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Spinning tornado wheel - circumference and radius

  1. Oct 28, 2009 #1
    Spinning "tornado" wheel - circumference and radius

    Another quote from my current lecture "elgant universe":

    Page 63, 3rd Paragraph
    This doesn't make much sense to me. While I can comprehend that the ruler must be shortened as it's layed out tagentially to the circumference, as it's moving in the direction of which it's layed out, I would assume that the circumference too is shortened, implying that Slim will get the same result for the circumference as we would, when the ride is standing still. In other words, let Slim start measuring the circumference when the ride is standing and then spin it up - assuming that his ruler is actually layed out exactly on the circumference, no discrepance should emerge.

    We however, perceive a contracted circumference due to Lorentz contraction but the same radius, meaning that in our system of observation, the ratio has changed.
     
  2. jcsd
  3. Oct 29, 2009 #2
    Re: Spinning "tornado" wheel - circumference and radius

    Not a single reply? Can it really be that hard to say whether this is true or false?
     
  4. Oct 29, 2009 #3
    Re: Spinning "tornado" wheel - circumference and radius

    To slim, his ruler is true, but to us, his ruler is contracted and no longer true. So we see him using a shortened ruler and are not surprised when he measures a very long circumference. We use a ruler which is in our inertial frame, and is therefore not contracted and appears longer than slim's ruler, so we measure a shorter circumference.
     
  5. Oct 29, 2009 #4
    Re: Spinning "tornado" wheel - circumference and radius

    In what context is the example given. Is the chapter or section from which it comes dealing with inertial frames or rotating frames.

    Matheinste.
     
  6. Oct 29, 2009 #5
    Re: Spinning "tornado" wheel - circumference and radius

    This is how it's presented in the book, but as I pointed out, this conclusion is incomplete and thus wrong. You did not address the issue I raised, just said what's wirtten in the book in your own words. Please read again and try to grasp my point:

    Yes, the ruler is shortened as we watch Slim measuring the rotating circumference while he spins with it, but so is the circumference itsself.

    Matheinste: The chapter brings about the introduction of general relavitivy, depicting the disortion of space in rotating frames.
     
  7. Oct 29, 2009 #6
    Re: Spinning "tornado" wheel - circumference and radius


    Its out of my league then.

    Matheinste
     
  8. Oct 29, 2009 #7
    Re: Spinning "tornado" wheel - circumference and radius

    It shouldn't be a different "league". All effects observed in rotating frames can allegedly be deduced from the concepts of spec. relativity (at least on the level this is being examined upon). The first quote I gave basically describes the whole idea, eventually concluding that hence space for Jim/Slim must be curved. I, however, object that - regardless of the outcome - the argument that Slim would go measure a longer circumference, is wrong.
     
  9. Oct 29, 2009 #8
    Re: Spinning "tornado" wheel - circumference and radius

    What makes you think we see a smaller circumference? The circle does not change shape as we look down on it.


    Also try not to assume I am misunderstanding you, or that the author's interpretation is wrong, just because you don't understand it.
     
    Last edited: Oct 29, 2009
  10. Oct 29, 2009 #9
    Re: Spinning "tornado" wheel - circumference and radius


    I know my own limitations but I am still learning. I have always found that the best way to learn is to trust the textbooks and if they do not make the subject clear, ask for help here. If I do that and disagree with both, then I assume there is a major lack of understanding on my part. I am always correct in this last assumption.

    Matheinste
     
  11. Oct 29, 2009 #10
    Re: Spinning "tornado" wheel - circumference and radius

    Mikey, regardless of on whose part the misunderstanding lies, your two contributions to this thread have in no kind of way helped resolving anything. I thus ask you to leave, instead of handing out smart advices such as "...just because you don't understand" and in the end not saying anything, until you find yourself apt to join me in the argumentative debate I tried to set off with my introductory post.

    Matheinste, the fact that I opened this thread suggests, that I'd rather believe the textbook's facts than my conclusions, which are in contrary to the former. But as it's popular literature, I think everyone qualifies for drawing his/her own conclusions and try to prove me/the book wrong.
     
  12. Oct 29, 2009 #11
    Re: Spinning "tornado" wheel - circumference and radius

    The circle you observe does not "shrink", special relativity analysis can only make conclusions about contractions along tangent lines which represent true inertial frames.


