On Saturday, February 8, 2025 at 12:29:55 AM UTC-7 Alan Grayson wrote:
The way I see it there are two frames f1 and f2, *one *rod located at the origin of f1, fixed in f1, but moving wrt f2. Of course, frames are coordinate systems and our single rod has coordinates in both frames, but only *EXISTS* in one frame, f1. Is "only exists in one frame" just a synonym for "only has one frame as its rest frame" or does it mean anything more? You agree the single rod can be assigned coordinates in both frames, so presumably you agree the observer in f2 is still able to see and measure the rod. So, using the formula from the pov of f2, the proper length of the rod in f1 is L, which is contracted to L'. Yes, that's what I said in point #1 above. Since the rod is fixed in f1, its length is* never* contracted from the pov of an observer in f1, Yes, that's what I said in point #2 above. but is always contracted from the pov of an observer in f2, which sees the rod moving with velocity v. Yes, that again agrees with point #1. The bottom line is the what is calculated by f2's observer is NEVER measured by f1's observer. Calculated using the length contraction equation, or calculated using the LT? "What is calculated by f2's observer" using the length contraction equation is just the length in the f2 frame (assuming the f2 observer inputs the velocity v of the rod in their own frame), not any prediction about what is measured in f1. *The simplist way to model this problem is to assume a symmetric situation, a rod in each frame of the same rest length, located at the respective origins, located at the center of each rod. Then, using the contraction formula, and assuming a relative velocity of v, observers within each frame, will measure the rod within each frame as having a non contracted length, whereas the calculated length of the moving rod it's observing will be calculated as contracted. Neither observer measures the rod in its own frame as contracted, but the rod from the perspective of either frame using the formula is calculated as contracted. For this reason I claim the contraction formula never predicts what it calculates as the measurement in the target or image frame. Also, if one of the frames does NOT have a rod, I mean that the contraction formula cannot be applied, since without a rod, the proper or rest frame length is undefined, or if you prefer equals zero. A frame without a rod presumably knows the coordinates of the rod in a frame which has a rod, and I think you're trying to show below that when the LT is applied to the coordinates of the rod in any moving frame, will show that my conclusion about contraction and measurement in the symmetric situation is mistaken. I need to further study your example using the LT to make an intelligent comment. AG* In contrast, the LT equations *can* be used to predict what is measured in f1 if you are given the coordinates of the rod in f2, and in this case the prediction will agree with my point #2 above that says the rod has no contraction in f1. *If so, then the LT and contraction formula disagree in their predictions since from the pov of f2, f1 is moving and must be contracted from the pov of f2. I'll have to study further what the LT predicts. AG * *My conjecture is that you might have lost the fact of motion between f1 and f2 when you used the rod's coordinates in f2 and the LT to calculate the measured value of rod in f1 and got its rest length. I suggest you review what you did, and if you don't find this error, I will study your results in detail. AG * You can, of course, reverse the frames which do the calculations, but you can't get symmetric results because *there is NO rod in f2's frame *for which the formula can be applied. If by "no rod in f2's frame" you just mean no rod *at rest* in f2's frame, and if "the formula" in your statement refers to the LT rather than the length contraction formula, then I would just point out that the LT equations do not *require* that an object be at rest in the input frame to predict the coordinates of the same object in the output frame. See my numerical example a few posts back, where I gave these coordinates for the front and back of a moving rod in the unprimed frame (corresponding to what you're calling the f2 frame): Back of the rod: x = 0.6c*t Front of the rod: x = 8 + 0.6c*t Do you AGREE/DISAGREE that these equations for position as a function of time describe a rod which is *not* at rest in f2? For example, at t=0 the equations tell you that the back of the rod is at x=0 and the front of the rod is at x=8, then 5 seconds later at t=5 the equations tell you the back of the rod is at x=3 and the front of the rod is at x=11, so the whole rod has moved forward by 3 light-seconds in those 5 seconds. So maybe the rod doesn't "exist" in this frame f2 according to your idiosyncratic phrasing, but it is clearly being *described* in terms of the coordinates of f2, and that description includes the fact that it is moving relative to f2. If we look at the set of spacetime coordinates that the front or back of rod will pass through in this frame, we can perfectly well use those unprimed coordinates as inputs for the x-->x' and t-->t' Lorentz transformation equations, and get the set of primed coordinates the front and back of the rod pass through in the f1 frame as output. For example if you take x=3, t=5 for a point the back end passes through, you can plug those values into the LT equations x' = 1.25*(x - 0.6*t) and t' = 1.25*(t - 0.6*x), and get x'=0, t'=4 as. output. Do you AGREE/DISAGREE that the LT can be used this way? Jesse -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion visit https://groups.google.com/d/msgid/everything-list/a03c8558-cee7-422e-9797-b5d3d0ec7e2dn%40googlegroups.com.
