Can a preferred frame of reference be identified?

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Offline David Cooper

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[Edit: This thread was resolved by page 5 - there is a hidden rotation involved when something is accelerated in one direction and is then accelerated sideways, and when that is taken into account all the issues go away, making it impossible to pin down a preferred frame of reference in the way I'd suggested. Beyond that point, the discussion turns to other matters, much of it concerning the business of whether SR can function without a preferred frame, and at the time of this edit it looks as it can't (unless you go for a model of SR in which objects take shortcuts into the future on some paths and get there sooner than other objects, at which point you have to handle event-meshing failures and need to bring Newtonian time back into the model to deal with that issue.]

I've found something I didn't think would ever be possible, but it looks as if there may be a way to pin down an absolute frame of reference.

Imagine a disc lying flat with four points marked on the circumference, N, E, S and W (for the four compass directions). We will move the disc northwards in a moment while rotating it clockwise, but let's first spin it up to speed without moving it along through space. I want to spin it until the edge is moving at 0.866c relative to the centre, a speed at which length contraction should act on the edge in such a way as to halve its length. If we also sandwich our rotating disc between two non-rotating discs of equal size we can eliminate all the non-Euclidean SR distractions by imposing a tight Euclidean metric upon our rotating disc in the middle of the sandwich and use that to lay down the law about how the rotating disc must behave in that space.

We can see that there is no longer enough material in our rotating disc to fill the whole space between the non-rotating discs, so it must stretch or break. Let's assume it splits and leaves us with gap in it, the gap being much wider the further out you go from the centre as the length contraction becomes more severe. It turns out then that we're going to  need to mend our disc once it's been spun up to the target speed so as to fill in the gap, and it's only after that that will we have a complete disc rotating at our target speed. This appears to go against some of the teachings of SR in relation to the behaviour of rotating discs, but it doesn't go against the rules as to how SR works for things moving in straight lines, and we can show that the two things are actually equivalent, which means that many of the existing ideas about how rotating discs behave are wrong.

Any rocket following a tangent to our rotating disc at 0.866c must display length contraction to half its rest length, and this must be matched by the material in the edge of the disc as they move side by side for a moment. That means that the edge of the disc must appear length contracted and cannot possibly fill the space all the way round the space demarcated for it by the two non-rotating discs. We can also eliminate most of the change in direction of the material in the disc's edge by using a disc of a diameter measured in billions of lightyears across, which means that the material in the disc's edge will be moving at the same speed and in the same direction as the material in the rocket flying past at a tangent to the disc not merely for an instant, but for many hours with the material in the disc edge and the rocket potentially being side by side and only a micron apart throughout that time - this is more than long enough to rule out any role for accelerations in breaking the normal rules of length contraction and time dilation. So, we can show that a rotating disc cannot behave the way that most SR experts claim it does: it turns out that they have been breaking some of the most fundamental rules of SR.

Our next step is to move the whole disc, and we want it to move at 0.866c northwards. By the way, our non rotating discs are transparent, so we can see the rotating disc through them, and our N, E, S, W markers are printed on the non-rotating discs, so N is always the leading point of the discs as they move through space, while S is the point most aft. Once we are moving our disc sandwich along at 0.866c, the material in our rotating disc starts to behave in unexpected ways, bunching up as it moves slowly past point W and whipping back past point E with all length contraction removed there. At point E the material is not moving in the frame of reference we're using as the base for all our measurements, but at point W it is moving northwards at 0.99c and the local length contraction is to 1/7. (To calculate this speed and length contraction at point W, I imagined a rocket moving north at 0.866c and firing a missile ahead at 0.866c from its point of view, and so in our reference frame that works out at 0.99c - that rocket must behave the same way as the material at the edge of the disc where the rocket may travel alongside it for a while as it follows a tangent to the disc at that point.) Our non-rotating discs have length-contraction applying across them exclusively in the NS direction, reducing all measurements running that way to 1/2 of their rest lengths, so the discs' shapes are now elliptical with the NS diameter half the length of the EW diameter. The rotating disc should match that shape if the idea of relativity is correct, but the length contractions on the material of the rotating disc and directions in which it contracts will be different in places, and it's in exploring this that I've found something that I thought couldn't happen.

The key thing is what happens at points N and S. The material there is moving at 0.968c (which can be broken down into two vectors: it's moving north at 0.866c, and it's moving sideways at 0.433c) which means that the length contraction will make the material sit four times as close together in its direction of travel as it would do at rest, and this contraction acts at an angle of 63.4 degrees forwards of the EW line. (I worked out the 0.433c figure by thinking about how a light clock aligned EW would work here: the light in it would actually move at 60 degrees ahead of sideways, and that reduces its progress between points E and W to half, so the same halving will apply to anything else moving from E to W and back.) The component of this contraction to 1/4 is greater in the NS direction than the length contraction in the non-rotating discs at points N and S (which is to 1/2), and that's the crucial thing here - this means it must pull the rotating disc in more at N and S than the non-rotating discs, so their shapes will no longer match up in the way they do when the apparatus is not moving along through space - the sandwich filling can no longer fill the whole space between the outer discs. On the non-rotating discs we have length contraction to 1/2 of the rest length all the way from N to S. On the rotating disc we only have that amount of length contraction at the very centre of the line NS: at all other points on the line NS we have more length contraction than that (running in the NS direction). That means that SR must have a theoretically identifiable preferred/absolute frame of referrence.

Again we can send a rocket at 0.968c over point N or S at the same angle as the material of the disc there is moving to illustrate that it must contract in exactly the same way in the disc as it does in the rocket, and by giving our disc a huge diameter measured in the billions of lightyears, we can reduce all the pesky accelerations caused by the rotation to such a low level that they can be ignored (while reducing the centrifugal forces to the point of irrelevance at the same time) - the material in the disc can now be thought of as moving in almost perfectly straight lines while we're comparing its behaviour with that of the material in the rockets which are temporarily co-moving with it.
« Last Edit: 04/09/2016 22:41:56 by David Cooper »

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #1 on: 04/08/2016 04:08:32 »
I've managed to come up with a related thought experiment which eliminates the rotation altogether, and it's now ridiculously simple! How has this been missed for a century?

Imagine a long, straight monorail floating in space with a train on it, both of them being as long as you like, but this is a special monorail in which the train runs on the side of the rail instead of the top - I'm making this change to its design to make it fit in with the previous thought experiment from which it was derived. We're looking down on the track and train from "above", and we see the rail as being a metre wide while the train is also a metre wide (and I chose that width for both just for simplicity with the numbers). The rail is aligned west-east and the train is to the north of it. We're going to move the whole caboodle northwards later on at 0.866c, just like we did with the discs before. First though, let's assume that the rail is stationary, and then we'll start the train rolling eastwards. We accelerate the train until it's moving east at 0.866c, and the length of each carriage must contract to half its rest length. The width of the train is unaffected though - it is still a metre wide, matching the width of the rail.

Now let's move the rail and train northwards at 0.866c. With the rail moving north at this speed, time dilation must kick in for it, so this will affect the train's actual speed of travel relative to the rail as measured from our stationary frame of reference: we will now measure it as going at 0.433c, though to anyone moving north with the rail at 0.866c the train will still appear to be doing 0.866c from their point of view. The length contraction on the rail will now make it half a metre wide, but what is the length contraction on the train, and in which direction is that contraction applied? The train is doing 0.866c northwards and 0.433c sideways (those are the two vectors), so the actual speed is the square root of 0.866^2 + 0.433^2, and that comes to 0.968c. The angle in which the material of the train is moving is 63.4 degrees ahead of the west-east line (a bit nearer to north than north-east), and the length contraction on it (which acts in that direction) will reduce its length to a quarter. We already know that the length contraction on the rail will reduce it to a half in the north-south direction, so we now measure the rail as being half a metre wide, but the train is now going to be quarter of a metre wide and a little contorted, the northern side now trailing the side adjacent to the rail. That is a major mismatch which doesn't occur when the rail is stationary in space with only the train moving. If you are travelling with the moving rail, you will still measure it as a metre wide from that frame of reference, but you will measure the width of the train as being half a metre instead of the whole metre you expected.

Unless there's an error in the above, it means that if you're in the frame of reference in which the rail is stationary, you can tell whether you're moving or not relative to an absolute frame of reference simply by measuring the width of the train. If we send a rocket at 0.968c in the same direction, 26.6 degrees round from north, the length contraction will be to 1/4 of the rocket's rest length, and the material of the train must undergo the same length contraction as that in the same direction. The rocket will also be seen to move with the train, staying above the same carriage all the time and keeping perfect pace with it.

So, have I made a mistake somewhere (and my brain's too far gone for me to spot it), or is this finally the death of relativity? And if the latter, can we turn this into a practical experiment to pin down how fast we're moving through the fabric of space?
« Last Edit: 04/08/2016 04:20:51 by David Cooper »

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Offline PmbPhy

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Re: Can a preferred frame of reference be identified?
« Reply #2 on: 04/08/2016 04:50:59 »
I've found something I didn't think would ever be possible, but it looks as if there may be a way to pin down an absolute frame of reference.
Hi David,

This subject touches on one of the things in relativity that I've been avoiding for years because thinking about it hurts my brain. Lol! What you've touched upon here is what's known as Ehrenfest's Paradox. It has to do with the circumference of a disk contracting while the radius remains constant which ends up leaving the value of π altered. You proposed to solve this problem by sandwiching the rotating disc between two non-rotating discs of equal size. It's your proposal that this will "eliminate all the non-Euclidean," correct? If that's the case then I believe that you made an error here by assuming that something like that can be done. Exactly what action will the two non-rotating disks have on the rotating disk? There's a great deal of physics to deal with here that you haven't mentioned. You can learn more about it in the physics literature if you're interested? You can first start with Wikipedia: https://en.wikipedia.org/wiki/Ehrenfest_paradox
You should also study Born rigidity while you're at it. I'd like to make it clear that I myself don't yet have a complete understanding of all the intricacies of Ehrenfest's paradox. It's something that I'm currently working to understand. It's tricky stuff and can be quite confusing.

Please let me know what you learn if you choose to study Ehrenfest's paradox. I'd be very interested.

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Offline jeffreyH

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Re: Can a preferred frame of reference be identified?
« Reply #3 on: 04/08/2016 08:26:47 »
If everything is in motion relative to everything else then there will be a positions that can be thought of as stationary with respect to everything else. The redshift of galaxies is a prime example. Any point in space from which all objects move away uniformly in all directions can be thought of as fixed relative to those galaxies. The problem for a preferential frame is that this point is not at infinity nor ever can be. So that some force is always acting upon it. Only if all external forces were exactly equal, creating  a perfect equilibrium point, could this stand in as a false preferential frame.

I have studied the concept of a fixed background for a while now. As Pete says it can make your head hurt. I look forward to reading more of this thread. It is a very pertinent point.
Fixation on the Einstein papers is a good definition of OCD.

