The Symmetric Twin Paradox
in [Relativity]
|
Prev: Black Hole is Black Day for Earth
Next: n-stars.
From: Daryl McCullough on 19 Jun 2010 23:09 colp says... > >On Jun 20, 12:32=A0pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) >wrote: >> colp says... > >> >The standard explanation for the classic twin paradox requires the >> >consideration of the role of acceleration and deceleration in the >> >experiment. Since you haven't considered this, your example is at odds >> >with that explanation in terms of resolving the paradox. >> >> The main point is that there *is* no paradox. There are many >> different ways to compute the age of each twin when they get >> back together, and they *all* give the same answer. > >In other words, you deny that there is a problem with your explanation >for the classic twin paradox, but you can't say why the standard >explanation claims that consideration of acceleration is necessary to >resolve the paradox when your own explanation of the paradox makes no >mention of it. There is no such thing as "the standard" explanation. There are many different ways of explaining it. In all cases, in the asymmetric case, the acceleration is what breaks the symmetry. But it is completely wrong to think that that means that "acceleration causes differential aging". The correct way to compute the differential ages is to use the rule: The elapsed time on a clock is given by T = t square-root(1-(v/c)^2) where t and v are measured in an inertial coordinate system. Acceleration does not enter into the calculation, except indirectly: it is impossible to have two different paths that start at the same point and end at the same point, unless one or the other is accelerated. -- Daryl McCullough Ithaca, NY
From: colp on 19 Jun 2010 23:33 On Jun 20, 12:32 pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > colp says... > > > > >On Jun 20, 3:57=A0am, stevendaryl3...(a)yahoo.com (Dary McCullough) wrote: > >> colp says... > > >> >Then what do you think the circumstances are in which SR predicts that > >> >a twin observes the other to age more quickly, and what mathematical > >> >relationship quantifies this? > > >> Sure. Let's assume the following set-up. One twin stays at > >> home throughout. The other zips away and comes back. > > >You are talking about the classic twin paradox, not the symmetric > >paradox. However, you haven't introduced acceleration or deceleration, > >so the issues are the same as in the symmetric case. > > The symmetric case works similarly: From the point of view > of each of the twins: > > Time period 1. Up until turnaround, the pulses from the other twin come > delayed (less than one pulse per second). > > Time period 2. After turnaround, and for a while, the pulses from > the other twin come at exactly one pulse per second. > > Time period 3. After the twin receives the last pulse from the other > twin sent before the other twin turned around, the pulses start > coming more frequently than one per second. > > The net effect in the symmetric case is that each twin sends and > receives the exact same number of pulses. Straw man. Your points are true, but the issue is what SR predicts will happen. What is different between the symmetric case and your worked example of the classic asymmetric case? You said: "So the traveling twin receives 400 pulses from the stay-at-home twin, while the stay-at-home twin receives only 200 pulses from the traveling twin." Yet there is nothing in your worked example which would make the stay- at-home twin any different to the travelling twin in terms of the math. IOW, your worked example is essentially the same as for the symmetric case. Saying that the classic case is asymmetric means nothing in your example unless that asymmetry is reflected in your math. > > >Also, you have given a worked example, rather than answering my > >questions directly. A direct answer would be that gamma is always > >larger than or equal to one, so according to SR, time dilation (or > >equivalent time if gamma is equal to one) is always observed, and time > >compression is never observed. > > That's not correct. O.K. it is a bit more complicated than that, but it doesn't affect my argument. > The Lorentz transformations do not > say that the relationship between the time coordinates > of the two observers is as simple as one time coordinate > being a multiple of the other time coordinate. > > If you look at the Lorentz transformation for time, you > find: > > t' = gamma (t - vx/c^2) OK. For our example the twins both have x as zero when the experiment starts, so for the outgoing leg: 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. 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. Let's call the velocity for the return leg r. r = -v t' = gamma (t - r.x/c^2) t' = gamma (t + v.vt/c^2) t' = gamma (t + t.v^2/c^2) t' = gamma t(1 + v^2/c^2) Since 1 + v^2/c^2 (and gamma) will always be greater than or equal to 1, the twins both see each other's time to be dilated here as well. > > 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'. > > You have to actually use the equations. > > >Thus the (previously contested) second > >point of the following argument is true: > > >1. SR predicts that each twin observes the other twin to age more > >slowly both on the outgoing leg and the return leg. > > False. I just explained it to you. No, you introduced signal transit time, which made it appear that time was passing more quickly for the remote twin when the incoming signal rate exceeded the rest rate. If you rework your example but with a zero signal transit time the paradox should become more apparent. > If by "observing the other > twin aging" you mean looking at the other twin through a powerful > telescope and seeing how old he *looks*, then what you would find > is: > > (1) During the outward journey, each twin will see the other twin > aging more slowly. > > (2) During the return journey, each twin will see the other twin > aging more rapidly. That isn't what I mean by observing the other twin. By including the effect of signal transit time, the measured signals differ from the remote time, which is what is being observed. > > Whether they end up the same age, or different ages depends on > how long the fast aging period lasts compared with the slow aging > period. At issue is not what actually happens, but what SR predicts will happen. > > >2. In no case does SR predict that a twin observes the other to > >age more quickly. > > That's not correct. Actually it is, as shown by my reduction of the Lorentz transformation. > > >A rate faster than the rest rate of pulse reception should not be > >confused with observed time compression, since the decreasing signal > >transit time results in the pulse rate increase. > > That's true, but you can reliably figure out how old each twin is > when they get back together by counting the number of pulses sent > by each twin that is received by the other twin. No, you can't figure it out reliably using SR because SR gives you different results depending on where you make your observations from.
From: Koobee Wublee on 20 Jun 2010 00:54 On Jun 19, 6:11 am, Tom Roberts wrote: > colp wrote: > > truth: SR predicts that each twin observes the other twin to age more > > slowly both on the outgoing leg and the return leg. One thumbs up. > > truth: In no case does SR predict that a twin observes the other to > > age more quickly. Two thumbs up. > > inference: SR predicts that each twin will younger than the other at > > the end of the experiment. Applaud. > All three of those are wrong. Huh! > You MUST learn what SR ACTUALLY says. <shaking my head> > That > requires STUDY, not wasting your time posting nonsense to the net. Self-styled physicists knows very little about the subjects in which they are supposed to be experts in. <shrug> Hint: You will become an Einstein Dingleberry if you continue to read books smeared with fermented diarrhea of Einstein the nitwit, the plagiarist, and the liar. In another words, accepting books written by Einstein Dingleberries will make you ever more mystified as if the academics are not mystified enough. <shrug> Truly unbelievable.
From: Koobee Wublee on 20 Jun 2010 02:02 On Jun 19, 5:07 pm, train <gehan.ameresek...(a)gmail.com> wrote: > My point is that the assertion that the twins show the same age is in > contradiction with the fact that the twins have moved relative to each > other. I cannot make it more simple than that Your point is very well taken among the truly scholars of physics. The simple logic is all in the yet also very simple mathematics of the Lorentz transform. <shrug> > The moving clock runs slow According to SR, all moving frames have slower time flow relative to an observer. Again, this is all in the Lorentz transform. <shrug> > The stay at home twin ages faster means that all stay at home twins > age faster than the all traveling twins that follow the exact same > flight profile. No, each twin should be observed to age slower according to the other twin regardless of the stay at home twin or not. Noticing the traveling twin has to accelerate away in the first place, Einstein the nitwit, the plagiarist, and the liar pulled out the nonsense that acceleration breaks the symmetry. Well, as you have proposed earlier, we can have the stay at home twin doing the traveling using the acceleration profile as the traveling twin. In doing so, if there is any effect of time dilation in would be nullify between these twins. The result unmistakably, still shows the twin's paradox. > oh maybe as AE said we have to give up common sense. And reason? Einstein the nobody was a nitwit, a plagiarist, and a liar. It is best not to listen to what this nitwit, this plagiarist, and this liar has to say. In his only book on relativity, Einstein the nitwit, the plagiarist, and the liar was able to show two equations of the Lorentz transform starting with two equations equating zero with zero. Any true scholars would brush this aside as nonsense but not the Einstein Dingleberries. The same Einstein Dingleberry known as PD said the following http://groups.google.com/group/sci.physics.relativity/msg/ec9892eae720ee0e?hl=en
From: colp on 20 Jun 2010 02:27
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... |