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Accelerating frame transformation

  1. Mar 21, 2016 #1


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    1. The problem statement, all variables and given/known data

    In Minkowski spacetime we are considering a (series of) frame(s), S', attached to a rocket with constant proper acceleration. The rocket's speed in S is v.

    We find with boundary conditions x = 0 at t = t' = 0 the relationships between S and S' (for x' = 0, i.e. at the rocket):

    $$t = \frac{1}{a}sinh(a't') $$
    $$ x = \frac{1}{a'}[cosh(a't') - 1] $$

    Let's use these results to construct a full coordinate transformation from the lab frame x,t, to the accelerating frame x',t'. Try

    $$t = Asinh(a't') + B$$
    $$ x= Acosh(a't') + C $$

    Prove that if
    i) surface of constant t' are surfaces of constant time in a frame moving instantaneously at v
    ii) t matches with t' at early times and small x', while x agrees with x' at early times

    then A, B and C are uniquely determined and that

    $$t = (\frac{1}{a} + x') sinh(a't'/c)$$
    $$ x = (\frac{1}{a} + x') cosh(a't'/c) - \frac{1}{a} $$

    2. Relevant equations

    3. The attempt at a solution
    For x to agree with x' at early times (cosh(a't')=1) we know:

    $$A = (k + x') $$
    $$C = -k$$

    Where k is a constant.

    Then t and t' to agree at early t' and small x' we can see k= 1/a.

    However I don't know what limit condition 1 implies and cannot see a way to make the proof "watertight".
  2. jcsd
  3. Mar 23, 2016 #2


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    Gold Member

    Condition 1 basically says you can set up a third frame S'' at the same speed as S' relative to S, and that t'' depends only on t' (not x').
    Sorry for the brief reply, I'll elaborate when I can.
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