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A Bell's spaceship paradox: after the thread breaks...

  1. May 24, 2017 #1

    Philip Wood

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    When Bell says that the thread in put under "intolerable stress" and breaks, what happens then? Suppose that instead of the thread there is a light rod, which breaks at the point of attachment to the back spacecraft, so it is left sticking out backwards from the front spacecraft..

    In the frame in which the spacecraft are at rest before they start to accelerate, would the rod become more and more contracted, while the distance between the spacecraft remains the same (fixed by hypothesis: [itex] x_A - x_B = (\frac{1}{2} at^2 + X) - \frac{1}{2} at^2 \ \text {with}\ X\ \text{being a constant} [/itex]), so a gap between the back end of the rod and the back spacecraft opens up and becomes wider and wider?

    I'm not trying to catch anyone out; I just want to know if this is correct.
     
  2. jcsd
  3. May 24, 2017 #2

    A.T.

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    Yes, just like the spacecraft themselves.
     
  4. May 24, 2017 #3

    Philip Wood

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    Thank you. So a gap between the broken back-end of the rod and the back spaceship will open up and widen? I just want to make sure that this is a valid deduction. [Incidentally, I'm taking the spacecraft as of negligible linear dimensions compared with the distance between them.]
     
  5. May 24, 2017 #4

    Ibix

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    Yes. This isn't a general truth about length contraction and how it happens under acceleration. For example one could construct a scenario where the join stays together and the ships get closer together. But in this case, I agree with your analysis.
     
  6. May 24, 2017 #5

    Philip Wood

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    Thank you. And I approve of your caveat! Perhaps one might say that the innocent-looking kinematics equations I quoted have unexpected consequences in SR ?
     
  7. May 24, 2017 #6

    pervect

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    If you mean:
    the equation is only valid in the limit for small t. You can find the relativistic version of ##\frac{1}{2} a\,t^2## <at this link>.

    $$x = \frac{c^2}{a} \, \left( \sqrt{1+\left( \frac{at}{c} \right)^2}-1 \right) $$

    which has a series expansion of

    $$
    \frac{1}{2}\,a{t}^{2}-\frac{1}{8}\,{\frac {{a}^{3}}{{c}^{2}}}{t}^{4}+\frac{1}{16}\,{\frac {{a
    }^{5}}{{c}^{4}}}{t}^{6}+O \left( {t}^{8} \right)
    $$

    Here t and x are the coordinates of the rocket in the inertial frame in which the rocket is initial at rest. It's worth looking at the velocity of the rocket as well, ##v=a\,t## gives the obviously non-relativistic result that the velocity of the rocket in the inertial frame exceeds the speed of light. The relativistic equation has the rocket approaching the speed of light asymptotically, as one would expect.
     
  8. May 24, 2017 #7

    Philip Wood

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    pervect: Thank you. My first thoughts are that the [itex]a[/itex] in the equations you quote must be the proper acceleration, [itex] \frac {d}{dt} \left( \frac{dx}{d \tau} \right) [/itex]. My [itex]a[/itex] is [itex]\frac {d}{dt} \left( \frac{dx}{dt} \right) [/itex].
     
  9. May 24, 2017 #8

    pervect

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    Yes, a is the proper acceleration. I should have mentioned that.
     
  10. May 24, 2017 #9

    PeterDonis

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    That's correct, ##a## is proper acceleration (i.e., a constant, independent of ##t##) in the equations pervect gave.

    That still doesn't make the equation ##x = \frac{1}{2} a t^2## correct. I suggest working through the details explicitly to see why. (All you have to do is take the second derivative of pervect's formula for ##x##, with respect to ##t##, to get the formula for your ##a##, i.e., for ##d^2 x / dt^2##, then rewrite the formula for ##x## using that. What do you get?)

