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Length contraction and accelerating frames

  1. Nov 16, 2011 #1
    We all know the definition of a rope between two identical accelerating spaceships and how that rope will break assuming they both accelerate with the identical same velocity (as defined from a observer on earth for example). And it makes sense, thinking of stress. But how do I define a Lorentz contraction inside a accelerating frame? It should be possible to do, as I think of it, as there should be a difference between the front and and the end of that spaceship accelerating (let's say uniformly).

    Or am I bicycling in the blue younder here?

    Maybe it's easier to see if not assuming a uniform constant acceleration, but to me they both should have it?
    Last edited: Nov 16, 2011
  2. jcsd
  3. Nov 16, 2011 #2


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    I'm not sure what exactly you mean by "Lorentz contraction inside a accelerating frame", but perhaps what you are looking for are Rindler coordinates, which are the "rest coordinates" of a "Born rigid" rocket undergoing constant proper acceleration. If (t,x) is the Minkowski coordinate system of an inertial observer (I'll ignore y and z), and (T,X) is the Rindler coordinate system in which the rocket is at rest, the coordinates are linked by[tex]
    x + \frac{c^2}{a} &= \left( X + \frac{c^2}{a} \right) \cosh \frac{aT}{c} \\
    ct &= \left( X + \frac{c^2}{a} \right) \sinh \frac{aT}{c}
    [/tex]Here a is the proper acceleration of the part of the spaceship located at X=0.

    Does that help?
  4. Nov 17, 2011 #3
    Thanks, the problem for me is how to think of it intuitively. By 'inside a accelerated frame' I meant that I started to think about it as if I split the spaceship into 'frames of reference', each one 'still' as in a instant, and wondered if they would notice a Lorentz contraction relative each other. In a uniformly moving spaceship it won't be noticeable as there is no 'shear stress', if I'm using the proper nomenclature there? But inside that accelerated frame? Shouldn't there be difference. If I assume that to all 'clocks' differing there is a reciprocal Lorentz contraction and if I think of it as taking 'snapshots', frozen in motion.

    And then I started to think of spinning a round plate close to light :) Wondering if that would help me see it. But there the 'cracks' come from them (like the tracks on a cd, sort of) not having a 'same speed' relative each other, if you wander from the spinning edge inwards towards the centre, well, as I assume. But it's bothering me as if it is right (which I think it is) then if I assume a extremely high acceleration, shouldn't that spaceship also be able to 'crack', as every 'acceleratory instant' will 'push' those 'frames of reference' a little more relative each other?

    On the other hand, the clocks go fastest at the front of the ship right, but the 'push' comes from the aft, not the front. So even thought the clocks goes faster at the front the clocks aft should get first updated?

    Awhhhh :)
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