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Configuration comparison.

  1. Nov 15, 2009 #1
    When we are taught relativity we never seem to be asked to consider how the configuration of objects in frames moving relative to ours actually got there. It just seems that we imply that is the way they always were and so an actual congruence measurement of direct comparison in the same frame was never made. If we physically transported the configuration of objects to our rest frame and measured them by direct comparison, could we guarantee the same result as applying the transformations of our measurements at a distance, or would the transportation profile make a difference. This leads on to the following question. If we transport a configuration of objects, presently at rest in our frame, to another inertial frame moving relative to us, and then return the configuration to rest in our frame, would the configuration return to its original dimensions in our frame no matter how the transportation was done. Or would the two transportations have to be exactly reciprocal with respect to each other.

    Matheinste.
     
  2. jcsd
  3. Nov 15, 2009 #2

    Dale

    Staff: Mentor

    This is not unique to relativity. All of the laws of physics are in the form of differential equations, and when you solve a differential equation you wind up with one or more undetermined values. These values cannot be derived from the laws of physics but by the boundary conditions of the specific problem at hand, such as the configuration of some set of objects. How the boundary conditions came about are not important in any branch of physics.
     
    Last edited: Nov 15, 2009
  4. Nov 16, 2009 #3
    If we accelerated a rocket of length d to a new inertial frame moving at positive v relative to us using Born rigid acceleration we could with sufficient care acclerate the rocket back to the initial frame using Born rigid acceleration and its length would still be d. I say "sufficient care" because once the rocket is accelerated using Born rigid acceleration the clocks at the back and front of the rocket will no longer be synchronised from the point of view of the launch frame OR from the point of view of the rocket observers, so you will have to take into account whose version of simultaneity you are going to use for initiating the return trip.

    If the rocket uses Born rigid acceleration on the outward leg and Bell acceleration on the return leg it will be longer than d on arriving back at the start frame. Also if we used Bell acceleration on the outward leg and Bell acceleration on the return leg the rocket would be even longer than in the previous case.

    It would seem that the configuration does not return to its original state, unless Born rigid acceleration is used and the rocket clocks are re-synchronised before the return journey. Even when the spatial configuration is the same upon return to the original frame, the transported clocks will be out out sync again.
     
    Last edited: Nov 16, 2009
  5. Nov 16, 2009 #4

    Dale

    Staff: Mentor

    So what? Again, this is not unique to relativity. In many different non-relativistic situations you also have to take great care in order to return a system to a previous state.
     
  6. Nov 16, 2009 #5
    I never claimed it was unique to relativity. I was simply answering matheinste's question "would the configuration return to its original dimensions in our frame no matter how the transportation was done" (?)
     
  7. Nov 16, 2009 #6
    Most people are aware that if one reference frame is accelerated to a velocity relative to another inertial frame, that observers in both frames would consider rulers in the other frame to shorter than their own and clocks in other frame to be slower than their own and that rulurs and clock rates in their own frame appear the same after acceleration as before acceleration. On the other hand, most people are not aware that after acceleration, the occupants of the accelerated frame would be able to tell that it was that it was their own frame that underwent acceleration, even if they slept through the acceleration phase, by observing that clocks in their own reference frame have gone out of sync. Because that is not well known, it is worth mentioning in the context of accelerating one frame relative to another.
     
  8. Nov 16, 2009 #7

    Dale

    Staff: Mentor

    Ah, good point. I answered an earlier part of the OP and not that part.

    @ matheinste, as I said above, the reason that we don't go into how the boundary conditions came to be is because it is irrelevant, this is the same with relativity as with any other branch of physics. If the state of the system is X then it doesn't matter how the state came to be X, the laws of physics then tell us how the system evolves to some other state Y. Once the system is in state Y there is not always an easy way back to state X, but if there is a way back to state X then it again does not matter how the system arrived at that state and you can again get back to Y in exactly the same manner as before. This has general applicability.

    Specifically for relativity, as kev mentioned simply accelerating and reversing the acceleration will not, in general, bring the system back to state X.
     
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