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Quote from: PhysBang on 12/08/2016 23:21:45You can assert all you like that I haven't included time in this, but when a diagram shows two space dimensions where every single point depicted in it has the same time coordinate as every other, it is unnecessary to give a specific value for it when they're all identical.The very first thing that Einstein proved when developing SR was that the time coordinates are not all the same. Please see the original paper, sections 1 and 2.http://www.fourmilab.ch/etexts/einstein/specrel/www/You are denying one of the most fundamental and important results of SR. It would be funny if it were not kind of sad.QuoteWhen someone insinuates that I'm a crank, I don't take kindly to it. If you don't want return fire, don't shoot insults at people.I'm sorry. I assumed, incorrectly, that you would be alarmed that you are using the same kind of reasoning that we see from physics cranks. I should not have used those words.QuoteQuoteYou are, by your own admission, only using part of SR. That means that you are using your own special form of relativity.I am using the relevant parts which are sufficent for the argument.Again, you asked for help, specifically what you needed to complete your argument and what you needed to get it published. Your argument needs to include the time coordinates because that it an important part of SR. Quoteyou can just make Time=0 for the whole diagram (by the clocks of Frame A). Why on Earth do you have to keep asking for a coordinate when you already know they're all the same?The problem is that t=0 for one frame does not translate to the same value at all locations in other frames. Again, I urge you to read up on this.QuoteNo, I expect you to be able to generate your own versions of the diagrams in your head in a matter of seconds and to recognise that they are right in the way that a real expert would.You expect to rule by fiat that your diagrams are correct without doing the work to show they are correct.QuoteWhich part of them do you take issue with? I told you specifically: you claim that the wheels of the trains leave the tracks, I claim that they do not because of where and when the tracks and the wheels are. Claiming that the wheels leave the tracks depends on the timing of where the wheels are and when the track is at certain locations.QuoteThere are no wheels. You should be able to picture the whole thing as a 2D setup with the train between the rails. The centre of the parallelogram is directly between the rails and it only sticks through them because the length contraction applied at 45 degrees (in this most recent example) requires that - as I've told you before, that is a crash, but you imagine it isn't because you wrongly think the parallelogram is somehow still all between the rails rather than sticking out.You are still not thinking about when the rail is at a given location. In order for part of the train to stick past a track, it depends on the track being in a certain place at a certain time. You have not done the work to establish where the track is at different times or where parts of the train are at different times.Quote So, what numers do you need here?I need you to demonstrate exactly where and when a part of the train goes outside of the track.Quote If you think the rails should be further apart and that they would accomodate the whole parallelogram, make the parallelogram a hundred squares long instead of four and see how long that idea lasts. You don't need any more numbers on this - you should be able to see it in your mind straight away.I see different things in my mind than you do, because I have been trained in using SR in applications. As I said many times, taking timing into account changes issues. See the pole-and-barn paradox, for example: http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/polebarn.html (This "argument" really is a modification of the pole-and-barn paradox.)QuoteAnd every time I tell you that you already have the time coordinate for the entire diagram, you demand it again and say you haven't got it.Yeah, because you haven't. You are so ignorant of the basics of SR that you do not understand why to include the time coordinate. Which is why I asked you to do the actual SR calculations, not the rough calculations you rely upon.QuoteI don't take kindly to being insulted by someone and then having that person threaten me with being banned from a forum for returning fire.I am sorry that you cannot see past your anger to actually try to use SR properly. I feel that in being aggressive in pointing out your errors I have done you psychological harm that will prevent you from ever learning SR. I hope that this is not the case and I urge you to try to actually use the full Lorentz transformations in working through your argument.
You can assert all you like that I haven't included time in this, but when a diagram shows two space dimensions where every single point depicted in it has the same time coordinate as every other, it is unnecessary to give a specific value for it when they're all identical.
When someone insinuates that I'm a crank, I don't take kindly to it. If you don't want return fire, don't shoot insults at people.
QuoteYou are, by your own admission, only using part of SR. That means that you are using your own special form of relativity.I am using the relevant parts which are sufficent for the argument.
You are, by your own admission, only using part of SR. That means that you are using your own special form of relativity.
you can just make Time=0 for the whole diagram (by the clocks of Frame A). Why on Earth do you have to keep asking for a coordinate when you already know they're all the same?
No, I expect you to be able to generate your own versions of the diagrams in your head in a matter of seconds and to recognise that they are right in the way that a real expert would.
Which part of them do you take issue with?
There are no wheels. You should be able to picture the whole thing as a 2D setup with the train between the rails. The centre of the parallelogram is directly between the rails and it only sticks through them because the length contraction applied at 45 degrees (in this most recent example) requires that - as I've told you before, that is a crash, but you imagine it isn't because you wrongly think the parallelogram is somehow still all between the rails rather than sticking out.
