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Quote from: Halc on 31/07/2020 17:40:17...Quote from: Jaaanosik on 31/07/2020 15:31:03Is the deformation real then?The deformation into an ellipse is frame dependent, but that doesn't make it not real. M-L seems to think otherwise.QuoteThe wheel is already accelerated and it has a constant rotation.Both wheels have identical proper angular velocity. That's not the same as constant rotation. The angular velocity of an object is frame dependent since it can be used as a clock, and time is dilated in a frame in which the object is moving.QuoteIf an A1 rim observer connects a string to B1 rim observer at the bottom then is it fair to expect the string to be broken at the top?If the two wheels are different wheels in the same frame, then the string breaks same as if I attach a string between a moving car and a parked one.I don't think you mean that, but I don't know what you mean by 'A1 rim' and 'B1 rim'. They seem to be references to different objects in relative motion, which probably breaks the string.If you mean a string from one side of a wheel to the other side of the same wheel, then no, that string will not break for either wheel so long as they keep spinning. The spokes already serve as such a string.QuoteIs the (a) axle observer going to see the string broken?I cannot figure out where you are putting your string. The (a) axle observer cannot see anything different than the (b) axle observer since, per principle of relativity, linear motion cannot be locally detected. The two wheels might be the same wheel, just considered in two different inertial frames.QuoteDo we have a multiverse here?? What brings this up?QuoteThe strings not broken for (a) but broken for (b)?Your description made it sound like one string between the two wheels, so if one sees it break, the other will see the same string break.Halc,This is the bottom of the cycloid as seen from (b).Please, ignore A0 and B0, it should say A1 and B1 because the speed is close to 0 in the (b) frame, the bottom part of the cycloid, therefore there is almost no gap between the A1 and B1 rim blocks from (b) point of view. The string is attached here at the bottom when there is very tiny gap between the A1 and B1 rim blocks.The wheel makes a half a turn.The gap grows. Is the string going to break?I agree both observers will see the same result, either it breaks or it does not.There is no multiverse, two different outcomes for two different observers. Jano
...Quote from: Jaaanosik on 31/07/2020 15:31:03Is the deformation real then?The deformation into an ellipse is frame dependent, but that doesn't make it not real. M-L seems to think otherwise.QuoteThe wheel is already accelerated and it has a constant rotation.Both wheels have identical proper angular velocity. That's not the same as constant rotation. The angular velocity of an object is frame dependent since it can be used as a clock, and time is dilated in a frame in which the object is moving.QuoteIf an A1 rim observer connects a string to B1 rim observer at the bottom then is it fair to expect the string to be broken at the top?If the two wheels are different wheels in the same frame, then the string breaks same as if I attach a string between a moving car and a parked one.I don't think you mean that, but I don't know what you mean by 'A1 rim' and 'B1 rim'. They seem to be references to different objects in relative motion, which probably breaks the string.If you mean a string from one side of a wheel to the other side of the same wheel, then no, that string will not break for either wheel so long as they keep spinning. The spokes already serve as such a string.QuoteIs the (a) axle observer going to see the string broken?I cannot figure out where you are putting your string. The (a) axle observer cannot see anything different than the (b) axle observer since, per principle of relativity, linear motion cannot be locally detected. The two wheels might be the same wheel, just considered in two different inertial frames.QuoteDo we have a multiverse here?? What brings this up?QuoteThe strings not broken for (a) but broken for (b)?Your description made it sound like one string between the two wheels, so if one sees it break, the other will see the same string break.
Is the deformation real then?
The wheel is already accelerated and it has a constant rotation.
If an A1 rim observer connects a string to B1 rim observer at the bottom then is it fair to expect the string to be broken at the top?
Is the (a) axle observer going to see the string broken?
Do we have a multiverse here?
The strings not broken for (a) but broken for (b)?
