock is going slower when it receives the signal than the sending clock was going, since they're both accelerating; that also contributes to the blue-shift, but only to the extent that the "sent" reading is behind the reading when the signal is "received". Thus, while both of these effects clearly result in blue-shift (and hence an accelerated tick rate on the observed clock), they both obviously can never cause the shift to be large enough, for long enough, such that the leading clock ever quite catches up to the trailing clock.

    This argument also applies equally well to the case of tandem accelerating ships which maintain a constant distance in the stationary frame.
     
  20. Aug 9, 2004 #19
    And if the tandem spaceships have constant separation in the stationary frame and the spaceships are accelerating, then their crews feel the same acceleration; that is, the acceleration is the same rate in the MCRFs.

    Agreed.

    Not sure what you mean here. Can you elaborate?

    I was thinking about this more this weekend. You are right. The solution to the pinwheel paradox is actually a nice proof of the excess circumference. With my last post I was thinking that if the pinwheel segments separated during acceleration then there isn’t an excess circumference because you can’t count the gaps. But you should as you pointed out. Plus, if the pinwheel was created in a rotating frame then those gaps would naturally be there to be filled in with extra segments. So if you came across such a rotating pinwheel then you’d know that it had more segments than it could if it were not rotating.

    Alas, I won’t have more time for this until this evening or tomorrow. I’ll digest your interesting twists then.
     
    Last edited: Aug 9, 2004
  21. Aug 9, 2004 #20

    sal

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    This is one of those statements that sounds deeper than it is...

    The fact that it's a ring forces the ships to be symmetric -- all are necessarily identical from the point of view of the central observer. With a line of ships, or when considering multiple floors on a single ship, that's not true.

    To put it another way, if we were dealing with a very small universe that had the topology of a cylinder, so that a "straight" line of ships could form a continuous ring, it would behave the same way our "curved" ring behaves, because there's no magic in the fact that the ring curves.
     
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