Can a preferred frame of reference be identified?

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #50 on: 09/08/2016 21:39:51 »
If you submit it, you have to do the math. That's it.

I've done all the maths that's needed to prove the case, assuming that it's read by someone who understands relativity.

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Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #51 on: 10/08/2016 14:52:47 »
Just out of curiosity: you have an education website, but you have never attended any university level education?

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #52 on: 10/08/2016 19:33:54 »
Just out of curiosity: you have an education website, but you have never attended any university level education?

I gave university a miss because my work on linguistics (I specialised in generative semantics) was already ahead of anything that was being taught in any university, and that led me on into work on AGI (artificial general intelligence). University would have held my work back by several years for no useful gain, wasting my time on such things as Chomsky's broken linguistcs (with its shallow analysis and ludicrous ideas about universal grammar).

My website exists because I am disgusted with education systems which spend most of their time pretending to teach while not actually teaching anything (and which can also provide abysmal teaching when they do finally get round to it, making it impossible to succeed in some subjects unless you can afford private tutoring to make up for the deficiencies), so I've tried to show how things could be done more efficiently. I have shown how children can learn most of the essentials in a fraction of the time they take to do so in school, but when an education system has a fixed learning schedule and doesn't reward faster learning (because it repeatedly puts children in holding pens until the rest catch up again), there is no benefit from using more efficient methods other than for children working outside the system through home-schooling. [You may want to read up on Unschooling to see what happens when children aren't systematically taught anything at all and are just left to play most of the time - on average they end up with the same level of qualifications at the same age as schooled children, so that illustrates just how bad schools are. Peter Gray's blog at Psychology Today is the best starting place for reading up on this: https://www.psychologytoday.com/blog/freedom-learn - I'm not a fan of Unschooling, but want to see something half way between that and schooling so that all children can have the best of both worlds and where they have the freedom to walk away from bad teaching and to find their own way through the work without wasting their childhood on the empty garbage that's currently inflicted on them.] I was hoping that other people might join in the effort to build the site properly, but no one has done so and the whole thing will soon be rendered completely redundant by AGI which will provide the same kind of teaching more directly while interacting intelligently with learners.

Now, what point were you trying to make by asking about this? One of the biggest problems afflicting society is that people worship status and qualifications and don't value reason. Dunning and Kruger have made this even worse by giving qualified "experts" further excuse not to bother checking that they've built their knowledge upon a sound base. Errors can occasionally be found which persist for decades or even centuries without any of the teachers or indoctrinated learners ever stopping to check. If someone does spot an error, they are simply shouted down and the establishment blunders blindly on regardless, overconfident about their rightness because they are "experts" who can't be wrong.

In this thread, I've identified such an error, and it's a monumental whopper of an error. How do you judge this though? You simply look to see if I'm saying something that goes against the "experts", and when you determine that I am, you decide that the "experts" must be right because they are "experts" and they couldn't possibly all have made the same mistake for over a hundred years. In this particular case though, they're failing to read diagrams correctly, not realising the relationship between a diagram showing two space dimensions at a fixed point in time (from the point of view of a specific frame of reference) and a Spacetime diagram which shows only one space dimension with time shown vertically, each slice showing a fixed point in time (from the point of view of a specific frame of reference). A horizontal slice of a Spacetime diagram is functionally identical to a straight line drawn through one of my diagrams: if two things are not in the same place in the diagram in one frame of reference, they cannot be in the same place in any frame of reference because they are at different Spacetime locations. When you see a parallelogram with sides that don't align with the rails in Frame A, any method of converting to a Frame B view of the same objects which shows the sides parallel with the rails is violating fundamental rules, and yet that is what the "experts" are clearly doing here. They are naively assuming that all frames of reference behave the same way, so when they shift from Frame A to Frame B they simply recalculate the positions, lengths and alignments of objects from scratch on the mistaken basis that all frames behave the same way, and in doing so they introduce distortions. They must see the misalignment in one frame, but they then carry out a translation to the other frame and the misalignment is magically gone, but they never stop to ask themselves how this can happen when they wouldn't accept the same kind of mismatch when handling Spacetime diagrams. The simple truth of it is that the translation method they're using is faulty because it's based on a false belief that all frames behave the same way.

The correct way to translate between frames is to stick with the original Frame A calculations and then convert using the ref-frame camera method where you change the clock synchronisations and run through events noting where things are when the local clock at any point hits the target time for the new frame. This is a much more computation-intense way of carrying out the translation from one frame to another, but the end result is that you get a correct translation instead of one that introduces distortions. It's clear to me now that no one has ever done these translations correctly because they've simply relied blindly on an incorrect assumption that all frames work identically. I am now writing a program to do such translations properly. I have attached a drawing showing the square on a rocket moving NW (north is up) at relativistic speed and the three different shapes it will have when viewed from frames A, B and B'. I haven't used accurate numbers for them, but it's the general shapes I want you to look at. The top row shows the kinds of shapes I calculate for them, while the bottom row shows the shapes that the "experts" would calculate for it, and it's with the middle one that you see the key difference. They simply recalculate the shape by assuming that Frame B works like Frame A, but I run the events in Frame A instead and convert by "taking a photograph" for Frame B as the local pixel clocks hit the target time for taking that picture with all clocks synchronised for Frame B.
« Last Edit: 10/08/2016 19:51:39 by David Cooper »

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Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #53 on: 10/08/2016 21:09:14 »
Now, what point were you trying to make by asking about this?
I was curious, given that you displayed ignorance of the form of journal articles.

Like you said, people often value things other than reason. Often, the form of journal articles are what they are for a reason. Sometimes these are good reasons, sometimes these are bad.

Similarly, the details of the transformations from one system of coordinates to another are what they are for a reason. You are free to believe what you want to believe, but you should know that you will be held to a high standard if you want to publish your work and that standard will include at least recognizing what trained physicists expect from a transformation from one system of coordinates to another. This will include including the time coordinate in all translations.

