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Relativity and Uncertainty

  1. Nov 9, 2009 #1
    Hi,

    Just had a quick question... Relativity seems to require the position and time of an event in one reference frame and the difference in velocity with a second reference frame. Given this starting information you're suppose to be able to calculate the position and time of the same event as seen from the second reference frame... Does the Heisenberg Uncertainty Principle prevent you from accurately knowing all the starting information for small scales... Thanks.

    Tegg
     
  2. jcsd
  3. Nov 9, 2009 #2

    Dale

    Staff: Mentor

    Hi Tegg, welcome to PF!

    Position and time are not canonically conjugate to each other so according to the uncertainty principle you can know both to arbitrary precision.
     
  4. Nov 10, 2009 #3
    Thanks for replying... I guess I'm thinking that in particle experiments the primary reference frame is normally some sort of sensor equipment and the secondary frame is anchored to a fast moving particle. Then if you know the velocity of the secondary reference frame you have trouble figuring out the it's position... But maybe that's not the way experiments are set up...
     
  5. Nov 11, 2009 #4

    Dale

    Staff: Mentor

    Remember, reference frames are not physical entities, they are coordinate systems. They are mathematical tools for keeping track of physics, they are in no way "anchored" to anything.

    So, let's say that you have a non-relativistic particle whose position you know to be x±∆x and whose velocity you know to be v±∆v such that (∆x)(m∆v) satisfies the HUP. Now, suppose further that you do a translation and a boost such that x'=0 and v'=0. Then in the primed reference frame the particle's position is 0±∆x and the particle's velocity is 0±∆v and (∆x)(m∆v) still satisfies the HUP.
     
  6. Nov 13, 2009 #5
    Thanks again for replying... Sorry for the delay. Had to try and get my head around the math. Not sure if I did... For a relativistic particle with position x±∆x and velocity v±∆v such that (∆x)(m∆v) satisfies the HUP, after the translation and boost I seem to get:

    ∆x' = ∆x / sqrt(1- v^2/c^2)

    ∆v' = ∆v / (1 - v^2/c^2 - (v)∆v/c^2)

    This seems to imply that ∆x'>∆x and ∆v'>∆v so that (∆x')(m∆v') > (∆x)(m∆v)...
     
  7. Nov 13, 2009 #6

    Dale

    Staff: Mentor

    Yes, I specified non-relativistic because I was too lazy to write out all of those terms.
     
  8. Nov 14, 2009 #7
    Ok. Last question... Things being relative. If you set (∆x')(m∆v') so that it satisfies the HUP then doesn't (∆x)(m∆v) violate the HUP...
     
  9. Nov 14, 2009 #8

    Dale

    Staff: Mentor

    No. Remember, mv is not the momentum of a relativistic particle.
     
  10. Nov 14, 2009 #9
    Thanks for your help.
     
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