Tim88 Posted October 1, 2016 Share Posted October 1, 2016 Again you are confusing me, because events that happen simultaneously cannot be seen. You should add the light lines that show what the red & blue car are seeing when at A. They do not see the bicycle through simultaneity. Maybe you mean that distant simultaneity is a matter of inference based on convention? Yes indeed, that's a basic understanding of SR - and therefore not explicitly shown. The observations of those events by the cars are also events, and they occur later. Link to comment Share on other sites More sharing options...
michel123456 Posted October 2, 2016 Author Share Posted October 2, 2016 Maybe you mean that distant simultaneity is a matter of inference based on convention? Yes indeed, that's a basic understanding of SR - and therefore not explicitly shown. The observations of those events by the cars are also events, and they occur later. I meant that at event A, logic says that the red & blue car should observe the same event, the bike "somewhere". But that is not shown on the diagram. If Relativity states that from event A, the blue car sees the bike here and the red car sees the bike there, then yes that would be a problem for me. Link to comment Share on other sites More sharing options...
Celeritas Posted October 2, 2016 Share Posted October 2, 2016 I meant that at event A, logic says that the red & blue car should observe the same event, the bike "somewhere". But that is not shown on the diagram. If Relativity states that from event A, the blue car sees the bike here and the red car sees the bike there, then yes that would be a problem for me. Not a problem. They see the bike at a specific bike location and corresponding bike time, although they then measure the range to the bike differently. However, they do see the car differently. For example, the blue-car driver sees the bike moving and length-contracted, while the red car driver sees the car stationary at its proper-length. Just extend a 45 deg light-path from the co-located cars into and thru the bottom right quadrant. Where it intersects the bike (an event) is the visual-image being received by the 2 cars at their co-location (another event). Best regards, Celeritas Link to comment Share on other sites More sharing options...
VandD Posted October 2, 2016 Share Posted October 2, 2016 Maybe you mean that distant simultaneity is a matter of inference based on convention? Whatever convention you use, the convention you use has to measure constant speed of light in all inertial frames. And that will tell you there is relativity of simultaneity between the two reference frames. Simultaneity is a basic concept in SR. Relativity of simultaneity even more. You want to refute simultaneity and relativity of simultaneity? Tell us, you still don't understand SR, or ... you do understand SR but you don't want to accept what SR tells you? The observations of those events by the cars are also events, and they occur later. Indeed. And based on their individual measuring ruler and wristwatch they will sort out which events did occur simultaneously and which not. And it will tell them relativity of simultaneity. Here the light from the events: events that happen simultaneously cannot be seen. Really? Below I added a dog chasing the orange car and bicycle. Watch how the light from simultaneous events reach a car driver's eye simultaneously. Verdict: relativity of simultaneity. Link to comment Share on other sites More sharing options...
michel123456 Posted October 2, 2016 Author Share Posted October 2, 2016 (edited) (...) Really? Below I added a dog chasing the orange car and bicycle. Watch how the light from simultaneous events reach a car driver's eye simultaneously. Verdict: relativity of simultaneity. Right. Wrong wording from me. instead of events that happen simultaneously cannot be seen. I should have stated: an observer cannot get instantaneous information from an event that happened at a distance. IOW the event and the information one gets of the event are not simultanate when the event takes place at a distance. Edited October 2, 2016 by michel123456 Link to comment Share on other sites More sharing options...
Tim88 Posted October 2, 2016 Share Posted October 2, 2016 Whatever convention you use, the convention you use has to measure constant speed of light in all inertial frames. And that will tell you there is relativity of simultaneity between the two reference frames. Simultaneity is a basic concept in SR. Relativity of simultaneity even more. You want to refute simultaneity and relativity of simultaneity? Tell us, you still don't understand SR, or ... you do understand SR but you don't want to accept what SR tells you?[..] I clarified relativity of simultaneity; if you don't understand that it's a convention, it means that you don't understand SR. Link to comment Share on other sites More sharing options...
VandD Posted October 2, 2016 Share Posted October 2, 2016 I clarified relativity of simultaneity; if you don't understand that it's a convention, it means that you don't understand SR. Feel free to think that I don't understand SR. Link to comment Share on other sites More sharing options...
michel123456 Posted October 3, 2016 Author Share Posted October 3, 2016 (edited) Another proof that there is a huge gap between making accurate calculations and extracting a conclusion. Please do not argue about who understands better. -------------------------- Dear VandD I took the liberty to write on your graph, I hope it will not raise any misunderstanding.See belowFrom what I understand:At event Q, the blue car and the bike shared a same spacetime coordinate: they were meeting together._While the blue car goes to the left, the bike goes to the right, they separate and go away from each other. _At event A, the blue car observes the bike at event P. That is: in its past._And at event A, the red car also observes the bike at event P. That is: also in its past.Both cars receive an image of the bike as it was in the past at event P. The 2 images may differ but both images emanate from event P.The blue car then asks his physicist to calculate where is the bike "today" (simultaneity with event A in blue car's frame). The physicist answers after some calculations that the bike "is today" at event B. The red car asks the same question to his physicist and gets the answer that the bike is "today" (simultaneity with event A in red car's frame) at event C. Is that correct so far? Edited October 3, 2016 by michel123456 Link to comment Share on other sites More sharing options...
