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Author Topic: Can a preferred frame of reference be identified?  (Read 6497 times)

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 »
 

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
 

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 »
 

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.
 

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!
 

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 »
 

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.
 

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?

Quote
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?
 

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 »
 

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.
 

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.
 

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.)

Quote
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.
 

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.
 

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.
 

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.
 

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.
 

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 »
 

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?
 

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.
 
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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.
Quote
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.
 

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!

Quote
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.

Quote
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.

Quote
Quote
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.

Quote
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.

Quote
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).

Quote
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 »
 

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 »
 

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.

Quote
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 »
 

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?
 

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.
 

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Re: Can a preferred frame of reference be identified?
« Reply #49 on: 09/08/2016 21:29:06 »

 

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