So, what numers do you need here?
If you think the rails should be further apart and that they would accomodate the whole parallelogram, make the parallelogram a hundred squares long instead of four and see how long that idea lasts. You don't need any more numbers on this - you should be able to see it in your mind straight away.
And every time I tell you that you already have the time coordinate for the entire diagram, you demand it again and say you haven't got it.
I don't take kindly to being insulted by someone and then having that person threaten me with being banned from a forum for returning fire.
The very first thing that Einstein proved when developing SR was that the time coordinates are not all the same.
You are denying one of the most fundamental and important results of SR. It would be funny if it were not kind of sad.
I'm sorry. I assumed, incorrectly, that you would be alarmed that you are using the same kind of reasoning that we see from physics cranks. I should not have used those words.
Again, you asked for help, specifically what you needed to complete your argument and what you needed to get it published. Your argument needs to include the time coordinates because that it an important part of SR.
The problem is that t=0 for one frame does not translate to the same value at all locations in other frames. Again, I urge you to read up on this.
You expect to rule by fiat that your diagrams are correct without doing the work to show they are correct.
I told you specifically: you claim that the wheels of the trains leave the tracks, I claim that they do not because of where and when the tracks and the wheels are. Claiming that the wheels leave the tracks depends on the timing of where the wheels are and when the track is at certain locations.
You are still not thinking about when the rail is at a given location. In order for part of the train to stick past a track, it depends on the track being in a certain place at a certain time. You have not done the work to establish where the track is at different times or where parts of the train are at different times.
I need you to demonstrate exactly where and when a part of the train goes outside of the track.
I see different things in my mind than you do, because I have been trained in using SR in applications.
As I said many times, taking timing into account changes issues.
See the pole-and-barn paradox, for example: http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/polebarn.html (This "argument" really is a modification of the pole-and-barn paradox.)
QuoteAnd every time I tell you that you already have the time coordinate for the entire diagram, you demand it again and say you haven't got it.Yeah, because you haven't. You are so ignorant of the basics of SR that you do not understand why to include the time coordinate. Which is why I asked you to do the actual SR calculations, not the rough calculations you rely upon.
I am sorry that you cannot see past your anger to actually try to use SR properly. I feel that in being aggressive in pointing out your errors I have done you psychological harm that will prevent you from ever learning SR. I hope that this is not the case and I urge you to try to actually use the full Lorentz transformations in working through your argument.
Quote from: PhysBang on 13/08/2016 03:05:37The very first thing that Einstein proved when developing SR was that the time coordinates are not all the same.If we are drawing a Frame A diagram showing what is where at a specific time by the clock of Frame A, the time coordinate is the same in that diagram for every single point depicted in the diagram. It is impossible for Einstein or anyone else to prove that any of those points in a diagram in which they are all specifically tied to the same Frame A time coordinate can be tied to a different Frame A time coordinate as well - they can only have one time coordinate and it is the same one for all those locations in the diagram. If you want to use a different time coordinate, you have to use a different diagram, redrawing all the content in such a way as to take into account how far objects will have moved in between the two diagrams.
I've shown that if you accelerate a square along rail A, two of its edges remain aligned parallel to that rail, but if you accelerate a square along rail B, the edges which were parallel to that rail initially drift off that alignment and do so more and more as it reaches higher speeds, and this behaviour will be fully visible to Frame B observers too.
I am using AGI-type reasoning - the application of reason is my speciality and I do it with much greater care than physicists.
I now know how to present the argument to them in such a way as to head off their invalid objections from the start by showing them where their beliefs generate contradictions.
If you pluck a violin instead of using the bow, you are still playing the violin. Where plucking the violin is sufficient to play a piece of music, you are fully capable of playing that piece of music. I use the parts of SR that are relevant to the case I'm proving and have absolutely no obligation to use the parts that aren't.
No, I invited people to point out any faults with the argument if they could find them, but the argument was complete from the outset. The time aspect has never been lacking from it.
No, I expect you to generate your own diagrams and try to generate ones that are incompatible with mine if you think I'm wrong.
The frame A diagram of this, which I attached to a post recently (the one with the points X, Y and Z marked in it), shows exactly where the parallelogram is relative to the track at a single point in time by the clock of Frame A. You are trying to play games with time where no such games are possible.
QuoteI need you to demonstrate exactly where and when a part of the train goes outside of the track.Already done - see the diagram I just referred to.
QuoteSee the pole-and-barn paradox, for example: http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/polebarn.html (This "argument" really is a modification of the pole-and-barn paradox.)The pole-and-barn "paradox" will not help you here. What it shows is that a length contracted ladder can be seen from a different frame as not being contracted while the timing of the opening and closing of doors is seen to change to accommodate this, but that is all it shows. If you apply what you should have learned from it to my argument, you will find that it fits in with my argument just fine: the contracted objects can be regarded as not-contracted when viewed from their own frame, but other things have to adjust to maintain compatibility, such as angles of rails changing relative to the edges of a square in order to maintain their misalignment. Your trouble is that you only half think things through and don't take them the full distance.