    However, your overall logic is still correct; the only thing you really need for that is that the equation for the worldlines of both spaceships takes the form ##x = f(t) + x_0##, where ##f(t)## is the same function of ##t## for both worldlines. That ensures that ##x_a - x_b## will be constant, independent of ##t##, regardless of the specific form of ##f(t).
     
  11. May 24, 2017 #10

    Philip Wood

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    PeterDonis. Thank you. I thought that [itex]x=\frac{1}{2}a t^2 [/itex] was an inescapable mathematical consequence, for constant [itex]a[/itex], of [itex]a= \frac{d}{dt} \left(\frac{dx}{dt} \right)[/itex], but I shall suppress incredulity and start with pervect's equation as you suggest. Thanks again.
     
  12. May 24, 2017 #11

    PeterDonis

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    But ##a## by your definition is not constant. It is by pervect's definition, but in relativity, pervect's definition and your definition are not the same. They are in non-relativistic Newtonian physics, but that's not what we're talking about here.
     
  13. May 24, 2017 #12

    DrGreg

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    Your equation is correct if coordinate acceleration ##\text{d}^2x/\text{d}t^2## is constant over time. PeterDonis and pervect are referring to the case when, instead, the proper acceleration is constant over time. But, as Peter pointed out, as far as the conclusion is concerned, it doesn't matter if acceleration is constant or not, as long as both ships accelerate in the same way as each other.
     
  14. May 24, 2017 #13

    PeterDonis

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    But if that is the case, the speed will exceed the speed of light in a finite time. Or, equivalently, ##\gamma \rightarrow \infty## in a finite time. Also, this motion requires constantly increasing rocket thrust, because the proper acceleration increases with time; and the rocket thrust also increases without bound in a finite time (it goes like ##\gamma^2##). So this motion cannot be maintained indefinitely.
     
  15. May 24, 2017 #14

    PeterDonis

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    Also, just for clarity, the motion that is standardly assumed in the Bell spaceship paradox is constant proper acceleration, with the same magnitude for both spaceships. In other words, each spaceship's motion is described by pervect's equation (with an appropriate offset in ##x## based on the initial positions of the spaceships) in the standard version of the paradox.
     
  16. May 24, 2017 #15

    Philip Wood

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    Yes, I've just derived "pervect's equation" for [itex]x[/itex] from constancy of proper acceleration.
    Yes indeed – if constant co-ordinate acceleration could be maintained – but it can't.

    But I'm being a bit of a devil's advocate; although I still believe that [itex]x=\frac{1}{2}at^2[/itex] isn't actually wrong for constant co-ordinate acceleration, I do agree that constant proper acceleration is a much more realistic thing to assume, and also that it doesn't matter which (if indeed either) we assume if we're concerned only with Bell's paradox.

    I've found this whole discussion very helpful. Thank you.
     
  17. May 24, 2017 #16

    PeterDonis

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    It isn't. But, as you appear to agree, that state of motion can't be maintained, and the assumption of constant proper acceleration is much more realistic--one reason being that that state of motion can be maintained indefinitely (at least in principle).
     
  18. May 25, 2017 #17

    Philip Wood

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    I do agree.

    Just a foonote… I find that the equation for [itex]x[/itex] under constant proper acceleration, [itex]a[/itex], can be cast into the form [tex]x=\frac{1}{2} a \left(t^2 - \left(\frac{x}{c}\right) ^2 \right).[/tex]
    Quite pretty, I thought.
     
  19. May 25, 2017 #18

    PeterDonis

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    Is the ##x## on both sides the same ##x##?
     
  20. May 25, 2017 #19

    Philip Wood

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    Yes – which means that this re-arrangement is not much use for calculation purposes, but I thought it bore an interesting resemblance to [itex]x=\frac{1}{2}a_{co-ord}\ t^2[/itex]. I'm thinking of [itex] t^2 - \left(\frac{x}{c} \right)^2 [/itex] as the space-time interval between (0. 0) and (ct, x).
     
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