All possible electromagnetic Lorentz invariants that can be built into the electromagnetic Lagrangian?

Given the electromagnetic Lagrangian density $$\mathcal{L}~=~-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}~=~\frac{1}{2}(E^2-B^2)$$ is a Lorentz invariant, how many other electromagnetic invariants exists that can be incorporated into the electromagnetic Lagrangian?

• One can show that all local, gauge and Lorentz invariant can be constructed from only two quantities $F_{\mu\nu}F^{\mu\nu}$ and $\tilde F_{\mu\nu}F^{\mu\nu}$ – Adam Mar 15 '14 at 16:46
• What is the meaning of the symbol "~" over "F"? – linuxfreebird Mar 15 '14 at 16:48
• $\tilde F_{\mu\nu}=\epsilon_{\mu\nu\sigma\lambda}F^{\sigma\lambda}$ and $\tilde F_{\mu\nu}F^{\mu\nu}\propto E^2-B^2$. – Adam Mar 15 '14 at 16:54
• Oh its the dual tensor. – linuxfreebird Mar 15 '14 at 16:55
• Related: physics.stackexchange.com/q/87817/2451 and links therein. – Qmechanic Mar 15 '14 at 22:45

To lowest order in $F$ the only non-zero invariants are: \begin{equation} F _{ \mu \nu } F ^{ \mu \nu} \quad , \quad \epsilon _{ \alpha \beta \gamma \delta } F ^{ \gamma \delta } F ^{ \alpha \beta } \end{equation} If we restrict ourselves to terms with mass dimension of $4$ or lower these are the only options (these terms are called renormalizable terms). However, one can also write down other invariants which have higher mass dimensions. One such example is the mass dimension six term, \begin{equation} \partial ^\mu F _{ \mu \nu } \partial ^\alpha F _\alpha ^{ \,\, \nu } \end{equation} Such terms are small at low energies and are often ignored. In general there are an infinite number of allowed (non-renormalizable) terms in the Lagrangian. Though it may not be trivial, such terms could be written in terms of the electric and magnetic fields to find the different combinations of $\bf E$ and $\bf B$ that form Lorentz invariants.
• One wonders if there are people doing $f(F)$ E&M in the same way there are people doing $f(R)$ gravity. – user10851 Mar 15 '14 at 17:57
• A Lorentz gauge term, $\partial_\mu A^\mu$ is not gauge invariant. If you want to fix the gauge such that this term is zero then you can add it in. But at that point there is no use to including the term as this term is zero. – JeffDror Mar 16 '14 at 17:14