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Author Topic: Can a preferred frame of reference be identified?  (Read 6571 times)

Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #75 on: 12/08/2016 20:07:04 »
Is there no real expert with qualifications here who's going to step in and let PhysBang know he's not up to this stuff? You could do it in a PM, but he needs to be told and he won't take it from me. The best way to defend physics isn't to dig in and defend things that are wrong, but to move on and make the subject right.
« Last Edit: 12/08/2016 20:11:31 by David Cooper »
 

Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #76 on: 12/08/2016 23:21:45 »
You are correct: your continued failure to actually perform the relevant calculations are examples of my avoiding issues.

Anyone competent can work out for themselves the shape that a square will take up when it moves through Frame A without needing to lift a calculator. You can't do it because you're incompetent to an extraordinary degree. How on Earth can you have qualifications in this if you can't even do that! Which university is responsible for this failure?
So you are refusing to do the calculations necessary to show you are correct? Need I remind you that this is your argument? That you are trying to convince me and others? That you came here, asking for flaws in your argument? It was only because you asked that I pointed out that you failed to take time into account and that you have not done the calculations to make your case.

You are free to believe that you don't have to be held to the same standard that physicists are held when making their arguments. You did, however, ask for help. You are behaving as if your requests for help are not genuine.

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I am using things that are 100% parts of SR. The length contraction, the way things appear in Frame A, the direction the contraction is applied, the degree to which it is applied, the coordinates for objects in Frame A (where north-south is one space dimension, east-west is another space dimension, and the whole diagram is a slice showing how things are located in that space at a specific Frame A time. To write it off as not being SR is one of the most ludicrous things you can do.
You are, by your own admission, only using part of SR. That means that you are using your own special form of relativity.

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Where is your problem? Look at the diagram and apply your own coordinates to it. A child could do it. For example, we could decide that a stationary square is centered on point (0,0) and with corners at (2,2), (2,-2), (-2,-2) and (-2,2) with north being the Y-axis and east being the X-axis. We can then move it up to relativistic speed moving NE and calculate its shape when it's still centered on (0,0). Two of the corners will retain the same coordinates: (-2, 2) and (2,-2). The other two will move towards (0,0) and you don't need to know exactly how close they will get to it because the effect we're looking at applies to any relativistic speed which takes all the edges of the shape off their original north-south and east-west alignments, so (1,1) and (-1,-1) will do fine. You should understand this intuitively without needing to reach for a calculator. Now string lots of these together to represent a rectangle undergoing the same contraction and you have a parallelogram which will clearly cut across the rails and lead to observers encountering parts of these objects passing them in the order I described.
Note what you refused to include there: the time coordinate. This is very important since you are claiming that there are times when the wheels of the train do not coincide with the track, yet you refuse to actually work out any of these times.

Were I not willing to grant you some charity, I would conclude that you do not know how to do this and you are attempting to hide this failure. As it stands, you simply seem to hard-headed to take the time to do your work properly.

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When something is so clearly proven with visual examples, there is no need to see precise numbers to know that the numbers must fit. If someone shows you a square and tells you it isn't a circle, you don't need to see any coordinates for the corners to know that it's not a circle.
Again, we have to take your word that your diagrams are correct because you will not justify them by using the correct mathematical physics.

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You have seen more than enough calculations to get well beyond the point where you should have recognised that I'm right, but you don't want to admit you're wrong, so your only face-saving tactic now is to demand an infinite number of wholly unnecessary numbers.
All I have demanded is to see the relevant time coordinates. That is one finite set of numbers. These are necessary numbers because you are speaking of events where the train wheels do not meet the train track; all events have a location in space and a time at which they occur.

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If you think your way of calculating things disproves my proof, it's your job to demonstrate that. I've told you what shape a moving square will be when it's travelling in a direction not aligned with its edges, but you are incapable of telling whether I'm right in saying that its edges will change their alignment with the grid. Any real expert would immediately confirm that I'm right about the shape, but you won't do that.
You are claiming, without working through the calculations, that they will verify your claims. You seem to be claiming to have a very wonderful precognitive ability. And yet you still seem unable to write a convincing argument. Shouldn't your precognitive abilities allow you to know in advance what will be a convincing argument?

