# Homework Help: What is the relative area of a rectangular field?

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1. Jan 18, 2016

### Julian102

1. The problem statement, all variables and given/known data
If the area of a rectangular field has length of 110 m and 80 m.If a spaceship is travelling 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?

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

3. 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

2. Jan 18, 2016

Why would it become a parallelogram?

3. Jan 18, 2016

### DaveC426913

Length contraction along one diagonal but not along the other would result in a distortion of the rectangle.

4. Jan 18, 2016

### Staff: Mentor

Contractraction occurs along the direction of travel, but not along direction perpendicular to the direction of travel.

Does this image help?

5. Jan 18, 2016

Ah yes the other diagonal.. totally forgot! Thanks

6. Jan 18, 2016

### Julian102

7. Jan 18, 2016

### Julian102

Then how will I find the area?

8. Jan 18, 2016

### haruspex

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.
No, not the other diagonal, exactly. It's a rectangle, not a square. See gneill's diagram.

9. Jan 18, 2016

### SteamKing

Staff Emeritus
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.