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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.
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?
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.
This subject touches on one of the things in relativity that I've been avoiding for years because thinking about it hurts my brain. Lol! What you've touched upon here is what's known as Ehrenfest's Paradox.
It has to do with the circumference of a disk contracting while the radius remains constant which ends up leaving the value of π altered. You proposed to solve this problem by sandwiching the rotating disc between two non-rotating discs of equal size. It's your proposal that this will "eliminate all the non-Euclidean," correct? If that's the case then I believe that you made an error here by assuming that something like that can be done. Exactly what action will the two non-rotating disks have on the rotating disk?
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.
Yeah, when people say this, it means that they are making a basic mistake.
QuoteNow 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.
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.
Quote from: David Cooper on 04/08/2016 19:52:30 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 fromthe 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.
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.
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.
Hi Pete,Yes, I read up on it a week or two back, and I wasn't impressed.
QuoteLet'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.
QuoteSo 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.
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.
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.
I will now provide a better description of how the length contraction must be applied to Train B as viewed from Frame A.