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I ask you to check each of my claims and to home in on specific ones that you take issue with, and if numbers are required to prove a specific point (which is so obvious that no expert should get stuck on it), I'll give you numbers for that point. What is not right is to demand an infinite supply of unnecessary numbers.
I asked you for a specific set of numbers, viz, the time coordinates and the associated calculations related to your claims. Lying about what I wrote will not help your case, and it may very well get you banned from these forums.

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You still appear to have no comprehension of how flimsy your grasp of this subject is. What distant events? The observer at X, Y or Z is right there at the place where the events he's observing are taking place!
Because the wheels of the train are not at the same spatial location, any events that happen there are distant from each other. Thus the order of events at the train wheels can differ in different systems of coordinates. This is very basic SR. I urge you to read about the relativity of simultaneity; it may save you a lot of embarrassment.

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If this involved observing two events which are both at a distance from the observer and a frame change could change the apparent order in which those events occur, you would have a point, but that is not the case here. All the events being observed happen at point X.
This cannot be the case, since you are speaking of separate wheels and separate points on two different train tracks. Again, carefully working out the actual transformations would help you realize the problems in your argument.
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I've done all the work necessary to prove the case and see no need in calculating irrelevant numbers for someone who doesn't understand the subject. If you can find a fault, show it. So far, all you've come up with is simultaneity issues which don't apply.
Sure, you can whine and repeat the same things over and over again. I understand your psychological pressure: you claim to be a self-taught expert on education and it might look bad if you fail to self-teach yourself the basics of a subject. However, everyone can make mistakes and it looks very bad if you become the example case that being self-taught means not being able to fix one's mistakes. By setting yourself up as an expert on education, you damage not only your character but also your expertise on education by refusing to actually consider a suggestion offered in good faith.

Is there no real expert with qualifications here who's going to step in and let PhysBang know he's not up to this stuff? You could do it in a PM, but he needs to be told and he won't take it from me. The best way to defend physics isn't to dig in and defend things that are wrong, but to move on and make the subject right.
I would be seriously surprised if there were an "expert" who would come forward and say that your argument is correct and SR is hopelessly inconsistent. I have no doubt that there are cranks around here who might PM you that I am incorrect because they too cannot answer the relevant questions I have offered them. If you want to side with these cranks, then so be it. You are free to believe what you want to believe. You can lie about the content of my posts and insult me if you would like, but do not be surprised if there are consequences.
 

Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #77 on: 12/08/2016 23:22:28 »
You are correct: your continued failure to actually perform the relevant calculations are examples of my avoiding issues.

Anyone competent can work out for themselves the shape that a square will take up when it moves through Frame A without needing to lift a calculator. You can't do it because you're incompetent to an extraordinary degree. How on Earth can you have qualifications in this if you can't even do that! Which university is responsible for this failure?
So you are refusing to do the calculations necessary to show you are correct? Need I remind you that this is your argument? That you are trying to convince me and others? That you came here, asking for flaws in your argument? It was only because you asked that I pointed out that you failed to take time into account and that you have not done the calculations to make your case.

You are free to believe that you don't have to be held to the same standard that physicists are held when making their arguments. You did, however, ask for help. You are behaving as if your requests for help are not genuine.

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You are claiming to discuss the basics of relativity theory. It is clear to me that you are using a relativity theory other than SR.

I am using things that are 100% parts of SR. The length contraction, the way things appear in Frame A, the direction the contraction is applied, the degree to which it is applied, the coordinates for objects in Frame A (where north-south is one space dimension, east-west is another space dimension, and the whole diagram is a slice showing how things are located in that space at a specific Frame A time. To write it off as not being SR is one of the most ludicrous things you can do.
You are, by your own admission, only using part of SR. That means that you are using your own special form of relativity.

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Where is your problem? Look at the diagram and apply your own coordinates to it. A child could do it. For example, we could decide that a stationary square is centered on point (0,0) and with corners at (2,2), (2,-2), (-2,-2) and (-2,2) with north being the Y-axis and east being the X-axis. We can then move it up to relativistic speed moving NE and calculate its shape when it's still centered on (0,0). Two of the corners will retain the same coordinates: (-2, 2) and (2,-2). The other two will move towards (0,0) and you don't need to know exactly how close they will get to it because the effect we're looking at applies to any relativistic speed which takes all the edges of the shape off their original north-south and east-west alignments, so (1,1) and (-1,-1) will do fine. You should understand this intuitively without needing to reach for a calculator. Now string lots of these together to represent a rectangle undergoing the same contraction and you have a parallelogram which will clearly cut across the rails and lead to observers encountering parts of these objects passing them in the order I described.
Note what you refused to include there: the time coordinate. This is very important since you are claiming that there are times when the wheels of the train do not coincide with the track, yet you refuse to actually work out any of these times.

