Janus 1123 Posted October 8, 2020 Share Posted October 8, 2020 21 hours ago, michel123456 said: Yes i am baffled. To me, geometry shows evidently that the relative velocity is almost null since it is perpendicular to the line of sight. The fact that the virtual observer is in the same FOR of earth & planet X has nothing to do with that. Geometry still counts. No, the geometry does not evidently show the relative velocity as be almost null. As the observer gets further away, the angular velocity he would measure would decrease, But since the actual relative velocity is w x d*, Where w is the observed angular velocity and d is the distance to b, the increase in d cancels out the decrease in w and gives a constant answer for the relative velocity no matter how far the distance d is. Also, the distance between E and X doesn't change with an increase of d, and the fact that the E- observer distance and the X-observer distance is the same, just makes its easier for the observer to determine that B traveled at 0.8c . For example, lets assume that the observer is 1000 ly away from both. B leaves E, and the light of this event heads off towards our observer, it takes 1000 yrs to reach him. It takes 1 1/4 hours for B to reach X, so 1 1/4 hr after the light of B leaving E is emitted, the light of B arriving at X is emitted. It also takes 1000 yrs to reach the observer. This means that the light telling the observer that B arrived at X arrives 1 1/4 hrs after the light telling the observer that B arrived at X. Since the observer knows that he is an equal distance from both, he know that the light from these events took equal times to reach him, so he knows that the 1 1/4 difference between his seeing these event is exactly the same as the time it actually took B to go from E to X. The distance between E and X is 1 light hour, so he knows be traveled at 0.8 relative to them, and since he is at rest with respect to them, 0.8c relative to himself. This is all pretty straight forward, and in no way supports your conclusion. * technically, this equation assumes that B is traveling along an arc of a circle with our observer at the center. B is actually traveling a straight line. This would have the effect of causing the measured angular velocity to change as B went from E to X. Starting at a one value, increasing to a maximum at the midpoint, then decreasing again. This variance in angular speed would be largest with small values of D and decrease ( but never totally vanish) at larger values of D. If we assume D is very large, we can safely ignore this without too much loss of accuracy.) While complicates things a bit, it does not change the fact that our observer will always get a value of 0.8c for the velocity of B Link to post Share on other sites

michel123456 547 Posted October 8, 2020 Author Share Posted October 8, 2020 7 hours ago, Eise said: Of course you can. Subtract the effect of the signal delay, and you should get the relativistic time dilation. But I am not so versed in such calculations anymore. OK, it really was so simple: Take the inverse Lorentz transformation for time (we are looking at how B sees X at the moment of passing Earth, not how Earth sees B's time): Whereby gamma: We take t = 0 for ease, so in our example we get: vx'/gamma =(0.8 x 0.6)/0.6 = 0.8. 1 Lh = 60 LMinutes, so 0.8 x 60 LMinutes makes 48 minutes. Now back to my databases... I expected a result like 60 min x 0.6 = 36 min. 1 hour ago, Janus said: No, the geometry does not evidently show the relative velocity as be almost null. As the observer gets further away, the angular velocity he would measure would decrease, But since the actual relative velocity is w x d*, Where w is the observed angular velocity and d is the distance to b, the increase in d cancels out the decrease in w and gives a constant answer for the relative velocity no matter how far the distance d is. Also, the distance between E and X doesn't change with an increase of d, and the fact that the E- observer distance and the X-observer distance is the same, just makes its easier for the observer to determine that B traveled at 0.8c . For example, lets assume that the observer is 1000 ly away from both. B leaves E, and the light of this event heads off towards our observer, it takes 1000 yrs to reach him. It takes 1 1/4 hours for B to reach X, so 1 1/4 hr after the light of B leaving E is emitted, the light of B arriving at X is emitted. It also takes 1000 yrs to reach the observer. This means that the light telling the observer that B arrived at X arrives 1 1/4 hrs after the light telling the observer that B arrived at X. Since the observer knows that he is an equal distance from both, he know that the light from these events took equal times to reach him, so he knows that the 1 1/4 difference between his seeing these event is exactly the same as the time it actually took B to go from E to X. The distance between E and X is 1 light hour, so he knows be traveled at 0.8 relative to them, and since he is at rest with respect to them, 0.8c relative to himself. This is all pretty straight forward, and in no way supports your conclusion. * technically, this equation assumes that B is traveling along an arc of a circle with our observer at the center. B is actually traveling a straight line. This would have the effect of causing the measured angular velocity to change as B went from E to X. Starting at a one value, increasing to a maximum at the midpoint, then decreasing again. This variance in angular speed would be largest with small values of D and decrease ( but never totally vanish) at larger values of D. If we assume D is very large, we can safely ignore this without too much loss of accuracy.) While complicates things a bit, it does not change the fact that our observer will always get a value of 0.8c for the velocity of B Pretty straight forward yes, except the bold part. I am naive to believe that velocity is measured by δ distance divided by δ time. Since distance between the observer & the traveling clock remains the same (between the observer & the traveling object), δ distance is zero. So relative velocity (between the observer & the traveling clock) is zero. And the observer should see no time dilation & no length contraction (in relation to him). Although the same observer will see (according to what everybody say here) exactly what diagram 1 shows: a clock traveling 1LH in 1h:15 and ticking only 45 minutes, contracted sideways. Link to post Share on other sites

