Re: relativity of simultaneity - real or perceived?

Thanks to all for their responses. It's as I thought. I was however
confused by some responses to an earlier post. (This is why I am
asking, Paul - thanks for your clarity.)

I'm trying to work out a simple thing in preparation for something more
complex.

Consider Inertial Frame of Reference 1: 2 photons (or particles if you
prefer) both travelling at c towards each other along the X-axis.
Photon A is at position zero on the X-axis, and photon B is 1.5
light-seconds away.
Frame 1:
A->........449688687 meters........<-B
O1(Observer)
x_A = 0
x_B = 1.5 light seconds
t_A = 0
t_B = 0

Now consider this SAME situation from another Inertial Frame of
Reference 2, in which the axes are the same as Frame 1 and photon A
is also at position zero on the X-axis, but the observer, at rest in
Frame 1, is travelling at 0.5c relative to Frame 1, in the same
direction as photon A:
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?
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. The distance however
between them would be calculated to be less. What is that distance?

.

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