Were I not willing to grant you some charity, I would conclude that you do not know how to do this and you are attempting to hide this failure. As it stands, you simply seem to hard-headed to take the time to do your work properly.

Quote
When something is so clearly proven with visual examples, there is no need to see precise numbers to know that the numbers must fit. If someone shows you a square and tells you it isn't a circle, you don't need to see any coordinates for the corners to know that it's not a circle.
Again, we have to take your word that your diagrams are correct because you will not justify them by using the correct mathematical physics.

Quote
You have seen more than enough calculations to get well beyond the point where you should have recognised that I'm right, but you don't want to admit you're wrong, so your only face-saving tactic now is to demand an infinite number of wholly unnecessary numbers.
All I have demanded is to see the relevant time coordinates. That is one finite set of numbers. These are necessary numbers because you are speaking of events where the train wheels do not meet the train track; all events have a location in space and a time at which they occur.

Quote
If you think your way of calculating things disproves my proof, it's your job to demonstrate that. I've told you what shape a moving square will be when it's travelling in a direction not aligned with its edges, but you are incapable of telling whether I'm right in saying that its edges will change their alignment with the grid. Any real expert would immediately confirm that I'm right about the shape, but you won't do that.
You are claiming, without working through the calculations, that they will verify your claims. You seem to be claiming to have a very wonderful precognitive ability. And yet you still seem unable to write a convincing argument. Shouldn't your precognitive abilities allow you to know in advance what will be a convincing argument?

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I ask you to check each of my claims and to home in on specific ones that you take issue with, and if numbers are required to prove a specific point (which is so obvious that no expert should get stuck on it), I'll give you numbers for that point. What is not right is to demand an infinite supply of unnecessary numbers.
I asked you for a specific set of numbers, viz, the time coordinates and the associated calculations related to your claims. Lying about what I wrote will not help your case, and it may very well get you banned from these forums.

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You still appear to have no comprehension of how flimsy your grasp of this subject is. What distant events? The observer at X, Y or Z is right there at the place where the events he's observing are taking place!
Because the wheels of the train are not at the same spatial location, any events that happen there are distant from each other. Thus the order of events at the train wheels can differ in different systems of coordinates. This is very basic SR. I urge you to read about the relativity of simultaneity; it may save you a lot of embarrassment.

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If this involved observing two events which are both at a distance from the observer and a frame change could change the apparent order in which those events occur, you would have a point, but that is not the case here. All the events being observed happen at point X.
This cannot be the case, since you are speaking of separate wheels and separate points on two different train tracks. Again, carefully working out the actual transformations would help you realize the problems in your argument.
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I've done all the work necessary to prove the case and see no need in calculating irrelevant numbers for someone who doesn't understand the subject. If you can find a fault, show it. So far, all you've come up with is simultaneity issues which don't apply.
Sure, you can whine and repeat the same things over and over again. I understand your psychological pressure: you claim to be a self-taught expert on education and it might look bad if you fail to self-teach yourself the basics of a subject. However, everyone can make mistakes and it looks very bad if you become the example case that being self-taught means not being able to fix one's mistakes. By setting yourself up as an expert on education, you damage not only your character but also your expertise on education by refusing to actually consider a suggestion offered in good faith.

Is there no real expert with qualifications here who's going to step in and let PhysBang know he's not up to this stuff? You could do it in a PM, but he needs to be told and he won't take it from me. The best way to defend physics isn't to dig in and defend things that are wrong, but to move on and make the subject right.
I would be seriously surprised if there were an "expert" who would come forward and say that your argument is correct and SR is hopelessly inconsistent. I have no doubt that there are cranks around here who might PM you that I am incorrect because they too cannot answer the relevant questions I have offered them. If you want to side with these cranks, then so be it. You are free to believe what you want to believe. You can lie about the content of my posts and insult me if you would like, but do not be surprised if there are consequences.
 

