The real twin paradox.
in [Relativity]
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From: Dirk Van de moortel on 21 Nov 2007 05:40 "colp" <colp(a)solder.ath.cx> wrote in message news:06b84031-18aa-4644-bfb7-43f49f46ae6a(a)i37g2000hsd.googlegroups.com... > This thought experiment is like the classic twin paradox, but in this > expirement both twins leave earth and travel symmetric return trips in > opposite directions. > > Since the paths taken by the twins in this experiment are symmetric, > they must be the same age when they meet on their return to earth. > > In this experiment the twins maintain constant observation of each > other's clocks, from when they depart until they return and find that > their clocks tell the same time. > > Special relativity says that each twin must observe that the other's > clock is running slow, and at no time does special relativity allow > for an observation which shows that the other clock is running fast. No, special relativity says much more precise than that "moving clocks" are running slow. It says something about intertial observers who measure times between ticks on remote, moving clocks. When your two clocks fly apart, each clock will measure this time to be longer and conclude that the other clock is "running slower". While clock A is coasting, according to clock A, each tick on clock A is simultaneous with some tick on clock B with a smaller time value. While clock B is coasting, according to clock B, each tick on clock B is simultaneous with some tick on clock A with a smaller time value. After clock A has made its turnaround, it has shifted to another inertial frame, in which according to clock A, each tick on clock A is simultaneous with some tick on clock B with a larger time value. After clock B has made its turnaround, it has shifted to another inertial frame, in which according to clock B, each tick on clock B is simultaneous with some tick on clock A with a larger time value. > > The paradox is that special relativity says that a twin will never see > the other twin's clock catch up, but the clocks must show the same > time at the end of the experiment because of symmetry. When they finally meet, for both clocks, this larger time reading of the simultaneous events on the other clock is compensated by the "more slowly running time" on that clock such that they read the same time when they are reunited. Dirk Vdm [copy and follow-up to sci.physics.relativity]
From: colp on 21 Nov 2007 11:37 On Nov 21, 11:40 pm, "Dirk Van de moortel" <dirkvandemoor...(a)ThankS-NO- SperM.hotmail.com> wrote: > "colp" <c...(a)solder.ath.cx> wrote in messagenews:06b84031-18aa-4644-bfb7-43f49f46ae6a(a)i37g2000hsd.googlegroups.com... > > This thought experiment is like the classic twin paradox, but in this > > expirement both twins leave earth and travel symmetric return trips in > > opposite directions. > > > Since the paths taken by the twins in this experiment are symmetric, > > they must be the same age when they meet on their return to earth. > > > In this experiment the twins maintain constant observation of each > > other's clocks, from when they depart until they return and find that > > their clocks tell the same time. > > > Special relativity says that each twin must observe that the other's > > clock is running slow, and at no time does special relativity allow > > for an observation which shows that the other clock is running fast. > > No, special relativity says much more precise than that > "moving clocks" are running slow. The Lorentz-Fitzgerald transform is more precise that my description, but that doesn't mean that my description is wrong. > > It says something about intertial observers who measure > times between ticks on remote, moving clocks. > > When your two clocks fly apart, each clock will measure > this time to be longer and conclude that the other clock > is "running slower". > While clock A is coasting, according to clock A, each > tick on clock A is simultaneous with some tick on clock B > with a smaller time value. > While clock B is coasting, according to clock B, each > tick on clock B is simultaneous with some tick on clock A > with a smaller time value. Yes, that is the standard theory. > > After clock A has made its turnaround, it has shifted to > another inertial frame, in which according to clock A, each > tick on clock A is simultaneous with some tick on clock B > with a larger time value. > After clock B has made its turnaround, it has shifted to > another inertial frame, in which according to clock B, each > tick on clock B is simultaneous with some tick on clock A > with a larger time value. Wrong. The other clock tick is still observed to have a smaller time value. This is because in the Lorentz-Fitzgerald transform the relative velocity term is squared, making the the issue of the clocks separating vs the clocks approaching irrelevant to the amount of time dilation. http://en.wikipedia.org/wiki/Time_dilation
