How Does Special Relativity Affect the Perceived Angle of a Sailboat Mast?

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In the discussion about how special relativity affects the perceived angle of a sailboat mast, participants explore the angle observed by a stationary observer as the boat moves at speed v. The initial calculation involves using the formula tan^-1(tan θ * γ), which raises concerns about the presence of tangent within the inverse tangent function. The correct transformation for the observed angle is confirmed as tan θ' = γ tan θ, highlighting the effects of relativistic length contraction. A shift in focus to a beam of light from a spotlight on the boat introduces complexity, as participants discuss the challenges of identifying an arbitrary point for analysis. The conversation concludes with an acknowledgment that the angles observed in both scenarios differ, prompting further curiosity about the implications of relativity.
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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.
 
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well

\arctan{(\tan{x})} = x
 
nicktacik said:
well

\arctan{(\tan{x})} = x

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.
 
Your answer, tan \theta ' = \gamma tan \theta is correct. I don't get what your concern is.
 
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.
 
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?
 
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