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Quote by James S Saint
 Quote by PeterDonis This version of the LT is only valid if the two frames have a common origin: in other words, if Einstein is just passing the station-master (assuming the station-master is at the spatial origin, x = 0, in the station frame) when both Einstein's and the station's clocks read 0. I don't think that's true of the OP's scenario, so all the LT formulas become more complicated since they have to include the offset of one frame's origin relative to the other. As far as I can see, all of the calculations in this thread need to be re-done because of this.
There you go. The presumption that when t=0, t'=0 corrupts the calculations.
What are you talking about? Peter just said that this version of the LT (he's talking about the standard configuration) is only valid if the two frames have a common origin which means t=0 and t'=0 (and the same for the x, y, and z coordinates as I mention on the previous thread). He then went on to suggest one way to do this but realized that wouldn't fit your scenario and then said there needed to be an offset between the two frame's origins. Well I accomplished the same thing by providing an offset between the station masters location and the station frame's origin. If you had provide details that you cared about when I asked for them, we could have avoided all these issues that you are now raising.

Please note that Peter is also talking about a station with a clock even though you never described the station, just a station master with no clock and two stop-clocks by the tracks.
 Quote by James S Saint As far as that 6 μls; If we take it that from the station's frame we have 6 between the clock and the train and thus must have 5.196 from the train's POV, then we have presumed that it is the train that is moving. If we take it that the station is moving and that from the train's POV the distance is 5.196, then we have to say that the station's POV is going to be 5.196*.866, yielding 4.5 μls for the station.
No, there is a distance in the station frame that you said was 6 μls. That distance when measured by Einstein is 5.196. It's actual length is frame dependent. In the station frame, it's 6, in Einstein's frame it's 5.196. When Einstein measures it in either frame (or any other frame), he gets 5.196. When the station master measures it in either frame (or any other frame), he gets 6. Nobody measures it as 4.5 although there are frames of reference in which it is 4.5 but not one of the ones we have considered so far.
 Quote by James S Saint There has to be a defined common state for both frames somewhere. In the typical scenario when [t,x] = [0,0] then [t',x'] = [0,0]. That was Lorenz' formulation presumption. But we have no idea of Einstein's clock readings and thus they cannot be used as a fact for calculation so directly using Lorenz. If we say that when the button is pressed, the train's Clock is at t'=0, we get different results.
I think this issue has already been dealt with. Just keep in mind that the times for events that got transformed are coordinate times on imaginary clocks and if you want to give Einstein a real clock, you can add or subtract an offset to relate it to the coordinate clocks.
 Quote by James S Saint In these scenarios we are really only concerned with the change in time readings thus we have to use something else as the common state. It shouldn't matter what Einstein's clocks read at any one moment as long as they remain consistent from there forward. In this scenario, the only thing common that we can use is that distance and if we don't use it, we have a conundrum of not knowing who is moving such as to indicate which clocks are slower.
We know who is moving. You told us. But we can transform your description based on the rest frame of the station into any other frame of our own choosing and it will be just as valid as the one you started with.