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

Deriving the poincare algebra

  1. Jul 3, 2014 #1
    Hey guys,

    as this is a basic QFT question, I wasn't sure to put it in the relativity or quantum section. Since this question specifically is about manipulating tensor expressions, i figured here would be appropriate.


    My question is about equating coefficients in tensor expressions, 2.4.10-11 in Weinberg's Quantum Theory of Fields (2005).

    [itex] i[ 1/2 \omega_{\mu \nu} J^{\mu \nu} - \epsilon_{mu}P^{\mu}, J^{\rho \sigma} ] = \omega_{\mu}^{\rho} J^{\mu \sigma} - \omega_{\nu}^{\sigma} J^{\rho \nu} - \epsilon^{\rho}P^{\sigma}+ \epsilon^{\sigma}P^{\rho}[/itex]

    [itex] i [ 1/2 \omega_{\mu \nu} J^{\mu \nu} - \epsilon_{\mu} P^{\mu} ,P^{\rho} ]=\omega_{\mu}^{\rho} P^{\mu} [/itex]

    The task is to equate coefficients on the epsilon and omega terms to find the commutators of the poincare algebra. I'm a bit confused because, for instance, the omega term on the LHS of the first equation has dummy indices, but on the RHS has real indices. I'm not sure what manipulations i can do besides raising and lowering with the metric.
     
  2. jcsd
  3. Jul 3, 2014 #2
    In order to equate coefficients, for example with the (2.4.11) expression, you can expand the commutator and collect all the coefficients together on the left hand side so that it looks like:
    [tex]
    -\frac{i}{2}\omega_{\mu\nu}\left[P^{\rho},J^{\mu\nu}\right] + i\epsilon_{\mu}\left[P^{\rho},P^{\mu}\right] = \omega_{\mu}^{\ \ \rho}P^{\mu}=\eta^{\rho\nu}\omega_{\mu\nu}P^{\mu}
    [/tex]
    This already gives you the (2.4.14) result because there is no εμ coefficient on the right-hand side. With the ω coefficients you have to do a little work because they are not in a form you can get (2.4.13) directly with. Notice that the left-hand side is completely symmetric under μ↔ν (this is because ωμν and Jμν are both completely antisymmetric). This means the right-hand side must also be symmetric under this exchange. You know that you can write the symmetric part of a tensor Aij (the whole part here) as (Aij+Aji)/2. Therefore you can do this with the right-hand side and write:
    [tex]
    -\frac{i}{2}\omega_{\mu\nu}\left[P^{\rho},J^{\mu\nu}\right] = \frac{1}{2}\left(\eta^{\rho\nu}\omega_{\mu\nu}P^{\mu} + \eta^{\rho\mu}\omega_{\nu\mu}P^{\nu}\right)
    [/tex]
    That cancels out the factors of 1/2 and you can get the ωμν and ωνμ to look the same by using the antisymmetry of ωνμ=-ωμν. That leaves you with:
    [tex]
    i\omega_{\mu\nu}\left[P^{\rho},J^{\mu\nu}\right] = \omega_{\mu\nu}\left(\eta^{\rho\mu}P^{\nu} - \eta^{\rho\nu}P^{\mu}\right)
    [/tex]
    which gives you what he has, only with different greek letters for the indices.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Deriving the poincare algebra
  1. Poincare' group (Replies: 5)

Loading...