swansont 7656 Posted October 8, 2020 Share Posted October 8, 2020 1 hour ago, michel123456 said: Pretty straight forward yes, except the bold part. I am naive to believe that velocity is measured by δ distance divided by δ time. 1LH divided by 1.25 hours is 0.8c 1 hour ago, michel123456 said: Since distance between the observer & the traveling clock remains the same (between the observer & the traveling object), δ distance is zero. But the object isn’t traveling in that direction, so that’s not the correct distance to use 1 hour ago, michel123456 said: So relative velocity (between the observer & the traveling clock) is zero. That’s not what you asked, though. You asked “what is the speed of B in the FOR of the virtual observer?” You did not ask for the speed component between the clock and the observer Link to post Share on other sites

Janus 1123 Posted October 8, 2020 Share Posted October 8, 2020 4 hours ago, michel123456 said: I expected a result like 60 min x 0.6 = 36 min. Pretty straight forward yes, except the bold part. I am naive to believe that velocity is measured by δ distance divided by δ time. Since distance between the observer & the traveling clock remains the same (between the observer & the traveling object), δ distance is zero. So relative velocity (between the observer & the traveling clock) is zero. And the observer should see no time dilation & no length contraction (in relation to him). Although the same observer will see (according to what everybody say here) exactly what diagram 1 shows: a clock traveling 1LH in 1h:15 and ticking only 45 minutes, contracted sideways. The distance between c and E and The distance between c and X are the same. So are you claiming that an observer at c will say an object going from E to X ( along the red arrow) traveled zero distance? Of course not. The same holds for an observer at d, or e, f and g. They all would say that an object moving from E to B traveled the same distance in the same time. It doesn't matter how far you put your observer away, the traveled distance doesn't converge to 0, but remains the same. Link to post Share on other sites

