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SR: Does observation change when changing reference frame

  1. Jul 22, 2017 #1
    1. The problem statement, all variables and given/known data
    Special relativity: A rod travelling with the relativistic velocity of 0.866c moving towards a garage. The length of the rod is L, that of the garage is L/2. From the reference frame of the garage, we find the value of Lorentz factor = 2 we say that rod would fit inside garage momentarily. Now we take this scenario from rod's frame and find the length of garage L/4. Since mechanical pressure can't travel faster than light, so we find by the time light(mechanical pressure) reaches the free end the rod has shrunk without any resistance to a size of L times (1-0.866) = 0.134L<0.25L, and therefore can easily fit inside garage.

    My question is - In garage's frame of reference rod barely fits inside while in rod's reference frame it can fit inside with margin to spare, is this difference in observation valid?

    Also if this is correct then what happens if I increase the size of rod and keep the size of garage fixed so that it can fit inside garage from rod's frame but not from garage's frame?

    2. The attempt at a solution
    I do understand that concept of simultaniety holds little value in SR because simultaneous event in one frame may not be so in other, but I think that observation like the one in this case should be same although explanation may differ, but I don't see how that would happen in the given question.

    Here by observation I mean the end result of whatever events occur.
     
  2. jcsd
  3. Jul 22, 2017 #2

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    This problem should not use anything about the speed of mechanical pressure. It's just a puzzle about the definition of "simultaneous" in each reference frame.
    This is not what you said above. Think about this again. It's always the other guy, who is moving, whose length seems to have shrunk.
     
  4. Jul 23, 2017 #3
    Actually, I just generalised the length of rod and garage in terms of L, the rest is given as an explanation in Relativity special, general and cosmological by Rindler (2006 edition - 3rd chapter), including the explanation about mechanical pressure travelling through the rod.

    I also read another similar example where there were 2 doors in and out of the garage and we find that rod's length was reduced to half by Lorentz factor(in a similar fashion as example in post #1) hence it could fit inside garage with both doors closed, but from rod's frame of reference we find that both doors did not close simultaneously and so rod was never completely inside the garage, when head of rod touched back door, the front door was open. Similarly when front door closed on back end of rod the head of rod was outside the garage because the back door was open.

    I think this is what you mean by
    But the example I quoted in my 1st post is different because there is no back door. Once the rod is inside, it's not getting out no matter how much force it applies (an assumption I did not state earlier).

    I did not understand what you meant by
    But now I think I do. When I said rod will shrink in its own reference frame I did not mean it will happen by using lorentz factor. It will shrink because back end (one touching the garage) would be pushed by garage and since mechanical pressure can't travel faster than c hence front end would not know that back end was being pushed so it stays at same place. End result being that rod gets shrinked.
    question.png
     
    Last edited: Jul 23, 2017
  5. Jul 23, 2017 #4

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    Your solution of the puzzle still should not have anything to do with mechanical pressure from an impact. Both reference frames should observe the same end result but for different reasons. Saying that the rod will not impact the rear of the garage is assuming some type of almost immediate slowing to 0 velocity. Making that assumption for one observer but not for the other is not valid. Suppose the rod does not change velocity. Both reference frames will observe that it will crash into the back of the garage. They will disagree on whether it briefly fit inside the garage because they can not agree on the "simultaneous" positions of both front and rear of the rod. In fact, they can not agree on what events are "simultaneous". But they will agree that the end result was a collision into the rear of the garage.
     
    Last edited: Jul 23, 2017
  6. Jul 23, 2017 #5
    What if we close the gate for the garage when the rod was inside the garage? Also, assume that once the door was closed the rod can't get out so either it bends or breaks.
     
  7. Jul 23, 2017 #6

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    The rod-based observer thinks that the garage is too short, but the door was closed long after the front of the rod hit the back of the garage. The garage-based observer thinks that the rod is short enough to fit and he closed the door before the rod hit the back of the garage. Both think that the rod hit the back of the garage. They only disagree on the relative timing of the collision and the door closing. Any assumption that the velocity of the rod changes is a completely different problem.
     
  8. Jul 23, 2017 #7
    I meant how can the door close from rod's frame of reference if garage was too small? rod can't get outside garage from behind. The door wont shut no matter how long rod based observer waits
     
  9. Jul 23, 2017 #8

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    This should be a thought experiment regarding moving reference frames and space-time positions in those frames. The garage and rod are just a way of referring to those space-time positions in two different reference frames. I advise you not to introduce issues involving collision forces and structural strength that are not part of a basic SR thought experiment.

    You are talking as though the rod will stop once it gets into the garage and so the door can never be closed. That is not part of the thought experiment. The space-time positions of the ends of the rod keep going at a constant velocity. So the rod front position hist the garage wall in both reference frames and the door can be closed as soon as the rod end position goes past.

    If you do assume that the rod reference frame slows to a stop as soon as the front of the rod reaches the back of the garage, than both reference frames become identical and they observe identical lengths and results.
     
    Last edited: Jul 23, 2017
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