michel123456 Posted October 23, 2020 Author Posted October 23, 2020 (edited) On 9/23/2020 at 5:39 PM, swansont said: Is the “length” moving? No. So it’s not contracted. Is that your objection? (still searching the quote) 26 minutes ago, Eise said: Uh? 'The distance'? Here you go again. Not mentioning which distance from which frame In E-X's FOR, the distance between E and X is 1Lh. So for B, moving with 0.8c, he sees the distance he must travel as 0.6Lh. There are no other relevant distances for this situation. Lengths of poles connected to B just play no role. Why, B could be a rocket of 1 Lh long. The only thing that interests us is the passing of the observer in B with his clock, passing E and X. No, I think that you are wrong there. The distance 0.6 LH is the one from the FOR of E-X. As in diag2. From the FOR of X-E, what B seem to experience is a change of scale. The units are changing. The meters are smaller & the seconds are smaller too (so that velocity remains the same) In its own frame, B does not travel 0.6 LH, it still travels 1LH. diag2 In the FOR of B, in 45 min B will travel & reach point Xb on the following: On 9/23/2020 at 7:10 PM, md65536 said: On 9/23/2020 at 5:39 PM, swansont said: Is the “length” moving? No. So it’s not contracted. Do you mean the lengths between the dice? Those lengths are contracted, see (b). Here's a thought experiment to show that the spaces between objects are contracted the same as objects themselves: Consider the dice in their rest frame. Put an enclosure around each die, and connect them with sticks. In the moving frame, everything contracts, and the dice never leave the enclosures. The distance between the dice must contract the same as an object of the same length. Edited October 23, 2020 by michel123456
Eise Posted October 23, 2020 Posted October 23, 2020 1 hour ago, michel123456 said: No, I think that you are wrong there. The distance 0.6 LH is the one from the FOR of E-X. As in diag2. The distance B must travel is given by E-X. This distance is 1Lh in the FOR of E-X, but 0.6Lh in the FOR of B. The distance B must travel is definitely not given by the length contraction of a rod attached to B, seen from the FOR of E-X. You see it yourself: B's rod does not reach X. 1 hour ago, michel123456 said: In the FOR of B, in 45 min B will travel & reach point Xb on the following Oh, c'mon. From the FOR of B the distance E-X is length contracted. Your diagrams do not show that.
swansont Posted October 23, 2020 Posted October 23, 2020 2 hours ago, michel123456 said: Is that your objection? (still searching the quote) You might notice that what I said is not what you had said. You said "If B is moving then the distance is contracted" Not the same thing at all. The length between E and X is at rest in E's frame. ergo, it is not length contracted. What md65536 had said was that the lengths were contracted in that example, because they were indeed moving.
joigus Posted October 23, 2020 Posted October 23, 2020 4 hours ago, Eise said: 'The distance'? Here you go again. Not mentioning which distance from which frame
michel123456 Posted October 23, 2020 Author Posted October 23, 2020 6 hours ago, Eise said: The distance B must travel is given by E-X. This distance is 1Lh in the FOR of E-X, but 0.6Lh in the FOR of B. Are you saying this? diag5.
