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Unequal accelerations among comoving objects

  1. Oct 1, 2013 #1
    http://en.wikipedia.org/wiki/Bell's_spaceship_paradox#Importance_of_length_contraction

    In Bell's spaceship paradox, the spaceships appear to maintain distance from one another from inertial frame S. Got it. But because of length contraction within each rigid body, when observed from the frame of one space ship, the distance between the ships would expand and the "contraction" isn't seen. Got it. How, though, is this due to unequal accelerations as claimed by apparently every great name attached to this? I see that from the perspective of one spaceship, the other ship will appear to accelerate away. But why? Where does this derive from? I'm failing to attain a visceral understanding of where the difference in velocity is arising.

    Bonus question: If this kind of expansion between bodies requires the frame to be moving relative to an inertial frame, is this effect inherently reliant on an inertial frame existing? See, from the perspective of the ships, they may see one another accelerate away inexplicably if the inertial frame is sufficiently unapparent. Forgive me, but this reminds me quite a bit of dark energy/cosmic expansion.

    Thanks to anyone with the patience to walk me through this.
     
    Last edited: Oct 1, 2013
  2. jcsd
  3. Oct 1, 2013 #2

    Nugatory

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    When we say that two spaceships have the same acceleration, we are saying that they are both changing their speed by the same amount at the same time.

    And, thanks to the relativity of simultaneity, "at the same time" is a frame-dependent concept. If in one frame the two spaceships are changing their speed by the same amount at the same time, then in other frames they are not.
     
  4. Oct 1, 2013 #3
    So does the speed of light and observation hold the explanation to this? What I'm thinking is that from frame S, both ships can be seen moving simultaneously because they are equally distance from an observer in S, thus light reaches the observer equally. But in S', each ship will observe a delay between its own movement and the movement of the other ship due to c?
     
  5. Oct 1, 2013 #4

    Nugatory

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    No. As I and others have said in many other threads, "relativity of simultaneity is what's left over after you've allowed for light travel time".

    As soon as the ships have started moving, they and the ground-based observer will have a different notion of simultaneity (Have you been through Einstein's train thought experiment of relativity of simultaneity yet? If not, you should get it down cold before you take on Bell's paradox). Therefore, the simultaneous (to the ground observer) speed changes required to maintain the same acceleration profile (for the ground observer) and therefore the same distance between the ships (for the ground observer) will not be simultaneous to the shipboard observers. And if one of them changes speed at a different time than the other, than the distance between them won't stay the same.
     
  6. Oct 1, 2013 #5
    I have, and that's actually what I was referencing but I may have worded it poorly. In the train paradox, the difference in simultaneity is due to the fact that c is constant relative to each observer. So in S (ground observer) for the case of the ships, both ships move equally and light reaches the observer in S simultaneously from both. But I also may note that as seen from S if the ships are moving in x+, the ship at x1 will perceive the movement of x2 (being x1 + some constant) before x2 perceives the movement of x1. Then from the frame of either ship, c is constant, but each (once acceleration starts) will only observe the other accelerate after some short delay, though x1 will observe a shorter delay than x2.

    Though I think I'm just reciting, this is actually helping. Thanks!
     
  7. Oct 3, 2013 #6
    MODERATORS NOTE: this post is factually incorrect, see DrGreg's reference below

    No, actually the real resolution to "Bell's Spaceship Paradox" is that there is no paradox at all ! This is because the thread, or string between the spaceships does not break according to Special Relativity. When Bell resurrected what was a 17 year old 'chestnut', he was analysing it using the pre-Einstein theory of Lorentz, Larmor & FitzGerald ! In Special Relativity there is no distinction between the length of the connecting thread and the distance between the identically moving spaceships located at the end of the thread. So the idea that the thread breaks is a fallacy, and identically accelerating spaceships, rockets or other bodies remain the same physical distance apart whether this is considered from the original launch frame, the co-moving frame, or any other frame ! There is therefore no thread breaking & no 'paradox'.
     
    Last edited by a moderator: Oct 3, 2013
  8. Oct 3, 2013 #7
    Interesting. Would you mind linking the sources for that?
     
  9. Oct 3, 2013 #8

    DrGreg

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    Almost all of that is incorrect. The string does break. See Bell's spacehip paradox.
     
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