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Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #4 on: 04/08/2016 13:16:27 »
I've managed to come up with a related thought experiment which eliminates the rotation altogether, and it's now ridiculously simple! How has this been missed for a century?
Yeah, when people say this, it means that they are making a basic mistake.
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Now let's move the rail and train northwards at 0.866c. With the rail moving north at this speed, time dilation must kick in for it, so this will affect the train's actual speed of travel relative to the rail as measured from our stationary frame of reference: we will now measure it as going at 0.433c, though to anyone moving north with the rail at 0.866c the train will still appear to be doing 0.866c from their point of view.
And there is your basic mistake: you introduced a contradiction. The speed relative to the track is given. This means that the track relative to the train is given as the same thing, there is no conversion for that speed. If you were doing some sort of calculation based on the operation of the engine, then you would have to do a conversion with time dilation taken into account.

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #5 on: 04/08/2016 18:59:35 »
This subject touches on one of the things in relativity that I've been avoiding for years because thinking about it hurts my brain. Lol! What you've touched upon here is what's known as Ehrenfest's Paradox.
Hi Pete,

Yes, I read up on it a week or two back, and I wasn't impressed. The point of sandwiching the rotating disc between two non-rotating ones is that they force you to measure the rotating disc properly using the normal Euclidean metric without being confused by the complications of what the rotating disc is imagined to be doing in non-Euclidean geometry. The rotating disc, whatever it may be doing in the minds of physicists, still has to interface with the Euclidean metric imposed on it by the two non-rotating discs. If its radius remains the same, the circumference must still be found directly between the two circumferences of the non-rotating discs. If the circumference length contracts, it cannot complete the circuit, so there will need to be a gap or gaps in it.

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It has to do with the circumference of a disk contracting while the radius remains constant which ends up leaving the value of π altered. You proposed to solve this problem by sandwiching the rotating disc between two non-rotating discs of equal size. It's your proposal that this will "eliminate all the non-Euclidean," correct? If that's the case then I believe that you made an error here by assuming that something like that can be done. Exactly what action will the two non-rotating disks have on the rotating disk?

The non-rotating discs simply force the observer to measure the rotating disc correctly without being put off by non-Euclidean complications. Whatever the rotating disc does, it must behave in accordance with the basic rules of SR. Let me show you an alternative scenario which is directly equivalent. Imagine a stationary non-rotating disc with 36 tangents to it touching at a series of points 10 degrees apart. Each tangent has a rocket moving along it, and each rocket is travelling at 0.866c, so it's been length-contracted to half its rest length by this movement. Each rocket touches the disc for a moment and they all do this simultaneously. Each rocket also hooks up to the one ahead of it at that same moment, so they become locked together into a ring, and from this point on that ring will rotate round the disc. For each rocket, this is no different from them just looping round in a circle - there will be some differences in the length contraction on them due to the change in direction which will introduce stresses, but they won't suddenly do anything weird: they will simply go round the non-rotating disc, remain hooked together, remain length contracted to half their rest length, and they will continue to fill the space available to them round the disc as measured by the Euclidean metric. If you then slow the rockets down to a halt, they will lose the length contraction and take up twice as much space round the disc as there is room for them, so there may be a bit of a pile up. This shows that the way SR normally tries to handle rotation is wrong. Furthermore, if you repeat this with the non-rotating disc moving along at 0.866c and have the rockets all approach it at the right speeds relative to it so that they still think they're running on tangents to a stationary disc, you will have them behaving in the way I described the material of a rotating disc moving along at 0.866c through space with the rocket at point E having no length contraction on it and the one at point W being shortened to 1/7 of its rest length. These thought experiments show that the normal way of attempting to handle rotation in SR is incompetent, because we're now doing it using more fundamental rules of SR where the rotation has been eliminated up to the point where the rockets hook together, and the direction changes on them that follow are no different from normal direction changes. When we look at a rotating disc a billion lightyears across and think how gentle those direction changes will be, we have the material moving practically in straight lines throughout, so the way the material in the disc's edge must match the behaviour of rockets moving past on tangents to the disc where they are moving at the same speed and in the same direction.

However, even if you still think there's some weird complication which can somehow override all of that, you have to look at the second post of this thread where I simplified the thought experiment by removing rotation from it altogether.
« Last Edit: 04/08/2016 19:04:40 by David Cooper »

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #6 on: 04/08/2016 19:13:20 »
If everything is in motion relative to everything else then there will be a positions that can be thought of as stationary with respect to everything else. The redshift of galaxies is a prime example. Any point in space from which all objects move away uniformly in all directions can be thought of as fixed relative to those galaxies. The problem for a preferential frame is that this point is not at infinity nor ever can be. So that some force is always acting upon it. Only if all external forces were exactly equal, creating  a perfect equilibrium point, could this stand in as a false preferential frame.

Hi Jeffrey,

It's important to realise that a preferred frame doesn't need to be the same thing as an absolute frame. A preferred frame at one point in the universe needn't be the same as the preferred frame at any other, but an absolute frame would have to be the same for them all. That might initially sound unlikely (or even impossible), 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.
« Last Edit: 04/08/2016 19:16:36 by David Cooper »

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #7 on: 04/08/2016 19:52:30 »
Hi PhysBang,

Yeah, when people say this, it means that they are making a basic mistake.

Well, let's see how long it takes you to find that basic mistake...

Quote
Quote
Now let's move the rail and train northwards at 0.866c. With the rail moving north at this speed, time dilation must kick in for it, so this will affect the train's actual speed of travel relative to the rail as measured from our stationary frame of reference: we will now measure it as going at 0.433c, though to anyone moving north with the rail at 0.866c the train will still appear to be doing 0.866c from their point of view.
And there is your basic mistake: you introduced a contradiction.

Where is that contradiction?

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The speed relative to the track is given. This means that the track relative to the train is given as the same thing, there is no conversion for that speed. If you were doing some sort of calculation based on the operation of the engine, then you would have to do a conversion with time dilation taken into account.

Do you understand the idea of picking a frame of reference for the analysis and sticking rigorously to it for all measurements? Let's call our chosen frame Frame A. If our rail is stationary in Frame A and our train is moving at 0.866c along the rail, there is no difficulty in labelling the two things with the speeds zero (for the rail) and 0.866c. Let's keep that rail and train so we can refer to them again, so we'll call them Rail A and Train A.

Let's now introduce Rail B (aligned west-east) and we'll have it moving northwards at 0.866c. We are still analysing it from Frame A, so a clock sitting on Rail B, and let's call this Clock B, will be ticking once for every two ticks of our clock, Clock A. If we have a distance marked out on the rail and watch train B (or one carriage of it) cover that distance, Train B may cover that distance along Rail B in one tick of Clock B, but that will be two ticks of our Clock A. We determine that in frame A, Train A is moving twice as fast along Rail A as Train B is moving along Rail B, so Train B's speed is 0.433c.

There is no error in that other than my rounded off figures. If you calculate the sine of 60 degrees you'll get a more accurate figure for Train A's speed eastwards, and it's the same figure for Rail B's speed northwards - this speed leads to length contraction to precisely a half and also slows clocks down to tick at exactly half their normal rate. Train B's speed eastwards will be half of that, but if anyone wanted to measure it using Frame B as the base for their measurements, they would calculate that it is doing 0.866c.

So where's this error/contradiction that you talk of? Even if there had been an error in the speed I'd chosen, it would have been irrelevant - any movement of Train B along rail B will necessarily lead to the material in Train B having more length contraction acting on it in the NS direction than there is on Rail B, and that is all you need to pin down a preferred frame of reference because the width of the train will be less than that of the rail (or the reverse if the rail is moving west faster than the train is moving east). It is only in the preferred frame that the train can move along the rail at any speed without it's width changing.
« Last Edit: 04/08/2016 22:40:49 by David Cooper »

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #8 on: 04/08/2016 22:55:08 »
There would be a small complication if you wanted to use this experiment for real to try to work out if you're moving or not. If the rail and train are the same width, that doesn't guarantee that you aren't moving north or south because the rail could be moving west at the same speed as the train is moving east, leading to them both being the same width because they'd both have the same extra length contraction acting on them, but changing the speed of either the rail or the train would show up a width difference and demonstrate that they are also moving sideways (either north or south). In such a case, it would be possible to identify a frame that's neither moving west nor east either by finding speeds for rail and train which generate a big width difference one way and then finding speeds which generate as big a difference the other way, and then you'd average the two. If no such width differences ever show up, it means you are already stationary in the north-south direction. This could then be repeated with the rail and train aligned north south and then up down to pin down the preferred frame.

In reality though, getting a train up to relativistic speed is impractical in the extreme, so how can this be turned into any kind of experiment that can actually be carried out?

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Offline alancalverd

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Re: Can a preferred frame of reference be identified?
« Reply #9 on: 04/08/2016 23:42:16 »
If we have a distance marked out on the rail and watch train B (or one carriage of it) cover that distance, Train B may cover that distance along Rail B in one tick of Clock B, but that will be two ticks of our Clock A. We determine that in frame A, Train A is moving twice as fast along Rail A as Train B is moving along Rail B, so Train B's speed is 0.433c.

There is no error in that other than my rounded off figures.

The error is that by the time train B has moved unit distance along the track, it has also receded from
the observer at A, so the information that it has reached its destination will be delayed. If you ignore half the informaton, you will obviously get a wrong result from your calculation.
helping to stem the tide of ignorance

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #10 on: 05/08/2016 00:21:20 »
Hi Alan,

If we have a distance marked out on the rail and watch train B (or one carriage of it) cover that distance, Train B may cover that distance along Rail B in one tick of Clock B, but that will be two ticks of our Clock A. We determine that in frame A, Train A is moving twice as fast along Rail A as Train B is moving along Rail B, so Train B's speed is 0.433c.

There is no error in that other than my rounded off figures.

The error is that by the time train B has moved unit distance along the track, it has also receded from
the observer at A, so the information that it has reached its destination will be delayed. If you ignore half the informaton, you will obviously get a wrong result from your calculation.

We can put an observer to the north and another to the south such that they see the action from different sides as the apparatus races towards the former observer and away from the latter, both our observers being stationary in Frame A. By doing this, we can cancel out the Doppler effect issue and determine that the train is moving relative to the track at 0.433c (as a Frame A measurement), and this should be undisputed stuff in any discussion of relativity where the standard rules are being applied (although in a case like this it's worth exploring all possible places where and error might lie because the implications of this are so extraordinary). We can also put an observer far above the apparatus at a lightyear's distance and have a light flash as the train passes each marker. The timing between the flashes for that observer will be pretty exact as Frame A timings of the event because they are travelling the same distance to reach that observer, and if we also have a middle marker with a flash when the train passes that, if our overhead observer is directly over the apparatus when that flash is sent out, the timings between the first and second flashes will be exactly the same as between the second and third, and his timing from the whole trip will be guaranteed free of any Doppler issues. My speed for the Train B's movement along Rail B must be 0.433c as measured in Frame A.