Offline jeffreyH

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Re: Can a preferred frame of reference be identified?
« Reply #78 on: 13/08/2016 00:14:24 »
Joking aside. Let's not resort to insults. Argue the case by all means but a little gentlemanly behaviour wouldn't go amiss.
 

Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #79 on: 13/08/2016 00:38:23 »
So you are refusing to do the calculations necessary to show you are correct? Need I remind you that this is your argument? That you are trying to convince me and others? That you came here, asking for flaws in your argument? It was only because you asked that I pointed out that you failed to take time into account and that you have not done the calculations to make your case.

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.

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You are free to believe that you don't have to be held to the same standard that physicists are held when making their arguments. You did, however, ask for help. You are behaving as if your requests for help are not genuine.

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.

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You 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.

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Note what you refused to include there: the time coordinate. This is very important since you are claiming that there are times when the wheels of the train do not coincide with the track, yet you refuse to actually work out any of these times.

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?

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Were I not willing to grant you some charity, I would conclude that you do not know how to do this and you are attempting to hide this failure. As it stands, you simply seem to hard-headed to take the time to do your work properly.

This again comes down to your lack of understanding. Any real expert on this can pin their own coordinates to everything I've said because it's all basic stuff that any expert should be able to do in his head while spreading his crackers.

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Again, we have to take your word that your diagrams are correct because you will not justify them by using the correct mathematical physics.

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? Where do you imagine you can produce anything different from my diagrams to show the things I've described? I could describe these things over the phone to any real expert and he'd have a diagram matching mine on his desk within seconds, all generated by him as he works through the same ideas. But for some reason, you can't do that.

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All I have demanded is to see the relevant time coordinates. That is one finite set of numbers. These are necessary numbers because you are speaking of events where the train wheels do not meet the train track; all events have a location in space and a time at which they occur.

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.

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You are claiming, without working through the calculations, that they will verify your claims. You seem to be claiming to have a very wonderful precognitive ability. And yet you still seem unable to write a convincing argument. Shouldn't your precognitive abilities allow you to know in advance what will be a convincing argument?

I have worked through all the calculations necessary to prove the point and to demonstrate it to anyone competent in this field. I should not be required to make people understand it who lack sufficient understanding of relativity to be able to get their heads around it.

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I asked you for a specific set of numbers, viz, the time coordinates and the associated calculations related to your claims. Lying about what I wrote will not help your case, and it may very well get you banned from these forums.

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. It's an irrelevant number because it's the same for every pixel of the picture. You have also had all the calculations necessary, and a whole host that you asked which you shouldn't have needed to ask for because they were too obvious for any expert to require.

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Because the wheels of the train are not at the same spatial location, any events that happen there are distant from each other. Thus the order of events at the train wheels can differ in different systems of coordinates. This is very basic SR. I urge you to read about the relativity of simultaneity; it may save you a lot of embarrassment.

The apparatus is on a 2D plane and there are no wheels. In the original train experiment it was a maglev, but considering connections to the rail is actually quite unnecessary (and was only necessary before when I was discussing a gap between the train and new rail). It is also totally unnecessary to consider the third space dimension other than for the observer's location which might be just over or under the apparatus, but he contacts it with the chalk, so his interaction with it is directly on the 2D plane. I had thought by this time that it would finally have dawned on you that you're wrong to bring simultaneity issues into this, but you're still trying to do it! You are the one who needs to read up on simultaneity - the order of events at X, Y or Z cannot change as you change frame, but you don't even recognise that!

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Sure, you can whine and repeat the same things over and over again. I understand your psychological pressure: you claim to be a self-taught expert on education and it might look bad if you fail to self-teach yourself the basics of a subject. However, everyone can make mistakes and it looks very bad if you become the example case that being self-taught means not being able to fix one's mistakes. By setting yourself up as an expert on education, you damage not only your character but also your expertise on education by refusing to actually consider a suggestion offered in good faith.

Suggestions offered in good faith don't come with insults. What you get back from me is a reflection of what you threw my way. But the key thing is that you aren't being co-operative at all - you're evasive, never answering any direct question about whether you agree with any point I ask you to commit yourself to a position on, and you keep telling me I haven't given you time coordinates for things where everything in the Frame A diagram shares the same moment in time and there is no need to have any particular number in mind for the value of t.

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I would be seriously surprised if there were an "expert" who would come forward and say that your argument is correct and SR is hopelessly inconsistent.

I wasn't asking one to. I was suggesting that one might give you a bit of help in recognising the many places where your understanding of relativity falls far below the level you think you're on so that you stop embarrassing them by association.

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I have no doubt that there are cranks around here who might PM you that I am incorrect because they too cannot answer the relevant questions I have offered them. If you want to side with these cranks, then so be it. You are free to believe what you want to believe. You can lie about the content of my posts and insult me if you would like, but do not be surprised if there are consequences.

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.
« Last Edit: 13/08/2016 01:12:04 by David Cooper »
 

Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #80 on: 13/08/2016 03:05:37 »
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.
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.

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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.
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.
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You 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.

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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?
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.

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

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Which 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.

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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.
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.

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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.

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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.)

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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.

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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.
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.
 

Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #81 on: 13/08/2016 21:11:57 »
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.

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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.

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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.

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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.

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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.

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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).

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

(8) 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 (8): 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.
 

Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #82 on: 13/08/2016 22:14:14 »
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.
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.

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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.
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.

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I 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.
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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.
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.

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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.
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.

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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.
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.

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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.
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.

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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.
If this is the case, then you should not have a problem completing your argument with the full Lorentz transformations.

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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 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.

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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.
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.

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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.
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.

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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.
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.

I'll have to go through this "point by point" argument later.
 

Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #83 on: 13/08/2016 23:11:00 »
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.

And if the entire proof can be carried out without changing frame at all, there are no such translations to be done.

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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.

The information from the observers at K, L and M (originally labelled as X, Y and Z) show us that the rails are not aligned with the edges of the rhombus/parallelogram. The Frame A diagram of this shape in relation to the rails shows the misalignment too, and every point in that diagram can have an observer placed at it to confirm that the diagram shows what they see.

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I 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.

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).

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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.

The observers at K and M are witnesses to the fact that there is a misalignment. Any transformation which changes the order of events which they witness is an illegal transformation.

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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.

That is certainly a good analogy for your method of reasoning.

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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.

On the contrary; I'm the one who's seeing this stuff clearly while you are tolerating contradiction. You imagine that a transformation which changes the order of events on a straight line running up a Spacetime diagram can be changed by changing frame of reference, but that is completely impossible. Address that before you accuse me of not understanding this stuff.

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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.

It is not my job to prove the case to anyone who is incapable of going through a proof and recognising it as the proof that it is. The proof is there and if you want to attack it, you need to try to pick apart specific parts of it and to show one of them to be unsound.

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If 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.

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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.

You should be able to generate your own diagrams of all these things in order to compare them with mine and see if they are compatible. If you can produce a diagram which is not compatible with mine but which meets the same requirements (such as centre of rhombus being midway between two rails), then I want to see where you're finding incompatibilities - there can't be any unless you're making mistakes.

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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.

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.

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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.

The proof doesn't depend on any frame change apart from showing why it is impossible to change the order of events on a line on a Spacetime diagram - these aren't distant events observed from that line, but events which occur directly on it.

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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.

If you go through it applying the rules of SR the way you want to, you will be able to see that every part of it stands up. However, you are not allowed to use illegal transformations which change the order of events taking place at a location (which becomes a straight line up a Spacetime diagram) because such an order change is impossible by the rules of SR.
« Last Edit: 13/08/2016 23:24:19 by David Cooper »
 

Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #84 on: 13/08/2016 23:38:27 »
Pay particular attention to the bit in my previous post relating to the attached diagram (the paragraph containing bold text). If the Frame A view of the shape moving NE makes it appear as a rhombus, an observer on that moving shape (which appears square to him) must see any square that isn't moving through Frame A as being a rhombus too, and by extension, all the grid lines of Frame A must appear to him to follow the same angles as the angles of the square that he is seeing as a rhombus (as must the rails). That is why he must see the gridlines aligned as they are in my diagram. A traveller on train A, however, will see no such change in the angle of the grid lines as they will remain parallel and perpendicular to his train.

