Accelerating frame transformation

  • Thread starter Fek
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  • #1
Fek
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Homework Statement


[/B]
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

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

Homework Equations




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".
 

Answers and Replies

  • #2
Twigg
Science Advisor
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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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