The Symmetric Twin Paradox
in [Relativity]
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From: Paul Stowe on 20 Jun 2010 19:10 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 :)....
From: Uncle Ben on 20 Jun 2010 19:20 On Jun 20, 6:29 pm, colp <c...(a)solder.ath.cx> wrote: > On Jun 21, 4:53 am, Uncle Ben <b...(a)greenba.com> 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? > > > > > <<Einstein's 1905 presentation of special relativity was soon > > > > supplemented, in 1907, by Hermann Minkowski, who showed that > > > > the relations had a very natural interpretation[C 5] in terms > > > > of a unified four-dimensional "spacetime" in which absolute > > > > intervals are seen to be given by an extension of the > > > > Pythagorean theorem.>>http://en.wikipedia.org/wiki/Lorentz_ether_theory#The_shift_to_relati... > > > > > << 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 > > > > > Sue...- Hide quoted text - > > > > - Show quoted text - > > > See the recent thread addressed to you: "colp, why did Einstein use > > the word 'relative'?" in sci.physics.relativity. > > I'm not interested unless it relates to the symmetric twin paradox. > It's up to you to show relevance.- Hide quoted text - > > - Show quoted text - It explains the contradiction.
From: colp on 20 Jun 2010 19:56 On Jun 21, 1:10 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > colp says... > > >On Jun 20, 1:11=A0am, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote: > >> You MUST learn what SR ACTUALLY says. > > >What do you think that SR actually says about the symmetric twin > >thought experiment? > > What SR says is that the laws of mechanics and electromagnetism > are invariant under the Lorentz transformations. That's the > physical content of SR. This was already true of Maxwell's > equations, but not of Newtonian mechanics. So part of the development > of SR involved finding a replacement to Newton's equations of motion > that were invariant under the Lorentz transformations. > > I know that's not what you meant---you don't actually want to > know about real physics, you want to know about thought experiments > involving clocks and twins and rockets. That's not true: I do want to know about real physics, however real physics is irrelevant when identifying a paradox, which is all I am interested in here. > > 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 = 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) So t' does not simply depend on t and gamma, it depends on v and x as well. If v is changing, then that will affect the relationship between t and t'. </quote> > 3. An extended object traveling at speed v will, after reaching its > equilibrium shape, be contracted in the direction of motion by a factor > of square-root(1-(v/c)^2). > 4. An object not under the influence of any forces will move at > constant speed. > > Those are all true in any inertial coordinate system. The final claim, > which is actually derivable from the first 4, (alternatively, > 2&3 can be derived from 5) is: > > 5. If C is an inertial coordinate system, and C' is obtained from C > by a translation, rotation or Lorentz transformation, then C' is > an inertial coordinate system. > > So anything you want to derive about twins moving around, sending > light signals, carrying clocks, etc. is derivable from 1-5. Pick > any inertial coordinate system you like to analyze the situation, > and compute away using 1-5. > > The usual twin paradox: A twin hops in a rocket at time t=0, > speeds away to a distance L, turns around rapidly, and speeds back > at speed v. How much older is he when he returns? By rule 2 above, > it's pretty simple: the elapsed time on the rocket for the full > trip will be 2 (L/v) square-root(1-(v/c)^2). > In contrast, the stay-at-home twin ages by an amount 2L/v. > Done. The traveling twin is younger. > > Now, if you want to compute the results from the point of view > of some other frame, that's fine. Go ahead, and use 1-5. > Let's compute it from the outgoing frame of the traveling twin. > > First, we have to figure out what happens in this frame, since > the description above was from the point of view of the stay-at-home > twin. So why should your approach for the traveling twin be any different to to that of the stay-at-home twin? > To get the corresponding description for the new frame, we > use the Lorentz transformations. But not the same one that you used previously. > > There are three major events involved: > > E1: The two twins depart. > In the stay-at-home frame, this has coordinates (x=0, t=0). > By the Lorentz transformations, in the new frame, it has the > same coordinates: (x'=0, t'=0). > > E2: The traveling twin turns around. > In the stay-at-home frame, this has coordinates (x=L, t=L/v). > By the Lorentz transformations, in the new frame, it has the > coordinates: x'=0, > t' = gamma (t-vx/c^2) > = gamma (L/v - vL/c^2) > = L/v gamma (1-v^2/c^2) > = L/v 1/gamma (since gamma = 1/square-root(1-(v/c)^2). What basis do you have for applying t' = gamma (t-vx/c^2) here, when you applied T = t / gamma previously? > > E3: The twins reunite. > In the stay-at-home frame, this has coordinates (x=0, t=2L/v). > In the new frame, it has coordinates > x' = gamma (x-vt) = -gamma 2L > t' = gamma (t - vx/c^2) = gamma 2L/v > > Okay, so in