Re: answer to YBM's bell problem

On Sep 10, 7:19�pm, YBM <ybm...@xxxxxxxx> wrote:
rbwinn a �crit :

one question at a time, let's simplify the point since you're clearly
lost from the very beginning :

n'=t(1-v/w)

Well. In frame A, I consider two light rays on the (Ox) line (does it
remind you something ?). One at speed w=c in A, the other one at speed
w=-c in A. How would an observer in B would compute, in the context of
you "theory", using n', the speeds of both light rays simultaneously ?

OK, say that the lights are emitted at -a and a in both frames of
reference when the origin of B is at the origin of A, each ray of
light directed at the origins. B is moving in the +x direction
relative to A at a velocity of v. The light ray emitted at -a has a
velocity of w=c in both frames of reference. The light ray emitted at
a has a velocity of w=-c in both frames of reference. The light ray
emitted at -a goes from x=-a to the origin of A in frame of reference
A in a time of t=a/c. The light ray emitted at -a goes from x'=-a to
the origin of B in a time of n'=a/c. The light ray emitted at a
travels from x=a to the origin of A in a time of t=a/c. The light ray
emitted at a travels from x=a to the origin of B in a time of n'=a/c.
The equation for n' is

n'=t(1-v/w)

You remember the equation for n', YBM.

Is there some kind of magic making its clock change whenever he consider
one light rays or the other one ?

It is not a transformation equation, YBM. It just tells how far light
has gone in B according to a clock that shows n'.
Robert B. Winn
.

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