Re: relativity of simultaneity - real or perceived?

In article <1116086135.741617.95740@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Curious <anthonyroseuk-curious@xxxxxxxxxxx> wrote:
>Frame 2:
> A->.............? meters...........<-B
> O2(Observer -> at 0.5c relative to O1)
>
>Question: Where is photon B calculated (not perceived) to be in Frame 2
>at the moment in Frame 2 when photon A in Frame 2 is at x'_A = x_A = 0?

I worked this out for you numerically a few days ago, using two
different methods, and got the value x'_B = 0.86603 light-seconds.

>To put it another way, If there was a 1.5 light-second long box between
>A and B in Frame 1, so that the photons were at each end of the box at
>this moment, then I would expect that in Frame 2 also the two photons
>would be at each end of the box simultaneously.

No, they would not arrive at the two ends of the box simultaneously in
frame 2. In frame 2, photon A arrives at the left end of the box at t'_A
= 0, whereas photon B arrives at the right end of the box at t'_B =
-0.86603 seconds. That is, photon B arrives at its end of the box first,
then, 0.86603 seconds later, photon A arrives at *its* end of the box.

>The distance however
>between them would be calculated to be less. What is that distance?

At which moment in frame 2? When photon B arrives at the right end of the
box, or when photon A arrives at the left end of the box?

--
Jon Bell <jtbell@xxxxxxxxxx> Presbyterian College
Dept. of Physics and Computer Science Clinton, South Carolina USA
.

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