The Symmetric Twin Paradox
in [Relativity]
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From: Inertial on 20 Jun 2010 22:40 "Sue..." <suzysewnshow(a)yahoo.com.au> wrote in message news:1a23b1f5-4139-4906-9464-0a92e5dc74e8(a)y11g2000yqm.googlegroups.com... > 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. SR is NOT optical illusion. Again , ,Sue speaks from ignorance >> 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. It may appear absurd .. but there is no contradiction or inconsistency >> 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 r�le as > the three space co-ordinates. >> > http://www.bartleby.com/173/17.html Sue does NOT think .. she pastes the same old quotes over and over in the hope they will be relevant > In other words, clock mechanisms can't > detect coordinate systems to cooperate > with either part of your thought experiment. That is certainl 'other words' .. it has NOTHING to do with what you quoted ... just your own word-salad > << 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. It doesn't say that. Again .. you don't understand what you copy and paste.
From: Daryl McCullough on 20 Jun 2010 22:42 colp says... > >On Jun 21, 1:10=A0am, stevendaryl3...(a)yahoo.com (Daryl McCullough) >wrote: >> What SR says for those sorts of thought experiments is that, as measured >> in any inertial coordinate system, >> >> 1. Light travels at constant velocity, at speed c, in all directions, >> independent of the motion of the source. >> 2. An ideal clock traveling at speed v for time period t will show >> an elapsed time of T =3D t square-root(1-(v/c)^2). > >You are saying that T = t / gamma, where gamma = 1 / square-root(1-(v/ >c)^2) > >How do you reconcile this with your earlier statement? > ><quote> >If you look at the Lorentz transformation for time, you >find: > >t' = gamma (t - vx/c^2) These are talking about two different things: (1) the elapsed time on a *single* clock, and (2) the time *coordinate*. They are related, but are not the same thing. To set up a coordinate system, you can't just use a single clock, you have to have a system of clocks, and you have to have a way of synchronizing them. The Lorentz transformation for the time coordinate takes into account two different things: (1) the relative rate of one clock as measured in a coordinate system in which that clock is not at rest, and (2) the fact that two different frames will use different synchronizations for their clocks. Let's see how that works for just two clocks. Suppose you have two clocks, A and B, that are lined up along the x-axis, and are at rest in some frame F'. Suppose that the distance between the clocks is X' as measured in frame F'. If you are an observer in frame F', then how do you go about making sure that your clocks are synchronized? Well, here's one way of doing it: You know the distance between the clocks, X'. You know that light has speed c. So it takes light a time X'/c to travel from clock A to clock B. So what you do is set clock A to show t'=0, and then immediately send a light signal to clock B. When B receives the signal, that clock is set to t'=X'/c. Voila! They are synchronized. Now, let's look at things from the point of view of frame F, the frame in which the two clocks are moving at speed v along the x-axis (with clock B ahead of clock A). Let's assume that clock A started at x=0 at time t=0, so the location of clock A at later times is given by x = vt. As measured in frame F, the distance between the clocks is not X', but is X'/gamma, because of length contraction. So the location of B is given by x = vt + X'/gamma. That's the first complication. The second complication is that when A sends a light signal to clock B, the light has to travel further than just X'/gamma, because B is moving away the whole time. To compute this effect, let t be the time it takes for the light to travel from A to B. During that time, light travels a distance ct, and clock B travels a distance vt. So the light has to travel a distance of X'/gamma + vt in time t. So we must have: ct = X'/gamma + vt or t = X'/gamma * 1/(c-v) Since 1/gamma = gamma * 1/gamma^2, and 1/gamma^2 = 1-(v/c)^2, we can rewrite this as follows: t = X' gamma (1-(v/c)^2)/(c-v) = X'/c^2 gamma (c^2 - v^2)/(c-v) = X'/c^2 gamma (c+v) So, from the point of view of frame F, it takes time X'/c^2 gamma (c+v) for the light signal to travel from A to B. During that time, clock A advances by 1/gamma * (that time), because moving clocks are slowed. So A advances by X'/c^2 (c+v) by the time the light signal reaches B. But, the synchronization scheme involved setting clock B to time X'/c. So as measured in frame F, the two clocks are out of synch; the difference in their setting is X'/c^2 (c+v) - X'/c = X' v/c^2. Clock B is behind clock A by an amount X' v/c^2. In addition, clock B is running slow by a factor of gamma. So the time t' shown on clock B is given by: t' = t/gamma + X' v/c^2 We said above that B's location is given by x = vt - X'/gamma. So we can express X' in terms of x as: X' = gamma (x-vt). Putting this into the equation for t' gives: t' = t/gamma - gamma (x-vt) v/c^2 = gamma (t/gamma^2 - xv/c^2 + v^2/c^2 t) = gamma (t (1-(v/c)^2) - xv/c^2 + (v/c)^2 t) = gamma (t - xv/c^2) So, in frame F, although each clock is running slower by a factor of gamma, the effect of the synchronization causes the coordinate t' to vary with x as predicted by the Lorentz transformations. -- Daryl McCullough Ithaca, NY
From: colp on 20 Jun 2010 23:10 On Jun 21, 12:10 pm, "Inertial" <relativ...(a)rest.com> wrote: > "colp" <c...(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 According to your logic you cannot be taken seriously. > > > You are the one claiming that turnaround compensates for the observed > > time dilation of the outgoing and return legs. > > It does According to your logic you cannot be taken seriously. > > > Using your own > > standards, you cannot be taken seriously until you show the math which [fixed typo] > > backs up your claim. > > I have No, you haven't. All you did was to pick a set of frames which didn't result in any paradox being apparent. There isn't any basis in SR for such selective application of reference frames.
From: Inertial on 20 Jun 2010 23:25 "colp" <colp(a)solder.ath.cx> wrote in message news:572cf302-7007-41ba-a08d-77cf2dde07a7(a)40g2000pry.googlegroups.com... > On Jun 21, 12:10 pm, "Inertial" <relativ...(a)rest.com> wrote: >> "colp" <c...(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 > > According to your logic you cannot be taken seriously. Of course I can .. by my own, and any reasonable logic. You are the one making claims about SR. . the burden of proof of these is on YOU. So far , you've failed >> > You are the one claiming that turnaround compensates for the observed >> > time dilation of the outgoing and return legs. >> >> It does > > According to your logic you cannot be taken seriously. Of course I can .. by my own, and any reasonable logic. You are the one making claims about SR. . the burden of proof of these is on YOU. So far , you've failed >> > Using your own >> > standards, you cannot be taken seriously until you show the math which >> > [fixed typo] >> > backs up your claim. >> >> I have > > No, you haven't. Yes .. I did. You are the one making claims about SR. . the burden of proof of these is on YOU. So far , you've failed > All you did was to pick a set of frames which didn't > result in any paradox being apparent. Because there *is* no paradox .. it doesn't matter *what* inertial frame you take. And note that you cannot just go changing the frame in which you are doing the analysis will-nilly .. that is an error called frame-jumping. You need to be very careful if you attempt to do that. > There isn't any basis in SR for > such selective application of reference frames. Tell me which inertial frame you want me to use. And why you expect it to give a paradox, and why an analysis in that frame is any better that the frame I chose to use ... which was TEH SAME FRAME that you chose (where x = 0 in the first leg for one of the twins).
From: colp on 20 Jun 2010 23:26
On Jun 21, 12:24 pm, "Sue..." <suzysewns...(a)yahoo.com.au> wrote: > 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. How do you think that works in the case of the symmetric twin paradox? > > > > > 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. No, that isn't what I'm referring to. Will you please answer my question? |