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Lorentz boosts and rotation matrices 
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#1
May2811, 11:37 PM

P: 30

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.
1. The problem statement, all variables and given/known data 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. 2. Relevant 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}} 3. The attempt at a solution Normally, to get a boostplusrotation 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. 


#2
May2911, 06:01 AM

P: 205

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
May2911, 10:57 AM

P: 30

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
May2911, 11:13 AM

P: 205

Lorentz boosts and rotation matrices
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
May2911, 11:40 AM

P: 30

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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