Re: answer to YBM's bell problem
- From: YBM <ybmess@xxxxxxxx>
- Date: Thu, 11 Sep 2008 23:12:49 +0200
rbwinn a écrit :
On Sep 11, 12:34�pm, YBM <ybm...@xxxxxxxx> wrote:....
Because of relativity of simultaneioty that the Lorentz equationThe rays of light willIs there something you don't understand above ?
not meet at the origins of both frames of reference. �So from A, an
observer will only observe the bell in A to ring, from B an observer
will only observe the bell in B to ring.
�With my equations, the bell in A will ring first, then the bell in
B. �You will hear both bells in both frames of reference.
Uh huh. �Now, YBM, explain the same events using the Lorentz equations.I did here :http://groups.google.com/group/sci.physics.relativity/msg/a39fe2523de...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))
But if you put a bell at the origin of each frame ofLook carefully at the coordinates of events I've provided
reference, what will happen?
Robert B. Winn
So it is quite certain that you didn't read.No, I read it, YBM. �So explain what happens if there are two bells,
one at the origin of each frame of reference.
above. Then figure out yourself :
- at what time in B does both light rays have been emitted ?
� �are they the same ?
require, the beams of light are not emitted at the same time in B.
right.
Therefore, according to you only the bell at the origin of A will ring
as observed by the observer in A.
- are the positions of the points of emission in frame B symetricThe points of emission are equal distances from the origin of B.
� �with respect to the origin of B ?
right.
- given that velocities of the light rays are c and -c in BI say they will, but the Lorentz equations say they will not.
� �are they going to meet at the origin of B ?
At least you should notice that the Lorentz equations are
perfectly coherent.
If you
have a bell at the origin of A and a bell at the origin of B, only one
of the bells will ring.
You mean according to LT, right ?
So, now you "theory" says : both bells will ring, right ?
(even if this not what your formulas implies).
This is sooooo utterly absurd that I suggest you to take
a rest and think a bit about it.
.
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