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

Dual tensors in Lagrangian

  1. May 26, 2014 #1
    Why is it the case that dual field tensors, e.g. [itex]\widetilde{F}^{\mu\nu}=\frac{1}{2}\epsilon^{\mu\nu\rho\sigma}F_{\rho \sigma}[/itex], aren't being included in the Lagrangian? For example, one doesn't encounter terms like [itex]-\frac{1}{4}\widetilde{F}^{\mu\nu}\widetilde{F}_{\mu\nu}[/itex] in QED or [itex]\widetilde{F}^{\mu\nu}W_{\mu}W_{\nu}^{\dagger}[/itex] in vector meson interactions (Phys. Rev. 58, 953). Has it something to do with (discrete) symmetries or..?
     
  2. jcsd
  3. May 26, 2014 #2
    In principle, there is no reason to exclude it. The point is that such a term would imply a strong CP violation which has not been observed experimentally so far. Understanding why such a term does not appear in the Lagrangian (a better way to say it is to understand why the parameter θ in front of it is so small) is one of the open problems in the Standard Model.
     
  4. May 26, 2014 #3

    Bill_K

    User Avatar
    Science Advisor

    This is equal to the same quantity without the tildes.

    The Lagrangian must be a scalar, and this is a pseudoscalar.
     
  5. May 26, 2014 #4
    Yes, that's true. If you want to include the dual tensor you have to write something like:
    $$
    F_{\mu\nu}\tilde{F}^{\mu\nu}
    $$
     
  6. May 26, 2014 #5
    Thanks, Bill_K, I didn't know that before. A quick proof for the eager:
    [itex]\begin{align*}\widetilde{F}^{\mu\nu}\widetilde{F}_{\mu\nu} &= \frac{1}{4}\epsilon^{\mu\nu\rho\sigma}\epsilon_{\mu\nu\alpha\beta}F_{ \rho \sigma}F^{\alpha\beta} = \frac{1}{2}\delta_{\alpha \beta}^{\rho\sigma}F_{\rho\sigma}F^{\alpha \beta} = \delta^{[\rho}_{\alpha}\delta^{\sigma]}_{\beta}F_{\rho\sigma}F^{\alpha\beta} = \frac{1}{2}(\delta^{\rho}_{\alpha}\delta^{\sigma}_{\beta} - \delta^{\sigma}_{\alpha}\delta^{\rho}_{\beta})F_{\rho\sigma}F^{\alpha \beta} =\\ &= \frac{1}{2}(F_{\rho\beta}F^{\rho\beta} - F_{\beta\sigma}F^{\sigma\beta}) = \frac{1}{2}(F_{\mu\nu}F^{\mu\nu}+F_{\mu\nu}F^{\mu\nu}) = F_{\mu\nu}F^{\mu\nu}\,.\end{align*}[/itex]
    The contracted term of dual and ... ugh... normal tensor is effectively zero (a 4-divergence, doesn't change the EOM):
    [itex]\begin{align*}F_{\mu\nu}\widetilde{F}^{\mu\nu} &= \epsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma} = \epsilon^{\mu\nu\rho\sigma}(\partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu})(\partial_{\rho}A_{ \sigma} - \partial_{\sigma}A_{\rho}) = 4\epsilon^{\mu\nu\rho\sigma}(\partial_{\mu}A_{\nu})(\partial_{\rho}A_{ \sigma}) =\\ &= 4\epsilon^{\mu\nu\rho\sigma}\partial_{\mu}(A_{\nu}\partial_{\rho}A_{ \sigma}) = 0\,,\end{align*}[/itex]
    since
    [itex]\begin{align*}\epsilon^{\mu\nu\rho\sigma}\partial_{\mu}(A_{\nu}\partial_{\rho}A_{\sigma}) = \epsilon^{\mu\nu\rho\sigma}[(\partial_{\mu}A_{\nu})(\partial_{\rho}A_{\sigma}) + A_{\nu}\partial_{\mu}\partial_{\rho}A_{\sigma}]\,.\end{align*}[/itex]

    Putting the calculus aside, any suggestions from the literature on the matter? I've copies of Peskin&Schröder, Zee, Srednicki and other classics.
     
    Last edited: May 26, 2014
  7. May 26, 2014 #6
    [itex]F\tilde{F}[/itex] is not zero in general. You are supposing that the theory is an abelian one. If you have a non-abelian theory then [itex]F_{\mu\nu}^a=\partial_\mu A^a_\nu-\partial_\nu A^a_\mu+gf^{abc}A_\mu^b A_\nu^c[/itex]. In this case it turns out that [itex]F\tilde{F}[/itex] is not zero but it's a total derivative which, however, is not irrelevant for the physics.

    You can take a look to the strong CP problem here: http://arxiv.org/pdf/hep-ph/0607268v1.pdf
     
  8. May 26, 2014 #7
    Yeah, I forgot to mention I'm interested in U(1) theory. Thanks for the link, will read it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook