- #1
dbkooper
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Clock transport can be used to compare one clock with another in an absolute sense. All we have to do is to transport a clock between two A-frame clocks that have been synchronized per Einstein's definition. Let's call the transported clock "T" and the left-hand and right-hand E-synch'd clocks "E1" and "E2" respectively. We will need to transport T twice - at different speeds wrt A - in order to have a comparison.
In both cases, T and E1 read zero at the start. It is most important to bear in mind the simple fact that the temporal relationship between E1 and E2 is constant. (This means "constant within the A frame," not "constant as seen by any outside observers.") (In other makes them actual experiments in anyone's view.)
In the first transport case, each frame* sees the other moving at 40/41c.
(*The T-clock frame and the E-clocks frame, aka, Frame A)
-------------[0]-->
---------<--[0]------Frame A------[?]-------------------------[0.30]-->
[?]------Frame A-----[1.10] difference = 0.80
In the 2nd case, each frame sees the other moving at 35/37c.
-------------[0]-->
---------<--[0]------Frame A------[?]-------------------------[0.34]-->
[?]------Frame A-----[1.06] difference = 0.74
If T and E1 always tick at the same physical rate (as special relativity demands), and given the fact that E1 and E2 always have the same temporal relationship, then T must faithfully carry or transport this relationship to E2.
In case one, we see that this transported relationship shows up as E2 reading 0.80 different from E1's transported time of 0.30.
But in case two, we see that T now differs from E2 by only 0.74.
Only if T read .26 at the end (of the 2nd case) would it have physically ticked at the same rate as E1 (because the latter is always - per the 1st case - 0.8 different from E2).
We see that simple clock transport tells us the very important fact that clocks do indeed run at physically different rates even with no acceleration or gravity.
This is important because it means that a clock's absolute motion affects its rate.
And this is important because it is evidence of absolute motion.
In both cases, T and E1 read zero at the start. It is most important to bear in mind the simple fact that the temporal relationship between E1 and E2 is constant. (This means "constant within the A frame," not "constant as seen by any outside observers.") (In other makes them actual experiments in anyone's view.)
In the first transport case, each frame* sees the other moving at 40/41c.
(*The T-clock frame and the E-clocks frame, aka, Frame A)
-------------[0]-->
---------<--[0]------Frame A------[?]-------------------------[0.30]-->
[?]------Frame A-----[1.10] difference = 0.80
In the 2nd case, each frame sees the other moving at 35/37c.
-------------[0]-->
---------<--[0]------Frame A------[?]-------------------------[0.34]-->
[?]------Frame A-----[1.06] difference = 0.74
If T and E1 always tick at the same physical rate (as special relativity demands), and given the fact that E1 and E2 always have the same temporal relationship, then T must faithfully carry or transport this relationship to E2.
In case one, we see that this transported relationship shows up as E2 reading 0.80 different from E1's transported time of 0.30.
But in case two, we see that T now differs from E2 by only 0.74.
Only if T read .26 at the end (of the 2nd case) would it have physically ticked at the same rate as E1 (because the latter is always - per the 1st case - 0.8 different from E2).
We see that simple clock transport tells us the very important fact that clocks do indeed run at physically different rates even with no acceleration or gravity.
This is important because it means that a clock's absolute motion affects its rate.
And this is important because it is evidence of absolute motion.