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Why light clock is flawed.

by gnomechompsky
Tags: flawed, light clock
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Passionflower
#19
Aug17-10, 10:16 AM
P: 1,555
Quote Quote by kev View Post
We can do the twins paradox in anther way that removes all acceleration from the problem. Consider one observer A that remains at location A. Another observer B is passing A and moving at constant velocity v relative to A. A and B synchronise clocks so that read the same time. B travels for some distance away from A until he passes another observer moving with velocity -v towards A. C adjusts here clock so that it reads the same as B's clock at the instant they pass each other. All observers agree that at the passing event when C synchronises her clock with B that C's clock reading is a good representation of the elapsed time on B's clock. When C passes A, C and A compare elapsed times at the passing event at agree that that the elapsed time on C's clock is less than the elapsed time on A's clock. No one has accelerated at all during this thought experiment, so the acceleration aspect has been completely removed.
Consider one observer A that remains at location A?
How is that relativity?
yuiop
#20
Aug17-10, 10:23 AM
P: 3,967
Quote Quote by Passionflower View Post
I do not see any problem with what you write Kev it is as you say: it all depends on the relative speed. But what causes a relative speed to change? Right, inertial and proper acceleration. So what drives a change in clock rates?
I agree that it is trivially true that the instantaneous time dilation factor is a function of the instantaneous velocity and in turn that the instantaneous velocity is a function of the acceleration, but this is an unhelpful distraction in analysing the twins paradox.

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.
yuiop
#21
Aug17-10, 10:25 AM
P: 3,967
Quote Quote by Passionflower View Post
Consider one observer A that remains at location A?
How is that relativity?
LOL. I meant that A is at rest in frame A and he measures the velocity of B to be 0.8c and the velocity of C to be -0.8c, but you were right to pick me up on my sloppy use of words there
yuiop
#22
Aug17-10, 10:27 AM
P: 3,967
Quote Quote by Mentz114 View Post
Kev, you said it in post #14, so my #15 is redundant.
Your contributions are never redundant Besides, your use of the term "wordline" is better than my vague "path through space time".
Passionflower
#23
Aug17-10, 11:12 AM
P: 1,555
Quote Quote by kev View Post
LOL. I meant that A is at rest in frame A and he measures the velocity of B to be 0.8c and the velocity of C to be -0.8c, but you were right to pick me up on my sloppy use of words there
Perhaps I am missing something but I cannot see how your example shows the twins paradox.

Quote Quote by kev View Post
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.
Can you show me the calculations?
yuiop
#24
Aug17-10, 01:56 PM
P: 3,967
Quote Quote by Passionflower View Post
Perhaps I am missing something but I cannot see how your example shows the twins paradox.
If you are being picky, you can argue that 3 clocks are involved rather than the usual two in the twins paradox and since none of the clocks (siblings) in this thought experiment were ever at rest wrt each other, they could never be born at the same time by a common parent without some considerable practical difficulties in the maternity ward That aside, some people find it an acceptable analogue of the twins paradox that removes all acceleration considerations, while others might feel it is not a satisfactory resolution. It is just one of many solutions, so it does not have to stand by itself.

Quote Quote by Passionflower View Post
Can you show me the calculations?
I was hoping you wouldn't ask, LOL.

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.
Passionflower
#25
Aug17-10, 07:40 PM
P: 1,555
Quote Quote by kev View Post
If you are being picky, you can argue that 3 clocks are involved rather than the usual two in the twins paradox and since none of the clocks (siblings) in this thought experiment were ever at rest wrt each other, they could never be born at the same time by a common parent without some considerable practical difficulties in the maternity ward That aside, some people find it an acceptable analogue of the twins paradox that removes all acceleration considerations, while others might feel it is not a satisfactory resolution. It is just one of many solutions, so it does not have to stand by itself.
In your example all the movement is inertial and there are no inertial accelerations thus how can you make definitive statements about one clock going slower than another?

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:
Quote Quote by kev View Post
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.
How so? Please explain how you come to that conclusion.