Eise 490 Posted October 9, 2020 Share Posted October 9, 2020 (edited) 13 hours ago, michel123456 said: gamma is 1.66666 in our example. 1/gamma is 0.6 Right, I mix up gamma and its inverse again and again. So instead of dividing by 0.6, one should multiply by 1.6. The end result is the same... Ususally I look at the result, and when it looks like 'time contraction' or 'length dilation', I know I made the wrong choice.. It seems to me that your not grasping of what a FOR is, goes back to your idea that in relativity everything is a question of perspective (different distances and angles). And as observers in the same FOR can have different perspectives, you seem to think that different observers in the same FOR must come to different conclusions about length contraction and time dilation. But that is not what relativity is based on: it is based on the speed of an object relative to a FOR, independent of which observer in that FOR is doing the observations. So as a first step: can you see that it is perfectly possible to define such a speed without any contradiction. Take just simple examples from daily life: e.g. can you say what the velocity of a car on a motorway is, even if your position is miles away, and you see the motorway under an angle (so not only when standing on a bridge over the motorway, or standing exactly perpendicular to the velocity of the car)? If you see this, then you can look at simple derivations of relativity (e.g.deducing time dilation from the example of the light clock). And there you will see that the only velocity figuring in the derivation is the velocity of one FOR to another, i.e. the velocity of one FOR to another. It just makes no sense to try to understand relativity if you are not clear about the first step. You will get more and more confused. And it is even worse if you involve what I would take only as the last step: Step 3: what different observers actually see. Now perspective plays a role, signal delays, angles, distances etc. E.g. a car driving away from you, will 'become' smaller and slower. But of course you know it does not become smaller and slower really. But in your considerations, you mix up all 3 steps! In short: First tackle the problem of the definition of the velocity between FORs. (Take Janus' previous post.) Do not involve anything that has to do with relativity! Follow a few derivations of relativity phenomena, and see that only the velocity between FORs plays a role in these derivations. Voluntary I would say, because it does not lead to physical effects: ponder about what different observers actually see. Just one tiny remark: Janus' animations with the clocks on Earth, X and B are not explanations. They show precisely what relativity says. However the later ones (with the 3 clocks and the light wave fronts) do explain relativity of simultaneity. Do not mix up these too: depictions of relativity and explanations of relativity. Edited October 9, 2020 by Eise Link to post Share on other sites

michel123456 547 Posted October 9, 2020 Author Share Posted October 9, 2020 (edited) 8 hours ago, Janus said: The distance between c and E and The distance between c and X are the same. So are you claiming that an observer at c will say an object going from E to X ( along the red arrow) traveled zero distance? Of course not. The same holds for an observer at d, or e, f and g. They all would say that an object moving from E to B traveled the same distance in the same time. It doesn't matter how far you put your observer away, the traveled distance doesn't converge to 0, but remains the same. Velocity is a vector: from wiki Quote Velocity is a physical vector quantity; both magnitude and direction are needed to define it. The scalar absolute value (magnitude) of velocity is called speed, being a coherent derived unit whose quantity is measured in the SI (metric system) as metres per second (m/s) or as the SI base unit of (m⋅s^{−1}). For example, "5 metres per second" is a scalar, whereas "5 metres per second east" is a vector. If there is a change in speed, direction or both, then the object has a changing velocity and is said to be undergoing an acceleration. For E, velocity is speed 0.8c going out. The input, for him, is Vout=0,8c for X, velocity is speed 0.8c going in. The input for him, is Vin=-0.8c The direction of Vin is opposite to the direction of Vout: it is a different vector, so it must be understood as a different velocity. Which one is the correct for observer g? Vin or Vout? Which one is the "good one"? ------------------------- And 13 hours ago, michel123456 said: 21 hours ago, Eise said: Of course you can. Subtract the effect of the signal delay, and you should get the relativistic time dilation. But I am not so versed in such calculations anymore. OK, it really was so simple: Take the inverse Lorentz transformation for time (we are looking at how B sees X at the moment of passing Earth, not how Earth sees B's time): Whereby gamma: We take t = 0 for ease, so in our example we get: vx'/gamma =(0.8 x 0.6)/0.6 = 0.8. 1 Lh = 60 LMinutes, so 0.8 x 60 LMinutes makes 48 minutes. Now back to my databases... I expected a result like 60 min x 0.6 = 36 min. I expected my calculation to be corrected by someone. (the one in the bottom that gives 36 min.) Edited October 9, 2020 by michel123456 Link to post Share on other sites