If you are determined to assign a wrong value to it though, feel free. Use any non-zero figure of your choice as the eastward vector to combine with the 0.866c nortward vector, then see what your actual speed is (but bin it and start again if it's greater than c) and calculate the direction the material of the train is actually moving in, then calculate the length contraction that must apply for that speed in that direction, and then tell me how wide the train must be. It will always be more contracted than the rail, though you will need a fair bit of speed before it shows up clearly as being a real width difference rather than just a calculator showing a 0.9999999 that actually means 1. The 0.433 that I chose shows it very well, but any other speed of reasonable magnitude will prove the point.
« Last Edit: 05/08/2016 00:27:46 by David Cooper »

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Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #11 on: 05/08/2016 02:27:41 »
Do you understand the idea of picking a frame of reference for the analysis and sticking rigorously to it for all measurements?
Yes. Unfortunately, I missed that you were using two trains moving orthogonal to each other.
Quote
Let's call our chosen frame Frame A. If our rail is stationary in Frame A and our train is moving at 0.866c along the rail, there is no difficulty in labelling the two things with the speeds zero (for the rail) and 0.866c. Let's keep that rail and train so we can refer to them again, so we'll call them Rail A and Train A.

Let's now introduce Rail B (aligned west-east) and we'll have it moving northwards at 0.866c. We are still analysing it from Frame A, so a clock sitting on Rail B, and let's call this Clock B, will be ticking once for every two ticks of our clock, Clock A. If we have a distance marked out on the rail and watch train B (or one carriage of it) cover that distance, Train B may cover that distance along Rail B in one tick of Clock B, but that will be two ticks of our Clock A. We determine that in frame A, Train A is moving twice as fast along Rail A as Train B is moving along Rail B, so Train B's speed is 0.433c.
You are already confusing things. What is "Clock A" is it any clock co-moving with the rails on which Train A is on or is it any clock co-moving with Train A? What is the frame of reference that you are actually using?

If we have a system of coordinates at rest with the tracks, then saying that the speed of the train is x in a given direction sets the speed of that train. It also sets the speed of the relevant track for a frame of reference co-moving with a given train.

If we want to consider a frame of reference co-moving with train A and think about the speed of train B, then we have a somewhat more complicated question. Especially since now train B is moving south-east in the frame co-moving with train A.
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So where's this error/contradiction that you talk of? Even if there had been an error in the speed I'd chosen, it would have been irrelevant - any movement of Train B along rail B will necessarily lead to the material in Train B having more length contraction acting on it in the NS direction than there is on Rail B, and that is all you need to pin down a preferred frame of reference because the width of the train will be less than that of the rail (or the reverse if the rail is moving west faster than the train is moving east). It is only in the preferred frame that the train can move along the rail at any speed without it's width changing.
Where is the preferred reference frame here? You are merely identifying that, when one combine two translations from frame to frame, one has to take both translations into account.

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Offline PmbPhy

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Re: Can a preferred frame of reference be identified?
« Reply #12 on: 05/08/2016 13:16:15 »
Quote from: David Cooper
Hi Pete,

Yes, I read up on it a week or two back, and I wasn't impressed.
The days of me trying to make an impression of members of forums or making attempts to correct errors they've made are long past. That last attempt that I made to correct a ridiculous error made by the newbie Lord Antares sealed if for me. Trying to correct the mistakes made by members who argue like he did in that thread was the worst waste of time that I've spent in a very long time. So when it comes to problems which have a solution such as the Ehrenfest paradox I'm only going to discuss it with those members who accept the solution, which is indeed correct. I can't see the point of rehashing physics that has already been done by first rate physicists and which is very clear and well presented.

I'm not saying that you're either right or wrong. I'm just letting you know what to expect from me on this point and in the future, that's all.

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #13 on: 05/08/2016 19:36:35 »
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Let's call our chosen frame Frame A. If our rail is stationary in Frame A and our train is moving at 0.866c along the rail, there is no difficulty in labelling the two things with the speeds zero (for the rail) and 0.866c. Let's keep that rail and train so we can refer to them again, so we'll call them Rail A and Train A.

Let's now introduce Rail B (aligned west-east) and we'll have it moving northwards at 0.866c. We are still analysing it from Frame A, so a clock sitting on Rail B, and let's call this Clock B, will be ticking once for every two ticks of our clock, Clock A. If we have a distance marked out on the rail and watch train B (or one carriage of it) cover that distance, Train B may cover that distance along Rail B in one tick of Clock B, but that will be two ticks of our Clock A. We determine that in frame A, Train A is moving twice as fast along Rail A as Train B is moving along Rail B, so Train B's speed is 0.433c.
You are already confusing things. What is "Clock A" is it any clock co-moving with the rails on which Train A is on or is it any clock co-moving with Train A? What is the frame of reference that you are actually using?

Clock A is a clock stationary in Frame A, like Rail A, so it can be imagined as sitting on Rail A just as Clock B is sitting on the moving Rail B.

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If we have a system of coordinates at rest with the tracks, then saying that the speed of the train is x in a given direction sets the speed of that train. It also sets the speed of the relevant track for a frame of reference co-moving with a given train.

If we want to consider a frame of reference co-moving with train A and think about the speed of train B, then we have a somewhat more complicated question. Especially since now train B is moving south-east in the frame co-moving with train A.

Up to now we haven't needed to do any analysis from the point of view of the frame of reference co-moving with Train A, but if we want to use it for any reason we could name it Frame A'. In the same way, we can name the frame in which Train B is stationary as Frame B', and this frame is moving at 0.968c through Frame A at an angle 26.6 degrees from north.

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So where's this error/contradiction that you talk of? Even if there had been an error in the speed I'd chosen, it would have been irrelevant - any movement of Train B along rail B will necessarily lead to the material in Train B having more length contraction acting on it in the NS direction than there is on Rail B, and that is all you need to pin down a preferred frame of reference because the width of the train will be less than that of the rail (or the reverse if the rail is moving west faster than the train is moving east). It is only in the preferred frame that the train can move along the rail at any speed without its width changing.
Where is the preferred reference frame here? You are merely identifying that, when one combine two translations from frame to frame, one has to take both translations into account.

Let's go through the whole thing looking at the four objects (two rails and two trains) from the point of view of the four different frames that we now have names for.

We start off by assuming that Frame A is the preferred frame, and if it happens to be the preferred frame, it doesn't matter what speed our train moves along the track, all length contraction on it must operate in the west-east direction and leave the train's width completely unchanged - it remains a metre wide.

If it was possible to carry out this experiment for real, we wouldn't initially know if Frame A is the preferred frame, so we could assert instead that Frame A' (the frame co-moving with Train A) is the preferred frame, and if we do that we can then assert that the rail is moving westwards at 0.866c. Again the length contraction would be acting west-east on the rail and its width will be unchanged, so we cannot tell at this stage whether Frame A or Frame A' is the preferred frame.

Let's now return to using Frame A as the base for our measurements and look at Rail B and Train B. We can see Rail B moving north at 0.866c, and we also see Train B moving along it at 0.433 relative to Rail B, but we measure all the material of Train B as moving through Frame A at 0.968c in a direction 26.6 degrees away from north. The length contraction that we measure on Rail B reduces its width to half a metre (and we're still measuring from Frame A). The length contraction on the material of the train will reduce its width to a quarter of a metre - although the length contraction is acting at an angle 26.6 degrees away from north (and 63.4 degrees away from east), this still leads to the width of the train reducing to a quarter in the north-south direction (and the same would happen with the length contracting acting at shallower angles with lower speeds because when you shorten a line drawn at an angle, both the vertical and horizontal vectors that are associated with that line will shorten by the exact same amount).

What we see then from Frame A is Rail B reduced to half a metre wide and Train B reduced to quarter of a metre wide. Before I go any further with this, I want to lock down the Euclidean metric on this in a similar way to what I did earlier with the disc. Let's put another rail in on the other side of Train B to sandwich it between the rails. This new rail is attached to Rail B by 3m poles, all aligned north-south, one metre of one end of each pole being welded across the top of Rail B, one metre of the other end of each pole being welded across the top of our new rail, and the middle metre of each pole runs over the train without touching it. We can have more poles underneath too, so the train is moving in a space a metre across between the two rails, and it's also running under the top set of our poles and over the bottom set. When we look at the apparatus now from Frame A, we see our first track contracted from a metre wide to half a metre wide, we see our poles lenth-contracted from 3 metres to 1.5m, and our new rail is contracted from one metre wide to half a metre wide. We see Train B contracted from a metre wide to quarter of a metre wide, and we see a gap quarter of a metre wide between Train B and our new rail.

We've now finished our Frame A analysis. We also looked at Rail A from the point of view of Frame A', but there's no great need to use Frame A' to analyse any of the B items (though we can do so later if anyone is keen to explore the irrelevant), so let's move on to analyse things from Frame B.

We are now co-moving with Rail B and using Frame B for our new analysis. When we measure the width of Rail B, we find it to be a metre wide. When we measure the length of the poles, we find them to be 3m long. When we measure the new rail, we find that it is a metre wide. When we measure the width of the gap between Rail B and the new rail, we find it to be a metre wide. When we measure the width of Train B, what do we find? If we find that it is a metre wide and that there is not a half-metre wide gap between it and the new rail, we are seeing something that doesn't tie in with what we saw from Frame A. If we are to see something compatible with what we saw from Frame A, we must now see a gap half a metre wide between the train and the new rail, and we must measure the train as being half a metre wide.

Bang! Relativity has just been shot dead.

Let me now spell out to you why the gap seen from Frame A cannot disappear when we look from Frame B.

How is our train attached to the rail? Let's make it some kind of mag-lev, but we'll use a T flange on each side of it to lock into the rails on either side, and the T is on its side with the sharp end attached to the train. There's a slot all the way along the side of Rail B, and the crossbar of the T is inside the rail while the stem of the T goes out through the slot and attaches to the train. The train is attached in the same manner to the new rail. (There's no actual contact, so we don't need to worry about friction or vapourisation of the material.)

What happens now? Let's return to what we see from Frame A. We are no longer going to see a gap between Train B and the new rail, so there are two options. (1) The two rails move closer together as the train accelerates up to speed and the poles bend to accommodate this, or perhaps we could make the poles telescopic so that they don't break - we can have markings on them which will show one part moving into the other and we'll know how much has become hidden. (2) The rails stay a metre apart and the train is forced to fill the whole space between them, but we can see that the length-contraction that should be acting on them is being prevented from doing so and that the train is artificially being stretched to fill the gap, to the point that it will be warping and breaking apart. We know that this must happen because we see the correct length-contraction operating on the rocket that's flying over the train, moving at 0.968c in a direction 26.6 degrees away from north, and this rocket is co-moving with the material of the train.

Either way, we have clear indications that something is up: with (1) we have poles telescoping or bending, and with (2) we have a train being ripped apart sideways by extreme forces.

