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Lorentz boosts and rotation matrices

  • Thread starter gnulinger
  • Start date
  • #1
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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.
 

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