Ghideon 477 Posted October 9, 2020 Share Posted October 9, 2020 I do not share your view on this. 25 minutes ago, michel123456 said: The direction of Vin is opposite to the direction of Vout: it is a different vector, so it must be understood as a different velocity. Ok, lets say it Vin and Vout are different velocities. What do you suggest the acceleration to be that changes the velocity from Vin to Vout? Link to post Share on other sites

Eise 490 Posted October 9, 2020 Share Posted October 9, 2020 1 hour ago, michel123456 said: The scalar absolute value (magnitude) of velocity is called speed, being a coherent derived unit whose quantity is measured in the SI (metric system) as metres per second (m/s) or as the SI base unit of (m⋅s^{−1}). And it is this scalar absolute value that is the base of relativity. And you can see that if you follow correct derivations of relativity, this speed is used, not its apparent speed for some observer. 1 hour ago, michel123456 said: I expected my calculations to be corrected by someone. Well, if you just show '60 min x 0.6 = 36 min' it is not clear to me how you got to these values. (60 min and 0.6). Which formula did you use? So I use the Lorentz transformation for time, and then fill in the values: \[ t = \frac{{t}' + \frac{v{x}'}{c^{2}}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} \] \[t = \frac{{0} + {0.8} . {0.6} }{\sqrt{1 - {0.8 ^{2}}}}\] You will end up with 0.8 hour, which is 0.8 x 60 minutes = 48 minutes. Link to post Share on other sites

michel123456 547 Posted October 9, 2020 Author Share Posted October 9, 2020 1 hour ago, Eise said: Well, if you just show '60 min x 0.6 = 36 min' it is not clear to me how you got to these values. (60 min and 0.6). Which formula did you use? Time dilation factor is 0.6 Janus calculated 1h & 15 min dilated into 45 min. That is 75min x 0.6= 45 min. Lenght contraction factor is 0.6: the 1LH distance contracted into 0.6 LH So I expected that the time lag 1 hour (the delay) would become 60min. x 0.6 = 36 min. (the time light need to travel 0,6 LH). Link to post Share on other sites

Eise 490 Posted October 9, 2020 Share Posted October 9, 2020 10 minutes ago, michel123456 said: Time dilation factor is 0.6 Janus calculated 1h & 15 min dilated into 45 min. That is 75min x 0.6= 45 min. Lenght contraction factor is 0.6: the 1LH distance contracted into 0.6 LH So I expected that the time lag 1 hour (the delay) would become 60min. x 0.6 = 36 min. (the time light need to travel 0,6 LH). This should have told you that your calculation is wrong. Did somebody already told you that relativity is not about delays? Do you still think that relativity is about signal delays? Did you already look up the derivation of time dilation by means of the light clock? Can you show us where the signal delay between light clock and observer is taken into account? Link to post Share on other sites

michel123456 547 Posted October 9, 2020 Author Share Posted October 9, 2020 (edited) On 10/3/2020 at 9:48 PM, Janus said: B is at rest while the Earth and planet X move to the left at 0.8c. the Earth, planet X, their clocks, and the distance between them is length contracted. B is not. It take 45 min by B's clock for planet X to travel from being 0.6 lh away to B at 0.8c Earth and planet X's clocks are time dilated and only accumulate 27 min during this time. However, due to Relativity of Simultaneity, the Planet X clock already reads 48 min later than Earth's clock when B and the Earth are next to each other. Thus the 27 minutes it advances brings it to 1 hr 15 min, as it passes B. The times on Earth's clock and B's clock when they are next to each other agree with the first animation, and the Times shown on B's clock and Planet X's clock when they are next to each other also agree with the first animation. @Eise here above. What is "Relativity of Simultaneity"? And what are the 48min that Planet X clock already reads when B and the Earth are next to each other ? Edited October 9, 2020 by michel123456 Link to post Share on other sites