Edit: actually, this bit isn't so clear after all because the rails are moving relative to the grid lines, and if we apply a grid co-moving with Frame B, it will produce similar angles for the Frame A' observer, so there may be a mirror image of events with those lines and it may not show up different physics at all. We need to know how things look with the actual rails, and that's harder to visualise, but we do know that the order of observed events for observers K and M show up something radically different for Train B which doesn't occur with Train A, and that's what we need to concentrate on.

Edit 2: Ah, I've got it - they are different right enough. The north-south aligned grid lines which are co-moving with Frame B don't veer off their alignment with the north-south grid lines co-moving with Frame A when viewed by observers riding Train A (although the east-west ones do), whereas for observers on Train B, Frame A 's north-south gridlines do veer off at an angle due to the length contraction on the train acting at 45 degrees, so the view is definitely not a mirror image. The computer program will make all of this much easier to see.
« Last Edit: 14/08/2016 00:37:02 by David Cooper »
 

Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #85 on: 14/08/2016 02:26:50 »
And 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.

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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).
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.

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If 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.
Again, the burden of proof is one you.

However, I am going to do this work. It's a nice exercise.


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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 just want to deny SR, then there is nothing that can be helped. It will place you in a category, however.

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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.
Sure, if we use David Cooper Relativity, then we get an inconsistency. I get that. But I and other people use SR.

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The proof doesn't depend on any frame change
Except 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?
 

Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #86 on: 14/08/2016 02:28:13 »
Edit: actually, this bit isn't so clear after all because the rails are moving relative to the grid lines
Ta da!
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Edit 2: Ah, I've got it
Nope, still don't got it.
 

Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #87 on: 14/08/2016 17:35:09 »
OK, so we start with a reference frame, A, that we will take to be at rest.

We will then imagine a reference frame , A', moving to the NE of our original reference frame at a speed of 0.866 of the speed of light. (We will just use units adjusted so that the speed of light in a frame is 1 in our units).

Let's make this a unit square, so that the SW corner of the square rests at the origin of A'. This means that the SW corner of the square is at A'(t,0,0) for all values of t. For al corners:

SW A'(t,0,0)
NE A'(t,1,1)
NW A'(t,0,1)
SE A'(t,1,0)

Let's make things easy for ourselves and arrange frames A and A' so that the SW corner of our square passes through A(0,0,0) as the square moves through A.

Then we have to ask where the other points of the square lie?

For ease of reference, I have used the boost matrix formulation available at wikipedia. https://en.wikipedia.org/wiki/Lorentz_transformation#Boost_matrix

Our velocity is 0.866, but we have to adjust this based on our angle of motion. We can do this by multiplying by the normals in each direction, n_x=0.707 and n_y = 0.707. Our gamma remains 2. (Please note that I am using reduced significant digits in this presentation.)

This means that to translate a position from A to A':
t'=2t + (-2)(0.866)(0.707)x + (-2)(0.866)(0.707)y
x'=(-2)(0.866)(0.707)t + (1 +(2-1)(0.707)^2)x + (2-1)(0.707)(0.707)y
y'=-2(0.866)(0.707)t  + (2-1)(0.707)(0.707)x + (1 +(2-1)(0.707)^2)y

To get the inverse, we have to essential run a translation at the same speed in the opposite direction. This means flipping the sign on the normals for each direction.

This means that to translate a position from A' to A:
t=2t' + (-2)(0.866)(-0.707)x' + (-2)(0.866)(-0.707)y'
x=(-2)(0.866)(-0.707)t' + (1 +(2-1)(-0.707)^2)x' + (2-1)(-0.707)(-0.707)y'
y=-2(0.866)(-0.707)t'  + (2-1)(-0.707)(-0.707)x' + (1 +(2-1)(-0.707)^2)y'

So let's look at the unit square at time A'(t=0):
SW A'(0,0,0)
NE A'(0,1,1)
NW A'(0,0,1)
SE A'(0,1,0)

Translated to frame A:
SW A(0,0,0)
NE A(2.45,2,2)
NW A(1.22,0.5,1.5)
SE A(1.22,1.5,0.5)

We see here a problem for determining the shape of our square in frame A: our points are now at different times! This isn't much of a problem, because we know that the parts of the square are all moving at a constant rate relative to frame A. We just have to trace their path back.