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #54 on: 10/08/2016 23:19:26 »
Now, what point were you trying to make by asking about this?
I was curious, given that you displayed ignorance of the form of journal articles.

I've looked up a few things and now see that having letters after your name isn't crucial as they don't tell the reviewers who you are (although you still have to get past the first wall of persuading them to send your paper to be reviewed). That makes the system a more fair than I'd expected it to be. It's also possible to include software along with the paper, or a link to a place from which it can be downloaded, and that makes it considerably easier to show them what's going on.

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Similarly, the details of the transformations from one system of coordinates to another are what they are for a reason.

Indeed, and it's a bad reason - it's based on a belief that different frames all work the same way, but I've shown that a person in Frame B sending things along Rail B will observe them to warp and to do so more the faster they go. Any system for doing the transformations which doesn't find that result is producing errors, hiding this warping with distortions which precisely cancel out the warping that should be there.

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You are free to believe what you want to believe, but you should know that you will be held to a high standard if you want to publish your work and that standard will include at least recognizing what trained physicists expect from a transformation from one system of coordinates to another. This will include including the time coordinate in all translations.

I am the one who has the high standards here, but there's never any guarantee that people with lower standards are capable of recognising that because they tend to reject anything they disagree with instead of checking it out carefully, and it's not surprising that they do this: they are bombarded with rubbish which they need to reject without wasting time on it, and yet it means they are likely to miss those extremely rare things that look wrong because they go so much against expectations but which happen to be right. There is a high chance that they'll simply reject my proof in the same way that you have without thinking it through.

I've attached a new diagram showing the asymmetry in the way different frames behave. For the top picture, you need to imagine Frame A being stationary while Frame B is racing up the screen, and the shape shown has no sides parallel with Rail B or perpendicular to it. This shape (not drawn very accurately this time because all you need to agree with is that none of its sides are parallel with Rail B or perpendicular to the track) is the one that the square on the rocket must have when viewed from Frame A, and it's the shape that squares on Train B must show too as they are co-moving with the rocket.

For the bottom picture, we are seeing an equivalent square on Train A while looking from Frame B, and two of the sides are parallel to Rail A. According to relativity (both Special Relativity and Lorentz Ether Theory [in their current form]) those shapes should be a mirror image of each other, but they aren't. It's easy to work out what the shape of the rocket square will look like from Frame A because you simply apply length contraction at an angle and all four edges end up tilting relative to the north-south and east-west lines. At no point have you dared to commit yourself to stating whether you agree with it or not, but you ought to.

It's harder to work out the shape of a Train A square as viewed from Frame B, but using the ref-frame camera method you do the following. First you find the Frame A shape which is a rectangle, the length-contraction applying in the east-west direction. The next step is to synchronise clocks for Frame B, pick a time to "take a photo", then run all the clocks as you imagine the shape moving east along the rail. Whenever a pixel clock hits the target time, you copy whatever's at that pixel to a new diagram, and that means you'll transfer the whole southern edge in one go (because the clocks for all those pixels read the same time as each other), then you move the shape east a bit, then you work your way up the sides, each bit being copied further to the right, and eventually you reach the top edge and transfer the whole of it in one go. The only other thing you have to do is fix the north-south width of this shape to make sure the correct length contraction is being applied in that direction, but it will certainly look like the picture I've drawn for it in regard to the direction the tilted edges tilt and in the fact that the other two edges must remain parallel to Rail A.

I don't know which mistake other people have been making with this, but they are producing shapes which are mirror images of each other, either by ignoring the necessity of the square on the rocket looking the same shape as the squares on Train B when viewed from Frame A (and therefore having a contradiction in their results which they failed to notice), or they're calculating the right shape for it and are then making the mistake of calculating the shape of a square on Train A as viewed from Frame B by treating Frame B like a preferred frame and applying length contraction to that square at an angle in the way the square on the rocket is treated when calculating it from Frame A (while also failing to notice the incompatible alignments when they change frame). Either way, they're making a mistake - the two shapes should not be a mirror image of each other because the way frames behave is not symmetrical. The only reason we've missed this for a hundred years is that we trusted Lorentz, Einstein and the other pioneers of relativity - they failed to explore this properly but gave the impression that they had, and everyone has just believed them ever since without bothering to check thoroughly. I believed them too: I only stumbled upon it by accident while having a conversation with an Einsteinist about rotating discs.
« Last Edit: 10/08/2016 23:58:30 by David Cooper »

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Offline jerrygg38

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Re: Can a preferred frame of reference be identified?
« Reply #55 on: 11/08/2016 12:03:08 »


Hi Jeffrey,

, but if you imagine the universe as being contained in the skin of an expanding bubble, the absolute frame of reference is tied to the centre of the bubble, which is a point not found inside the universe, and no frame of reference inside the universe can be the absolute frame. At every point inside the universe there is a preferred frame of reference which is different from the preferred frame at any other point, but they are all preferred frames of reference regardless, being the frame at that point which matches up closest to the absolute frame. On the local scale though, such as within our solar system, you can consider that all points in that local space have the same frame as their preferred frame of reference, even if that isn't quite true, because the errors will be too small to have any relevance.
   As I see it there is an absolute frame of reference at the center of the universe where the big bang took place. The expanding bubble of energy exploded along the bubble simultaneously billions of times to form the center of the galaxies.  the gravitational field from all the galaxies reach the center which is 13.78 billion light years from the surface of the universe. At the same time a sphere of 27.56 billion years is the outer sphere of the universe. This is a perfect sphere as well.
   So you are correct in my opinion that we have one absolutely stationary point which is not within our visible universe. It seems to me that everything else is distorted common mode but relative frame of references depends upon the gravitational field. The sun is a relative reference. The earth is another relative reference. Everything within a preferred frame of reference is distorted equally but we see a sphere as a sphere because we are distorted as well.
   

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Offline jerrygg38

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Re: Can a preferred frame of reference be identified?
« Reply #56 on: 11/08/2016 12:11:58 »
If you submit it, you have to do the math. That's it.

I've done all the maths that's needed to prove the case, assuming that it's read by someone who understands relativity.
   Relativity is a best fit approximation to reality. Absolute reality has to take into account the distortions produced by all the speeds relative to the absolute reference point. This includes how fast the universe is rotating.
  The beauty of relativity is that so many of the distortions are common mode that a perfect sphere appears to us  as a perfect sphere but in truth it is an ellipsoid.  Fortunately the gravitational field of the Earth tends to equalize the distortions and we cannot readily measure them. So the real world from an absolute sense is different but upon our reference plane if seems to us that a sphere is a sphere. So you can write equations and feel you have solved a particular problem but you are dealing with a best fit approximation.

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Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #57 on: 11/08/2016 14:21:48 »
Indeed, and it's a bad reason - it's based on a belief that different frames all work the same way, but I've shown that a person in Frame B sending things along Rail B will observe them to warp and to do so more the faster they go. Any system for doing the transformations which doesn't find that result is producing errors, hiding this warping with distortions which precisely cancel out the warping that should be there.
I know that you believe that. But any physicist who looks at your argument will reject it because you have not actually discussed the reference frames. All reference frames used in the Special Theory of Relativity have their own time coordinate and you do not include this in your reasoning. Until you do, no physicist will take your argument seriously.

You are free to believe that the time coordinate is not important. However, since the people who work with the theory have all gone through training that demonstrates to them that the theory does not work properly without taking the time coordinate into account, they have a reason for rejecting your argument.


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At no point have you dared to commit yourself to stating whether you agree with it or not, but you ought to.
As someone trained in the use of the Special Theory of Relativity, I can't recognize your claims as an intelligible part of that theory, since they do not use the theory properly. Since your claims do not include transformations to the time coordinate, they do not meet the standard I have been trained to expect for such work. Because of this, I can't recognize your argument as one that is about the Special Theory of Relativity, instead it is an argument about David Cooper's Theory of Relativity.

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The next step is to synchronise clocks for Frame B, pick a time to "take a photo", then run all the clocks as you imagine the shape moving east along the rail.
According to the Special Theory of Relativity, one cannot have a frame without a set definition of synchronized clocks. These clocks will not be synchronized to the clocks in the other frames in your example.

Again, you are free to use David Cooper's Theory of Relativity. However, since you show that it is not consistent, most people will continue to use the Special Theory of Relativity.

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The only reason we've missed this for a hundred years is that we trusted Lorentz, Einstein and the other pioneers of relativity - they failed to explore this properly but gave the impression that they had, and everyone has just believed them ever since without bothering to check thoroughly.
Given the vast literature on relativity theory, including the prevalence of homework problems combining reference frames and the decades of crank attempts to deny relativity theory, it is extremely unlikely that someone would miss this kind of combination.

Anyone who would work through such a scenario would be trained to work from the actual Lorentz transformations, not merely use purely spatial length contractions. This means that they would include the time coordinate in their work and when considering how something looks at a certain time, they would have to consider where everything looks at the time of each frame. Since transformation the time from one frame to another depends on position, this can change the shape of objects from one frame to another.
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I believed them too: I only stumbled upon it by accident while having a conversation with an Einsteinist about rotating discs.
That one person you had a conversation with may have been mistaken. Reading your original post, it is not clear what you believe is the position on rotation taken by the Special Theory of Relativity. Talking to one person is a poor form of education. Did you read about rotation in any textbook on the Special Theory of Relativity?

How were you educated on the Special Theory of Relativity? You keep saying that you have done the work to convince, "someone who understands relativity,"  but you do not want to use the Special Theory of Relativity, you only want to use length contraction. Someone who understands the theory would like to see the theory applied in an argument that purports to be demonstrating a contradiction in the theory.
« Last Edit: 11/08/2016 14:27:53 by PhysBang »

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #58 on: 11/08/2016 20:43:57 »
I know that you believe that. But any physicist who looks at your argument will reject it because you have not actually discussed the reference frames.

Sure! I haven't discussed Frames A, A', B and B' and haven't described the speeds at which they move relative to each other and the direction they're moving in relative to each other, so no one could possibly make sense of what I've said, apart from people who understand relativity and recognise that I have actually discussed those frames and defined how they move relative to each other such that they can visualise the entire setup. I don't know why you're incapable of doing so, and I don't know why you imagine that you're qualified to take part in this conversation when your understanding of the subject is so lacking.

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All reference frames used in the Special Theory of Relativity have their own time coordinate and you do not include this in your reasoning. Until you do, no physicist will take your argument seriously.

When I describe the scene from Frame A, I'm describing Frame A at a single point in time for Frame A - that is a time coordinate, and you can give it any number you like. Let's call it 0 (zero). If we then run events, we redraw the scene for time=1 and move all the things that are moving through Frame A to their new postitions. We can then do it again for time=2 and move all the objects that are moving through Frame A again. We can run all these pictures as a video too, and time will tick up as we do so. I don't know how you were incapable of working out where time comes into this because it should be obvious to anyone with even a modicum of education in the subject. With Spacetime diagrams it's the same, but they normally (due to the limitations of 2D paper) only show one space dimension, each horizontal slice through the diagram having a different time coordinate from the one above and the one below (in the time of the selected frame), while all parts of that horizontal slice have the same time coordinate by that frame. The difference with my diagrams is that they show two space dimensions, so the running of time would be shown by replacing one diagram (one layer of a Spacetime diagram) with another diagram (the next layer up), and the time coordinate is incremented as you run through the diagrams. You can imagine stacking these diagrams in a pile with time running upwards, and then they are orientated the same way as Spacetime diagrams, only with two space dimensions set out in each layer instead of one. When you switch to another frame, you are simply taking a slice through that pile at an angle, cutting through each layer at those points where the times for the new frame are identical, then you warp everything to make the new slice horizontal while keeping the time dimension straight up and down (just like changing frame in the interactive diagram half way down my webpage on relativity). It is astonishing that anyone should find it hard to understand this given that it is all bog standard stuff that anyone who knows the basics of relativity should be able to follow with ease. Look at the interactive diagrams on my webpage - every single one of them have a counter underneath which represents the time coordinate.

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You are free to believe that the time coordinate is not important.

Even Lewis Carroll would have struggled to make up a conversation this surreal! Time coordinate not important? Who the blazes thinks it's not important? What do you imagine my diagrams show if there's no time coordinate tied to them? This is getting into deeper and deeper farce now. Do you really have a qualification in this stuff? Who trained you? You attack me for discussing something I understand but don't have a qualification in, but there you are apparently with a qualification in something that you manifestly don't understand! How could that have happened?

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However, since the people who work with the theory have all gone through training that demonstrates to them that the theory does not work properly without taking the time coordinate into account, they have a reason for rejecting your argument.

I would hope they actually have qualifications that genuinely relate to the level of their understanding and that they can see exactly how time is tied up in my descriptions of things.

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At no point have you dared to commit yourself to stating whether you agree with it or not, but you ought to.
As someone trained in the use of the Special Theory of Relativity, I can't recognize your claims as an intelligible part of that theory, since they do not use the theory properly. Since your claims do not include transformations to the time coordinate, they do not meet the standard I have been trained to expect for such work. Because of this, I can't recognize your argument as one that is about the Special Theory of Relativity, instead it is an argument about David Cooper's Theory of Relativity.

Whoever trained you, I would strongly recommend that you go back and demand a refund from them. It is absolutely appalling that they can award a qualification that leads you into imagining that you have been trained in the use of the Special Theory of Relativity when you have such a shallow grasp of the subject. It is outrageous that my argument is being attacked by someone who is trying to pull rank on the basis of qualifications when he can't follow the simplest of descriptions of how objects appear in a single frame of reference at a single point in time (by the time of that frame) as they move at different speeds in different directions, to the point that he can't even commit himself to saying how a square (when at rest in the frame) will appear once it is moving at relativistic speed in a dircection not aligned with its edges. That is the most astonishing failure I have ever encountered when dealing with someone who claims to be qualified in this subject.

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The next step is to synchronise clocks for Frame B, pick a time to "take a photo", then run all the clocks as you imagine the shape moving east along the rail.
According to the Special Theory of Relativity, one cannot have a frame without a set definition of synchronized clocks. These clocks will not be synchronized to the clocks in the other frames in your example.

And your problem is? The clocks in my ref-frame camera can be set to different synchronisations depending on which frame you want to capture a picture for, and there is a different pattern of syncronisation for every frame of reference. If you set them to by synchronised for frame B, you can then run events until a target time for Frame B appears on a pixel's clock, and when it does, you copy the local content of that part of space to the "photograph" that you're trying to take. Once all the clocks have reached that point (which they don't all do at once, but progressively as you run the events by the rules of Frame A), out comes a perfect Frame B photo of the action at a specific Frame B time.

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Again, you are free to use David Cooper's Theory of Relativity. However, since you show that it is not consistent, most people will continue to use the Special Theory of Relativity.

Everything I'm describing relates directly to SR and to LET. Your failure to recognise that shows that you are not qualified to discuss this stuff.

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Given the vast literature on relativity theory, including the prevalence of homework problems combining reference frames and the decades of crank attempts to deny relativity theory, it is extremely unlikely that someone would miss this kind of combination.

And yet they have all missed it, and when you realise how little understanding of the subject most people with qualifications in the subject actually have, you begin to understand how this sorry state of affairs has come about.

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Anyone who would work through such a scenario would be trained to work from the actual Lorentz transformations, not merely use purely spatial length contractions.

Anyone who knows their stuff knows that all that's needed for analysing this are the simple tools of LET. If a problem shows up when applying length contraction in the way that I have demonstrated happens, that problem will not go away for SR, no matter how many irrelevant things you try to throw at it to try to hide the problem.

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This means that they would include the time coordinate in their work and when considering how something looks at a certain time, they would have to consider where everything looks at the time of each frame.

How do you imagine the ref-frame camera is supposed to work if it doesn't handle time? The program (which I've now finished designing and will get on with writing the code after I've finished responding to your ludicrous objections) will not only show up the length contraction asymmetries, but it will also show up any similar issues with time dilation if there are any. It is possible though that the time dilation numbers will always be right even if the length contractions are calculated incorrectly by people using the wrong method (my initial investigations into this suggest that no difference will show up there), but we'll see what happens once it's up and running.

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Since transformation the time from one frame to another depends on position, this can change the shape of objects from one frame to another.

Which is exactly why you see the changes in the observed shapes of a square as observed from different frames, as I've shown you in the last two diagrams I attached to posts here.

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That one person you had a conversation with may have been mistaken.

He was mistaken alright - he was arguing your side of things, but we were discussing a different issue. I just happened to think about the vectors at point N on the disk and the length contractions not being related to the length contraction for the line that the vectors describe in the same way as the vectors relte to that line. The combined length contraction is much stronger. The anomaly there then transferred through to cases with no rotation, and it's come as a massive surprise to me that the effect is so easy to find when you look for it.

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Reading your original post, it is not clear what you believe is the position on rotation taken by the Special Theory of Relativity. Talking to one person is a poor form of education. Did you read about rotation in any textbook on the Special Theory of Relativity?

I was the one educating him - he was just robotiaclly spouting the standard piffle while being incapable of visualising anything (which is par for the course in this field).

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How were you educated on the Special Theory of Relativity?