Eise 490 Posted October 9, 2020 Share Posted October 9, 2020 (edited) 1 hour ago, michel123456 said: What is "Relativity of Simultaneity"? Janus explained that here. If you do not understand his explanation, why did't you simply ask? 1 hour ago, michel123456 said: And what are the 48min that Planet X clock already reads when B and the Earth are next to each other ? As Janus already said, when two clocks on different locations are synchronous in their common FOR, then they are not synchronised in a FOR moving relative to the first. But this was already said to you. I just used the Lorentz transformations, i.e. the coordinate transformations between 2 FORs. Do not forget: B is already moving with 0.8c. As you see from my calculation example, t also depends on x', not just on t' alone. I hope you know what the Lorentz transformations are, if not, look them up in Wikipedia. Edited October 9, 2020 by Eise Link to post Share on other sites

michel123456 547 Posted October 9, 2020 Author Share Posted October 9, 2020 (edited) 4 hours ago, Eise said: Janus explained that here. If you do not understand his explanation, why did't you simply ask? As Janus already said, when two clocks on different locations are synchronous in their common FOR, then they are not synchronised in a FOR moving relative to the first. But this was already said to you. I just used the Lorentz transformations, i.e. the coordinate transformations between 2 FORs. Do not forget: B is already moving with 0.8c. As you see from my calculation example, t also depends on x', not just on t' alone. I hope you know what the Lorentz transformations are, if not, look them up in Wikipedia. I am sorry. I am continuously messing the FOR's. B is at rest. Earth & X are moving. B is calculating with Lorentz transform what clocks on Earth & X are showing. At 12:00 the travel begins (that's the only thing I understand so far!!). By your calculations you say that when the departure bell rings at 12:00, clock on X shows 12:48. (according to B). Is that it? Edited October 9, 2020 by michel123456 Link to post Share on other sites

Eise 490 Posted October 10, 2020 Share Posted October 10, 2020 18 hours ago, michel123456 said: By your calculations you say that when the departure bell rings at 12:00, clock on X shows 12:48. (according to B). Is that it? Yes. But that is not the time that B sees, it is what B concludes, given the signal delay. Just take care that you do not mix up what observers see, and what they can conclude about what really is going on. Your 'show' is ambiguous. Link to post Share on other sites

michel123456 547 Posted October 10, 2020 Author Share Posted October 10, 2020 8 hours ago, Eise said: Yes. But that is not the time that B sees, it is what B concludes, given the signal delay. Just take care that you do not mix up what observers see, and what they can conclude about what really is going on. Your 'show' is ambiguous. Thanks. Other question, on the grounds of what has been explained previously. I understand that there are at least 3 stages of calculations: Stage 1, Earth & Planet X are at rest, B is moving, contracted & time delayed : you insert the opening statements (1LH of distance, velocity 0.8c) and extract the result= Time of travel 75 min. in the FOR of E & X Stage 2, B is at rest, Earth & Planet X are moving: the distance between E & X is contracted by a factor of 0.6, so the distance becomes 0.6LH & velocity 0.8c give Time of travel 45 min. in the FOR of B Stage 3, B is at rest again, (quoting from Janus) “B is at rest while the Earth and planet X move to the left at 0.8c. the Earth, planet X, their clocks, and the distance between them is length contracted. B is not. It take 45 min by B's clock for planet X to travel from being 0.6 lh away to B at 0.8c Earth and planet X's clocks are time dilated and only accumulate 27 min during this time. However, due to Relativity of Simultaneity, the Planet X clock already reads 48 min later than Earth's clock when B and the Earth are next to each other. Thus the 27 minutes it advances brings it to 1 hr 15 min, as it passes B. The times on Earth's clock and B's clock when they are next to each other agree with the first animation, and the Times shown on B's clock and Planet X's clock when they are next to each other also agree with the first animation.” I have put these 3 stages in this sketch below: (this is not a spacetime diagram, simply a sketch that puts my ideas all together) Relativity of Simultaneity (RoS) is used in stage 3 in such a way that after the calculation pirouette, everything comes back in place: stage 3 corresponds to stage 1 with 75 min. recorded Time of Travel. That is fine. My questions are thus the followings, 1. why is there no use of RoS in stage 2? (the question mark on the sketch) 2.I am confused about the calculation procedure generally: why does it go one way (here from up to down) instead of going forth & back? Link to post Share on other sites

