# Accelerating frame transformation

1. Mar 21, 2016

### Fek

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. Mar 23, 2016

### Twigg

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.