I started out by trying to work everything out for myself, which is how I found a different way of calculating length contraction and time dilation which produces identical numbers. I simply looked to see where the light was actually going, and I found that when the MMX moves at 0.866c, the light following the arm perpendicular to the motion of the MMX is actually moving at an angle of 60 degrees forward of that through space, and time dilation comes directly out of this because it takes twice as long to complete the trip along that arm and back. For light to take the same length of time to complete the journey on the other arm of the MMX, the length had to be shortened to half (which I had heard about before), so the length contraction and time dilation are directly related. I then found out that I had rediscovered LET, and I read up on SR after that and saw that it uses exactly the same rules of length contraction and time dilation in relation to how things present themselves in Euclidean planes. If an asymmetry is found in the way things behave in different frames in those Euclidean planes, it necessarily applies both to LET and SR and there is no getting out of that - I have found exactly such an asymmetry, and that dictates that there is a preferred frame of reference which we should be able to identify.

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You keep saying that you have done the work to convince, "someone who understands relativity,"  but you do not want to use the Special Theory of Relativity, you only want to use length contraction. Someone who understands the theory would like to see the theory applied in an argument that purports to be demonstrating a contradiction in the theory.

To see the asymmetry in the way frames behave, is is sufficient to use length contraction. If you want to prove that a boat is three metres long, it is not important to consider its mass, colour, manufacturer, crew requirements etc. - what you do is you stick to the minimum amount of stuff necessary to show that the boat is three metres long. Length contraction is a key part of SR, and what I have shown is that that this key part is not compatible with the idea that all frames of reference behave the same way.

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Offline jeffreyH

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Re: Can a preferred frame of reference be identified?
« Reply #59 on: 11/08/2016 21:11:59 »
We can say that

L = cos(asin(v/c))
T = 1/cos(asin(v/c))

Where L is a multiplier for length and T is a multiplier for time.

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Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #60 on: 11/08/2016 21:32:02 »
I don't know why you're incapable of doing so, and I don't know why you imagine that you're qualified to take part in this conversation when your understanding of the subject is so lacking.
I imagine that I am qualified because I took several courses in grad school on relativity theory. It might be that I, all my professors, and all my fellow students were under some sort of delusion. You are free to believe that.

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When you switch to another frame, you are simply taking a slice through that pile at an angle, cutting through each layer at those points where the times for the new frame are identical, then you warp everything to make the new slice horizontal while keeping the time dimension straight up and down (just like changing frame in the interactive diagram half way down my webpage on relativity). It is astonishing that anyone should find it hard to understand this given that it is all bog standard stuff that anyone who knows the basics of relativity should be able to follow with ease. Look at the interactive diagrams on my webpage - every single one of them have a counter underneath which represents the time coordinate.
It is not clear that your representation of how things appear is taking the correct "slice through that pile at an angle", since you at no point do any translation of time coordinates.

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Time coordinate not important? Who the blazes thinks it's not important? What do you imagine my diagrams show if there's no time coordinate tied to them? This is getting into deeper and deeper farce now. Do you really have a qualification in this stuff? Who trained you? You attack me for discussing something I understand but don't have a qualification in, but there you are apparently with a qualification in something that you manifestly don't understand! How could that have happened?
I'm not attacking you, I'm merely pointing out that you never do a translation that includes a time coordinate. You are free to believe that this is not important.

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Whoever trained you, I would strongly recommend that you go back and demand a refund from them. It is absolutely appalling that they can award a qualification that leads you into imagining that you have been trained in the use of the Special Theory of Relativity when you have such a shallow grasp of the subject. It is outrageous that my argument is being attacked by someone who is trying to pull rank on the basis of qualifications when he can't follow the simplest of descriptions of how objects appear in a single frame of reference at a single point in time (by the time of that frame) as they move at different speeds in different directions, to the point that he can't even commit himself to saying how a square (when at rest in the frame) will appear once it is moving at relativistic speed in a dircection not aligned with its edges. That is the most astonishing failure I have ever encountered when dealing with someone who claims to be qualified in this subject.
You are being very defensive and I am sorry that you feel that you are being attacked. I am merely trying to get you to produce the most rigorous version of your argument. It is true that I believe that your argument will disappear when rigor is applied. It is also true that I do not think that you will believe me if I work out the details.
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Everything I'm describing relates directly to SR and to LET. Your failure to recognise that shows that you are not qualified to discuss this stuff.
You are free to believe this. I believe that you are missing something crucial. I hope that it is not defensiveness that prevents you from working out your argument in all the required details.

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Anyone who would work through such a scenario would be trained to work from the actual Lorentz transformations, not merely use purely spatial length contractions.

Anyone who knows their stuff knows that all that's needed for analysing this are the simple tools of LET. If a problem shows up when applying length contraction in the way that I have demonstrated happens, that problem will not go away for SR, no matter how many irrelevant things you try to throw at it to try to hide the problem.
If one looks at the history of the so-called "paradoxes" of the Special Theory of Relativity, you will find that most of them do not appear problematic when the relativity of simultaneity is properly taken into account. In all cases, properly applying the Lorentz transformations is crucial to understanding what happens in the Special Theory of Relativity, since one can produce incorrect results if one applies only a part of the transformations and not all the transformations.

Again, you are free to believe what you want. I am only trying to report how your argument will be received by someone who has training in the relevant physics.
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How do you imagine the ref-frame camera is supposed to work if it doesn't handle time? The program (which I've now finished designing and will get on with writing the code after I've finished responding to your ludicrous objections) will not only show up the length contraction asymmetries, but it will also show up any similar issues with time dilation if there are any. It is possible though that the time dilation numbers will always be right even if the length contractions are calculated incorrectly by people using the wrong method (my initial investigations into this suggest that no difference will show up there), but we'll see what happens once it's up and running.
If your calculations do not recognize that translations to the time coordinate depend upon the location, then you will be presenting incorrect pictures, since you will not be presenting simultaneous points.

