The real twin paradox.
in [Relativity]
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From: Sue... on 29 Nov 2007 05:54 On Nov 29, 4:40 am, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote: > "Sue..." <suzysewns...(a)yahoo.com.au> wrote in news:2141a9a7-70f6-4527-a1fa- > e931094c4...(a)j44g2000hsj.googlegroups.com: > > > The page shows a young twin beside an old twin after > > one twin has traveled. > > Does one of the twins have a medical disorder or is the page > > it offering argument against the principle of relativity? > > > << The general principle of relativity states that physical > > laws are the same in all reference frames -- inertial or non- > > inertial.>> > >http://en.wikipedia.org/wiki/Principle_of_relativity > > The laws of physics being the same does NOT make measurements BETWEEN > frames identical. Yes... but if light is the tool for comparing the frames and we use different models of light propagation we might honestly disagree about how the frames differ. > > It just says that WITHIN a single frame, you will get the same results as > someone within another frame would get UNDER IDENTICAL CONDITIONS. > > I am in frame A and travel from New York to Boston at 60 MPH. > You are in frame B and travel from San Francisco to Boston at 60 MPH. > > By your reasoning, we would each travel the same number of miles and spend > the same on gas. Yes indeed... that is my interpretation because neither time, energy nor lunch is free. << invariance with respect to time translation gives the well known law of conservation of energy >> http://en.wikipedia.org/wiki/Noether's_theorem > > Physics doesn't work that way. We travel different distances because we > took different paths. Even if we BOTH go from SF to Boston, one of us can > go by route 66 and the other can take an airplane. Same trip. different > miles traveled and different time to take the trip. We start at the same > time and place, we arrive at the same time and place but take a different > number of hours to make the trip and take a different number of miles to > make the trip. > [quote from AE's 1905 paper] > From this there ensues the following peculiar consequence. If at the points > A and B of K there are stationary clocks which, viewed in the stationary > system, are synchronous; and if the clock at A is moved with the velocity v > along the line AB to B, then on its arrival at B the two clocks no longer > synchronize, but the clock moved from A to B lags behind the other which > has remained at B by 1/2 t v^2/c^2 (up to magnitudes of fourth and higher > order), t being the time occupied in the journey from A to B. > It is at once apparent that this result still holds good if the clock moves > from A to B in any polygonal line, and also when the points A and B > coincide. > [unquote] If you will use the 1920 paper you won't have to unlearn so many of the concepts that come with Newton's inertial ether, tho his particle light still remains in the "relativity of simultaniety" boondoggle. There is probably no possibility you will trust an author from Brussels any more than one from Russia? http://arxiv.org/abs/physics/0606233 http://www.ulb.ac.be/homepage_uk.html > > If one twin travels a greater distance through time than the other, that > twin will be older. But that was never shown with rigourous transformations from the spactime displacements. Y. Pierseaux's analysis seems consistant with J.D Jackson's paper. http://arxiv.org/abs/physics/0204034 ....with respect to temporal and spatial properties he attributes to Coulomb and Lorenz gauge. > > Your problem is that you still think that t being plotted on the i axis > makes the time it takes to make a trip somehow 'unreal'. Your problem is that; I don't believe spacetime displacememts can become temporal displacements just because BZ and H.G Wells would like it to be true. :-) Sue... > > -- > bz
From: Bryan Olson on 29 Nov 2007 05:58 colp wrote: > Bryan Olson wrote: >> 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. > > The only way for that to happen is if B's clock does not appear to be > dilated to A, and vice-versa. Wrong yet again, as SR will show if you decide to learn. > From A's local frame: > The distances from the starting point to the turnaround points are > equal. True. > A sees B receeding at twice the velocity that the starting point > appears to be receeding. False, according to SR. In your theory, up to you. In SR, velocities do not combine by simple addition: http://www.bartleby.com/173/13.html > Thus A knows from t=d/v that he will arrive at the turnaround point at > the same time that B arrives at his respective turnaround point. Once again, colp-theory fails. > Only If A does not see B's time as being dilated will he see B's clock > tick the same number of times as his own. Why stick to your theory after it has failed so consistently? Try Einstein's. -- --Bryan
From: Sue... on 29 Nov 2007 06:11 On Nov 29, 5:54 am, "Sue..." <suzysewns...(a)yahoo.com.au> wrote: > > > > > It just says that WITHIN a single frame, you will get the same results as > > someone within another frame would get UNDER IDENTICAL CONDITIONS. > > > I am in frame A and travel from New York to Boston at 60 MPH. > > You are in frame B and travel from San Francisco to Boston at 60 MPH. > > > By your reasoning, we would each travel the same number of miles and spend > > the same on gas. A quick geography refresher suggest we would not travel the same distance but we would use the same fuel. Zero. Sue... > Sue... >
From: Daryl McCullough on 29 Nov 2007 09:28 colp says: > >On Nov 29, 6:03 pm, Bryan Olson <fakeaddr...(a)nowhere.org> wrote: >> 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. > >The only way for that to happen is if B's clock does not appear to be >dilated to A, and vice-versa. Colp, if you want to complain about what Special Relativity says about the situation, you *MUST* understand what Special Relativity says. You can't just make stuff up and then complain that it doesn't make sense. Look at what SR says about this case! This case involves three inertial frames: Frame 1, in which A and B are both traveling at velocity v. Frame 2, in which A is initially at rest, and Frame 3, in which B is initially at rest. Here is the way things are computed in these three frames: For definiteness, let v = 4/5 c, and let each twin travel for 100 seconds, as measured by Earth clocks, before turning around. So as measured by Earth clocks, the entire trip takes 200 seconds. With that speed, the time dilation factor is 5/3. Assume A travels to the left initially (-x direction) and B travels to the right (+x direction). In Frame 1: A travels in the -x direction for 100 seconds. Because of time dilation, his clock only advances 60 seconds. Then A travels in the +x direction for 100 seconds. Again, his clock advances 60 seconds. The total elapsed time for A's clock is 120 seconds. B travels in the +x direction for 100 seconds. Because of time dilation, his clock only advances 60 seconds. Then B travels in the -x direction for 100 seconds. Again, his clock advances 60 seconds. The total elapsed time for B's clock is 120 seconds. In Frame 2: A stays at rest for 60 seconds. His clock advances 60 seconds. During this time, B travels in the +x direction. Because of time dilation, his clock advances only about 13 seconds. (The speed of B in frame 2 is greater than his speed in frame 1, so the time dilation factor is 41/9 instead of 5/3.) Next, A turns around so that he is traveling at the same speed as B for the next 213 seconds. During this time, because of time dilation, both A's clock and B's clock advance by only about 47 seconds. Next, B turns around so that he is at rest in Frame 2 for the next 60 seconds. His clock advances 60 seconds, while A's clock advances only 13 seconds (due to time dilation). Total elapsed time for A = 60 + 47 + 13 = 120 Total elapsed time for B = 13 + 47 + 60 = 120 In Frame 3: B stays at rest for 60 seconds. His clock advances 60 seconds. During this time, A travels in the -x direction. Because of time dilation, his clock advances only about 13 seconds. (The speed of A in Frame 3 is greater than his speed in Frame 1, so the time dilation factor is 41/9 instead of 5/3.) Next, B turns around so that he is traveling at the same speed as A for the next 213 seconds. During this time, because of time dilation, both A's clock and B's clock advance by only about 47 seconds. Next, A turns around so that he is at rest in Frame 3 for the next 60 seconds. His clock advances 60 seconds, while B's clock advances only 13 seconds (due to time dilation). Total elapsed time for A = 13 + 47 + 60 = 120 Total elapsed time for B = 60 + 47 + 13 = 120 ------------------------- Note: All three frames agree about the following facts: e1. A's clock shows time 60 seconds when he turns around. e2. B's clock shows time 60 seconds when he turns around. e3. Both clocks shows time 120 seconds when the two twins get back together. In all three frames, clocks at rest advance faster than clocks that are moving. Where the three frames disagree about are: 1. What speed is each clock traveling? In Frame 1: both clocks travel at 4/5 c throughout. In Frame 2: Clock A travels at speed 0 and clock B travels at speed 0.976c for 60 seconds. Then both clocks travel at speed 0.976c for 213 seconds. Then B travels at speed 0 and A travels at speed 0.976c for the last 60 seconds. In Frame 3: Clock B travels at speed 0 and clock A travels at speed 0.976c for 60 seconds. Then both clocks travel at speed 0.976c for 213 seconds. Then A travels at speed 0 and B travels at speed 0.976c for the last 60 seconds. 2. Does event e1 take place before, after, or the same time as event e2? In Frame 1: They take place at the same time. In Frame 2: e1 takes place 213 seconds before e2. In Frame 3: e2 takes place 213 seconds after e1. In any *given* coordinate system, the facts are perfectly consistent. But if you try to mix and match facts from one coordinate system with facts from another coordinate system, you get nonsense, just as if you switched from using inches to using centimeters in the middle of a calculation. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 29 Nov 2007 09:41
colp says... > >On Nov 29, 1:36 pm, stevendaryl3...(a)yahoo.com (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): There is no such thing as "the first twin's frame of reference"! There are three inertial frames of reference in this problem: In Frame 1: A and B move in opposite directions at the same speed, turn around and return at the same speed. In Frame 2: A is initially at rest, and B is moving away from A. Then A accelerates and starts moving in the same direction as B at the same speed as B. Then (later, as measured in this frame) B decelerates until he is at rest in this frame. In Frame 3: B is initially at rest, and A is moving away from B. Then B accelerates and starts moving in the same direction as A at the same speed as A. Then (later, as measured in this frame) A decelerates until he is at rest in this frame. So what do you mean by "A's frame"? Do you mean Frame 2 or Frame 3? >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. What SR says is that, from the point of view of Frame 1, A and B have equal time dilation throughout. Conclusion: Their clocks show the same time when they get back together. From the point of view of Frame 2, B's time dilation is greater than A's on the first part of the trip, and A's time dilation is greater than B's on the last part of the trip. Conclusion: Their clocks show the same time when they get back together. From the point of view of Frame 3, A's time dilation is greater than B's during the first part of the trip, and B's time dilation is greater than A's during the last part of the trip. Conclusion: Their clocks show the same time when they get back together. When you talk about "A's frame" I don't know what you mean. Do you mean Frame 1, Frame 2, or Frame 3? Each of those frames gives the same answer: the clocks show the same time when the twins get back together. The only way to get some different answer is to mix up facts from one frame with facts from another frame. But that's a *stupid* thing to do. It's like doing a measurement problem in which you change from inches to centimeters in the middle of the problem. -- Daryl McCullough Ithaca, NY |