Speed of Time
in [Relativity]
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From: Ste on 28 Dec 2009 23:16 On 28 Dec, 14:16, "papar...(a)gmail.com" <papar...(a)gmail.com> wrote: > On 28 dic, 09:19, Ste <ste_ro...(a)hotmail.com> wrote: > > > On 28 Dec, 11:03, "Dirk Van de moortel" > > Let me be clear I didn't actually disagree with your model within its > > own mathematical terms. What I disagree with is the approach which you > > say is mathemetically more simple and yet is in fact less intuitive. > > The fact that this supposed paradox has not been laid to bed yet > > surely proves that's it's in want of a simple explanation. > > > And indeed the explanation does seem to be simple: the astronaut > > accelerates more. If the universe *had* accelerated around the > > astronaut, then the astronaut would end up *older* than his twin - but > > in fact, a simple accelerometer *proves* that the universe did not in > > fact accelerate around the astronaut - the astronaut accelerated > > around the universe, and that is why the astronaut returns to Earth > > younger. > > I always cite the clear and simple explanation of Landau and Lifshitz > for this "symmetric" behavior, which goes like this: > > "Suppose some clocks are moving in uniform rectilinear motion, > relative to an inertial system K. A reference frame K' linked to the > latter is also inertial. Then from the point of view of an observer in > the K system the clocks in the K' system fall behind. And conversely, > from the point of view of the K' system, the clocks in K lag. To > convince ourselves that there is no contradiction, let us note the > following. In order to establish that the clocks in the K' system lag > behind those in the K system, we must proceed in the following > fashion. Suppose that at a certain moment the clock in K' passes by > the clock in K, and at that moment the readings of the two clocks > coincide. To compare the rates of the two clocks in K and K', we must > once more compare the readings of the same moving clock in K' with the > clocks in K. But now we compare this clock with different clocks in K > with those past, which the clock in K' goes at this new time. Then we > find that the clock in K' lags behind the clocks in K with which it is > being compared. We see that to compare the rates of clocks in two > reference frames we require several clocks in one frame and one in the > other, and that therefore this process is not symmetric with respect > to the two systems. The clock that appears to lag is always the one > which is being compared with different clocks in the other system. If > we have two clocks, one of which describes a closed path returning to > the starting point (the position of the clock which remained at rest), > then clearly the moving clock appears to lag relative to the one at > rest. The converse reasoning, in which the moving clock would be > considered to be at rest (and vice versa) is now impossible, since the > clock describing a closed trajectory does not carry out a uniform > rectilinear motion, so that a coordinate system linked to it will not > be inertial. > Since the laws of nature are the same only for inertial reference > frames, the frames linked to the clock at rest (inertial frame) and to > the moving clock (non-inertial) have different properties, and the > argument which leads to the result that the clock at rest must lag is > not valid." I appreciate your reply, but it really does go to show how out of touch Landau and Lifshitz are when they consider this a "simple and clear" explanation. The simple and clear explanation, in my mind, is this: the astronaut accelerates more. And anyone who resides on Earth should immediately reply "of course!".
