What is the relative area of a rectangular field?

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Julian102
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Homework Statement


If the area of a rectangular field has length of 110 m and 80 m.If a spaceship is traveling with 0.9c velocity along the diagonal of the field.Then what is the area of the field seen by the astronaut in the spaceship?

Homework Equations


L=L(initial) Root over (1- (v/c)^2)

The Attempt at a Solution


Guys,this will change to a non-rectangular parallelogram.I used resolved part theory at first to find Lx(along length) and Ly(along breadth). Then, I used the formula of relativity for length contraction on Lx(along length) and Ly(along breadth) . Now I get the area A= Lx*Ly
 
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Why would it become a parallelogram?
 
faradayscat said:
Why would it become a parallelogram?
Length contraction along one diagonal but not along the other would result in a distortion of the rectangle.
 
Contractraction occurs along the direction of travel, but not along direction perpendicular to the direction of travel.

Does this image help?

Fig2.PNG
 
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DaveC426913 said:
Length contraction along one diagonal but not along the other would result in a distortion of the rectangle.

Ah yes the other diagonal.. totally forgot! Thanks
 
gneill said:
Contractraction occurs along the direction of travel, but not along direction perpendicular to the direction of travel.

Does this image help?
Well now how will I solve the problem?

View attachment 94471
 
Then how will I find the area?
 
Julian102 said:
Then how will I find the area?
Consider a small element dxdy, using coordinates in the direction of and at right angles to the velocity of the spaceship. What is the apparent area of that element?
Sum.
faradayscat said:
Ah yes the other diagonal.. totally forgot! Thanks
No, not the other diagonal, exactly. It's a rectangle, not a square. See gneill's diagram.
 
Julian102 said:
Then how will I find the area?
If you study gneill's diagram, you have two triangles created by the major diagonal. Since there is no length contraction perpendicular to the line of travel, the heights of the triangles are the same as for the original field layout. You have computed the contracted length of the diagonal, so it's a matter of calculating the areas of these triangles and adding them together. If you do a little algebra, there is a simple answer as to what the contracted area of the field is.