Now let's return to Frame B and see what it looks like there. Do we see the poles bending or telescoping into each other? Do we see the train being ripped apart? Clearly we must see one or other of these things - it is impossible to see the train happily zipping along at a metre wide in a space a metre wide with straight, untelescoped poles and no sideways stresses acting on the train.

Relativity has broken down and the game's well and truly up.

Do I need to look at things from Frame B'? There's no need to bother, but I'll do it later if anyone thinks it's necessary - there may be a limit on how long this post can be, so it's best to stop here and post it now.

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #14 on: 05/08/2016 19:52:23 »
The days of me trying to make an impression of members of forums or making attempts to correct errors they've made are long past. That last attempt that I made to correct a ridiculous error made by the newbie Lord Antares sealed if for me. Trying to correct the mistakes made by members who argue like he did in that thread was the worst waste of time that I've spent in a very long time. So when it comes to problems which have a solution such as the Ehrenfest paradox I'm only going to discuss it with those members who accept the solution, which is indeed correct. I can't see the point of rehashing physics that has already been done by first rate physicists and which is very clear and well presented.

That's fine, but I did provide you with a precise analysis of how things must work when you form a rotating ring by sending rockets in on tangents to a circle, and they must show how a rotating ring or disc would actually behave. I also pointed out that if you make the ring a billion lightyears across, the turning forces become so small as to be irrelevant. It was a very precise, correct analysis. If what I said isn't compatible with the "correct" solution to the Ehrenfest paradox, then the "correct" solution to Ehernfest paradox needs to be looked at again. If it's compatible with it though, then what are we arguing about? I haven't studied it as deeply as you have because I was put off by Einstein's non-Euclidean voodoo at the start. Non-Euclidean voodoo does not give you permission to fail to have a rotating ring interface correctly with a Euclidean metric and to conform to the basic laws of length contraction and time dilation.

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I'm not saying that you're either right or wrong. I'm just letting you know what to expect from me on this point and in the future, that's all.

That's fine - I don't know if we even have an argument with each other here, but it would be more to the point if you moved on to discussing the thought experiment outlined in the second post of this thread where I found a way of removing all the rotation aspects from things altogether. We are now dealing with things that move exclusively in perfectly straight lines (once our trains have reached their target speeds along the rails).

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #15 on: 05/08/2016 20:01:52 »
One correction to make from earlier: when I was describing flashing lights and an observer a lightyear above the experiment, the time between the first two flashes would actually be shorter than the time between the second and third flash, so that didn't work. What works better is to bring the Frame A observer in close, but still above the plane on which the action plays out. The middle flash should occur directly under him, so a line from the point where the flash is emitted running up to this observer will be perpendicular to the plane.

There are other ways to make the measurements though, and one would be to set up a camera of pixels all over the plane on which the action takes place. Each pixel would simply record what's happening on the plane right under it, and they would all record the action by taking pictures in accordance with their own clocks, all synchronised for Frame A. To synchronise two clocks, you simply position yourself halfway between them and have one adjusted until you see them both tick at the same time. Once you've done this for trillions of clocks governing trillions of pixels, you can take photographs of the action from the Frame A perspective with no Doppler effect complications interfering. When you look at the pictures and compare the ones taken at different times, you can then calculate the speeds of all items through Frame A as measured from Frame A.

[Edit: During the course of this thread, I wrote a piece of reference-frame camera software ( www.magicschoolbook.com/science/ref-frame-camera ) which illustrates some of the thought experiments discussed here and allows them to be explored properly. It runs in JavaScript straight off the web page, so there is nothing to install. You can load the example objects or program in your own, then change the frame of reference to see how they appear from that perspective.]
« Last Edit: 04/09/2016 22:50:22 by David Cooper »

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Offline jeffreyH

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Re: Can a preferred frame of reference be identified?
« Reply #16 on: 06/08/2016 12:51:12 »
Let us consider a disc trravelling flat in the x direction along the x/y plane. Let us also consider a projectile that is fired vertically upwards. From a frame of reference considered stationary with respect to the plane the projectile takes a non vertical path with its direction at an angle to the moving disc that is less that 90 degrees in the direction of motion. If the system were moving at relativistic speeds length contraction would be in both the x and z dimensions. If the projectile had a greater velocity than the moving disc then David's point becomes apparent.
Fixation on the Einstein papers is a good definition of OCD.

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Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #17 on: 06/08/2016 15:55:23 »
Why would a train on B experience any additional length contraction in the direction of motion of A'?

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #18 on: 06/08/2016 21:43:44 »
I will now provide a better description of how the length contraction must be applied to Train B as viewed from Frame A.

First, let's consider the rocket flying over the train, moving at 0.968c in a direction 26.6 degrees off north. At this speed of travel, the rocket will be length-contracted in that direction to a quarter of its rest length. If, before launching the rocket, we paint a square on its top surface (this being a flat surface parallel to the plane we're running all the apparatus through) and we also align the rocket to point 26.6 degrees off north before we paint the square onto it, we can paint this square so that its edges (each a metre long) are aligned north-south and west east. Once we have sent the rocket off and it is moving at the target speed of 0.968c, the length contraction on it will act at the angle shown in the diagram below. The arrow shows the direction for the contraction, the amount of contraction will be to 0.25 of its rest length, and the end result is shown on the right: we see how the square will now appear as viewed from Frame A. I drew the diagram and it is not quite accurate as I don't have software capable of rotating by 26.6 degrees, contracting to 1/4 of the height and then rotating back by 26.6 degrees, so it may be a bit wider vertically than it should be, but it's good enough.

If we now draw lots of squares on both the trains, each a metre by a metre and with edges aligned north-south and west-east, all of the squares on Train B should look (when viewed from Frame A) exactly like the distorted, contracted square painted on the rocket flying overhead. Clearly they cannot be that shape though, because our rails force the two long sides to be horizontal, so the train will be crushed into a different shape to make those sides horizontal, and we'll be left with a buckled train a quarter of its rest width. In addition to that, it will either pull the two rails closer together to match, or if the poles holding the rails apart don't buckle or telescope to a shorter length, the train will be further buckled by being pulled apart to twice the width that the material of the train is comfortable with.

All of this destruction of the train is serious stuff which will render it a write-off, so what will it look like from the point of view of a Frame B observer standing on Rail B? Is the train running along happily in the metre-wide gap between the rails without its material being twisted and torn? Is that compatible with the devastated ruin of shattered metal that we see from Frame A?
« Last Edit: 06/08/2016 21:46:35 by David Cooper »

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Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #19 on: 06/08/2016 21:58:34 »
I will now provide a better description of how the length contraction must be applied to Train B as viewed from Frame A.
Except that you didn't do this. Just show us the translations that you are using. Don't introduce some new object, just show us the work.

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #20 on: 06/08/2016 22:14:35 »
Why would a train on B experience any additional length contraction in the direction of motion of A'?

The length contraction on each square drawn on the train (see my previous post) should be exactly the same as for the square painted on the rocket co-moving with it, but they are prevented from taking up that shape by the rails. Once they have warped to conform to the restrictions imposed on them by the rails, they may have lost all length-contraction in the west-east direction, and even without the rails causing the train to warp, for the train to align itself comfortably it would have to change its alignment away from west-east, making it impossible to align with the rail it's supposed to be travelling along.

The problem we have here, which Lorentz, Einstein and co. were apparently too lazy to explore (or at least to do so properly by forcing things to interface with Frame A's Euclidean metric), is simply that the length contraction acting north-south and west-east vectors are not compatible with the length contraction on the angled line which the vectors add up to. I originally spotted that problem when I was looking at the rotating disc and wondered how the material at point N could retain the required length contraction in the north-south direction without having more length contraction applied to it in that direction as a result of its extra movement eastwards, and it turned out that it couldn't - it was clearly impossible for the material of the disc to spread out north-south as much as the non-rotating discs. Having also found that I could minimise the accelerations to trivial levels simply by using bigger and bigger discs, I realised that the effect must show up in a straight-line experiment too with no rotation in it whatsoever, and that indeed turned out to be the case.

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #21 on: 06/08/2016 22:24:06 »
I will now provide a better description of how the length contraction must be applied to Train B as viewed from Frame A.
Except that you didn't do this. Just show us the translations that you are using. Don't introduce some new object, just show us the work.

I did precisely what I said. What new object did I introduce? The rocket co-moving with Train B was already there - I put it there precisely to make the point that the material in the train has to behave exactly like the material in the rocket, which means that what happens to the square drawn on the rocket should happen to the squares drawn on the train, except that the squares on the train can't behave like that because the're constrained by the rails and are warped away from that shape as a result, buckling the train.

If you want to see what length contraction to 0.25 of the original length does to a square, you can look at the diagram I provided. If you don't belive the diagram, you can make your own in Microsoft Paint in the way that I did. I drew a square angled so that its sides were about 26.6 degrees off the vertical/horizontal, then I squished the picture to a quarter of the original height. I then redrew what I saw, turning it through apx. 26.6 degrees to get a close approximation of what the square on the rocket would look like from Frame A.

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Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #22 on: 06/08/2016 22:29:34 »
The length contraction on each square drawn on the train (see my previous post) should be exactly the same as for the square painted on the rocket co-moving with it, but they are prevented from taking up that shape by the rails.
And the reason for this is? If you would actually work this out, you would not find a problem here.
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The problem we have here, which Lorentz, Einstein and co. were apparently too lazy to explore (or at least to do so properly by forcing things to interface with Frame A's Euclidean metric), is simply that the length contraction acting north-south and west-east vectors are not compatible with the length contraction on the angled line which the vectors add up to.
Who should I believe here? On the one side, I have a number of very smart individuals who have had their work checked for over a century, work that I have gone through myself at various times, and work that has been instrumental in some of the highest precision applications in all of human history. On the other hand, there is a person who will not actually work out all the details of their example.

Who should I expect is "too lazy" to present this subject properly?
« Last Edit: 07/08/2016 14:17:53 by PhysBang »

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #23 on: 07/08/2016 01:33:39 »
How have I not worked it out properly? Do you seriously imagine that the edges of the length-contracted square will still be aligned north-south and west-east after the contraction has been applied at 26.6 degrees away from north so that it will sit comfortably between the rails? Likewise, do you think the north-south component of the contraction will only be to 0.5 of the rest width? You should be able to visualise it in your head without reaching for a calculator. However, let's go through all the numbers and see if the result conforms to the diagram I drew:-

Let's name the corners of the square s top left, t top right, u bottom left and v bottom right. Before we apply the contraction, lines su and tv are aligned north-south while lines st and uv are aligned west-east. The contraction is applied at 26.6 degrees away from north (to the east of it) and reduces the length in that direction to 0.25. What angle do the lines st and uv now lie at?