VandD Posted October 3, 2016 Share Posted October 3, 2016 Another proof that there is a huge gap between making accurate calculations and extracting a conclusion. Blue car and orange car calculate what happens 'now', i.o.w. simultaneous with event A. This leads to relativity of simultaneity. What else do you need? Please do not argue about who understands better. -------------------------- Dear VandD I took the liberty to write on your graph, I hope it will not raise any misunderstanding. See below sMAB0xO-editedbyMichel.jpg From what I understand: At event Q, the blue car and the bike shared a same spacetime coordinate: they were meeting together. _While the blue car goes to the left, the bike goes to the right, they separate and go away from each other. _At event A, the blue car observes the bike at event P. That is: in its past. _And at event A, the red car also observes the bike at event P. That is: also in its past. Both cars receive an image of the bike as it was in the past at event P. The 2 images may differ but Differ? Impossible. The image from event P is what it is. If event P = bike with clock with hands pointing to f.ex. 5:05 o'clock, then both cars at event A see the same image of the bike and the bike's clock showing 5:05. both images emanate from event P. The blue car then asks his physicist to calculate where is the bike "today" (simultaneity with event A in blue car's frame). The physicist answers after some calculations that the bike "is today" at event B. The red car asks the same question to his physicist and gets the answer that the bike is "today" (simultaneity with event A in red car's frame) at event C. Is that correct so far? O.K. Link to comment Share on other sites More sharing options...
michel123456 Posted October 3, 2016 Author Share Posted October 3, 2016 Differ? Impossible. The image from event P is what it is. If event P = bike with clock with hands pointing to f.ex. 5:05 o'clock, then both cars at event A see the same image of the bike and the bike's clock showing 5:05. ? The 2 cars are moving WRT the bike at different velocities, so I expected that the length contraction of the bike would not be the same as observed by the 2 cars. IOW that the image would not be the same. Link to comment Share on other sites More sharing options...
VandD Posted October 3, 2016 Share Posted October 3, 2016 (edited) ? The 2 cars are moving WRT the bike at different velocities, so I expected that the length contraction of the bike would not be the same as observed by the 2 cars. IOW that the image would not be the same. Strictly speaking you are correct, BUT... when I show the bike at the flag pole, the event means what it says: the bike at the flag pole. To be fully exact I should have said: event B = (f.ex) "front of bike is at the left surface paint of the pole". Hence both cars are seeing one and the same image: front of bike at pole paint with its clock at 5:05. An event has no 'length'. But on a larger scale, for practical reasons, we say: at event A the bike is at the pole, at event C the bike is at the tower, a.s.o. Length measurement: Length is measured between events: there would be a length between a front event of the bike and a rear event of the bike. For showing length contraction of the bike of the diagram I should zoom in a lot. Therefore, lets consider the length between the dog and the bike (or make it a long train from dog to bike). Orange car measures the length between events D and B. It's the spatial distance an orange ruler at rest (the orange ruler in a frame is also a set of simultaneous events in orange frame) measures between simultaneous events of the dog and the bike (or rear and front of train) in his orange 3D space frame of simultaneous events. How does blue car measure the distance between dog and bike (rear and front of train) present in his 3D space of simultaneous events? Not between events D and B because those events are not simultaneously occurring in his blue 3D space. Watch which events of the dog and bike (rear and front of train) are in his blue 3D space: events F and B ! Blue car driver measures the distance between events F and B. That length is shorter than what orange measures between dog and bike (events D and B) simultaneously present in orange 3D space of simultaneous events. Length "contraction" occurs because different observers measure distance/length between a DIFFERENT set of simultaneous events. This is to be kept in mind especially when proper length changes over time, see sketch 9 here: http://www.scienceforums.net/topic/98501-lost-in-langevins-language/page-3#entry944651 If blue car driver would measure the distance between (non-simultaneous in blue frame) events B and D on his blue ruler (ruler at rest in his blue 3D space of simultaneous events), the driver will measure a LONGER distance/length than proper orange length between D and B. Because on his ruler he will first tick off one event (event D) and then after some time the other event (event B). You are interested in what blue and orange car drivers SEE of all of this? I hope you understand the proper train and the shorter train are two different set of simultaneous events. All you have to do is draw 45 degree lines from the (train) events you are interested in, and see where and when they hit the blue driver's eye and orange driver's eye. Edited October 3, 2016 by VandD Link to comment Share on other sites More sharing options...
