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bernhard.rothenstein is offline
Jun16-08, 11:00 PM
P: 997

absolute simultaneity with Einstein synchronized clocks?

Quote Quote by RandallB View Post
The set up seems extremely confused or at least incomplete.

What is “a”
“M(a(xa) “ looks like a typo.

Can we assume x’=0 is where the observer R’ is at on I’
When t=? & t’=? did R’ crossed x=0 sometime before the light flashed.
The light flashes at x=0 t=0; where and when does it go off in the I’ frame.
Since the source could be a point in the I’ frame. Where and when is that I’ frame point, x=? & t= ? when R’ was at x=0.

I don’t know what the question “Is there some flow?” means;
But I don’t see any prospects for setting any form of “Absolute Simultaneity” in this description.
Thanks for your answer. Consider please the following one space dimensions detected from I. A particle moves with constant speed V in the positive direction of the OX axis. x defines its position at t=0 when a light signal is emitted from the origin O (APPARENT POSITION) X defining its position when the light signal arrives at its location. (ACTUAL POSITION). We have
X=x+Vx/c=x(1+V/c) (1)
Let t=x/c and T=X/c be the times when the light signal arrives at the apparent and at the actual positions respectively. The mentioned light signal performs the synchronization of the clocks of I. Performing the Lorentz transformations to the rest frame of the moving particle I'
we obtain
T'=g(T-Vx/c^2)=T[(1-V/c)/(1+V/c)]^1/2=t/g (2)
X'=g(X-VT)=X[(1-V/c)/(1+V/c)]=x/g (3)
Do you consider that (2) and (3) are the transformation equations for the events e(x,x/c) and E(X,X/c) for the space-time coordinates of events associated with the apparent and the actual positions of the same particle? Is (2) an expression for absolute simultaneity (t=0, T'=0)
The inertial reference frames I and I' are in the standard arrangement.