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My point was that the one second lag in the onset of acceleration, due to the distance between them

SP1 was convinced by all evidence that SP2 was increasingly lagging behind and that SP2’s clock was running slow. Not just one second behind, but increasingly behind. SP2 was convinced by all evidence that SP1 was going too slow and SP2 was gaining on SP1 and that SP1’s clock was increasingly behind. Not until they got back into a common inertial reference frame was the illusion broken. The distance between the ships is the same as at the start and the string is unbroken.

As we will see, the rod is not always stationary in SP2’s frame.

Quote from: Halc QuoteThe point of my exercise was that at the end of the story, the two ships are found to be in a common inertial frame at their original separation and with an unbroken string between them despite their interprettion of what they are seeing.That is your goal, but you need to perform invalid mathematics to get that story. SR predicts otherwise.My math was completely valid.

QuoteThe point of my exercise was that at the end of the story, the two ships are found to be in a common inertial frame at their original separation and with an unbroken string between them despite their interprettion of what they are seeing.That is your goal, but you need to perform invalid mathematics to get that story. SR predicts otherwise.

The point of my exercise was that at the end of the story, the two ships are found to be in a common inertial frame at their original separation and with an unbroken string between them despite their interprettion of what they are seeing.

If after coming back into a common inertial frame, they each emailed a copy of their acceleration record to each other, would either one see a difference between the other’s record and his own?

Quote from: Halc link=topic=80177.msg610442#msg610442 date=1596631551I’ll do the brutal method since it illustrates the concepts you’re trying to avoid by keeping the speed low.Two ship 10 light days apart each accelerate at 100g proper acceleration for exactly 4 days as measured on ship clock. That’s today's scenario.SP1 initiates acceleration at event SP1a and shuts the engine off at SP1b. We’ve chosen SP1a event as the origin of both frames.SP2 initiates acceleration at event SP2a and shuts the engine off at SP2b.In the original frame:SP1a is at x=0,t=0 at the start (or 0,0). SP2a is at (-10, 0) Units in (light days, days)After end of acceleration,SP1b coordinates are (2.5123, 4.9082) andSP2b coordinates are (-7.4877, 4.9082). Both ships are moving at .8112c which has a Lorentz factor λ of 1.71008. The SP1 clock reads 4 at SP1b and the SP2 clock reads 4 at SP2b.At time 4.9082, The rod is still accelerating at various rates. It is stationary relative to SP2 only in SP2’s frame, so SP1 observes a rod moving outside his window at event SP1b.[/color In the post-acceleration frame:Both ships are initially moving at -.8112c at come to a halt after 4 subjective days of acceleration.The easy part. By symmetry, SP1 begins acceleration SP1a at (0, 0) (our chosen common event) and SP1b coordinates are (-2.5123, 4.9082). If you don’t use proper acceleration, the symmetry is lost and it takes much more work to compute that.For SP2 we need to do a Relativity of Simultaneity calculation to compute how far out of sync their clocks will be in this new frame.ΔT = λ * separation distance (in original frame) * Δv = 1.71008 * 10 * .8112 = 13.8721Therefore:event SP2a coordinates are (-17.1008, 13.8721) t = 0 on SP2 clock, v = -.8112event SP2b coordinates are (- 19.6131, 18.7803) t = 4 on SP2 clock, v = 0and while we’re at it, we can computewhere SP2 is at time zero: (-5.8477, 0) t=-8.112 on SP2 clock, v = -.8112At time 18.78 in this frame, both ships become stationary. SP2 clock reads 4, and SP1 clock reads 17.8721, and only at this time does the rod outside his window stop moving. When it does, he reads 17.1008 on it (the difference between their x coordinates). If there was a 10 light-day string between them, it would have broken long ago since it cannot span the 17+ light-day gap.If you see an error in the above numbers where they conflict with SR, point it out. If however you feel the numbers do not correspond to your naive beliefs, then keep it to yourself.Notice that at no point did I bother to have either ship ‘observe’ the other. The only observation done was SP1 looking at the markings on the rod in his presence.For clarity, the rod is attached to SP2 who is pushing it and it is not attached to SP as per your comments above.

I’ll do the brutal method since it illustrates the concepts you’re trying to avoid by keeping the speed low.Two ship 10 light days apart each accelerate at 100g proper acceleration for exactly 4 days as measured on ship clock. That’s today's scenario.SP1 initiates acceleration at event SP1a and shuts the engine off at SP1b. We’ve chosen SP1a event as the origin of both frames.SP2 initiates acceleration at event SP2a and shuts the engine off at SP2b.In the original frame:SP1a is at x=0,t=0 at the start (or 0,0). SP2a is at (-10, 0) Units in (light days, days)After end of acceleration,SP1b coordinates are (2.5123, 4.9082) andSP2b coordinates are (-7.4877, 4.9082). Both ships are moving at .8112c which has a Lorentz factor λ of 1.71008. The SP1 clock reads 4 at SP1b and the SP2 clock reads 4 at SP2b.At time 4.9082, The rod is still accelerating at various rates. It is stationary relative to SP2 only in SP2’s frame, so SP1 observes a rod moving outside his window at event SP1b.[/color In the post-acceleration frame:Both ships are initially moving at -.8112c at come to a halt after 4 subjective days of acceleration.The easy part. By symmetry, SP1 begins acceleration SP1a at (0, 0) (our chosen common event) and SP1b coordinates are (-2.5123, 4.9082). If you don’t use proper acceleration, the symmetry is lost and it takes much more work to compute that.For SP2 we need to do a Relativity of Simultaneity calculation to compute how far out of sync their clocks will be in this new frame.ΔT = λ * separation distance (in original frame) * Δv = 1.71008 * 10 * .8112 = 13.8721Therefore:event SP2a coordinates are (-17.1008, 13.8721) t = 0 on SP2 clock, v = -.8112event SP2b coordinates are (- 19.6131, 18.7803) t = 4 on SP2 clock, v = 0and while we’re at it, we can computewhere SP2 is at time zero: (-5.8477, 0) t=-8.112 on SP2 clock, v = -.8112At time 18.78 in this frame, both ships become stationary. SP2 clock reads 4, and SP1 clock reads 17.8721, and only at this time does the rod outside his window stop moving. When it does, he reads 17.1008 on it (the difference between their x coordinates). If there was a 10 light-day string between them, it would have broken long ago since it cannot span the 17+ light-day gap.If you see an error in the above numbers where they conflict with SR, point it out. If however you feel the numbers do not correspond to your naive beliefs, then keep it to yourself.Notice that at no point did I bother to have either ship ‘observe’ the other. The only observation done was SP1 looking at the markings on the rod in his presence.

In that case, you are wrong. You are assuming that the end of the rod will stay stationary outside SP1’s window until he stops accelerating. It will not.

The acceleration force cannot move through the rod at greater than the speed of sound in the material.

So let us assume that the rod is as strong as it needs to be and that the speed of sound in the material is the speed of light.

SP1 saw SP2 start accelerating ten days after himself.

He will also see SP2 stop accelerating ten days after himself.