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I then found out that I had rediscovered LET, and I read up on SR after that and saw that it uses exactly the same rules of length contraction and time dilation in relation to how things present themselves in Euclidean planes.
Have you consulted textbooks on the Special Theory of Relativity?

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To see the asymmetry in the way frames behave, is is sufficient to use length contraction. If you want to prove that a boat is three metres long, it is not important to consider its mass, colour, manufacturer, crew requirements etc. - what you do is you stick to the minimum amount of stuff necessary to show that the boat is three metres long. Length contraction is a key part of SR, and what I have shown is that that this key part is not compatible with the idea that all frames of reference behave the same way.
You are free to believe this. As I said before, anyone with training in SR will assume, even if they do not immediately recognize your error, that you are presenting an artefact of ignoring the change in the time coordinate. If you do not go through the rigor of actually doing the transformations to justify your argument, then your argument will be justifiably rejected on the basis that you have not done the work to establish your case.

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Offline jeffreyH

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Re: Can a preferred frame of reference be identified?
« Reply #61 on: 11/08/2016 21:35:17 »
Then 2*pi*L will show a contraction of the circumference of the unit circle due to gravitation during relativistic rotation. Quite a neat way to do this.

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Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #62 on: 11/08/2016 22:16:20 »

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Offline jeffreyH

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Re: Can a preferred frame of reference be identified?
« Reply #63 on: 11/08/2016 22:34:43 »
Since the radial direction is constantly changing and itself undergoing acceleration then there can be an argument for reduction in radial length as well as reduction in circumference during relativistic rotation. Rebuttals on a postcard to ...

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Offline jeffreyH

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Re: Can a preferred frame of reference be identified?
« Reply #64 on: 11/08/2016 22:42:29 »
The question is how fast does an object have to be rotating so that to the rest of the universe the radius appears to have contracted so that the mass is within its own Schwarzschild radius?

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #65 on: 11/08/2016 23:00:05 »
This conversation reminds me of this recent article: https://aeon.co/ideas/what-i-learned-as-a-hired-consultant-for-autodidact-physicists

This is just the Dunning Kruger thing making you overconfident again. You may have done a lot of learning, but you're incapable of applying it unless the argument is presented to you in exactly the form you've been taught to understand it in. As soon as it's expressed in a clearer form designed to enable untrained people to understand it too, for some bizarre reason, you can't hack it.

You're still too scared to state whether you agree that a square which was initially at rest in Frame A with its edges aligned north-south and east-west and which has subsequently been accelerated up to relativistic speed in the direction NE (without at any stage being rotated) will now be contracted in such a way that none of its edges are aligned north-south or east-west. What is your training worth if you can't even commit yourself to an answer on that simple point? Woeful!

You're also too scared to state whether you agree that squares on the roof of Train B should, if under no stress, take up the same shape as a square painted on a rocket that is co-moving with the train. Woeful again - what an embarrassment to the people who gave you your qualificatiions!

You're also too scared to discuss the shape of an unstressed square on the roof of Train A as viewed from a Frame B observer. That's a tougher task as it involves a frame change, but you claim to be capable of doing this. I've shown you the shape that I get, but you won't say anything at all about the shape that you would get, if you really knew how to calculate it. Not quite so woeful this time, as it's a harder test, but you should be able to give it a go at least. But no - you won't, and you won't because you're out of your depth.

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #66 on: 11/08/2016 23:25:56 »
Here's another clear way to show that I'm right. The square moving NE has none of its edges aligned north-south or east-west, so how is it going to fit nicely between Rail B and B2 which are aligned precisely east-west? Every point of the rhombus that is the Frame A view of this moving shape is shown in its correct position in the two space dimensions for the same instant in time by Frame A's clock, and only someone who doesn't understand relativity could say it's a simultaneity issue that prevents the edges of this square from lining up with the rails. If you're moving with that square, you still think the edges are aligned north-south and west-east, but you'll see that they are not alligned parallel to and perpendicular to the rails in that frame: the alignment of the rails in that frame is not east-west, and the same applies to any observer travelling on a square aligned at the same angle as the rocket and dangled between the rails, if the gap between the rails has been adjusted to allow it to fit there. That observer will then see two of the corners touching the rails at either side, but he will be able to put his arm in the gaps between the edges and the rails without being harmed while you would claim he couldn't do this because the difference in alignment is an error in his judgement of simultaneity. You (physbang) still won't get it though, because you've been miseducated and you have never learned how to think for yourself.
« Last Edit: 11/08/2016 23:28:03 by David Cooper »

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Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #67 on: 12/08/2016 02:13:28 »
This is just the Dunning Kruger thing making you overconfident again. You may have done a lot of learning, but you're incapable of applying it unless the argument is presented to you in exactly the form you've been taught to understand it in. As soon as it's expressed in a clearer form designed to enable untrained people to understand it too, for some bizarre reason, you can't hack it.
I'm glad you are being so pleasant and attempting to diagnose me. It speaks very highly of your character.
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You're still too scared to state whether you agree that a square which was initially at rest in Frame A with its edges aligned north-south and east-west and which has subsequently been accelerated up to relativistic speed in the direction NE (without at any stage being rotated) will now be contracted in such a way that none of its edges are aligned north-south or east-west. What is your training worth if you can't even commit yourself to an answer on that simple point? Woeful!
I am so glad that you have made the decision to insult me rather than actually work through your argument in detail. It again speaks very highly of your character.
Every point of the rhombus that is the Frame A view of this moving shape is shown in its correct position in the two space dimensions for the same instant in time by Frame A's clock, and only someone who doesn't understand relativity could say it's a simultaneity issue that prevents the edges of this square from lining up with the rails.
It is rather the reverse: the relativity of simultaneity ensures that the train wheels continue to stay lined up with the rails.

