The Symmetric Twin Paradox
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From: Paul Stowe on 20 Jun 2010 11:08 On Jun 20, 2:07 am, "Inertial" <relativ...(a)rest.com> wrote: > "PaulStowe" <theaether...(a)gmail.com> wrote in message > > news:0a3a7ed7-17b1-41ee-8393-b1285126b234(a)k25g2000prh.googlegroups.com... > > > > > > > On Jun 19, 9:23 am, "Inertial" <relativ...(a)rest.com> wrote: > >> "Dary McCullough" <stevendaryl3...(a)yahoo.com> wrote in message > > >>news:hvipdt01aeo(a)drn.newsguy.com... > > >> > colp says... > > >> >>Then what do you think the circumstances are in which SR predicts that > >> >>a twin observes the other to age more quickly, and what mathematical > >> >>relationship quantifies this? > > >> > Sure. Let's assume the following set-up. One twin stays at > >> > home throughout. The other zips away and comes back. > >> > Each twin sends out a radio pulse once per second (as > >> > measured by his own clock). > > >> > From the point of view of the stay-at-home twin, the traveling twin > >> > travels away at 0.866 c for 200 seconds, turns around rapidly, > >> > and comes back at 0.866 c for 200 seconds. The traveling twin > >> > experiences time dilation of a factor of two, so the trip takes > >> > 400 seconds, as measured by the stay-at-home twin, and only > >> > 200 seconds, as measured by the traveling twin. According to > >> > the stay-at-home twin, the traveling twin sends out pulses at > >> > the rate of one pulse every two seconds. > > >> > What about the pulses? The stay-at-home twin will receive signals > >> > from the traveling twin at the rate of one signal every 3.73 seconds > >> > for the first 373 seconds (for a total of 100 pulses). Then, the > >> > stay-at-home twin will receive signals at the rate of one signal > >> > every 0.27 seconds for the next 27 seconds, for a total of 100 more > >> > pulses. > > >> > Why these numbers? On the way out, each successive pulse from > >> > the traveling twin is sent from farther and farther away. Since > >> > the traveling twin travels 1.732 light seconds between sending > >> > any pulses. That means that the second pulse takes an additional > >> > 1.732 seconds to travel back to Earth. Since it is sent 2 seconds > >> > later, that means it will arrive at Earth 3.732 seconds later. > > >> > When the traveling twin is on his way back, each pulse is sent > >> > from a closer and closer distance. One pulse is sent. Then the > >> > next pulse is sent 2 seconds later. But since it is sent from > >> > closer in, it takes 1.732 seconds *less* time to travel the > >> > distance back to Earth. So the second pulse arrives only > >> > 2 - 1.732 = 0.268 seconds later. > > >> > So the stay-at-home twin sees pulses arrive at rate once per > >> > 3.732 seconds for part of the time, and sees pulses arrive at > >> > the rate of once per 0.268 seconds for the rest of the time. > >> > When does the changeover happen? Well, it happens when the > >> > last pulse from the traveling twin's outward journey is sent. > >> > Since the traveling twin travels outward for 200 seconds, he > >> > is 200*0.866 = 173.2 light-seconds away. So it takes another > >> > 173.2 seconds for that pulse to reach the Earth. So the > >> > earth only gets that pulse at time 200 + 173.2 = 373.2 seconds. > > >> > So the earth receives at the rate of one per 3.732 seconds for > >> > 373.2 seconds, for a total of 100 pulses, and then receives at > >> > the rate of one per 0.268 seconds for the next 26.8 seconds, > >> > for a total of 100 more pulses. So the Earth twin receives > >> > 200 pulses from the traveling twin. > > >> > Now, let's look at the situation from the point of view > >> > of the traveling twin: > > >> > The two rates are the same (by relativity): The traveling > >> > twin receives pulses from the Earth twin at the rate of one > >> > pulse per 3.732 seconds during his outward trip, which lasts > >> > 100 seconds (according to his clock) for a total of about 27 > >> > pulses received. In his return trip, he receives pulses from the > >> > Earth at the rate of one pulse per 0.268 seconds for the > >> > next 100 seconds, for a total of 373 more pulses. So altogether, > >> > the traveling twin receives 373 + 27 = 400 pulses. > > >> > So the traveling twin receives 400 pulses from the stay-at-home > >> > twin, while the stay-at-home twin receives only 200 pulses from > >> > the traveling twin. By counting pulses, they both agree that > >> > the traveling twin is younger. > > >> What would be instructive is to do the same analysis for the so-called > >> symmetrical twins paradox. > > > Too bad you can't do it... > > Too bad you wouldn't understand it if I did. THAT! I don't have to worry about since you are incapable of providing the detailed analysis and explanation to begin with. Since it was not 'I' to whom you were responding. ahahah... Paul Stowe
