The Symmetric Twin Paradox
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From: Koobee Wublee on 16 Jun 2010 19:49 On Jun 16, 10:12 am, hagman <goo...(a)von-eitzen.de> wrote: > On 16 Jun., 09:21, Koobee Wublee wrote: > > Let me chime in. There have been no experiments showing that > > accelerating does indeed exhibit any time dilation. So, the classical > > resolution as proposed by Einstein the nitwit, the plagiarist, and the > > liar is totally bullshit in the first place. <shrug> > > Of course, such experiments have been made with fast-moving atomic > clocks, say.. No experiments can support the existence of a paradox. In fact, there are not a single experiment that shows so. <shrug> > The time differences were more subtle than with a twin moving > at almost c for a long time, but fully consistent with Einstein's > theory. Einstein was a nitwit, a plagiarist, and a liar. The nitwit came up with nothing original and innovative. Under Larmors transform, it manifests no twins paradox. Only under Poincares work now called the Lorentz transform that it does. Below is a link explaining the differences. http://groups.google.com/group/sci.physics.relativity/msg/c5a0a3c587fd4df4?hl=en > > That is true. Not to mention that twins with the same acceleration > > profile can also coast away without any acceleration for some random > > time. This will enter into the time dilation in which there is > > absolutely no mathematical remedy or resolution for that one. <shrug> > > > Don't expect the self-styled physicists to understand that one. They > > are indeed morons who cannot even understand or comprehend the most > > basic of logics. <shrug> > > In order to level out the effects of the intermediate period of > acceleration > each twin will be better off, calculation-wise, to resort to some > inertial > system. Why not the point they started from and meet again? Go ahead and start the calculation then for the time where each twin coast away or towards each other without any acceleration and with non- zero speed. It should be very easy. In fact, intelligent ones would not even attempt to because the mutual time dilation can be built up fact depending on the time of coasting (with no acceleration applied). <shrug>
From: Inertial on 16 Jun 2010 20:10 "colp" <colp(a)solder.ath.cx> wrote in message news:16fe04c0-8313-4242-87de-4ca11a04e750(a)42g2000prb.googlegroups.com... > On Jun 17, 9:24 am, "Peter K" <pe...(a)parcelvej.dk> wrote: >> "colp" <c...(a)solder.ath.cx> wrote in message >> >> news:267c724a-a11c-4cfe-ae6d-b5b9395cf382(a)a39g2000prb.googlegroups.com... >> >> >> >> > The classic twin paradox is asymmetric in that one twin remains on >> > Earth while the other leaves (i.e. only one of them accelerates and >> > deaccelerates). In the symmetric twin paradox both twins leave Earth, >> > setting out in opposite directions and returning to Earth at the same >> > time. The conventional explanation for the classic twin paradox is >> > since only one twin accelerates, the ages of the twins will be >> > different. In the symmetric case this argument cannot be applied. >> >> > The paradox of the symmetric twins is that according to special >> > relativity (SR) each twin observes the other twin to age more slowly >> > both on the outgoing leg >> > and the return leg, so SR paradoxically predicts that each twin will >> > be younger than >> > the other when they return to Earth. >> >> > The symmetric twin paradox is described more fully in the following >> > paper: >> >> > The Twin Paradox Revisited and Reformulated -- On the Possibility of >> > Detecting Absolute Motion >> > Authors: G. G. Nyambuya, M. D. Ngobeni >> >> >http://adsabs.harvard.edu/abs/2008arXiv0804.2008N >> >> > "We introduce a symmetric twin paradox whose solution can not be found >> > within the currently accepted provinces of the STR if one adopts the >> > currently accepted philosophy of the STR namely that it is impossible >> > for an inertial observer to determine their state of motion." >> >> The only way to check this, is to send a couple of watches out on a >> journey - say one to paris and back, and one to New York and back. Then >> when >> they get back to NZ we can check the time on each of them! Sheesh, how >> hard >> was that? > > Your proposal does not test the paradox because the paradox involves > observations of time dilation of non-local frames of reference. Your > watches only record the passage of time for their own local frame of > reference. No .. its exactly the same (other than we aren't talking purely inertial frame of reference) But would need to make the trip north-south from the equator .. east west doesn't work for that - its not symmetric.
