# The Naked Scientists Forum

### Author Topic: Can a preferred frame of reference be identified?  (Read 6475 times)

#### David Cooper

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##### Can a preferred frame of reference be identified?
« on: 03/08/2016 22:43:41 »
[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 »

#### 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 »

#### 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,

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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#### 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.

#### 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.
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. 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.

#### 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.

Quote
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 »

#### 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 »

#### 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.

Quote
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 »

#### 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?

#### 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.

#### 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 »

#### 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.
Quote
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.

#### 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.

#### 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.

#### 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).

#### 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 »

#### 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.

#### 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'?

#### 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 »

#### 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.

#### 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.

#### 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.

#### 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 »

#### 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.

#### 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.

#### The Naked Scientists Forum

##### Re: Can a preferred frame of reference be identified?
« Reply #24 on: 07/08/2016 01:45:43 »