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

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Homework Help Overview

The discussion revolves around the effects of special relativity on the perceived angle of a sailboat mast as observed from a stationary point. The problem involves understanding how motion affects the observation of angles, particularly in the context of relativistic effects such as length contraction.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between the angle of the mast and the observer's perception, with one participant expressing confusion over the use of inverse tangent in their calculations. Others clarify properties of the tangent function and discuss the implications of replacing the mast with a beam of light.

Discussion Status

The discussion is active, with participants providing insights and clarifications regarding the mathematical relationships involved. There is an ongoing exploration of different scenarios, including the transition from a physical mast to a beam of light, which raises further questions about the application of special relativity.

Contextual Notes

Participants note the challenge of applying concepts of length contraction to different types of objects (a mast versus a beam of light) and question the assumptions made in their approaches. There is an acknowledgment of the complexity introduced by relativistic effects in these scenarios.

ehrenfest
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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

[tex]\arctan{(\tan{x})} = x[/tex]
 
nicktacik said:
well

[tex]\arctan{(\tan{x})} = x[/tex]

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, [itex]tan \theta ' = \gamma tan \theta[/itex] 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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