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Sailboat and special relativity

  1. Aug 20, 2007 #1
    In a sailboat, the mast leans at an angle of theta with respect to the deck. An observer stainging on a dock watches the boat go by at speed v. Assume the boats is perpendicular to the observers line of sight and that its motion is also perpendicular to the observers line of sight. What angle does the observer observe the mast at.

    I got tan^-1( tan theta * gamma). However, that seems very wrong because I have never seen a tangent inside a tan^-1.
  2. jcsd
  3. Aug 20, 2007 #2

    [tex]\arctan{(\tan{x})} = x[/tex]
  4. Aug 20, 2007 #3
    if x is between pi/2 and -pi/2. :biggrin:

    However, I cannot really make any simplifications like when the tan(x) is only one factor in arctan.
  5. Aug 20, 2007 #4


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    Your answer, [itex]tan \theta ' = \gamma tan \theta [/itex] is correct. I don't get what your concern is.
  6. Aug 28, 2007 #5
    How would you do the problem if the mast were replaced with a beam of light from a spotlight mounted on the boat?

    In the original case, I took an arbitrary point on the mast (x,y) with tan(theta) = y/x and performed a length contraction on the y.

    You cannot really do that now that it is a beam of light, so I am having trouble finding an "arbitrary" point in the new problem.
  7. Nov 27, 2011 #6
    You didn't take an arbitrary point. You picked the end of the mast. For the beam of light, consider two events in the boats frame: the emission of a photon and the detection of said photon at the end of the mast. Transform these events to the stationary frame and find the angle made. Curious that they aren't the same isn't it?
    Last edited: Nov 27, 2011

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