Re: answer to YBM's bell problem
- From: YBM <ybmess@xxxxxxxx>
- Date: Thu, 11 Sep 2008 22:46:45 +0200
rbwinn a écrit :
On Sep 11, 11:24�am, YBM <ybm...@xxxxxxxx> wrote:....
They are both correct. One shows where light from x'= -a is asWrong : t(1-v/c) and t(1+v/c) are different except at t=0 (or if v=0).Well, if you are talking about an event in B, the n' values would beRight. The observer in A has one clock that shows t, and the observerWhich one : n'=t(1-v/c) or n'=t(1+v/c) ?- Hide quoted text -
in B has one clock that shows n'.
the same, and the two t times wouled be different. �
AGAIN : which n'=t(1-v/c) or n'=t(1+v/c) do you choose ?
compared to a clock in B which shows light to be traveling at a speed
of c, and the other shows where light from x'=a is as compared to a
clock in B which shows light to be traveling at a speed of c if both
beams of light are traveling toward the origin of B.
If they are "both correct" it means that from the point of view
of B, the events "left light ray arrives at the origin of A" and
"right light ray arrives at the origine of A" have different
time coordinates
n'=a/c(1-v/c) for the right one
n'=a/c(1+v/c) for the left one
so they don't arrive on A at the same instant (let's forget
for a moment how stupid is to compare events with coordinates
build from clocks being broken first, and broken on a different
way then), so for B, the bell in A don't ring.
Sorry you don't
like where this light is, YBM. The light will meet at the origin of B
at a time of n'=a/c.
Now getting back to the two bells that ring when light from both
directions reaches them, you will notice that from frame of reference
A, the Lorentz equations show only the bell at the origin of A
ringing. According to your rules, the bell at the origin of B will
not ring.
I say both bells will ring.
????!!!!! You'd better think before writing such absurdities...
.
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