How Is the Area of a Relativistically Contracted Football Field Calculated?

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Homework Help Overview

The problem involves calculating the area of a football field that is relativistically contracted as observed by a moving observer. The field is described as a square with semi-circular arcs, and the observer is moving at a significant fraction of the speed of light.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of length contraction on the dimensions of the football field, particularly how the semi-circular arcs may be perceived by the observer. Questions arise regarding the transformation of shapes under relativistic effects and the completeness of the problem statement.

Discussion Status

The discussion is ongoing, with participants exploring various interpretations of how the shapes change due to relativistic effects. Some guidance is offered regarding the general principles of area reduction when dimensions are altered, but no consensus has been reached on a specific approach to the problem.

Contextual Notes

There are concerns about missing information in the problem statement, particularly regarding the observer's speed and the proper dimensions of the field. Participants are encouraged to clarify the original problem as it was presented.

Turhan
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Homework Statement



A football field is given in the following shape, where, ABCD is a square of side-length and AEB, CFD are semi-circular arcs. If an observer is moving with uniform velocity .along AB, what is the area of the football-field measured by the observer? ( is the velocity of light in free-space.)

EDIT: The side of the square is 1m and the speed of the observer is 0.8c.

Homework Equations



L=L0√1-v2/c2

The Attempt at a Solution


Okay so I know I have to use the length contraction equation. As the observer is moving along AB, the lengths of AB and CD will contract. The relativistic length is .6 meters. The square becomes a rectangle. What do I do with the semi circular regions?
P.S- Not any homework problem. Just a question from a previous olympiad.
 
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What happens to a circle when you squash it in one direction?
 
Simon Bridge said:
What happens to a circle when you squash it in one direction?
Ellipse. Thanks a lot captain.
 
I think the semi circular arcs will be viewed as semi-elliptic arcs from the moving observer.
 
The problem statement seems to be missing information regarding speeds and lengths. How did you find the relativistic length of 0.6 meters unless you know the observer speed and proper dimensions of the field? Can you provide a clear, complete statement of the problem as it was given to you?
 
gneill said:
The problem statement seems to be missing information regarding speeds and lengths. How did you find the relativistic length of 0.6 meters unless you know the observer speed and proper dimensions of the field? Can you provide a clear, complete statement of the problem as it was given to you?[/QU
The side of the square is 1m and the speed of the observer is 0.8c. Dont know ehre the info went and can't even edit it right now.
 
Turhan said:
The side of the square is 1m and the speed of the observer is 0.8c. Dont know ehre the info went and can't even edit it right now.
Thanks. There's a short window of time wherein one can edit their own posts, after which they're closed to alteration. I will take the information and add it for you.
 
Turhan said:
Ellipse. Thanks a lot captain.
There is a more general principle that may be of more use.

You know that shrinking a rectangle's width by a factor of k reduces its area by a factor of k. You know that shrinking a rectangle's length by a factor of k reduces its area by a factor of k. Can you generalize this to other shapes or other directions? If you know integral calculus, can you justify this generalization?

Edit: If you do not know integral calculus, what if you start by tiling the interior of an arbitrary planar shape with a bunch of small rectangles and squares of various sizes all lined up along the "shrink axis".
 

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