This paper might be of interest to you: http://arxiv.org/ftp/physics/papers/0610/0610226.pdf
Austin0
#26
Aug18-10, 08:35 AM
P: 1,162
Quote Quote by kev View Post
We can do the twins paradox in anther way that removes all acceleration from the problem. Consider one observer A that remains at location A (EDIT: A remains at rest in frame A). Another observer B is passing A and moving at constant velocity v relative to A. A and B synchronise clocks so that they read the same time. B travels for some distance away from A, until he passes another observer C moving with velocity -v towards A. C adjusts here clock so that it reads the same as B's clock at the instant they pass each other. All observers agree that at the passing event, when C synchronises her clock with B, that C's clock reading is a good representation of the elapsed time on B's clock. When C passes A, C and A compare elapsed times at the passing event and agree that the elapsed time on C's clock is less than the elapsed time on A's clock. No one has accelerated at all during this thought experiment, so the acceleration aspect has been completely removed.
Yes you have successfully removed acceleration from the question but can you now calculate quantitative elapsed times for the frames involved????
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
Austin0
#27
Aug18-10, 08:43 AM
P: 1,162
Quote Quote by Passionflower View Post
In your example all the movement is inertial and there are no inertial accelerations thus how can you make definitive statements about one clock going slower than another?

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.

This paper might be of interest to you: http://arxiv.org/ftp/physics/papers/0610/0610226.pdf
I think your logic is solid here. We can't make any evaluation of whether a change in velocity is acceleration or deceleration and we can't attribute time dilation to acceleration outside the instantaneous velocities dilation but logically there is some change of velocity and a resulting change in relative clock rate.
Mentz114
#28
Aug18-10, 08:52 AM
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Quote Quote by Austin0
Yes you have successfully removed acceleration from the question but can you now calculate quantitative elapsed times for the frames involved????
Dead easy. Just calculate the proper time. If you persist in ignoring proper time you will never understand this. You also persist in talking about 'time dilation' which has nothing to do with the times elapsed on clocks.
Passionflower
#29
Aug18-10, 09:04 AM
P: 1,555
Quote Quote by Austin0 View Post
We can't make any evaluation of whether a change in velocity is acceleration or deceleration
You are right.
Austin0
#30
Aug18-10, 10:08 AM
P: 1,162
Quote Quote by Mentz114 View Post
Dead easy. Just calculate the proper time. If you persist in ignoring proper time you will never understand this. You also persist in talking about 'time dilation' which has nothing to do with the times elapsed on clocks.
OK How do you distinguish between two inertial frames where the worldlines are totally reciprocal mirror images of each other???
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???
Mentz114
#31
Aug18-10, 12:03 PM
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Quote Quote by Austin0 View Post
OK How do you distinguish between two inertial frames where the worldlines are totally reciprocal mirror images of each other???
Each worldline taken between two events, has a unique proper time associated with it.
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???
I don't follow you. Any segment of any worldline has a proper time.

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??
How do you explain the length of a piece of string ?


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???
You're making a puzzle where there isn't one. Observing a moving clock may give you the impression that it is running slower than yours, but that's an instantaneous velocity dependent measurement. To get the proper time you have to integrate ( sum the infinitesimals) of this factor over the whole journey. It's a postulate of SR that this gives the correct elapsed time on the local clock.

The Wiki page is not bad, I recommend a quick look.

http://en.wikipedia.org/wiki/Proper_time
CPL.Luke
#32
Aug18-10, 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...y-theor&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 Hafele-Keating experiment, which experimentally tested the resolution of the twin paradox.

http://en.wikipedia.org/wiki/Hafele-Keating_experiment
yuiop
#33
Aug18-10, 03:01 PM
P: 3,967
Quote Quote by Passionflower View Post
In your example all the movement is inertial and there are no inertial accelerations thus how can you make definitive statements about one clock going slower than another?
In my example you are right that I can not make definitive statements about the relative clock rates of spatially separated, beyond the trivial observation that any two ideal clocks that are at rest wrt each other tick at the same rate even when they are they spatially separated and moving relative to the observer. However, I can make a definitive statement about the elapsed proper time between any two timelike events.

