Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook