The real twin paradox.
in [Relativity]
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From: G on 18 Dec 2007 12:45 On Dec 17, 7:02 am, "Jeckyl" <no...(a)nowhere.com> wrote: > "Sue..." <suzysewns...(a)yahoo.com.au> wrote in message > > news:92b09294-5ed5-438b-b06f-d56cd0c03b02(a)i72g2000hsd.googlegroups.com... > > > > > On Dec 16, 3:02 pm, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote: > > [Apparent power snipped but not forgotten. > > Thermocouple ammeters on todo list] > > >> Right. So admit that you were wrong and get on with life. > > >> > Twins can't disagree on the number of orbits > >> > Jupiter's moon makes. PERIOD! > >> > No appeal to authority can change that. > > >> Right. But they can disagree upon how long the orbits took by their > >> clocks. > > > Clocks aren't invented yet. > > We traveled back in time. :o) > > There are candles bound to conform to PoR. > > And they will confirm that less time has elapsed for the travelling twin .. > just as predicted by the application of: > > > << All inertial frames are totally equivalent > > for the performance of all physical experiments >> > >http://farside.ph.utexas.edu/teaching/jk1/lectures/node7.html > > "c" is the speed of light. > > PoR does not say experiments should allow for > > the speed of light. > > That's right. But that the speed of light is 'c' is a law of physics that > the PoR says is the same in every frame. Didn't they teach you that in > school? > > > The candles were laid side by side. > > They could have even been weighed. > > Doesn't matter .. you still get less elapsed tome for the travelling twin > > >> Einstein et al, and tested by many experiments since). > > Where was such experiment conducted free from > > local gravitational influence? > > That was obviosuly taken into account. Further, locally the speed of light > is still 'c' regardless, and experiments at one gravitational potential are > still valid. > > >> > Neither Newton nor Einstein used a theory > >> > of inertia but is clear from GR > >> > that Einstein rejected the absoulte > >> > space that you and Jeckly need to > >> > cause a *real* clock to slow with motion. > > >> The clocks on the GPS satellites have rates that are consistent with the > >> predictions of Einstein. > >> Those clocks are real enough for me. > > > The planet they orbit is real enough for me. > > And you get the same number of orbits .. as I've been saying for months now Assume the case where two spaceships leave the earth at exactly the same time, accelerate at 1g for 10 hours, and pass by two "space buoys" SpaceBouy 1 and 2 SpaceBuoy 1 Ship1<<< Earth >>>Ship2 SpaceBuoy2 at the time ship 1 and 2 pass the space buoys 1 and 2, a signal is sent to space bouy 1 and 2 giving the exact time shown by the on board clocks of each spaceship. Each clock will be dutifully running slower than the earth clock, and will indicate Ship1t1 and Ship2t1 Ship1t1 = Ship2t1 and << Spacebouy and earth clock times However since Ship1 is moving relative to Ship2 Ship1_t1 <> Ship2t_1 One may raise the objection that Ship1_t1 = Ship2_t1 only in the "stationary" earth frame of reference. However the times are communicated to the space bouys at the exact same instant for all the bodies concerend and cannot be denied. Although time aboard the spacecraft is slowed, by symmetry the times on board have to be exactly the same. And they see the same number of orbits. Is this the real twin paradox? G
From: Jeckyl on 18 Dec 2007 18:18 <paparios(a)gmail.com> wrote in message news:a56c3459-a27c-4735-96ce-5b025b5c1df6(a)i3g2000hsf.googlegroups.com... > On 18 dic, 14:45, G <gehan_ameresek...(a)hotmail.com> wrote: >> On Dec 17, 7:02 am, "Jeckyl" <no...