The real twin paradox.
in [Relativity]
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From: harry on 26 Nov 2007 07:05 One last attempt ... "colp" <colp(a)solder.ath.cx> wrote in message news:81a29c49-6048-4f2d-87fd-b59380b5dd98(a)b40g2000prf.googlegroups.com... > On Nov 25, 5:54 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) > wrote: >> colp says... >> >> >The point is that a paradox exists due to the time dilation expected >> >by SR. >> >> No, there is no paradox in the sense of contradiction. > > The contradiction between SR prediction ant reality is described > below: > > This thought experiment is like the classic twin paradox, but in this > experiment both twins leave earth and travel symmetric return trips in > opposite directions. > > Since the paths taken by the twins in this experiment are symmetric, > they must be the same age when they meet on their return to earth. > In this experiment the twins maintain constant observation of each > other's clocks, from when they depart until they return and find that > their clocks tell the same time. Sure. > Special relativity says that each twin must observe that the other's > clock is running slow, and at no time does special relativity allow > for an observation which shows that the other clock is running fast. There is no need for that, and this has been explained to you from the very start. We also gave you links to articles that explain it. But I'll try to explain it with an example - see below! > The paradox is that special relativity says that a twin will never see > the other twin's clock catch up, but the clocks must show the same > time at the end of the experiment because of symmetry. -> Here is another paradox, specially prepared for you! :-) Suppose you warm up a container with water while in Europe, from room temperature (20 degrees) to 40 degrees and you keep it at that temperature in a thermos flask while you travel to the USA (technically difficult and even more so with the new regulations, but never mind). Then you buy a thermometer and you cool the water down to room temperature - back to its original state. Now you want to verify the temperature. You have read somewhere that EU degrees are dilated compared to US degrees: 1 EU degree = 9/5 US degree. The trip to the USA did not change its temperature. Thus the end reading should be 40 - 9/5*20 = 4 degrees. However, the thermometer indicates 68 degrees. This is a paradox, for you simply cooled it down to its original state; at no time during the trip did you see the temperature rise! Harald
From: Daryl McCullough on 26 Nov 2007 10:17 colp says... > >On Nov 25, 5:50 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) >wrote: >> No, you haven't. As I said, you have to look at >> what relativity *actually* predicts, not your >> own distorted version of relativity. > >What do you think the difference is between my version of relativity >and your version of relativity? The biggest single difference is that you seem to believe that time dilation is a relationship between two *clocks*. It isn't. It's a relationship between *one* clock and *one* inertial coordinate system. You CANNOT apply the time dilation formula to compare distant accelerated clocks. Also, the Lorentz transformations is a relationship between two *inertial* coordinate systems. It cannot be applied to accelerated observers. The elapsed time tau shown on a clock is given by the following formula: tau = integral of square-root(1-(v(t)/c)^2) dt where v(t) is the velocity of the clock at time t, and where v and t are both measured in a single inertial coordinate system. This formula gives the same answer in *every* inertial coordinate system. Let's go through the case in which a rocket starts at rest in some inertial coordinate system, accelerates rapidly to speed v, travels outward for a time T, turns around quickly and comes back at speed v. Let x,t refer to the initial rest frame, and let x',t' refer to the frame in which the rocket is at rest on the outward trip. (There *is* no inertial coordinate system in which the rocket is always at rest, because the rocket is not traveling inertially). Let e1 be the event at which the rocket blasts off, let e2 be the event at which the rocket turns around, and let e3 be the event at which the rocket returns. The coordinates of e1 are x1 = 0, t1 = 0 x1' = 0, t1' = 0 The coordinates of e2 are x2 = vT, t2 = T, x2' = gamma (x2 - v t2) = 0 t2' = gamma (t2 - v/c^2 x2) = gamma (T - v^2/c^2 T) = T/gamma (where I used (1-(v/c)^2) = 1/gamma^2) The coordinates of e3 are x3 = 0 t3 = 2T x3' = gamma (x3 - v t3) = - 2 gamma vT t3' = gamma (t3 - v/c^2 x3) = 2 gamma T Okay, now let's compute the elapsed time tau in both frames: Unprimed frame: The first leg of the journey, the rocket travels from x=0 to x=vT in time T. So the velocity is v. So the elapsed time for the first leg is T square-root(1-v^2/c^2). The return journey also has velocity v and lasts for a time T. So the total trip has elapsed time 2T square-root(1-v^2/c^2). Now, compute tau in the primed frame: The first leg of the journey, the rocket travels from x'=0 to x'=0 in time T/gamma. So the rocket's speed is 0 as measured in this frame. So the elapsed time for the first leg is just Integral of square-root(1-0) dt' = t2' - t1' = T/gamma = T square-root(1-(v/c)^2). On the return journey, the rocket travels from x'= 0 at time t' = T/gamma to x' = -2gamma vT at time t' = 2 gamma T. So the speed for the second leg, as measured in this frame, is v' = (change in x')/(change in t') = 2gamma vT/(2gamma T - T/gamma) = (2v)/(2 - 1/gamma^2) = 2v/(1 + (v/c)^2) So the elapsed time on the second leg of the journey is Integral of square-root(1 - (2v/(1 + (v/c)^2))^2) dt' = (2 gamma T - T/gamma) * square-root(1 - (2v/(1 + (v/c)^2))^2/c^2) We can simplify the expression inside the square-root: 1 - (2v/(1 + (v/c)^2))^2/c^2 = ((1+(v/c)^2) - 4 v^2/c^2)/(1+(v/c)^2)^2 = (1+2(v/c)^2 + (v/c)^4 - 4(v/c)^2)/(1+(v/c)^2)^2 = (1-2(v/c)^2 + (v/c)^4)/(1+(v/c)^2)^2 = (1-(v/c)^2)^2/(1+(v/c)^2)^2 So the elapsed time on the second leg is (2 gamma T - T/gamma) * (1-(v/c)^2)/(1+(v/c)^2) Now, using gamma = 1/square-root(1-(v/c)^2), we have T (2/square-root(1-(v/c)^2) - square-root(1-(v/c)^2)) * (1-(v/c)^2)/(1+(v/c)^2) which simplifies to T square-root(1-(v/c)^2) So the elapsed time, as measured in this frame, is *also* 2T square-root(1-(v/c)^2) (adding the elapsed times for outward and return trips). So the predictions of SR are perfectly consistent. -- Daryl McCullough Ithaca, NY
From: Sue... on 26 Nov 2007 11:12 On Nov 26, 10:17 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > colp says... > > > > >On Nov 25, 5:50 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) > >wrote: > >> No, you haven't. As I said, you have to look at > >> what relativity *actually* predicts, not your > >> own distorted version of relativity. > > >What do you think the difference is between my version of relativity > >and your version of relativity? > > The biggest single difference is that you seem to believe > that time dilation is a relationship between two *clocks*. It can be the difference in two clocks. If one is a light-clock. It is also the appearance of the stayhome clock when receeding. (corrrectable with doppler) It is also the appearance of a clock dropped at the turnaround point when returning home. (correctable with doppler) Except for parlor tricks, what good is "elapsed proper time" ? It is not mentioned here: http://farside.ph.utexas.edu/teaching/em/lectures/node114.html Sue... > Daryl McCullough > Ithaca, NY
From: Daryl McCullough on 26 Nov 2007 11:45 colp says... > >On Nov 25, 5:54 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) >wrote: >> colp says... >> >> >The point is that a paradox exists due to the time dilation expected >> >by SR. >> >> No, there is no paradox in the sense of contradiction. > >The contradiction between SR prediction ant reality is described >below: > >This thought experiment is like the classic twin paradox, but in this >experiment both twins leave earth and travel symmetric return trips in >opposite directions. > >Since the paths taken by the twins in this experiment are symmetric, >they must be the same age when they meet on their return to earth. That's correct. And that's exactly what SR predicts. >In this experiment the twins maintain constant observation of each >other's clocks, from when they depart until they return and find that >their clocks tell the same time. > >Special relativity says that each twin must observe that the other's >clock is running slow No, it doesn't say that. That's what I mean when I say that you have to look at what SR *actually* says, not your own parody of it. SR does not say that each clock observes the other clock running slow. What it says is that any *inertial coordinate system* measures any moving clock to be running slow. Time dilation is *not* a relationship between two clocks, but is a relationship between *one* clock and one inertial coordinate system. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 26 Nov 2007 11:48
Sue... says... >I won't dispute you have stated one of many refutations of >what I call the "parlor trick" interpretation of SR. What you call a "parlor trick" is just taking the axioms of Special Relativity and applying them, using ordinary high-school level algebra. -- Daryl McCullough Ithaca, NY |