[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.

Best Answer

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 $\hbar = c = 1$.

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