But don't trust me: work it out for yourself, you are the expert here.

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Offline jeffreyH

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Re: Can a preferred frame of reference be identified?
« Reply #68 on: 12/08/2016 13:40:44 »
Relativistic rotational speed at the circumference of a disc implies a gradient of relativistic mass increase from the centre of the disc outwards. With length contraction of the radius this indicates a longer Schwarzschild radius due to increase in mass. Therefore a larger mass black hole.

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #69 on: 12/08/2016 17:55:24 »
I'm glad you are being so pleasant and attempting to diagnose me. It speaks very highly of your character.

Nice - you can throw words like "crank" around, but I'm not allowed to tell you what your problem is.

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You're still too scared to state whether you agree that a square which was initially at rest in Frame A with its edges aligned north-south and east-west and which has subsequently been accelerated up to relativistic speed in the direction NE (without at any stage being rotated) will now be contracted in such a way that none of its edges are aligned north-south or east-west. What is your training worth if you can't even commit yourself to an answer on that simple point? Woeful!
I am so glad that you have made the decision to insult me rather than actually work through your argument in detail. It again speaks very highly of your character.

Where's the insult in there? Your tactics of avoiding the issues aren't the most attractive thing on show here.

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It is rather the reverse: the relativity of simultaneity ensures that the train wheels continue to stay lined up with the rails.

Is it going to be like this with the peer review people too? Do I have to spend many pages teaching them the basics too before I can introduce them to my proof? How did physics get itself into this mess! I've told you how these diagrams work: they show points plotted out in two space dimensions with the time being identical for each point (by the clock of that frame), but it just doesn't register with you. How far do I have to go before it dawns on you that your interpretation of them is ridiculous? Let me put you into the diagram (see the picture attached below) so that you can experience the events at specific points as objects pass you by. Here we have a parallelogram moving NE at relativistic speed which has contracted it in that direction. When it was at rest, before it was accelerated to this speed, it was a rectangle (made of four squares stuck together), the long side aligned east-west. It has not been rotated at any stage. The parallelogram shape that we see in the diagram is its Frame A shape when it's moving NE through Frame A at relativistic speed. The rails are like Rail B and B2, moving directly north at the same speed as the north vector for the movement of the parallelogram, so they move up the diagram at the same speed. Note: I have made the rails much narrower in order to keep the size of the diagram down - this allows me to use a much shorter rectangle.

Now, I want you to stand at point X and imagine this shape and the rails passing you. The parts of the parallelogram which the blue arrow passes through must all pass through point X, so you will meet them there. The parts of the track that pass you though will be the ones that are directly to the south of point X. In what order do you think you'll meet the different objects? Well, the first thing to reach you will be the longer leading edge of the parallelogram. The second point of sigificance to pass you will be the longer trailing edge of the parallelogram. Soon after that you will meet Rail B2, and lastly you will meet Rail B.

Now do the same thing at point Y: Rail B2 passes you first, then the longer leading edge of the parallelogram, then the longer trailing edge, and then rail B. Now do the same at point Z: Rail B2 passes you first, then Rail B, then the shorter leading edge of the parallelogram, and lastly the longer trailing edge of the parallelogram.

Now, can you see the problem with your stance on all this? You imagine that if you jump to a Frame B or Frame B' view of things, the whole rectangle will magically fit between the rails, but that would mean that a Frame B or Frame B' observer would see the observer at X or Z experiencing events in a different order from the one they experienced them in when viewed from Frame A. That kind of reordering of events is impossible.

To spell this out more clearly for you (because you clearly need all the help you can get), I want you to imagine that when you're standing at point X, Y or Z you are holding a piece of chalk which you're going to use to mark the objects as they pass you, but it's a special stick of chalk which is partly coloured, the top part being red and the rest being white. So, there you are at point X, marking the paralellogram with the chalk. The parallelogram wears away most of your stick of chalk, so it has a red line across it which turns white about a quarter of the way across. By the time the rails reach you, they can obviously only get marked in white. Now do the same at point Y (starting with a new piece of chalk): you mark the first rail with red, then about a quarter of the parallelogram in red too before that line turnes white, then you mark the second rail with white. Now do the same from point Z (again starting with a new piece of chalk): you mark both rails with red chalk, then you draw a line across the parallelogram which starts red and turns white before it reachs the far side.

Now let's switch to Frame B or B' and look at the events from there. I will be the observer from here. By your magical version of physics, I must see you (the X observer) mark Rail B first with white chalk, and you're holding a well-worn piece of chalk with all the red part missing. Next, I see you mark the parallelogram with red chalk, and your stick of chalk has magically grown such that all the red part is back. I watch the red erode away on the parallelogram and see the line turn white, then the stick of chalk suddenly loses a little chunk of its length while not touching anything before it marks Rail B with white. I am watching magic in action.

Everything will work fine with you working from Y, so let's jump to you as observer Z and see what you look like to Frame B and B' observers. Again, I will be Frame B/B' observer. I see the first rail reach you at point Z first, and you mark it with red. Next, I see some of the red chalk vanish without touching anything. Next, I see the parallelogram reach you and you mark it in a line that turns from red to white. Lastly, I see the chalk sprout most of its length back again so that you can mark Rail B with red, then the chalk magically shortens back down so that you aren't left with any of the chalk which must end up on the parallelogram.

That is how ludicrous your position is. How are you going to fix it now? Perhaps you'll assert that the length contraction can't be applied in the first place and that it should be a rectangle moving NE through Frame A rather than a parallelogram, but if you do that you've just eliminated length contraction from physics.

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But don't trust me: work it out for yourself, you are the expert here.

Yes - I am the expert here and I have worked it out. You're the one who's doing magic and who has been awarded physics qualifications from Hogwarts University.
« Last Edit: 12/08/2016 18:04:11 by David Cooper »

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Offline jeffreyH

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Re: Can a preferred frame of reference be identified?
« Reply #70 on: 12/08/2016 18:05:33 »
By accelerating chocolate to relativistic speeds we may well be able to make a Heston Blumenthal chocolate fountain to die for.

