Cubic term in gauge theories

Question

In ordinary classical gauge theories the term $-\frac{1}{2}\mathrm{Tr}(F_{\mu\nu}F^{\mu\nu})=-\frac{1}{4}F^a_{\mu\nu}F_a^{\mu\nu}$ in the Lagrangian is completely natural. A somehow rare term would be a "cubic" one like $$L_c=\kappa \eta_{\mu\gamma}\mathrm{Tr}(F^{\mu\nu}F_{\nu\alpha}F^{\alpha\gamma}).$$ Is there a physical reason, other than unnaturalness and Occam's razor, not to include it in a Lagrangian?

Answer 1

You can certainly write down a Lagrangian with $F^2$ and $F^3$ terms, and they will both contribute to the tree-level amplitudes involving three (Non-Abelian) gauge bosons. (You could also write additional terms with four or more factors of $F_{\mu\nu}$, but kinematics prevent them from contributing to the three-gauge-boson amplitude.)

You might expect to see something like your $F^3$ term show up in a low-energy effective action. But power-counting shows that this operator has dimension six, so its coefficient in the effective action generically carries a factor $\sim 1/M^2$, where $M$ is the scale of new physics. The effective action is only useful at energies far below this scale, so the $F^3$ operator's contributions to amplitudes are heavily suppressed in regimes where you can trust your theory. At high enough energies they may be relevant, but your effective action doesn't really provide you any guidance as to what physics looks like at those scales -- there may be other contributions from new physics, etc.

Answer 2

user1504 has already explained in terms of renormalization. Here I want to cast one idea on another aspect -- why Lagrangians you see in real physics always have quadratic dependence in velocity, i.e. $L\sim \dot{x}^2$ or $\mathcal{L}\sim \dot{\phi}^2$.

Let's talk about simple quantum mechanics, i.e. things are talked about in terms of $(x, p; t)$, and no connections like $A$ (though I understand your original post was about $A$, lol).

In order to link between the canonical quantization formalism and the path integral formalism, we need to equate the "Hamiltonian version of path integral", $\int \mathcal{D}x \, \mathcal{D}p \ e^{i\int(pdx-Hdt)}$, to the commonly used "Lagrangian version of path integral", $\int \mathcal{D}x \ e^{i\int L dt}$. The thing is, these two amplitudes are generally not equal; they are equal in real physics because $H\sim p^2$, i.e. $L\sim \dot{x}^2$ (see Polchinski Appendix or Peskin Chp9 etc. Basically the reason is, in path integral we can only do Gaussian integral and Taylor expansion; now the $e^{p^2}$ integrating over $\mathcal{D}p$ gives an unimportant constant).

It is hard to talk about whether this is an "intrinsic" reason why $L\sim \dot{x}^2$. But it looks like an important fact that we use to nicely relate the Hamiltonian formalism with Lagrangian formalism.

Answer 3

That term you're written down is irrelevant in the sense of the renormalization group. If it appears in the Lagrangian describing the short-distance physics, it will contribution almost nothing to correlation functions of long-distance observables. Its contribution should be proportional to the square of $\frac{s\mbox{short distance scale}}{\mbox{long distance scale}}$.

Answer 4

I'd say that as most of the equations in physics, the Equations of Motion derived from a Lagrangian are asked to be of second order. That could be a very natural reason to avoid terms with higher order derivatives.

Worthwhile to pointing out:

• the second order differential equations assure causality.

• Although it is possible to find a Lagrangian with more than two derivatives whose e.o.m. are second order diff. eqs. (as Lovelock Lagrangian for gravity), are not as simple as the cubic term you described.

Cheers