# [Physics] Units of the Scalar field theory Lagrangian density

dimensional analysisfield-theorylagrangian-formalism

The Lagrangian (with units $$J$$) connects to the Lagrangian density (with units $$J/m^3$$) as:

$$L=\iiint_V \mathcal{L}d^3x$$

Let $$\mathcal{L}$$ be the classical Lagrangian density of the scalar free theory (see https://en.wikipedia.org/wiki/Scalar_field_theory):

$$\mathcal{L}=\frac{1}{2} \eta^{\mu\nu} \partial_\mu \phi \partial_\mu \phi – \frac{1}{2}m^2 \phi^2$$

What are the units for each term?

Let me have a go:

I am assuming that $$\phi$$ has no units.

$$\mathcal{L}=\underbrace{\frac{1}{2} \eta^{\mu\nu} \partial_\mu \phi \partial_\mu \phi}_\text{no units} – \underbrace{\frac{1}{2}m^2 \phi^2}_{kg^2}$$

Clearly, a non-sensical equation.

In field theory, it is assumed that action is dimensionless and so is the speed of light ($$\hbar=c=1$$). So, we can write everything in terms of the mass unit. From the speed of light being unitless, we get, $$[L]= [T]$$ and from action being unitless we get, $$[M]=[L]^{-1}$$. In these units, the Lagrangian density has the dimension, $$[\mathcal{L}] = [M]^4$$. From the mass term in the Lagrangian density, we can find the units of the field $$\phi$$, $$[\phi] = [M].$$ To check whether the first term is consistent we should have the dimension of the first term as $$[M]^4$$. $$[\partial_\mu \phi] = \frac{[\phi]}{[x]} = \frac{[M]}{[L]} = [M]^2.$$ Remember that the partial derivative has $$\partial x^\mu$$ in the denominator which has the dimension of length. So, finally the dimension of the first term is $$[(\partial_\mu\phi)^2] = [M]^4$$, which is the correct dimension as $$[\mathcal{L}] =[M]^4$$.
So, there are two misconceptions in your assumptions. $$\phi$$ is not unitless and the form of Lagrangian density is not written in SI units, it is written in what is called natural units with $$\hbar = c = 1$$.