The real twin paradox.
in [Relativity]
|
From: Bryan Olson on 24 Nov 2007 00:08 kenseto wrote: > "Bryan Olson" wrote: >> Ken Seto wrote: >>> [...] I said that a clock second for a moving clock >>> contains a larger amount of absolute time. This corresponds to the SR >>> assertion that the rate on the moving clock is running slow compared to > the >>> observer's clock. >> On the other hand, a second observer, in the frame of the >> 'moving' clock, sees the first observer's clock running slow. > > No that's not observed experimentally. Look up special relativity. > From the ground clock point of view > the SR effect on the GPS clock is 7 us/day running slow compared to the > ground clock. But from the GPS clock point of view the SR effect on the > ground clock is approx. 7 us/day running fast. Relativity works again, but but situation in this thought experiment is symmetric. >> So what's the deal on this "absolute time"? > > It's a BIG deal. It died about a hundred years ago. -- --Bryan
From: colp on 24 Nov 2007 00:41 On Nov 24, 5:01 pm, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote: > colp <c...(a)solder.ath.cx> wrote innews:93dca4c9-c5c5-4708-9d0c-258e521aeb9b(a)s36g2000prg.googlegroups.com: > > > O.K. I see what you are saying now. But saying that they see a 1 GHz > > signal for "a while" is a bit misleading because they only see that > > signal for one instant in the entire journey. They start out seeing a > > redshifted signal and end up seeing a blue shifted signal so of course > > they've got to see the signal at it's natural frequency at some point. > > No! It is much longer than a single instant. The reason that it is only an instant is that redshift/blueshift would be expected to occur for the receiver as well as the sender, based on symmetry. Typically earth-bound receivers don't have to deal with this issue, so it's no suprise that it didn't occur to you. Your example said that the maximum signal transit time was longer than the "a while" that it took for turnaround to occur. So by the time the signal sent during turnaround got to the other twin, that twin would be accelerating or cruising, so the 1GHz signal would be blueshifted during observation and thus observed to be more than 1GHz.
From: bz on 24 Nov 2007 04:59 colp <colp(a)solder.ath.cx> wrote in news:2145a8d3-287f-4ac0-a232-46a77fb32680(a)s19g2000prg.googlegroups.com: > On Nov 24, 5:01 pm, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote: >> colp <c...(a)solder.ath.cx> wrote >> innews:93dca4c9-c5c5-4708-9d0c-258e521aeb9b(a)s36g2000prg.googlegroups.com >> : >> >> > O.K. I see what you are saying now. But saying that they see a 1 GHz >> > signal for "a while" is a bit misleading because they only see that >> > signal for one instant in the entire journey. They start out seeing a >> > redshifted signal and end up seeing a blue shifted signal so of >> > course they've got to see the signal at it's natural frequency at >> > some point. >> >> No! It is much longer than a single instant. > > The reason that it is only an instant is that redshift/blueshift would > be expected to occur for the receiver as well as the sender, based on > symmetry. Typically earth-bound receivers don't have to deal with this > issue, so it's no suprise that it didn't occur to you. > > Your example said that the maximum signal transit time was longer than > the "a while" that it took for turnaround to occur. So by the time the > signal sent during turnaround got to the other twin, that twin would > be accelerating or cruising, so the 1GHz signal would be blueshifted > during observation and thus observed to be more than 1GHz. > There is NO doppler shift when both ships are moving in the same direction at the same velocity. Ship A can not know that B has turned around for a period of time. Let us see how long that will take. At .5 c it will take our ship 1 year to return to earth. By our clock on the ship, that will be 316 days. 365 days if relativity had no effect on our clock. Signals we transmitted at turn around took 182.5 days to reach earth. That was 158 days by our clock. We will be receiving signals from ship B at 1 GHz until we 'see' his ship turn around. During our mutual, 1 year trips away from earth, before turn around at exactly 1 ly from earth, we will both have transmitted 2.731x10^7 one second ticks. we will both have only received 9.104x10^6 'one second ticks' from the other ship up to the time we turn around. We assume instant turn around so no ticks at that instant. We need to pick up 1.821x10^7 more ticks at 1 tick per second before we see B turn around. That is 210.733 days by our clock. So we will see 1 GHz signals from B for 210.733 days. But it will take us 316 days to reach earth, so we will be left with 105.366 days during which we will see signals transmitted by B after turn around. That means we will receive 2.731x10^7 1 second ticks (take the number of seconds in 105.366 days multiplied by 3[remember the doppler shift]) And guess what? That turns out to be the exact number of ticks that each of us transmitted during our trip out and that will be the same number of ticks transmitted during our trip back. No ticks are missing at all. Amazing, isn't it. No paradox. Not even a single doc. -- 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: Sue... on 24 Nov 2007 05:47 On Nov 24, 4:59 am, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote: > colp <c...