Let's give the corners initial coordinates (using what I hope are convenient numbers [so you can divide any value by 4 to convert to metres if you wish]), s (0,4), t (4,4), u (0,0) and v (4,0). The contraction acts along the line from (3,4) to (1,0), and we can put a line in perpendicular to that to use as the midline for both the original square and for the new resulting shape, this midline running from (0,3) to (4,1).

To find the coordinates for point s', we can draw a line through s perpendicular to our midline, so cos(26.6)=d/1 gives us the length of this perpendicular line from point s to where it hits our midline (so d=0.923), then the vertical distance back up to the same altitude as s is calculated by cos(26.6)=a/d, so a=0.852, then we take that away from 1 and add it to 3 to get the y-coordinate for the intersection point (which is 3.147). For the x-coordinate of the intersection point, we multiply sine(26.6) by d, so the intersection point is at (-0.413,3.147). Point s' is 1/4 of the way from the intersection point to s, so s' is at (-0.31,3.36). Point v' can be calculated using the same offsets in the opposite direction, so it's (4.31,0.64).

To find the coordinates for point t', we draw a line through t perpendicular to our midline, so cos(26.6)=d/3 gives us the distance to our new intersection point, and this time d=2.682. The vertical distance back up to t is cos(26.6)=a/d, so a=2.4, then we take that away from 4 to get the y-coordinate for the intersection point (which is 1.6). For the x-coordinate of the intersection point, we multiply sin(26.6) by d and subtract from 4, so the intersection point is at (2.8,1.6). Point t' will be a quarter of the way from there to t, so t' is at (3.1,2.2). Point u' can be calculated using the same offsets in the opposite direction, so it's (0.9,1.8). Plot this out and you'll get the shape in the diagram below, which isn't far off the shape I drew before - it's a little bigger, and it's rotated a bit more.

Armed with these coordinates, we can now calculate the angles of its sides. Horizontal separation between u' and v' = 4.31 - 0.9 = 3.41; vertical separation = 1.8 - 0.64 = 1.16, so tan x = 1.16/3.41, and that means u'v' slopes down at 18.8 degrees to the horizontal. The length of u'v' is, using Pythagoras, is 3.6, but we have to divide by 4 to convert to metres, so that's 90cm. This is the correct length contraction for something moving at 0.433c, but crucially the angle is wrong and does not fit the alignment of the track. Also, The actual length west-east component of the contraction that we have on this line is to 85cm.

The horizontal separation between t' and v' is 4.31 - 3.1 = 1.21, and the vertical separation is 2.2 - 0.64 = 1.56, so the angle of t'v' is at 37.8 degrees off vertical. The length of t'v' = 1.974, which is close to the 2 required for north-south length contraction and may be out due to rounding errors (as I kept ditching digits beyond the ones I wrote down), but again the angle is wrong and the actual north-south component of the contraction here is to 39cm. Each square drawn on the train should appear the same shape as in my diagram when viewed from Frame A, but it is actually going to be forced to warp to fit the space between the two rails with enormous stresses applying to it. In the course of adapting to that space (which will destroy the structural integrity of the train), it must maintain the same area if the material isn't to be stretched overall, so simply warping it until the lines are horizontal and vertical won't do. Rotating it until the long sides are aligned with the track would give a good approximate guide as to how wide it would actually end up being once it's adapted to the space, but it will clearly be less than a third of a metre. Because we have to warp the material (and destroy its structural integrity) in order to make it conform to the space, we can argue about what the resulting width and length will end up being, but if we contrive to make the length of our buckled train 0.9m, the width will be less than a third of a metre, while if we contrive to make it half a metre wide, the length will be much shorter than the required 0.9m.

Now, you told me I'd find no problem by this point, but I think a train which has to be severely buckled to try to make it conform to the required north-south and west-east length contractions and which still fails to come close to meeting both those requirements even after you've destroyed it is more than a small problem.

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Offline jeffreyH

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Re: Can a preferred frame of reference be identified?
« Reply #24 on: 07/08/2016 01:45:43 »
Objects traveling at speeds very close to c can be considered to be approaching a Rindler horizon. So destruction of such objects would be similar to the destruction of objects due to the tidal forces near to small dense masses.
Fixation on the Einstein papers is a good definition of OCD.

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #25 on: 07/08/2016 03:53:50 »
Objects traveling at speeds very close to c can be considered to be approaching a Rindler horizon. So destruction of such objects would be similar to the destruction of objects due to the tidal forces near to small dense masses.

It's not the same: we have a rocket moving at 0.968c in the direction 26.6 degrees away from north with a square painted on it which takes up the form and alignment shown below (north being up in the picture), and it is not being put under any stresses. If we take a carriage of the train and send it like a rocket at 0.968c in the same direction (with it aligned west-east when we launch it, and without rotating it at all as we accelerate it up to speed), the squares on that carriage will also look just like the square on the rocket, again without any stresses acting on them. However, the squares on Train B which is trapped between the rails cannot take up that same shape (the shape at which the material would sit naturally without stresses being applied to it), and that's the key thing here - the train has to buckle severely to fit into the space available to it, and that will destroy its structural integrity.

The camera won't lie here either: I described a camera earlier which could take Frame A pictures to show snapshots of where everything is at given points in time by Frame A's clock. These photos is required to show the square painted on the rocket as having the shape shown below on the left. The squares on the train would, if they could spread themselves into the shape where they are not under any stresses, be that same shape too, but they cannot be if they are to appear between Rail B and the new rail. There will be extreme stesses running through the material of the train which will buckle it out of shape (and wreck it), and those same forces must act on the train even if you're viewing it while standing on Rail B and measuring everything by Frame Bs Euclidean metric instead of As [I'd normally put apostrophes in those, but I don't want them to be mistaken for the other frames A' and B' tied to the two trains] - those stresses cannot magically be absent now with Train B magically being unbuckled and in perfect condition. If it's wrecked in Frame A, it must also be wrecked in Frame B, and that means the buckling would still have to occur and it would astonish the Frame B observer, unless he has read this thread and realises that he is moving through the fabric of space at relativistic speed.
« Last Edit: 07/08/2016 04:53:30 by David Cooper »

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #26 on: 07/08/2016 06:19:40 »
I'm now going to restate the proof that a preferred frame is theoretically detectable, but this time with numbers to make it easier for people to state which parts they agree with or object to. If they agree with every point on this list, they are logically required to accept that the proof is valid. (Also, if they disagree with points 6 to 8 but agree with point 9, they are logically required to accept that the proof is valid and that they're wrong about points 6 to 8.) This time, the contraction is described more accurately.

1. We start using Frame A as the basis for our measurements. If it it happens to be the preferred frame, it doesn't matter what speed our train moves along the track - all length contraction on it must operate in the west-east direction and leave the train's width completely unchanged, so it remains a metre wide. No one should disagree with this.

2. When we look at Rail B from Frame A, we see it moving north at 0.866c (sideways) and its width is contracted to half a metre. We know that its rest width is one metre, so we are seeing it contracted to half that, and that is the correct amount of contraction for that speed. Again, no one should disagree with this.

3. From Frame A, we also see Train B moving along Rail B at 0.433 relative to Rail B, but we measure all the material of Train B as moving through Frame A at 0.968c in a direction 26.6 degrees away from north. We can see that Clock B (which counts out the time of Frame B) is ticking at half the rate of our own clock, so we can tell that the Frame B observer will measure Train B as passing through Frame B at 0.866c, even though we measure it as 0.433c. No one should disagree with any of this.

4. The length contraction that we see (from Frame A) on the rocket sent at an angle 26.6 degrees from north at a speed of 0.968c will reduce its length in that direction to a quarter. A square painted on it when it was at rest (with the rocket already aligned in the direction it was due to go in and with the square aligned with its sides north-south and west-east) will appear the shape shown in the picture attached at the end of this post. This rocket will, once it's up to speed, travel over the train, maintaining position over the same carriage at all times because the train is actually moving in the same direction and at the same speed as the rocket. No one should disagree with any of this.

5. If there are squares painted on the train too (covering the roof from side to side so that they are a metre across, these painted when the train was at rest in Frame A before the experiment began), then once they are moving through Frame A at 0.968c in the same direction as the rocket, they should appear to be the same shape as the square on the rocket if they are not under any stress to distort them away from that shape. Again, no one should disagree with that: the material must behave identically whether it's in the rocket or in the train when it is not being warped by stresses.

6. We see Rail B as being half a metre wide when we measure it from Frame A. We also see a gap half a metre wide between it and the new rail which we added to mark out the space in which the train operates. The required shape for the squares does not fit in the space between the rails and the angles are wrong. If all the material of a carriage ten metres long was to sit together without stresses on it, it would actually have to burst out through the rails to either side at an angle. No one should disagree with that.

7. In order to make the train fit the space between the rails, we have to warp it, and that will put stresses on it which will be so severe that they will destroy the structural integrity of the train. No one should disagree with that.

8. We are switching now to Frame B to analyse things again, so we are co-moving with Rail B. When we measure the width of Rail B, we find it to be a metre wide. When we measure the length of the poles, we find them to be 3m long. When we measure the new rail, we find that it is a metre wide. When we measure the width of the gap between Rail B and the new rail, we find it to be a metre wide. When we measure the width of Train B, we find that it has become warped and broken to fit in the space, and the material is under a weird stress that seems to have no cause. It is likely that the train is substantially less than a metre wide, though not impossible that it is that wide if it has been buckled sufficiently to force it to that width. No one should disagree with any of that because the buckling that we saw happen from Frame A must also happen when viewed from Frame B, though that damage will be much more puzzling when viewed from Frame B.

9. When we go up a ladder and look down on the rocket as it flies above the train (it will pass under us at a speed we measure as 0.866c), it's showing weird length contraction which doesn't fit in with what we have been taught about relativity. We know that there's a square painted on it with the edges running north-south and west-east when it was at rest, and we know that it hasn't rotated since. What we've been taught about relativity tells us that it should have been contracted into a rectangle half as long as it is wide, just like we expected to see the squares on the train, but what does the square on the rocket look like in Frame B? This is much better than looking at the train, because the train's been warped by stresses which mask the shape the squares on it should naturally be. The material of the rocket is unstressed, so it can't be warped and buckled, and that makes it perfect for proving my argument. Frame B can use our special camera too, but we'll have to change the synchronisation of the pixel clocks to make them operate correctly for taking Frame B pictures, although the camera itself remains stationary in Frame A. What happens now when it takes a photograph (having resynchronised it for Frame B) is that the northern pixels take their part of the picture after their southern neighbours in order to stretch out all objects that are co-moving with Frame B, but all the pixels in the west-east direction still take a picture at the same time as the relative coordination of those is unchanged. Crucially, what does the camera now do to the square on the rocket which is also moving sideways? It takes a picture of the most southern point first (the lowest part of the picture attached to this post below), then it takes the next horizontal line up from there, although our shape will have moved slightly to the right by then, and then it'll take then next horizontal line up from there, with our shape again having moved a little to the right, and it will keep doing this until the photo is complete. This will steepen the angle of all the lines, making two of them closer to the vertical not only by stretching out the image, but by moving the higher parts of it progressively further to the right. For the other two lines though, rather than heading for being horizontal, they actually become even more tilted, and in that we have our proof that Lorentz, Einstein and the rest failed to study this properly, because they would have had you believe that the shape should come out as a rectangle. It won't though. I'll try to calculate the actual shape tomorrow and produce accurate numbers for it, but it's too late to do it tonight (i.e. at six in the morning). With the picture it's easy though: it should be sufficient to stretch the picture vertically to twice the height, then push each row of pixels along to the right to a greater degree the higher you go, so if you do that in your mind now, you'll be able to see straight away that I'm right without having to wait for the numbers and exact shape: the lines that would supposedly end up aligned west-east will tilt further away from that angle than they do in the image we're starting with.

