Using the transformation (from your post):
x' = x
y' = y
z' = (z + (c^2)/g)*cosh(g*t/c) - (c^2)/g
t' = (z/c + c/g)*sinh(g*t/c) ==>
The third equation says that an event at z = -(c^2)/g has the primed coordinate z' = -(c^2)/g, a constant in time.
This seems to contradict the second statement of :
"Note (from the equation for t') a clock at rest w.r.t. the spaceship at z = -(c^2)/g is infinitely dilated and stopped relative to frame K'.
Such a clock is moving at speed c in the K' frame."
What am I missing?
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
. . .
x' = x
y' = y
z' = (z + (c^2)/g)*cosh(g*t/c) - (c^2)/g
t' = (z/c + c/g)*sinh(g*t/c) .
The event of O and O' being momentarily coincident marks the zero of both time-like coordinates, t & t' for each corresponding frame. At the moment t = t' = 0 all the spatial coordinates of both frames coincide, and both frames are momentarily at rest w.r.t. each other.
The inverse of the transformation above is given by:
x = x'
y = y'
z = (z' + (c^2)/g)*sqrt(1 - (t'/(z'/c + c/g))^2) - (c^2)/g
t = (c/g)*arctanh(t'/(z'/c+ c/g)) .
. . .