Change in area due to relative motion

[Amith2006]

Homework Statement

1) A square of area 100 cm^2 is at rest in the reference frame of O. Observer O’ moves relative to O at 0.8c along the diagonal of the square. What is the area measured by O’?

Homework Equations

The Attempt at a Solution

I solved it the following way:
I resolved each side of the square into 2 components- One along the diagonal and the other perpendicular to the diagonal.
Given that side of square(a)=10 cm
Component of side a along the diagonal=a(cos(45)) cm
Component of side a perpendicular to the diagonal=a(sin(45)) cm
Due to the motion of O’ only the component of a along the diagonal gets
contracted.
Contracted length= a(cos(45))[1 – v^2/c^2]^(1/2)
= 6/(2^(1/2))
Let a1 be the side of the new square w.r.t O’.
By applying Pythagoras theorem,
a1=[ {6/(2^(1/2))}^2 + { a(sin(45))}^2]^(1/2)
= sqrt(68) cm
Area of new square= (a1)^2
= 68 cm^2
But the answer the given in my book is 60 cm^2.

 

[PhanthomJay]

Homework Statement

1) A square of area 100 cm^2 is at rest in the reference frame of O. Observer O’ moves relative to O at 0.8c along the diagonal of the square. What is the area measured by O’?

Homework Equations

The Attempt at a Solution

I solved it the following way:
I resolved each side of the square into 2 components- One along the diagonal and the other perpendicular to the diagonal.
Given that side of square(a)=10 cm
Component of side a along the diagonal=a(cos(45)) cm
Component of side a perpendicular to the diagonal=a(sin(45)) cm
Due to the motion of O’ only the component of a along the diagonal gets
contracted.
Contracted length= a(cos(45))[1 – v^2/c^2]^(1/2)
= 6/(2^(1/2))
Let a1 be the side of the new square w.r.t O’.
By applying Pythagoras theorem,
a1=[ {6/(2^(1/2))}^2 + { a(sin(45))}^2]^(1/2)
= sqrt(68) cm
Area of new square= (a1)^2
= 68 cm^2
But the answer the given in my book is 60 cm^2.

You calculated the sides of the shape correctly, but why did you assume that the new shape as viewed by the observer was still a square?

 

[Amith2006]

You calculated the sides of the shape correctly, but why did you assume that the new shape as viewed by the observer was still a square?

That is because all the sides are contracted to the same extent. So I thought that it would still remain a square.

 

[PhanthomJay]

That is because all the sides are contracted to the same extent. So I thought that it would still remain a square.

When you calculated the side a1 using Pythagorus, you correctly took the sq rt of the sum of the squares of the triangle legs. The vertical leg is contracted
along the diagonal of the square in the direction of the motion. The horizontal leg (and horizontal diagonal of the square) is not contracted, since it is perpendicular to the motion. Draw a quick sketch . You no longer will have a square.

 

[Amith2006]

So the resultant figure is a rhombus, isn’t it?
Area = ½ x (diagonal1)x(diagonal2)
= ½ x [6(2^(1/2))]x[10(2^(1/2)]
= 60 cm^2
Am I right now?

 

[PhanthomJay]

So the resultant figure is a rhombus, isn’t it?
Area = ½ x (diagonal1)x(diagonal2)
= ½ x [6(2^(1/2))]x[10(2^(1/2)]
= 60 cm^2
Am I right now?

Yes, you've got it.

 

[Amith2006]

Thanx buddy.