From: Inertial on 29 Dec 2009 02:33 "Ste" <ste_rose0(a)hotmail.com> wrote in message news:d01da796-4522-4253-a8d1-3dc294523d4c(a)n16g2000yqm.googlegroups.com... > On 29 Dec, 04:58, "Inertial" <relativ...(a)rest.com> wrote: [snip] http://users.telenet.be/vdmoortel/dirk/Physics/ImmortalFumbles.html > > That's a very funny site. One line that caught my eye was: "assertion > carries no weight". Another cracker was: "The simultaneity occurs at a > different time interval." There's some hum-dingers in there :)
From: paparios on 29 Dec 2009 06:12 On 29 dic, 01:16, Ste <ste_ro...(a)hotmail.com> wrote: > On 28 Dec, 14:16, "papar...(a)gmail.com" <papar...(a)gmail.com> wrote: > > > > > > > On 28 dic, 09:19, Ste <ste_ro...(a)hotmail.com> wrote: > > > > On 28 Dec, 11:03, "Dirk Van de moortel" > > > Let me be clear I didn't actually disagree with your model within its > > > own mathematical terms. What I disagree with is the approach which you > > > say is mathemetically more simple and yet is in fact less intuitive. > > > The fact that this supposed paradox has not been laid to bed yet > > > surely proves that's it's in want of a simple explanation. > > > > And indeed the explanation does seem to be simple: the astronaut > > > accelerates more. If the universe *had* accelerated around the > > > astronaut, then the astronaut would end up *older* than his twin - but > > > in fact, a simple accelerometer *proves* that the universe did not in > > > fact accelerate around the astronaut - the astronaut accelerated > > > around the universe, and that is why the astronaut returns to Earth > > > younger. > > > I always cite the clear and simple explanation of Landau and Lifshitz > > for this "symmetric" behavior, which goes like this: > > > "Suppose some clocks are moving in uniform rectilinear motion, > > relative to an inertial system K. A reference frame K' linked to the > > latter is also inertial. Then from the point of view of an observer in > > the K system the clocks in the K' system fall behind. And conversely, > > from the point of view of the K' system, the clocks in K lag. To > > convince ourselves that there is no contradiction, let us note the > > following. In order to establish that the clocks in the K' system lag > > behind those in the K system, we must proceed in the following > > fashion. Suppose that at a certain moment the clock in K' passes by > > the clock in K, and at that moment the readings of the two clocks > > coincide. To compare the rates of the two clocks in K and K', we must > > once more compare the readings of the same moving clock in K' with the > > clocks in K. But now we compare this clock with different clocks in K > > with those past, which the clock in K' goes at this new time. Then we > > find that the clock in K' lags behind the clocks in K with which it is > > being compared. We see that to compare the rates of clocks in two > > reference frames we require several clocks in one frame and one in the > > other, and that therefore this process is not symmetric with respect > > to the two systems. The clock that appears to lag is always the one > > which is being compared with different clocks in the other system. If > > we have two clocks, one of which describes a closed path returning to > > the starting point (the position of the clock which remained at rest), > > then clearly the moving clock appears to lag relative to the one at > > rest. The converse reasoning, in which the moving clock would be > > considered to be at rest (and vice versa) is now impossible, since the > > clock describing a closed trajectory does not carry out a uniform > > rectilinear motion, so that a coordinate system linked to it will not > > be inertial. > > Since the laws of nature are the same only for inertial reference > > frames, the frames linked to the clock at rest (inertial frame) and to > > the moving clock (non-inertial) have different properties, and the > > argument which leads to the result that the clock at rest must lag is > > not valid." > > I appreciate your reply, but it really does go to show how out of > touch Landau and Lifshitz are when they consider this a "simple and > clear" explanation. The simple and clear explanation, in my mind, is > this: the astronaut accelerates more. And anyone who resides on Earth > should immediately reply "of course!". It is clear enough to demonstrate that playing with relativistic clocks is quite more tricky than most people here think. Einstein also recognized the problem in its 1905 paper, where he writes: "...From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by (1/2)t(v^2/c^2)(up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B. It is at once apparent that this result still holds good if the clock moves from A to B in any polygonal line, and also when the points A and B coincide. If we assume that the result proved for a polygonal line is also valid for a continuously curved line, we arrive at this result: If one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which has remained at rest the travelled clock on its arrival at A will be (1/2)t(v^2/ c^2) seconds slow. Thence we conclude that a balance-clock at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions..." Miguel Rios
