Why "light clock" is flawed.

gnomechompsky

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

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.

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

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

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 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 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

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

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.

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'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

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

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.

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

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.

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

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

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

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

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

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

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

   

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

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

 

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?

Same with the graph above, the blue curved line could equally be Earth from the frame of reference of the Traveller.