This means looking at delta(t) and then figuring out the change in each coordinate and applying that (taking a negative value of delta(t) to go backwards in time). So:
delta(x) = delta(t)(v)n_x = delta(t)(-0.866)(0.707)
delta(y) = delta(t)(v)n_y = delta(t)(-0.866)(0.707)

So, looking back, we can find all the points of the square at A(t=0):
SW A(0,0,0)
NE A(0,0.5,0.5)
NW A(0,-0.25,0.75)
SE A(0,0.75,-0.25)

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.

So what if we imagine that this square was hovering over rails? Does the distortion of the box mean that it goes over the rails?

Well, since the box was literally placed over the rails, then we should expect the rails to be in exactly the same place under the box: they undergo the same translations!

These tracks are 1 unit apart in A', but only 0.707 units apart in frame A, which we know because of 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.

But let's move the box along, say 10 units in frame A. Then the x position and the y position of every point changes by 10(0.866)(0.707)=6.123 and the t of every point becomes 10.

Now we're looking at these points:
SW A(10,6.123,6.123)
NE A(10,6.623,6.623)
NW A(10,5.873,6.873)
SE A(10,6.873,5.873)

Since the rail is moving along with the square, the square stays nicely with the rail. But can we translate back?

Here are the translated positions (3 significant figures):
SW A'(5,0,0)
NE A'(3.78,1,1)
NW A'(4.39,0,1)
SE A'(4.39,1,0)

As we can see, these are all points on the unit square in Frame A', which means that they are points on the rail.

Let's imagine a rail stationary in frame A, over which the square (rhombus) is moving, touching at the NW and SE corners. This means that the rail runs SW to NE, is sqrt(2) wide and one rail passes through A(0,-0.25,0.75) and the other through A(0,0.75,-0.25).

We already know that the square stays on this track in frame A', since the exact same translations that we used above for the other track apply for this track on the corner.

I think that's enough for now.
 
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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #88 on: 14/08/2016 19:42:31 »
And 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.

I have established the order of events in which an observer at K or M (who is at rest in Frame A) sees objects move past him, and that tells you everything you need to know - the order is incompatible with the rhombus being contained between the rails.

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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.

You tolerate contradictions, but I don't. That's a big difference between us.

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If you just want to deny SR, then there is nothing that can be helped.

An illegal transformation (which generates a contradiction) has no place in SR because it breaks more fundamental rules of SR - it is being used by your lot in error.

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Sure, if we use David Cooper Relativity, then we get an inconsistency. I get that. But I and other people use SR.

The error you make is that when you change frame, you also switch from the frame that's stationary to one that isn't and then calculate on the basis that the new one's stationary, and that's why you generate contradictions which you have yet to recognise that you are doing.

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The proof doesn't depend on any frame change
Except 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]

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.
 

Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #89 on: 14/08/2016 19:45:05 »
Edit: actually, this bit isn't so clear after all because the rails are moving relative to the grid lines
Ta da!
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Edit 2: Ah, I've got it
Nope, still don't got it.

You're the one who hasn't got it. The angles are different - wait till you see the computer program.
 

Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #90 on: 14/08/2016 19:47:58 »
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.
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.
 

Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #91 on: 14/08/2016 20:12:34 »
Thanks for going through the calculations and showing them to me. Your Frame A' is the one I called B' in my example, so that's something readers need to avoid tripping over - I will use your A' label the same way you have in this post.

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.

So we agree on the same shape for the square moving NE at 0.866c through Frame A and will draw it the same way as each other in a diagram in which all points have t=0.

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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.

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?

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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.

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.
 

Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #92 on: 14/08/2016 20:21:02 »
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?
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.

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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.
Given that you admit that they are not moving together, who cares if they intersect or not? Where is the paradox?
 

Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #93 on: 14/08/2016 20:24:17 »
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.
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.

Again, what are you on about? How does there being a frame of reference in which the square is at rest require any frame change when analysing events entirely from that frame?

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Yet you refuse to actually consider that you might be missing something and you lash out with insults rather than doing work.

What insults? Is "what are you on about" an insult? If you think it was, it was no bigger an insult than the "do you even read what you write" which it was replying to and which you had said when making an extremely wayward point.

