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Re: [Phys-L] another DIY relativity experiment



David,
Thanks more! This will keep me occupied for a while!

But, right now, please clarify this ==>

In the equation (c^2)*(dτ)^2 = ((c + g*z/c)^2)*(dt)^2 -((dx)^2 + (dy)^2 + (dz)^2) , does the "z" in the first term of the right hand side refer to the z coordinate of the K origin as measured in K' ?
Thanks

Bob Sciamanda
Physics, Edinboro Univ of PA (Em)
treborsci@verizon.net www.sciamanda.com
-----Original Message----- From: David Bowman Sent: Sunday, May 22, 2016 6:41 PM To: Phys-L@Phys-L.org Subject: Re: [Phys-L] another DIY relativity experiment
David,
Thanks for the helpful details.

For those who are interested, here are some even more details concerning the uniformly accelerating spacecraft problem.

Suppose we have a uniformly accelerating intergalactic spaceship whose acceleration w.r.t. a sequence of instantaneously coincident inertial frames is g along the +z direction. Suppose we make our lab frame K to be at rest w.r.t. the spaceship (because the lab is on board the ship). This is a *non*inertial frame. Let's also consider a fully inertial frame K' whose velocity relative to frame K is momentarily at rest after which the K frame accelerates away from the the inertial Minkowski K' frame along the + z' direction. In this case the invariant time-like square space-time (square proper time) interval between a pair of two infinitesimally time-like separated space-time events as seen in both the inertial Minkowski K' frame and the rocket's non-inertial frame K is given by:

(c^2)*(dτ)^2 = (c^2)*(dt')^2 -((dx')^2 + (dy')^2 + (dz')^2) &

(c^2)*(dτ)^2 = ((c + g*z/c)^2)*(dt)^2 -((dx)^2 + (dy)^2 + (dz)^2) .