
#19
Aug1710, 10:16 AM

P: 1,555

How is that relativity? 



#20
Aug1710, 10:23 AM

P: 3,967

Let us say that A and B are initially at rest wrt each other. Now let us say that B accelerates to 0.8c and that B's clock rate is now slower than A's clock by a factor of 0.6 is some meaningful absolute way. After some time t, A accelerates to 0.8c in the same direction as B and both clocks are now at rest wrt each other. We might conclude that B has been at lower clock rate than A for a longer time until at the last minute A boosted his speed to match that of B and his clock rate slowed down to match that of B. However, when we calculate the elapsed proper times in the new mutual rest frame of A and B, we find that less time has elapsed on A's clock, contrary to our expectations when assuming B's clock physically slowed down when B accelerated. To an observer that remained in the original frame of A and B before they accelerated, the elapsed time of A's clock is greater than that of B's clock so it can be seen there is no way to obtain universal agreement about elapsed times of spatially separated clocks. It only when the two clocks are initially at the same location and finally at the same location (not necessarily the same location they both started from) that all observers can agree on the differences in elapsed proper time. However, when two spatially separated clocks are at rest wrt each other, some people might attach more significance to the elapsed time as measured in the final rest frame of the clocks. The fact is, that when one object accelerates relative to another, some change in relative clock rates does come about, but it is impossible to quantify in any meaningful way, whether that change is an increase or a decrease in the clock rate of the accelerated object relative to the unaccelerated object. It is only when the two objects return to a common rest frame (not necessarily the original one) that we can say anything meaningful in a physical sense about the instantaneous relative clock rates and it is only when two clocks return to a common location (but not necessarily at rest wrt each other) that we can say anything meaningful about comparative elapsed times. 



#21
Aug1710, 10:25 AM

P: 3,967





#22
Aug1710, 10:27 AM

P: 3,967





#23
Aug1710, 11:12 AM

P: 1,555





#24
Aug1710, 01:56 PM

P: 3,967

I know it is true intuitively, but the maths might get rather involved. Here is my shot at it. Let us say that rocket B accelerates to 0.8c in one second and then cruises, and exactly 10 years after B departed, A accelerates to 0.8c in one second (as measured by an observer C that remains at rest in the frame that A and B were originally at rest) and now both A and B are rest in new frame that has velocity 0.8c relative to the original frame. From the accelerating rocket equations of Baez, that I gave earlier, the proper acceleration of a rocket with terminal velocity 0.8c (as measured in the unaccelerated frame) after a time of 1 second (as measured in the unaccelerated frame) is given as: a = (v/t)/sqrt[1(v/c)^2] = 0.8*0.6 = 0.48 The proper elapsed time (T) of the rocket during the acceleration phase is given by Baez as: T = (c/a)*asinh(at/c) = (1/0.48)*asinh(0.48) = 0.9651 seconds. using units of c=1. Note that the proper elapsed time is not much less than the 1 second measured by the unaccelerated observer C. During the cruise phase of B's journey the elapsed proper time of B's clock according to C is 10years*0.6 = 6 years so the total elapsed proper time of B's clock is 6years + 0.9651 seconds in C's frame. The total elapsed proper time of rocket A from the time B took of to the time A joined B in the new rest frame is 10 years + 0.9651s seconds according to C. In frame C the elapsed time of A is obviously much greater than the elapsed time of B. Now we look at the times measured by an observer that was at rest in a frame (D) that was always moving with velocity 0.8c relative to frame C. i.e frame D is the final rest frame of observers A and B. In frame C the elapsed time from B taking off, to A joining B in frame D, was 10 years + 1 second, so in frame D the elapsed time between the two events is (10y + 1s)/0.6 = 16.6667 years + 1.66667 seconds. Other than the initial 1.6666 seconds that rocket B initially took to accelerate to rest in frame D, rocket B has been at rest in frame D for 16.6667 years so the total elapsed proper time of B's clock, according to C is 16.6667 years (plus the 0.9651 seconds of proper time B spent accelerating). The elapsed proper time of rocket A, according to observer D is 10 years (plus the 0.9651 seconds of proper time A spent accelerating). So in frame D, the time that elapses between B taking off and A joining B in frame D is 10 years + 0.9651 seconds proper time as measured by clock A and 16.6667 years + 0.9651 seconds proper time as measured by clock B. In frame D much more proper time has elapsed on clock B while in the original frame C, much more proper time elapses on clock A. It can also be seen that I have taken the time dilation due to acceleration into account and it is insignificant compared to the velocity time dilation and an unnecessary complication. The differences in elapsed proper times in frames C and D comes about because the two rockets are spatially separated and frames C and D have different notions of what is simultaneous. Note that in the final instance when both rockets A and B are rest in frame D, we could get A and B to move very slowly towards each other so that they meet in the middle and confirm that there is less elapsed proper time on A's clock than on B's clock when they are again at a common location, this despite the fact that B accelerated earlier than A and was presumably time dilating for a longer duration than A. 



