James S Saint
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What did I say about "distractive argumentation"?
At some point from Einstein's POV;
Clock1 = 11.072
Clock2 = 15.690So the question becomes;
For what time t1 and other clock time t1+4.6184 separated by 6.928 μls would there be an x location from which to observe the difference of 4.6184 μs?
x' = γ(x-tβ)
x' = 1.1547(x - t1/2) ;for clock1
x’ = 1.1547((x+6.928) - (t1+4.6184)/2) ;for clock2
(x-t1/2) = ((x+6.928) - ( t1+4.6184)/2)
(x) = x + 6.928 - ( t1+4.6184)/2 + t1/2
at this point x drops out meaning that any distance will do.
6.928 = ( t1+4.6184)/2 + t1/2
2*6.928 = ( t1+4.6184) + t1
2*t1 = 2*6.928 - 4.6184
t1 = 6.928 - 4.6184/2
= 4.6188
So at any location x, as long as t1 = 4.6188, the proper difference in time can be observed by Einstein at the button press.
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The distance of 6 μls was irrelevant.
But you could never convince me that because I said that the train was traveling rather than saying that the station was traveling, the distance between the station and train would be different. Again if it had been 2 rockets coming toward each other, due to symmetry, neither can claim ownership of the distance.
I am curious of one detail though. Is length contraction the same for objects moving away as it is for object moving toward? The equations would imply that it is, but sometimes that sort of thing just gets left out as presumption much like the t=0=t' concern.
{..and I meant that the original had 4 timing paths, not 4 clocks.}
At some point from Einstein's POV;
Clock1 = 11.072
Clock2 = 15.690So the question becomes;
For what time t1 and other clock time t1+4.6184 separated by 6.928 μls would there be an x location from which to observe the difference of 4.6184 μs?
x' = γ(x-tβ)
x' = 1.1547(x - t1/2) ;for clock1
x’ = 1.1547((x+6.928) - (t1+4.6184)/2) ;for clock2
(x-t1/2) = ((x+6.928) - ( t1+4.6184)/2)
(x) = x + 6.928 - ( t1+4.6184)/2 + t1/2
at this point x drops out meaning that any distance will do.
6.928 = ( t1+4.6184)/2 + t1/2
2*6.928 = ( t1+4.6184) + t1
2*t1 = 2*6.928 - 4.6184
t1 = 6.928 - 4.6184/2
= 4.6188
So at any location x, as long as t1 = 4.6188, the proper difference in time can be observed by Einstein at the button press.
-----------------------------------
The distance of 6 μls was irrelevant.
But you could never convince me that because I said that the train was traveling rather than saying that the station was traveling, the distance between the station and train would be different. Again if it had been 2 rockets coming toward each other, due to symmetry, neither can claim ownership of the distance.
I am curious of one detail though. Is length contraction the same for objects moving away as it is for object moving toward? The equations would imply that it is, but sometimes that sort of thing just gets left out as presumption much like the t=0=t' concern.
{..and I meant that the original had 4 timing paths, not 4 clocks.}
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