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Relativity - Light At An Angle

  1. May 17, 2009 #1
    1. The problem statement, all variables and given/known data
    A light beam is emitted at an angle [tex]\theta_o[/tex] with respect to the x' axis in S'

    a) Find the angle [tex]\theta[/tex] the beam makes with respect to the x axis in S.
    - Ans. : [tex] cos\theta = (cos\theta_o + \frac{v}{c})(1 + \frac{v}{c} cos\theta_o)[/tex]


    3. The attempt at a solution
    From an example problem in our book, we know that with a ruler at an angle in the same situation has angle in laboratory frame:

    [tex] \theta = arctan(tan(\theta_o)\gamma)[/tex], where [tex]\gamma = \frac{1}{\sqrt{1-frac{v^2}{c^2}}}[/tex]

    if you just take cos of both sides, then you have

    [tex]cos\theta = cos(arctan((tan(\theta_o)\gamma)))[/tex]

    Drawing a triangle with the right leg as [tex]tan(\theta_o)[/tex] and the bottom leg as [tex]\frac{1}{\gamma}[/tex] you get a hypotenuse of [tex]\sqrt{(\frac{1}{\gamma})^2 + tan(\theta_o)^2}[/tex]

    from this i narrowed it down to

    [tex]cos\theta = \frac{\frac{1}{\gamma}}{\sqrt{(\frac{1}{\gamma})^2 + tan(\theta_o)^2}}[/tex][tex] = \frac{1}{\sqrt{1 + frac{\gamma*2}{cos{\theta_o)^2} - \gamma^2}}[/tex]

    I'm not seeing how this can reduce to my answer I'm suppose to get yet?
     
  2. jcsd
  3. May 17, 2009 #2

    Doc Al

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    Staff: Mentor

    That answer doesn't look right. It should be:

    [tex] cos\theta = \frac{(cos\theta_o + \frac{v}{c})}{(1 + \frac{v}{c} cos\theta_o)}[/tex]

    The easy way to derive it is to use the velocity transformations.
     
  4. May 17, 2009 #3
    oops I missed the divide sign in my book - your answer is the right one.

    I'll give that a shot
     
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