# Why don't we have logarithms or exponentials of the fields in the Lagrangians?

All the Lagrangian densities I have seen have always been polynomials of the fields. Is this a coincidence or is there a reason which forbids, say, Lagrangians with logarithms or exponentials of the fields?

• – innisfree Apr 21 '15 at 14:52
• Actually, people do sometimes use functions in the potentials in scalar-field inflation e.g. natural inflation $V(\phi) \propto cos(\phi/M)$ (arxiv.org/abs/hep-ph/0404012) and I think Starobinsky is equivalent to $V(\phi) \propto 1-exp(\phi/M)$ – innisfree Apr 21 '15 at 14:56

Let us use units where action is dimensionless, i.e., $\hbar = 1$, and also $c = 1$. Then the action is $$S = \int d^4 x \, \mathcal L.$$ and $d^4x$ has the units of $\text{mass}^{-4}$; it's mass dimension is $-4$. Hence the Lagrangian $\mathcal L$ must have units of $\text{mass}^4$, i.e., mass dimension $4$. In these units, the electromagnetic field tensor $F_{\mu\nu}$ has mass dimension $2$ and a Dirac field $\psi$ mass dimension $\frac 3 2$.
Now it turns out that in a renormalizable theory, any coupling constants must have mass dimension $d \ge 0$ -- the proof is in Weinberg, Ch. 12. Hence a term like $\frac{1}{M} \bar \psi \gamma^\mu\gamma^\nu F_{\mu\nu} \psi$ or $\frac{1}{M^4}(F_{\mu\nu}F^{\mu\nu})^2$ cannot appear in a renormalizable Lagrangian. A forteriori, this rules out functions like $\exp(F_{\mu\nu}F^{\mu\nu})$. [This is also why general relativity and QFT don't mix: the curvature has dimension of $\text{length}^{-2} = \text{mass}^2$, so $G$ has mass dimension $-2$.]
However, if we drop the requirement of renormalizabilty, then in an effective field theory there is no limit to what we can put in our Lagrangian. In fact a term like $\frac{1}{M} \bar \psi \gamma^\mu\gamma^\nu F_{\mu\nu} \psi$ appears in the effective theory of electrons in external fields -- it's the famous anomalous magnetic moment. While this is still a polynomial, Heisenberg and Euler calculated the effective Lagrangian for QED in a strong background field already in the 30s. It is $${\displaystyle {\mathcal {L}}=-{\mathcal {F}}-{\frac {1}{8\pi ^{2}}}\int _{0}^{\infty }\exp \left(-m^{2}s\right)\left[(es)^{2}{\frac {\operatorname {Re} \cosh \left(es{\sqrt {2\left({\mathcal {F}}+i{\mathcal {G}}\right)}}\right)}{\operatorname {Im} \cosh \left(es{\sqrt {2\left({\mathcal {F}}+i{\mathcal {G}}\right)}}\right)}}{\mathcal {G}}-{\frac {2}{3}}(es)^{2}{\mathcal {F}}-1\right]{\frac {ds}{s^{3}}}}$$ where $\mathcal F = \frac{1}{4}F_{\mu\nu} F^{\mu\nu}$ and $\mathcal G = \frac{1}{4}\epsilon_{\mu\nu\rho\sigma} F^{\mu\nu}F^{\rho\sigma}$. Quite the complicated function of the fields!