There you go again - you're demanding that I add time coordinates to things that already have them, and it's your ignorance that's the problem here. The worst of it is that you don't learn when someone shows you you're wrong.
If I'm annoyed, it's because when a proof of something is presented, the job of those who comment on it is to address the points made within the proof and to state which ones they take issue with and to say which ones they agree with so that the discussion can home in on the points of conflict and resolve them for the person who's got it wrong. Instead of doing that, you're asking me to provide a different proof to prove the same thing, and that's not on. I want you to judge the proof that I have provided and not some other proof.
If you are relying on events from another frame and translating these events to frame A, then you need to take time change into account.
If you want to claim that they drift out of alignment, then you have to show where the rails are at different times. Heck, the shape has to deform in certain reference frames in order to match where the rails are, since the rails are moving and the train is moving.
QuoteI am using AGI-type reasoning - the application of reason is my speciality and I do it with much greater care than physicists.That is an incredibly bizarre claim, unless you are an artificial intelligence. In which case, this is a either a case of GIGO in terms of someone supplying you the rules of SR or a mistake in assigning the Bayesian updating protocols on this subject.
OK, but in this case, SR has no contradiction because the train never overlaps the rails because of where and when the rails are. The rails change position in different frames as well as the train.
A better analogy would be that you have violin held upside down and have the strings of the violin pressed against your shoulder. You are complaining that the violin doesn't make the sound you expect.
You have big blinders on because you were never trained in SR. For whatever reason, you refuse to even consider that you have these blinders and you refuse to do any of the rigorous work that would establish your case.
This is your argument. The burden of proof is on you, especially if you want to disprove a theory with scads of highly accurate applications over the last century.
If this is the case, then you should not have a problem completing your argument with the full Lorentz transformations.
I cannot trust your diagrams because you haven't done the mathematical work to justify them. All you do is repeatedly apply the same length contraction without ever showing when the parts of the train match the parts of the rail.
No angles have to change, merely when the parts of the train align with the parts of the rail. This is almost exactly the same as the pole-and-barn paradox.
You can claim that you have time coordinates, but until you show how you translate the time coordinates from one frame to the other, your claim is baseless.
OK: your proof is hopeless lacking. What it lacks is an application of SR. An application of SR uses the full Lorentz transformations, including time coordinates, which is important since checking where things are requires knowing when those things are.
And if the entire proof can be carried out without changing frame at all, there are no such translations to be done.
It's not a bizarre claim at all - I'm simply applying the rules of reasoning in the way that proper mathematicians do instead of the AGS way of doing things which mirrors the way which physicists do it (where contradictions are tolerated).
QuoteIf this is the case, then you should not have a problem completing your argument with the full Lorentz transformations.If it is not the case, you should have no problem showing it to be wrong by using whatever kind of calculations you like, and so long as you don't use any illegal transformations, your work should confirm my proof.
If you use a transfromation that makes the rhombus fit between the rails without the corners sticking through them, you are using an illegal transformation which can be shown to be illegal by the way it changes the order of events for an observer taking place where he is on a straight line up a Spacetime diagram. If you expect me to apply your illegal transformation and to trust its result, then that is not acceptable.
If you jump to the frame of the rhombus (B'), it becomes a square from your point of view. If you are moving with that square, you will think its edges are aligned north-south and east-west, but you'd be wrong. I've attached a diagram showing what happens to the north-south and east-west lines of the original grid in relation to this square. Observing from this frame puts you in the same frame as an observer on the train, and the rails follow the blue lines (the pair which are closer to horizontal than the other pair, except that the rails would actually cut through the corners of the square). In the equivalent case of being an observer on Train A and looking at the alignment of your train and its rails, the scene is quite different, again proving that different frames produce different physics.
The proof doesn't depend on any frame change
Edit: actually, this bit isn't so clear after all because the rails are moving relative to the grid lines
Edit 2: Ah, I've got it
Quote from: David Cooper on 13/08/2016 23:11:00And if the entire proof can be carried out without changing frame at all, there are no such translations to be done.Obviously not, since you are claiming that what is seen in one frame is not seen in another. You have to establish what is seen in each frame.
You admit that you are not trained in physics, yet you are making claims about how physicists reason. You also claim to reason like an artificial intelligence. Both of these claims are extraordinary and require extraordinary evidence.
If you just want to deny SR, then there is nothing that can be helped.
Sure, if we use David Cooper Relativity, then we get an inconsistency. I get that. But I and other people use SR.