QED

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #27 on: 07/08/2016 18:45:20 »
It turns out that you have to move each pixel row to the right more than the row below in such a way that two of the lines end up vertical, but the other two are still sloping. I haven't crunched the numbers though, so how do I know this? Well, it's all about my ref-frame camera:-

I described earlier a camera which consists of billions of pixels spread out across a plane to photograph the action. Each pixel has its own clock, but they all tick at the same rate as each other because they're all stationary in the same frame. We initially assume our camera's in the preferred frame and synchronise all the clocks on that basis.

While writing my previous post, I realised that by changing the synchronisation of the clocks in the north-south direction, I can make my camera take Frame B pictures instead of Frame A pictures, but it's actually much better than that. If I change the synchronisation in the west-east direction, I can make it take Frame A' pictures, Frame B' pictures, and indeed pictures revealing how things look as observed from any other imaginable frame (of the non-rotating variety).

Does my camera actually work properly though? Yes. If I set it for Frame B', this being the frame co-moving with the rocket at 0.968c through Frame A, lo and behold it takes pictures of the rocket which show no length contraction on it and which show the square that we painted on it as square. When we set the camera for frame A', this being the frame co-moving with Train A, it takes perfect pictures of the squares on the roof of Train A, while turning any squares on Rail A into rectangles. When we set the camera for Frame B, we see squares of Rail B as square and squares on Rail A as rectangles, but we see squares on Train A as parallelograms with two of their sides aligned parallel to Rail A and the other two sides aligned at an angle rather than perpendicular to Rail A. This is very different from what we see of Train B with the camera set to Frame A, because in that situation we see that the squares on Train B should be the same shape as the square on the rocket, as shown in the picture attached to the end of my previous post, but they can't look like that because Rail B and the new rail prevent them from taking up that form, buckling the material of the train and writing it off.

My ref-frame camera also works as a 3D camera - you just have a matrix of pixels, each of which detects whatever is at that same point and which records it, then pictures of anything can be generated from the data at any angle showing how they should appear from the reference frame of your choice - all you'd need to do is set the clock synchronisation in the three directions, north-south, east-west and up down, and you'd simply set them for the speed at which the required frame is moving in those directions.

I wonder if anyone's already written such ref-frame camera software? Perhaps no one has, because if they had they should have spotted that there's a problem with relativity: when you set the camera to Frame B and look at the square on the rocket, it should have two sides arranged perpendicular to the track while the other two are at an angle to it. I'm guessing that any existing software used to calculate the Frame B view of the square on the rocket would make an error and show it as a rectangle, because if it did the job properly someone should have found all this out long ago.

Now, how long is it going to be before anyone dares to stick their neck out and say they think I'm right...
« Last Edit: 07/08/2016 19:13:18 by David Cooper »

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Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #28 on: 07/08/2016 18:50:57 »
Can you actually show us the math rather than merely say things by fiat. Then we, you included, can see exactly where your errors are.

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #29 on: 07/08/2016 19:15:51 »
You should be able to see that I'm right without the numbers: the argument is more than clear enough. I've given you numbers for the shape of the contracted square on the rocket, and that should be enough in itself - the squares on Train B should look the same as that when measured from Frame A, but they can't take up that shape because the rails get in the way and warp them. What's stopping you seeing that!

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #30 on: 07/08/2016 19:31:45 »
Here are three simple things you can do to see that I'm right.

(a) Calculate the shape of a square co-moving with Train A as observed by a Frame B observer.

(b) Calculate the shape of a square co-moving with Train B as observed by a Frame A observer.

(c) Calculate the shape of the square on the rocket as observed by a Frame A observer.

Then ask yourself, do your shapes for (b) and (c) match. If not, you've got a contradiction. Then ask if your shape for (c) matches mine. If it doesn't, you're breaking the rules of length contraction. Once you've recognised that my shape for (c) is correct and that (b) must have the same shape, your final task is to compare (a) with (c) and to ask if they are equivalent. With (a) we have a parallelogram with 2 edges aligned parallel to Rail A, but with (b) and (c) we have a parallelogram with no edges aligned parallel to the track.

That should be enough for anyone competent at maths/physics to recognise that I'm right.

Is there really no software that physics experts can use where they put in a shape, define its frame of reference, then define a frame of reference to observe from, and then see at a glance what happens to that shape?
« Last Edit: 07/08/2016 23:27:25 by David Cooper »

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Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #31 on: 07/08/2016 23:04:21 »
You should be able to see that I'm right without the numbers: the argument is more than clear enough.
See, that's crank reasoning. You are claiming that everyone in the world, up to you, has been missing a simple problem that invalidates all of the reasoning in relativity theory. Yet you do not want to take the time to go carefully through your example to show that your numbers work out.

If you won't do the work, then I can relax and trust in the history of science since 1905. I can draw the entirely reasonable conclusion that you are making a basic error of reasoning. I won't be alone in drawing this conclusion, either. That you call everyone who every worked on relativity theory "lazy" for missing this supposed problem but you won't bother to work through the details yourself speaks to your character, not theirs.

If you want your Nobel prize, then work through the details.

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #32 on: 07/08/2016 23:46:11 »
I don't understand your attitude. I've told you exactly what to look at in post #30, but no, you can't be bothered to look. I've shown you the shape of the square on the rocket as seen from Frame A (and given you the numbers that I calculated for it), and I've explained that any unstressed square co-moving with Train B must look the same as that from Frame A. Do you at any stage get to the point where you agree with those two things? No. Why not?

There is only one more thing to do, and that is to calculate what unstressed squares travelling with Train A look like from Frame B. I have told you what shape these will be (two of their sides must be parallel to Rail A) and I have told you that this is not equivalent to the supposedly equivalent case in which we look at the shape of unstressed squares co-moving with Train B as viewed from Frame A. Are you prepared to accept that this would prove the case. No. Why ever not?

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Yet you do not want to take the time to go carefully through your example to show that your numbers work out.

There are ways to prove things that use reasoning which don't require numbers on every single irrelevant issue. What counts here is extremely simple: are two edges of the shapes parallel or not parallel with rails. I've shown you a picture of the distorted square on the rocket (and shown you the numbers I used to calculate its shape) and it is manifestly obvious that none of its edges are parallel with rail B as viewed from Rail A. No one sane should need any more numbers on that point to recognise that this is the case.

The only part I haven't put numbers on is the shape of squares co-moving with Train A as viewed from Frame B. If I do that for you, will you accept the proof, and if not, why not?

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #33 on: 08/08/2016 00:22:42 »
Key point: in all cases, the squares talked of in this thought experiment have initially been aligned with their edges lined up north-south and west-east, and a no time subsequently have any of them ever been rotated.

Those with agile minds should realise already that it's impossible for two of the edges of squares co-moving with Train A to be anything other than parallel to Rail A as seen by a Frame B observer. What will have surprised them (and it certainly surprised me) is that the reverse is not the case; that squares co-moving with Train B do not have any of their sides aligned parallel to Rail B as seen by a Frame A observer. My analysis of the appearance of the square on the rocket settles the latter point entirely, and it shows up the mistake that Lorentz and Einstein made in their analysis of that, because they failed to realise that if you analyse the Frame A shape for a square moving in the direction that the one in my example does and at that speed, it must have the same shape as a square sitting on a train moving along a path like Rail B. They would jump frame to Frame B instead, then naively apply length contraction from there on the train, then jump back to Frame A, thereby producing a parallelogram with two of its sides parallel to Rail B (a mirror image of the one that a Frame B observer will see when he looks at a square on Train A), and they'd fail to recognise that it didn't match up with the rocket's square even though it is required to do so.

Absolutely shocking that such big names could make such a fundamental error and for no one to spot it until now!
« Last Edit: 08/08/2016 00:52:39 by David Cooper »

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Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #34 on: 08/08/2016 02:04:57 »
I don't understand your attitude. I've told you exactly what to look at in post #30, but no, you can't be bothered to look.
Dude, you are either outright lying or you have no clue what is going on.

Let's see the actual functions you are using along with your actual numbers. Walk us through the calculations.

Absolutely shocking that such big names could make such a fundamental error and for no one to spot it until now!
Exactly. The only reasonable conclusion is that you are making a mistake. So walk through your example with actual numbers and calculations instead of fudging things with numbers you are cutting and pasting from other sources.

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #35 on: 08/08/2016 07:26:17 »
Absolute incompetence! I don't think they can ever have looked at this at all, because if they had, they'd have found another problem:-

What is the length contraction on one of our squares on the roof of Train A as measured from Frame A (through which the train is moving east at 0.866c)? This part's easy - it contracts from one metre to half a metre. But what is the east-west length contraction seen on this square by a Frame B observer (whose frame is moving north at 0.866c relative to Frame A)? The answer is that it must still be half a metre wide east-west, even though from Frame B the train is measured as moving at 0.433c, a speed which should only be able to reduce the square's width in that direction to 90cm. The observer in Frame B will also see the square contracted to half a metre in the north-south direction because of his speed of travel in that direction, but the east and west edges of the square will run at an angle.

For relativity to hold, it should be the case that the Frame A observer will see squares on the top of Train B showing the same amount of east-west length contraction on them as the Frame B observer will see on the squares on the top of Train A. The photographs they take of these squares should be exact mirror images of each other, but they aren't. Observer A 's photo of a square co-moving with Train B shows a parallelogram with no sides parallel to Rail B. Observer B's photo of a square co-moving with Train A shows a paralellogram of much shorter length and with two of its sides aligned parallel to Rail A.

I have yet to work out one detail, and that's the angle at which the east and west sides of squares on Train A will slope in Observer B 's photo, but it is clear now that no one else has ever done this either. If they had, they'd have realised that it's impossible for the strong east-west contraction on Train A to be hidden from Observer B.

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #36 on: 08/08/2016 07:52:31 »
I don't understand your attitude. I've told you exactly what to look at in post #30, but no, you can't be bothered to look.
Dude, you are either outright lying or you have no clue what is going on.

Or, I understand this stuff and you don't.

Quote
Let's see the actual functions you are using along with your actual numbers. Walk us through the calculations.

