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Bell spaceship paradox

  1. Jun 20, 2012 #1
    Hello,

    Suppose, A and B spaceships are moving with constant acceleration. There is a string tied to spaceships center to center. A is on left of B and B is on right of A. C is outside observer.

    As speed increase, A sees that B going further ahead, and B sees A going further behind. So, from ships PoV, string would be broken.

    But, from C's PoV, distance between A and B remains constant.

    So, I cannot understand, how string would be broken from C's PoV?

    Thanks.
     
  2. jcsd
  3. Jun 20, 2012 #2

    Dale

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    From C's PoV the string gets length-contracted. So the (constant) distance between A and B is greater than the (contracting) length of the string. Therefore the string snaps.
     
  4. Jun 20, 2012 #3

    Doc Al

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    The only way that the length of the moving string could remain constant is if it were being stretched apart. (Consider length contraction.)

    This--Bell's Spaceship paradox--has been discussed many, many times here. Do a search to find the relevant threads.
     
  5. Jun 20, 2012 #4
    Here is a very good explanation, complete with detailed math.
     
  6. Jun 20, 2012 #5
    DaleSpam, Doc Al, GAsahi

    The A and B starts simultaneously and continue with simultaneous constant acceleration in C's frame. But, not in spaceship's frame. The simultaneous changes of speed in C's frame remains spaceships at constant distance in C's frame. The distance remains equals for all journey even the distance is equal to the distance before journey.

    From spaceships' frame, distance between them increased than distance before journey. Because, B starts and changes speed before A in spaceships' frame.

    Length contraction cannot applied to only string. It can also apply to the space. If space between ships not contracted than how only string can?
     
  7. Jun 20, 2012 #6
    It is impossible to understand what you are writing (at least, for me). Did you read the Analysis? Do you realize that there are two different frames (launcher vs. the frame comoving with the two rockets)? Do you realize that relativity of simultaneity plays a big role in solving the problem?

    Because, as the problem statement says, IN THE FRAME OF THE LAUNCHER, the rockets accelerate in such a way that they KEEP THE DISTANCE BETWEEN THEM CONSTANT. Whereas the length of the string appears contracted due to Lorentz contraction.
     
  8. Jun 20, 2012 #7

    Doc Al

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    Why do you think the space between the ships is not contracted?
     
  9. Jun 20, 2012 #8

    A.T.

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    The distance between the ships stays constant in the launch frame. But the atoms of the string are contracting in the launch frame. At some point they cannot span the constant distance anymore.

    You can replace the string with a chain, and consider the individual chain links as rigid elements, which are all contracting as the chain accelerates:

    102oxg6.png
     
  10. Jun 20, 2012 #9

    Dale

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    Yes, I said that.

    I have no idea what you are talking about. What is "space's" rest frame, and what is its length in that frame? To use the length contraction formula you have to have an object with a rest frame and a length in that rest frame. I don't know how you would apply that to "space".
     
  11. Jun 20, 2012 #10

    Doc Al

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    Good point. Here's how I interpreted his statement about 'space' contracting: Imagine you have two ships traveling at high speed with respect to another observer (arranged as in the Bell set up, once the acceleration is completed). According to the ship frame, the distance between them is L. According to that other observer, the distance is L/γ.
     
  12. Jun 21, 2012 #11

    A.T.

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    If something happens to "space" then "elongation". Moving objects occupy less space then they do while at rest. This can be interpreted as:

    a) Moving object is shorter than at rest (the usual interpretation)
    b) Space is elongated for a moving object

    Interpretation b) sounds a bit weird. But when you go into a rotating frame, rulers at rest measure a circumference greater than 2πr. You cannot claim that the rulers are contracted, because they are at rest. You must conclude that the space along the circumference is elongated.
     
  13. Jun 21, 2012 #12

    Dale

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    A rotating frame is non inertial so rulers at rest can contract. You cannot simply apply the formulas for inertial frames.
     
  14. Jun 21, 2012 #13

    Dale

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    Yes, that is reasonable. Then the condition for snapping is that the distance between the ships is greater than the length of the string. In the ships frame that condition is met because the length stays the same and the distance increases. In the launch frame that condition is met because the distance stays the same and the length contracts.
     
  15. Jun 21, 2012 #14

    Doc Al

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    Right!
     
  16. Jun 22, 2012 #15

    A.T.

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    But it is not the length contraction from movement, because they are at rest. It is the apparent contraction that we get, when observing rulers resting further down in a gravitational field from a distance. This is however usually not interpreted as "contraction of the rulers", but rather "distortion of space".
     
  17. Jun 22, 2012 #16
    Ok.... Thanks guys.. for detailed explanation.
     
  18. Jun 25, 2012 #17

    Meir Achuz

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    I usually recommend <http://arxiv.org/abs/0906.1919> [Broken] for this.
     
    Last edited by a moderator: May 6, 2017
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