the new frame, the story looks like this: > The stay-at-home twin travels at constant speed v in the > -x direction the whole time. He moves from x'=0 to x'=-gamma 2L. > The total time for the trip is gamma 2L/v. > So using our rule 2 above, if the stay-at-home twin travels > at speed v for a length of time t, then he will age by > an amount t square-root(1-(v/c)^2). What basis do you have for applying T = t / gamma here, when you applied t' = gamma (t-vx/c^2) previously? > In this case, the > trip took time gamma 2L/v as measured in this frame, so > the stay-at-home twin ages by an amount: > > gamma 2L/v * square-root(1-(v/c)^2) > = 2L/v (since gamma = 1/square-root(1-(v/c)^2) > > So, in this frame, the stay-at-home twin is *also* 2L/v, > just like in the first frame. > > Now, let's calculate the age of the traveling twin in this > frame: > > As viewed in this frame, the traveling twin stays put at > x'=0 between the times t'=0 and t'= L/v 1/gamma. His > velocity during this period is 0 in this new frame, so > during the first part of the trip, he ages by an amount > L/v 1/gamma. > > What about the second part of the trip (after turnaround)? > During this part of the trip, the traveling twin goes from > x'=0 at time t'=L/v 1/gamma to x'=-gamma 2L at time > t' = gamma 2L/v. > > So let's figure out what the traveling twin's speed was > during this part of the trip. The total elapsed time was: > > t_end - t_start > = gamma 2L/v - L/v 1/gamma > = gamma L/v (2 - 1/gamma^2) > = gamma L/v (1 + (v/c)^2) > > (where I used 1/gamma^2 = 1 - (v/c)^2) > > The total distance traveled was: > > |x_end - x_start| > = gamma 2L > > So the velocity during this part of the journey is > v' = gamma 2L/(gamma L/v (1 + (v/c)^2)) > = 2v/(1+(v/c)^2) > (This is what would have been obtained using the velocity > addition formula) > > Now, let's calculate square-root(1-(v'/c)^2) for this > part of the trip: > > (v'/c)^2 = 4(v/c)^2/(1 + 2(v/c)^2 + (v/c)^4) > > 1 - (v'/c)^2 = (1 - 2(v/c)^2 + (v/c)^4)/(1+2(v/c)^2 + (v/c)^4) > square-root(1-(v'/c)^2) = (1-(v/c)^2)/(1+(v/c)^2) > > To find out how much the traveling twin ages during the > last part of the journey, we multiply this factor by > the elapsed time, which we calculated as gamma L/v (1 + (v/c)^2) > > Aging during last part = gamma L/v (1 + (v/c)^2) * (1-(v/c)^2)/(1+(v/c)^2) > = gamma L/v (1-(v/c)^2) > = L/v 1/gamma (where again I used gamma = 1/square-root(1-(v/c)^2). > > So the total aging done by the traveling twin turns out to be > L/v 1/gamma during both legs of the journey. So when the journey > is over, he has aged by > > 2L/v 1/gamma > > while the stay-at-home twin has aged by > 2L/v > > So once again, the traveling twin is younger when they get back together. > > So if you use the actual equations of SR (as opposed to guessing), then > the results are completely consistent: In *every* inertial coordinate > system, the answer is the same: The stay-at-home twin ages more than > the traveling twin. > > We could go through the symmetric case, if you want, but the results > will be this: > > During the outward trip, each twin ages by an amount > L/v square-root(1-(v/c)^2). > During the return trip, each twin ages by the same amount. > So when the twins get back together, they have both aged > by an amount 2L/v square-root(1-(v/c)^2). > > They are the same age when they get back together. > Every inertial coordinate system agrees on this answer. > > There is no paradox, in the sense of two different ways to > compute things leading to two different results for the > ages. The paradox comes up when you *fail* to correctly > use SR. Unless you can show a rational basis for an apparently arbitrary application of two different formula for time dilation, your conclusion is invalid.
From: train on 20 Jun 2010 19:56 On Jun 20, 10:52 pm, "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. > > http://chimera.roma1.infn.it/apehdoc/apemille/grid.gif > > Build the above grid, climb inside and shoot a wabbit. > Have a friend reposition the grid. > Shoot the wabbit again. > > Is it clear to you that the X,Y,Z descriptions of your > first and second shots will differ even tho the > shots are identical? > > Your response to Daryl indicates you are confused > about this. It is pointless to discuss the > 4D description until you are sure about the > 3D version. > > Are you quite sure the coordinate systems > used to describe experiments, don't in any way > affect an experiment, only its description in > terms of the coordinates? > > ??? > > Sue... > > > > > > <<Einstein's 1905 presentation of special relativity was soon > > > supplemented, in 1907, by Hermann Minkowski, who showed that > > > the relations had a very natural interpretation[C 5] in terms > > > of a unified four-dimensional "spacetime" in which absolute > > > intervals are seen to be given by an extension of the > > > Pythagorean theorem.>> > > http://en.wikipedia.org/wiki/Lorentz_ether_theory > > > > > > > > << 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. >> > I think I like this - its intuitive. > http://www.bartleby.com/173/17.html > > > > > > Sue...
From: colp on 20 Jun 2010 20:03
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. 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? If so, what do you think that SR does predict in this case? |