1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Special relativity- Uniform circular motion.

  1. Apr 3, 2013 #1
    1. The problem statement, all variables and given/known data
    Imagine two clocks that both perform uniform circular motion of radius a in the x-y plane, but in opposite directions: xμ(u) ={t, a cos(ωt),±a sin(ωt), 0}. Suppose these clocks are synchronized to agree when they are coincident at x = a at t = 0. How much time elapses until the next time the clocks are at x = a, as seen by each clock as well as by the inertial observer whose time is labelled by t?

    2. Relevant equations
    xμ(u) ={t, a cos(ωt),±a sin(ωt), 0}
    xσ(u)= {t,0,0,0}


    3. The attempt at a solution
    The trajectories are symmetrical, so I only need to calculate the proper time elapsed for in one of the moving reference frames and it should be the same for the other.
    First I converted each trajectory into its proper time. I set t=γτ where τ=tau

    xμ(u) ={γτ, a cos(ωγτ),±a sin(ωγτ), 0}

    Then I wanted the 4-velocities so I could determine how the opposite observer looks in one of the moving reference frames.

    dxμ(u)/dτ =γ{1, -aωsin(ωγτ),±aωcos(ωγτ), 0}. The γ and τ are different depending on the trajectory.
    I then subtracted the velocity vectors from each other.

    xΩ+-{1, -aωsin(ωγ+τ+)-aωsin(ωγ-τ-),aωcos(ωγ+τ+)-aωcos(ωγ-τ-), 0}

    So that equation is what the observer moving in the +sin direction would see in his reference frame for the observer moving in -sin direction. Now I need to determine how much time elapses between t=0 and the next time they meet.

    I think this is at points where τ+-? This is where I am confused. Is my reasoning good so far?
     
  2. jcsd
  3. Apr 4, 2013 #2
    Sherejg,

    Welcome to Physics Forums.

    It seems to me that this problem is a lot simpler than the way you are approaching it. From the problem statement, it appears that the motion of the two clocks is being described as reckoned from the inertial (stationary) frame of reference. It is easiest to do all the calculations as reckoned from this frame of reference. All you need to do is determine the proper time interval (of the clocks) between successive instances that each clock passes through the point a,0 (in the inertial frame). The differential of proper time for the traveling clocks is given by

    [tex](d\tau)^2= (dt)^2(1-(\frac{dx}{cdt})^2-(\frac{dy}{cdt})^2)=(\frac{dt}{\gamma})^2[/tex]

    The two clocks and and observers in the stationary frame of reference all agree on the simultaneity of the events when the two clocks meet up.

    Chet
     
  4. Apr 5, 2013 #3
    Okay, so I think I misread the problem. Your equation gives the proper time elapsed for a circular moving clock of uniform speed in reference to an inertial observer. I thought I also needed to use one of the clocks as an observer with a uniform circular motion reference frame, and try to discover the time elapsed in his frame of the other clock. But its really asking for the time in inertial observer frame compared to proper time. Thanks for clearing that up! So much easier now haha.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Special relativity- Uniform circular motion.
Loading...