Janus Posted October 23, 2020 Posted October 23, 2020 On 10/17/2020 at 2:33 AM, michel123456 said: I have read this thread again and again However, all the answers from valuable members here are based on the assumption that I don't understand relativity. Exactly, it is not so difficult to grasp. Here below some new diagrams that I hope will make you think again. The basis has been stolen from the animations of @Janus on page 9. Find the error: diag 1 E and X are at rest, B is moving. I have inserted a solid rod 1LH long in order to evade the comments like "space does not move & thus do not contract" Since B is moving together with its rod, both are time dilated & space contracted. The rod of B is 1LH long in the frame of B (see below diag 4). The notation 1LH(B) means 1 Light Hour from the FOR of B. diag 2 Exactly the same diagram, with the position of 0,6LH which is the contraction of B's rod from the FOR of Earth and planet X diag 3 Exactly the same diagram with the position of 0,6LH in the rod of B (from the FOR of E & X). I have labeled the point as Xb, that is 0,6LH from B. diag 4 Now the things as from the FOR of B: E and X are moving together with their rod, thus they are time dilated & length contracted. diag 5. Exactly the same diagram with annotated the 0,6 LH traveled by E & X. diag 6 Exactly the same diagram with annotated the point Xb that is on the rod of B at 0.6 LH from him. X will reach this point in 45 min from the FOR of B. It is the same point Xb that is shown in diag 3. Xa is the label for the point at the end of the rod of B AND NOW THE QUESTION: diag 7 Return to Earth and planet X FOR. Where is the correct contraction? Is it along diagonal 1 or along diagonal 2? We can check out: Here with diagonal 1, the upper part has been stretched along the direction of movement in order to make correspond point Xb and planet X (because that is the result of calculations that say the travel of B will last 45 min in the FOR of B) diag 8 And here below with diagonal 2. diag 9 Which one is the correct one, where is the error? Your error is that you again are not taking relativity of simultaneity into account. To illustrate, we can put clocks at the end of rod attached to B and at the point's along the rods. Thus from The E-X rest frame you get this: The green boxes indicate two clocks passing each other and their respective readings as they do. In the rest frame of B, you get this: Comparing the the times shown in the green boxes as clocks pass each other agree with those in the first image. In each image, there is a an instance when two pairs of clocks pass at the same time, but they are different pairs of clocks in each image. This just illustrates that, with the Relativity of simultaneity, events that are simultaneous for one frame of reference are often not simultaneous for another. 2
md65536 Posted October 23, 2020 Posted October 23, 2020 (edited) 16 hours ago, michel123456 said: Is that your objection? (still searching the quote) "If it's not moving, it's not Lorentz contracted" seems like a good rule to me. "If it's moving, it's length-contracted" could be made into a rule of thumb, but it's problematic (point particles, c, distances between relatively moving points, I think don't easily fit). "If B is moving, then (something else) is length contracted" is not a rule as you seem to think. I don't remember stating a rule though, so you might not find it. I did ask you which of certain objects (not distances) were moving and which were length-contracted. Misinterpreting things like that, and misinterpreting what SR says, and assuming it says something that you've invented, is a recurring problem here. I wrote: Quote The distance between the dice must contract the same as an object of the same length. Not that it'll matter for you, but it would have been helpful if I'd instead said something like that the "rest or proper distance between the dice is contracted in the frame where they're moving." If you're talking about a length being contracted, it's only relative to the length measured in another frame. Here the frames of reference are implied, but if you want to think about rules, it'd be better to be explicit about frames. If you want to think of a distance being contracted, think of a ruler that is measuring the distance. If that ruler is moving (in frame F) then distances being measured using that ruler are contracted (in frame F). If you're talking about measuring a distance to B, as measured by X, using X's ruler (that is not moving relative to X), then distances as measured using that (relatively) stationary ruler are not contracted according to X. I expect you to either ignore or twist this idea. Also, this is just my attempt to explain things as far as they make sense to me. It's not an "official rule of SR", and if it disagrees with SR or can be so easily misinterpreted, it's not a good rule. Certainly there are clearer and/or more precise ways to explain it. Edited October 24, 2020 by md65536
Eise Posted October 24, 2020 Posted October 24, 2020 15 hours ago, michel123456 said: Are you saying this? Yep. If I see it correctly, this is Janus 4th diagram in the series 'In the rest frame of B, you get this:'. I also had the idea that you would run into troubles because of the relativity of simultaneity. Janus worked it out into the last detail. It might look complicated, but you did it to yourself, by introducing rods that add nothing to the original situation. Really, Markus already said it clearly: it is not possible to derive some inconsistency in the framework of relativity. You behave like somebody who still thinks he can trisect an angle.
