> But maybe the catch is that it doesn't matter how slowly he moves the clocks into position, they'll always be skewed by time dilation.
I dealt with that by moving the clocks with identical velocity profiles, so time dilation should be the same...
Unless time dilation is anisotropic. I dealt with that by sending multiple clocks, with some sent on triangular routes and some direct. In more detail:
C
A B
D
If I send a clock from A to C to B, and a clock from A to D to B, and the two clocks arrive with the same time, then I have evidence that time dilation is anisotropic (for at least those two routes). I don't necessarily expect that they have the same time as a clock sent direct from A to B - they have an additional acceleration, from the change of direction at C or D, and they have more time at velocity, because of traveling the longer distance. I think I said that very badly in my first post.
But the point is, if I can show that time dilation is anisotropic, then the clocks that went direct from A to B, and the clocks that went the same distance in exactly the opposite direction, should have the same time on them.
> If I send a clock from A to C to B, and a clock from A to D to B, and the two clocks arrive with the same time, then I have evidence that time dilation is anisotropic (for at least those two routes).
You mean isotropic, and you don't really. D->B is the same as A->C and C->B is the same as A->D; whatever clever path you come up with, a clock going from A to B will end up having had vertical movements that sum up to 0. If moving up induces some extra time dilation and moving down reduces it, or vice versa, you'll never be able to detect it; ultimately you can only ever make measurements when you and your clocks (and/or signals) have moved in closed loops, however squiggly.
OK, so that diagram was intended to be horizontal, not vertical. Obviously you want to minimize vertical movement as much as possible.
But I see what you mean about the sides (as drawn) being parallel.
And, yes, I meant isotropic, not anisotropic. Embarrassing.
OK, how about this: I have an equilateral triangle, with vertices A, B, and C. I synchronize clocks four clocks at A. I send one clock to B directly, and one to C and then B. I send one clock to C directly, and one to B and then C. I do the same from points B and C. Then, I can look at the difference between clocks that came direct and clocks that came the long way. If all the differences are the same, then I can say that going A-to-B-to-C has the same effect as going A-to-C-to-B or B-to-A-to-C or any other route. Doesn't that show isotropy?
> Then, I can look at the difference between clocks that came direct and clocks that came the long way. If all the differences are the same, then I can say that going A-to-B-to-C has the same effect as going A-to-C-to-B or B-to-A-to-C or any other route. Doesn't that show isotropy?
Again, no, because you can only measure around the full loop. Any clock you can compare has gone just as far east as it has gone west, just as far north as south, and just as far up as down. You can rule out some particular kinds of anisotropy, but there are possible patterns that just wouldn't show up.
I dealt with that by moving the clocks with identical velocity profiles, so time dilation should be the same...
Unless time dilation is anisotropic. I dealt with that by sending multiple clocks, with some sent on triangular routes and some direct. In more detail:
If I send a clock from A to C to B, and a clock from A to D to B, and the two clocks arrive with the same time, then I have evidence that time dilation is anisotropic (for at least those two routes). I don't necessarily expect that they have the same time as a clock sent direct from A to B - they have an additional acceleration, from the change of direction at C or D, and they have more time at velocity, because of traveling the longer distance. I think I said that very badly in my first post.But the point is, if I can show that time dilation is anisotropic, then the clocks that went direct from A to B, and the clocks that went the same distance in exactly the opposite direction, should have the same time on them.