md65536 383 Posted October 10, 2020 Share Posted October 10, 2020 (edited) 5 hours ago, michel123456 said: That is fine. My questions are thus the followings, 1. why is there no use of RoS in stage 2? (the question mark on the sketch) 2.I am confused about the calculation procedure generally: why does it go one way (here from up to down) instead of going forth & back? The 3 "stages" sound good to me... Please don't go backward from this point, where the details are no longer fine! Answers: 1. RoS concerns the simultaneity of separated clocks or events, in this case the time at Earth relative to the time at X. It's not showing up yet because you're not comparing the times of things at E and X. However, your stages 2 and 3 are describing the same calculations. They're just "what is measured in B's inertial frame." RoS can be used in stage 2 to explain why, when B is arriving at X, B can find that X's clock is at 1:15 while E's clock is only at 0:27. Both have recorded 27 minutes passing (according to inertial B) during B's trip, but X started with a clock 48 minutes ahead (according to inertial B). Meanwhile, in the E+X frame, these same clocks are synchronized. That's relativity of simultaneity. 2. There's no reason, the "stages" are just calculations from different frames. You can calculate from any frame in any order that you want, and you don't have to do them all. Any one frame of reference can make all the measurements necessary, assuming all the clocks are accessible to them, to calculate the proper times mentioned here (B arrives at X at 1:15 on X's clock and 0:45 on B's clock, all frames of reference can show that on their own, using the initial conditions described above. 0:27 on E's clock is a coordinate time (measured by B from a distance) and is a frame dependent measure, differing in different frames again due to RoS). It doesn't make sense to line up all the stages as you did in the diagram. They're times measured by different clocks. By analogy, imagine if you made a map lining up all the countries in the world and attaching them vertically. It would show relative lengths of countries, but it wouldn't show how they're actually connected. Edited October 10, 2020 by md65536 Link to post Share on other sites

md65536 383 Posted October 11, 2020 Share Posted October 11, 2020 9 hours ago, michel123456 said: 1. why is there no use of RoS in stage 2? (the question mark on the sketch) Another answer based only on the sketch: 27+48=75 basically says "The amount of time that X's clock ticks while B travels, plus the time on X's clock when B begins, equals the arrival time on X's clock of 1:15" 27+48=75 expresses the sentence from B's frame. 75+0=75 expresses the exact same sentence, from the E+X frame of reference. Those are sensible statements because they're adding times measured by the same clock. 45+?=75 says "The amount of time that B's clock ticks while B travels, plus ??? equals the time on X's clock then B arrives." That doesn't make sense, because those times are measured by different clocks. Link to post Share on other sites

michel123456 547 Posted October 12, 2020 Author Share Posted October 12, 2020 (edited) On 10/11/2020 at 7:37 AM, md65536 said: 45+?=75 says "The amount of time that B's clock ticks while B travels, plus ??? equals the time on X's clock then B arrives." That doesn't make sense, because those times are measured by different clocks. Taking from your statements, reversed, that should give: 45+?=75 says "The amount of time that B's clock ticks while B travels, plus the time on B's clock when X begins equals the time on X's clock then B arrives." And we can assume that the result should be: ?=30min. Edited October 12, 2020 by michel123456 Link to post Share on other sites