michel123456 Posted October 24, 2020 Author Posted October 24, 2020 (edited) On 10/23/2020 at 11:12 PM, Janus said: Your error is that you again are not taking relativity of simultaneity into account. To illustrate, we can put clocks at the end of rod attached to B and at the point's along the rods. Thus from The E-X rest frame you get this: The green boxes indicate two clocks passing each other and their respective readings as they do. In the rest frame of B, you get this: Comparing the the times shown in the green boxes as clocks pass each other agree with those in the first image. In each image, there is a an instance when two pairs of clocks pass at the same time, but they are different pairs of clocks in each image. This just illustrates that, with the Relativity of simultaneity, events that are simultaneous for one frame of reference are often not simultaneous for another. I am very attentive to your analysis. You have time on the vertical axis, distance on the horizontal, and...time again on the horizontal? Ticking at a different rate for the moving FOR? And not ticking for the FOR at rest? 13 hours ago, Eise said: Yep. If I see it correctly, this is Janus 4th diagram in the series 'In the rest frame of B, you get this:'. I also had the idea that you would run into troubles because of the relativity of simultaneity. Janus worked it out into the last detail. It might look complicated, but you did it to yourself, by introducing rods that add nothing to the original situation. Really, Markus already said it clearly: it is not possible to derive some inconsistency in the framework of relativity. You behave like somebody who still thinks he can trisect an angle. Here below the diagram, that as you said is the same with one of Janus (it cannot be totally wrong): The problem with it is that I believe there is a small mistake in it: the vertical line above X. Explanation: the length contraction factor makes a direct link between the 1LH in the FOR at rest and the 1LH in the moving FOR. It remains 1LH length but measured differently. The link is diagonal, like this below: The contraction link goes from Xa to X. It goes from 0,6 LH above to 0,6 LH below. It goes along the thin red lines. There is no vertical link. And there is a way to corroborate, as a posted previously. Take attention at point Xb. This point lies at 0,6 LH as measured by B (at rest). It correspons to the 45min calculated by @Eise, when it is supposed that B reaches X (or X reaches B) as measured by B. See here below: Do you spot point Xb on the diagram? Do you really believe that B will reach X at point Xb? Where is the correct length contraction? Is it between Xb & X (Line 1) or between Xa & B (Line 2)? Don't you see that point Xb is the result of a (erroneous) double contraction (0.6 of a contracted line)? I let you ponder carefully. Edited October 24, 2020 by michel123456
Janus Posted October 25, 2020 Posted October 25, 2020 17 hours ago, michel123456 said: I am very attentive to your analysis. You have time on the vertical axis, distance on the horizontal, and...time again on the horizontal? Ticking at a different rate for the moving FOR? And not ticking for the FOR at rest? Each set of images shows a series of chosen moments for each frame. Those moments are chosen as they represent when a pair of clocks pass each other. The numbers tell us what clocks at those positions read at that moment according to the chosen frame. So each successive diagram in the image represents a single moment in time. Thus in the first image of the first set, All the clocks on the E-X rod read the same time(12:00) at that moment, while the clocks along the B rod do not ( reading 12:00, 11:31:12, and 11:12 as you go from left to right) at that same moment.