Do you know how to apply length contraction? Are you able to take a speed like 0.866c and calculate the time dilation and length contraction that is associated with it? If you understand relativity, this should be dead easy for you and you should be able to see that the numbers I use are correct. When I say that 0.866c contracts things to half their rest length, or that 0.968c contracts them to a quarter, or that 0.433c contracts things to 0.9, or that 0.99c contracts them to 1/7, you should be able to check that without anyone holding your hand and you should immediately recognise that I know how to calculate these numbers, even though I use a different method to do so that the one you would use. (I use arcsine of the speed to calculate an angle, then cosine of that angle to calculate the time dilation and length contraction. You can try that out and compare it with the formula Lorentz came up with, and then you can ask yourself how the heck I worked out a different way of doing it that produces the same results so easily, and you can wonder what the angle half way through the calculation represents and what it's relevance is.)

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The only reasonable conclusion is that you are making a mistake. So walk through your example with actual numbers and calculations instead of fudging things with numbers you are cutting and pasting from other sources.

Why would I need to cut and paste numbers when they're so ridiculously easy to compute? I have given you the numbers and they're easy to calculate - you should be able to check any part of what I've said with ease, but you don't appear to have a clue how to. You don't know how to work out how to contract a square to 1/4 of its rest length at an angle of 26.6 degrees, but I've done it here and given you numbers for coordinates to pin down its precise shape. You could check that with ease if you were able to hack the maths of it, but you don't even need to - I told you how I got my approximate drawing before that, and you could to the same thing to do a quick check without having to do any maths: I told you to draw a square tilted at about 26.6 degrees in Microsoft Paint (which is free with Windows) and to use the stretch function to reduce the height to 25%, then you'll see the same shape that I provided, and all you have to do is put the 26.6 degree rotation in to get the correct alignment for it. But no, you can't even do that. The truth of it is, you're working outside of your knowledge and pretending to understand a subject which you manifestly don't.

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Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #37 on: 08/08/2016 14:13:47 »
Dude, you are either outright lying or you have no clue what is going on.

Or, I understand this stuff and you don't.[/quote]
Yes, you understand it so well you can't do a few Lorentz transformations to justify your numbers.

Quote
Quote
Let's see the actual functions you are using along with your actual numbers. Walk us through the calculations.

Do you know how to apply length contraction? Are you able to take a speed like 0.866c and calculate the time dilation and length contraction that is associated with it? If you understand relativity, this should be dead easy for you and you should be able to see that the numbers I use are correct.
Yes, see, here's the problem: you always use the same numbers in the same sloppy fashion. You won't show your work... have you done the work?
Quote
Why would I need to cut and paste numbers when they're so ridiculously easy to compute?
Because a) you haven't done the work, b) the work you've done so far is incorrect, or c) if you were to actually go carefully through this example you would see that you were wrong.

Could if be that you know c) and this is why you refuse to do the work? Is this why you say ridiculous things like that you have discovered a mathematical error that nobody has found in over 100 years?

Quote
I told you to draw a square tilted at about 26.6 degrees in Microsoft Paint (which is free with Windows) and to use the stretch function to reduce the height to 25%, then you'll see the same shape that I provided, and all you have to do is put the 26.6 degree rotation in to get the correct alignment for it. But no, you can't even do that. The truth of it is, you're working outside of your knowledge and pretending to understand a subject which you manifestly don't.
Cranks think that physics is done in MS Paint. But that's not how physics is done.

Since I have a fair bit of home repair and other work to do today, I'll probably get back on this tomorrow.

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Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #38 on: 08/08/2016 17:12:56 »
Well, after a little messing around with the numbers, the obvious problem with the parallelogram shape came up: the relativity of simultaneity.

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #39 on: 08/08/2016 17:55:21 »
You quite clearly haven't the foggiest idea what you're talking about. In our basic scenario, we don't even need to change frame to see the problem as it shows up entirely from Frame A. Train B has different length-contraction acting on it than Rail B as soon as it starts moving relative to Rail B, and it can no longer sit in the space provided for it without stresses building up which will warp it more and more the faster it goes. You have no answer for that, and nor do the deities which you're defending in your role as part of the clergy.

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #40 on: 08/08/2016 18:46:30 »
Pretty awful that I should have to do trivial stuff this for a physics expert who can't handle the numbers in the blink of an eye, but here's a link to a page about length contraction that he might trust: https://en.wikipedia.org/wiki/Length_contraction

The formula used there is: Length = RestLength times root(1 - v^2 / c^2)

If we make c=1 and RestLength=1, this simplifies to root(1-v^2)

Let's apply this to the values I've used and see what length contraction we get on a meter-long object moving at different speeds:-

0.866c --> root(1-0.866^2) = root(1-0.75) = root(0.25) = 0.5

[Or, arcsine(0.866) = 60 degrees; cosine(60) = 0.5]

(Note: whenever I say 0.866, the more accurate figure of sin(60) can be used instead, but it won't make any practical difference to the results.)

0.9682458366 --> root(1-0.968^2) = root(1-0.9375) = root(0.0625) = 0.25

[Or, arcsine(0.9682458366 = 75.522 degrees; cosine(75.522) - 0.25]

0.433c --> root(1-0.433^2) = root(1-0.187489) = root(0.812511) = 0.901

[Or, arcsine(0.433) = 25.658 degrees; cosine(25.658) = 0.901]

There is no need to justify the speed of Frame B relative to Frame A or the speed of Train B relative to Rail B as the values I've chosen are fully possible (in a thought experiment, at least, though accelerating the material to such speeds would be costly). However, to calculate the speed of travel of the movement of the material of Train B through Frame A, we have to combine the two vectors which are 0.866 and 0.433. We do this through Pythagoras:-

v = root(0.866^2 + 0.433^2) = root(0.75+0.187489) = root(0.937489) = 0.968

Now, anyone competent who reads this thread will have taken a minute with a calculator to check that my speeds match the amount of length contraction that I've stated for them. They will also have worked out the angle at which the material moving at 0.968c through Frame A is moving in, and again this will only have taken a moment for them to do: you simply use the two vectors and a bit of standard trig:-

tan(x) = 0.433/0.866

tan(x) = 0.5

x = arctan(0.5)

x = 26.565 degrees

The next simple step is to take a square with its edges aligned north-south and east-west, then calculate its new shape once it's been length contracted to a quarter of its rest length in the direction 26.6 degrees away from north. I have done that for you, and I've given you a simple way to check my result without you having to do the maths yourself, so it takes quite some effort for an expert reading this to fail to recognise that my numbers all stack up. Again, I have attached a picture of that shape below.

That shape is the one that any square of Train B should take up because the material of the train is all moving at 26.6 degrees to north at 0.968c, and it will not fit comfortably between the rails as a result. The only way the train can remain between the rails is by having stresses applied to it to force it to remain in that space, and those forces will destroy its structural integrity.

There is nothing there that anyone sane should dispute. It is clear that Lorentz and Einstein never looked at this at all.

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #41 on: 08/08/2016 18:56:58 »
Well, after a little messing around with the numbers, the obvious problem with the parallelogram shape came up: the relativity of simultaneity.

If we take a Frame A photo of Rail B and the new rail using my ref-frame camera, it will give us a picture of two rails aligned east-west, each with its width length-contracted to half a metre and with a half metre wide space between them. The squares on the train, if not under any stress to warp them, must come out in the image looking like the picture I attached to the end of my previous post. There is no magical adjustment to its shape that can make it fit neatly between the rails because the shape I've shown is the Frame A shape for it. If a carriage is ten metres long, for its material to be unstressed it will have to have both ends embedded deep into the rails, which is technically known as a crash.

In post #23 I gave you the angle for the long side of the parallelogram, and it was 18.8 degrees to the east-west angle. If the whole train is made in one piece, and if its material is under no warping stress imposed on it by the rails, it would have to be angled through the picture at 18.8 degrees to the east-west line all the way across the image while the rails would be aligned perfectly east-west. The train would only be seen as passing through the rails and the space between the rails at one place in the picture, while in the rest it would be further north or south of that. A person standing on the rail at any point where the train is not seen to be in the right place will be able to jump into the gap between the rails without being hit by the train no matter how long he stays there - you are not going to see him bounce off the train or be smashed to pieces by it because it is manifestly not there.

Let me just take a moment to explain to anyone slow witted that if there is an impact between any object moving at any speed through Frame A and another object, both things will always appear in the same place in any photograph taken of the event, and they will do so regardless of which Frame the photographer is working from. If the train is actually between the rails the whole way along, it will appear in all pictures to be between the rails the whole way along, and it will be seen to be there in all pictures of it taken from all frames. The only way the train can be in that space though is if it is under high stress and has been destructively warped to keep it between the rails.
« Last Edit: 08/08/2016 19:19:55 by David Cooper »

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #42 on: 08/08/2016 19:25:17 »
How many times do I have to prove the case before people are prepared to come forward and say they recognise it as correct, or that they can't find any fault in it? Where are you? Where are the powerful, rational minds? We have one person so far who appears to have accepted that I have a point. Do I have to take this to the mathematicians to get some action on this? It should be on the news. Relativity has been blown out of the water and all you can say is nothing?

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Offline jeffreyH

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Re: Can a preferred frame of reference be identified?
« Reply #43 on: 08/08/2016 20:14:50 »
I need to go through this thread again but it is important to do so as this is a very important idea.
Fixation on the Einstein papers is a good definition of OCD.

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Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #44 on: 08/08/2016 20:49:43 »
Pretty awful that I should have to do trivial stuff this for a physics expert who can't handle the numbers in the blink of an eye, but here's a link to a page about length contraction that he might trust: https://en.wikipedia.org/wiki/Length_contraction
There are quite a few numbers here, and it pays to be careful, since you obviously have not been careful.

You are half-assing your efforts by just looking at length contraction. You are ignoring when elements on the train match up with the tracks. And because of this, you have a distorted picture of the relevant physics.

Probably 99% of the time that someone thinks they have a problem with relativity theory it is because they haven't taken relativity of simultaneity into account.
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There is nothing there that anyone sane should dispute. It is clear that Lorentz and Einstein never looked at this at all.
Nope, it is clear that you think you can get by with just adding in length contraction without considering the actual coordinates involved. Which is why I asked to see your calculations and you have revealed to us that you haven't actually done them.
Well, after a little messing around with the numbers, the obvious problem with the parallelogram shape came up: the relativity of simultaneity.
If we take a Frame A photo of Rail B and the new rail using my ref-frame camera, it will give us a picture of two rails aligned east-west, each with its width length-contracted to half a metre and with a half metre wide space between them. The squares on the train, if not under any stress to warp them, must come out in the image looking like the picture I attached to the end of my previous post. There is no magical adjustment to its shape that can make it fit neatly between the rails because the shape I've shown is the Frame A shape for it. If a carriage is ten metres long, for its material to be unstressed it will have to have both ends embedded deep into the rails, which is technically known as a crash.
You might be surprised to learn how no object is absolutely rigid. However, you are also ignoring when the different parts of the train are in contact with the different parts of the track. If you were to actually go through your scenario and work out the coordinates, you would find that you are misrepresenting what is "seen" in an instant of Frame A.
How many times do I have to prove the case before people are prepared to come forward and say they recognise it as correct, or that they can't find any fault in it? Where are you? Where are the powerful, rational minds? We have one person so far who appears to have accepted that I have a point. Do I have to take this to the mathematicians to get some action on this? It should be on the news. Relativity has been blown out of the water and all you can say is nothing?
All you have to do is make your case once. Just once. Yet you just haven't done the work. You think that you can get by in a relativity example by just incorporating length contraction and it doesn't work like that. I'm not going to use words like, "slow witted," but you might want to think twice before you do.

