The Symmetric Twin Paradox
in [Relativity]
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From: Inertial on 20 Jun 2010 20:10 "colp" <colp(a)solder.ath.cx> wrote in message news:48810f79-2ef0-495b-bb2b-cbd27fd317ce(a)11g2000prv.googlegroups.com... > On Jun 20, 9:09 pm, "Inertial" <relativ...(a)rest.com> wrote: >> "colp" <c...(a)solder.ath.cx> wrote in message >> >> news:d909afc3-3c9b-4e50-80e2-e1a97793fbad(a)23g2000pre.googlegroups.com... >> >> > 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. >> >> you are the one ignoring what SR predict and making up your own nonsense. > > Now you are contradicting yourself. Nope > Later in your post, you agree 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, Yes > and that this observation applies both on the > outgoing and return legs, and it applies for both twins. Yes The turnaround event is VERY important where the rest frame of each twin changes dramatically. >> >> > but SR does not >> >> > predict that result >> >> >> WRONG >> >> > No, not wrong. >> >> Yes .. WRONG .. show the math is you think otherwise >> >> So far nothing from you at all > > Wrong. The following is an excerpt from an earlier post mine: > >> t' = gamma (t - vx/c^2) That's just writing down just one of the Lorentz transforms equations .. that's not showing the SR math that shows how you arrive at your conclusion > OK. For our example the twins both have x as zero when the experiment > starts, so for the outgoing leg: > > t' = gamma t OK . .so we have two frames here S the frame of one of the twin, in which that twin remains at rest for the initial leg and the second twin moves away, and t being the time in that frame S' the frame of the second twin, in which that twin remains at rest for the initial leg and the first twin moves away in the opposite direction, and t' being the time in that frame If we look at the first twin in frame S for the initial leg.. the time there is always t. The correspond time in the second twin's frame is t' = gamma t. > and since gamma is greater than one, t' > t and the twins both see > each other's time to be dilated like I said in my previous post. And as I've agree. > Let's call the distance between the twins at turnaround x. > > x = vt, where v is 0.866c and t = 200 seconds per your example. The particular values don't matter. I didn't have a particular example. > Let's call the velocity for the return leg r. > > r = -v [snip rest of math following mistake as it doesn't matter] Nope. There's your mistake. You've ignored that BOTH twins have change their frames of reference at turn around. Indeed, the first clock changes from being at rest in S to being at rest in S'. The second clock changes from being at rest in S' to being at rest in S The second leg starts when a given twin reaches a time T_ret on their own clock. Time dilation means that, as seen from frame S, the second twin's clock is ticking slower, so it reaches time T_ret after the first twin (at rest in S) does. So in frame S the series of events are (in order): 1) twin B moves away from A (who remains at rest at x=0) 2) when the clock on twin A shows time T_ret, A moves away in the same direction and speed as twin B did 3) when the clock on twin B shows time T_ret, B stops and is now at rest some distance from the origin 4) twin A stops and is now at rest Here's a diagram in S of the motion of the twins (A and B) from A's initial frame of reference S: (you'll need a fixed space font to see it) B B B B B B B B B B A B A B A B A B A B A A A A A A A A A A As you can see, both twins do the same motion .. just in the opposite order, and so the same time is elapsed > There is nothing that occurs at turnaround which compensates for the > time dilation of the outgoing and return legs. If you disagree and > claim that it is so,then the burden of proof is yours. I just showed your math was wrong
From: Inertial on 20 Jun 2010 20:10 "colp" <colp(a)solder.ath.cx> wrote in message news:73c42da8-03e8-4f07-acbf-92c78718d7ba(a)j36g2000prj.googlegroups.com... > On Jun 20, 9:14 pm, "Inertial" <relativ...(a)rest.com> wrote: >> "colp" <c...(a)solder.ath.cx> wrote in message > >> > What do you think that SR actually says about the symmetric twin >> > thought experiment? >> >> You are the one making claims .. you'd been asked repeatedly to show the >> math backing up your claim. you refuse to do so. Until you do, you >> cannot >> be taken seriously > > I have already shown the math, and I've also reposted it in response > to an earlier post of yours. I've shown you are wrong > You are the one claiming that turnaround compensates for the observed > time dilation of the outgoing and return legs. It does > Using your own > standards, you cannot be taken seriously until you show the math with > backs up your claim. I have
From: Inertial on 20 Jun 2010 20:11 "Paul Stowe" <theaetherist(a)gmail.com> wrote in message news:80cea935-b8cc-4cb3-bf21-d38bb0d201c5(a)u20g2000pru.googlegroups.com... > On Jun 20, 3:24 pm, colp <c...(a)solder.ath.cx> wrote: >> On Jun 20, 9:14 pm, "Inertial" <relativ...(a)rest.com> wrote: >> >> > "colp" <c...(a)solder.ath.cx> wrote in message >> > > What do you think that SR actually says about the symmetric twin >> > > thought experiment? >> >> > You are the one making claims .. you'd been asked repeatedly to show >> > the >> > math backing up your claim. you refuse to do so. Until you do, you >> > cannot >> > be taken seriously >> >> I have already shown the math, and I've also reposted it in response >> to an earlier post of yours. >> >> You are the one claiming that turnaround compensates for the observed >> time dilation of the outgoing and return legs. Using your own >> standards, you cannot be taken seriously until you show the math with >> backs up your claim. > > Don't hold your breath, I've never seen him respond with substantive > analyses. Maybe you'll get 'lucky' heheheh... > > Not likely :).... You wouldn't know .. perhaps you are confusing me with someone else
From: Androcles on 20 Jun 2010 20:19 "train" <gehan.ameresekere(a)gmail.com> wrote in message news:94dda074-b06f-4afe-8874-af3abc04e720(a)s6g2000prf.googlegroups.com... On Jun 20, 10:52 pm, "Sue..." <suzysewns...(a)yahoo.com.au> wrote: > > > > > << 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 r�le as > > > the three space co-ordinates. >> > I think I like this - its intuitive. =========================== That's because you don't know what a vector space is. http://en.wikipedia.org/wiki/Vector_space This is vector addition: http://upload.wikimedia.org/wikipedia/commons/thumb/e/e6/Vector_addition3.svg/180px-Vector_addition3.svg.png This is utter bullshit, and certainly not intuitive: http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img71.gif Intuition fails in the face of mathematical rigour. You only think you like it.
From: Sue... on 20 Jun 2010 20:24
On Jun 20, 8:03 pm, colp <c...(a)solder.ath.cx> wrote: > On Jun 21, 5:52 am, "Sue..." <suzysewns...(a)yahoo.com.au> wrote: > > > 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. > > Additional rigor does not invalidate the usual interpretation of SR. It certainly does. Apply rigorous maths to an optical illusion and your interpretation will change. > > In the symmetric twin paradox, do you deny that SR predicts that each > twin observes the other twin to age more slowly in the outgoing leg? I suppose you refer to the absurdity in the 1905 paper pointed out by Paul Langevin, among others. Well, two wrongs don't make a right. > If so, what do you think that SR does predict in this case? << 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 In other words, clock mechanisms can't detect coordinate systems to cooperate with either part of your thought experiment. << Einstein's relativity principle states that: All inertial frames are totally equivalent for the performance of all physical experiments. >> http://farside.ph.utexas.edu/teaching/em/lectures/node108.html That means candles, torsion pendulum, guns and even Rolex watches can't operate differently as the result of relative motion. Sue... |