Rotation and Boost Commutating on the Same Axis

  • Thread starter JordanGo
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  • #1
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I want to prove that:
[tex] [J_1,G_1] = 0 [/tex]
Where J is the rotation operator and G is the boost operator (subscript refers to the axis).

I am using the Jacobi identity:
[tex] [[J_1,J_2],G_3] = [[G_3,J_2],J_1] +[[J_1,G_3],J_2] [/tex]

Using other identities, I got:
[tex] [J_3,G_3] = [G_2,J_2] - [G_1,J_1] [/tex]

Now I'm not sure what to do, can someone help?
 

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  • #2
Bill_K
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You will need to use information specifically about the Lorentz group and these two generators. For example you could write their action as 4 x 4 matrices on (x, y, z, t), and show that these matrices commute.
 
  • #3
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I would start from the Poincare algebra:
[tex]\frac{ 1 }{ i }[M_{\mu\nu}, M_{\rho\sigma}] = \eta_{\mu\rho} M_{\nu\sigma} - \eta_{\mu\sigma} M_{\nu\rho} - \eta_{\nu\rho} M_{\mu\sigma} + \eta_{\nu\sigma} M_{\mu\rho}[/tex]

Then the boosts are: (μ = 1, 2, 3)
[tex]G_{\mu} = M_{0\mu}[/tex]

And the rotations: (μ, ρ, σ = 1, 2, 3; ρ ≠ σ; ρ ≠ μ; σ ≠ μ)
[tex]J_{\mu} = M_{\rho\sigma}[/tex]

You will prove the commutativity knowing that certain components of the metric tensor are zero and the generator of the Lorentz transformations is antisymmetric.
 
  • #4
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I want to prove that:
[tex] [J_1,G_1] = 0 [/tex]
Where J is the rotation operator and G is the boost operator (subscript refers to the axis).
That is pretty simple,if you are allowed to use the commutation relation [Ji,Gj]=iεijkGk,for i=j=1.you have ε11k term,which is of course zero.(you can get this commutation using above post)
 

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