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What is the shape of length-contracted space?

  1. Apr 21, 2014 #1

    BitWiz

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    What is the shape of an "isodistance" volume from the point of view of an observer traveling at relativistic speeds? (EDIT: ... compared to that of an outside observer)

    If it is not spherical, how do the distorted (?) distances of all mass objects surrounding the observer affect the observer's path through spacetime? Is there more gravity parallel to the direction of travel? If a rocket points at a massive destination object, will that object tend to align the relativistic rocket in that direction as velocity increases?

    Thanks!
     
    Last edited: Apr 21, 2014
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  3. Apr 21, 2014 #2

    phinds

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    I'm not sure I really understand your question, based on the wording.

    You, right now as you read this, are traveling at a speed of .99999c relative to SOMETHING in the universe but your observable universe, for example is still a perfect sphere. If you define a sphere, say 1,000,000 miles in diameter, centered on you, and there is an observer 1LY away traveling at relativistic speeds not directly towards or away from you, then objects at the edge of that sphere will not be appear to be arrange in a perfect sphere based on you but will be deformed a bit, depending on both his angle of travel and his speed.

    A rocket traveling at relativistic speeds would pass by a massive object so fast that the amount of time it spent in the serious part of the gravity well, even if it came fairly close, would be too small to make much of a course change unless it was a black hole and you were going really close so no, a rocket traveling at relativistic speeds and far enough away to not be there yet would not be much affected.

    EDIT: and by the way, space does not appear length-contracted since space is not a thing. Stuff IN space appears length-contracted to someone moving at relativistic speeds relative to the stuff.
     
    Last edited: Apr 21, 2014
  4. Apr 21, 2014 #3

    BitWiz

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    Thanks for the reply, phinds. I agree I didn't word this very well. Let's try this:

    Two observers are co-located in space at the same point. One of the observers is at rest with respect to the stars, and the other is traveling at relativistic speed. Each observer is at the center of an isodistance sphere of the same arbitrary radius. With respect to either observer, what would the other's "sphere" look like? What shape would it be?

    Again, a badly worded question. Let me try this: Let's say that an observer manages to accelerate very close to the speed of light toward the core of the Andromeda Galaxy over a very short distance. Length contraction will bring Andromeda arbitrarily closer. After the acceleration, will the observer feel a stronger gravity pull from Andromeda?

    Thanks!!
     
  5. Apr 21, 2014 #4

    pervect

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    A stationary sphere , observed by a observer in relative motion, would appear as an ellipsoid. The easiest way to visualize it is to visualize a cube around the observer, apply the lorentz transformation to shrink the sides of the cube parallel to to the relative motion (while not affecting the transverse sides), then embedding an ellipsoid in the resulting box.

    More on the second question later...
     
  6. Apr 21, 2014 #5

    pervect

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    OK - for the second question. I'm going to cheat and give the answer for electromagnetism, because the notion of an electromagnetic force is better defined than the notion of a gravitational force, and because the math is easier.

    If andromeda was charged, and you measured the electric field (by measuring the relative acceleration between a charged and noncharged particle, for instance), and the moving observer also measured the electric field in a similar manner, you'd both get the same answer if the moving observer was headed directly towards andromeda.

    The math looks like this (from wiki)

    0f4d89f56e8e782c7062ae9e054183e5.png

    where the subscripts indicate fields parallel and transverse to the direciton of motion, repsectively.

    If you sketched out the electric field from Andromeda, and compared it from relatively stationary and relatively moving points of view, you'd get something like:

    attachment.php?attachmentid=37949&d=1313150679.png
     
  7. Apr 21, 2014 #6

    PeterDonis

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    It's worth pointing out that the word "appear" here is not to be taken literally; what you're describing is the shape of a spacelike slice through the sphere's world tube, taken orthogonally to the observer's worldline. This is *not* the same as the way the sphere would literally appear to the observer's eye, because of light travel time delays; to the observer's eye, the sphere will actually appear spherical, but rotated.

    For more gory details, see, for example, here:

    http://math.ucr.edu/home/baez/physics/Relativity/SR/penrose.html
     
  8. Apr 22, 2014 #7

    pervect

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    Right, I'm not talking about the visual appearance, but what happens after one corrects for the propagation delay of light.

    I'd agree that that is what I'm describing, but while it is accurate, I wouldn't say it's a "beginner friendly" way to describe it :(.
     
  9. Apr 22, 2014 #8

    BitWiz

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    Thank you, pervect and PeterDonis.

    And thanks for trying to make the concepts accessible. ;-)

    ... But I still feel I have been unable to convey my question ... or I may have the wrong idea entirely based on what phinds said:

    I don't understand this. If a rigid rod connects me and a distant object, and (instantaneously if necessary) the distance between me and the object closes relativistically, does the space(time) inhabited by the rod collapse or not collapse with the rod?

    Perhaps I need the answer to this before I can reattempt the other question.

    Thank you, sirs!
     
  10. Apr 22, 2014 #9

    phinds

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    Your problem here, I think, is with your concept of a rigid rod. It is something that does not exist in the sense that I think you mean it in. A rod cannot collapse at relativistic speeds, the way your example would require, by simple having the object at the other end push on it and move itself and the end of the rod towards you at relativistic speeds. That would require compressive forces to move in the rod faster than the speed of sound in the rod and that can't happen. On the other hand, if you think of a rod that COULD compress at relativitic speeds (I can't imagine what that would be made of) it will just get shorter as the object moves towards you. The END of the rod will get shorter and the rest of the rod will be unaffected and the object will appear length contracted.
     
