Some Inertial Frames may be more Equivalent than Others Silly proposition - but take the peculiar results predicted by Einstein in Part IV of the 1905 paper with a little extra added. We place two clocks at locations A and B separated by a great distance L. We identify A clock with an S' frame and B clock with an S frame. A third clock is added at point P in the S frame on a line that connects A and B. All clocks are initially synchronized and then A clock is accelerated toward B to 0.9c as measured in the S frame. A clock reaches its terminal velocity as it passes P, thereafter it continues at a uniform velocity 0.9c until it reaches B. The distance required to reach 0.9c is small compared to L. Einstein says the A clock will lag behind the B clock when A arrives at B. This is not a reciprocal observation - this is a permanent difference in the amount of time logged by the A clock and the B clock...the one-way time difference confirmed by all experiments with high speed particles (no turn around required). During the acceleration phase, A clock will lose time as measured by P due to the acceleration relative to P (It can't lose a different amount of time because it is also being monitored by B). This provides an offset, but thereafter the S and S' frames should be regarded as equivalent - yet the final age difference is based upon the time traveled from P to B (plus whatever is lost in getting up to speed). As A passes P, from P's perspective, the S' frame is simply a frame with relative velocity v and therefore both frames should be equivalent (in transit from P to B we can no longer consider the history that included the acceleration phase). So the primary contribution to the final age difference between A and B depends upon travel time of A as it moves at a uniform velocity from P to B. So, does the initial acceleration create asymmetry between the S and S' frames that leads to the final age difference?