# Transformation for accelerating observers

1. Jun 15, 2012

### ralqs

Suppose two observers A and B are each accelerating at the same rate wrt an inertial reference frame, but are moving relative to one another at some uniform velocity. Will the transformation between the coordinates of an event as measured by A and B be the Lorentz transformation?

2. Jun 15, 2012

### Staff: Mentor

I'd say yes, since it doesn't matter that A and B are accelerating wrt to a third frame of reference. The key is that they are moving relative to each other at uniform velocity and you're using the lorentz transformation between them.

3. Jun 15, 2012

### Staff: Mentor

It's not that simple. Check out the Bell Spaceship Paradox:

http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html [Broken]

In addition to the Bell Spaceship Paradox, it's also good to check out Rindler coordinates and the Rindler horizon:

http://en.wikipedia.org/wiki/Rindler_coordinates

http://gregegan.customer.netspace.net.au/SCIENCE/Rindler/RindlerHorizon.html

As the above links should show you, there are a number of subtleties lurking in the question you've posed. One is that the term "accelerating at the same rate" is ambiguous. Another is that "the coordinates of an event as measured by A and B" is also ambiguous. So the answer to your question could be "yes" or "no" depending on how you choose to tighten up the specifications of your scenario to resolve the ambiguities. I suspect that the answer to the question you were trying to ask, once the ambiguities are resolved, is "no".

Edit: I should mention that "moving relative to one another at uniform velocity" is also ambiguous in this scenario.

Last edited by a moderator: May 6, 2017
4. Jun 16, 2012

### Staff: Mentor

The transformation between uniformly accelerating and inertial observers is given here:

http://en.wikipedia.org/wiki/Rindler_coordinates

If you start with an observer "at rest" in the accelerating frame you have a world line$$(t,x,y,z)=(t,x_0,0,0)$$

Transforming that to the inertial coordinates you get: $$(T,X,Y,Z)=(x_0 \sinh(g t), x_0 \cosh(g t),0,0)$$

So a similarly accelerating observer, but with different initial velocity will have the worldline: $$(T,X,Y,Z)=(x_0 \sinh(g t), x_0 \cosh(g t)+v_0 t,0,0)$$

Transforming this back to the accelerating frame gives: $$(t,x,y,z)=\left( \frac{1}{g} atan\left( \frac{x_0 \sinh(g t)}{v_0 t+x_0 \cosh(g t)} \right),\sqrt{v_0^2 t^2+x_0^2+2v_0 x_0 t \cosh(g t)},0,0\right)$$