# General scalar Lagrangian

I have a doubt, I usually heard that the most general Lagrangian for a scalar field, up to two fields and two derivatives is $$\mathscr{L}=\partial_\mu \phi \partial^\mu \phi + c_2 \phi^2$$ My question is;

First, why don't include a term like $$c_1\phi$$?

Second, why is usually assumed that $$c_1$$ and $$c_2$$ are constant, wouldn't be possible a Lagrangian like $$\mathscr{L}=c_0(x)\partial_\mu\phi\partial^\mu\phi + c_1(x)\phi + c_2(x)\phi^2~?$$

## 2 Answers

Let's see what we can get out of this. Our derivatives are the 4-vector $$\partial_\mu \phi$$. The actual most general Lagrangian with your constraints is

$$$$\mathcal{L} = M^{\mu\nu}(x) \partial_\mu \phi \partial_\nu \phi + A^\mu(x) \partial_\mu \phi + B^\mu(x)\phi\partial_\mu \phi + C(x) \phi^2 + D(x) \phi + F(x)$$$$

First, let's do ourselves a favor. For an action principle, we have that two Lagrangians have the same dynamic if they only differ by a total divergence. If $$F$$ is well-behaved enough, it can be expressed by a divergence. We can also deal with $$A_\mu$$ via the integration by parts (The boundary of $$\phi$$ vanishes at infinity)

$$\begin{eqnarray} \int A^\mu(x)\partial_\mu \phi dx &=& - \int \phi \partial_\mu A^\mu(x) \end{eqnarray}$$

So that we can combine $$A$$ and $$D$$.

You can also use this trick to change the first term to

$$\begin{eqnarray} M^{\mu\nu}(x) \partial_\mu \phi \partial_\nu \phi = -\phi \left[(\partial^\mu M) (\partial_\mu \phi) + M(x) \partial_\mu \partial^\mu \phi\right] \end{eqnarray}$$

We are left with just

$$$$\mathcal{L} = M^{\mu\nu}(x) \partial_\mu \phi \partial_\nu \phi + B^\mu(x)\phi\partial_\mu \phi + C(x) \phi^2 + D(x) \phi$$$$

An easy way to see that we can't just transform what remains away is that those are the types of terms we get for a scalar field in a curved spacetime with a source term $$D$$.

We want our Lagrangian to be Poincaré-invariant. Just considering translation invariance, we have

$$\begin{eqnarray} x^\mu &\to& x^\mu + a^\mu \end{eqnarray}$$

Our field transforms as

$$\begin{eqnarray} \phi(x) \to \phi(x) + a^\mu \partial_\mu \phi(x) \end{eqnarray}$$

And the same goes for our various factors. The total variation of the linear part is

$$$$a^\mu (\partial_\mu D(x)) \phi + D(x) a^\mu (\partial_\mu \phi) + a^\mu (\partial_\mu D(x)) a^\nu (\partial_\nu\phi) = \partial_\mu (a^\mu D (x)\phi) + a^\mu (\partial_\mu D(x)) a^\nu (\partial_\nu\phi)$$$$

We want it to be equal to a total divergence. As this needs to be true for any function $$\phi$$, this implies that $$D$$ is a constant. Similarly, Poincaré invariance will constrain $$M = \eta^{\mu\nu}$$, and $$C$$ and $$B$$ are constant as well.

The term in $$B$$ goes away by

$$\begin{eqnarray} \int b^\mu \phi \partial_\mu \phi &=& [\frac{\phi^2}{2}]_\infty = 0 \end{eqnarray}$$

We just need to get rid of the term linear in $$\phi$$. For the same reason as $$F$$, this term can be removed by an appropriate divergence.

It all has to be with how we construct our theories, they have to be physically sound. One of the most important facts is that a theory has to be lorentz invariant. With a scalar theory the only terms with which you can build up a lagrangian are $$\partial_\mu \phi\qquad\phi$$ and higher order product of $$\phi$$. Now, since we want lorentz invariance we cannot put only the term$$\partial_\mu\phi$$ alone, since it's clearly not Lorentz invariant, so we put $$\partial_\mu\phi\partial^\mu\phi$$. $$\phi$$ alone is lorentz invariant since it's a scalar. You of course cannot put terms like $$c(x)$$ since they are not Lorentz invariant. You could say then that we could put a termlike $$c(\phi^2)$$ but that term then will just be absorbed in the part of the lagrangian that depends on $$\phi^2$$. Or you could say that we could use a term like $$c(x^\mu x_\mu)$$ but that it's just a number, again, it does not depend on anything.

The reason behind why you cannot put a linear term, nor a cubic term for that matter, is that a linear term makes the hamiltonian not lower bounded. If the hamiltonian is not lower bounded you have negative energies, which you don't want from a physical theory.

In reality there's even the concept of renormalizability of a theory. In fact the most general scalar lagrangian contains even a $$\phi^4$$ term but not terms of higher order. Terms with order bigger than $$4$$ make the theory not renormalizable.

By using some given prescriptions, we can construct any lagrangian just by using the basic building blocks that a specific theory gives us. The prescriptions are dictated through physical soundness like Lorentz invariance, rinormalizability and lower boundedness of the hamiltonian.