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A question regarding length contraction

  1. Mar 24, 2008 #1

    aaj

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    I am not a physics student (my background is that of an engineer + MBA) but have read a lot about relativity and have built up a fair level of understanding.

    I just thought up a situation regarding Lorentz contraction that has kind of confused my understanding of the same. Consider the following scenario:

    A is an earth based observer. B is an observer 10 LY away from the earth and is at rest with respect to observer A. C is an observer situated on the line between A and B and is situated 5 LY from both A and B. In other words, C is located midway between A and B. C is also at rest with respect to A and B.

    Hence in the above scenario, A measures B and C to be located a distances 10 LY and 5 LY respectively from himself.

    Now suppose that for some reason C starts moving towards B with a velocity 0.866c. This means that the relative velocity between A and C is also 0.866c (in the opposite direction). Therefore, A would now measure the distance between himself and C to be only 2.5 LY (due to Lorentz contraction). However, A still measures the distance to B to be 10 LY.

    The above would imply that A would now measure the distance between B and C to be 7.5 LY.

    Is the above conclusion correct? It appeared very weird to me that despite C attaining a velocity in the direction of B, i.e. away from A, A still measures a decrease in the distance between himself and C and an increase in the distance between B and C.
     
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  3. Mar 24, 2008 #2

    Doc Al

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    Staff: Mentor

    No, A and B will still measure their distance to C (as C passes the midpoint between A & B) to be 5 LY. It's C, not A or B, who will measure the distances as being contracted.
     
  4. Mar 24, 2008 #3

    aaj

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    This confuses me. I thought that if two observers have a velocity between themselves, each observes the distance to the other to be length contracted. In this discussion, the relative velocity between A and C is 0.866c. From the standpoint of either of them, the other observer is moving away at 0.866c. When then do you say that it is only C who would observe the distance between themselves to be Lorentz contracted? The situation is symmetrical for A, so why would he not observed the Lorentz contraction of his distance from C?
     
  5. Mar 24, 2008 #4

    Doc Al

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    Staff: Mentor

    No.
    True.
    Because the distance from A to C was already measured to be 5 LY according to A. As far as A is concerned, he's just sitting there--why should his distance measurements change because C is moving?

    What would be symmetric is something like this: Say C carries along with him a giant stick that's 10 LY long (according to C) and aligned along the direction of travel. Then A will measure the length of that moving stick to be only 5 LY long according to him.
     
  6. Mar 24, 2008 #5

    aaj

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    So you mean that it is possible for two observers in relative motion to each measure a different value for the distance between them?

    For instance, if a planet revolves in a precise circular orbit around a star, will an observer on the planet measure a different value for the distance between the planet and the star, compared the to the value measured by an observer located on the surface of the star?
     
  7. Mar 24, 2008 #6

    Doc Al

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    Absolutely!

    That's not a good example since the motion is perpendicular to the distance between the star and planet. (Distance only "contracts" in the direction of motion.)

    Taking your previous example, imagine two planets A and B that are at rest with respect to each other and are 10 LY apart (according to their own measurements). An astronaut in spaceship C is zooming by planet A on her way to planet B. She's moving at 0.866c with respect to the planets. The moment she passes planet A, how far is she from planet B?

    According to planets A & B, she is 10 LY away from planet B. But according to her own measurements, she is only 5 LY away.
     
  8. Mar 24, 2008 #7

    JesseM

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    As with almost all confusions about relativity, it comes down to the relativity of simultaneity. Suppose C is traveling along with a rod that is 5 ly long in his own frame, and is sitting in the middle, so that each end of the rod is 2.5 ly away from him. Then in C's frame, the event of one end of the rod passing by A is simultaneous with the event of the other end passing by B, which is also simultaneous with the event of C passing a buoy that is at the midpoint of the line between A and B. So obviously in C's frame, the distance between A and B must by 5 ly. In A and B's frame, the rod is shrunk to only 2.5 ly long total, so each end is only 1.25 ly away from C. But in A and B's frame, the event of one end of the rod passing A happens long before the event of C passing the buoy at the midpoint, which happens long before the event of the other end of the rod reaching B. In A and B's frame the distance between them is of course going to stay 10 ly...nothing short of A or B accelerating could change that.
     
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