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Why light clock is flawed. 
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#37
Aug1810, 05:45 PM

P: 11




#38
Aug1810, 06:13 PM

P: 3,967

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. 


#39
Aug1810, 07:50 PM

P: 1,555




#40
Aug1810, 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 noninertial parts of the spacetime diagram causes the calculation to fail from the spaceships perspective. 


#41
Aug1910, 03:01 AM

P: 1,162

Explanation??? 


#42
Aug1910, 03:15 AM

P: 11

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? 


#43
Aug1910, 07:20 AM

P: 1,162

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 t_{0} 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 = Tt_{0} =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?? 


#44
Aug1910, 10:08 AM

PF Gold
P: 4,087

I don't understand why the differential ageing of the twins is seen as some sort of paradox. Usually the proparadox 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. 


#45
Aug1910, 10:42 AM

P: 11

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


#46
Aug1910, 12:09 PM

PF Gold
P: 4,087

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


#47
Aug1910, 12:22 PM

P: 11

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? 


#48
Aug1910, 01:19 PM

PF Gold
P: 4,087

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. 


#49
Aug1910, 01:40 PM

P: 11

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?



#50
Aug1910, 01:59 PM

PF Gold
P: 4,087

This graph shows a traveller doing a smooth trip out and back. The worldline is given by
x =0.01 (t^{2}  1 )^{4} using numerical integration the elapsed time on the traveller's clock is 1.9998564 compared with 2.0000 on the earth clock. 


#51
Aug1910, 02:41 PM

P: 11

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


#52
Aug1910, 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. 


#53
Aug1910, 03:10 PM

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P: 4,087




#54
Aug1910, 03:30 PM

P: 11

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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