From: spudnik on 16 Jun 2010 20:13 doc Atlas, there is no paradox, if you accept that there is no phenomenon, including sub-atomic angular momenta, that"goes" faster than light. see, it was only a "twin paradox" til explained via Einstein et al's extension of Galilean relativity -- a strawman, really. are you going to argue Ole Roemner's dyscovery of the "retardation" of light, way back, when ever? your proposed "balancing" is almost cute, but iff they accelerate at the same average rate, there clocks'll be in synch at the rendezvous; so, you've described a Twins Miming Each Other "experiment" of no account. just get rid of the useless notion of Minkowski's phase-space, and you won't have to think too hard about it. > Go ahead and start the calculation then for the time where each twin > coast away or towards each other without any acceleration and with non- > zero speed. It should be very easy. In fact, intelligent ones would > not even attempt to because the mutual time dilation can be built up > fact depending on the time of coasting (with no acceleration > applied). --Stop BP's Waxman's arbitrageurs' CAP&TRADE Last Bail-out of Wall Street, the City of London and George Soros et al ad vomitorium! --Fermat's next theorem! http://wlym.com
From: Peter K on 16 Jun 2010 20:13 "colp" <colp(a)solder.ath.cx> wrote in message news:16fe04c0-8313-4242-87de-4ca11a04e750(a)42g2000prb.googlegroups.com... > On Jun 17, 9:24 am, "Peter K" <pe...(a)parcelvej.dk> wrote: >> "colp" <c...(a)solder.ath.cx> wrote in message >> >> news:267c724a-a11c-4cfe-ae6d-b5b9395cf382(a)a39g2000prb.googlegroups.com... >> >> >> >> > The classic twin paradox is asymmetric in that one twin remains on >> > Earth while the other leaves (i.e. only one of them accelerates and >> > deaccelerates). In the symmetric twin paradox both twins leave Earth, >> > setting out in opposite directions and returning to Earth at the same >> > time. The conventional explanation for the classic twin paradox is >> > since only one twin accelerates, the ages of the twins will be >> > different. In the symmetric case this argument cannot be applied. >> >> > The paradox of the symmetric twins is that according to special >> > relativity (SR) each twin observes the other twin to age more slowly >> > both on the outgoing leg >> > and the return leg, so SR paradoxically predicts that each twin will >> > be younger than >> > the other when they return to Earth. >> >> > The symmetric twin paradox is described more fully in the following >> > paper: >> >> > The Twin Paradox Revisited and Reformulated -- On the Possibility of >> > Detecting Absolute Motion >> > Authors: G. G. Nyambuya, M. D. Ngobeni >> >> >http://adsabs.harvard.edu/abs/2008arXiv0804.2008N >> >> > "We introduce a symmetric twin paradox whose solution can not be found >> > within the currently accepted provinces of the STR if one adopts the >> > currently accepted philosophy of the STR namely that it is impossible >> > for an inertial observer to determine their state of motion." >> >> The only way to check this, is to send a couple of watches out on a >> journey - say one to paris and back, and one to New York and back. Then >> when >> they get back to NZ we can check the time on each of them! Sheesh, how >> hard >> was that? > > Your proposal does not test the paradox because the paradox involves > observations of time dilation of non-local frames of reference. Your > watches only record the passage of time for their own local frame of > reference. Oh. But when they get back can't I see what they measured? Then I've at least observed the effects on a "non local frame of reference". Otherwise, could we get the watches to continually send small signals back to me when the travel? The could send a "tick" and a "tock" for each of their seconds - and I could compare them to the "ticks" and "tocks" on a third watch I'm holding. Would that work?
From: Androcles on 16 Jun 2010 20:22
"Peter K" <peter(a)parcelvej.dk> wrote in message news:4c1968bf$1(a)news.xnet.co.nz... | "colp" <colp(a)solder.ath.cx> wrote in message | news:16fe04c0-8313-4242-87de-4ca11a04e750(a)42g2000prb.googlegroups.com... | > On Jun 17, 9:24 am, "Peter K" <pe...(a)parcelvej.dk> wrote: | >> "colp" <c...(a)solder.ath.cx> wrote in message | >> | >> news:267c724a-a11c-4cfe-ae6d-b5b9395cf382(a)a39g2000prb.googlegroups.com... | >> | >> | >> | >> > The classic twin paradox is asymmetric in that one twin remains on | >> > Earth while the other leaves (i.e. only one of them accelerates and | >> > deaccelerates). In the symmetric twin paradox both twins leave Earth, | >> > setting out in opposite directions and returning to Earth at the same | >> > time. The conventional explanation for the classic twin paradox is | >> > since only one twin accelerates, the ages of the twins will be | >> > different. In the symmetric case this argument cannot be applied. | >> | >> > The paradox of the symmetric twins is that according to special | >> > relativity (SR) each twin observes the other twin to age more slowly | >> > both on the outgoing leg | >> > and the return leg, so SR paradoxically predicts that each twin will | >> > be younger than | >> > the other when they return to Earth. | >> | >> > The symmetric twin paradox is described more fully in the following | >> > paper: | >> | >> > The Twin Paradox Revisited and Reformulated -- On the Possibility of | >> > Detecting Absolute Motion | >> > Authors: G. G. Nyambuya, M. D. Ngobeni | >> | >> >http://adsabs.harvard.edu/abs/2008arXiv0804.2008N | >> | >> > "We introduce a symmetric twin paradox whose solution can not be found | >> > within the currently accepted provinces of the STR if one adopts the | >> > currently accepted philosophy of the STR namely that it is impossible | >> > for an inertial observer to determine their state of motion." | >> | >> The only way to check this, is to send a couple of watches out on a | >> journey - say one to paris and back, and one to New York and back. Then | >> when | >> they get back to NZ we can check the time on each of them! Sheesh, how | >> hard | >> was that? | > | > Your proposal does not test the paradox because the paradox involves | > observations of time dilation of non-local frames of reference. Your | > watches only record the passage of time for their own local frame of | > reference. | | Oh. But when they get back can't I see what they measured? Then I've at | least observed the effects on a "non local frame of reference". | | Otherwise, could we get the watches to continually send small signals back | to me when the travel? The could send a "tick" and a "tock" for each of | their seconds - and I could compare them to the "ticks" and "tocks" on a | third watch I'm holding. Would that work? | A devout relativist will argue black is white in support of his religion. |