Eise 490 Posted October 12, 2020 Share Posted October 12, 2020 On 10/10/2020 at 8:16 PM, michel123456 said: I understand that there are at least 3 stages of calculations: I might repeat a few things that md65536 might already have said, but repeating in other words might help... So, no, there are no 3 stages. The 3 different calculations show the events from different FORs, and also different things (the numbers correspond to your 'stages') Calculates the travel duration of B according to an observer in FOR 'E-X'. Calculates the travel duration of the travel of X, according an observer in FOR 'B', from the moment it passes E. Calculates the time of the clock at X, seen from the viewpoint from B at the moment B passes E and at the moment when B passes X. The specialty of the 3^{rd} calculation is that it is not only about durations, but about points in time. I had some troubles here too, because from the view of B the clock at X is also time dilated. From Bs perspective it takes 45 minutes before X reaches him. Taking time dilation into account this makes 0.6 x 45 = 27 minutes. But when X arrives at B, X's clock shows 75 minutes. This apparent contradiction however disappears when you take the RoS in account: according B, X's clock stood at 48 minutes when E was passing. Some authors take the RoS as the essence of relativity. Measurements of lengths and points in time should always refer to correct measurements. E.g. when a stick is moving in the direction of its length, and one measures it by measuring the begin point and the end point at the same time, i.e. simultaneously, one get the 'correct length' (the proper length). However, observers in different FORs from the stick's FOR do not agree on 'the same time'. Therefore from the FOR of the stick other FOR's do not do that. They measure at different times, and so they get the length of the stick wrong. Link to post Share on other sites

md65536 383 Posted October 12, 2020 Share Posted October 12, 2020 (edited) 8 hours ago, michel123456 said: Taking from your statements, reversed, that should give: 45+?=75 says "The amount of time that B's clock ticks while B travels, plus the time on B's clock when X begins equals the time on X's clock then B arrives." And we can assume that the result should be: ?=30min. But that's incorrect. Times on B's clock don't add up to times on X's clock. Do you at least understand that you're talking about 2 different clocks? I know it's only page 12 but do you understand that much so far? 5 hours ago, Eise said: I might repeat a few things that md65536 might already have said, but repeating in other words might help... Agreed, there are so many ways to describe the concepts, and different people "get it" different ways, plus I often make mistakes. It's too bad this isn't in the relativity forum and might be read by others who'll get it. There's always something that makes more sense with someone else's explanation. 5 hours ago, Eise said: E.g. when a stick is moving in the direction of its length, and one measures it by measuring the begin point and the end point at the same time, i.e. simultaneously, one get the 'correct length' (the proper length). However, observers in different FORs from the stick's FOR do not agree on 'the same time'. Therefore from the FOR of the stick other FOR's do not do that. They measure at different times, and so they get the length of the stick wrong. I don't agree, as worded. Different observers do measure the ends of the stick simultaneously in their respective frames, and they get the correct relative length of the stick. But it's true those measurement times are different for different observers. The proper length is measured when the stick's at rest, it's not enough to measure at a single time, because any observer can do the latter. (Not mentioned here, but also it's no good to measure the length at times that are simultaneous in the stick's rest frame, because in other frames the stick moves between those two times.) Edited October 12, 2020 by md65536 Link to post Share on other sites

michel123456 547 Posted October 13, 2020 Author Share Posted October 13, 2020 16 hours ago, md65536 said: On 10/12/2020 at 1:02 PM, michel123456 said: Taking from your statements, reversed, that should give: 45+?=75 says "The amount of time that B's clock ticks while B travels, plus the time on B's clock when X begins equals the time on X's clock then B arrives." And we can assume that the result should be: ?=30min. But that's incorrect. Times on B's clock don't add up to times on X's clock. Do you at least understand that you're talking about 2 different clocks? I know it's only page 12 but do you understand that much so far? You should read again the statement. It is about B's clock. It states that you must add to B's clock the RoS caused by the fact that E and X are not experienced in simultaneity by B. Link to post Share on other sites