QuoteThe proof doesn't depend on any frame changeExcept that you are talking about the deformation of a square from a frame in which it is at rest. Do you even read what you write?
Quote from: David Cooper on 13/08/2016 23:38:27Edit: actually, this bit isn't so clear after all because the rails are moving relative to the grid linesTa da!QuoteEdit 2: Ah, I've got itNope, still don't got it.
What are you on about? A square at rest in Frame A is accelerated to relativistic speed in Frame A and we see its shape change in Frame A. There is no switch there.
This gives us a nice rhombus shape, with a distance between NW and SE of sqrt(2), which is what we expect and a distance between SW and NE of 0.707, which is also what we expect from length contraction.
The tracks run W-E where they are at rest, but they are at rest in A', which means that they are moving NE in A.
This also means that the tracks are farther north the farther east one goes and are farther south the farther west one goes. This is a product of the relativity of simultaneity: where each frame assigns the track to be at different times. It depends on when frame A assigns the track to cross certain points. In A', we have a track sitting still, running W to E, but in A we have a track moving NE, oriented NW to SE.
No. The tracks are moving directly north in Frame A and are not at rest in Frame A', so you're turning them into objects co-moving with the rhombus through Frame A, and that will automatically give them the same alignment as the edge of the rhombus. Have you just worded this wrongly or have you actually treated them as co-moving with the rhombus through Frame A?
So you are indeed moving the rails incorrectly and it wasn't just a mistake in the wording. In Frame A, the rails are moving directly north and appear on the Frame A diagrams aligned perpendicular to their direction of travel precisely because every point of them shown in a single diagram shows them at the same Frame A time. If they were co-moving with Frame A', they would appear in Frame A diagrams with the same alignment as two of the edges of the rhombus. Your calculations are for that case and not for the one in my thought experiment where the rails are not co-moving with the rhombus.
Quote from: David Cooper on 14/08/2016 19:42:31What are you on about? A square at rest in Frame A is accelerated to relativistic speed in Frame A and we see its shape change in Frame A. There is no switch there.And yet, there is a frame of reference in which the square is at rest! These is another very basic fact of SR of which you are completely ignorant.
Yet you refuse to actually consider that you might be missing something and you lash out with insults rather than doing work.
I await your reply to my worked out example with actual coordinates and actual transformations. Or rather, I expect you to ignore it or blatantly deny the application of SR.
I do feel sorry to you, since it is apparent that you have had some learning problems in your life and may continue to have these problems. However, this does not excuse your attitude.
If you're just going to have the tracks not move with the object, then who cares? Of course a train moving in a different direction from a track can go over a track.
I tried to follow your setup from post #81 as much as possible. But I get it now, you want to see if you can move the square along the tracks.
Given that you admit that they are not moving together, who cares if they intersect or not? Where is the paradox?
If you calculate the angle of the track in Frame A based on it co-moving with the train, it will have length contraction applied to it in the NE-SW direction and will appear in Frame A diagrams at an angle to the east-west line, just like two of the edges of the rhombus. If you have the rails moving directly north, it is aligned directly east-west on the diagram. That is a radical difference: your calculations deviated from the events described in my proof and are invalid as commentary upon it.
QuoteGiven that you admit that they are not moving together, who cares if they intersect or not? Where is the paradox?Who cares? How slapdash do you want to be? This is shocking, and I thought the biscuit had already been taken lang syne. If I am moving north at 0.7c (or any other relativistic speed) through Frame A along with a pair of rails aligned perpendicular to my direction of travel, when I send my mag-lev squares along between them at relativistic speeds relative to me, I expect (if I have been misinformed by physicists) that square to stay happily between the rails no matter how fast it goes, just as if it is in the preferred frame, but no - it either warps and breaks or it buckles the rails because the physics of the frame I'm in doesn't work like the preferred frame.
So stop being slapdash and do the work properly.
I assumed that you were trying to create a contradiction so I did my best to recreate consistency across descriptions based on post #81. So you were pointing out that things deform differently under different transformations. How is this a contradiction for anyone?
If you send it at the right speed, it should be fine. Is your problem that you are not combining the speeds correctly?
QuoteSo stop being slapdash and do the work properly.Since I'm the only person here actually using SR, you seem like a complete jerk to be saying that I'm slapdash. Again, your learning problems are not an excuse for your poor behavior.
I showed that at least the deformation relative to one kind of track is consistent. If you want to show something else, then do what I did, use some actual Lorentz transformations, and actually make your case.
I'll see when I have enough time. I'm sorry that you have learning problems, but not sorry that you're a real ____.
In the end, you are a crank, so there is not much I can do to help you. I wish you the best of luck and I hope you don't waste too much time with your education website. And I really hope you have some support system to pay for your needs.