Revisiting the Flaws of the Light Clock in Special and General Relativity

[gnomechompsky]

Why "light clock" is flawed.

I have been considering this thought experiment for some time, and the principle seems "broken". I'm wondering if by playing devil's advocate you can show me what is going wrong.

-In SR, the light clock was used to predict time dilation in a "moving" object.

-According to the Twin Paradox, two distinct frames of reference in different inertial motion would see time dilating in the other one, because all motion is relative.

-The counter argument to this is that the frames of reference are not identical, because in the Twin Paradox, the Twin flying away from Earth accelerates and is not in constant inertial motion.

-We therefore can't use SR to answer the Twin Paradox, we have to use GR.

-Using a GR thought experiment, the Light Clock would no longer function if acceleration is great enough. The acceleration would cause the beam of light to bend outside the light clock, even from the light clock's frame of reference.

 

[HallsofIvy]

You can use SR even when acceleration is involved by doing a different "SR" calculation for each instant, treating the speed at that instant as constant.

 

[Mentz114]

Special relativity is the theory of Minkowski spacetime, which includes accelerating observers, see Rindler chart,en.wikipedia.org/wiki/Rindler_coordinates[/url] and also this is very interesting [url]http://gregegan.customer.netspace.net.au/SCIENCE/Rindler/RindlerHorizon.html[/URL]

 

[AJ Bentley]

-We therefore can't use SR to answer the Twin Paradox, we have to use GR.

Simply not true.

Acceleration plays no part in the resolution of the paradox.
In fact, the whole point of the paradox from a tuition point of view is to get student's to realize that there is no paradox.

The point is to illustrate the absurdity of our ordinary concept of simultaneity - which is what the 'paradox' relies on for it's puzzle effect.

 

[yuiop]

-Using a GR thought experiment, the Light Clock would no longer function if acceleration is great enough. The acceleration would cause the beam of light to bend outside the light clock, even from the light clock's frame of reference.

As others have mentioned SR can handle acceleration of the observers and so it not a GR thought experiment, but it is related to GR via the equivalence principle. It is true that the light beam will bend during acceleration and miss the mirror or the receiver on the way back in a traditional transverse light clock which is not ideal for the accelerating case. If you wish to analyse time dilation using a light clock in the accelerating case it would be best to have a light clock that is orientated parallel to the acceleration. One obvious limitation of such a clock is that when the signal leaves the source the velocity of the source is v, and when the signal returns the velocity is v+a*t, so the light clock is not measuring the time dilation corresponding to velocity v except in the case the the light clock and the a*t term is vanishingly small, so the accuracy of the clock is size dependent. We could design the clock so that it has one source, but two mirrors in opposite directions so that when the clock is at rest the signals return simultaneously. When the light clock is accelerating parallel to its length, I don't think the signals will return simultaneously (but I have not done the calculations yet) and so you could obtain some sort of average time from the two oppositely orientated clocks.

 

[gnomechompsky]

Simply not true.

Acceleration plays no part in the resolution of the paradox.
In fact, the whole point of the paradox from a tuition point of view is to get student's to realize that there is no paradox.

The point is to illustrate the absurdity of our ordinary concept of simultaneity - which is what the 'paradox' relies on for it's puzzle effect.

Please explain this because every resolution of the Twin Paradox I have seen refers to acceleration. If both frames of reference are otherwise in relative inertial motion what is the difference between them qualitatively?

 

[Mentz114]

Please explain this because every resolution of the Twin Paradox I have seen refers to acceleration. If both frames of reference are otherwise in relative inertial motion what is the difference between them qualitatively?

The twin 'paradox' requires the twins to start in the same place and meet up again later. The elapsed time on their clocks is equal to the proper-time of their worldlines. The proper time can be calculated simply and has nothing to do with 'time dilation' effects observed during the trips. The proper time takes into account the 4-D path and so any changes in velocity will be reflected in the proper time. See http://en.wikipedia.org/wiki/Proper_time

 

[Rasalhague]

Acceleration plays no part in the resolution of the paradox.