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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 await your corrections to your work.

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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.

What learning problems? I've had problems getting access to decent maths and physics teaching at school, but I just worked out my own ways of doing everything instead, and when I can produce the same results or better ones (that don't produce the contradictions that your methods do), I think that puts me some way ahead. As for my attitude, I am a mirror - whatever you fling at me, it will come back.
 

Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #94 on: 14/08/2016 20:41:37 »
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.

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.

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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.

If you want to use the right coordinates for the track, it has to be moving through Frame A in the right direction so that you don't warp it off its correct alignment.

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Given 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.
 

Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #95 on: 14/08/2016 21:06:58 »

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.
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?

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Given 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.
If you send it at the right speed, it should be fine. Is your problem that you are not combining the speeds correctly?
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So 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.
 

Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #96 on: 14/08/2016 21:37:36 »
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 were working by post #81, the wording of point (4) was as follows:-

(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.

That, plus all previous examples, gives the direction of travel of the rails as northwards, and that is why they appear in Frame A diagrams perpendicular to their direction of travel. As you have discovered, if the rails are moving NE instead, they will appear in Frame A diagrams at a different angle. My proof depends on them moving north and you must conform to that if your calculations are to be valid. Once you have worked through the numbers, you will find that the rhombus does not fit between the rails, as I've told you all along. The Frame A diagrams do not lie and cannot lie: what you see with them is what you get.

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If you send it at the right speed, it should be fine. Is your problem that you are not combining the speeds correctly?

If you cheat by having the rails co-moving with the rhombus, you will hide the effect you're meant to be looking for because you are literally looking in the wrong place.

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So 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 am the one applying SR correctly - the Frame A diagrams show what's going on perfectly, but you dismiss them out of ignorance of how they work and what they mean. I tried to educate you about this, but you wouldn't listen, and now you've made a right royal fool of yourself over four pages.

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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.

Are you playing games of avoidance? Did you deliberately use rails co-moving with the rhombus in order to hide something? Well, I hope not. The big question though is, will you be man enough to post your results for the rails moving in the right direction when it finally dawns on you that I've been right throughout? If you do post them though, that will be an admirable act for which you will deserve respect.
 

Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #97 on: 14/08/2016 21:43:17 »
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.
 
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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #98 on: 14/08/2016 22:10:59 »
I'll see when I have enough time. I'm sorry that you have learning problems, but not sorry that you're a real ____.

Charming! I'm not the one with learning problems, as you'll find if you do the work properly instead of being slapdash.

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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.

I'm not the one that needs help - I don't tolerate contradictions in my model of reality, but you do, and so do hordes of other qualified "experts" who make a mockery of mathematics. The rhombus does not fit between the rails, and Frame A diagrams show that. The Frame A diagram for the numbers you provided would show the rails running at an angle like the edges of the rhombus, so of course you couldn't find the issue there - the Frame A diagram of that doesn't lie, but shows the rhombus sitting neatly between the rails. The Frame A diagram of my setup though shows the rails aligned perpendicular to their 100% northward direction of travel, and it doesn't lie about the placing of the rhombus over it where two of the corners stick outside of the rails. Just by drawing the diagram and looking at it for a few seconds, a real expert would have realised that the argument I've presented is correct: different frames of reference produce different physics. In the preferred frame you can move the squares and rails east or west at any speed and there is no problem, but for the Frame B physicist testing how the same setup works in his frame, he will confirm that Lorentz and Einstein called it wrong. Now we need to design a practical experiment to make use of this new understanding so that we can finally tell how fast we're moving through the preferred frame, and in which direction.

[Details of my ref-frame camera program will follow at some point, and then the program itself will be made available for all to use: I'll do a JavaScript version that anyone can run straight off a webpage, so watch this space.]
« Last Edit: 14/08/2016 22:28:37 by David Cooper »
 

Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #99 on: 15/08/2016 14:04:30 »
Mr. Cooper stumbled upon an interesting feature of relativity theory. One day, if he learns to do SR, he may actually be able to incorporate this into his JavaScript program. I wish him the best of luck and health on this.

https://en.wikipedia.org/wiki/Wigner_rotation
 

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Re: Can a preferred frame of reference be identified?
« Reply #99 on: 15/08/2016 14:04:30 »

 

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