michel123456 Posted October 4, 2016 Author Share Posted October 4, 2016 I half agree and half disagree.I understand and agree with Length "contraction" occurs because different observers measure distance/length between a DIFFERENT set of simultaneous events. I also understand what you wrote Strictly speaking you are correct, BUT... when I show the bike at the flag pole, the event means what it says: the bike at the flag pole. To be fully exact I should have said: event B = (f.ex) "front of bike is at the left surface paint of the pole". Hence both cars are seeing one and the same image: front of bike at pole paint with its clock at 5:05.An event has no 'length'. But on a larger scale, for practical reasons, we say: at event A the bike is at the pole, at event C the bike is at the tower, a.s.o.Length measurement:Length is measured between events: there would be a length between a front event of the bike and a rear event of the bike. For showing length contraction of the bike of the diagram I should zoom in a lot. Then you write this: Therefore, lets consider the length between the dog and the bike (or make it a long train from dog to bike).Orange car measures the length between events D and B. It's the spatial distance an orange ruler at rest (the orange ruler in a frame is also a set of simultaneous events in orange frame) measures between simultaneous events of the dog and the bike (or rear and front of train) in his orange 3D space frame of simultaneous events. Ok that is the train's proper length. An orange train of length from D to B And then you continue saying: How does blue car measure the distance between dog and bike (rear and front of train) present in his 3D space of simultaneous events? Not between events D and B because those events are not simultaneously occurring in his blue 3D space. Watch which events of the dog and bike (rear and front of train) are in his blue 3D space: events F and B ! Blue car driver measures the distance between events F and B. That length is shorter than what orange measures between dog and bike (events D and B) simultaneously present in orange 3D space of simultaneous events. Let's say I am OK with that. The blue car measures the orange train of length F to B, which is shorter than DB. But then please make the same explanation beginning with a blue train of proper length FB. How will you get a shorter train as seen from the orange frame? Link to comment Share on other sites More sharing options...
VandD Posted October 4, 2016 Share Posted October 4, 2016 (edited) But then please make the same explanation beginning with a blue train of proper length FB. How will you get a shorter train as seen from the orange frame? Be carefull when you write "as SEEN" ... You probably mean how the blue train is shorter in the orange frame? If you want to know what an observer SEES, then you have to add the 45 degrees light paths and see when they hit an observer's eye. I leave that for you to add in following diagram... I give you the set of 'train events' (the big blue dots) that make the 4D blue train. Light blue zones indicates the set of train events in the blue 3D space of simultaneous events.These are the proper train at different blue instant of time. Orange zones indicates the set of train events in the orange 3D space of simultaneous events. These trains are shorter than proper trains in blue spaces. The shorter train in orange 3D space is made of a different set of train events than the (proper) train in blue 3D space. SR is about realtivity of simultaneous events. Edited October 4, 2016 by VandD Link to comment Share on other sites More sharing options...
michel123456 Posted October 5, 2016 Author Share Posted October 5, 2016 (edited) Be carefull when you write "as SEEN" ... You probably mean how the blue train is shorter in the orange frame? If you want to know what an observer SEES, then you have to add the 45 degrees light paths and see when they hit an observer's eye. I leave that for you to add in following diagram... I give you the set of 'train events' (the big blue dots) that make the 4D blue train. Light blue zones indicates the set of train events in the blue 3D space of simultaneous events.These are the proper train at different blue instant of time. Orange zones indicates the set of train events in the orange 3D space of simultaneous events. These trains are shorter than proper trains in blue spaces. The shorter train in orange 3D space is made of a different set of train events than the (proper) train in blue 3D space. SR is about realtivity of simultaneous events. Thank you. Maybe you have hit the nail that block my understanding. One of the multiple nails. You wrote: Be carefull when you write "as SEEN" ... You probably mean how the blue train is shorter in the orange frame? If you want to know what an observer SEES, then you have to add the 45 degrees light paths and see when they hit an observer's eye. I leave that for you to add in following diagram... And SR is about realtivity of simultaneous events. I thought SR was about observation. Let me explain: when you take a ruler in your hand, you can see the entire ruler. The ruler is at rest. It has a length. It exists in spacetime. On the diagram, the ruler at rest is represented by a segment that lies upon the 45 degrees, because if it were not on the 45 degrees, it would not be directly observable. It is the same as if you replace the ruler with the distance to a far away star: the star is observed as it was in the past (a few years ago). So, the end of the ruler (the star) is in the past, the other end of the ruler (in your hand) is at present time. The entire ruler (the distance from your hand to the star) is entirely visible: it lies upon the 45 degrees. The "proper length" of the ruler, the ruler that lies upon simultaneity line, is not directly observable: you cannot see the end of the ruler (the star) "now". You can calculate where it is in spacetime, but it is not directly observable. You must calculate. And the input in the equations are measurements. And the source of measurements is directly observable (I suppose) From this above follows a confusion (in my head) about what are the conclusions to get from SR. But is the above correct? Edited October 5, 2016 by michel123456 Link to comment Share on other sites More sharing options...
