The 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.

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 am the one applying the basic rules of SR rigorously and not allowing them to generate contradictions through the application of illegal transformations. The funny/sad thing is that so many experts are happy to allow those contradictions to be generated and think it's not a problem, but I think it's most likely because they've never stopped to look carefully at what they're doing because they've been programmed to believe that all frames of reference produce identical physics. They don't though: 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'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.

I am using AGI-type reasoning - the application of reason is my speciality and I do it with much greater care than physicists. I must apologise to you too though for using the word "timewasting" in my previous post, because you've actually been very helpful in showing me how hard it is likely to be to get through to the peer review experts who may have similar misunderstandings about relativity to yours. 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.

You are, by your own admission, only using part of SR. That means that you are using your own special form of relativity.

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.

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.

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.

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.

I have proved the case without needing to change frame at all (see below).

You expect to rule by fiat that your diagrams are correct without doing the work to show they are correct.

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.

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.

The Frame A view of the parallelogram shows very clearly that its long sides are no longer aligned with the tracks and you can see that an observer at point X will meet both rails after the parallelogram has gone by, while an observer at point Z will meet both rails before the parallelogram has gone by. If you draw that out on a standard Spacetime diagram, you'll find that no transformation of it can change the order of events at those points (which are straight lines in the Spacetime diagram). That proves that the edges of the parallelogram are not parallel to the rails.

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.

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.

I 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.

I see different things in my mind than you do, because I have been trained in using SR in applications.

If you're seeing different things from what happens in my diagrams, you're doing it wrong.

As I said many times, taking timing into account changes issues.

I've taken timing fully into account. The problem here is that you haven't managed to get your head around the diagrams that have been put before you and to understand the idea of every point in a diagram having the same time coordinate - you're determined to misunderstand them.

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.)

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.

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.

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.

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.

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.

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.

So, here again is the proof with each point numbered. If you agree with any of the points, please say so. If you disagree with any, please point to them and explain what your problem is with them. [The first one (0) is not a point, but sets the scene.]

(0) In a frame of reference called Frame A (which we will treat as if it is the preferred frame), we have a square with its edges aligned north-south and east-west. Our coordinate grid is also aligned with the Y-axis running north-south and the X-axis running east-west. This frame is the scene for all the action, so every diagram involved in it has the same time coordinate for every point shown in that diagram, though different diagrams may depict different times.

(1) If the square is then moved (without rotating it at any point) such that it is now moving through Frame A at 0.866c in the direction NE, it will become a rhombus shape with its NE and SW corners twice as close together as its NW and SW corners, while the NW and SW corner will retain their original separation (distance).

[You are free to use your own numbers for the coordinates of the corners of the square/rhombus, and you can also make the square any size you like because whatever values you use, your result will be 100% compatible with my description. That means you don't need any numbers from me other than the 0.0866c figure and you should not be demanding them. You can work out for yourself that the contraction is to 0.5x the rest length. I also don't need to tell you where the rhombus shape is in Frame A as its location will make no difference to its shape. The only thing we're concerned with here is what shape it will be in Frame A when moving NE at 0.866c, and any expert in relativity will produce the exact same shape, so no one should be disagreeing with this point.]

(2) If the square, when it was at rest, had sides one metre long, the north-south component of the separation between the NW and SE corners of the rhombus must still be one metre, as will the east-west component of their separation.

[Again this will fit whatever coordinates you have used, so you don't need to demand any from me.]

(3) The speed of travel of the rhombus is 0.866c, so the north-south and east-west components of this movement, v, can be calculated from v^2 + v^2 = 0.866^2, and that means v = 0.612. The length contraction acting on things moving at this speed reduces them to 0.791 of their rest length.

[You can do the simple maths to check this for yourself - I've shown you how to do it before, and experts in relativity don't need to be told.]

(4) If we have two rails aligned east-west separated by one metre in the north-south direction when at rest in Frame A (and attached together by one-metre long poles to hold them in place relative to each other), when we move them at 0.612c northwards through Frame A they will contract closer together to a separation of 79.1cm.

(5) If we arrange these elements such that the centre of our rhombus shares the same Y-coordinate as a point midway between our rails, the NW and SE corners of our rhombus project out of the space between the rails, the NW corner being further north than the northern rail and the SE corner being further south than the southern rail.

[This is the case for any T-coordinate, and as the rails can be considered to be infinitely long, the X-coordinates for the rhombus are unimportant - any value will do.]

(6) If we take an identical square at rest in Frame A and then move it northwards at 0.612c, it will become a rectangle with its north-south length reduced to 79.1cm while its east-west length remains one metre.

(7) If we then maintain that rectangle's northwards speed of travel and add an eastwards component of movement to it, by the time that eastward component reaches 0.612c (still as measured from Frame A), its shape must match that of our rhombus as it's now co-moving with it and neither of them have been rotated at any stage.

(

If we place an observer at a frame A location L further north of the rails and the rhombus, we can arrange things in such a way that when we run through a series of diagrams in which we increment T for each, we will see the rails and centre of the rhombus pass through point L. One of the rails will reach point L first, then the rhombus will reach it, and the other rail will be the last to reach it.

(9) If we place more observers at points K and M to either side of L, careful placement of these can enable the objects passing through them to do so in a different order than in (

: at K we can have the NW corner of the rhombus arrive first followed by the rails, and at M we can have both rails pass through this point before the SE corner of the rhombus.

[Note: I have used letters K, L and M this time because the X, Y and Z that I used in the past could be muddled up with the idea of coordinates for some readers.]

(10) We can also have three observers called K', L' and M' who are going to do the equivalent with a square (again identical to our original square when at rest in Frame A) which we're going to send along another set of rails. These rails are aligned east-west and are stationary in Frame A. The square was sitting between them (with two of its edges touching them) while it was stationary. We now move that square eastwards at any relatistic speed you care to use and the contraction on its east-west length turns it into a rectangle. Our observers, K', L' and M' are moving northwards through Frame A at 0.612c, but no matter where we place them and no matter what speed we move our square/rectangle at along this track, we cannot find any way for our observers to encounter these objects in any order other than southern rail first, then square/rectangle (which they will see as a parallelogram with two of its edges running parallel to the rails), and then the northern rail last.

(11) We have just looked at two cases of observers encountering a pair of rails and a shape moving along between them (and in one case partly through them). Those two cases should be directly equivalent if all frames are to behave the same way, but that is not what we've found. For our L observer, the order of events, rail-shape-rail, matches up with the order of events for our K', L' and M' observers as they encounter different parts of the shape that they meet between two rails. However, for our K and M observers, we get different orders of events: shape-rail-rail; and rail-rail-shape. There is no valid transformation between frames that can change these orders, so the case it proven.

[To process this action, we simply work through a series of Frame A diagrams for a series of different time coordinates, moving our shape and observers each time we increment the time coordinate. I have given you all the information you need to do this - an expert in relativity needs nothing else and should be embarrassed if he needs to ask for more.]

That is a proof, and it's a sound one. If you want to disprove it, all you have to do is find a counterexample to any of the points made in it, but no such counterexamples can exist. Now, I'm going to get on with writing my ref-frame camera software so that I can find out what the Frame B view of our rhombus shape actually looks like (and I now think that none of its sides will be parallel to our grid lines at all). I will then post some key details of the program so that other people can write their own version of it without having to work out all the details for themselves.