On Wed, Feb 12, 2025 at 3:23 AM Alan Grayson <agrayson2...@gmail.com> wrote:
> > > On Saturday, February 8, 2025 at 9:36:31 PM UTC-7 Jesse Mazer wrote: > > On Sat, Feb 8, 2025 at 9:59 PM Alan Grayson <agrays...@gmail.com> wrote: > > > > On Saturday, February 8, 2025 at 7:26:18 PM UTC-7 Jesse Mazer wrote: > > On Sat, Feb 8, 2025 at 8:45 PM Alan Grayson <agrays...@gmail.com> wrote: > > > > On Saturday, February 8, 2025 at 5:46:02 PM UTC-7 Jesse Mazer wrote: > > On Sat, Feb 8, 2025 at 7:17 PM Alan Grayson <agrays...@gmail.com> wrote: > > > > 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* > > > I do try to read what you write carefully and imagine different possible > meanings to see if I can make sense of statements that seem ambiguous, but > if even after that your statements seem equivocal, you can't reasonably ask > me to just take it on faith that they make sense. And you use these > statements to derive conclusions I'm confident are wrong (both because I've > done the calculations to check and because I have never seen any physicist > suggest a disagreement between predictions and measurements), so naturally > I'm going to suspect that the apparent equivocation is connected to your > wrong conclusions, especially when you are so evasive when I ask you to > clarify some point about your wording, refusing to give me a straight > answer even when I repeat the same question many times. > > > > > *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 * > > > Both the LT and the length contraction formula predict the same thing, > that if the rod R1 is moving relative to frame f2 and at rest relative to > frame f1, then the rod will be contracted as measured in frame f2, and > non-contracted as measured in f1. When you say "congratulations to us", do > you think *I* ever said anything different than this? You already agreed > that the length contraction equation predicts both of these things when I > stated them as points #1 and #2 earlier. As for the LT equation, the whole > point of my numerical example is to show that if you start with the > coordinates of the rod in f2 and use the LT to predict the rod's > coordinates in f1, you get the conclusion that THE ROD IS NON-CONTRACTED IN > F1, which is exactly what was predicted when we used the length contraction > equation to find the length of the rod in f1 (point #2 which you agreed > with). > > > *Please refresh my memory; what was point #2. * > > > I wrote: > 2. If the observer O1 has a velocity of v=0 relative to the rod, and uses > the length contraction formula to predict the length in her own frame, then > the output of the formula is that the rod has its non-contracted length L' > = L. > > You replied: > Ok. > > Do you want to change your answer? > > > *No. Of course not. I posted the same thing. But the formula is generally > used to predict CONTRACTION from the pov of a frame moving relative to f1, > namely f2, and it never measures or predicts the rest length of a rod in > f1. AG* > > > It is permissible to use the formula to predict the length L' in any > frame, including the frame f1 where the rod has velocity v=0. You can't > deny this is an acceptable use of the formula without contradicting > yourself, since this is exactly what point #2 above was saying (plugging > v=0 into the length contraction formula to get L' = L in O1's frame f1), > and you said "Ok." > > > > > > *Regardless, I can read what the contraction formula states or implies; it > always implies contraction, never NON-CONTRACTION. It will only imply the > rest length of rod in f1 if v is set to zero, but that's not the scenario > we're discussing, which is frames in relative motion. If you think > otherwise, prove it. AG* > > > We are discussing a scenario where we have two frames f1 and f2 which have > a relative velocity of 0.6c, and a rod R1 which has a velocity of 0 > relative to f1 (which implies the rod has a velocity of 0.6c relative to > f2). But if you want to know what the length contraction equation says > about the length of the ROD as observed by frame f1, THE VELOCITY OF F2 > RELATIVE TO F1 HAS NOTHING TO DO WITH IT! The velocity v you put in the > length contraction equation is always the velocity of the OBJECT relative > to the observing frame, and in this case the object is the rod R1 which has > velocity v=0 relative to the observing frame f1. > > > *Maybe that's how it's done on Mars. Here on Earth the formula is used > from a frame for which the rod is observed in motion, namely from f2, where > f1 and f2 are in relative motion. Using your method, contraction would > never be calculated as non-zero. AG * > > > Why do you say "Using your method, contraction would *never* be calculated > as non-zero"? Are you under the impression that "my method" says the length > contraction formula can *only* be used in the frame f1 where the rod has > v=0? If so this is another example of your bad reading > comprehension/memory, I never said any such thing and my point #1 > explicitly applied the formula in the other frame f2 where both the > velocity of the rod and its contraction were non-zero. "My method" is the > standard one that says the length contraction formula can be used in *any* > inertial frame, and in each frame the value of v that you should plug into > the formula is the velocity of the object (e.g. the rod) in that frame. So > if you use to calculate the length of the rod in f1 you'll plug in v=0 and > get L' = L as the output of the equation, if you use it to calculate the > length of the rod in f2 you'll plug in v=0.6c and get L' = 0.8*L as the > output of the equation, just as you already agreed to in point #2 and point > #1. > > And this agrees perfectly with the numerical example I gave of the LT, > where I considered a rod with rest length L = 10, I wrote the equations of > motion for the front and back of the rod in the unprimed frame f2 where it > had a length of L' = 8 and was moving at a velocity of 0.6c, then I applied > the LT to these equations to output the corresponding equations of motion > for front and back in the primed frame f1, equations which showed that in > the primed coordinates of f1 the rod has a length of 10 and a velocity of 0. > > Jesse > > > I understand and agree in full with your above paragraphs. AG > Sounds good--does that mean you agree that both the length contraction formula *and* the LT would make the following predictions about the length of the rod in frame f1 (where it's at rest) and f2 (where it's moving)? 1. When measured using f1's ruler/clock system, the rod is predicted to have its non-contracted length L 2. When measured using f2's ruler/clock system, the rod is predicted to have a contracted length L' (equal to 0.8*L in my numerical example where the rod has velocity 0.6c relative to f2) Jesse > > > > If you think the math of my numerical example is wrong feel free to find > the flaw, but it's completely incoherent to say that *my* results support > your notion of a conflict between what is predicted and what is measured in > f1, since NON-contraction of the rod R1 in f1 is what I said was predicted > by both the length contraction equation and the LT, and it's what we both > agree is measured by O1. > > Jesse > > > > > > > > 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-li...@googlegroups.com. > > To view this discussion visit > https://groups.google.com/d/msgid/everything-list/302bf988-a364-45bf-8277-9ef1569068efn%40googlegroups.com > <https://groups.google.com/d/msgid/everything-list/302bf988-a364-45bf-8277-9ef1569068efn%40googlegroups.com?utm_medium=email&utm_source=footer> > . > > -- > 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/d7f880d7-3bdd-4c89-8924-07dd688f10a5n%40googlegroups.com > <https://groups.google.com/d/msgid/everything-list/d7f880d7-3bdd-4c89-8924-07dd688f10a5n%40googlegroups.com?utm_medium=email&utm_source=footer> > . > -- You received this message because you are subscribed to the Google Groups "Everything List" group. 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