From: Josef Matz on 21 Nov 2007 11:48 "Dirk Van de moortel" <dirkvandemoortel(a)ThankS-NO-SperM.hotmail.com> schrieb im Newsbeitrag news:tWT0j.217820$gM5.12435682(a)phobos.telenet-ops.be... > > "colp" <colp(a)solder.ath.cx> wrote in message news:06b84031-18aa-4644-bfb7-43f49f46ae6a(a)i37g2000hsd.googlegroups.com... > > This thought experiment is like the classic twin paradox, but in this > > expirement both twins leave earth and travel symmetric return trips in > > opposite directions. > > > > Since the paths taken by the twins in this experiment are symmetric, > > they must be the same age when they meet on their return to earth. > > > > In this experiment the twins maintain constant observation of each > > other's clocks, from when they depart until they return and find that > > their clocks tell the same time. > > > > Special relativity says that each twin must observe that the other's > > clock is running slow, and at no time does special relativity allow > > for an observation which shows that the other clock is running fast. > > No, special relativity says much more precise than that > "moving clocks" are running slow. > > It says something about intertial observers who measure > times between ticks on remote, moving clocks. > > When your two clocks fly apart, each clock will measure > this time to be longer and conclude that the other clock > is "running slower". > While clock A is coasting, according to clock A, each > tick on clock A is simultaneous with some tick on clock B > with a smaller time value. > While clock B is coasting, according to clock B, each > tick on clock B is simultaneous with some tick on clock A > with a smaller time value. > > After clock A has made its turnaround, it has shifted to > another inertial frame, in which according to clock A, each > tick on clock A is simultaneous with some tick on clock B > with a larger time value. > After clock B has made its turnaround, it has shifted to > another inertial frame, in which according to clock B, each > tick on clock B is simultaneous with some tick on clock A > with a larger time value. > > > > > The paradox is that special relativity says that a twin will never see > > the other twin's clock catch up, but the clocks must show the same > > time at the end of the experiment because of symmetry. > > When they finally meet, for both clocks, this larger time reading of > the simultaneous events on the other clock is compensated by the > "more slowly running time" on that clock such that they read the > same time when they are reunited. > Hello Dirk If you could mathematically demonstrate that the time delays of the symmetric clock A as viewed by B can be compensated somehow you have solved the paradox ! Would you tell us idiots how this runs in SR ? And please: SR says that both clocks go physically physically different (slower) than the other. Josef > Dirk Vdm > [copy and follow-up to sci.physics.relativity] >
From: mluttgens on 21 Nov 2007 13:02 On Nov 21, 11:40 am, "Dirk Van de moortel" <dirkvandemoor...(a)ThankS-NO- SperM.hotmail.com> wrote: > "colp" <c...(a)solder.ath.cx> wrote in messagenews:06b84031-18aa-4644-bfb7-43f49f46ae6a(a)i37g2000hsd.googlegroups.com... > > This thought experiment is like the classic twin paradox, but in this > > expirement both twins leave earth and travel symmetric return trips in > > opposite directions. > > > Since the paths taken by the twins in this experiment are symmetric, > > they must be the same age when they meet on their return to earth. > > > In this experiment the twins maintain constant observation of each > > other's clocks, from when they depart until they return and find that > > their clocks tell the same time. > > > Special relativity says that each twin must observe that the other's > > clock is running slow, and at no time does special relativity allow > > for an observation which shows that the other clock is running fast. > > No, special relativity says much more precise than that > "moving clocks" are running slow. > > It says something about intertial observers who measure > times between ticks on remote, moving clocks. > > When your two clocks fly apart, each clock will measure > this time to be longer and conclude that the other clock > is "running slower". > While clock A is coasting, according to clock A, each > tick on clock A is simultaneous with some tick on clock B > with a smaller time value. > While clock B is coasting, according to clock B, each > tick on clock B is simultaneous with some tick on clock A > with a smaller time value. > > After clock A has made its turnaround, it has shifted to > another inertial frame, in which according to clock A, each > tick on clock A is simultaneous with some tick on clock B > with a larger time value. > After clock B has made its turnaround, it has shifted to > another inertial frame, in which according to clock B, each > tick on clock B is simultaneous with some tick on clock A > with a larger time value. > > > > > The paradox is that special relativity says that a twin will never see > > the other twin's clock catch up, but the clocks must show the same > > time at the end of the experiment because of symmetry. > > When they finally meet, for both clocks, this larger time reading of > the simultaneous events on the other clock is compensated by the > "more slowly running time" on that clock such that they read the > same time when they are reunited. > > Dirk Vdm > [copy and follow-up to sci.physics.relativity] Blah Balh Blah You are not even able to solve an elementary problem about a relativistic rocket! Marcel Luttgens
From: Sue... on 21 Nov 2007 13:21
On Nov 21, 11:48 am, "Josef Matz" <josefm...(a)arcor.de> wrote: > "Dirk Van de moortel" <dirkvandemoor...(a)ThankS-NO-SperM.hotmail.com> schrieb > im Newsbeitragnews:tWT0j.217820$gM5.12435682(a)phobos.telenet-ops.be... > > > "colp" <c...(a)solder.ath.cx> wrote in message > > news:06b84031-18aa-4644-bfb7-43f49f46ae6a(a)i37g2000hsd.googlegroups.com... > > > > > > > > This thought experiment is like the classic twin paradox, but in this > > > expirement both twins leave earth and travel symmetric return trips in > > > opposite directions. > > > > Since the paths taken by the twins in this experiment are symmetric, > > > they must be the same age when they meet on their return to earth. > > > > In this experiment the twins maintain constant observation of each > > > other's clocks, from when they depart until they return and find that > > > their clocks tell the same time. > > > > Special relativity says that each twin must observe that the other's > > > clock is running slow, and at no time does special relativity allow > > > for an observation which shows that the other clock is running fast. > > > No, special relativity says much more precise than that > > "moving clocks" are running slow. > > > It says something about intertial observers who measure > > times between ticks on remote, moving clocks. > > > When your two clocks fly apart, each clock will measure > > this time to be longer and conclude that the other clock > > is "running slower". > > While clock A is coasting, according to clock A, each > > tick on clock A is simultaneous with some tick on clock B > > with a smaller time value. > > While clock B is coasting, according to clock B, each > > tick on clock B is simultaneous with some tick on clock A > > with a smaller time value. > > > After clock A has made its turnaround, it has shifted to > > another inertial frame, in which according to clock A, each > > tick on clock A is simultaneous with some tick on clock B > > with a larger time value. > > After clock B has made its turnaround, it has shifted to > > another inertial frame, in which according to clock B, each > > tick on clock B is simultaneous with some tick on clock A > > with a larger time value. > > > > The paradox is that special relativity says that a twin will never see > > > the other twin's clock catch up, but the clocks must show the same > > > time at the end of the experiment because of symmetry. > > > When they finally meet, for both clocks, this larger time reading of > > the simultaneous events on the other clock is compensated by the > > "more slowly running time" on that clock such that they read the > > same time when they are reunited. > > Hello Dirk > > If you could mathematically demonstrate that the time delays of the > symmetric clock A as viewed by B can be > compensated somehow you have solved the paradox ! > > Would you tell us idiots how this runs in SR ? > > And please: SR says that both clocks go physically physically different > (slower) than the other. You have to use a light-clock moving in stationary media to get the *physical* behavior because light does not move inertially. http://www.anselm.edu/homepage/dbanach/st9.jpg <<A Lorentz transformation or any other coordinate transformation will convert electric or magnetic fields into mixtures of electric and magnetic fields, but no transformation mixes them with the gravitational field. >> http://www.aip.org/pt/vol-58/iss-11/p31.html If you want to relate the light-clock's rate to inertial frames or kinetic energy, mass/energy equivalence is used to couple to an inertial field or pseudo-ether. http://www.bartleby.com/173/15.html What the travelers see each other's light-clocks or normal clock doing is irrelevant. They all have to agree on the total number of ticks at experiment's conclusion and normal clocks that are not responsive to motion through a dielectic shouldn't change rate. Sue... > > Josef > > > |