Why "light clock" is flawed.


by gnomechompsky
Tags: flawed, light clock
gnomechompsky
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#37
Aug18-10, 05:45 PM
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Quote Quote by kev View Post
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.
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
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#38
Aug18-10, 06:13 PM
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Quote Quote by gnomechompsky View Post
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
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#39
Aug18-10, 07:50 PM
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Quote Quote by kev View Post
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.
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
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#40
Aug18-10, 10:46 PM
P: 444
Mentz,

I think this is the time where you have to demonstrate how the observer in the space ship 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
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#41
Aug19-10, 03:01 AM
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Quote Quote by kev View Post
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
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#42
Aug19-10, 03:15 AM
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Quote Quote by kev View Post
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.

Quote Quote by kev View Post
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?

Quote Quote by kev View Post
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
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#43
Aug19-10, 07:20 AM
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Quote Quote by kev View Post
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
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#44
Aug19-10, 10:08 AM
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Quote Quote by CPL.Luke View Post
Mentz,
I think this is the time where you have to demonstrate how the observer in the space ship 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
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#45
Aug19-10, 10:42 AM
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Quote Quote by Mentz114 View Post
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
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#46
Aug19-10, 12:09 PM
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Quote Quote by gnomechompsky View Post
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

[tex]
c^2d\tau^2=c^2dt^2-dx^2-dy^2-dz^2
[/tex]

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

I don't see that the light clock comes into at all.
gnomechompsky
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#47
Aug19-10, 12:22 PM
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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
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#48
Aug19-10, 01:19 PM
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Quote Quote by gnomechompsky View Post
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.
Attached Thumbnails
earth view.png   first leg.png  
gnomechompsky
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#49
Aug19-10, 01:40 PM
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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
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#50
Aug19-10, 01:59 PM
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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.
Attached Thumbnails
non-inertial-trip.png  
gnomechompsky
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#51
Aug19-10, 02:41 PM
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Quote Quote by Mentz114 View Post
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.
I think you misunderstood my question. Referring back to the original two graphs, why can we not just switch the reference frames so it is Earth that is moving away and then doing a return journey?

Same with the graph above, the blue curved line could equally be Earth from the frame of reference of the Traveller.
CPL.Luke
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#52
Aug19-10, 02:52 PM
P: 444
Mentz you are assuming that it is minkowski space for the traveler, if that was the case then the equations you show would have both parties disagreeing on who was younger.

The problem is solved from the travelers frame when you realise that the traveler passes through rindler space each time he accelerates, the earth is always in minkowski space.

To illustrate this, in post 50 you performed the calculation from the earths perspective, can you perform it from the travelers perspective? your solution is correct, but for the wrong reasons.
Mentz114
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#53
Aug19-10, 03:10 PM
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Quote Quote by gnomechompsky
I think you misunderstood my question. Referring back to the original two graphs, why can we not just switch the reference frames so it is Earth that is moving away and then doing a return journey?
Because finding rocket motors powerful enough to accelerate the earth is a problem. But if you could do it, then the earth's worldline would be curved and its proper time shorter.

Quote Quote by CPL.Luke
Mentz you are assuming that it is minkowski space for the traveler, if that was the case then the equations you show would have both parties disagreeing on who was younger.
Proper time is invariant under Lorentz transformation, and any instant on the curved world line is connected to the earth worldline by a Lorentz transformation.

The problem is solved from the travelers frame when you realise that the traveler passes through rindler space each time he accelerates, the earth is always in minkowski space.
It is always Minkowski space. There's no 'Rindler' spacetime, only Rindler coords in Minkowski spacetime.

To illustrate this, in post 50 you performed the calculation from the earths perspective, can you perform it from the travelers perspective? your solution is correct, but for the wrong reasons.
Please explain why the calculation is wrong. Seriously, if you think that the elapsed times are frame dependent for the trip under question then demonstrate it.
gnomechompsky
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#54
Aug19-10, 03:30 PM
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Quote Quote by Mentz114 View Post
Because finding rocket motors powerful enough to accelerate the earth is a problem. But if you could do it, then the earth's worldline would be curved and its proper time shorter.
Ah, so we are back to the point in my original post, it is the acceleration "felt" by the object that generates the asymmetry in the Twins Paradox.

However, even if we know that it is the Space traveller accelerating, according to the formulas for proper time and SR time dilation, acceleration is not a factor.


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