Lagrangian density

Remarks

The action is an integral functional with the integrand $\mathcal{L}$. Its smooth critical points under variations with compact support in $U$ are the solutions of the Euler–Lagrange equations $$\frac{\partial \mathcal{L}}{\partial y^a} - \partial_\mu \left( \frac{\partial \mathcal{L}}{\partial z^a_\mu} \right) = 0, \qquad a = 1, \dots, k,$$ with summation over $\mu$ and with all terms evaluated along $\varphi$.

Example: let $n = 4$, $k = 1$, $\eta = \mathrm{diag}(1, -1, -1, -1)$ and $m \ge 0$. The density $\mathcal{L} = \frac{1}{2} \eta^{\mu\nu} \partial_\mu \varphi \, \partial_\nu \varphi - \frac{1}{2} m^2 \varphi^2$ has $\partial \mathcal{L} / \partial y = -m^2 \varphi$ and $\partial \mathcal{L} / \partial z_\mu = \eta^{\mu\nu} \partial_\nu \varphi$. So its Euler–Lagrange equation is $\eta^{\mu\nu} \partial_\mu \partial_\nu \varphi + m^2 \varphi = 0$, the Klein–Gordon equation.

Non-example: the expression $\int_U \int_U K(x, x') \varphi(x) \varphi(x') \, d^n x \, d^n x'$ with a kernel $K$ is not the integral of a Lagrangian density, because the integrand at $x$ depends on the values of $\varphi$ at other points.

Densities that also depend on higher derivatives of $\varphi$ are used too; then the Euler–Lagrange equations contain more terms. On a manifold, a Lagrangian density is a horizontal $n$-form on the first jet bundle, written $\mathcal{L} \, d^n x$ in coordinates.