At the end of the ten days, he will look out the window and see that the rod he left behind ten days earlier has now caught up with him and is where it started.

Everything is symmetric and everything will end up in the same relative positions, although all equally Lorentz contracted as per an inertial observer.

You assumed that SP1a and SP2a were simultaneous and that the rod would start moving at SP1a.

A basic lesson of Relativity Theory, which you should learn someday, is that at any non-trivial distance, the idea of even approximate simultaneity has no meaning.

Quote from: Malamute Lover on 06/08/2020 00:52:24Acceleration is absolute in that it is experienced by the accelerating observer. The magnitude of the acceleration is not absolute in that time dilation disassociates proper acceleration from that measured by an inertial frame observer.The acceleration is experience by the accelerating observer and observered by everyone else. The magnitude of the proper acceleration is absolute. The magnitude of the acceleration is indeed frame dependent.

Acceleration is absolute in that it is experienced by the accelerating observer. The magnitude of the acceleration is not absolute in that time dilation disassociates proper acceleration from that measured by an inertial frame observer.

QuoteThe traveling twin thinks he is continuing to accelerate at the same rate – which he can feel - and therefore exceeds cc is a speed. Acceleration is not. Different units. You’re equivocating the two here. It is meaningless to say that acceleration exceeds some speed.

The traveling twin thinks he is continuing to accelerate at the same rate – which he can feel - and therefore exceeds c

QuoteThis is why the pilot of the spaceship and the inertial frame observer disagree on how fast the ship is going.They disagree because the pilot always says he is stationary. One cannot have a nonzero velocity relative to ones self.

This is why the pilot of the spaceship and the inertial frame observer disagree on how fast the ship is going.

QuoteIn the roller coaster thing, the cars are being held against centripetal force by an outside track which is not rotating with them. As the cars accelerate, they push the track in the opposite direction. (Newton’s Third Law.) That is the acceleration will not be as much as expected, half of the energy going into the track. No energy goes into the track. We’re assuming lack of friction here. No work is being done by or on the track. It just guides the cars. The cars are not thrusting against the track, so the track remains stationary.

In the roller coaster thing, the cars are being held against centripetal force by an outside track which is not rotating with them. As the cars accelerate, they push the track in the opposite direction. (Newton’s Third Law.) That is the acceleration will not be as much as expected, half of the energy going into the track.

QuoteWhat is holding them in a circular orbit rather than flying out in a straight line and crashing into the track?Wheels

What is holding them in a circular orbit rather than flying out in a straight line and crashing into the track?

QuoteAs it turns out, the track is going to contract with the cars. How? Lense-Thirring frame dragging. (The names are Austrian and are pronounced LENsuh TEERing).Congrats, you managed to find yet another way to add obfuscation to this topic. This is an SR example, so you can assume negligible mass of the moving parts. Nobody considers the mass of the space ship in the twins scenario, despite the fact that that mass will affect the dilation of the ship clock. It also is assumed to be negligible.

As it turns out, the track is going to contract with the cars. How? Lense-Thirring frame dragging. (The names are Austrian and are pronounced LENsuh TEERing).

QuoteIf you do not know the rules, do not try to play the game....I know SR coldSays the guy who cannot compute the coordinates of some spaceships except using Newtonian physics.

If you do not know the rules, do not try to play the game....I know SR cold

QuoteAcceleration puts different observers in different reference frames. Time dilation is observer dependent.It makes them stationary in different reference frames. It doesn’t put them in different reference frames since you can’t enter or exit a reference frame under SR.

Acceleration puts different observers in different reference frames. Time dilation is observer dependent.

QuoteBringing two clocks into a common inertial frame will allow comparison, with the slower clock the one that experienced more acceleration. Not necessarily true. I can sync a pair of clocks, separate them and have clock A undergo 10x the acceleration of clock B, and yet clock B be the slower of the two when they’re compared in each other’s presence. It’s the temporal length of their respective worldlines that counts, not the amount of acceleration.

Bringing two clocks into a common inertial frame will allow comparison, with the slower clock the one that experienced more acceleration.

Quote from: Malamute Lover on 06/08/2020 00:59:22To an observer on the surface (and therefore to the mechanical properties of the cylinder) the same number of revolutions a minute are taking placeNow you’ve contradicted yourself again. We have a stationary marker by which a rotation can be measured. If their clocks are not running at the same pace, they necessarily measure a different time for one revolution.

To an observer on the surface (and therefore to the mechanical properties of the cylinder) the same number of revolutions a minute are taking place

QuoteNope. The reduced circumference circle is spinning faster because tangential speed is still the same but the increased time dilation makes an observer on the circumference think a revolution takes as long as if the circumference were still longer. You’re failing simple grade school arithmetic now. Angular rate is measured in say RPM which isn’t a function of radius or linear speed of any of the parts. If a rotation takes 12 seconds as measured by the central clock, then regardless of the radius and rim speed, that object is rotating at 5 RPM. If the clock at the rim runs slower and measures 10 seconds, then the object rotates at 6 RPM. It is only a function of the measured time to do one revolution, so if the two measured times are different, the measured number of revolutions a minute must be different.

Nope. The reduced circumference circle is spinning faster because tangential speed is still the same but the increased time dilation makes an observer on the circumference think a revolution takes as long as if the circumference were still longer.

QuoteAnd “a pilot always experiences being stopped”? No, the pilot experiences being accelerated to 0.99 c from his original reference frame. If he decelerates and finds himself in a matching inertial frame that is actually accelerated from the starting frame, is he stopped?A pilot is an observer observing from a position of rest in his current reference frame. That means he experiences his clock running at normal rate, and experiences his proper mass, not something 7x his prior mass.What you seem to be referring to is proper velocity, which is indeed 7c. Proper velocity is the result of integration of past acceleration, and there’s no limit to that, which is why, with a fast enough ship, I can get to the far side of the galaxy before I die.

And “a pilot always experiences being stopped”? No, the pilot experiences being accelerated to 0.99 c from his original reference frame. If he decelerates and finds himself in a matching inertial frame that is actually accelerated from the starting frame, is he stopped?

QuoteIf you were able to see the GPS clock it would be going at the same rate as your own clock because it was intentionally set to run slower.Looking at a clock known to be designed to run slower and seeing it run the same rate as my normal clock is a direct observation of objective time dilation between those clocks.

If you were able to see the GPS clock it would be going at the same rate as your own clock because it was intentionally set to run slower.

QuoteAcceleration is in the domain of General Relativity.Gravity actually

Acceleration is in the domain of General Relativity.

QuoteEinstein was not believed by everyone for some time. There are relativity deniers today, to use your phrase.If you’re not a relativity denier, then why have you not shown any relativistic mathematics when computing the coordinates of the ships in the string example? That exercise is pretty trivial, and yet all I get is low speed Newtonian math from you, which indicates a fear to expose the inconsistencies in your view. Stop it with these repetitive assertions and give workable numbers.Choose a different scenario preferably, but not one from a website where somebody else has done it. Notice that I don’t use the same one twice. You must use a scenario with relativistic dilation. Ideally, the situation should become evident with a couple digits of precision, but I did mine with around 5 digits. A speed of 100 m/sec would require more digits than my calculator has. It demonstrates nothing.I’ve put up numbers. You’ve not found any inconsistency in them except disagreements with your personal vision, which I do not accept as evidence that I did it wrong. If it’s wrong, then somewhere there will be a pair of events that don’t relate properly with fixed light speed. I can’t do that with yours because you’ve given no relativistic example to work with. You apparently don’t know your physics at all because you continue to decline to do this.

Einstein was not believed by everyone for some time. There are relativity deniers today, to use your phrase.

The traveling twin thinks he is continuing to accelerate at the same rate – which he can feel - and therefore exceeds cThe driver thinks he is continuing to accelerate at the same rate – which he can feel - and therefore exceeds the speed limit.Where is the problem?

Quote from: HalcQuoteWhat is holding them in a circular orbit rather than flying out in a straight line and crashing into the track?WheelsThe track is holding them. If the track disappeared, they would fly off in straight lines despite having wheels.

The twins paradox has no answer in SR because acceleration is involved. SR cannot deal with acceleration. There is centripetal acceleration even with a constant speed. SR cannot handle this problem. You need GR and that is what GR says will happen. If you insist on staying in SR, then you have just created another fake twins paradox with no way to answer it.

Since the situations are symmetric at the beginning and in the acceleration applied according to synchronized clocks points directly at the end results having the same appearance to the participants. Clear concepts first. No need for math in this.

like the fact that the rod attached to SP2 will not yet move at time SP1a.

Quote from: HalcQuoteBringing two clocks into a common inertial frame will allow comparison, with the slower clock the one that experienced more acceleration. Not necessarily true. I can sync a pair of clocks, separate them and have clock A undergo 10x the acceleration of clock B, and yet clock B be the slower of the two when they’re compared in each other’s presence. It’s the temporal length of their respective worldlines that counts, not the amount of acceleration.… with the one having experienced the most acceleration having less elapsed time. If you think you can set up a situation where that is not the case, show me. Don’t need to do complicated calculations. Keep it simple.

I am reminded of Poul Andersen’s novel Tau Zero from years back. A ship with a gigantic fuel supply and a hyper-efficient engine suffers a problem. The engine won’t turn off.

Not objective. Another observer in a different reference frame will observe a different a time dilation factor between you and the GPS satellite.

Even if you exclude other observers from consideration, the GPS satellite will consider you time dilated since you are moving relative to it and are deeper in a gravity well and your clock is running slower.

I have shown the inconsistency with your concepts.

...The graph shows an accelerating observer and a non-accelerating observer. What does this have to do with the subject, which is two observers in a common inertial reference frame accelerating at the same rate in the same direction beginning at an agreed on moment on synchronized clocks?In that scenario, the entire ensemble will not be aware of any contraction. An external observer in an inertial frame would observe increasing Lorentz contraction on the entire ensemble. Are you still imagining that there is an Absolute Space in which the shrinkage is happening? There isn't. Contraction is relative, not objective.

Sloppy.

In the original frame [M]:SP1a is at x=0,t=0 at the start (or 0,0). SP2a is at (-10, 0) Units in (light days, days)After end of acceleration,SP1b coordinates are (2.5123, 4.9082) andSP2b coordinates are (-7.4877, 4.9082). Both ships are moving at .8112c which has a Lorentz factor λ of 1.71008. The SP1 clock reads 4 at SP1b and the SP2 clock reads 4 at SP2b.At time 4.9082, The rod is still accelerating at various rates. It is stationary relative to SP2 only in SP2’s frame, so SP1 observes a rod moving outside his window at event SP1b.

If both clocks read 4 and they are initially synchronized and undergo identical acceleration histories, then both engines have stopped. Why should the rod still be accelerating?

You are comparing the 4.9082 elapsed time that an unaccelerated observer would see (unaccelerated with respect to the original frame of SP1 and SP2) with the 4 elapsed time that both SP1 and SP2 see.

The unaccelerated observer sees 4.9082 for both.

The accelerated observers (SP1 and SP2) both see 4. The rod is stopped relative to both SP1 and SP2 in all frames of reference.Sloppy.

QuoteIn the post-acceleration frame:Both ships are initially moving at -.8112c at come to a halt after 4 subjective days of acceleration.The easy part. By symmetry, SP1 begins acceleration SP1a at (0, 0) (our chosen common event) and SP1b coordinates are (-2.5123, 4.9082). If you don’t use proper acceleration, the symmetry is lost and it takes much more work to compute that.If you use proper acceleration the elapsed time for both is 4.

In the post-acceleration frame:Both ships are initially moving at -.8112c at come to a halt after 4 subjective days of acceleration.The easy part. By symmetry, SP1 begins acceleration SP1a at (0, 0) (our chosen common event) and SP1b coordinates are (-2.5123, 4.9082). If you don’t use proper acceleration, the symmetry is lost and it takes much more work to compute that.

If you do not use proper acceleration, that is, if you use an unaccelerated frame then the elapsed time for both is 4.9082.

QuoteFor SP2 we need to do a Relativity of Simultaneity calculation to compute how far out of sync their clocks will be in this new frame.The clocks are not out of sync at all. You are mixing two different reference frames. That does not work.

For SP2 we need to do a Relativity of Simultaneity calculation to compute how far out of sync their clocks will be in this new frame.

Quote from: Malamute Lover on 07/08/2020 14:41:36Sloppy.Ooh, I hit a nerve.

QuoteIn the original frame [M]:SP1a is at x=0,t=0 at the start (or 0,0). SP2a is at (-10, 0) Units in (light days, days)After end of acceleration,SP1b coordinates are (2.5123, 4.9082) andSP2b coordinates are (-7.4877, 4.9082). Both ships are moving at .8112c which has a Lorentz factor λ of 1.71008. The SP1 clock reads 4 at SP1b and the SP2 clock reads 4 at SP2b.At time 4.9082, The rod is still accelerating at various rates. It is stationary relative to SP2 only in SP2’s frame, so SP1 observes a rod moving outside his window at event SP1b.For brevity, I will call the two frames M and N, conveniently M (maroon) for the original frame, and N (navy) for the frame of either ship after the end of acceleration. This corresponds to the colors I used for the text.QuoteIf both clocks read 4 and they are initially synchronized and undergo identical acceleration histories, then both engines have stopped. Why should the rod still be accelerating?Your question again displays a lack of awareness of relativity of simultaneity. The event of the halting of the 17.1 LD mark on the rod (which I shall call event eW for where SP1 sees the rod stop outside his Window) is simultaneous with distant event SP2b in N, so it cannot be simultaneous with event SP2b in M. In fact, the coordinates of eW in M is (21.7559, 28.6306), so it actually takes 23.7224 days to stop in frame M after SP2 shuts down. If the rod is longer than this, extending well past SP1, then those parts have not yet stopped at that time in M.I chose my numbers (high acceleration, separation greater than distance to Rindler horizon) to highlight such effects.QuoteYou are comparing the 4.9082 elapsed time that an unaccelerated observer would see (unaccelerated with respect to the original frame of SP1 and SP2) with the 4 elapsed time that both SP1 and SP2 see.The coordinate system of M is what your stationary observer in M would see, yes. That coordinate system is unaffected by what one pair of ships happen to be doing.[QuoteThe unaccelerated observer sees 4.9082 for both.He can work out that time for both. He’s probably not present at either event, so what he sees of those events comes some time later. I expended no effort computing what anybody sees except things happening in their presence. But yes, if there was a series of unaccelerated observers with clocks synced in M, that just happen to be present at SP1b and SP2b, their respective clocks would read 4.9082 at those events.QuoteThe accelerated observers (SP1 and SP2) both see 4. The rod is stopped relative to both SP1 and SP2 in all frames of reference.Sloppy.I said no such thing, and if you look at my numbers in M, the rod is very much still moving relative to SP1 at event SP1b.The rod is stopped by definition in frame N when SP2 stops in that frame. Outside SP1’s window, event eW thus happens immediately in N, but takes over 3 weeks to happen in M. Again, your statement suggests a complete lack of awareness of relativity of simultaneity, which for somebody claiming to know even the most basic fundamentals of relativity, is not so much sloppy as plain uneducated. If two separated events (SP2b and eW) happen simultaneously in one frame, they’re not going to be simultaneous in the other. There’s exceptions to this, but not in a 1D example.QuoteQuoteIn the post-acceleration frame:Both ships are initially moving at -.8112c at come to a halt after 4 subjective days of acceleration.The easy part. By symmetry, SP1 begins acceleration SP1a at (0, 0) (our chosen common event) and SP1b coordinates are (-2.5123, 4.9082). If you don’t use proper acceleration, the symmetry is lost and it takes much more work to compute that.If you use proper acceleration the elapsed time for both is 4.The elapsed ship proper time is 4. Elapsed time time in frame N is 4.9082. I explicitly said these are N coordinates (or rather the post-acceleration frame which I’m now calling N).QuoteIf you do not use proper acceleration, that is, if you use an unaccelerated frame then the elapsed time for both is 4.9082. Frame N is an inertial reference frame, as is M. Neither are accelerated frames. The use of proper acceleration or not affects the flight plan and the symmetry of the mathematics, but has no effect on what has been defined as an inertial reference frame.If the flight plan is constant acceleration relative to M, then proper acceleration increases over time resulting in a lower average speed in the M frame than the same phase in the N frame where most of the acceleration (reduction of speed) is at the end, resulting in a higher average speed. The symmetry is lost and the elapsed time of accelration in frame M would be different than in frame N. I did not care to attempt to compute that.Also, 100g of acceleration for 4 days from a stop relative to some frame is impossible. It involves infinite proper acceleration before 4 days.QuoteQuoteFor SP2 we need to do a Relativity of Simultaneity calculation to compute how far out of sync their clocks will be in this new frame.The clocks are not out of sync at all. You are mixing two different reference frames. That does not work.Your baseless assertion, not mine. You’ve shown no self-inconsistency with my numbers, only asserting inconsistency with your fictional ones where there is no RoS, which is of consequence to me only if I can see your alternate version of all the numbers.

Quote from: Malamute Lover on 07/08/2020 03:07:20The traveling twin thinks he is continuing to accelerate at the same rate – which he can feel - and therefore exceeds cThe driver thinks he is continuing to accelerate at the same rate – which he can feel - and therefore exceeds the speed limit.Where is the problem?No problem with that. Proper velocity has no upper limit and can exceed c, as I explained above.

QuoteQuote from: HalcQuoteWhat is holding them in a circular orbit rather than flying out in a straight line and crashing into the track?WheelsThe track is holding them. If the track disappeared, they would fly off in straight lines despite having wheels.That they would. Same thing if the spokes disappeared in the spoke example.

QuoteThe twins paradox has no answer in SR because acceleration is involved. SR cannot deal with acceleration. There is centripetal acceleration even with a constant speed. SR cannot handle this problem. You need GR and that is what GR says will happen. If you insist on staying in SR, then you have just created another fake twins paradox with no way to answer it.You’re welcome to invoke GR if that’s the only way you can figure it out. SR contains Lorentz transformations between frames, and that’s all I used in my example.

QuoteSince the situations are symmetric at the beginning and in the acceleration applied according to synchronized clocks points directly at the end results having the same appearance to the participants. Clear concepts first. No need for math in this.Until you actually do the math as I did and it shows otherwise. Do the trivial math then. I gave coordinates of the four important events (SP1a/b and SP2a/b). If I gave incorrect coordinates, then correct them. Because with my numbers, a 10-LD rod length-contracted down to size 5.8477 isn’t going to bridge the gap between SP1b and SP2b which are still 10 LD apart. My string has broken. You have to have some kind of different numbers to have the string not break, but you’ve shown nothing but Newtonian math, examples with near stationary motion, plus a lot of assertions that don’t add up.

Quotelike the fact that the rod attached to SP2 will not yet move at time SP1a.Sp1a is an event, not a time. The 10 LD mark on the rod happens to be present at that event, and it indeed begins acceleration at that event, but since it has zero velocity at the moment it begins to accelerate, I’ll have to agree that it does not yet move at that event.

Quote from: HalcQuoteQuoteBringing two clocks into a common inertial frame will allow comparison, with the slower clock the one that experienced more acceleration. Not necessarily true. I can sync a pair of clocks, separate them and have clock A undergo 10x the acceleration of clock B, and yet clock B be the slower of the two when they’re compared in each other’s presence. It’s the temporal length of their respective worldlines that counts, not the amount of acceleration.… with the one having experienced the most acceleration having less elapsed time. If you think you can set up a situation where that is not the case, show me. Don’t need to do complicated calculations. Keep it simple.Clock A accelerates at 1g for however long it takes it to get to the nearest star and back. Clock B meanwhile gets put into a cyclotron on Earth and is subjected to 10g of continuous acceleration the whole time Clock A is gone. Clock A will have logged less time when they reunite, despite having accelerated far less. I didn’t do the numbers, so I can’t say how much. Can do if you think it matters. Clock B will age negligibly slower than one not in the cyclotron, depending on the radius of the thing.

QuoteQuoteBringing two clocks into a common inertial frame will allow comparison, with the slower clock the one that experienced more acceleration. Not necessarily true. I can sync a pair of clocks, separate them and have clock A undergo 10x the acceleration of clock B, and yet clock B be the slower of the two when they’re compared in each other’s presence. It’s the temporal length of their respective worldlines that counts, not the amount of acceleration.… with the one having experienced the most acceleration having less elapsed time. If you think you can set up a situation where that is not the case, show me. Don’t need to do complicated calculations. Keep it simple.

QuoteI am reminded of Poul Andersen’s novel Tau Zero from years back. A ship with a gigantic fuel supply and a hyper-efficient engine suffers a problem. The engine won’t turn off.And nobody thinks to just put the ship into a slow turn so the speed never gets stupid.

QuoteNot objective. Another observer in a different reference frame will observe a different time dilation factor between you and the GPS satellite.Oopsie. Math shows otherwise.

Not objective. Another observer in a different reference frame will observe a different time dilation factor between you and the GPS satellite.

QuoteEven if you exclude other observers from consideration, the GPS satellite will consider you time dilated since you are moving relative to it and are deeper in a gravity well and your clock is running slower.Relative to the ISS, my clock runs objectively faster, despite my continuous speed relative to it, and despite my greater gravity well depth. I’m only slower relative to GPS because the gravity difference is enough to outweigh the motion difference. GPS satellites don’t move all that fast.

QuoteQuoteQuoteQuoteAcceleration is in the domain of General Relativity.Gravity actuallyGravity and acceleration are dealt with by the same mathematical framework. Mathematically gravity is acceleration and vice versa. SR cannot handle acceleration. GR can. I have shown the inconsistency with your concepts.You’ve shown my numbers to be inconsistent with your assertions is all. Without showing light speed to be not c somewhere, or to show that events seen in one frame differ from the observations made from the same events in the other frame, or that 1+2 for one observer is different than 1+2 for the other, you’ve not demonstrated any inconsistency at all. You’ve just make empty assertions with no numbers to back them. And that’s all you’ll continue to do, because if you post ‘corrected’ numbers for my scenario, I’ll show exactly those inconsistencies.Use my scenario: 100g for 4 days. You say there’s no need for mathematics, so it should be trivial for you. Show the numbers in both frames, as I did. But you can’t.

QuoteQuoteQuoteAcceleration is in the domain of General Relativity.Gravity actuallyGravity and acceleration are dealt with by the same mathematical framework. Mathematically gravity is acceleration and vice versa. SR cannot handle acceleration. GR can. I have shown the inconsistency with your concepts.

QuoteQuoteAcceleration is in the domain of General Relativity.Gravity actuallyGravity and acceleration are dealt with by the same mathematical framework. Mathematically gravity is acceleration and vice versa. SR cannot handle acceleration. GR can.

The numbers done the right way.

SP1a is at x=0,t=0 at the start (or 0,0). SP2a is at (-10, 0) Units in (light days, days)After end of acceleration,SP1b coordinates are (2.5123, 4.9082) relative to an unaccelerated observer in the original inertial frame pf SP1 and SP2SP2b coordinates are (-7.4877, 4.9082) relative to an unaccelerated observer in the original inertial frame of SP1 and SP2Both ships are moving at .8112c which has a Lorentz factor λ of 1.71008 relative to an unaccelerated observer in the original inertial frame of SP1 and SP2

But since both ships have been accelerating at the same rate for the same time in the same direction beginning at an agreed time on synchronized clocks, they are in the same inertial frame and there is no difference in their λ.

Neither is there any difference in their proper separation from the start, although that unaccelerated observer sees the whole ensemble contracted.

The SP1 clock reads 4 at SP1b and the SP2 clock reads 4 at SP2b.

At time 4.9082 relative to an unaccelerated observer in the original inertial frame of SP1 and SP2,

which is time 4 according to synchronized clocks on the two ships, both ships and the rod are stationary relative to each other.

Since both ships have stopped accelerating at the same time, they are in a common inertial frame

Relativity of Simultaneity refers to the difference in apparent time separation between two events as viewed by two observers not in the same inertial frame.

A light is turned on at point A and at Point B at Midnight UTC (synchronized clocks). Observer Alice is on a point midway on a straight line between A and B. Alice sees the light from both A and B at the same time.

Bob is on a fast motorcycle on that straight line heading toward A and passes Alice at Midnight UTC on Alice’s watch. Bob will see the light from A earlier than the light from B. A and B send each other information about the time they turned the lights on and when they saw the light from the other. Since they have synchronized clocks, A and B will then agree that they turned their lights on at the same time, even though each saw the other’s light go on later than his own.QuoteIf they agree on the time delay value, they know they are in a common inertial frame and can even calculate the proper separation.They already know they were stationary in the one frame, so the subsequent timings just confirm again that those clocks are indeed synced in that frame, yes.

If they agree on the time delay value, they know they are in a common inertial frame and can even calculate the proper separation.

I see that you have given up on the first picture you presented that attempted to show changing real lengths after I explained how it was wrong.

Quote from: Malamute Lover on 07/08/2020 03:07:20… and therefore exceeds cAs I showed, I was not equating speed and acceleration. You were wrong about what you said but rather than admit such a thing, you switch to another subject.

… and therefore exceeds c

The subject was that the track would necessarily move in a direction opposite to the cars courtesy of Newton #3. But you skip that part because you cannot deal with the consequences.

I have addressed your math earlier and shown just how wrong it is.

The contraction of the entire ensemble, ships and rod, is only as seen by the inertial observer.

To the ships traveling at the same speeds because of identical acceleration histories, contraction does not exist because there is no relative motion.

Quote from: Halc on 07/08/2020 06:37:12Sp1a is an event, not a time.SP1a is the time on SP1’s clock when SP1 acceleration begins.

Sp1a is an event, not a time.

Quote from: HalcClock A accelerates at 1g for however long it takes it to get to the nearest star and back. Clock B meanwhile gets put into a cyclotron on Earth and is subjected to 10g of continuous acceleration the whole time Clock A is gone. Clock A will have logged less time when they reunite, despite having accelerated far less. I didn’t do the numbers, so I can’t say how much. Can do if you think it matters. Clock B will age negligibly slower than one not in the cyclotron, depending on the radius of the thing.Wrong.

Clock A accelerates at 1g for however long it takes it to get to the nearest star and back. Clock B meanwhile gets put into a cyclotron on Earth and is subjected to 10g of continuous acceleration the whole time Clock A is gone. Clock A will have logged less time when they reunite, despite having accelerated far less. I didn’t do the numbers, so I can’t say how much. Can do if you think it matters. Clock B will age negligibly slower than one not in the cyclotron, depending on the radius of the thing.

Clock B will have gone much faster and been time dilated much more than Clock A and will be way behind Clock A. It will be obvious which one had experienced more acceleration. Where they went or did not go is irrelevant as long as they end up side by side in a common inertial frame. What makes you think otherwise?

No, you showed mixed numbers, partly from the accelerated frame and partly from an inertial frame and assigned different numbers to SP1 and SP2 despite identical acceleration histories.

I have posted corrected numbers.

The observers at the end of the ruler never see any change in the length of the ruler. SP1 and SP2 never see any change in their separation and the position of the rod.

Quote from: Malamute Lover on 09/08/2020 20:58:06SP1 and SP2 never see any change in their separation and the position of the rod.This is not so straight forward.Are you suggesting a rod between the two spaceships?

SP1 and SP2 never see any change in their separation and the position of the rod.

If yes, then you need to understand/explain the post #101.Do you see where is the simultaneity line?

