The real twin paradox.
in [Relativity]
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From: paparios on 28 Nov 2007 21:26 On 28 nov, 21:36, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > colp says... > > > > >On Nov 29, 10:44 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) > >wrote: > >> No, they aren't. The Wiki example, as you noticed, is about > >> a case in which one of the twins remains *inertial*. In that > >> case, you can talk about time dilation for that twin's inertial > >> coordinate system. If *both* twins are accelerating, then you > >> *can't* use the simple time dilation formula. > > >The paradox does not rely on observations that are made from > >accelerating frames. > > If you calculate elapsed times using only *inertial* coordinates, > then you don't get any contradictions. So you are completely > wrong. > > >The essential part of the paradox is that when the two twins are > >travelling away from each other in inertial frames they expect to > >receive fewer clock ticks than they send. > > I don't care what they expect. The question is: what does > Special Relativity predict? It predicts that in the > symmetric case, the number of signals each twin receives > from the other is the same. > > -- > Daryl McCullough > Ithaca, NY My last effort in this subject. In the following web page (http:// www.phys.unsw.edu.au/einsteinlight/jw/module4_twin_paradox.htm), there is a good and detailed graphical explanation of the asymmetrical twin paradox. The explanation I previously gave in a post, I will try to put it into simpler words, using the diagrams of that page. Recall that Rocket 1 is departing from Earth along the +x axis at a speed v=0.6c, while Rocket 2 is going the opposite way. Let us assume both rockets depart from Earth, where a third twin is located. The rockets are moving away to a location 6 light years from Earth and then they will turn around and come back to Earth. The outway trip will last 10 years for every rocket and gamma=1.25 Every year, according to their local calendars they sent light signals to every one. The following tables show the local clock at all the twins and the signals as received from the travelling twins. In the tables 0 means no light signal is received. Local clock R1: 1-2-3-4-5-6-7-8-9-10-11-12-13-14-15-16 Reception from Earth:0-1-0-2-0-3-0-4-6-8-10-12-14-16-18-20 Reception from R2: 0-0-0-1-0-0-0-2-3-4-5-6-7-8-12-16 Local clock R2: 1-2-3-4-5-6-7-8-9-10-11-12-13-14-15-16 Reception from Earth: 0-1-0-2-0-3-0-4-6-8-10-12-14-16-18-20 Reception from R1: 0-0-0-1-0-0-0-2-3-4-5-6-7-8-12-16 Local clock Earth: 1-2-3-4-5-6-7-8-9-10-11-12-13-14-15-16-17-18-19-20 Reception from R1: 0-1-0-2-0-3-0-4-0-5-0-6-0-7-0-8-10-12-14-16 Reception from R2: 0-1-0-2-0-3-0-4-0-5-0-6-0-7-0-8-10-12-14-16 OK, now what is happening according to these numbers. a) From the frame of reference of R1, in the time interval (1,8) of its local clock, both Earth and R2 clocks are running slow, according to received light signals. So Earth first light signal arrives to R1 when R1's local clock is showing two years have passed. Similarly, R2 first light signal arrives to R1 when R1's local clock is showing four years have passed. In the time interval (9,14), R1 has turn around and is coming back to Earth, and R1 is receiving light signals sent by R2 at a rate of one per year. Thus R2 light signal sent at R2's local clock = 8 years, arrives to R1 when its local clock reads 14 years. Things speed up significantly in the time interval (15,16). There in two years of R1's local clock, eight light signals from R2 are received. b) From the frame of reference of R2, in the time interval (1,8) of its local clock, both Earth and R1 clocks are running slow. So Earth first light signal arrives to R2 when R2's local clock is showing two years have passed. Similarly, R1 first light signal arrives to R2 when R2's local clock is showing four years have passed. In the time interval (9,14), R2 has turn around and it is coming back to Earth and it is receiving light signals sent by R1 at a rate of one per year. Thus R2 light signal sent at R2's local clock = 8 years, arrives to R1 when its local clock reads 14 years. Things speed up significantly in the time interval (15,16). There in two years of R2's local clock, eight light signals from R1 are received. So we see both R1 and R2 clocks read the same 16 years at the end of the trip, while the twin who remained at Earth is reading 20 years. Both R1 and R2 have experienced time dilation of their clocks with respect to the Earth twin and no paradox is present. Miguel Rios
