The real twin paradox.
in [Relativity]
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From: Dono on 29 Nov 2007 01:18 On Nov 28, 10:13 pm, colp <c...(a)solder.ath.cx> wrote: > On Nov 29, 2:22 pm, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote: > > The paradox remains. You have not rebutted anything that I said. > http://www.dkimages.com/discover/previews/855/35092155.JPG
From: colp on 29 Nov 2007 01:39 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. From A's local frame: The distances from the starting point to the turnaround points are equal. A sees B receeding at twice the velocity that the starting point appears to be receeding. 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. 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.
From: Sue... on 29 Nov 2007 01:56 On Nov 28, 9:35 pm, colp <c...(a)solder.ath.cx> wrote: > On Nov 29, 1:36 pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) > wrote: > > > > > > > colp says... > > > >On Nov 29, 10:44 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) > > >wrote: > > >> No, they aren't. The Wiki example, as you noticed, is about > > >> a case in which one of the twins remains *inertial*. In that > > >> case, you can talk about time dilation for that twin's inertial > > >> coordinate system. If *both* twins are accelerating, then you > > >> *can't* use the simple time dilation formula. > > > >The paradox does not rely on observations that are made from > > >accelerating frames. > > > 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. > The number of clock ticks of the other twin that are counted before > the other twin reaches the turnaround point will be less than then > number of local clock ticks counted at the turnaround point of the > first twin. > > Since the example is symmetric the difference in clock ticks is > paradoxical. At the university teaching level, there are two interprtations of this statement: << As judged from K, the clock is moving with the velocity v; as judged from this reference-body, the time which elapses between two strokes of the clock is not one second, but http://www.bartleby.com/173/M5.GIF seconds, i.e. a somewhat larger time. As a consequence of its motion the clock goes more slowly than when at rest. Here also the velocity c plays the part of an unattainable limiting velocity. >> --Albert Einstein http://www.bartleby.com/173/12.html Authoritative support exist for both views and both have been offered in this thread without ever putting teams on the field to decide which is the better school. (That may be why Caltech doesn't have a football team) Until the semantic operators and metaphysical propagation is removed, rigourous resolution seems unlikely. That hard work is done for you because each term is thoroughly described and rigoursly derived here: Time-independent Maxwell equations Time-dependent Maxwell's equations Relativity and electromagnetism http://farside.ph.utexas.edu/teaching/em/lectures/lectures.html Unless you can convince the H.G. Wells fans that only one interpretation is logically correct it is fruitless to present yet another conundrum just like the one they have tolerated for over 100 years. Sue...
From: Sue... on 29 Nov 2007 02:23 On Nov 28, 9:26 pm, "papar...(a)gmail.com" <papar...(a)gmail.com> wrote: > On 28 nov, 21:36, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > > > > > > > colp says... > > > >On Nov 29, 10:44 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) > > >wrote: > > >> No, they aren't. The Wiki example, as you noticed, is about > > >> a case in which one of the twins remains *inertial*. In that > > >> case, you can talk about time dilation for that twin's inertial > > >> coordinate system. If *both* twins are accelerating, then you > > >> *can't* use the simple time dilation formula. > > > >The paradox does not rely on observations that are made from > > >accelerating frames. > > > If you calculate elapsed times using only *inertial* coordinates, > > then you don't get any contradictions. So you are completely > > wrong. > > > >The essential part of the paradox is that when the two twins are > > >travelling away from each other in inertial frames they expect to > > >receive fewer clock ticks than they send. > > > I don't care what they expect. The question is: what does > > Special Relativity predict? It predicts that in the > > symmetric case, the number of signals each twin receives > > from the other is the same. > > > -- > > Daryl McCullough > > Ithaca, NY > > My last effort in this subject. In the following web page ( http://www.phys.unsw.edu.au/einsteinlight/jw/module4_twin_paradox.htm ), there > is a good and detailed graphical explanation of the asymmetrical twin > paradox. 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 Sue...
From: bz on 29 Nov 2007 04:40
"Sue..." <suzysewnshow(a)yahoo.com.au> wrote in news:2141a9a7-70f6-4527-a1fa- e931094c4626(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. 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. 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 one twin travels a greater distance through time than the other, that twin will be older. 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'. -- 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 |