Lorentz boosts and rotation matrices

  • Thread starter gnulinger
  • Start date
  • #1
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I also posted this in the homework help for introductory physics, but it wasn't getting any responses, so I guess it's slightly more advanced.

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.
 

Answers and Replies

  • #2
213
8
L(-a)B(0)L(a) tells you to start with a vector in the xy plane rotate it so that it is along the x axis boost it and then rotate it back to its original angle in the xy plane. L(a)B(0) tells you to start with a vector along the x axis boost and then rotate. So defining v to be a vector in the xy plane and L(a)v the corresponding vector along the x axis. Then the boosted v along the x axis is L(a)B(v)v = B(0)L(a)v
 
  • #3
30
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Thanks for the reply. I don't follow the last part of your post though. If v is a vector in the xy plane, wouldn't L(-a)v be the corresponding vector along the x axis? I also dont understand how L(a)B(v)v = B(0)L(a)v.
 
  • #4
213
8
it is L(a)v since that is what you gave me anyway it doesn't matter as long as we have opposites in L(-a)B(0)L(a). B(v)v = v' is a boosted vector along v which we can rotate to get a vector along x i.e L(a)v'. However this is the same as getting the original vector v rotating it so that it is along the x axis and then boosting i.e. B(0)L(a)v
 
  • #5
30
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I guess that makes sense, but my linear algebra skills aren't so great. I thought that you would normally get the rotated and boosted vector B(a) by L(a)*B(0).

I think my confusion is coming from the difference between a pure boost and something like L(a)*B(0). What is the difference?

Is it that when you have L(a)*B(0), or a boost followed by a rotation, that you get Thomas precession, and the pure boost avoids that?
 

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