From: colp on 28 Nov 2007 21:35 On Nov 29, 1:36 pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > colp says... > > > > >On Nov 29, 10:44 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) > >wrote: > >> No, they aren't. The Wiki example, as you noticed, is about > >> a case in which one of the twins remains *inertial*. In that > >> case, you can talk about time dilation for that twin's inertial > >> coordinate system. If *both* twins are accelerating, then you > >> *can't* use the simple time dilation formula. > > >The paradox does not rely on observations that are made from > >accelerating frames. > > If you calculate elapsed times using only *inertial* coordinates, > then you don't get any contradictions. So you are completely > wrong. The contradiction is as follows (the following points ar made with respect to the first twin's frame of reference): A twin is travelling in an inertial frame, and observes the clock of the other twin, who is also in an inertial frame. SR predicts that time dilation of the other twin will be observed regardless of whether the other twin is approaching or retreating. When the twin gets to his turnaround point his proper time will be different to other twins apparent time when the other twin is at his respective turnaround point. The number of clock ticks of the other twin that are counted before the other twin reaches the turnaround point will be less than then number of local clock ticks counted at the turnaround point of the first twin. Since the example is symmetric the difference in clock ticks is paradoxical.
From: bz on 28 Nov 2007 20:22 colp <colp(a)solder.ath.cx> wrote in news:ba52b16d-97b9-4823-a1c0- beb3511293b6(a)e6g2000prf.googlegroups.com: > Yes, but the paradox is that the proper times are different to the > observed times, not that the proper times are the same. No. Proper time is the time on his clock as measured by the observer in his own reference frame. The Proper Times are identical in the symetric case. -- bz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set. bz+spr(a)ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap
From: Bryan Olson on 29 Nov 2007 00:03 colp wrote: > Daryl McCullough wrote: >> If you calculate elapsed times using only *inertial* coordinates, >> then you don't get any contradictions. So you are completely >> wrong. > > The contradiction is as follows (the following points ar made with > respect to the first twin's frame of reference): > > A twin is travelling in an inertial frame, and observes the clock of > the other twin, who is also in an inertial frame. > SR predicts that time dilation of the other twin will be observed > regardless of whether the other twin is approaching or retreating. > When the twin gets to his turnaround point his proper time will be > different to other twins apparent time when the other twin is at his > respective turnaround point. Not so. In all frames, the clock traveling with A ticks the same number of times before A turns around as B's local clock ticks before B's turnaround. Two observers, one in each of the outgoing frames, disagree on which turnaround happened first, but the number of times a clock ticks before that same clocks turns around is the same for both clocks, observed from any frame. A clock reaching turnaround and that same clock showing its tick count at turnaround are, by definition, co-located events. The simultaneity of co-located events is *not* frame-relative. The simultaneity of distant events, such as the two turnarounds, is frame-relative. The explanation is still there, should you decide you want to understand SR: http://www.bartleby.com/173/9.html > The number of clock ticks of the other twin that are counted before > the other twin reaches the turnaround point will be less than then > number of local clock ticks counted at the turnaround point of the > first twin. That may be your theory, but it's not what SR says. > Since the example is symmetric the difference in clock ticks is > paradoxical. The example is symmetric about the Earth, not about either twin. Twin A reaching his turnaround while finding B still far from B's turnaround is not a violation of symmetry. Symmetry does imply that B's experience will be symmetric with A's: B finds himself about to reach turnaround while A is still a ways from A's turnaround. In the Earth's frame, the turns-around happen at the same time. SR is self-consistent, whether or not one believes the universe actually works that way. -- --Bryan
From: colp on 29 Nov 2007 01:13
On Nov 29, 2:22 pm, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote: > colp <c...(a)solder.ath.cx> wrote in news:ba52b16d-97b9-4823-a1c0- > beb351129...(a)e6g2000prf.googlegroups.com: > > > Yes, but the paradox is that the proper times are different to the > > observed times, not that the proper times are the same. > > No. The paradox remains. You have not rebutted anything that I said. > Proper time is the time on his clock as measured by the observer in his own reference frame. Yes, that's right. > > The Proper Times are identical in the symetric case. Yes, that is also true. |