Please explain this because every resolution of the Twin Paradox I have seen refers to acceleration. If both frames of reference are otherwise in relative inertial motion what is the difference between them qualitatively?

There must be some point on the path through spacetime of the twin who's younger when they meet where this twin's velocity changes. In practice, this would entail acceleration. I think what AJ Bentley means is that you can get some intuition about the scenario by considering a simplified, not physically realistic, case where, instead of the path having a curved segment, representing acceleration, you consider the outward journey and the journey home as two different straight line segments, and you treat the change in velocity where the twin reverses direction in space as an abrupt corner connecting these line segments. So in this simplified model, there's not really acceleration in the sense of a derivative of velocity, because the velocity isn't continuous, and therefore not differentiable, at the corner, although you might hear some people refer to it informally as an "infinite acceleration".

 

[Bussani]

Please explain this because every resolution of the Twin Paradox I have seen refers to acceleration. If both frames of reference are otherwise in relative inertial motion what is the difference between them qualitatively?

I got some good replies when I asked a question about this myself. It might be helpful to you.

https://www.physicsforums.com/showthread.php?p=2549150

 

[yuiop]

Let us say for the sake of argument that acceleration does affect time dilation and see how that affects the twins paradox resolution.

Consider one twin accelerating from 0 to 0.8c in one second and then traveling at a constant velocity for 10 years as measured in the Earth frame) before turning around in the space of 2 seconds and returning at 0.8c and finally coming to a stop by decelerating to zero in 1 second and re-uniting with his twin. The total journey time is 20 years (+4 seconds) and the SR time dilation factor at 0.8c is 0.6 so the traveling twin has aged roughly 12 years in the time the stay at home twin has aged 20 years. Now the acceleration takes place in a total of 4 seconds. If time stops still during the acceleration (worst case) then the largest error that is introduced by ignoring the time dilation during the acceleration phase in this example is 4 seconds and that fails to account for why the traveling twin has lost a total of 8 years compared to his stay at home twin.


DrGreg also produced a very nice diagram that makes it very intuitively clear why acceleration is not the explanation in the twins paradox here: https://www.physicsforums.com/showpost.php?p=1747855&postcount=4

Lastly, as I mentioned before, analysis of a light clock parallel to the axis of motion would give us an informative insight into how an ideal clock performs during acceleration. This should not be too difficult to calculate for constant acceleration. Are you interested in that aspect or just acceleration as it applies to the twins paradox?

 

[gnomechompsky]

Let us say for the sake of argument that acceleration does affect time dilation and see how that affects the twins paradox resolution.

Consider one twin accelerating from 0 to 0.8c in one second and then traveling at a constant velocity for 10 years as measured in the Earth frame) before turning around in the space of 2 seconds and returning at 0.8c and finally coming to a stop by decelerating to zero in 1 second and re-uniting with his twin. The total journey time is 20 years (+4 seconds) and the SR time dilation factor at 0.8c is 0.6 so the traveling twin has aged roughly 12 years in the time the stay at home twin has aged 20 years. Now the acceleration takes place in a total of 4 seconds. If time stops still during the acceleration (worst case) then the largest error that is introduced by ignoring the time dilation during the acceleration phase in this example is 4 seconds and that fails to account for why the traveling twin has lost a total of 8 years compared to his stay at home twin.


DrGreg also produced a very nice diagram that makes it very intuitively clear why acceleration is not the explanation in the twins paradox here: https://www.physicsforums.com/showpost.php?p=1747855&postcount=4

Lastly, as I mentioned before, analysis of a light clock parallel to the axis of motion would give us an informative insight into how an ideal clock performs during acceleration. This should not be too difficult to calculate for constant acceleration. Are you interested in that aspect or just acceleration as it applies to the twins paradox?

-This doesn't solve the twin paradox. If we pretend acceleration plays a negligible impact, using the light clock thought experiment then both observers will see the same amount of time dilation in the other observer.

-You are stating that time dilation over the course of the 10 years is caused by this inertial motion, but the formula from this is derived from the light clock thought experiment which we have resolved on the basis of the twin paradox yet.