    Anyway I'm not rising to this, I've tried to help and my advice is free and educated, I'm sorry you don't want it.
     
  13. Oct 29, 2009 #12

    Ich

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    Re: Spinning "tornado" wheel - circumference and radius

    Then I htink it's time for you to start again from the beginning: Why do you thing the circumference would shrink? What exactly do you mean by a "shrinking circumference"? Can you try to imagine the situation in spacetime, with this helical simultaneity?
    You have to rid yourself of that misconception, as the book's (and MikeyW's) arguments are quite clear.
     
  14. Oct 29, 2009 #13

    DrGreg

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    Re: Spinning "tornado" wheel - circumference and radius

    Instead of a solid disk, consider a thin rod bent to form a circle of radius r, initially not spinning. The two ends of the rod are not welded together to form a complete circle; there is a tiny 1 mm gap between the two ends. The length of the rod, measured by the bird's eye observer Jim stationary relative to it, is 2πr.

    Now start the rod rotating around the centre of the circle. Relative to the bird's eye observer Jim, the rod contracts to a length of 2πr/γ. This means the rod no longer forms a complete circle; the gap has expanded to 2πr(1 − γ−1). However, if an observer, Slim, standing on the rotating rod measures its length by slowing walking around it with a tape measure, the length will be an uncontracted 2πr.

    Slim has with him some rods and welding equipment and welds more rods into place to fill the gap and complete the circle. The complete circle has a circumference of 2πr as measured by the bird's eye inertial observer Jim. But this length is contracted from the length of 2πrγ measured by the rotating observer Slim. Both observers agree that the radius of the circle is r as the radial direction is at right angles to the motion.

    If you start off with a solid, non-spinning disk, then start spinning it, then either parts of the disk will stretch "to fill the gap", or else the disk will shatter into pieces. There's no such thing as a truly rigid, unbreakable disk.
     
  15. Oct 29, 2009 #14

    DaveC426913

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    Re: Spinning "tornado" wheel - circumference and radius

    I thnk what the OP is struggling with is this:

    Why would the ruler not shorten exactly in proportion to the circumference of the Tornado? They're both going the same speed and in the same direction.

    If the circumference is relativistically shortened by half, then the ruler will be shortened by half as well, meaning his measurement of the circumference does not change.
     
  16. Oct 29, 2009 #15

    DrGreg

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    Re: Spinning "tornado" wheel - circumference and radius

    In my example, the measurement of the bent rod, initially at rest and then made to rotate, does not change, according to Slim. It gets longer only when he welds in the extra rods to complete the circle.
     
  17. Oct 29, 2009 #16
    Re: Spinning "tornado" wheel - circumference and radius

    You say it "no longer" forms a complete circle. But the previous paragraph seems to say that it never did form a complete circle.
     
  18. Oct 29, 2009 #17

    DrGreg

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    Re: Spinning "tornado" wheel - circumference and radius

    Well it almost did apart from a tiny gap (I said 1 mm but it can be as small as you like). The point is the two ends aren't stuck together, they are free to move apart.
     
  19. Oct 29, 2009 #18
    Re: Spinning "tornado" wheel - circumference and radius

    Ah, right, so they're effectively touching?
     
  20. Oct 29, 2009 #19

    DrGreg

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    Re: Spinning "tornado" wheel - circumference and radius

    Yes. Touching but free to move apart.
     
  21. Oct 30, 2009 #20
    Re: Spinning "tornado" wheel - circumference and radius

    Why would the gap expand? "It" too is moving along "its" direction. It's like you were argueing that all objects appear smaller at a distance and thus looking at Emmental cheese at a 10 meters would render its holes smaller and hence create more cheese.

    Gap or no gap is not the question. The predictions made my s/r (based upon which, we are trying to deduce the effects of g/r) do not concern objects but they concern space and time itsself. That said, it's irrelevant what we are looking it, whether there is vacuum (a gap) - even air if you like - or rod at the place. Since we regard space and not matter, everything transforms homogeneously.

    Indeed, Slim will take the exact same measure of the rod's lengh as we did, when it was standing still.

    Don't get me wrong: I'm aware of argueing for a lost cause, but so far, I don't see any convincing line of thought - although the results concluded are apparently true.
     
    Last edited: Oct 30, 2009
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