What Is the Length of a Meter Stick Moving at an Angle?

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Homework Help Overview

The problem involves a meter stick moving at a significant fraction of the speed of light (0.8C) and at an angle (60 degrees) relative to its direction of motion. The discussion centers around the concept of length contraction in special relativity and how to apply it to determine the observed length of the stick from a specific frame of reference.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the Lorentz factor and its application to length contraction, noting that contraction occurs only in the direction of motion. There is an exploration of visualizing the meter stick as part of a triangle and considering coordinate transformations to simplify the problem.

Discussion Status

The discussion is active, with participants providing insights and affirmations regarding the approach to the problem. There is a focus on visualizing the problem correctly and ensuring that one axis aligns with the direction of motion to avoid complications in calculations.

Contextual Notes

Participants express uncertainty about visualizing the problem and the implications of changing coordinate systems. The original poster references a solution from the textbook, which adds a layer of complexity to their understanding.

SpaceTrekkie
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Homework Statement


A meter stick is moving with 0.8C relative to frame S. What is the sticks length when measured by an observer in S if the stick is 60 degrees to v, as seen in the rest frame?


Homework Equations


Length contraction = proper length /lorentz factor


The Attempt at a Solution


Okay for the speed I got the lorentz factor = 1.66667. So if the meter stick was moving directly parallel the length would be contracted to .6 meters. I know that length contraction only occurs in the direction of motion, so NOT perpendicular. And I know that I can solve it by thinking about the meter stick as the hypotenuse of a 30-60-90 triangle. I think I am just mis visualizing something, but I really can't seem to work it out. I looked in the back of the book and the answer is .917 meters. I used that to try to work backwards and still could not figure it out.

Any direction would be awesome...
 
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Hi SpaceTrekkie! :smile:
SpaceTrekkie said:
Okay for the speed I got the lorentz factor = 1.66667.
So if the meter stick was moving directly parallel the length would be contracted to .6 meters. I know that length contraction only occurs in the direction of motion, so NOT perpendicular.
And I know that I can solve it by thinking about the meter stick as the hypotenuse of a 30-60-90 triangle.

All correct. :wink:

Draw the triangle in x,y coordinates.

Now draw it again in x',y' coordinates.

y' = y and x' = 0.6x, so … ? :smile:
 
Hmm, Does it work to change the x and y-axis so that they x-axis is in like with the direction of motion? I think that might make it easier, or am I over thinking the problem?
 
SpaceTrekkie said:
Hmm, Does it work to change the x and y-axis so that they x-axis is in like with the direction of motion? I think that might make it easier, or am I over thinking the problem?

No, you're completely correct :smile:

always have one of the axes in the direction of motion …

otherwise the equations get too complicated! :rolleyes:
 

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