Quote from: Jaaanosik on 10/08/2020 12:19:19Quote from: Malamute Lover on 09/08/2020 20:58:06SP1 and SP2 never see any change in their separation and the position of the rod.This is not so straight forward.Are you suggesting a rod between the two spaceships?Yes, the rod was first suggested in post 3 and defined exactly in post 8 where I affixed it to the front of the rear ship rather than having it trail behind the lead ship, which leads to it breaking even if not attached to anything.So our scenario since post 14 has been a pair of negligible length ships (shorter than about 10 light minutes, the precision of the calculations) initially separated by 10 light days, which accelerate at 100g for 4 days ship time. The rod in front of the rear ship is 20 light days long, so it extends past the front ship, who therefore at any time can determine the separation between the two by just reading the markings on it. No need to send time-consuming light signals between the ships.QuoteIf yes, then you need to understand/explain the post #101.Do you see where is the simultaneity line?Why did you post the same picture twice in that post?It is sort of appropriate, except it depicts continuous acceleration, whereas our scenario has them stop accelerating after 4 subjective days. M-L claims that the proper distance between the ships will somehow revert to the original 10 light-day separation after they stop accelerating. Your picture shows continuous acceleration, and thus perhaps doesn't clearly show it. It also lacks numbers on the ruler, making it difficult to assess the separation in a frame not depicted.The picture BTW seems entirely accurate to me, even if it doesn't depict our exact scenario.Based on the most recent post, M-L seems in total denial of the existence of the frame in which both ships are subsequently stopped, as he refuses to compute any numbers for it. It would of course show that length contraction is real, so I suppose that's why he feels the need to take that stance.I can think of three possibilities why he would do that:1) He actually has as little clue about relativity as he presents himself to have. The only relativistic math I've seen in any post is a chart of speed-to-Lorentz-factors. Never any math applied to a specific scenario except very slow speed examples using Newtonian mathematics.2) He has a pathological fear of being shown wrong, and refuses to show math that would illustrate it.I personally love being shown to be wrong, because I'm here to learn, not to be correct.My numbers could very well have mistakes in them. I did them quickly without double-check. They need interval checks for instance, the first thing I'd do with somebody else's numbers.3) He knows his relativity quite well, but is a troll getting his jollies from yanking our collective chains.

Halc,I see your point about acceleration and stopping.

Having said that, this is not a uniform acceleration between the two spaceships as per the textbook definition.

Quote from: Jaaanosik on 10/08/2020 17:01:20Halc,I see your point about acceleration and stopping.No, not stop moving, just cease accelerating. The ships continue at the same inertial speed off to the right, not slowing back to the original frame.You picture also shows one rigid object accelerating up and down to a stop, not two separate objects accelerating identically.There's seemingly strain being put on the object in your picture. If it was brittle (not bendy), then it would perhaps break given what being done to it. Hard to tell since the speed never gets high enough to make it obvious.QuoteHaving said that, this is not a uniform acceleration between the two spaceships as per the textbook definition.Not sure what you consider uniform. Our ships accelerate at a constant proper acceleration of 100g for 4 days ship time. That's identical acceleration, not necessarily uniform, depending on your definition of uniform. If it means the object is always stationary along its length in its own frame, then yes, it illustrates that sort of motion.Your prior picture I think showed uniform acceleration, where the left side of the ruler accelerates a higher g force than the right side. Those two endpoints are not representative of the motion of our two ships.

the proper distance between the ships will somehow revert to the original 10 light-day separation after they stop accelerating.

SP1a [0,0]SP1 proper acceleration is 100gSP1 duration of proper acceleration is 4 daysSP1 final proper speed is 1.130 cSP1 proper acceleration has been continuous, so average proper speed is 0.565 cSP1 proper distance covered is 2.261 LDSP1b proper [2.261,4]

SP2a [-10,0]SP2 proper acceleration is 100gSP2 duration of proper acceleration is 4 daysSP2 final proper speed is 1.130 cSP2 proper acceleration has been continuous, so average proper speed is 0.565 cSP2 proper distance covered is 2.261 LDSP2b proper [-7.739,4]2.261 -(-7.739) = 10 LD

Same proper separation between SP1 and SP2 at end of acceleration as at start of acceleration.

The rod has been defined as having constant acceleration along its length.

SP1 end [0,0]

SP1 end proper acceleration is 100gSP1 end duration of proper acceleration is 4 daysSP1 end final proper speed is 1.130 cSP1 end proper acceleration has been continuous, so average proper speed is 0.565 cSP1 end proper distance covered is 2.261 LDSP1 end at SP1b proper [2.261,4]SP2 end [0,-10]SP2 end proper acceleration is 100gSP2 end duration of proper acceleration is 4 daysSP2 end final proper speed is 1.130 cSP2 end proper acceleration has been continuous, so average proper speed is 0.565 cSP2 end proper distance covered is 2.261 LDSP2 end at SP2b proper [-7.739,4]2.261 -(-7.739) = 10 LD

Same proper separation between SP1 end of rod and SP2 end of rod at end of acceleration as at start of acceleration. The rod is the same length at the beginning and at the end.

Let us define SP1x as Time = 2 on the SP1 clock

SP1 proper acceleration is 100gSP1 duration of proper acceleration is 2 daysSP1 proper speed is 0.565 cSP1 proper acceleration has been continuous, so average proper speed is 0.2825 cSP1 at SP1x proper distance covered is 0.565 LDSP1 at SP1x proper [0.565,2]Let us define SP2x as Time = 2 on the SP2 clockSP2 proper acceleration is 100gSP2 duration of proper acceleration is 2 daysSP2 proper speed is 0.565 cSP2 proper acceleration has been continuous, so average proper speed is 0.2825 cSP2 at SP2x proper distance covered is 0.565 LDSP2 at SP2x proper [-9.435,2]

Now the rodSP1 end proper acceleration is 100g

SP1 end duration of proper acceleration is 2 daysSP1 end final proper speed is 0.565 cSP1 end proper acceleration has been continuous, so average proper speed is 0.2825 cSP1 end at SP1x proper distance covered is 0.565 LDSP1 end at SP1x proper [0.565,2]The difference between SP1’s proper distance covered and the SP1 end of the rod say SP1x is 0.565 - 0.565 = 0

The SP1 end of the rod is motionless outside SP1’s window for any SP1 time you choose.

Let Rh be the point on the rod 5 LD forward of SP2 and 5 LD aft of SP1 at t=0 before any acceleration begins. That is, at the mid-point of the rod. Coordinates are [-5,0]

How long is the rod at the 2 day mark?

Rh proper acceleration is 100g

The proper length of the rod does not change at any point in time.

Since both ships have been accelerating at the same rate for the same time in the same direction beginning at an agreed time on synchronized clocks, they are in the same inertial frame and there will be no difference in proper lengths.

If both spaceship pause acceleration then the distance between the spaceships will not revert to the original 10 light-day interval according to the SR.

Quote from: Malamute Lover on 09/08/2020 20:09:35The numbers done the right way.All the numbers are the same as mine, if only the frame M coordinates. You offer no frame N coordinates at all.If you are correcting something, I expected different numbers. It appears you just copied my stuff, which contradicts your assertion that I had not done them correctly.We each say the other is wrong, so it is a he-said/she-said sort of situation. But numbers don’t lie. If my numbers are wrong, you need to produce different ones. Producing the same ones means you are in agreement with my assessment of the situation. My numbers show the string breaking as there is over 17 light-days of proper separation between the ships after acceleration. You’ve not shown otherwise since you don’t compute the proper distance between the ships at all.

