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Length contraction

  1. Sep 12, 2010 #1
    http://cas.web.cern.ch/cas/Baden/PDF/Relativity.pdf
    on pg 16, it says that a muon travelling at a relativistic velocity towards earth would see the distance to earth shorted by the gamma factor... i thought the length of the muon is all that is shortened by the factor of gamma? am i wrong here? :s
     
  2. jcsd
  3. Sep 13, 2010 #2

    Doc Al

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    They are talking about an observer moving along with the muon. To that observer, the distance is shortened.
     
  4. Sep 13, 2010 #3
    In other words, each observes the OTHER as being shorter than the other measures locally. It's a reciprical effect.

    Likewise, each would observe the passage of time of the OTHER as passing more slowly than their own time....Neither measures are absolute but are depend on relative observer velocities.
     
  5. Sep 13, 2010 #4
    okay, so if a muon is travelling towards to earth at 0.9c, lets say we travel alongside it - we are the muon's eyes.. what we see is what the muon would "see"

    we will see the earth as being closer because all the space in between us and the earth is moving relative to us if we take our own coordinate system to be at rest, therefore every bit of every atom appears to be moving fast to us and everywhere in between the earth and us appears to be contracted in the direction in which we are travelling

    is this corre
     
  6. Sep 13, 2010 #5

    Doc Al

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    That's correct. Either viewpoint--from the muon's frame or from the earth's--is equally valid.
     
  7. Sep 14, 2010 #6
    What would the muon see as the shape of the earth? Would it too be flattened in the direction of travel, sort of a disk-ish shape?
     
  8. Sep 14, 2010 #7

    Doc Al

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    Yes. If the muon observer measured the dimensions of the earth, he would find it to be squashed along the direction of motion.
     
  9. Sep 16, 2010 #8
    And if there were an object to the opposite direction of the earth, 'behind' the muon; that object would seem farther away right?
     
  10. Sep 17, 2010 #9
    No, it would also be contracted.
     
  11. Sep 17, 2010 #10
    Are this true? An object moving away from the observer will appear closer? Would it not appear farther away? (Like red shift and blue shift essentially.)
     
  12. Sep 17, 2010 #11
    For a constant velocity the distance to the observer will increase but there will still be length contraction between them. However if an object accelerates in a certain way not even that is true, an object could accelerate away from an observer and get increasingly closer to him.
     
  13. Sep 17, 2010 #12

    DaveC426913

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    Essentially, the entire universe - every measurement you might care to make (except c) - would be contracted along the axis of travel.
     
  14. Sep 18, 2010 #13
    I have difficulty accepting this. Velocity is relative. There is no means for an observer to tell their velocity in space, only relative to other objects. If your statement were true, there would always be contraction in all directions even when the observer is at rest compared to his surroundings (e.g. the observer is in a galaxy, but the galaxy is moving very fast compared to other galaxies, so there would be length contraction on that axis). This does not work. You have to look at the 'special' case of referring to one object and it's relative velocity to determine if there is relative length contraction. It has no 'general' length contraction that is innate to it's reference frame.

    Tell me this, for the photon, are not all distances in front of it zero, and all distances behind it infinite? That would mean that approaching the speed of light, all distances behind become relatively longer while distances in front become relatively shorter--that is, with reference to objects on that axis of travel whose difference in velocity is nearly c. (Whereas, in reference to objects co-moving with the reference frame will not see this effect when compared to one another--because their relative difference in velocity is zero.)
     
    Last edited: Sep 18, 2010
  15. Sep 18, 2010 #14
    Nevertheless it is so.

    For instance when someone travels at 0.6 times the speed of light everything in front and behind him will be contracted by a factor 1.25. So one meter in front of him and one meter behind him will be 80 centimeters.
     
  16. Sep 18, 2010 #15

    JesseM

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    Yes, that's right.
    I think DaveC426913 was assuming for simplicity that everything else that makes up "the universe" shares a common rest frame, and that the observer is moving relative to this frame. If different galaxies are moving at different velocities, then any given observer should measure different galaxies to have different length contraction factors in that observer's rest frame.
     
  17. Sep 18, 2010 #16
    You cannot have a velocity that is simply 60% the speed of light, because you cannot have velocity relative to light--velocity relative to light is always c. Thus, you can only have velocity to other objects. So, if you have velocity of 0.6c relative to another object, then only with reference to that object would there be a length contraction of a factor of 1.25. If you looked a different object along the same axis which was at a different relative velocity, the factor of length contraction would be different.

    Now, separately from this, I see from special relativity that there is always length contraction in the direction of relative motion for measurement of length. However, this does not account for if the observer in the moving frame of reference is looking forward or backward--for it would then be his view of spacetime intervals that seemed shortened in front and lengthened behind. Correct?
     
  18. Sep 18, 2010 #17
    Obviously.

    No, first of all I suppose you want to say distance instead of spacetime interval. If you actually mean spacetime interval you have another conceptual misunderstanding about special relativity.

    Several people have now told you that length contraction applies to the axis of travel, that means both in front and behind, why do you keep asking the same question? The answer will not change.
     
  19. Sep 18, 2010 #18

    JesseM

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    Visually it's true that the apparent length of the same object would depend on whether it was traveling towards you or away from you, but length contraction has nothing to do with visual appearances, it's what remains after you correct for light delays and figure out where the front and back of an object were at the same moment of coordinate time in your frame (even though light from events at the front and back which occurred simultaneously in your frame does not actually reach your eyes simultaneously) So, length will contract by the same factor regardless of the object's direction of travel.

    See here for more on the difference between visual changes in length and "actual" length contraction of an object which is moving relative to your rest frame:

    http://math.ucr.edu/home/baez/physics/Relativity/SR/penrose.html

    And this page has a good discussion of differences in apparent visual length for objects moving towards you or away from you (turns out objects moving towards you actually appear stretched visually, not contracted):

    http://www.spacetimetravel.org/bewegung/bewegung3.html
     
    Last edited: Sep 18, 2010
  20. Sep 18, 2010 #19

    DaveC426913

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    As far as the observer is concerned, he is stationary and every other object in he universe is moving at a sizeable fraction of c. Thus, they are all flattened along that axis.

    Note: there is no mention of any absolute velocity here (since we all know there's no such thing); it is all relative.

    Yes, an observer moving at a sizeable fraction of c through the universe could indeed tell that he is moving at that speed. The average of multiple objects in his view would be observed as moving to his read at a sizeable fraction of c. He can presume its really him who's moving.

    This does not follow.
     
  21. Sep 19, 2010 #20
    Yes, but this is still velocity relative to the content of the universe, a bit like the often used and useful concept of the fixed stars forming a background against which all local motion takes place. Although it is true that in your scenario it is a reasonable assumption that it would be the observer who is in motion, it would not be a valid proof that the observer is "moving."

    I know from the quality of your previous posts that you are aware of this all this: I pointed it out for the benefit of absolute/relative beginners to who it might be misleading.

    Matheinste.
     
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