md65536 383 Posted October 13, 2020 Share Posted October 13, 2020 6 hours ago, michel123456 said: You should read again the statement. It is about B's clock. Bold emphasis mine: On 10/12/2020 at 4:02 AM, michel123456 said: 45+?=75 says "The amount of time that B's clock ticks while B travels, plus the time on B's clock when X begins equals the time on X's clock then B arrives." That's your statement, it's about two clocks. 6 hours ago, michel123456 said: It states that you must add to B's clock the RoS caused by the fact that E and X are not experienced in simultaneity by B. I can't imagine how to explain why this is wrong if you don't understand that you're talking about different clocks. Are you purposefully making statements that you know are nonsense? (A strawman to defeat) Or do you think your statement makes sense and is true? Link to post Share on other sites

michel123456 547 Posted October 13, 2020 Author Share Posted October 13, 2020 1 hour ago, md65536 said: Bold emphasis mine: That's your statement, it's about two clocks. I can't imagine how to explain why this is wrong if you don't understand that you're talking about different clocks. Are you purposefully making statements that you know are nonsense? (A strawman to defeat) Or do you think your statement makes sense and is true? I believe that in the FOR of B, X does not start to move instantaneously. Link to post Share on other sites

md65536 383 Posted October 13, 2020 Share Posted October 13, 2020 1 hour ago, michel123456 said: I believe that in the FOR of B, X does not start to move instantaneously. I see. That kind of makes sense... B measures a shorter trip but a delayed start and ends up with the same time that X has. That's not special relativity. There's no point in discussing what special relativity predicts any further, if we're talking SR while you're talking about your own ideas in the language of SR. I could demonstrate why "when B starts moving relative to X, X is delayed before moving relative to B" is inconsistent, but if you have no problem picking aspects of relativity that you like while rejecting others, you'll continue finding ways to make the numbers add up to whatever you want, with no regard for consistency. You won't understand relativity while ignoring what it predicts. I don't think that's a problem, and I don't think you do either. Not everyone needs to understand it. Sorry it didn't work out. Link to post Share on other sites

michel123456 547 Posted October 14, 2020 Author Share Posted October 14, 2020 (edited) 13 hours ago, md65536 said: I see. That kind of makes sense... B measures a shorter trip but a delayed start and ends up with the same time that X has. That's not special relativity. There's no point in discussing what special relativity predicts any further, if we're talking SR while you're talking about your own ideas in the language of SR. I could demonstrate why "when B starts moving relative to X, X is delayed before moving relative to B" is inconsistent, but if you have no problem picking aspects of relativity that you like while rejecting others, you'll continue finding ways to make the numbers add up to whatever you want, with no regard for consistency. You won't understand relativity while ignoring what it predicts. I don't think that's a problem, and I don't think you do either. Not everyone needs to understand it. Sorry it didn't work out. But when @Eise quoted @Janus & explained the "However" part, that made sense to you, and it was about Relativity. As a reminder: On 10/3/2020 at 9:48 PM, Janus said: B is at rest while the Earth and planet X move to the left at 0.8c. the Earth, planet X, their clocks, and the distance between them is length contracted. B is not. It take 45 min by B's clock for planet X to travel from being 0.6 lh away to B at 0.8c Earth and planet X's clocks are time dilated and only accumulate 27 min during this time. However, due to Relativity of Simultaneity, the Planet X clock already reads 48 min later than Earth's clock when B and the Earth are next to each other. Thus the 27 minutes it advances brings it to 1 hr 15 min, as it passes B. The times on Earth's clock and B's clock when they are next to each other agree with the first animation, and the Times shown on B's clock and Planet X's clock when they are next to each other also agree with the first animation. 13 hours ago, md65536 said: I see. That kind of makes sense... B measures a shorter trip but a delayed start and ends up with the same time that X has. Yes. B measures 75 min. Not 45 min. And there is no paradox. Edited October 14, 2020 by michel123456 Link to post Share on other sites

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