QuoteSP1a is at x=0,t=0 at the start (or 0,0). SP2a is at (-10, 0) Units in (light days, days)After end of acceleration,SP1b coordinates are (2.5123, 4.9082) relative to an unaccelerated observer in the original inertial frame pf SP1 and SP2SP2b coordinates are (-7.4877, 4.9082) relative to an unaccelerated observer in the original inertial frame of SP1 and SP2Both ships are moving at .8112c which has a Lorentz factor λ of 1.71008 relative to an unaccelerated observer in the original inertial frame of SP1 and SP2All in agreement, and seemingly copied from my post, except for some of the rearangement of the wording.

QuoteBut since both ships have been accelerating at the same rate for the same time in the same direction beginning at an agreed time on synchronized clocks, they are in the same inertial frame and there is no difference in their λ.Agree. They both are moving at the same speed, and thus are stationary relative to each other after both have finished accelerating.

QuoteThe SP1 clock reads 4 at SP1b and the SP2 clock reads 4 at SP2b.Yes, kind of by definition, since that’s exactly the flight plan.QuoteAt time 4.9082 relative to an unaccelerated observer in the original inertial frame of SP1 and SP2,That took a whole line to say “At 4.9082 in M”. That’s why I gave the frames names, so I didn’t have to type all that each time.

Quotewhich is time 4 according to synchronized clocks on the two ships, both ships and the rod are stationary relative to each other.They’re both moving at the same speed in M at that time in M. That’s different than saying they’re stationary relative to each other, because you have to use the ship frame to state that they’re stationary at that time. I go so far as to agree that relative to the ships at event SP2b, both ships and the rod are stationary relative to each other, but I cannot agree with your statement above. Again, without numbers in N, your assertion is just words. I have no coordinates in N to work with, so I cannot show you the contradictions that result from your assertions.

QuoteSince both ships have stopped accelerating at the same time, they are in a common inertial frameThey’re ‘in’ all the frames, but each is stationary in only one. I think you meant that, but being clear is nice. It’s because they both accelerate to the same speed that makes them stationary in the same frame after acceleration. When they do it or how long they take to do it makes no difference, so long as net acceleration from M is .8112c. We’re more controlled than that of course, but them being stationary relative to each other has nothing to do with when they stopped accelerating, especially since the ‘when’ part is completely frame dependent.

QuoteRelativity of Simultaneity refers to the difference in apparent time separation between two events as viewed by two observers not in the same inertial frame.It has to do with the frame, and not the observers, and again, all observers are in all frames. But yes, something like how you word it.

QuoteA light is turned on at point A and at Point B at Midnight UTC (synchronized clocks). Observer Alice is on a point midway on a straight line between A and B. Alice sees the light from both A and B at the same time.No frame reference. I presume the clocks are stationary in the frame in which they are in sync, and Alice is also stationary in this same frame. Just making sure.

QuoteBob is on a fast motorcycle on that straight line heading toward A and passes Alice at Midnight UTC on Alice’s watch. Bob will see the light from A earlier than the light from B. A and B send each other information about the time they turned the lights on and when they saw the light from the other. Since they have synchronized clocks, A and B will then agree that they turned their lights on at the same time, even though each saw the other’s light go on later than his own.If they agree on the time delay value, they know they are in a common inertial frame and can even calculate the proper separation.They already know they were stationary in the one frame, so the subsequent timings just confirm again that those clocks are indeed synced in that frame, yes.

Bob is on a fast motorcycle on that straight line heading toward A and passes Alice at Midnight UTC on Alice’s watch. Bob will see the light from A earlier than the light from B. A and B send each other information about the time they turned the lights on and when they saw the light from the other. Since they have synchronized clocks, A and B will then agree that they turned their lights on at the same time, even though each saw the other’s light go on later than his own.If they agree on the time delay value, they know they are in a common inertial frame and can even calculate the proper separation.

You didn’t mention what Bob concludes since he doesn’t know where in his frame the light pulses originated, and therefore seems unable to assess which one flashed first. The one might have been before the other, or just closer. So what’s the purpose of Bob here? A stationary observer off to one side of Alice would similarly observe one pulse before the other. RoS seems not to have been well demonstrated with this example.

You didn’t label any events, but there are the two events where the light pulses were emitted, and various observation events. RoS says that the two emission events, if simultaneous in Alice’s frame are not simultaneous in Bob’s frame, even if that wasn’t demonstrated by your example.

QuoteNeither is there any difference in their proper separation from the start, although that unaccelerated observer sees the whole ensemble contracted.Words, which lie. Show numbers, which don’t. Without that, all I can do is assert that you’re wrong, and my numbers show it, and your lack of different numbers lends zero support to your assertion. I see no computation of their proper separation after acceleration.

QuoteI see that you have given up on the first picture you presented that attempted to show changing real lengths after I explained how it was wrong.I don’t even remember posting a picture. What post is this? Have you now switched to replies to Jaaanosik without indication?

His picture shows a continuously accelerating rigid object (a ruler apparently), much like SP2 at 100g and the first ~5 light days of rod in front of it. It shows some lines of proper simultaneity for the object, but doesn’t bother with putting units on anything. The picture seems accurate enough.I haven’t much been paying attention to what Jaaanosik has been trying to convey with this picture, but he used it in post 85 to illustrate length contraction, which it does.Notice that SP2 would be the left end of the ruler (@) accelerating at 100g, and the right end of the ruler (@') corresponds approximately to the 5 LD mark on SP2’s rod. It is not accelerating at 100g as evidenced by the different hyperbolic curves of the two lines. So the two worldlines, accelerating differently, converge in the original frame. SP1 is not like that since it follows the same curve as the left side of the ruler, so there is no contraction of the distance between them, but the rod in front of SP2 does contract, so 10 LD is insufficient to reach SP1 once they’re moving.

Quote from: Malamute Lover on 09/08/2020 23:19:07Quote from: Malamute Lover on 07/08/2020 03:07:20… and therefore exceeds cAs I showed, I was not equating speed and acceleration. You were wrong about what you said but rather than admit such a thing, you switch to another subject. I was commenting on the bit I quote just above. You didn’t say what exceeded c. Turns out you were talking about proper velocity, but apparently without knowing the term, and once that was clarified, I agreed that yes, that can exceed c.

QuoteThe subject was that the track would necessarily move in a direction opposite to the cars courtesy of Newton #3. But you skip that part because you cannot deal with the consequences.It would not since there is not just the one car, but a series of them equally spaced. The vector sum of the forces on the track is zero and it stays put. Surely you know this, and are just trying to nitpick. I was started to back off the troll assessment when I thought you were putting out numbers, but no, you just copied mine and put out comments like this one above.

QuoteI have addressed your math earlier and shown just how wrong it is.You quoted a subset of the exact same numbers, either showing how much you agree with what I’ve said, or you are unable to do the math so you just copied them. Numbers don’t lie.Then you added a bunch of words unbacked by different numbers.