 

[Passionflower]

Let us say for the sake of argument that acceleration does affect time dilation and see how that affects the twins paradox resolution.

Acceleration does effect the rate a clock ticks with respect to one that does not accelerate. So the longer the time interval the greater the time differential.

 

[yuiop]

Acceleration does effect the rate a clock ticks with respect to one that does not accelerate. So the longer the time interval the greater the time differential.

That may be true, but I have shown in the last post, that the time dilation due to acceleration in the twins paradox can be reduced to a negligible error, e.g less than 4 seconds "lost" due to acceleration time dilation compared to 8 years "lost" due to constant velocity.

It can also be shown that if an object is accelerating away from us, its instantaneous clock rate relative to our clock, is a function of its instantaneous velocity relative to us (i.e t' = t*sqrt{1-v^2/c^2) where v is its instantaneous velocity, and is independent of its acceleration.

 

[yuiop]

-You are stating that time dilation over the course of the 10 years is caused by this inertial motion, but the formula from this is derived from the light clock thought experiment which we have resolved on the basis of the twin paradox yet.

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 traveling 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 traveling twin returns home, and the two clocks are alongside each other again, there is no ambiguity. The path of the traveling 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.

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.

 

[Mentz114]

-This doesn't solve the twin paradox. If we pretend acceleration plays a negligible impact, using the light clock thought experiment then both observers will see the same amount of time dilation in the other observer.

Time dilation has nothing to do with what the clocks will read when they meet !
There's no way to understand this generally by looking at bits of journeys and talking about 'changing frame'. The fact is that what is shown on a clock is the proper time of the worldline. The proper time takes into account the details of the trip in 4 dimensions and always gives the correct answer. It's got nothing to do with whether one traveller is inertial or not, although it's perhaps a little simpler. Talking about acceleration doesn't help much eaither. The proper time takes into account all changes of velocity on the worldline.

 

[yuiop]

It can also be shown that if an object is accelerating away from us, its instantaneous clock rate relative to our clock, is a function of its instantaneous velocity relative to us (i.e t' = t*sqrt{1-v^2/c^2) where v is its instantaneous velocity, and is independent of its acceleration.

I thought I would elaborate a little on this unsupported statement.

First let's take the equation for the instantaneous velocity on a accelerating rocket with constant proper acceleration (a) at time (t) in the non accelerating frame:

v = at / sqrt[1 + (at/c)2]

as given by Baez http://www.xs4all.nl/~johanw/PhysFAQ/Relativity/SR/rocket.html

We can solve this equation for (at) to obtain:

at = v/sqrt[1-(v/c)^2]

Baez also gives the instantaneous gamma factor of the accelerating rocket at time t as:

γ = sqrt[1 + (at/c)^2]

and if we substitute the symbolic value for (at) obtained earlier into the equation for gamma immediately above we get:

γ = sqrt[1+((v/c)/sqrt[1-(v/c)^2]] = 1/sqrt[1-(v/c)^2]

demonstrating that the instantaneous time dilation of an accelerating particle can be formulated in such a way that it is only a function of the instantaneous velocity at any given instant and independent of the acceleration.

 

[Passionflower]

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?

 

[Mentz114]

Kev, you said it in post #14, so my #15 is redundant.

 

[Passionflower]

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]

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]

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]

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]

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.

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]

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.

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]

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:

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]

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]

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]

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]

We can't make any evaluation of whether a change in velocity is acceleration or deceleration

You are right.

 

[Austin0]

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]

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]

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 spaceship went on the trip, according to the observer on the spaceship the Earth went on a trip away from him.http://www.scientificamerican.com/article.cfm?id=how-does-relativity-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]

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.

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.

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_speed_of_light#Slow_transport or google the terms.

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]

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]

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 traveling 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 traveling twin returns home, and the two clocks are alongside each other again, there is no ambiguity. The path of the traveling 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]

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 traveling 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 traveled 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.

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.

 

[gnomechompsky]

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 traveling 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 traveled a longer path through spacetime and than A and all will agree that B has less elapsed proper time than A.