  11. Apr 22, 2014 #10

    BitWiz

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    Thanks, phinds. What I'm trying to separate is the reduction in distance I get from the derived velocity I integrate from acceleration, i.e. "speed," and the reduction in distance due to relativistic effects. Let me modify the example:

    A rigid rod extends from a distant object to a point in space one meter in front of me. I accelerate for one meter. At the point that I just touch the rod, has the length of the rod changed from the time that I was at rest with respect to the rod? If so, what has happened to the space(time) that the rod occupied?

    Thanks! ;-)
     
  12. Apr 22, 2014 #11

    timmdeeg

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    In my opinion nothing happens to the space-time. As long as there is relative velocity you will "see" the rod length contracted parallel to the direction of the motion. As soon as you stop in front of the rod you are in it's rest-frame and you will see it's length accordingly (not contracted).
     
  13. Apr 22, 2014 #12

    BitWiz

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    Hi, Timmdeeg. I don't stop. In fact, I'm still accelerating. I'm just at that instantaneous point in my travel. What would you say then?

    Thanks for the reply!
     
  14. Apr 22, 2014 #13

    PeterDonis

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    It depends on what you mean by the "length" of the rod. Length is frame-dependent; the rod does not have a single unique "length". As far as any invariants describing the rod, they do not change at all; physically, nothing about the rod changes, only your state of motion changes.
     
  15. Apr 22, 2014 #14

    phinds

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    No, the length of the rod has not changed. Nothing happened to the rod at all. Your PERCEPTION of the rod will have changed and you will think it is shorter but that's just an effect of your motion and length contraction and has nothing to do with the rod.

    You, right now as you read this, are traveling at a speed of .99999c relative to SOMETHING in the universe and relative to that object you are massively length contracted. Do you feel any different?
     
  16. Apr 22, 2014 #15

    timmdeeg

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    Well, then the rod is still length contracted in your frame. Do you suspect that something physical happens to the rod? As already mentioned that's not the case. Something else must be the background of your reasoning. What could that be?
     
  17. Apr 22, 2014 #16

    BitWiz

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    Good question. I think I must be missing something very fundamental.

    Say I have a series of rigid cylinders with their long axes aligned along a single line in space, and alternating between those cylinders (their ends) is an equal amount of empty space. I now accelerate in a direction parallel to, and adjacent to, the cylinder axes. As my speed increases, the length of the cylinders will *appear* to contract. Correct? And because of the angle, cylinders further away from me will appear to length contract more than those adjacent to me. Correct? If so, then:

    1) Is this merely a visual illusion, similar to a Fun House mirror?

    2) If the cylinder lengths contract, does the space in between contract, widen, or stay the same?

    3) If I double my relative velocity, will I see *more* than twice as many cylinders go past me?

    4) If #4 is true, what the heck is going on? ;-)

    Thanks, Timmdeeg.

    Bit
     
  18. Apr 22, 2014 #17

    PeterDonis

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    In the special sense of "appear" that pervect used in his earlier post, yes. But as I noted in response to that post, this sense of "appear" does not describe what you will actually *see*; it is what you will *calculate* the length of the cylinders to be, after correcting for light travel time delays.

    Not in the sense of "appear" used above, no. Length contraction depends only on relative velocity; at any given instant of your proper time (i.e., time by your clock), your velocity relative to all the cylinders is the same, so they are all length contracted the same.

    It is true that, since different cylinders are at different distances from you, what you will actually see at a given instant of your proper time will vary from cylinder to cylinder. But in that sense of "appear", the cylinders do not appear length contracted; they appear rotated, as I pointed out in my earlier post.

    I would say yes, because I think the key error you are making here is thinking that length contraction is something that actually happens to the object, instead of something that's purely due to you changing your state of motion. The fact that you are accelerating does not change the cylinders at all. Attach whatever sensors, strain gauges, accelerometers, etc., etc. you like to the cylinders: their readings will *not* change at all as you accelerate. Attach a laser and reflector to each cylinder to precisely measure its length in its own rest frame; that won't change either as you accelerate. The only thing that changes as you accelerate is you.

    It depends on what you mean by "space in between". But in any case, you should not be thinking in terms of space; you should be thinking in terms of spacetime. Spacetime does not change just because you accelerate. The only thing that changes as you accelerate is you.
     
  19. Apr 22, 2014 #18

    WannabeNewton

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    It is not about perception in any sense of the word. Length is simply a frame-dependent quantity. Saying "the length of the rod has not changed" has no meaning in SR. If you want to qualify length using "rest length" then the statement is fine but in doing so, you are singling out a specific Lorentz frame in order to do this even though the rest length isn't any more special than the length of the rod as measured in an arbitrary Lorentz frame.

    As such, length contraction is not an illusion. It is simply a consequence of making temporal and spatial measurements functions of space-time foliations.
     
  20. Apr 22, 2014 #19

    phinds

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    Fair enough. I should have said "proper length" and explained it more fully, to be as accurate as you are being.
     
  21. Apr 22, 2014 #20

    Bill_K

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    I'd say the proper length is the maximum of the lengths as measured in various Lorentz frames.
     
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