Re: answer to YBM's bell problem

rbwinn a écrit :
On Sep 11, 6:40�am, YBM <ybm
Funny : the same light rays have a different behaviour in A and in B !
Here is what you pretend:
They meet at the origin of A from the point of view of A (so the bell
ring)
They meet at the origin of B from the point of view of B (so the bell,
which is NOT at the origin of B at this time, doesn't ring)
AGAIN: This is a direct conclusion OF WHAT YOU JUST WROTE, not what
I pretend, not what GT or LT pretend.

A bell at the origin of A does not depend on light meeting in B. It
depends on light meeting at the origin of A.

These sentences are insanely meaningless... The bell at the origin of A
depends on light meeting or not at the origin of A regardless of the
frame you consider this event. It's why any decent theory could not
predict, as yours, that it rings in a frame and not in another one.

By your definition of
the problem, light meets at the origin of A because light is emitted
at -a and a on the x axis in frame of reference A at the same time.
The light meets at the origin of A in a time of t=a/c.

Right.

The bell rings
in all frames of reference when that happens.

It's true, BUT it's not enough to write it down to make your "theory"
says that. As a matter of fact your "theory" predict it won't in
B frame.

Now you want to talk about frame of reference B. To make this easier
to see, we will put another bell at the origin of B. The bells are
right next to each other when the light is emitted. Bell A rings when
a time of t=a/c has elapsed in A. The bell in B is now a distance of
vt from the bell in A.


Scientists have determined by experiment that
the clock in B is slower than the clock in A.

Nope. What experiment shows is :
- from the point of view of frame B, clocks in A are running slower
- from the point of view of frame A, clocks in B are running slower

Note that this is as well what LTs predicts.

The clock in B
continues to move away from the clock in A. When the clock in B reads
n'=a/c, the bell in B rings.

You're going more insane every day : what you just write implies this :
- For an observer in A, the bell at rest in A rings, the one at rest in
B don't
- For an observer in B, the bell at rest in B rings. the one at rest in
A don't

Now, YBM, explain the same events using the Lorentz equations.

I did here :
http://groups.google.com/group/sci.physics.relativity/msg/a39fe2523de79ced
There is no such absurdities in SR : for SR, the bell
rings in both frames... It just happens that in frame
B they were emitted at coordinates
(-a/sqrt(1-v^2/c^2),0,0) at time va/(c^2*sqrt(1-v^2/c^2))
for the "left" light ray, and :
(a/sqrt(1-v^2/c^2),0,0) at time -va/(c^2*sqrt(1-v^2/c^2))

So it is quite certain that you didn't read.

� � � 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'.
You way you evade questions is ridiculous...

FIRST : they is two light rays in the experiment, SO there is two "n'" :
� � � � � �n'=t(1-v/c) (for light ray coming from (-a,0,0))
� � � � � �n'=t(1+v/c) (for light ray coming from (a,0,0))
� � � � �Why should the observer in B use two clocks, both of them
� � � � �being broken ?

Well, according to you, the observer in B only had one clock, which
showed light to be traveling at a speed of c in B. The observer in B
knows that an observer in A has a clock that shows light to be
traveling at a speed of c in A. You specified in the beginning that B
was moving with a velocity of v relative to A. That makes A a
preferred frame of reference because it is not moving.

There is nothing "preferred" for frame A in the problem I described.
There is indeed something special about A : the bell is at rest in A !

To transform
coordinates, you use a clock in A. To determine where light appears
to be according to the n' clock, you use the n' clock. In B light was
emitted at x'=-a and x'= a at n'=0. The light will meet at the origin
of B at a time of n'=a/c, and the bell in B will ring when that
happens. It will ring in all frames of reference when it rings, not
just in B.

There is always the problem you refuse to adress : there is two discinct
"n'" clocks in B !

SECOND : You have a double language : You claim that n' give
� � � � �a time coordinate for obervers of events in B, so the
� � � � �equation for n' is part of a transformation (the fact
� � � � �I've proven is that this transformation is absurd), then
� � � � �you claim that this equation is NOT a part of a transformation,
� � � � �but then you just support Galilean Transformations were
� � � � �speed of light rays in frame B is no more c.

The speed of light in B is definitely c according to the n' clock.

There is always the problem you refuse to adress : there is two discinct
"n'" clocks in B !
.

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