From: Uncle Ben on 20 Jun 2010 12:53 On Jun 20, 2:27 am, colp <c...(a)solder.ath.cx> wrote: > On Jun 20, 11:35 am, "Sue..." <suzysewns...(a)yahoo.com.au> wrote: > > > On Jun 19, 7:17 pm, colp <c...(a)solder.ath.cx> wrote: > > > 1. SR predicts that each twin observes the other twin to age more > > > slowly both on the outgoing leg and the return leg. > > > No... > > How does a four dimensional model of spacetime provide for an > alternative interpretation of the symmetric twin thought experiment? > > > > > > > <<Einstein's 1905 presentation of special relativity was soon > > supplemented, in 1907, by Hermann Minkowski, who showed that > > the relations had a very natural interpretation[C 5] in terms > > of a unified four-dimensional "spacetime" in which absolute > > intervals are seen to be given by an extension of the > > Pythagorean theorem.>>http://en.wikipedia.org/wiki/Lorentz_ether_theory#The_shift_to_relati... > > > << the four-dimensional space-time continuum of the > > theory of relativity, in its most essential formal > > properties, shows a pronounced relationship to the > > three-dimensional continuum of Euclidean geometrical space. > > In order to give due prominence to this relationship, > > however, we must replace the usual time co-ordinate t by > > an imaginary magnitude > > > sqrt(-1) > > > ct proportional to it. Under these conditions, the > > natural laws satisfying the demands of the (special) > > theory of relativity assume mathematical forms, in which > > the time co-ordinate plays exactly the same rôle as > > the three space co-ordinates. >>http://www.bartleby.com/173/17.html > > > Sue...- Hide quoted text - > > - Show quoted text - See the recent thread addressed to you: "colp, why did Einstein use the word 'relative'?" in sci.physics.relativity. Uncle Ben
From: Sue... on 20 Jun 2010 13:52 On Jun 20, 2:27 am, colp <c...(a)solder.ath.cx> wrote: > On Jun 20, 11:35 am, "Sue..." <suzysewns...(a)yahoo.com.au> wrote: > > > On Jun 19, 7:17 pm, colp <c...(a)solder.ath.cx> wrote: > > > 1. SR predicts that each twin observes the other twin to age more > > > slowly both on the outgoing leg and the return leg. > > > No... > ================= > How does a four dimensional model of spacetime provide for an > alternative interpretation of the symmetric twin thought experiment? It contributes a bit of mathematical rigour. http://chimera.roma1.infn.it/apehdoc/apemille/grid.gif Build the above grid, climb inside and shoot a wabbit. Have a friend reposition the grid. Shoot the wabbit again. Is it clear to you that the X,Y,Z descriptions of your first and second shots will differ even tho the shots are identical? Your response to Daryl indicates you are confused about this. It is pointless to discuss the 4D description until you are sure about the 3D version. Are you quite sure the coordinate systems used to describe experiments, don't in any way affect an experiment, only its description in terms of the coordinates? ??? Sue... > > > > > <<Einstein's 1905 presentation of special relativity was soon > > supplemented, in 1907, by Hermann Minkowski, who showed that > > the relations had a very natural interpretation[C 5] in terms > > of a unified four-dimensional "spacetime" in which absolute > > intervals are seen to be given by an extension of the > > Pythagorean theorem.>> http://en.wikipedia.org/wiki/Lorentz_ether_theory > > > << the four-dimensional space-time continuum of the > > theory of relativity, in its most essential formal > > properties, shows a pronounced relationship to the > > three-dimensional continuum of Euclidean geometrical space. > > In order to give due prominence to this relationship, > > however, we must replace the usual time co-ordinate t by > > an imaginary magnitude > > > sqrt(-1) > > > ct proportional to it. Under these conditions, the > > natural laws satisfying the demands of the (special) > > theory of relativity assume mathematical forms, in which > > the time co-ordinate plays exactly the same rôle as > > the three space co-ordinates. >> http://www.bartleby.com/173/17.html > > > Sue... > >
From: Daryl McCullough on 20 Jun 2010 13:52 colp says... > >On Jun 20, 12:32=A0pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) >wrote: >> If by "observing the other >> twin aging" you mean looking at the other twin through a powerful >> telescope and seeing how old he *looks*, then what you would find >> is: >> >> (1) During the outward journey, each twin will see the other twin >> aging more slowly. >> >> (2) During the return journey, each twin will see the other twin >> aging more rapidly. > >That isn't what I mean by observing the other twin. By including the >effect of signal transit time, the measured signals differ from the >remote time, which is what is being observed. Okay, let's take into account the transit time. Remember the set-up: the traveling twin is traveling at 0.866c, which means a time dilation factor of 2. He travels away from Earth for 200 seconds, earth time, which means 100 seconds, his time, and then comes back. The earth twin sends signals at a rate of 1 per second (according to Earth coordinates). The traveling twin doesn't receive them at that rate, for two reasons (1) the transit time for light keeps changing, and (2) according to the traveling twin, the earth twin's clock is running slow, by a factor of 2. Taking both factors into account, we find for the traveling twin that: Before turn-around, he sees one signal coming from the Earth every 3.732 seconds. After turn-around, he sees one signal every 0.268 seconds. Right