VandD Posted October 5, 2016 Share Posted October 5, 2016 (edited) Thank you. Maybe you have hit the nail that block my understanding. One of the multiple nails. I bet you would never see the light without spacetime diagrams. They show more than only pages of 'shut up and calculate'. You wrote: And I thought SR was about observation. Let me explain: when you take a ruler in your hand, you can see the entire ruler. Correct, but you don't SEE that proper 'at rest' ruler. Because the light from the simultaneous events of the atoms of that ruler do not reach your eyes simultaneously. The ruler is at rest. It has a length. It exists in spacetime. On the diagram, the ruler at rest is represented by a segment that lies upon the 45 degrees,No, it's rather 22.2 degrees. Only light paths are at 45 degrees. because if it were not on the 45 degrees, it would not be directly observable.How can a ruler be 'on' the 45 degree line??? Only light is represented by a 45 degre line. It is the same as if you replace the ruler with the distance to a far away star: the star is observed as it was in the past (a few years ago).correct So, the end of the ruler (the star) is in the past, the other end of the ruler (in your hand) is at present time. The entire ruler (the distance from your hand to the star) is entirely visible: it lies upon the 45 degrees.I don't understand what you mean by that. If you keep the ruler at rest in your hand (you hold the ruler at the left end), then the other end of that ruler at rest is out there. You don't SEE that other 'now' end of the ruler. But you do see a full ruler, but not light coming from the simultaneous ruler events. What you see of the ruler 'at rest' is light coming from successive 'proper ruler' in time. (I'll add that to the diagram) The "proper length" of the ruler, the ruler that lies upon simultaneity line, is not directly observable: you cannot see the end of the ruler (the star) "now".Correct You can calculate where it is in spacetime, but it is not directly observable.Nothing is directly observable. Even the light of a fly sitting on your eye takes a split second to reach the retina, and from there to your brain... You must calculate.Science doesn't get far without calculating. And the input in the equations are measurements.The input in the Lorentz Transformations are time and spatial coordinates. And the source of measurements is directly observable (I suppose)I don't know what you mean by that. You don't observe a length. You observe events. Light coming from events. And events can be: leaving a mark on a ruler, etc. From this above follows a confusion (in my head) about what are the conclusions to get from SR. But is the above correct? Here visualizing what an observer SEES. Edited October 5, 2016 by VandD Link to comment Share on other sites More sharing options...
michel123456 Posted October 5, 2016 Author Share Posted October 5, 2016 Here visualizing what an observer SEES. The black dots compose the image of the ruler. If the ruler was the Universe then all that we are observing of the Universe are the black dots. All the rest of the diagram comes from our calculations. The measurements that we are doing are taken from the black dots. Link to comment Share on other sites More sharing options...
VandD Posted October 5, 2016 Share Posted October 5, 2016 The black dots compose the image of the ruler. If the ruler was the Universe then all that we are observing of the Universe are the black dots. All the rest of the diagram comes from our calculations. The measurements that we are doing are taken from the black dots. I don't see what you are getting at. Every split second you see another set of events black dots. Don't you take them into account to make your "measurements" of space and time? Link to comment Share on other sites More sharing options...
VandD Posted October 6, 2016 Share Posted October 6, 2016 (edited) =============================================================================================Below I show you how you feel the shorter contracted car between hands of stretched arms:. ...============================================================================================Have you tried to read symmetry of time dilation in the 4D spacetime diagram? (If you want to know when one of the observers SEES the wristwatch.... add the 45 degree light paths! )Relative v = 0.7cHence time dilation factor is about 1.4 . .Michel, I can do no better -and I guess you can not find better on the internet- to show you howthe relativity of simultaneity of events leads to time dilation and length contraction.Now it's up to you to draw spacetime diagrams... Edited October 6, 2016 by VandD Link to comment Share on other sites More sharing options...
michel123456 Posted October 6, 2016 Author Share Posted October 6, 2016 (edited) ============================================================================================= Below I show . . . Michel, I can do no better -and I guess you can not find better on the internet- to show you how the relativity of simultaneity of events leads to time dilation and length contraction. Now it's up to you to draw spacetime diagrams... You guess correctly. Edited October 6, 2016 by michel123456 Link to comment Share on other sites More sharing options...
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