The Symmetric Twin Paradox
in [Relativity]
|
Prev: Black Hole is Black Day for Earth
Next: n-stars.
From: Dary McCullough on 19 Jun 2010 11:57 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. -- Daryl McCullough Ithaca, NY
From: Inertial on 19 Jun 2010 12:23 "Dary McCullough" <stevendaryl3016(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.
From: Paul Stowe on 19 Jun 2010 16:48 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... Paul Stowe
From: paparios on 19 Jun 2010 17:20 On 19 jun, 02:56, colp <c...(a)solder.ath.cx> wrote: > On Jun 19, 2:33 pm, "papar...(a)gmail.com" <papar...(a)gmail.com> wrote: > > > > > First typical mistake of students. > > > > ... is believing that they have been told without proving it for > > > themselves. > > > So you are guilty as charged!!! > > How do you do you figure that? > > > You have been defending these young > > south african students who do not understand a bit about SRT, > > What are you talking about? Say no more...clearly you know nothing about physics...period. Just having fun trolling right? Miguel Rios
From: colp on 19 Jun 2010 17:57
On Jun 19, 8:08 pm, "Inertial" <relativ...(a)rest.com> wrote: > "colp" <c...(a)solder.ath.cx> wrote in message > > news:97c45dd7-e152-4e3d-8197-42bc43980300(a)y18g2000prn.googlegroups.com... > > In reality the twins age the same as each other, > > As SR predicts .... if you ignore what SR predicts that each twin will individually observe throughout the entire experiment. > > > but SR does not > > predict that result > > WRONG No, not wrong. > > > if you examine the experiment from the point of > > view of either twin. > > WRONG Nope. You can hack my statement anyway you like, but the fact is that SR predicts that an observer observing a non-local clock moving in a inertial frame at a relativistic velocity will observe that clock to be running slow. This observation applies both on the outgoing and return legs, and it applies for both twins. > > Show the supposed SR analysis you claim does this. And don't just look at > the two legs individually .. So you don't want me to talk about the essential element of the paradox, right? > as in the usualy twins paradox, the main point > is the change in rest reference frame at hte turnaround. ignore that, and > its NOT SR. I haven't ignored the turnaround. I have previously posted an explanation of why turnaround can't eliminate the paradox with an example of pulsed radio waves that are measured by a third observer on Earth. |