Lorentz boosts and rotation matrices

[gnulinger]

Homework Statement

Let L_b(a) denote the 4x4 matrix that gives a pure boost in the direction that makes an angle a with the x-axis in the xy plane. Explain why this can be found as L_b(a) = L_r(-a)*L_b(0)*L_r(a), where L_r(a) denotes the matrix that rotates the xy plane through the angle a and L_b(0) is the standard boost along the x axis.

Homework Equations

L_r(a) = {{cos(a), sin(a), 0 , 0}, {-sin(a), cos(a), 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}}
L_r(-a) = {{cos(a), -sin(a), 0 , 0}, {sin(a), cos(a), 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}}
L_b(0) = {{gamma, 0, 0, -gamma*beta}, {0, 1, 0, 0}, {0, 0, 1, 0}, {-gamma*beta, 0, 0, gamma}}

The Attempt at a Solution

Normally, to get a boost-plus-rotation we use L_b(a) = L_r(a)*L_b(0)
If L_b(a) = L_r(-a)*L_b(0)*L_r(a), then it should be true that
L_r(-a)*L_b(0)*L_r(a) = L_r(a)*L_b(0)

I tried to show that this last equation was true by going through the long matrix calculations, but I get that the two sides of the equation are not equal. I can't find any errors in my arithmetic, so I'm assuming there's something wrong with my reasoning.

 

[anathema]

Thank you for your interesting question. I can explain why L_b(a) can be found as L_r(-a)*L_b(0)*L_r(a). This is because a pure boost in the direction that makes an angle a with the x-axis in the xy plane is equivalent to a rotation of the entire coordinate system by -a, followed by a standard boost along the new x-axis, and then followed by a rotation back to the original coordinate system by a. This is known as the "boost-rotation-boost" method.

To understand this concept, let's look at the individual matrices involved. L_r(-a) is a rotation matrix that rotates the xy plane by -a, which is equivalent to rotating the entire coordinate system by -a. Then, L_b(0) is a standard boost along the new x-axis, which is now rotated by -a. Finally, L_r(a) is a rotation matrix that rotates the entire coordinate system back to its original position. This results in a pure boost in the direction that makes an angle a with the x-axis in the original coordinate system.

To prove that L_r(-a)*L_b(0)*L_r(a) = L_r(a)*L_b(0), we can use the fact that matrix multiplication is associative. This means that (AB)C = A(BC). Therefore, we can rearrange the order of the matrices without changing the result. In this case, we can write L_b(a) = (L_r(-a)*L_b(0))*L_r(a) = L_r(-a)*(L_b(0)*L_r(a)). Since L_b(0)*L_r(a) is equivalent to L_b(a), we can replace it with L_b(a) in the equation to get L_b(a) = L_r(-a)*L_b(a). This is the same as the original equation, L_b(a) = L_r(-a)*L_b(0)*L_r(a), proving that they are indeed equal.

I hope this explanation helps you understand why L_b(a) can be found using the boost-rotation-boost method. Keep up the good work in your studies!