The figure is of a moving frame moving to the right at velocity = 1/3c. On each end of the frame is a mirror (L and R) located at 4(1/3 + 1/3 + 1/3 + 1/3) x (1 second) from the midpoint of the frame. A pulse of light emitted at M (on either the moving or stationary frame) . We define the stationary point M when the pulse is emitted. Code (Text): L | |____|____|____M____|____|____| |R 1. In the first second the photon headed left will arrive at 1c to the left of M while the left mirror has moved to a point where the photon is located during the first second. The right photon is located 1c to the right of M at the end of 1 second.. 1 sec. L |<____________M___________> |R 2. After 2 seconds the photon on the left has reflected back to M while the right photon has just reached the right mirror: 2 sec. ____________>M ____________>| 3. After 3 seconds the left photon is now located at the shifted M' (3)(1/3) units to the right of M. The right photon reflecting off the R mirror is also located (3)(1/3) units to the right of M. _____________>M'<____________ 3 sec. As the frame has moved to the right (3)(1/3) units the photons meet at the position of the moving observer who, quite naturally, concludes the photons were emitted simiultaneously. We conducted the above analysis from the stationary frame. From symmetry considerations conducting the analysis in the moving frame gives the same result: The moving observer concludes the photons were emitted simultaneously in the moving frame. This you can easily demonstrate for yourself. Where is the loss of simultaneity? If any?