I am hopeful someone can give me a quick lesson here. I have an idea that time does not slow as one's velocity increases (bear with me, please). I'll state this in familiar terms with a person on a train vs platform and the light beam traveling vertically from the ceiling (P1) to the floor of the train (P2). Let's just say that it stops at that point (P2). X is a person on the train and Y is a person on the platform. Both X and Y have the same sort of clock (measuring distance, by the way, not time, but save that for another day). They each have an identical clock, however you wish to see it. The STR argument is of course very familiar and of course X's clock runs slow blah blah blah so enough said. But I wonder, is it true? (Again, please humor me for a minute) Here is X looking at the light beam from P1 to P2. We'll say for the sake of argument that the light has travelled four units and X's clock has travelled four (different) units, so X determines c to be 1 (4/4). Note, in all diagrams please ignore the "o" symbols as it is the only way I could get the diagrams to format correctly. P1 . . . . P2 X And here is Y looking at the light beam from P1 to P2. P1 . o . ooo . ooooo . ooooo P2 Y And since the light has travelled more than four units (diagonal distance), then Y's clock has travelled more than four (different) units in order for c to be constant. Hence Y's clock is faster than X. Or as more commonly reported, X's clock is slower than Y's. Even though the clocks were identical. But wait a minute. Put X and Y in together now. The beam of light was emitted the moment the train passed Y (moving from left to right as you can see) P1 o . ooo . ooooo . ooooooo . ooooooo P2 Y oooooo X In the diagram above, Y has not yet seen the light as it terminates at point P2. This 'information' must still traverse the distance from P2 to Y before Y stops his clock and reports his time. And while this happens, the train still travels. So we have: P1 . ooo . ooooo . ooooooo . P3 . . . . P2 . . . . P4 Y oooooooooooo X So Y clearly has to stop his clock AFTER X has stopped her clock, since X stopped her clock at P2 and is already at P4 when Y stops his clock. So of course Y's clock has 'ticked' longer to report the same information (or, if you prefer, X's clock 'ticking' less). But then it is not really a case of X's clock "running slow", is it? Thanks in advance for your feedback.