Eise Posted October 26, 2020 Posted October 26, 2020 (edited) On 10/24/2020 at 11:24 PM, michel123456 said: I am very attentive to your analysis. You have time on the vertical axis, distance on the horizontal, and...time again on the horizontal? Ticking at a different rate for the moving FOR? And not ticking for the FOR at rest? Wow. You want to argue that there is an error in relativity, but you did not understand Janus' diagrams? The message of the diagrams is, that you cannot account for the situation you created, without taking the relativity of simultaneity into account. As you don't in any of your diagrams, you will fail. It is obvious for everybody here that you argue against something that is over your head. To be honest, as you probably noticed, it is also partially above my head. But my mindset is different than yours: understanding the basic principles, knowing that special relativity is fully integrated in every basic physics theory (EM, QED, general relativity), that special relativity is tested through and through since more than 100 years, I completely trust the real physicist saying something about relativity that I might not understand fully, but makes sense against that background. OTOH you go even that far that you insult one of our outstanding experts on relativity (Markus), and you invest not enough time in what another one is showing you in animations and extend diagrams (Janus). In the meantime, do you understand Janus' latest diagrams by now? If you do, you should see that your diagrams with all these connection lines have next to nothing to do with the real situation. When I said: On 10/24/2020 at 10:09 AM, Eise said: Yep. If I see it correctly, this is Janus 4th diagram in the series 'In the rest frame of B, you get this:'. I meant of course without all the superfluous rods and connection lines. We started the example with an observer with a clock (B), and 2 planets, with their clocks synchronised, and B traveling from planet E to planet X. Now you complicated the example by making B a rocket (B + its rod) with the same length as the distance it must travel (seen from their respective FORs). Now your diagrams are worthless, because they do not take the relativity of simultaneity into account. (Janus did that for you, so now you can try to setup the same argument in his diagrams. Does your argument still work)? Please explain to me why muons that live too short to travel from the place where they are created high in the atmosphere to the earth's surface. Let my see how your ideas work in this simple, empirically tested situation. Then I think you can learn a lot by looking at the 'ladder in the barn' thought experiment. It is also called a 'paradox', but just as the twin 'paradox', it is not a real paradox. This situation is very similar to the one you created with the rods (the 'ladder in the barn paradox' is also know as the 'pole in the barn paradox'). But it can only be solved if you take the relativity of simultaneity in account. Good luck pondering carefully these examples. Edited October 26, 2020 by Eise
studiot Posted October 26, 2020 Posted October 26, 2020 2 minutes ago, Eise said: Wow. You want to argue that there is an error in relativity, but you did not understand Janus' diagrams? It's that Thermopylae spirit.
michel123456 Posted November 8, 2020 Author Posted November 8, 2020 On 10/26/2020 at 1:18 PM, Eise said: you should see that your diagrams with all these connection lines have next to nothing to do with the real situation. About the "connection lines": Here below an ANALOGY You remember this picture showing foreshortening. Below my own hand from one side % the reverse: Very roughly, let's assume that, as a real measurement, the red distance is the same lenght as the the green one. Because of the effect of perspective, they look different depending on the position of the camera; the "apparent length" is observer dependent. Apparently, length 1 (red) and length 2 (green) are linked by "connection lines" in white color. The situation is perfectly symmetric. The length of line 1 (red) in the first picture corresponds to line 2 (green) in the second, and the length of line 2 (green) in the first corresponds to line 1 (red)in the second. If someone links lines 1 & 2 with a bold red line that "jumps" somehow vertically between lines 1 & 2, it is possible to calculate some result of the operation but the result will correspond to exactly nothing. If you want to "jump" from line 1 to line 2, you must follow the white lines. Otherwise the result is wrong.
Eise Posted November 8, 2020 Posted November 8, 2020 I have no idea what different perspectives of objects in the same FOR have to do with relativity which is about how observers in different FOR measure time and space. And I remember you have serious problems already with perspective... "Analogies are like cars: if you take them too far, they break down." (Don't remember where I saw that, but it applies very well here). Please test your ideas at the examples of time dilation/length contraction of muons. Can you explain these with your ideas? And did you look up the 'paradox of the pole in the barn'? If you work this through, you would realise how essential the relativity of simultaneity is, and that your diagrams do not suffice to account for that. Therefore you should understand the message of Janus' diagrams, which show the different time readings. One of the best understandable explanations I found in Chapter 2.3 Paradoxes in David J. Griffith, Revolutions in Twentieth-Century Physics. I think it is a very useful introduction for laypeople in modern physics, if your are not afraid for a bit of mathematics. But a little understanding of simple algebra is enough (which also is enough to have a basic understanding of special relativity).
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