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #71 on: 12/08/2016 18:44:05 »
By accelerating chocolate to relativistic speeds we may well be able to make a Heston Blumenthal chocolate fountain to die for.

You'd think the experiment would already have been done with squares of Galaxy or entire Mars Bars.

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Offline jeffreyH

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Re: Can a preferred frame of reference be identified?
« Reply #72 on: 12/08/2016 19:00:41 »
I am off to search for scientific papers on the matter!

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Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #73 on: 12/08/2016 19:09:23 »
Where's the insult in there? Your tactics of avoiding the issues aren't the most attractive thing on show here.
You are correct: your continued failure to actually perform the relevant calculations are examples of my avoiding issues.

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It is rather the reverse: the relativity of simultaneity ensures that the train wheels continue to stay lined up with the rails.

Is it going to be like this with the peer review people too? Do I have to spend many pages teaching them the basics too before I can introduce them to my proof?
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. If you want to claim to be discussing SR, then you are going to have to go over the basics. This means actually performing the calculations rather than taking the shortcuts that you have been doing so far.

You are free to believe whatever you wish. You can even attack me for my refusal to simply abandon rigor and adopt your rough methods.

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How did physics get itself into this mess! I've told you how these diagrams work: they show points plotted out in two space dimensions with the time being identical for each point (by the clock of that frame), but it just doesn't register with you.
You have made claims about how your diagrams work. Yet since you do not actually show us the relevant calculations, we must take your word that your diagrams are correct. On the one hand we have you: one person who has not done all the calculations despite being asked, who claims to have found something that everyone who uses the full calculations has missed. On the other hand we have dozens of textbooks, thousands of articles, and thousands of academics who have reviewed and worked with SR. You ask us to take your authority that your diagrams are correct, but the scientific world does not work like that, it requires the demonstration mathematically.
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Now, can you see the problem with your stance on all this? You imagine that if you jump to a Frame B or Frame B' view of things, the whole rectangle will magically fit between the rails, but that would mean that a Frame B or Frame B' observer would see the observer at X or Z experiencing events in a different order from the one they experienced them in when viewed from Frame A. That kind of reordering of events is impossible.
You are actually contradicting one of the fundamental results of SR, that the order of distant events depends on the system of coordinates used. As I said, most of the time that someone thinks that they have a problem with SR, they really have a problem with the relativity of simultaneity. You have just demonstrated that you reject the relativity of simultaneity, so you are firmly with the majority of those who mistakenly think they have discovered a problem for SR.

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But don't trust me: work it out for yourself, you are the expert here.

Yes - I am the expert here and I have worked it out. You're the one who's doing magic and who has been awarded physics qualifications from Hogwarts University.
And yet you actually haven't bothered to use the Lorentz transformations.

I know that it might be painful to own up to the truth, but perhaps you have made a mistake? Won't you even consider actually using the transformations that, supposedly, you derived?

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #74 on: 12/08/2016 20:03:33 »
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?

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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.

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If you want to claim to be discussing SR, then you are going to have to go over the basics. This means actually performing the calculations rather than taking the shortcuts that you have been doing so far.

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.

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You are free to believe whatever you wish. You can even attack me for my refusal to simply abandon rigor and adopt your rough methods.

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.

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You have made claims about how your diagrams work. Yet since you do not actually show us the relevant calculations, we must take your word that your diagrams are correct.

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.

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On the one hand we have you: one person who has not done all the calculations despite being asked,

I've done all the calculations necessary to prove the case.

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...who claims to have found something that everyone who uses the full calculations has missed.

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.

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On the other hand we have dozens of textbooks, thousands of articles, and thousands of academics who have reviewed and worked with SR. You ask us to take your authority that your diagrams are correct, but the scientific world does not work like that, it requires the demonstration mathematically.

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.

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You are actually contradicting one of the fundamental results of SR, that the order of distant events depends on the system of coordinates used.

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! The other observer is also right there at the place where the events are taking place. The chalk on the objects is witness to the events taking place where they are too and to their order. The events at the single point X take place in a fixed order, and there is no frame in the universe which you can switch to to show those events taking place in a different order.

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As I said, most of the time that someone thinks that they have a problem with SR, they really have a problem with the relativity of simultaneity. You have just demonstrated that you reject the relativity of simultaneity, so you are firmly with the majority of those who mistakenly think they have discovered a problem for SR.

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. In a Spacetime diagram, my point X becomes a line running upwards, vertically in one frame and at an angle in others, the the events that occur in it happen in a fixed order with the first one appearing lowest on that line in every frame and the last appearing highest on that line in every frame. By trying to make the parallelogram fit between the rails when you switch frame, you are changing the order of events on that line, and that is a massive breach of the rules of SR. No one qualified in this stuff should be making such a shocking error.

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And yet you actually haven't bothered to use the Lorentz transformations.

I have used a method of calculating length contraction which I have demonstrated produces the exact same numbers and in demonstrating that I used Lorentz's formula too. If you have a problem with my results, you need to show me how you get a different shape for objects or different alignments of them from the ones I've claimed they will have (and which I've backed with more than enough numbers to prove the point). At no point have you managed to do so in any case, and the reason for that is that you have nothing to offer: you can't find any fault other than imagined ones based on your lack of understanding of SR.

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I know that it might be painful to own up to the truth, but perhaps you have made a mistake? Won't you even consider actually using the transformations that, supposedly, you derived?

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.

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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 »

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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.

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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.

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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.

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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.

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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 »

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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.

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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.

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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.

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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 »

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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 »

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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?

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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.

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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?
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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.

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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.

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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.

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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.

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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?

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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.

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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.

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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.

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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.

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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 »

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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