Quote from: Malamute Lover on 30/07/2020 22:31:36Quote from: Jaaanosik on 30/07/2020 22:27:05I see the problem this way. The frame (a) sees the rim of the wheel symmetrically. See the figures 13.14 and 13.15.The frame (b) sees the rim of the wheel asymmetrically. See the figure 2 of the paper.Both are the inertial frame observers.The (a) and (b) observers are not on the rim itself though.If there is an accelerated observer on the rim of the wheel then this local observer will measure either symmetrical centripetal acceleration as predicted by (a) frame or asymmetrical acceleration where the spacing between 'the rim blocks' changes as predicted by (b) or ... completely something else that neither reference frame predicted.JanoAs already stated, the outside observer who sees (b) is not looking at an inertial reference frame. The spokes are going faster on top and slower on the bottom relative to overall motion.My apologies, I introduced a new scenario when the axle is accelerated. Having said that, my last couple of posts are about the textbook and the paper. There is no acceleration of the axle here, just to make it clear.Both, (a) and (b) are inertial observers looking at the rotating wheel.The question stands, is the accelerated wheel rim observer going to see/observe/measure the deformation or not.What prediction/observation wins for the accelerated observer on the wheel rim? Is it (a) or (b)?Jano
Quote from: Jaaanosik on 30/07/2020 22:27:05I see the problem this way. The frame (a) sees the rim of the wheel symmetrically. See the figures 13.14 and 13.15.The frame (b) sees the rim of the wheel asymmetrically. See the figure 2 of the paper.Both are the inertial frame observers.The (a) and (b) observers are not on the rim itself though.If there is an accelerated observer on the rim of the wheel then this local observer will measure either symmetrical centripetal acceleration as predicted by (a) frame or asymmetrical acceleration where the spacing between 'the rim blocks' changes as predicted by (b) or ... completely something else that neither reference frame predicted.JanoAs already stated, the outside observer who sees (b) is not looking at an inertial reference frame. The spokes are going faster on top and slower on the bottom relative to overall motion.
I see the problem this way. The frame (a) sees the rim of the wheel symmetrically. See the figures 13.14 and 13.15.The frame (b) sees the rim of the wheel asymmetrically. See the figure 2 of the paper.Both are the inertial frame observers.The (a) and (b) observers are not on the rim itself though.If there is an accelerated observer on the rim of the wheel then this local observer will measure either symmetrical centripetal acceleration as predicted by (a) frame or asymmetrical acceleration where the spacing between 'the rim blocks' changes as predicted by (b) or ... completely something else that neither reference frame predicted.Jano
Lorentz contraction is relative, not real.
Quote from: Malamute Lover on 31/07/2020 23:54:22Lorentz contraction is relative, not real.What about it being relative makes it not real?
If the Special Relativity is reciprocal then the Lorentz Contraction is not real.Please, let us discuss the statement above. Do we agree it is a true statement?Jano
Quote from: Jaaanosik on 01/08/2020 20:37:11If the Special Relativity is reciprocal then the Lorentz Contraction is not real.Please, let us discuss the statement above. Do we agree it is a true statement?JanoWhat do you mean by special relativity being "reciprocal"?
How about different observers seeing different things? Who is right?
Quote from: Malamute Lover on 02/08/2020 04:18:03How about different observers seeing different things? Who is right?They both are right. Someone travelling near the speed of light might experience only a few hours of the passage of time if they traveled to Alpha Centauri, but someone on Earth watching them through a telescope will see that it took them over 4 years to get there. You can't say that one is right and one is wrong in that case either. They are both correct in their own reference frames. It's the same with length contraction.
Malamute, you forgot that the energy of acceleration cannot propagate faster than the speed of light. There is a delay between the front and the back as soon as you apply a force on an object. Thus the front and the back cannot be calculated as if they were in the same frame. SR and GR are classical theories. In classical physics, any object is inside space and time, it is embedded. Space and time are represented by a continuum, it is not divisible in chunk. The only way to represent an object which is not embedded in space is to picture it by infinitely small particles separated by space so the space is still a continuum. In Newtonian physics, you can considered objects as having a fixed length because space and time are not interwoven, but space is still a continuum and object are also embedded in space, it is just that you can make abstraction of the continuum to solve the problem when the object is solid.
Does anyone see the string break?
The traveler watches a couple of hours on the Earth to take 4 years on the ship?Correct?
Quote from: Kryptid on 01/08/2020 00:42:19Quote from: Malamute Lover on 31/07/2020 23:54:22Lorentz contraction is relative, not real.What about it being relative makes it not real?How about different observers seeing different things? Who is right?