QuoteThe contraction of the entire ensemble, ships and rod, is only as seen by the inertial observer.Wrong. It is contracted only relative to a frame in which it is moving. SP2 is inertial after it shuts down, and yet it is not then contracted in the SP2 frame (N), so that contradicts your statement above. The statement also implies that any accelerating observer will not see SP2 contracted, which is wrong as well. Some observer could be accelerating the other way for instance. SP2 will be contracted relative to him.

QuoteTo the ships traveling at the same speeds because of identical acceleration histories, contraction does not exist because there is no relative motion.Agree if you add ‘Relative’ in front of that statement. For instance, I claim no length contraction of the rod in frame N once SP2 has stopped accelerating. Look at the numbers. Don’t take my word for it. If the rod were contracted, a length 17.1 rod would not span from N coordinates -19.6131 to -2.5123 where SP1 is parked. It needs its full uncontracted length to reach between them. Numbers don’t lie.QuoteQuote from: Halc on 07/08/2020 06:37:12Sp1a is an event, not a time.SP1a is the time on SP1’s clock when SP1 acceleration begins.OK, so you don’t know what an event is. An event is an objective point in spacetime, a frame-invariant fact. The location and time of an event is relation with a coordinate system, typically an inertial frame.

QuoteQuote from: HalcClock A accelerates at 1g for however long it takes it to get to the nearest star and back. Clock B meanwhile gets put into a cyclotron on Earth and is subjected to 10g of continuous acceleration the whole time Clock A is gone. Clock A will have logged less time when they reunite, despite having accelerated far less. I didn’t do the numbers, so I can’t say how much. Can do if you think it matters. Clock B will age negligibly slower than one not in the cyclotron, depending on the radius of the thing.Wrong.Thank you. I’d question my statement if you agreed with me. You’re getting predictable.QuoteClock B will have gone much faster and been time dilated much more than Clock A and will be way behind Clock A. It will be obvious which one had experienced more acceleration. Where they went or did not go is irrelevant as long as they end up side by side in a common inertial frame. What makes you think otherwise? Want numbers? You first this time. Pick a real distance to a star and say a cyclotron with a 20m radius, which seems reasonable. Otherwise, just more words from you. You seem to have the mistaken impression that the magnitude of time dilation is a function of the magnitude of acceleration, so I picked a counterexample as requested.

QuoteNo, you showed mixed numbers, partly from the accelerated frame and partly from an inertial frame and assigned different numbers to SP1 and SP2 despite identical acceleration histories. They’re not mixed. They’re carefully separated in discussions relative to each frame, just like Einstein talks about the frame of the train and the frame of the platform, but with no statement mixing the two frames.All events are consistent between the two frames. That shows I did it correctly.If thinking in terms of observers helps you, just consider the N stuff to be relative to some inertial observer stationary the whole time in frame N, watching the ships go from -.8112c to a halt over the course of 4.9082 days each. All the N coordinates are relative to that observer. So if my numbers are wrong relative to that unaccelerated observer (who is at location 0 the whole time), then give your own numbers. I never once referenced an accelerated reference frame, so I used nothing but inertial SR mathematics to arrive at my numbers. Feel free to do it your own way.

QuoteI have posted corrected numbers.You mean correct numbers, not corrected, since they were my numbers.

You asked for proper numbers, as seen by SP1 and SP2.

As can be clearly seen from those numbers, at all times the proper lengths and location of SP1, the rod and SP2 never change.

You are still missing the point. The λ of SP1, the rod and SP2 are identical as seen by an unaccelerated observer.

I have provided lots of numbers about proper separation. It is identical at all times.

I was emphasizing that SP1, rod and SP2 are all in one reference frame and the observer is in another

You are insisting on a change in proper separation as seen by SP1 and SP2. So who is right, M or Z?

I have provided numbers.

They are stationary in the common inertial frame that they (SP1, rod, SP2) are all in

An observer is in the frame he is in. He is not in any other frame.

If he were in all frames, then he would have all length contractions

Not to mention that he would be traveling in all directions at all speeds.

If you want to see the other picture Jano posted here it is.It relates a fixed observer and a moving observer.

Consider the ruler to be the rod with observers at each end (SP1 and SP2).

BTW can anyone explain why the shape of the colored area first grows then shrinks as time goes on?

There is no Newtonian Absolute Space and no Newtonian Absolute Time. There are only coordinates of convenience relative to some observer.

Quote from: HalcClock A accelerates at 1g for however long it takes it to get to the nearest star and back. Clock B meanwhile gets put into a cyclotron on Earth and is subjected to 10g of continuous acceleration the whole time Clock A is gone. Clock A will have logged less time when they reunite, despite having accelerated far less.Wrong. Clock B will have gone much faster and been time dilated much more than Clock A and will be way behind Clock A.

Clock A accelerates at 1g for however long it takes it to get to the nearest star and back. Clock B meanwhile gets put into a cyclotron on Earth and is subjected to 10g of continuous acceleration the whole time Clock A is gone. Clock A will have logged less time when they reunite, despite having accelerated far less.

with the slower clock the one that experienced more acceleration.

It will be obvious which one had experienced more acceleration.

Where they went or did not go is irrelevant as long as they end up side by side in a common inertial frame.

Quote from: HalcWant numbers? You first this time. Pick a real distance to a star and say a cyclotron with a 20m radius, which seems reasonable. Otherwise, just more words from you. You seem to have the mistaken impression that the magnitude of time dilation is a function of the magnitude of acceleration, so I picked a counterexample as requested.Clocks can only be compared when they are nearby in a common inertial frame.

Want numbers? You first this time. Pick a real distance to a star and say a cyclotron with a 20m radius, which seems reasonable. Otherwise, just more words from you. You seem to have the mistaken impression that the magnitude of time dilation is a function of the magnitude of acceleration, so I picked a counterexample as requested.

To get to a given point and back at a constant g factor means accelerating to the midpoint then decelerating, then accelerating again in the other direction to the midpoint then decelerating again ending up in the place and inertial frame where the trip started. Constant acceleration for half the trip and constant deceleration for half the trip.

The acceleration will slow the onboard clock relative to an inertial observer (in a common inertial frame with Earth) as the relative speed between them increases.

The deceleration will speed up the clock relative to that inertial observer as the relative speed between them decreases eventually bringing the relative clock rate back to that of the original inertial frame. In the following scenarios, all accelerations are proper, as felt by the accelerating entity, and unless specified otherwise all times are as per an unaccelerated clock on Earth.Let us have a clock accelerate at 1 g for 1 year then decelerate at 1 g for 1 year . The clock will then reverse direction, accelerate at 1 g for 1 year then decelerate at 1 g for 1 year. To make the situation symmetrical, let us have a clock in a cyclotron accelerate at 10 g for 1 year

Let us have a clock in a cyclotron accelerate at 10 g for 1 year and decelerate at 10 g for 1 year.

The Earth clock elapsed time is 4 years.

The 1 g acceleration clock elapsed time is 3.50 years

The 10 g acceleration clock elapsed time is 1.17 years

Nope. You applied Lorentz time dilation to SP2 but not to SP1. Mixed numbers.

I corrected your erroneous mismatch in time values for SP1 and SP2. You were mixing numbers from different reference frames.