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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?

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Its not clear to me by what you mean by 'unnaturalness' here. Its also not clear to me that this term in gauge invariant. –  DJBunk Nov 9 '12 at 18:08
    
The trace makes it gauge invariant. It's a three glueball vertex. –  user1504 Nov 9 '12 at 18:18
    
@DJBunk by unnatural I mean strange, mathematically. One doesn't see kinetic terms like $\dot{x}^3$ in point-particle mechanics, for instance. As for the gauge invarance, it's true and follows from the cyclicity of the trace: $L_c=$Tr$(F F F) \mapsto$ Tr$(U^{-1}FU \,U^{-1}F U \,U^{-1}F U )=L_c$. –  c.p. Nov 9 '12 at 18:19
    
Oh yeah, thanks. –  DJBunk Nov 9 '12 at 18:22
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4 Answers

up vote 3 down vote accepted

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.

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That's nicely explained. I'm confused though (my fault): when would it make sense to consider dim-$6$ operators? cannot they precisely say something about "new physis"? –  c.p. Nov 9 '12 at 22:02
    
Adding F^3 to the standard YM Lagrangian gives a non-renormalizable theory; you should only consider it as part of a low-dimensional effective theory. F^3 is the "least irrelevant" gauge-invariant operator for pure YM. So the low-energy expansion of some theory that holds at higher energies might have F^2 as the leading term and F^3 term as the first "correction". –  Robert McNees Nov 12 '12 at 20:11
    
Since the dimension of a field is read off from the kinetic term, what if we take $F^3$ to be the kinetic term and say $A$ (where $F=dA$) is of dimension $[m]^{1/3}$? –  Jingyuan Chen Nov 12 '12 at 21:24
    
I'm not sure why you'd do that. Do you know of some way of building a sensible F^3 theory? –  Robert McNees Nov 13 '12 at 18:59
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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.

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Comment to the answer(v1): A convenient ansatz may make calculations easier and the theory simpler, but this is by itself not a rigorous argument to forbid certain terms in the first place. –  Qmechanic Nov 9 '12 at 21:28
    
... sometimes only a linear dependence, as in the action for the Dirac equation. –  Arnold Neumaier Nov 12 '12 at 19:00
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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 $\frac{s\mbox{short distance scale}}{\mbox{long distance scale}}$.

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I was askink for its classical meaning, but now I see your answer is much more interesting :) Could you give a deatail, why is that so? Or you made dimensional analysis? –  c.p. Nov 9 '12 at 18:21
    
The terms doesn't appear in physically relevant classical gauge theories because these are approximating quantum gauge theories, where it doesn't appear because it's irrelevant. I don't think there's an independent classical reason it doesn't appear. The naturalness arguments seem weak to me, but there is some justification for preferring quadratic kinetic terms. See physics.stackexchange.com/a/43743/1504 for some elaboration. The scaling behavior is essentially dimensional analysis, plus an assumption on the normalization of the kinetic term. –  user1504 Nov 9 '12 at 18:33
    
yes, I admitted that. The question is now: why is that term suppresed by $(E/\Lambda)^2$? –  c.p. Nov 9 '12 at 18:51
    
It's a standard renormalization scaling argument, like I said, dimensional analysis plus normalization of the kinetic term. If you want details, you're going to need to learn some of the theory of renormalization. See physics.stackexchange.com/q/743/1504 for a reading list. –  user1504 Nov 9 '12 at 19:01
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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

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