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.

Quote Quote by Passionflower View Post
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.
You make a nice observation here, but we are left with the problem that even when we know the proper acceleration of a given clock, there is still ambiguity about its clock rate relative to other clocks with relative motion, depending upon which reference frame the comparisons are made from. As you quite rightly pointed out, what looks like a acceleration of a clock with resultant slowing down down of the clock in one frame looks like deceleration of the clock and resultant speeding up of the clock in another frame.

Quote Quote by Passionflower View Post
How so? Please explain how you come to that conclusion.
I was talking about an old fashioned alternative method of clock synchronisation or comparison of elapsed clock times by "slow transport" of clocks. This is probably an unnecessary distraction in this thread and I probably shouldn't have mentioned it. If you are still curious about slow clock transport synchronization see http://en.wikipedia.org/wiki/One-way...Slow_transport or google the terms.

Quote Quote by Passionflower View Post
This paper might be of interest to you: http://arxiv.org/ftp/physics/papers/0610/0610226.pdf
Yes, thanks, it is of interest. The authors are trying yet another way to make it clear to students that acceleration is not the resolving factor in the twins paradox, but unfortunately their approach seems even less intuitive (to me anyway) than the well known demonstrations.
Mentz114
#34
Aug18-10, 03:13 PM
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P: 4,087
Quote Quote by CPL.Luke
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.
No. General relativity has nothing to do with it. Nor is the question whether the spaceship is a 'valid frame' (?).

It's the proper time !
gnomechompsky
#35
Aug18-10, 03:45 PM
P: 11
Quote Quote by kev View Post
Yes, I am assuming the time dilation equations of SR are correct. What is important in the twins paradox is the path length through spacetime. On the outward trip of one of the the twins, there is some ambiguity of the time dilation of the twins relative to each other. When the paths are plotted on a spacetime diagram, it is true that the path of the Earth twin looks longer from the travelling twins frame and vice versa. There is always some ambiguity in clock rates when the clocks are not at rest with one another. However, after the travelling twin returns home, and the two clocks are alongside each other again, there is no ambiguity. The path of the travelling twin through spacetime is ALWAYS longer than the Earth twins path, from the point of view of ANY inertial observer. If we define relative time dilation in terms of spacetime path lengths, then the ambiguities can be made to disappear when the two clocks come to rest wrt to each other even if they are spatially separated.
I don't understand this. An inertial observer moving at half the velocity of the traveller could legitimately see the traveller's and the Earth twin's path lengths as being the same.

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 earth-bound twin, because there are no special frames of reference?

Happy to ignore acceleration completely.
yuiop
#36
Aug18-10, 05:20 PM
P: 3,967
Quote Quote by gnomechompsky View Post
I don't understand this. An inertial observer moving at half the velocity of the traveller could legitimately see the traveller's and the Earth twin's path lengths as being the same.
This is true on the travellers outward leg, when the twins are spatially separated. The third observer (C) with intermediate velocity says equal proper times have elapsed for the Home twin (A) and the travelling twin (B), when B arrives at the turnaround point, while from B's point of view, A has the longer path and the shortest elapsed proper time and A says B has the longest path and shortest elapsed proper time. Everyone has a different opinion and ambiguity reigns. Once B turns around and arrives back home, observers A, B and C will all agree that B has travelled a longer path through spacetime and than A and all will agree that B has less elapsed proper time than A. All ambiguity is removed once the clocks return to a common location and the differences are unambiguously accounted for in all frames in terms of differences in path lengths.

Quote Quote by gnomechompsky View Post
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 earth-bound twin, because there are no special frames of reference?

Happy to ignore acceleration completely.
As mentioned in other posts by Mentz, it is better to think in terms of proper time between events, which can be unambiguously defined, rather than in terms of relative time dilation of spatially separated clocks with relative motion, which is ambiguous. When two clocks have non-zero relative motion, there is by definition no single reference frame in which they are both at rest and no way to unambiguously define their relative instantaneous clock rates.

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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