Consider Lagrangian Mechanics. In the same way that we have Lagrangians and Euler-Lagrange equations for particles or collections of particles, we have Lagrangians for these infinite sets (continuum) of particles: the fields.
According to Wikipedia:
Lagrangian field theory is a formalism in Classical Field Theory. It is the field-theoretic analogue of Lagrangian Mechanics. Lagrangian mechanics is used to analyze the motion of a system of discrete particles each with a finite number of degrees of freedom. Lagrangian field theory applies to continua and fields, which have an infinite number of degrees of freedom.
One motivation for the development of the Lagrangian formalism on fields, and more generally, for Classical Field Theory, is to provide a clean mathematical foundation for Quantum Field Theory, which is infamously beset by formal difficulties that make it unacceptable as a mathematical theory.
A kind of dual approach is that of Hamiltonian field theory.
From the mathematical point of view, classical fields may be described by sections $\phi$ of a bundle $E\to M$ (where $M$ represents spacetime). Physics is incorporated by introducing an horizontal $n$-form (where $n=dim(M)$) on the first order jet bundle $J^1(E)$, the Lagrangian density
$$ \lambda=L(x_i,u^{\alpha},u^{\alpha}_i)dx_1\wedge \cdots \wedge dx_n. $$See the annotation variational bicomplex.
The important sections $\phi$ are those which extremize the functional
$$ S[\phi]=\int_R (j^1\phi)^*(\lambda) $$being $R$ a compact region of $M$, where $j^1\phi$ is the prolongation to the first jet bundle.
The condition of $\phi$ extremizing this functional is equivalent to satisfying Euler-Lagrange equations, but in this case is a system of PDEs, not ODEs.
But I think there is a more geometrical object called Poincare-Cartan form.
See Noether's theorem#In Lagrangian field theory to understand what it is said in No-nonsense qft, pag155. Apply it in two cases:
... under construction...
We are dealing with a spacetime symmetry if the Lagrangian density, denoted as $L(\phi(x_{\mu}), \partial_\mu \phi(x_{\mu}), x_{\mu})$, remains unchanged under a transformation from $x_{\mu}$ to $x'_{\mu}$. Mathematically, this implies:
$$ \delta L = L(\phi(x_{\mu}), \partial_\mu \phi(x_{\mu}), x_{\mu}) - L(\phi'(x'_{\mu}), \partial_\mu \phi'(x'_{\mu}), x'_{\mu}) = 0. $$We have that:
$$ \begin{aligned} x_\mu & \rightarrow x_\mu^{\prime}=x_\mu+\delta x_\mu \\ \phi & \rightarrow \phi^{\prime}=\phi+\delta \phi \\ \left(\partial_\mu \phi\right) & \rightarrow\left(\partial_\mu \phi\right)^{\prime}=\left(\partial_\mu \phi\right)+\delta\left(\partial_\mu \phi\right) \end{aligned} $$and then, the transformed Lagrangian becomes, up to first order:
$$ \begin{aligned} L\left(\left(\phi^{\prime}\left(x_\mu^{\prime}\right), \partial_\mu \phi^{\prime}\left(x_\mu^{\prime}\right), x_\mu^{\prime}\right)=\right. & L\left(\left(\phi\left(x_\mu\right), \partial_\mu \phi\left(x_\mu\right), x_\mu\right)\right. \\ & +\frac{\partial L}{\partial \phi} \delta \phi+\frac{\partial L}{\partial\left(\partial_\mu \phi\right)} \delta\left(\partial_\mu \phi\right)+\frac{\partial L}{\partial x_\mu} \delta x_\mu . \end{aligned} $$and then
$$ \delta L=\frac{\partial L}{\partial \phi} \delta \phi+\frac{\partial L}{\partial\left(\partial_\mu \phi\right)} \delta\left(\partial_\mu \phi\right)+\frac{\partial L}{\partial x_\mu} \delta x_\mu . $$Suppose we have, for example, translational symmetry $x_{\mu} \rightarrow x_{\mu}^{\prime}=x_{\mu}+a_{\mu}$. Then
$$ 0=\frac{\partial L}{\partial \phi} \delta \phi+\frac{\partial L}{\partial\left(\partial_\mu \phi\right)} \delta\left(\partial_\mu \phi\right)+\frac{\partial L}{\partial x_\mu} \delta x_\mu . $$...to be continued...
...to be continued...
________________________________________
________________________________________
________________________________________
Author of the notes: Antonio J. Pan-Collantes
INDEX: