# Understanding Hamilton's equations in classical field theory in a rigorous way

So, I'm in a quest of understanding classical field theory on my own, and I'm interested in its rigorous construction. Here's the link for a previous post of mine on mathoverflow. The interesting discussions there led me to this new post here, which I will use some of the notations I used there.

Notation: If $${\bf{x}} = (x_{1},...,x_{n}) \in \mathbb{R}^{n}$$ and $$f=f({\bf{x}})$$ is real-valued and differentiable, I'll denote: $$\frac{\partial f}{\partial \bf{x}} := \bigg{(}\frac{\partial f}{\partial x_{1}},...,\frac{\partial f}{\partial x_{n}}\bigg{)} \equiv \nabla f.$$

This notation is useful since, if $$f$$ is a function of more than one variable, e.g. $$f=f(\bf{x},\bf{y},\bf{z})$$, then $$\partial f/{\partial \bf{x}}$$ means the gradient with respect to the $$\bf{x}$$ variable.

## Legendre Transforms for many variable functions

Here, I'm following Arnold. Let $$f: \mathbb{R}^{n}\to \mathbb{R}$$ be a twice-differentiable function such that its Hessian $$\nabla^{2}f$$ is positive-definite (so $$f$$ is strictly convex). Let $$G=G({\bf{p}},{\bf{x}}) := \langle {\bf{p}},{\bf{x}}\rangle - f({\bf{x}})$$, where $$\langle \cdot, \cdot \rangle$$ is the usual inner product on $$\mathbb{R}^{n}$$. Then, the Legendre transform of $$f$$ is defined to be the function $$g=g({\bf{p}}) := \max_{{\bf{x}}}G({\bf{p}},{\bf{x}})$$. Notice that $$G$$ attains its maximum iff $$\frac{\partial G}{\partial \bf{x}} = 0$$, so that the vector $$\bf{x}$$ which maximizes $$G$$ for a fixed $$\bf{p}$$ is the solution of: $$\begin{eqnarray} \frac{\partial f}{\partial \bf{x}} = \bf{p} \tag{1}\label{1} \end{eqnarray}$$

Let $$L=L(t,{\bf{x}},\dot{{\bf{x}}})$$ be a Lagragian on the phase space as studied in classical mechanics. Because the Hamiltonian $$H=H(t,{\bf{x}},{\bf{p}})$$ is the Legendre transform of $$L$$, equation (\ref{1}) becomes: $$\begin{eqnarray} \frac{\partial L}{\partial \dot{{\bf{x}}}} = {\bf{p}} \tag{2}\label{2} \end{eqnarray}$$ which is one of the Hamilton's equations usually found in textbooks.

## Classical Field Theory

As discussed in my previous question linked above, the Lagrangian and the Hamiltonian now become functions of fields, which are infinite-dimensional vectors indexed by space-time coordinates $$(t,{\bf{x}})\in \mathbb{R}^{4}$$. Let us denote $$\mathcal{F}$$ the space of fields, which we assume to be sufficiently smooth and regular at infinity so that the following integrals are always finite.

In textbooks, the Hamiltonian for a classical field theory is given by: $$\begin{eqnarray} H(t, \phi, \partial_{{\bf{x}}}\phi,\pi) := \int \pi(t,{\bf{x}})\dot{\phi}(t,{\bf{x}})d{\bf{x}} - L(t, \phi, \partial_{\mu}\phi) \tag{3}\label{3} \end{eqnarray}$$

Question 1: How does one define the Legendre transform in such infinite-dimensional space such that the Hamiltonian becomes (\ref{3})?

Question 2: Once question 1 is answered and the Hamiltonian is defined in this infinite-dimensional space, there should be an identity similar to (\ref{2}) so that the usual formula: $$\begin{eqnarray} \pi(t,{\bf{x}}) = \frac{\partial \mathscr{L}}{\partial \dot{\phi}(t,{\bf{x}})} \tag{4}\label{4} \end{eqnarray}$$ holds. What is the meaning of the derivative in the right hand side of (\ref{4})? I'm assuming the space of fields $$\mathcal{F}$$ is a Banach space (actually, probably an inner product space) so that the above derivative is Fréchet?

ADD: As I stressed before, in classical mechanics one can define the Hamiltonian as: $$\begin{eqnarray} H(t,{\bf{p}},{\bf{x}}) = \langle {\bf{p}}, \dot{{\bf{x}}}\rangle - L(t,{\bf{x}},\dot{{\bf{x}}}) \tag{5}\label{5} \end{eqnarray}$$ where, in (\ref{5}) it is understood that $$\dot{{\bf{x}}}$$ should be considered as a function of $${\bf{p}}$$ by means of the solution of (\ref{2}). Thus, in classical field theory, we can define the Hamiltonian following the same recipe, by setting: $$\begin{eqnarray} H(t,\phi, \partial_{{\bf{x}}}\phi, \pi) := \int \pi(t,{\bf{x}})\dot{\phi}(t,{\bf{x}})d{\bf{x}} - L(t, \phi, \partial_{\mu}\phi). \tag{6}\label{6} \end{eqnarray}$$

However, in classical mechanics, the Hamiltonian (\ref{5}) is the Legendre transform of $$L$$ and (\ref{2}) follows naturally. So, the objective of my question is to check wether the infinite-dimensional case can also be defined by means of an appropriate infinite-dimensional Legendre transform analogous to the finite-dimensional case, so that the conjugate variable $$\pi$$ as defined by (\ref{4}) is naturally enherited from the maximality of this Legendre transform as it is the case for the finite-dimensional case.

• Comment to the post (v2): Eq. (2) is not one of the Hamilton's equations per se. Feb 2 at 0:55
• Related question in point mechanics: physics.stackexchange.com/q/105912/2451 Feb 2 at 1:05
• Off the top of my head, I do not recall Arnold discussing classical field theory, but Abraham, Marsden, and Ratiu do. And Giaccheta and Sardanshvili wrote two monographs on mathematical field theory. One of these three references should a rigorous definition of a Legendre transformation of the "configurations space". Feb 2 at 1:14
• @DanielC yes, Arnold do not discuss it sadly. I followed Arnold only in the section about Legendre transforms for finitely many variables. Feb 2 at 1:16
• @Qmechanic thanks for the comments. About eq. (2), you mean that this is not one of Hamilton's equations because it is usually stated as the definition of ${\bf{p}}$, right? This is what I meant, but it sounds wrong in my post. You are right. About the post linked: this is exactly what I'm looking for but for a field theory instead of point mechanics! Feb 2 at 1:18