I still don't think you are justifying this. If all movement is relative, and A,B and C were all in inertial frames of reference, why has B gone through the longer path when he returns to A? Is it not equally valid to say that A (Earth) moved away from B then returned to B?

 

[yuiop]

I still don't think you are justifying this. If all movement is relative, and A,B and C were all in inertial frames of reference, why has B gone through the longer path when he returns to A? Is it not equally valid to say that A (Earth) moved away from B then returned to B?

Your third observer C would certainly never see A (Earth) moving away from B and then returning to B. Observer B would feel the acceleration as he changed direction, while observer A never feels a change in direction, so the two situations are certainly not equivalent or symmetrical from that point of view. The observer that feels or meausres proper acceleration is the one that has really changed direction and acceleration is absolute in that sense. I can only suggest you get some graph paper and try drawing the the space versus time diagrams yourself. That is one of the best ways to get an intuitive understanding of what is happening.

Note: You said "If all movement is relative, and A,B and C were all in inertial frames of reference, why has B gone through the longer path when he returns to A? the important word is "were". If B has returned to A, B has not remained in an inertial frame.

 

[Passionflower]

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_speed_of_light#Slow_transport or google the terms.

I am familiar with the term but I still like to understand what you mean, do you have the calculations? Note that if two clocks run with a different rate and they want to meet halfway they will disagree about where halfway actually is, slow transport or not.

 

[CPL.Luke]

Mentz,

I think this is the time where you have to demonstrate how the observer in the spaceship would go about calculating the proper time of the observer on the earth, from their reference frame.

you are correct that the proper times will differ, however with just plane SR, you can't calculate the proper time of the Earth correctly (from the space ships perspective). If you attempt to do this in SR you will find that the earthbound twin has aged less than the spacebound twin.

EDIT: by valid frame I meant inertial frame, the space ships frame is non inertial at various points, thus in order to properly calculate the proper time you would have to sum over these areas, while you could do this for certain special examples (such as rindler space) the solution is not so simple as to look at the proper time.

Ignoring the non-inertial parts of the spacetime diagram causes the calculation to fail from the spaceships perspective.

 

[Austin0]

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.

It would seem that you have demonstrated that frame C was actually moving relative to frame A as the inertial clock located in C and moving toward A had less elapsed proper time than inertial clock A moving toward clock C at the same relative v.

Explanation?

 

[gnomechompsky]

Your third observer C would certainly never see A (Earth) moving away from B and then returning to B.

Disagree. If C was at 1/2 the velocity of B (and also began his return to C at the same time) he would see both moving away from him (and then towards him) at the same speed.

Observer B would feel the acceleration as he changed direction, while observer A never feels a change in direction, so the two situations are certainly not equivalent or symmetrical from that point of view.

Ok, I can agree on this, but as I said in my first post, the only asymmetry in the Twins Paradox comes from acceleration.

However, what is it about the light clock thought experiment that makes it valid on B from A's frame of reference, but not valid on A from B's frame of reference? The acceleration doesn't matter, because B will still see A's light clock as "dilating", on both the outward and return journey. Is there another way to derive the time dilation equation without this paradox?

I can only suggest you get some graph paper and try drawing the the space versus time diagrams yourself. That is one of the best ways to get an intuitive understanding of what is happening.

I can't do this yet, because it would seem to be easily possible to reverse the space/time graph from the frame of reference for B and get the same result.

 

[Austin0]

Originally Posted by kev
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.

Suppose there is a frame D such that A and C are traveling relative to it, at 0.5c and -0.5c
respectively. At event (B,C) D.. t'''=0 and the spatial interval between A and C is dx''' with observed clock times in A and C of t₀ and t''₀

As observed in D ...A and C meet at dx'''/2 with observed clock times in A and C of T and T''
In both frames dx'''/ 0.5 = dt , dt'' from this it would seem to follow that dt = T-t₀ =T''-t''₀= dt'' Wouldn't this be equal elapsed proper time observed in both frames between events (B,C) and (C,A) ?

Do you see some reason why this would not apply??

