Why is the lagrangian polynomial in fields and derivatives

[IRobot]

I started to answer this question, and I have quite a bit an answer, but still not complete, let's say that we write a Lagrangian in QFT, which an unknown function of the scalar field \phi and its derivative \partial \phi. We can always Taylor-expand it and get: L(\phi,\partial\phi) = a + b \phi + c \partial\phi + d \phi^2 + e \partial\phi^2 + f \phi\partial\phi + g \phi^3 + h \phi^2\partial\phi + ...
But than, since we are in QFT, we need the Lagrangian to be renormalizable so we cut this expansion at the fourth level (couting \partial\phi as a term of order 2). We then get a Lagrangian polynomial, with finite terms, and assuming that in the limit quantum to classical, the "shape" of Lagrangian doesn't change, this applies to classical field theory. But it is still incomplete because I can't get rid of some terms like f \phi\partial\phi. Maybe knows how to finish this demonstration.

 

[The_Duck]

The Lagrangian is a Lorentz scalar. But a term like \phi (\partial_\mu \phi) is a Lorentz vector, so can't appear in the Lagrangian. All vector indices have to be contracted, as in, say, \phi (\partial_\mu \phi) (\partial^\mu \phi).

 

[IRobot]

Yes, I thought but we could have f^\mu\phi\partial_\mu\phi we break isotropy in space-time so it can't correspond to the most general lagrangian, thank you ;)