To MacM:
Ok I have skimmed the rest of it now. Not much there except your often repeated claims that those who disagree with you are spouting "BS" - I.e. most (at least 95%) of the well educated physics community and all the designers of particles accelerators, many pratical engineers working with accelerators etc. are full of BS as MacM knows better, based on his ??? - you tell me.
There was, however, one thing that did surprize me as I recall you had the opposite possition only a few months ago. Here you said:
MacM said:
Simultaneity is nothing more than signal delays and dilation combined. Both of which is calculable and can be offset, corrected and taken into account so as to time events in real universal time and when you understand that and do that all your BS vanishes.
The first post of this thread carefully avoids all "signal delays" by having a multitude of observers standing "shoulder-to-shoulder" adjacent to the fast moving train with their noses just being "grazed" by the passing bombs that are the "start" and "stop events." So first part of your first sentence is irelivant. - the are no "signal delays" in my proof that time is not universal.
Also, I make no use of "dialation" in it, so that is irrelivant also. What I clearly show in that proof is the the temporal separation between the "start event" and the "stop event" is different in the two frames. I use the constant /same speed of light to do this. The flash
at the center of the train, which setoffs the two explosions takes (L/2c) time to travel to each in the train's frame, where L is the length of the train. (In the train's frame, they go off simultaneously.) But, in the station frame, this flash of light reaches the bomb at the end of the train in less than (L/2c) as the train is moving forward to meet the on coming light. (I don't want to worry about the "contraction of L" in the station frame. - That would just make the time in the station frames even less, or be exactly compensated by the opposite effect at the front of the train. I don't want to take time to think it thru carefull or argue about the effect of contraction, which obviously cancel out in the computation of the total time interval between the two explosions.) Thus, in the station frame the end of train explosion preceeds the one at the front. - The explosions are not simultaneous. I agreed that the explosion at the rear of the train could be, by mutual agreement, taken as the same t = 0 time.
The explosion at the front of the train is then not at the same t = x time in the station frame as it is in the train frame. (Because the forward going light front from the flash must travel, in the station frame, the extra distance the train has moved while the light was rushing toward the front bomb.) That was the essence of the proof. I invite you to look at the numerical example also given.
The simple fact is that, yes it is possible to make any pair of good clocks agree in two different frames, for example the one at the end of the train and the one on the ground both had t = 0 when the end of train explosion occurred. If you want to keeep them in agreement a "correction factor" must be added. But only one pair can be in agreement with the same correction factor. Other clocks at different locations will require their own different correction factors - I.e. not keep local time in their frame correctly.
What surprized me was you basically agreed with this months ago and even stated that you did not assume time was universal, but now your state:
MacM said:
...Simultaneity is nothing more than signal delays and dilation combined. Both of which is calculable and can be offset, corrected and taken into account so as to time events in real universal time and when you understand that and do that all your BS vanishes.
The "so as to time events in real universal time" appears to be a change in your position.
I have also explainded in words (not math) trying to get thru to you the fundamental reason why only a pair of clocks can keep both local time and be sychronized to one in another frame. When one tranforms to another moving coordinate system , there is a most one point with only "time like separtation" at any time. That is for this one point (which constantlychanges) point (x,y,z) = point (X,Y,Z) where x = X, y = Y & z = Z. For all other points the transform between the two coordinated systems mixes the space separation of the two points into some time separation in the the other frame. That is why only the end of the train and the adjacent station frame clock can be set to both have t = 0 simulatinuously but the ones at the front of the train and adjacent one on the ground must have different times if the are "good clock" in their own frames. (keeping local time correctly.)
Hard to state this clearly without any errors in words. Hope the basic idea gets thru to you. If not look at the numetical example given earlier in this thread (train going at 0.8c and 20 unitis long - where a "length unit" is the distance light travels in one "time unit" - you can work results out for your self and see the interval numerically between "start clocks end of train explosion event" until the "stop clocks front of train explosion event" are not the same in the two frames. (In the "flash at center of train" case this interval is zero in the train frame and greater than zero on the ground. I have also worked out, in earlier posts, the case where the flash is one length unit closer to the end of the train and thus not simultaneous in either frame.)
That is the "ticks" accumulated will obviously differ in the two frames and neither observer will agree that the other stoped his accumlating clock correctly (simultaneously with his). Each will think the other should have stopped accumulating differently. If the other had stopped "correctly" (simulatneously with his "now") then ther other would have accumulated less "ticks" - both think this. - it is not your "obviously impossible." They think this because they do not agree on what is "simulataneous."
You, in your simple minded view, (no insult intened - only stating it is a simple view, without the necessary understanding), which repeated denys that there is any need to discuss how they start and stop their accumulation of "ticks" correctly ignore the heart of the problem and falsely conclude it is impossible for both to accumulate less time. Each accumulates between the start and stop evnets as is correct in his frame, but these proceedures are not the same as their concepts of when the other should have stopped disagree. There is no univesal time. They can not, therefore, agree to start accumulating at noon and stop at 1PM. etc. It is a complex problem inherently involving a "start event" and a "stop event" when you "accumulated ticks". If you do not like my exposions, you could have the observers start when the are just passing and call that tick the "start event" or "tick zero" and agree that the 1,000,000th tick later will be the "stop event" then of course both will accumulate a million ticks, but the will not have accumulated for the same time duration because their millionth ticks are not simultaneous. (I assume you want them to accumulate for "the same time," do you not?)
I am just trying, in various ways, to show you that "tick accumulations comparisons" is not the extremely simple minded procedure you think it is because universal time does not exist.