On Saturday, February 8, 2025 at 3:20:43 PM UTC-7 Alan Grayson wrote:
On Saturday, February 8, 2025 at 9:23:11 AM UTC-7 Jesse Mazer wrote: On Sat, Feb 8, 2025 at 9:35 AM Alan Grayson <agrays...@gmail.com> wrote: 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.* You can introduce a second rod if you like, but then I would request that you have different names for the two rods--say the rod at rest in O1's frame (f1) is called R1, and the rod at rest in O2's frame (f2) is called R2--and that when you make a statement about what any observer predicts about the length of a rod, you specify which rod you are talking about by name. Note that if we describe the rods this way, the previous discussion was only about the rod R1 at rest in f1, there was no rod R2 in that scenario. I think it would be simpler to stick to that scenario with one rod viewed in two different frames, but we can also talk about the second rod R2 if you feel it's essential. * Neither observer measures the rod in its own frame as contracted,* Yes, that agrees with the previous discussion where we just had the rod R1, and the observer O1 did not predict it as contracted according to the length contraction formula (this was point #2 on my list of 4 points which you agreed with. BTW, you didn't respond to my earlier argument that you should have no reason to object to point #3, see the argument I made in the two paragraphs beginning with my comment "I've asked you a bunch of times before about what you mean when you say there is 'no rod'." Do you intend to respond to that?) * but the rod from the perspective of either frame using the formula is calculated as contracted.* This is the sort of phrase I would want to avoid because it doesn't specify which rod is being predicted in which frame using which formula. Using the length contraction equation, we get the prediction that the rod R1 is contracted on O2's frame, but that the rod R2 has no contraction in O2's frame; and the length contraction equation also gives us the symmetrical prediction that the rod R2 is contracted in O1's frame, but that the rod R1 has no contraction in O1's frame. And I claim that if we start with the equations of motion for both R1 and R2 in one frame and use the LT formula to predict R1 and R2 in the other frame, we get the same predictions as above; for example if we take the coordinates of both rods in O1's frame as input, and use the LT to translate to O2's frame as output, we again get the prediction that the rod R1 is contracted in O2's frame, but that the rod R2 has no contraction in O2's frame. * For this reason I claim the contraction formula never predicts what it calculates as the measurement in the target or image frame.* If by "target or image frame" you mean the same thing as what I call the "output" of the LT (you seemed to deny this is what you meant by 'target frame' in an earlier post, but you never explained what I got wrong), then I would deny that there's any disagreement between the LT prediction and the measurement, see the paragraph above. Do you disagree with my claim in the last sentence that "if we take the coordinates of both rods in O1's frame as input, and use the LT to translate to O2's frame as output, we again get the prediction that the rod R1 is contracted in O2's frame, but that the rod R2 has no contraction in O2's frame"? Or do you agree with that, but think that this prediction differs from what O2 actually measures for R1 and R2? * 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.* But you still aren't defining what you mean by vague phrases like "the rod doesn't exist in one frame" or "one of the frames does not have a rod", I keep asking you over and over and over and over again if you're just saying there is no rod *at rest in* the frame you're referring to, or if you mean something different, but you never answer. In our previous scenario where there was just one rod R1 at rest in O1's frame, would you say that O2 doesn't "have a rod" because there is no second rod at rest in O2's frame? But if that's all you mean, surely you can't be claiming that there is anything wrong with O2 applying the length contraction formula to find the length in his own frame of that rod R1 despite the fact that R1 isn't at rest in O2? You had no problem with this when I stated it as point #1 in my list of 4 points from before. It seems like you're just exploiting the verbal ambiguity between O2 "having a rod" in the sense of there being a rod at rest in O2, vs. "having a rod" in the sense of having the values of L and v associated with some rod to plug into the length contraction formula. But these are totally different meanings, and this ambiguity would never arise if you would just give me a straight answer to my question asking you to define what you mean by such phrases. * 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* No, I didn't lose that. The equations of motion in the unprimed f2 frame described a rod R1 which is moving at 0.6c in the +x direction of the f2 frame, then I used the LT equations to translate R1's coordinates to a primed f1 frame which is also moving at 0.6c in the +x direction relative to the f2 frame. In this case the LT equations for x-->x' and t-->t' look like this (using units of light-seconds for length and seconds for time so that c=1): x' = 1.25*(x - 0.6*t) t' = 1.25*(t - 0.6*x) The 0.6 in these equations represents the relative velocity of 0.6c between f1 and f2, and the 1.25 is the gamma factor which is also based on that relative velocity (since 1/sqrt[1 - 0.6c^2/c^] = 1/(1 - 0.36) = 1/0.64 = 1.25). So yes, please study that numerical example and see if you agree with all the steps, and if not tell me where you disagree. Jesse *Let me make a constructive criticism; don't parse my comments. Read them in their entirety before responding. And make better use of your imagination in understanding my comments, which are quite clear and not at all ambiguous as you claim. AG* *Congratulations to us! We've proven that SR has a fatal irreparable flaw; specifically, that the LT, using coordinate transformations, predicts the rest length of a rod in f1 as calculated from f2, where the frames have non zero relative velocity v; whereas the contraction formula derived from the LT, always predicts a contracted length under the same conditions, and never the rest length. Where shall we publish? 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 everything-list+unsubscr...@googlegroups.com. To view this discussion visit https://groups.google.com/d/msgid/everything-list/49ff3541-2e6f-4f0d-91f8-d5077889fe3an%40googlegroups.com.