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #45 on: 08/08/2016 22:36:12 »
There are quite a few numbers here, and it pays to be careful, since you obviously have not been careful.
Obviously!

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You are half-assing your efforts by just looking at length contraction. You are ignoring when elements on the train match up with the tracks. And because of this, you have a distorted picture of the relevant physics.

Allow me to let you in on a secret: I know what I'm talking about. I also know, as do most people who are reading this thread [thanks, by the way, to the people who have now made comments in PMs], that you're just winging it and that you're horribly out of your depth. Take a look at the second interactive diagram on this page: www.magicschoolbook.com/science/relativity . This diagram shows the MMX apparatus with length contraction applied to it, and it gives you the Frame A view of something moving along at 0.866c. Notice that the vertical arm is straight and perpendicular to the direction of travel, just like Rail B. The only distortion is the length contraction, and it applies solely in the direction of travel. There are no complications - the pulses of light that you see running thorugh the apparatus move across the screen at all times at the same speed (until they are captured by the detector). This displays how things behave in the Euclidean metric of a frame of reference. If you click the "stop" button, you will get a "photograph" at a point in time where it is the exact same time at every single point in the picture (based on all pixels having clocks synchronised for Frame A, which is fully possible to do with the Frame A camera in my thought experiment). If there was a square in that diagram moving at 26.6 degrees down from the direction of the MMX apparatus at 0.968c, and if it's rest shape had its edges aligned up-down and left-right and it hasn't subsequently rotated) it would be the shape shown in the picture I keep attaching to posts (only rotated by 90 degrees), and each point of that shape would be exactly where it would appear on the screen, its coordinates in space and time being locked down precisely. If one of these snapshots showed it near to the vertical arm of the MMX, it would be clear to anyone who isn't blind that none of its edges are parallel to that arm. We're talking here about fundamental rules as to where and when things appear in a Euclidean metric for a single frame of reference, and there is no room for messing with what the diagram shows - things are exactly where they appear to be and not somewhere else. There is no leeway for any funny business to go on: objects always appear where they are supposed to be and the rules of length contraction dictate the separation of their component parts and dictates their effective shape as a consequence.

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Probably 99% of the time that someone thinks they have a problem with relativity theory it is because they haven't taken relativity of simultaneity into account.

That's wonderful - what you need to do is identify them carefully and tell them where they've made that mistake. When you're dealing with someone in the other 1%, it would be a good idea not to offer the same unthinking "solution" on the basis that you'll be right 99% of the time.

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There is nothing there that anyone sane should dispute. It is clear that Lorentz and Einstein never looked at this at all.
Nope, it is clear that you think you can get by with just adding in length contraction without considering the actual coordinates involved. Which is why I asked to see your calculations and you have revealed to us that you haven't actually done them.

Which calculations are you asking for now? What is wrong with you that you need to have more stuff spelt out to you that anyone competent should be able to understand already? Do you need a multitude of coordinates to understand the idea of a rail aligned east-west running north through Frame A at 0.866c and to visualise it? Do you need a multitude of coordinates to understand the idea of a train moving eastwards along that rail at 0.433c and to visualise it? Do you suffer from some kind of disability that I need to take into account? This is one of the simplest thought experiments you're likely to encounter, so I'm struggling to understand why it's causing you so much difficulty to get your head round it.

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You might be surprised to learn how no object is absolutely rigid.

There are tight constraints on how far you can flex most materials before they become damaged. The train is not made of rubber.

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However, you are also ignoring when the different parts of the train are in contact with the different parts of the track. If you were to actually go through your scenario and work out the coordinates, you would find that you are misrepresenting what is "seen" in an instant of Frame A.

That is an incorrect assertion. The shape of the warped square on the rocket is the shape that a square on the train must take if it has no warping stresses applied to it, and all parts of it are where that shape shows them to be at a single point in time in Frame A's Euclidean metric - no one competent should be arguing that any of those points are representations of different times for the same snapshot of events. These snapshots specifically show a single time throughout (by Frame A 's time).

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All you have to do is make your case once. Just once. Yet you just haven't done the work. You think that you can get by in a relativity example by just incorporating length contraction and it doesn't work like that. I'm not going to use words like, "slow witted," but you might want to think twice before you do.

No, it seems that I have to make it again and again, and every time I'm asked for numbers it's not good enough for you because you don't accept numbers as numbers and try to make out you haven't been given any! You don't understand this stuff and all you've done here is throw your own misunderstandings at me. You don't understand the functionality of a Euclidean metric and don't understand that the rules of length contraction and time dilation are related specifically to that kind of metric.
« Last Edit: 09/08/2016 00:21:19 by David Cooper »

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Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #46 on: 09/08/2016 05:51:33 »
Dude, I had a look at your webpage. I'm sorry.

You believe what you want to believe, I'm not going to pressure you.
« Last Edit: 09/08/2016 06:00:02 by PhysBang »

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #47 on: 09/08/2016 20:08:48 »
Dude, I had a look at your webpage. I'm sorry.

Well, it's a wee bit out of date now, but then so is every other page on the Net, and mine will need the least modification to correct it.

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You believe what you want to believe, I'm not going to pressure you.

Balls. Let's do it with balls (or circles in the diagrams, so I may use the words interchangeably) instead of squares, because that will reveal something beautifully clearly, and they don't have those troublesome corners on them to catch on anything. Importantly, at no point will these balls ever rotate - they'll retain their original alignment throughout, and each of them will have N, E, S and W points marked on them.

Let's use a circle of diameter 1m, and we'll draw one of these on our rocket while it's at rest in Frame A before sending it off on its journey. Instead of trains, we're going to send balls along between the rails, and crucially we're going to see which parts of them touch the rails and how far apart the rails are. We'll give Rail A a second rail to go with it (just like the one Rail B already has), so Rail A is a metre wide and runs east-west across the plane on which all the action takes place. The second rail, Rail A2, is also a meter wide, and the gap between the two rails is again one metre. The balls will be fired along in the gap between the rails. Whatever speed we send the balls at between rails A and A2, the points marked N and S on them will always be scraping against the rails no matter how much length contraction is acting on them, and that's because the length contraction will act exactly in the east-west direction as that's the way they're moving through space.

As before, we're going to use Rail B and B2, but we'll move them through Frame A at a new speed so that I can illustrate my method for cutting and pasting these numbers from random web pages. I want to apply length contraction to 1/8 of the rest length this time, and to do it at 45 degrees to north. So. I start with 0.125, then use arccos to get an angle, so that's 82.81924422 degrees [readers of my website will know what that angle means for the direction of light moving through the MMX on the arm that's perpendicular to the direction of travel of the apparatus], then I use sine to get the speed that I want the rocket to travel at, and this comes out at 0.9921567416c. So, having copied and pasted that into here, I can now collect the vectors (from a randomly selected Japanese website selling "Hello Kitty" knickers) by using Pythagoras: v^2 + v^2 = 0.9921^2, so I have to square the 0.9921, which gives me 0.984, then I halve that and find the square root, which is 0.70156076c. That means I want to move Rail B and B2 at 0.70156c northwards, and then I'll fire balls along it at 0.70156c (which will appear to Frame B observers as a much higher speed, but still a fully viable one, as it's the exact same speed as the Rocket which is going in the same direction as the material of those balls). The rocket will of course travel at 45 degrees off north at 0.9921c, so the circle painted on its top surface must match the shape of the balls fired along between Rail B and B2: the balls and the rocket are co-moving.

What is the Frame A shape for the circle on the rocket? See the diagram attached below (and note that I've used Microsoft Paint again to carry out the contraction to save time - it does a perfect mathematical job of this and it would be extremely unwise for anyone to criticise this method). At the top, we see a circle at rest (contained inside a square which I drew first both to ensure that the circle was a circle rather than an ellipse and to make it easy to draw in the north-south and east-west lines which mark out the points N, E, S and W on the ball). Underneath it, we see the contracted version, and because the balls running along between the rails must be the same shape and share the same alignment, I've added in the rails too as orange lines. The purple line XY running through the middle of the circle goes from one rail to the other, and it is aligned perpendicular to the rails. We can compare its length with the diameter of the uncompressed ball. If it's to fit between Rail B and B2, it needs to fit into 0.7126 of the length of the diameter of the uncompressed ball (because the rails and gap between them have each been contracted to 71cm wide by their speed of travel through Frame A), and if you measure it on your screen 64mm vs 92mm, so 92/64 = 0.695, so that's smaller, but close enough to say that the numbers need to be crunched properly to find out if it might actually be a good fit. We may have to move the rails closer together for the balls to touch them, so that is something that Frame B physicists may need have to do with Rail B and B2, and they won't understand why unless they've read this thread - that is something I haven't checked yet. Of course, if we use anything longer than the ball in the east-west direction, it will stick out into the rails at both sides, so this question is a side issue. More to the point is where the rails contact the balls.

The circle on the rocket and the balls running in the gap between Rail B and B2 have been contracted to an eighth of their rest width in the NE-SW direction, but are not contracted at all in the NW-SE direction. Take careful notice of the points N and S marked on the contracted ball. Will those points on the balls be in contact with the rails in the way they are on balls moving between Rail A and A2? The answer is no - the points of contact have migrated far away from there. In the preferred frame of reference, points N and S on each ball contact the rails, but in frames moving at relativistic speeds, points N and S are not the points on the balls' surfaces nearest to the rails, so they cannot touch the sides.

What we see here again, very clearly indeed, is that different frames of reference are not equivalent: objects behave differently in different frames.
« Last Edit: 09/08/2016 21:10:52 by David Cooper »

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Offline David Cooper

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Re: Can a preferred frame of reference be identified?
« Reply #48 on: 09/08/2016 20:32:59 »
So, now I need some advice. (1) Which Journal should I send this to? (2) How do I know they're going to look at it if I don't have a string of letters after my name? And (3) Do I have to put it into an impenetrable form in order to make it impossible for ordinary mortals to follow it, or is it socially acceptable to submit a paper written in normal language?

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Offline PhysBang

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Re: Can a preferred frame of reference be identified?
« Reply #49 on: 09/08/2016 21:29:06 »
If you submit it, you have to do the math. That's it.