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Rotation and Boost Commutating on the Same Axis

  1. Oct 22, 2013 #1
    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?
     
  2. jcsd
  3. Oct 22, 2013 #2

    Bill_K

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    Science Advisor

    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.
     
  4. Oct 23, 2013 #3
    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.
     
  5. Oct 24, 2013 #4
    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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