(a)nowhere.com> wrote: >> >> > >> > > The planet they orbit is real enough for me. >> >> > And you get the same number of orbits .. as I've been saying for months >> > now >> >> Assume the case where two spaceships leave the earth at exactly the >> same time, accelerate at 1g for 10 hours, and pass by two "space >> buoys" SpaceBouy 1 and 2 >> >> SpaceBuoy 1 Ship1<<< Earth >>>Ship2 SpaceBuoy2 >> >> at the time ship 1 and 2 pass the space buoys 1 and 2, a signal is >> sent to space bouy 1 and 2 giving the exact time shown by the on board >> clocks of each spaceship. Each clock will be dutifully running slower >> than the earth clock, and will indicate Ship1t1 and Ship2t1 >> >> Ship1t1 = Ship2t1 and << Spacebouy and earth clock times >> >> However since Ship1 is moving relative to Ship2 >> >> Ship1_t1 <> Ship2t_1 >> >> One may raise the objection that Ship1_t1 = Ship2_t1 only in the >> "stationary" earth frame of reference. >> >> However the times are communicated to the space bouys at the exact >> same instant for all the bodies concerend and cannot be denied. >> Although time aboard the spacecraft is slowed, by symmetry the times >> on board have to be exactly the same. And they see the same number of >> orbits. >> >> Is this the real twin paradox? >> >> G > > While the three frames are moving away from each other (inertially), > everyone of them will see every other clock running slow. So Ship1 > sees Ship2 clock running slow and Ship2 sees Ship1 running slow and > Earth also sees both of them clocks running slow. Yes .. so far so good > The reason is that > the signals sent after passing the space bouys take a long time to > reach the other ship (and also Earth), when they are moving at > relativistic speeds. No .. that is not the reason at all .. its not just some optical illusion due to delays in the arrival of signals.
From: Jeckyl on 18 Dec 2007 18:16 "G" <gehan_ameresekere(a)hotmail.com> wrote in message news:b27f63a7-4e95-4425-805b-3c71f932e8b8(a)d4g2000prg.googlegroups.com... > Assume the case where two spaceships leave the earth at exactly the > same time, accelerate at 1g for 10 hours, and pass by two "space > buoys" SpaceBouy 1 and 2 Putting acceleration into account really makes it unnecessarily complicated. For the sake of what you're saying, you can just have the ships moving at constant velocities and that at some time the both pass the earth, and then some time later they go past the space buoy. Then you can regard the buoys, earth, spaceship etc as all inertial frames. > SpaceBuoy 1 Ship1<<< Earth >>>Ship2 SpaceBuoy2 > > at the time ship 1 and 2 pass the space buoys 1 and 2, a signal is > sent to space buoy 1 and 2 giving the exact time shown by the on board > clocks of each spaceship. Each clock will be dutifully running slower > than the earth clock, and will indicate Ship1t1 and Ship2t1 > > Ship1t1 = Ship2t1 and << Spacebouy and earth clock times Yeup .. relative to the earth .. as both have the same time dilation effect in the earth iFoR > However since Ship1 is moving relative to Ship2 > > Ship1_t1 <> Ship2t_1 Yes .. for the iFoR of each of the ships > One may raise the objection that Ship1_t1 = Ship2_t1 only in the > "stationary" earth frame of reference. Yes.. that is exactly what one argues. > However the times are communicated to the space buoys at the exact > same instant for all the bodies concerned Only in the iFoR of the earth (and probably the buoys, assuming they are stationary wrt the earth) > and cannot be denied. In the iFoR of the earth. . no you cannot deny that. > Although time aboard the spacecraft is slowed, by symmetry the times > on board have to be exactly the same. In the iFoR of the earth .. yes. > And they see the same number of > orbits. Yes .. noone is saying there are different numbers of orbits (other than my gaff earlier on where I wrote the wrong thing) > Is this the real twin paradox? No .. this is a different situation, and there is no paradox when you consider relativity of simultaneity
From: paparios on 18 Dec 2007 15:01 On 18 dic, 14:45, G <gehan_ameresek...(a)hotmail.com> wrote: > On Dec 17, 7:02 am, "Jeckyl" <no...(a)nowhere.com> wrote: > > > > > > The planet they orbit is real enough for me. > > > And you get the same number of orbits .. as I've been saying for months now > > Assume the case where two spaceships leave the earth at exactly the > same time, accelerate at 1g for 10 hours, and pass by two "space > buoys" SpaceBouy 1 and 2 > > SpaceBuoy 1 Ship1<<< Earth >>>Ship2 SpaceBuoy2 > > at the time ship 1 and 2 pass the space buoys 1 and 2, a signal is > sent to space bouy 1 and 2 giving the exact time shown by the on board > clocks of each spaceship. Each clock will be dutifully running slower > than the earth clock, and will indicate Ship1t1 and Ship2t1 > > Ship1t1 = Ship2t1 and << Spacebouy and earth clock times > > However since Ship1 is moving relative to Ship2 > > Ship1_t1 <> Ship2t_1 > > One may raise the objection that Ship1_t1 = Ship2_t1 only in the > "stationary" earth frame of reference. > > However the times are communicated to the space bouys at the exact > same instant for all the bodies concerend and cannot be denied. > Although time aboard the spacecraft is slowed, by symmetry the times > on board have to be exactly the same. And they see the same number of > orbits. > > Is this the real twin paradox? > > G While the three frames are moving away from each other (inertially), everyone of them will see every other clock running slow. So Ship1 sees Ship2 clock running slow and Ship2 sees Ship1 running slow and Earth also sees both of them clocks running slow. The reason is that the signals sent after passing the space bouys take a long time to reach the other ship (and also Earth), when they are moving at relativistic speeds. Miguel Rios
From: G on 18 Dec 2007 19:54
On Dec 19, 4:16 am, "Jeckyl" <no...(a)nowhere.com> wrote: > "G" <gehan_ameresek...(a)hotmail.com> wrote in message > > news:b27f63a7-4e95-4425-805b-3c71f932e8b8(a)d4g2000prg.googlegroups.com... > > > Assume the case where two spaceships leave the earth at exactly the > > same time, accelerate at 1g for 10 hours, and pass by two "space > > buoys" SpaceBouy 1 and 2 > > Putting acceleration into account really makes it unnecessarily complicated. > For the sake of what you're saying, you can just have the ships moving at > constant velocities and that at some time the both pass the earth, and then > some time later they go past the space buoy. Then you can regard the buoys, > earth, spaceship etc as all inertial frames. Yes, I thought of that, but I thought it unnecessarily complicated :) > > > SpaceBuoy 1 Ship1<<< Earth >>>Ship2 SpaceBuoy2 > > > at the time ship 1 and 2 pass the space buoys 1 and 2, a signal is > > sent to space buoy 1 and 2 giving the exact time shown by the on board > > clocks of each spaceship. Each clock will be dutifully running slower > > than the earth clock, and will indicate Ship1t1 and Ship2t1 > > > Ship1t1 = Ship2t1 and << Spacebouy and earth clock times > > Yeup .. relative to the earth .. as both have the same time dilation effect > in the earth iFoR > > > However since Ship1 is moving relative to Ship2 > > > Ship1_t1 <> Ship2t_1 > > Yes .. for the iFoR of each of the ships > > > One may raise the objection that Ship1_t1 = Ship2_t1 only in the > > "stationary" earth frame of reference. > > Yes.. that is exactly what one argues. > > > However the times are communicated to the space buoys at the exact > > same instant for all the bodies concerned No : the point of the experiment is that the time that the ships reach the bouys in their own respective reference frames can be calcuated using the time dilation equation. This will be the actual times on each spaceship in their own reference frame. Is there an instant that is common to all frames? If not, is there a time value that is equivalent in magnitude in each frame? In which case you will still have the inequality. Why doesnt someone calculate the values? You cannot have two different times in space ship1 and sapceship2 Also, EA hinted at "memorylessness" which may mean that when a historical trace of all events is taken and compared, there are contradictions. G > > Only in the iFoR of the earth (and probably the buoys, assuming they are > stationary wrt the earth) > > > and cannot be denied. > > In the iFoR of the earth. . no you cannot deny that. > > > Although time aboard the spacecraft is slowed, by symmetry the times > > on board have to be exactly the same. > > In the iFoR of the earth .. yes. > > > And they see the same number of > > orbits. > > Yes .. noone is saying there are different numbers of orbits (other than my > gaff earlier on where I wrote the wrong thing) > > > Is this the real twin paradox? > > No .. this is a different situation, and there is no paradox when you > consider relativity of simultaneity Simultaneity beleiving that "Information/ reality is limited by the speed of light" . When you take historical tracking of physical phenomena this disappears, I think. G |