Pentcho Valev
2013-12-14 21:23:36 UTC
http://www.feynmanlectures.caltech.edu/I_16.html#Ch16-S2
Richard Feynman: "Just as the mu-mesons last longer when they are moving, so also will Paul last longer when he is moving. This is called a "paradox" only by the people who believe that the principle of relativity means that all motion is relative; they say, "Heh, heh, heh, from the point of view of Paul, can't we say that Peter was moving and should therefore appear to age more slowly? By symmetry, the only possible result is that both should be the same age when they meet." But in order for them to come back together and make the comparison, Paul must either stop at the end of the trip and make a comparison of clocks or, more simply, he has to come back, and the one who comes back must be the man who was moving, and he knows this, because he had to turn around. When he turned around, all kinds of unusual things happened in his space ship - the rockets went off, things jammed up against one wall, and so on - while Peter felt nothing. So the way to state the rule is to say that the man who has felt the accelerations, who has seen things fall against the walls, and so on, is the one who would be the younger; that is the difference between them in an "absolute" sense, and it is certainly correct. When we discussed the fact that moving mu-mesons live longer, we used as an example their straight-line motion in the atmosphere. But we can also make mu-mesons in a laboratory and cause them to go in a curve with a magnet, and even under this accelerated motion, they last exactly as much longer as they do when they are moving in a straight line."
It takes severe megalomania to teach that the acceleration suffered by the traveller is both responsible (in Paul's case) and not responsible (in the mu-meson case) for lasting longer. Less megalomaniac Einsteinians teach either the one or the other but never both:
http://www.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/spacetime_tachyon/index.html
John Norton: "Then, at the end of the outward leg, the traveler abruptly changes motion, accelerating sharply to adopt a new inertial motion directed back to earth. What comes now is the key part of the analysis. The effect of the change of motion is to alter completely the traveler's judgment of simultaneity. The traveler's hypersurfaces of simultaneity now flip up dramatically. Moments after the turn-around, when the travelers clock reads just after 2 days, the traveler will judge the stay-at-home twin's clock to read just after 7 days. That is, the traveler will judge the stay-at-home twin's clock to have jumped suddenly from reading 1 day to reading 7 days. This huge jump puts the stay-at-home twin's clock so far ahead of the traveler's that it is now possible for the stay-at-home twin's clock to be ahead of the travelers when they reunite."
http://www.damtp.cam.ac.uk/research/gr/members/gibbons/gwgPartI_SpecialRelativity2010.pdf
Gary W. Gibbons FRS: "In other words, by simply staying at home Jack has aged relative to Jill. There is no paradox because the lives of the twins are not strictly symmetrical. This might lead one to suspect that the accelerations suffered by Jill might be responsible for the effect. However this is simply not plausible because using identical accelerating phases of her trip, she could have travelled twice as far. This would give twice the amount of time gained."
Twin paradox without any acceleration at all:
http://www.people.fas.harvard.edu/~djmorin/chap11.pdf
Introduction to Classical Mechanics With Problems and Solutions, David Morin, Cambridge University Press, Chapter 11, p. 44: "Modified twin paradox *** Consider the following variation of the twin paradox. A, B, and C each have a clock. In A's reference frame, B flies past A with speed v to the right. When B passes A, they both set their clocks to zero. Also, in A's reference frame, C starts far to the right and moves to the left with speed v. When B and C pass each other, C sets his clock to read the same as B's. Finally, when C passes A, they compare the readings on their clocks."
Pentcho Valev
Richard Feynman: "Just as the mu-mesons last longer when they are moving, so also will Paul last longer when he is moving. This is called a "paradox" only by the people who believe that the principle of relativity means that all motion is relative; they say, "Heh, heh, heh, from the point of view of Paul, can't we say that Peter was moving and should therefore appear to age more slowly? By symmetry, the only possible result is that both should be the same age when they meet." But in order for them to come back together and make the comparison, Paul must either stop at the end of the trip and make a comparison of clocks or, more simply, he has to come back, and the one who comes back must be the man who was moving, and he knows this, because he had to turn around. When he turned around, all kinds of unusual things happened in his space ship - the rockets went off, things jammed up against one wall, and so on - while Peter felt nothing. So the way to state the rule is to say that the man who has felt the accelerations, who has seen things fall against the walls, and so on, is the one who would be the younger; that is the difference between them in an "absolute" sense, and it is certainly correct. When we discussed the fact that moving mu-mesons live longer, we used as an example their straight-line motion in the atmosphere. But we can also make mu-mesons in a laboratory and cause them to go in a curve with a magnet, and even under this accelerated motion, they last exactly as much longer as they do when they are moving in a straight line."
It takes severe megalomania to teach that the acceleration suffered by the traveller is both responsible (in Paul's case) and not responsible (in the mu-meson case) for lasting longer. Less megalomaniac Einsteinians teach either the one or the other but never both:
http://www.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/spacetime_tachyon/index.html
John Norton: "Then, at the end of the outward leg, the traveler abruptly changes motion, accelerating sharply to adopt a new inertial motion directed back to earth. What comes now is the key part of the analysis. The effect of the change of motion is to alter completely the traveler's judgment of simultaneity. The traveler's hypersurfaces of simultaneity now flip up dramatically. Moments after the turn-around, when the travelers clock reads just after 2 days, the traveler will judge the stay-at-home twin's clock to read just after 7 days. That is, the traveler will judge the stay-at-home twin's clock to have jumped suddenly from reading 1 day to reading 7 days. This huge jump puts the stay-at-home twin's clock so far ahead of the traveler's that it is now possible for the stay-at-home twin's clock to be ahead of the travelers when they reunite."
http://www.damtp.cam.ac.uk/research/gr/members/gibbons/gwgPartI_SpecialRelativity2010.pdf
Gary W. Gibbons FRS: "In other words, by simply staying at home Jack has aged relative to Jill. There is no paradox because the lives of the twins are not strictly symmetrical. This might lead one to suspect that the accelerations suffered by Jill might be responsible for the effect. However this is simply not plausible because using identical accelerating phases of her trip, she could have travelled twice as far. This would give twice the amount of time gained."
Twin paradox without any acceleration at all:
http://www.people.fas.harvard.edu/~djmorin/chap11.pdf
Introduction to Classical Mechanics With Problems and Solutions, David Morin, Cambridge University Press, Chapter 11, p. 44: "Modified twin paradox *** Consider the following variation of the twin paradox. A, B, and C each have a clock. In A's reference frame, B flies past A with speed v to the right. When B passes A, they both set their clocks to zero. Also, in A's reference frame, C starts far to the right and moves to the left with speed v. When B and C pass each other, C sets his clock to read the same as B's. Finally, when C passes A, they compare the readings on their clocks."
Pentcho Valev