before turn around, he's looking at signal number 27 coming from the Earth. That means that it was sent when the Earth twin was 27 seconds older than when they left. To figure out how old the Earth twin is right now, you have to add in the transit time for the signal. How do you compute that? Well, let L' be the current distance to Earth. Since, according to the traveling twin, it has been 100 seconds since departing, and the Earth has been receding at 0.866 c, the Earth must now be 86.6 light-seconds away. Suppose that signal number 27 was sent at time T in the past. Since Earth is moving away at speed 0.866 c, that means that in the past, Earth was at location L - vT. Since light travels at speed c, that means that light traveled distance L-vT in time T. That means L-vT = cT. So we can solve for T: T = L/(c+v) For our numbers, L = 86.6 light-seconds, v = 0.866 c, so we have: T = 86.6/1.866 = 46.4 seconds So signal number 27 was sent 46.4 seconds ago. The Earth twin ages at half the normal rate, so he has aged 237 seconds since the signal was sent. So his age at the moment right before turn-around is about 27 + 23 = 50. So immediately before turnaround, the traveling twin concludes that the Earth twin has aged only 50 seconds since the beginning of the journey, while the traveling twin has aged 100 seconds. So the traveling twin thinks that he is older. Now, he does the turn-around instantaneously. Immediately afterwards, he's still looking at signal number 27. But now, the traveling twin is in a different frame, in which the Earth is moving *towards* the traveling twin at speed 0.866 c, instead of away from him. Now, let's redo the computation to figure out how old the Earth twin is, as measured in the new frame after turnaround. Once again, we assume that the signal was sent at some time T1 ago. Since in this new frame, the Earth is moving *towards* the traveling twin, that means that in the past, he was *farther* away from the traveling twin. If the Earth is currently a distance L away, then in the past, it was at a location L + v T1 away---farther away by an amount v T1. Since light travels at speed c, that means that light traveled distance L+vT1 in time T1. That means L+vT1 = cT1. So we can solve for T1: T1 = L/(c-v) For our numbers, L = 86.6 light-seconds, v = 0.866 c, so we have: T = 86.6/0.134 = 646 seconds So, as measured in the *new* frame, the transit time for signal number 27 was 646 seconds! Since the Earth twin ages at half the normal rate, it means that he has aged 323 seconds since sending signal number 27. That means that Earth twin has now aged 323 + 27 = 350 seconds since the beginning of the journey! So, when the traveling twin switches frames, the new frame disagrees with the old frame about how old the Earth twin is, by a whopping 300 seconds! That's disconcerting, but if you are discontinuously changing coordinate systems, you have to expect the coordinates of things to change discontinuously. So, the traveling twin would compute the age of the Earth twin to change in the following way: 1. Between time t'=0 and time t'=100, the Earth twin ages 50 seconds. 2. During turnaround, the Earth twin's age jumps ahead 300 seconds. 3. Between time t'=100 and time t'=200, the Earth twin ages another 50 seconds. Total aging for Earth twin = 400 seconds. Total aging for traveling twin = 200 seconds. So you are right, that after accounting for transit time, the traveling twin would compute the Earth twin to age at half the normal rate during the first half of the journey, and also to age at half the normal rate during the second half of the journey. But during turn-around, the traveling twin would have to conclude that the Earth twin aged 300 seconds. This principle is actually important for the transition to General Relativity: In an accelerated coordinate system (which is what the traveling twin is using), clocks that are "higher up" (that is, in the same direction the traveling twin is accelerating) run faster. This effect is actually measurable: If you take a clock to the top of Mount Everest, it will run faster than the same clock on the ground (after accounting for any differences in speed of the two clocks). -- Daryl McCullough Ithaca, NY
From: Sue... on 20 Jun 2010 14:16
On Jun 20, 1:52 pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > colp says... > > > > > > >On Jun 20, 12:32=A0pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) > >wrote: > >> If by "observing the other > >> twin aging" you mean looking at the other twin through a powerful > >> telescope and seeing how old he *looks*, then what you would find > >> is: > > >> (1) During the outward journey, each twin will see the other twin > >> aging more slowly. > > >> (2) During the return journey, each twin will see the other twin > >> aging more rapidly. > > >That isn't what I mean by observing the other twin. By including the > >effect of signal transit time, the measured signals differ from the > >remote time, which is what is being observed. > > Okay, let's take into account the transit time. > > Remember the set-up: the traveling twin is traveling at 0.866c, > which means a time dilation factor of 2. He travels away from > Earth for 200 