# Dual tensors in Lagrangian

1. May 26, 2014

### guest1234

Why is it the case that dual field tensors, e.g. $\widetilde{F}^{\mu\nu}=\frac{1}{2}\epsilon^{\mu\nu\rho\sigma}F_{\rho \sigma}$, aren't being included in the Lagrangian? For example, one doesn't encounter terms like $-\frac{1}{4}\widetilde{F}^{\mu\nu}\widetilde{F}_{\mu\nu}$ in QED or $\widetilde{F}^{\mu\nu}W_{\mu}W_{\nu}^{\dagger}$ in vector meson interactions (Phys. Rev. 58, 953). Has it something to do with (discrete) symmetries or..?

2. May 26, 2014

### Einj

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.

3. May 26, 2014

### Bill_K

This is equal to the same quantity without the tildes.

The Lagrangian must be a scalar, and this is a pseudoscalar.

4. May 26, 2014

### Einj

Yes, that's true. If you want to include the dual tensor you have to write something like:
$$F_{\mu\nu}\tilde{F}^{\mu\nu}$$

5. May 26, 2014

### guest1234

Thanks, Bill_K, I didn't know that before. A quick proof for the eager:
\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*}
The contracted term of dual and ... ugh... normal tensor is effectively zero (a 4-divergence, doesn't change the EOM):
\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*}
since
\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*}

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
6. May 26, 2014

### Einj

$F\tilde{F}$ is not zero in general. You are supposing that the theory is an abelian one. If you have a non-abelian theory then $F_{\mu\nu}^a=\partial_\mu A^a_\nu-\partial_\nu A^a_\mu+gf^{abc}A_\mu^b A_\nu^c$. In this case it turns out that $F\tilde{F}$ 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

7. May 26, 2014

### guest1234

Yeah, I forgot to mention I'm interested in U(1) theory. Thanks for the link, will read it.