From: Inertial on 29 Dec 2009 06:42 <paparios(a)gmail.com> wrote in message news:0a2d5a76-4d35-4e4c-9045-27581b39c22a(a)21g2000yqj.googlegroups.com... > On 29 dic, 01:16, Ste <ste_ro...(a)hotmail.com> wrote: >> On 28 Dec, 14:16, "papar...(a)gmail.com" <papar...(a)gmail.com> wrote: >> >> >> >> >> >> > On 28 dic, 09:19, Ste <ste_ro...(a)hotmail.com> wrote: >> >> > > On 28 Dec, 11:03, "Dirk Van de moortel" >> > > Let me be clear I didn't actually disagree with your model within its >> > > own mathematical terms. What I disagree with is the approach which >> > > you >> > > say is mathemetically more simple and yet is in fact less intuitive. >> > > The fact that this supposed paradox has not been laid to bed yet >> > > surely proves that's it's in want of a simple explanation. >> >> > > And indeed the explanation does seem to be simple: the astronaut >> > > accelerates more. If the universe *had* accelerated around the >> > > astronaut, then the astronaut would end up *older* than his twin - >> > > but >> > > in fact, a simple accelerometer *proves* that the universe did not in >> > > fact accelerate around the astronaut - the astronaut accelerated >> > > around the universe, and that is why the astronaut returns to Earth >> > > younger. >> >> > I always cite the clear and simple explanation of Landau and Lifshitz >> > for this "symmetric" behavior, which goes like this: >> >> > "Suppose some clocks are moving in uniform rectilinear motion, >> > relative to an inertial system K. A reference frame K' linked to the >> > latter is also inertial. Then from the point of view of an observer in >> > the K system the clocks in the K' system fall behind. And conversely, >> > from the point of view of the K' system, the clocks in K lag. To >> > convince ourselves that there is no contradiction, let us note the >> > following. In order to establish that the clocks in the K' system lag >> > behind those in the K system, we must proceed in the following >> > fashion. Suppose that at a certain moment the clock in K' passes by >> > the clock in K, and at that moment the readings of the two clocks >> > coincide. To compare the rates of the two clocks in K and K', we must >> > once more compare the readings of the same moving clock in K' with the >> > clocks in K. But now we compare this clock with different clocks in K >> > with those past, which the clock in K' goes at this new time. Then we >> > find that the clock in K' lags behind the clocks in K with which it is >> > being compared. We see that to compare the rates of clocks in two >> > reference frames we require several clocks in one frame and one in the >> > other, and that therefore this process is not symmetric with respect >> > to the two systems. The clock that appears to lag is always the one >> > which is being compared with different clocks in the other system. If >> > we have two clocks, one of which describes a closed path returning to >> > the starting point (the position of the clock which remained at rest), >> > then clearly the moving clock appears to lag relative to the one at >> > rest. The converse reasoning, in which the moving clock would be >> > considered to be at rest (and vice versa) is now impossible, since the >> > clock describing a closed trajectory does not carry out a uniform >> > rectilinear motion, so that a coordinate system linked to it will not >> > be inertial. >> > Since the laws of nature are the same only for inertial reference >> > frames, the frames linked to the clock at rest (inertial frame) and to >> > the moving clock (non-inertial) have different properties, and the >> > argument which leads to the result that the clock at rest must lag is >> > not valid." >> >> I appreciate your reply, but it really does go to show how out of >> touch Landau and Lifshitz are when they consider this a "simple and >> clear" explanation. The simple and clear explanation, in my mind, is >> this: the astronaut accelerates more. And anyone who resides on Earth >> should immediately reply "of course!". > > It is clear enough to demonstrate that playing with relativistic > clocks is quite more tricky than most people here think. Not those who have a decent understanding of SR > Einstein also > recognized the problem in its 1905 paper, where he writes: It's not a problem .. but is surprising when first presented. > "...From this there ensues the following peculiar consequence. If at > the points A and B of K there are stationary clocks which, viewed in > the stationary system, are synchronous; and if the clock at A is moved > with the velocity v along the line AB to B, then on its arrival at B > the two clocks no longer synchronize, but the clock moved from A to B > lags behind the other which has remained at B by (1/2)t(v^2/c^2)(up to > magnitudes of fourth and higher order), t being the time occupied in > the journey from A to B. It is at once apparent that this result still > holds good if the clock moves from A to B in any polygonal line, and > also when the points A and B coincide. If we assume that the result > proved for a polygonal line is also valid for a continuously curved > line, we arrive at this result: If one of two synchronous clocks at A > is moved in a closed curve with constant velocity until it returns to > A, the journey lasting t seconds, then by the clock which has remained > at rest the travelled clock on its arrival at A will be (1/2)t(v^2/ > c^2) seconds slow. Thence we conclude that a balance-clock at the > equator must go more slowly, by a very small amount, than a precisely > similar clock situated at one of the poles under otherwise identical > conditions..." This is simply the famous twins paradox. > Miguel Rios
From: Inertial on 30 Dec 2009 18:54
"Ste" <ste_rose0(a)hotmail.com> wrote in message news:d5f88bd1-1462-43bc-87d5-4fe81af4ae37(a)34g2000yqp.googlegroups.com... > Again, any sensible person can see the legitimacy of this > interpretation. The fact that you sling insults again just proves that > you're a stubborn fool. No .. it shows that I was dealing with one |