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Special Relativity Diagram Problem

  1. Feb 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Frame S' has velocity v relative to frame S. At time t = 0, a light ray leaves the origin of S, traveling at a 45 degree angle with the x-axis.

    (a) What angle does the light ray make with respect to the x' axis in frame S'?

    (b) Repeat part (a), replacing the light ray with a particle of mass m and speed u.

    (c) Repeat part (a), replacing the light ray with a rod that is stationary in frame S.


    2. Relevant equations

    Minkowski diagram

    [tex]v'_x = \frac{v_x - v}{1-vv_x/c^2}[/tex]
    [tex]v'_y = \frac{v_y}{\gamma(1-vv_x/c^2)}[/tex]
    [tex]\theta = \tan^{-1}(v/c)[/tex]

    3. The attempt at a solution

    (a) In frame S', the axes are both rotated inwards (if v is positive) by angle [itex]\theta[/itex]. However, I am not sure if the angle wanted in part (a) is still 45 degrees in frame S' (even though it is clearly [itex]45 - \theta[/itex] looking from frame S).

    (b) The particle has identical components of velocity [itex]v_x = v_y[/itex] so I should just be able to use the above equations to solve for [itex]v'_x[/itex] and [itex]v'_y[/itex]. However, the same question from part (a) remains; is angle measured differently in the S' frame as opposed to the S frame? In other words, in the S' frame, is the angle between x' and ct' still considered 90 degrees?

    (c) You can treat S' as the stationary frame and S as the moving frame, in which case the axes of S will rotate outwards assuming it is moving with negative velocity relative to S'. In that case, it is a simple matter of determining the angle from the above formula and calculating the apparent angle with the x' axis.

    Thanks!
     
  2. jcsd
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