The real twin paradox.
in [Relativity]
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From: colp on 23 Nov 2007 18:04 On Nov 24, 11:45 am, "Dirk Van de moortel" <dirkvandemoortel(a)ThankS-NO- SperM.hotmail.com> wrote: > "colp" <colp(a)solder.ath.cx> wrote in messagenews:f00a630f-91a1-49ea-93ad-37a674fd0fcf(a)i29g2000prf.googlegroups.com... > > On Nov 24, 6:18 am, "Dirk Van de moortel" <dirkvandemoor...(a)ThankS-NO- > > SperM.hotmail.com> wrote: > >> "colp" <c...(a)solder.ath.cx> wrote in messagenews:efba2671-a734-4801-a8ec-64c62f428ff5(a)d4g2000prg.googlegroups.com... > >> > On Nov 23, 9:53 am, "Dirk Van de moortel" <dirkvandemoor...(a)ThankS-NO- > >> > SperM.hotmail.com> wrote: > > >> [snip] > > >> >> Have you tried drawing that diagram? > > >> > There's not much point unless we can both see the diagram and talk > >> > about what it represents. That would involve getting the diagram onto > >> > the internet, which is more work that I am willing to contemplate > >> > right now. > > >> Well, it has become clear from this (and from your other responses) > >> that, for some unknown reason, you act like a person who is too stupid > >> to understand the basics > > > Coming from someone what can't event get SR time dilation right for > > decreasing relative distance in an inertial frame, your opinion isn't > > very credible. > > (unsnip) > > >> or to even *try* to understand them, so I will > >> stop wasting your time. If - and only if - you are ready to reply directly > >> to my explanation with the spacetime diagram, feel free to do so. > > "If and only if" I said. So what? > You obviously also act like a person who is too much of a coward > to even reply to the only explanations that could really help him out > of his self-inflicted misery. Like your explanation? According to you, observed time not only dialates, it compresses in inertial frames. From your opening post: When your two clocks fly apart, each clock will measure this time to be longer and conclude that the other clock is "running slower". While clock A is coasting, according to clock A, each tick on clock A is simultaneous with some tick on clock B with a smaller time value. While clock B is coasting, according to clock B, each tick on clock B is simultaneous with some tick on clock A with a smaller time value. After clock A has made its turnaround, it has shifted to another inertial frame, in which according to clock A, each tick on clock A is simultaneous with some tick on clock B with a larger time value. After clock B has made its turnaround, it has shifted to another inertial frame, in which according to clock B, each tick on clock B is simultaneous with some tick on clock A with a larger time value. > So you now are in the "autistic imbecile cowards" category. > But don't this put you off - you are not alone. This very place was > actually created to suck in people like you. Enjoy the attention. LOL Where is your solution to the paradox, Dirk? As another poster said: Would you tell us idiots how this runs in SR ?
From: Dono on 23 Nov 2007 18:09 On Nov 23, 1:04 pm, colp <c...(a)solder.ath.cx> wrote: <...> You are still: http://eldoradoclub.net/images/wacko-lg_1_.gif
From: Dono on 23 Nov 2007 18:25 On Nov 23, 3:04 pm, colp <c...(a)solder.ath.cx> wrote: > > Would you tell us idiots how this runs in SR ? Riiiight :-)
From: Bryan Olson on 23 Nov 2007 18:24 colp wrote: > Coming from someone what can't event get SR time dilation right for > decreasing relative distance in an inertial frame, your opinion isn't > very credible. I think you misunderstood Dirk on that. He wrote: After clock A has made its turnaround, it has shifted to another inertial frame, in which according to clock A, each tick on clock A is simultaneous with some tick on clock B with a larger time value. He's saying that in the returning frame of A, clock B is now ahead of clock A; running slower due to time dilation, but ahead. -- --Bryan
From: colp on 23 Nov 2007 18:48
On Nov 24, 11:01 am, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote: > colp <c...(a)solder.ath.cx> wrote innews:ce4ba04e-546a-4468-a814-90c056b77de4(a)b40g2000prf.googlegroups.com: > > >> You 'see' your twins clock ticking faster than yours. > >> Once the signals from his ship, after his turn around, reach you, the > >> doppler shifts are doubled. > > > Wrong. The cumulative effect is nil. > > > f' = f + fv / c > > > v is the velocity of the transmitter relative to the receiver in > > meters/second: positive when moving towards one another, negative when > > moving away > > You are forgetting the fact that the signals from his ship do not reach > you the instant they are transmitted. No I'm not. > > You don't see the effect of his turn around until signals transmitted at > the time of his turn around reach you. But at the instant you turn around > you DO see a change in the frequency of the signal Possibly, if he was decelerating when he sent the signal that you saw at that instant. > > Let us say you both start from earth at the same time, each traveling at > .5c away from earth. Your relativistic velocities away from each other > will be 0.8 c (see the composition of velocities formula). > > Let us say each ship transmits on 1 GHz (near some cell phone bands). > As the ships go away from earth, receivers on earth will receive both > ships on 0.577 GHz. Each ship will receive the other ship on 0.333 GHz, as > they are separating at 0.8 c relative. > Sounds reasonable, I'm assuming that your maths is O.K. > After the ships turn around, They will receive each others signals at 1 GHz > for a while because each will be > receiving signals sent while the other was moving in the same direction > and at the same velocity that they are now moving. They see the other > ship's relative velocity as ZERO. That doesn't work. > When they get closer together and 'see' > the other ship turn around, they will pick up the other ship's signals at > 3 GHz, while the earth bound sister receives them both at 1.732 GHz as they > are approaching earth at .5 c each. > > All frequencies were computed using the relativistic Doppler shift > formula. Relative velocity computed using the composition of velocity > formula. You haven't addressed the the issue of the clock tick count that was measured by either of the twins, which is an essential element of the paradox. The issue could also be addressed by integrating the frequency of the signal received by a twin over the duration of the experiment. |