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Angular contractions?

  1. May 24, 2004 #1
    Say, if you had a ruler attached tangent to the outside of a wheel, and this wheel is rotating near c. Does the ruler still contract?
     
  2. jcsd
  3. May 24, 2004 #2

    selfAdjoint

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    Relative to what?
     
  4. May 25, 2004 #3
    The ruler will have the same speed from one instant in time to the next relative to a stationary object and will then have a constant length contraction.Relative to an object in motion the ruler will have a different length from one instant in time to the next.
     
  5. May 25, 2004 #4

    selfAdjoint

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    What stationary object? I am trying to get the point across that you are assuming we have a fixed frame of reference where everything is either in motion or at rest with no problems. And this is not so.
     
  6. May 25, 2004 #5

    turin

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    You say that the wheel is rotating near c. One must assume that you mean the outer edge of the wheel has a tangential velocity near c (because rotation is unambiguously characterised by an angular velocity with units of rad/s, not m/s). In this frame of reference, the ruler would be observed to contract. Note: the geometry of the wheel isn't flat.
     
  7. May 25, 2004 #6
    Then does the circumferance of the wheel also contract(relative to a stationary observer)?

    I don't even know enough about this to ask a proper question so please bear with me.
     
  8. May 27, 2004 #7

    turin

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    Well, there are two ways that you can consider it. Let us assume a flat space against which the wheel rotates and call it the background.

    In the background, Euclid's postulates hold, so the circumference of any circle would measure 2πr. To think about this operationally, imagine the background containing a canvas or sheet of paper that can be marked on. Then, imagine that the wheel is parallel to this sheet with just enough clearance to rotate freely. If you send a brief blast of paint or some other marking agent straight down from above the sheet, then the "shadow" will give you a circle with a circumference C = 2πr.

    In the space that rotates with the wheel, all Euclidean bets are off. This space is not flat (because it is absolutely rotating wrt the flat space). Operationally, you could imagine a robotic probe on the wheel that is programed to travel the diameter of the wheel and then the circumference. It records the distance on an odometer. Of course, it is important to assume that the probe has exceptional traction on the wheel, so this is far more quasi operational than even the paint shadow measurement from above, but, maybe you can allow your imagination to take you there (and remember, this is only a model anyway). The probe would show a relationship of C > 2πr assuming only special relativistic effects. In other words, the wheel would have a saddle shape geometry as far as the probe readings are concerned, with the inflection point in the middle.

    If the wheel were to stop rotating wrt the background, then the shape would rinkle up and it really would be a saddle shape. Inversely, if you start out with a wheel that is not rotating and then make it rotate wrt the background, then it would "curl up" into a paraboloid and cease to be a wheel wrt the background.

    Keep in mind, this is all based on the model of special relativity that does not really deal well with curved geometries. Also, the distortions suggested in the previous paragraph are rather speculative and I am making a liberal extension of the model for the purpose of visualization.
     
    Last edited: May 27, 2004
  9. May 27, 2004 #8
    No. I don't believe so. However this is related to something which has been the subject of much debate in the relativity literature. To learn more about it, look up "Ehrenfest's Paradox"

    PEte
     
  10. Aug 5, 2004 #9

    sal

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    The assertion that the rule won't contract can't be correct. If it's taped in place at both ends with very strong tape, it may stretch and appear unchanged in the stationary frame, but it must be under tension in that case.

    Consider a simple "reduction" argument:

    Make the wheel very large -- say, a lightyear in radius. Now, to maintain the path at a velocity near C, the acceleration will be on the order of 1G (give or take a factor of 3 or so), and the curvature of the path will be quite small.
    Make the radius even larger, and the acceleration and curvature of the path will be smaller still, with a straight path with no acceleration in the limit (angular velocity drops linearly as the radius increases, assuming fixed tangential velocity at the rim).

    Surely, when the radius is very, very large and the acceleration is very, very small, the contraction must be almost exactly what it would be for straight-line motion.

    So, since the radius can be varied continuously, the ruler must contract no matter what the radius is.

    Of course, this argument assumes the ruler is short relative to the radius. If it's long relative to the radius, its center will still contract, but what's going on out near the ends will be a lot more complicated, and if you try to spin the rim of the wheel up near C in that case the ends of the ruler will break off (or break C, which won't happen).
     
  11. Aug 5, 2004 #10

    turin

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    The neat thing about this "paradox" is what happens if you have a tape-measure that goes all the way around at a fixed distance. I like the way that you have just posed the issue, sal, because it prevents people from hiding behind "the complications." In fact, the complication that you mentioned and then obviated is precisely the heart of the issue: General relativity. Call it the equivalence principle, if you will. The acceleration is 1g at a certain distance, less at further distances, and greater at closer distances. There is an "acceleration field." This is a gravitational field. Of course, the acceleration can be entirely transformed away in General Relativity by transforming to the non-rotating frame (and thus eliminating the acceleration field), but in so doing, then you have Special Relativity effects to consder. The bottom line is that the geometry of the rotating frame is fundamentally non-Euclidean.
     
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