 

[Mentz114]

Mentz,
I think this is the time where you have to demonstrate how the observer in the spaceship would go about calculating the proper time of the observer on the earth, from their reference frame.

Given any worldline, one can calculate the elapsed time between two events on it. Furthermore, all observers will agree on this. So the elapsed time as defined above will be the same from all IFRs.

..., however with just plain SR, you can't calculate the proper time of the Earth correctly (from the space ships perspective). If you attempt to do this in SR you will find that the earthbound twin has aged less than the spacebound twin.

I don't agree with this. As I've said, the proper time is the same in all frames.

EDIT: by valid frame I meant inertial frame, the space ships frame is non inertial at various points, thus in order to properly calculate the proper time you would have to sum over these areas, while you could do this for certain special examples (such as rindler space) the solution is not so simple as to look at the proper time.

In the classic twins 'paradox', we don't need Rindler coords, and the resolution is to ignore the instantaneous velocity dependent time-dilation and use the proper time to calculate what the elapsed times are.

Ignoring the non-inertial parts of the spacetime diagram causes the calculation to fail from the spaceships perspective.

The proper time interval is found by integrating along the whole worldline. It doesn't matter if parts are non-inertial.

I don't understand why the differential ageing of the twins is seen as some sort of paradox. Usually the pro-paradox arguments are

1. symmetry - both twins see the other as moving.
That's irrelevant, it's the proper times that give the elapsed times
2. each sees the others clock running slowly
also irrelevant.

 

[gnomechompsky]

1. symmetry - both twins see the other as moving.
That's irrelevant, it's the proper times that give the elapsed times.

How do we derive the formula for proper time? Is it from our geometric understanding of the light clock thought experiment? We can't use proper time to resolve the Twin Paradox if we have not yet justified the premise on which it is based.

Is it possible to derive proper time without using the light clock?

 

[Mentz114]

How do we derive the formula for proper time? Is it from our geometric understanding of the light clock thought experiment? We can't use proper time to resolve the Twin Paradox if we have not yet justified the premise on which it is based.

Is it possible to derive proper time without using the light clock?

In Minkowski spacetime, the infinitesimal line element is given by this formula

<br /> c^2d\tau^2=c^2dt^2-dx^2-dy^2-dz^2<br />

and this forms the basis of the calculation. It is the geometry of Minkowski spacetime, and a postulate of SR is that \tau is the time measured by a clock on the worldline.

I don't see that the light clock comes into at all.

 

[gnomechompsky]

Well Minkowski must have derived the formula somehow. What experiment did he do to get this?

Also, it would seem that the formula above doesn't address the paradox because either side of the equation can equally be applied to either observer. Why is the twin on Earth not on a world line moving away from the Traveller?

 

[Mentz114]

Well Minkowski must have derived the formula somehow. What experiment did he do to get this?

Good question. There is experimental support for SR, but I think Minkowski just intuited the line element.

Also, it would seem that the formula above doesn't address the paradox because either side of the equation can equally be applied to either observer.

The formula applies to whatever segment of worldline you want it to, and it will give a number.

Why is the twin on Earth not on a world line moving away from the Traveller?

From the travellers point of view the Earth is moving away.

I don't know how you are with spacetime diagrams but here's two that show a traveller from the Earth frame, and the Earth from the travellers frame on his outbound trip.

Reading the propertimes off the graphs we find

earth frame : Earth elapsed time ~13
first leg frame : Earth elapsed time sqrt[(14.5*14.5)-(5.5*5.5)] ~ 13

The point is that applying the proper time formula, you get same elapsed time on a segment from any frame.

 

[gnomechompsky]

On those graphs why is Earth always on the straight line? Why can't we make a graph where the traveller is on the straight line and Earth reverses direction?

 

[Mentz114]

Elapsed time on a non-inertial worldline

This graph shows a traveller doing a smooth trip out and back. The worldline is given by

x =0.01 (t² - 1 )⁴

using numerical integration the elapsed time on the traveller's clock is 1.9998564 compared with 2.0000 on the Earth clock.