(a)solder.ath.cx> wrote innews:2145a8d3-287f-4ac0-a232-46a77fb32680(a)s19g2000prg.googlegroups.com: > > > > > > > On Nov 24, 5:01 pm, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote: > >> colp <c...(a)solder.ath.cx> wrote > >> innews:93dca4c9-c5c5-4708-9d0c-258e521aeb9b(a)s36g2000prg.googlegroups.com > >> : > > >> > O.K. I see what you are saying now. But saying that they see a 1 GHz > >> > signal for "a while" is a bit misleading because they only see that > >> > signal for one instant in the entire journey. They start out seeing a > >> > redshifted signal and end up seeing a blue shifted signal so of > >> > course they've got to see the signal at it's natural frequency at > >> > some point. > > >> No! It is much longer than a single instant. > > > The reason that it is only an instant is that redshift/blueshift would > > be expected to occur for the receiver as well as the sender, based on > > symmetry. Typically earth-bound receivers don't have to deal with this > > issue, so it's no suprise that it didn't occur to you. > > > Your example said that the maximum signal transit time was longer than > > the "a while" that it took for turnaround to occur. So by the time the > > signal sent during turnaround got to the other twin, that twin would > > be accelerating or cruising, so the 1GHz signal would be blueshifted > > during observation and thus observed to be more than 1GHz. > > There is NO doppler shift when both ships are moving in the same direction > at the same velocity. > > Ship A can not know that B has turned around for a period of time. > Let us see how long that will take. > > At .5 c it will take our ship 1 year to return to earth. > By our clock on the ship, that will be 316 days. > 365 days if relativity had no effect on our clock. > > Signals we transmitted at turn around took 182.5 days to reach earth. > That was 158 days by our clock. > > We will be receiving signals from ship B at 1 GHz until we 'see' his ship > turn around. > > During our mutual, 1 year trips away from earth, before turn around at > exactly 1 ly from earth, we will both have transmitted 2.731x10^7 one > second ticks. > > we will both have only received 9.104x10^6 'one second ticks' from the > other ship up to the time we turn around. > > We assume instant turn around so no ticks at that instant. > > We need to pick up 1.821x10^7 more ticks at 1 tick per second before we see > B turn around. > > That is 210.733 days by our clock. So we will see 1 GHz signals from B for > 210.733 days. > > But it will take us 316 days to reach earth, so we will be left with > 105.366 days during which we will see signals transmitted by B after turn > around. > > That means we will receive 2.731x10^7 1 second ticks (take the number of > seconds in 105.366 days multiplied by 3[remember the doppler shift]) > > And guess what? > > That turns out to be the exact number of ticks that each of us transmitted > during our trip out and that will be the same number of ticks transmitted > during our trip back. > No ticks are missing at all. > > Amazing, isn't it. No paradox. Not even a single doc. Now strap a light clock on the traveler and see if you get the same result. <<We consider the two cases where the clock is stationary, Fig 10.1(a), and when it travels at velocity v relative to the aether, [dielectric] Fig 10.1(b), at right angles to the clock axis. Given that light will always travel at speed c relative to the aether we have the following clock times (1 tick) for stationary and moving cases: >> http://www.esotericscience.com/Relativity.aspx http://en.wikipedia.org/wiki/Free_space http://www-ssg.sr.unh.edu/ism/what.html Sue... > > -- > bz > > please pardon my infinite ignorance, the set-of-things-I-do-not-know is an > infinite set. > > bz+...(a)ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap- Hide quoted text - > > - Show quoted text -
From: bz on 24 Nov 2007 06:53
"Sue..." <suzysewnshow(a)yahoo.com.au> wrote in news:b2e46734-0b8a-4637-99d5-7f3d2b9d822e(a)w28g2000hsf.googlegroups.com: > On Nov 24, 4:59 am, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote: >> colp <c...(a)solder.ath.cx> wrote >> innews:2145a8d3-287f-4ac0-a232-46a77fb32680(a)s19g2000prg.googlegroups.com >> : >> >> >> >> >> >> > On Nov 24, 5:01 pm, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote: >> >> colp <c...(a)solder.ath.cx> wrote >> >> innews:93dca4c9-c5c5-4708-9d0c-258e521aeb9b(a)s36g2000prg.googlegroups. >> >> com >> >> : >> >> >> > O.K. I see what you are saying now. But saying that they see a 1 >> >> > GHz signal for "a while" is a bit misleading because they only see >> >> > that signal for one instant in the entire journey. They start out >> >> > seeing a redshifted signal and end up seeing a blue shifted signal >> >> > so of course they've got to see the signal at it's natural >> >> > frequency at some point. >> >> >> No! It is much longer than a single instant. >> >> > The reason that it is only an instant is that redshift/blueshift >> > would be expected to occur for the receiver as well as the sender, >> > based on symmetry. Typically earth-bound receivers don't have to deal >> > with this issue, so it's no suprise that it didn't occur to you. >> >> > Your example said that the maximum signal transit time was longer >> > than the "a while" that it took for turnaround to occur. So by the >> > time the signal sent during turnaround got to the other twin, that >> > twin would be accelerating or cruising, so the 1GHz signal would be >> > blueshifted during observation and thus observed to be more than >> > 1GHz. >> >> There is NO doppler shift when both ships are moving in the same >> direction at the same velocity. >> >> Ship A can not know that B has turned around for a period of time. >> Let us see how long that will take. >> >> At .5 c it will take our ship 1 year to return to earth. >> By our clock on the ship, that will be 316 days. >> 365 days if relativity had no effect on our clock. >> >> Signals we transmitted at turn around took 182.5 days to reach earth. >> That was 158 days by our clock. >> >> We will be receiving signals from ship B at 1 GHz until we 'see' his >> ship turn around. >> >> During our mutual, 1 year trips away from earth, before turn around at >> exactly 1 ly from earth, we will both have transmitted 2.731x10^7 one >> second ticks. >> >> we will both have only received 9.104x10^6 'one second ticks' from the >> other ship up to the time we turn around. >> >> We assume instant turn around so no ticks at that instant. >> >> We need to pick up 1.821x10^7 more ticks at 1 tick per second before we >> see B turn around. >> >> That is 210.733 days by our clock. So we will see 1 GHz signals from B >> for 210.733 days. >> >> But it will take us 316 days to reach earth, so we will be left with >> 105.366 days during which we will see signals transmitted by B after >> turn around. >> >> That means we will receive 2.731x10^7 1 second ticks (take the number >> of seconds in 105.366 days multiplied by 3[remember the doppler shift]) >> >> And guess what? >> >> That turns out to be the exact number of ticks that each of us >> transmitted during our trip out and that will be the same number of >> ticks transmitted during our trip back. >> No ticks are missing at all. >> >> Amazing, isn't it. No paradox. Not even a single doc. > > Now strap a light clock on the traveler and see if you > get the same result. I used the relativistic formula for my calculations, Einstein et al. Everything works just fine. No paradox at all. In your terms, I used a light clock. In my terms I used a clock. > <<We consider the two cases where the clock is > stationary, Fig 10.1(a), and when it travels at > velocity v relative to the aether, [dielectric] > Fig 10.1(b), at right angles to the clock axis. > Given that light will always travel at speed c > relative to the aether we have the following > clock times (1 tick) for stationary > and moving cases: >> > http://www.esotericscience.com/Relativity.aspx the web site should be called kookscience. Registrant: Esoteric Science .... Melbourne, Vic 3079 Australia ..... You seem to like fringe science sources. Quote from their home page: Combining Science, Spirituality and Esoteric knowledge You lose credibility when you cite such sites. http://www.esotericscience.com [quote] In Astrology we give an esoteric or spiritual perspective on how and why astrology works. A free natal chart report is available based on channeled intepretations of the planets. In the Channeling section we present information channeled by Olgaa Fienco, that covers a range of topics. You can also order a personal reading from your guides and helpers. The Physics section presents an alternative view of our universe based on the concept of an underlying fluid like hyperdimensional aether. Including the application of this to the concept of free energy, antigravity and teleportation. [unquote] The sections on Astrology, and Channeling explain where you have been getting your ideas about science. You must think you have been channeling Einstein. > http://en.wikipedia.org/wiki/Free_space The fact that we don't have a perfect vacuum may make travel at .5 c somewhat hazardous, but experiments performed in the best vacuums we can obtain on earth are consistent with relativity and the medium out to 1 ly should be a better vacuum than anything we can obtain. > http://www-ssg.sr.unh.edu/ism/what.html intersteller media is unimportant. All measurments were relative to EITHER earth or the other ship, as appropriate. The whole system (two ships, and solar system) could be [and are] moving at .9 c wrt some reference point and it would make no difference to anybody. Heck, pick an incoming OMG particle, moving at .99999c as a reference point. It still makes no difference. Put that particle 2 light years out from earth and perpendicular to the paths our ships will be taking. Run ALL calculations using the arbitrary direction relativistic doppler shift formula. And just as our two brothers are shaking hand with each other and their stay at home sister, the particle arrives. Of course, it has only take 3.264651 days for all this to happen, by the clock the OMG particle is carrying. -- 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 |