# Why does the Lagrangian Density have to be a polynomial of the field?

In a lecture, a professor appeared to have said that the Lagrangian can only contain terms that have powers of $$\phi$$ and a term with $$\partial_\mu \partial^\mu \phi$$ . I imagine this would make any physically possible Lagrangian of the form

$$\mathcal{L}(\phi, \partial_\mu \phi, t) = k\partial_\mu \phi \partial^\mu \phi + \sum_{i\in I} c_i \phi^i$$ For arbitrary real numbers $$k$$ and $$c_i$$ and index set $$I$$.

Is this truly the case? If so, why would it be impossible to have a Lagrangian that has a term with, say, $$\cos(\phi)$$ or even $$(\partial_\mu \partial^\mu \phi)^2$$?

• As a first comment, the first term in your Lagrangian density really should be $k\partial_{\mu}\phi\partial^{\mu}\phi$, as this is the canonical kinetic term of a scalar field theory (you could also see this from the following facts -- 1. A lagrangian density shouldn't contain any linear terms, for you could simply complete the square to get rid of them and 2. The kinetic term you wrote down has a free Lorentz index.). – Bob Knighton Jan 3 '19 at 19:03
• @BobKnighton Ah yes of course, my bad. Edited accordingly. – John K Jan 3 '19 at 19:04
• The $\mu$ index cannot be free like this. It needs to be contracted on another similar index to get a Lorentz and coordinates scalar. So something like $(\partial_{\mu} \phi)(\partial^{\mu} \phi)$, with some arbitrary exponent. – Cham Jan 3 '19 at 19:05
• You could add things like $\sum_n c_n \, \big( (\partial_{\mu} \phi)(\partial^{\mu} \phi) \big)^n$ – Cham Jan 3 '19 at 19:06
• the chiral lagrangian, something that has been experimentally verified over and over again, it's definitely not of the form the OP has described. In fact, any theory of Goldstone Bosons is not of that form. – TwoBs Jan 4 '19 at 10:14

The Lagrangian density of any field theory does not need to by polynomial in the field. The polynomial form of a Lagrangian density is typically taken to be an approximation in the spirit of effective field theory. Indeed, one could easy write down a field theory whose Lagrangian density takes the form

$$\mathcal{L}=\frac{1}{2}\partial\varphi\cdot\partial\varphi+a^2m^2(\cos{\frac{\varphi}{a}}-1).$$

This theory is known as Sine-Gordon theory (for obvious reasons). In $$d=2$$ dimensions Sine-Gordon theory is actually incredibly interesting and has many applications in the study of duality.

Of course, I could simply Taylor expand the cosine and write

$$\mathcal{L}=\frac{1}{2}\partial\varphi\cdot\partial\varphi+\frac{1}{2}m^2\,\varphi^2-\frac{1}{4!}\frac{m^2}{a^2}\varphi^4+\frac{1}{6!}\frac{m^2}{a^4}\,\varphi^6+\cdots,$$

which resembles the form in which you wrote your Lagrangian density.

The polynomial approximation is typically taken because one cannot do traditional perturbation theory without it (polynomial terms lead to $$n$$-valent graphs in the Feynman diagrammatic expansion of the partition function in a field theory) and because of the fact that terms in the Lagrangian with high powers of $$\varphi$$ typically are less important in certain approximation regimes (this is the basis of effective field theory).

I hope this helps!