seconds, earth time, which means 100 seconds, > his time, and then comes back. The earth twin sends signals > at a rate of 1 per second (according to Earth coordinates). > The traveling twin doesn't receive them at that rate, for > two reasons (1) the transit time for light keeps changing, > and (2) according to the traveling twin, the earth twin's > clock is running slow, by a factor of 2. Taking both factors > into account, we find for the traveling twin that: > > Before turn-around, he sees one signal coming from the > Earth every 3.732 seconds. After turn-around, he sees > one signal every 0.268 seconds. > > Right before turn around, he's looking at signal number 27 > coming from the Earth. That means that it was sent when the > Earth twin was 27 seconds older than when they left. To > figure out how old the Earth twin is right now, you have > to add in the transit time for the signal. How do you compute > that? > > Well, let L' be the current distance to Earth. Since, > according to the traveling twin, it has > been 100 seconds since departing, and the Earth has been > receding at 0.866 c, the Earth must now be 86.6 light-seconds > away. Suppose that signal number 27 was sent at time T in the > past. Since Earth is moving away at speed 0.866 c, that means > that in the past, Earth was at location L - vT. Since light > travels at speed c, that means that light traveled distance > L-vT in time T. That means L-vT = cT. So we can solve for T: > > T = L/(c+v) > > For our numbers, L = 86.6 light-seconds, v = 0.866 c, so > we have: > > T = 86.6/1.866 = 46.4 seconds > > So signal number 27 was sent 46.4 seconds ago. The Earth > twin ages at half the normal rate, so he has aged 237 seconds > since the signal was sent. So his age at the moment right > before turn-around is about 27 + 23 = 50. So immediately > before turnaround, the traveling twin concludes that the > Earth twin has aged only 50 seconds since the beginning > of the journey, while the traveling twin has aged 100 > seconds. So the traveling twin thinks that he is older. > > Now, he does the turn-around instantaneously. Immediately > afterwards, he's still looking at signal number 27. But > now, the traveling twin is in a different frame, in which > the Earth is moving *towards* the traveling twin at speed > 0.866 c, instead of away from him. Now, let's redo the > computation to figure out how old the Earth twin is, as > measured in the new frame after turnaround. > > Once again, we assume that the signal was sent at some > time T1 ago. Since in this new frame, the Earth is moving > *towards* the traveling twin, that means that in the past, > he was *farther* away from the traveling twin. If the > Earth is currently a distance L away, then in the past, > it was at a location L + v T1 away---farther away by > an amount v T1. Since light travels at speed c, that > means that light traveled distance > L+vT1 in time T1. That means L+vT1 = cT1. > So we can solve for T1: > > T1 = L/(c-v) > > For our numbers, L = 86.6 light-seconds, v = 0.866 c, so > we have: > > T = 86.6/0.134 = 646 seconds > > So, as measured in the *new* frame, the transit time for > signal number 27 was 646 seconds! Since the Earth twin ages > at half the normal rate, it means that he has aged 323 seconds > since sending signal number 27. That means that Earth twin > has now aged 323 + 27 = 350 seconds since the beginning > of the journey! > > So, when the traveling twin switches frames, the new frame > disagrees with the old frame about how old the Earth twin > is, by a whopping 300 seconds! That's disconcerting, but > if you are discontinuously changing coordinate systems, > you have to expect the coordinates of things to change > discontinuously. > > So, the traveling twin would compute the age of the Earth > twin to change in the following way: > > 1. Between time t'=0 and time t'=100, the Earth twin ages 50 seconds. > 2. During turnaround, the Earth twin's age jumps ahead 300 seconds. > 3. Between time t'=100 and time t'=200, the Earth twin ages another > 50 seconds. ============= > > Total aging for Earth twin = 400 seconds. > Total aging for traveling twin = 200 seconds. > > So you are right, that after accounting for transit time, > the traveling twin would compute the Earth twin to age at > half the normal rate during the first half of the journey, > and also to age at half the normal rate during the second > half of the journey. But during turn-around, the traveling > twin would have to conclude that the Earth twin aged 300 > seconds. > > This principle is actually important for the transition > to General Relativity: In an accelerated coordinate system > (which is what the traveling twin is using), clocks that > are "higher up" (that is, in the same direction the > traveling twin is accelerating) run faster. This effect > is actually measurable: If you take a clock to the top > of Mount Everest, it will run faster than the same clock > on the ground (after accounting for any differences in > speed of the two clocks). The higher clock runs faster because its moving masses follow a flatter trajectory. When did Biology become the basis for fundamental physcs? Sue... > > -- > Daryl McCullough > Ithaca, NY |