The real twin paradox.
in [Relativity]
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From: Cosmik de Bris on 23 Nov 2007 19:17 Dirk Van de moortel wrote: > > "colp" <colp(a)solder.ath.cx> wrote in message > news: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, 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. > > Dirk Vdm Thanks for trying Dirk, I have been trying to explain it to him in a different newsgroup but everybody but him is wrong. He just doesn't want to learn anything. -- Posted via a free Usenet account from http://www.teranews.com
From: bz on 23 Nov 2007 19:10 colp <colp(a)solder.ath.cx> wrote in news:9b0f8062-9892-4501-9eba-7c48834920c1(a)d21g2000prf.googlegroups.com: > 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. But it does. Lets draw a few pictures. A and B are ships and E is earth ( and ) are transmitted signals phase 1 ships going away from earth <A )))))) E ((((((( B> Now we just look at things from A's Viewpoint and ignore A's transmitted signal <A ((((((( E (notice the signals that B transmitted have not yet caught A. Other signals have but these haven't. I didn't bother to draw all the signals. phase 2, A turns back toward earth ((A>(((( E As SOON as A turns back, he starts encountering signals from B at a higher rate. In fact those signals were transmitted by B when B was moving at .5 c to the right. NOW, A is also moving at .5 c to the right. The received signal is 1 GHz. From B's viewpoint, the same thing is happening. Both see the other's clocks as ticking at the same rate because they don't yet see signals transmitted by the other AFTER turn around. phase 3, A and B close on earth and each other A>)) E ((<B They NOW see signals that the other transmitted after turn around. Those signals are doppler shifted to 3 GHz. > >> 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. Try it. You will find that all ticks are accounted for. -- 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: colp on 23 Nov 2007 22:06 On Nov 24, 1:10 pm, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote: > colp <c...(a)solder.ath.cx> wrote innews:9b0f8062-9892-4501-9eba-7c48834920c1(a)d21g2000prf.googlegroups.com: > > > > > 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. > > But it does. 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 beecause 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. <snip description of signals> > > > >> 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. > > Try it. You will find that all ticks are accounted for. Here's how it works. During the experiment both ships send the same number of clock ticks. Redshifts and blueshifts do not change the numbers of ticks that are sent. Signal propogation time doesn't affect that either, it just adds a delay between when they are generated and when they are received at by the other twin. At the beginning and end of the experiment there is no delay for the signals, and there is no way that the signal can lose or gain clock ticks. The number of ticks that will be expected by a twin is a product of the tick rate, the proper time duration of the experiment, and the mean time dilation factor for the other twin. The number of ticks that are sent by a twin is equal to the product of the tick rate and the proper time. So the only way the expected tick count will equal the number of sent ticks is if the mean time dilation factor is one. The problem is that the time dilation factor can never exceed one in special relativity, and it must be less than one during the time spent in inertial frames during both the outbound and inbound legs, meaning than the mean value must be less than one. Since the number of ticks expected by SR does not equal the number of ticks sent, and every tick that is sent will be eventually recieved by the other twin, SR fails.
From: Sue... on 23 Nov 2007 23:25 On Nov 23, 10:06 pm, colp <c...(a)solder.ath.cx> wrote: > > Since the number of ticks expected by SR does not equal the number of > ticks sent, and every tick that is sent will be eventually recieved by > the other twin, SR fails. This is where you need the light-clock or a method of reading the four-space intervals as Timo was explaining. The light clock will indeed give an index that correctly calculates the kinetic energy between traveler and stayhome if their mass is known. It is the argument that a twins age anomalously or some other process is affectecd by the motion that fails. If some process were affected by the motion it would be a violation of the principle of relativity. SR certainly does not claim the PoR is invalid. << in reality there is not the least incompatibility between the principle of relativity and the law of propagation of light, >> http://www.bartleby.com/173/7.html Sue... - Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -
From: bz on 23 Nov 2007 23:01
colp <colp(a)solder.ath.cx> wrote in news: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. Say they are 1 light year apart when they turn back toward earth, This implies they each traveled 1/2 light-year away from earth and turned around. Since they traveled at .5 c, this implies that they each traveled for 1 year (earth time) before turning toward home. I will leave it as an exercise for you to work out the time they will have seen on their clock when they turned for home. Let us call A's location at that instant A_b (for A was here when he turned for home) Let us call B's location at that instant B_b (for B was here when he turned for home) It will take 1 year for the signals from A emitted at A_b to reach B_b. And it will take the signals emitted by B at B_b to reach the point A_b. Of course, A won't have to wait an entire year to receive the signal B emitted at B_b because A is traveling in that direction at .8 c wrt B. However it should be clear to you that A will be receiving B's signals at 1 GHz from the moment that A turns back toward earth until A encounters the signal that B sent when he was at B_b AND this will take several months but less than a year. It is NOT an instant. It is much longer than that. -- 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 |