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Showing that Lorentz transformations are the only ones possible |
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| Nov27-12, 02:17 AM | #120 |
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Showing that Lorentz transformations are the only ones possibleAFAIK RoS has always been explained with the usual linear Lorentz transformations. |
| Nov27-12, 02:47 AM | #121 |
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We are looking for theories in which there's a set K of curves in M (i.e. in spacetime) such that each member of K represents a possible motion of an accelerometer that measures 0. A global inertial coordinate system should be a bijection from M into ℝ4 that takes every one of those curves to a straight line. But we can't take this as the definition of a global inertial coordinate system, because we know that in SR, those curves are all timelike, and a global inertial coordinate system in SR also takes spacelike geodesics to straight lines. I think we need to leave the term "global inertial coordinate system" partially undefined at this point. We can define it properly after we have found a group of inertial coordinate transformations. The partial definition of "global inertial coordinate system" doesn't imply that inertial coordinate transformations take *all* straight lines to straight lines. It just implies that there's a set L of straight lines such that each member of L is taken to a member of L. It does seem natural to also require that every inertial coordinate transformation takes all constant-velocity motions to constant-velocity motions, but this assumption doesn't pin down what an inertial coordinate transformation does to an infinite-speed straight line. |
| Nov27-12, 02:55 AM | #122 |
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| Nov27-12, 04:43 AM | #123 |
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| Nov27-12, 05:56 AM | #124 |
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I'm just saying that I haven't yet seen a satisfactory way to explain why inertial coordinate transformations should take *all* straight lines to straight lines. For example, why should the straight line with (t,x,y,z) coordinates t=0, x=0, y=z be taken to a straight line? |
| Nov27-12, 07:32 AM | #125 |
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But of course if you took as starting point a flat spacetime that would trivially come from the fact that all geodesic lines in such a space are straight lines by definition. Still, this begs the question why should one choose such a space for SR in the first place. And the only answer is the postulates wich are arbitrary to some extent. |
| Nov27-12, 01:40 PM | #126 |
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I think I have an argument that works. Consider the set of "vertical" lines (t variable, x,y,x constant) through a segment on that line. They would represent the motion of the component parts of a thin non-accelerating rod, in a comoving inertial coordinate system. Their union is the "world sheet" of the rod. We have already assumed that all the finite-speed lines in the world sheet, including the ones with arbitrarily high speeds, are taken to straight lines.
Suppose that the world sheet's intersection with the t=x=0 plane (i.e. the line I described), isn't taken to a straight line, then the images of the vertical lines under the inertial coordinate transformation have discontinuities at the t=x=0 plane, and a straight line with one point removed and put somewhere else isn't really a straight line. So this contradicts the assumption that finite-velocity straight lines are taken to straight lines, and this means that the line I described must be taken to a straight line. My explanation may not be perfectly clear, but I have a spacetime diagram in my head that's seems clear enough, so I think this idea works even if I didn't explain it well enough. |
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