#25
Aug1710, 07:40 PM

P: 1,555

In relation to your example with A, B, C, and D I follow all that you say but I do not see how that contradicts that acceleration changes the rate that a clock ticks with respect to one that does not accelerate. For C, when A and B accelerate to increase the relative velocity with respect to C, the relative clock rates diverge. For D however A and B accelerate (decelerate) to decrease the relative velocity with respect to D, thus the clock rates converge. But then you write: This paper might be of interest to you: http://arxiv.org/ftp/physics/papers/0610/0610226.pdf 



#26
Aug1810, 08:35 AM

P: 1,162

I don't think so. You have now created a perfectly reciprocal situation without any basis for a comparative analysis between A and C's clocks. Equivalent worldlines and diagrams. If you think there is some basis for a comparison I would really be interested in seeing your calculations. Thanks 



#27
Aug1810, 08:43 AM

P: 1,162





#28
Aug1810, 08:52 AM

PF Gold
P: 4,081





#29
Aug1810, 09:04 AM

P: 1,555





#30
Aug1810, 10:08 AM

P: 1,162

WHere the passage of proper time is equivalent in each diagram as depicted by the length of the worldline how do you decide which one is accurate??? Given kev's three inertial frames; if in fact you derive a difference in elapsed time between frames how do you explain this without an implication of actual motion on the part of some frame?? Out of curiosity what are you talking about " my persisting in ignoring proper time" ANd what do you mean time dilation has nothing to do with elapsed time on clocks?? Do you think time dilation has nothing to do with the slope of worldlines and vice versa??? 



#31
Aug1810, 12:03 PM

PF Gold
P: 4,081

The Wiki page is not bad, I recommend a quick look. http://en.wikipedia.org/wiki/Proper_time 



#32
Aug1810, 02:40 PM

P: 444

I think the issue here is that there is no basis for deciding who actually went on the trip and who did not.
according to the observer on earth the twin on the space ship went on the trip, according to the observer on the spaceship the earth went on a trip away from him. http://www.scientificamerican.com/ar...ytheor&page=2 the above article goes through the full paradox including a nice space time diagram, I disagree with parts of the resolution, namely the distinction between which object left the reference frame, Ialso need to do the math to see if it works out from the spaceships perspective I still believe that a key point is that the spaceship is not a valid reference frame for the entire journey, because it does undergo acceleration at 3 points, meaning that without very careful general relativity work, you would not be able to calculate the proper time of the planet earth. also illuminating is the HafeleKeating experiment, which experimentally tested the resolution of the twin paradox. http://en.wikipedia.org/wiki/HafeleKeating_experiment 



#33
Aug1810, 03:01 PM

P: 3,967

Let us say A remains at rest in frame A. All references to coordinate measurements will mean measurements made by observers at rest in frame A. B passes A at coordinate time zero at a coordinate velocity of +0.8c. After a coordinate time of 10 years, B passes C who is going in the opposite direction with a coordinate velocity of 0.8c. The coordinate distance between event (B passing A) = event(B,A) and event(B,C) is 8 lightyears. Other observers in different reference frames will disagree with the coordinate times, distance and velocities measured by A and will also disagree on what clock A reads at event(B,C), but all observers will agree that 6 years of proper time elapses on clock B between events (A,B) and (B,C). (Definitive statement 1.) Eventually clock C passes A at event(C,A). All observers agree that 6 years of proper time passes on clock C between events (B,C) and (C,A). (Definitive statement 2). All observers will agree that 20 years of proper time elapses on clock A between events (B,A) and (C,A). (Definitive statement 3). All observers agree that the combined elapsed proper time between the 3 events (B,A), (B,C) and (C,A) is 12 years (Definitive statement 4) and that this proper time interval is less that the proper time interval between events (B,A) and (C,A). (Definitive statement 5). Here I am defining the proper time between any two timelike events (the invariant time interval) as being the time measured by an inertially moving ideal clock that is coincident at both events. I am also using the term "all observers" to mean inertial observers that are not necessarily at rest in frame A. 



#34
Aug1810, 03:13 PM

PF Gold
P: 4,081

It's the proper time ! 



#35
Aug1810, 03:45 PM

P: 11

Whilst this thread has gotten quite big, the point I am actually trying to bring out (as you are probably aware), is how do we know the time dilation SR equations are valid if the light clock thought experiment can be applied equally to the "traveller" and the earthbound twin, because there are no special frames of reference? Happy to ignore acceleration completely. 



#36
Aug1810, 05:20 PM

P: 3,967

Even when B has turned around and is on his way home, it is still possible to find reference frames in which more time has elapsed for B than for A. It is not the turn around (or the acceleration involved) that removes the ambiguity. It is the arrival of the two clocks at common location that removes the ambiguity about the elapsed proper times, but if the twins do not come to rest wrt each other, they will still disagree about their instantaneous relative clock rates at the final passing event. 


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