This is a second part of my previous question. I'm trying to figure it out by myself how to deduce Hamilton's equations in classical field theory as it is usually obtained in physics books.

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}

Classical Field Theory

I know that the more general setting of classical field theory (as well as classical mechanics) is defined in terms of manifolds and tangent bundles, but for our purposes I will work in the usual spacetime $\mathbb{R}^{4}$. In the present context, a field is a function $\phi: \mathbb{R}^{4} \to \mathbb{R}^{n}$. A point ${\bf{x}} = (x_{1},x_{2},x_{3},x_{4}) \in \mathbb{R}^{4}$ is represented by space coordinates $x_{1},x_{2},x_{3}$ and a time coordinate $x_{4} = t$. It is useful to write the equations that follow in Einstein's notation: $x_{\mu}$ denotes any coordinate of ${\bf{x}}$ and $\partial_{\mu}$ denotes the partial derivative with respect to the $\mu$-th entry, $\mu \in \{1,2,3,4\}$.

The action $S = S(\phi)$ is defined by: \begin{eqnarray} S({\bf{\phi}}) := \int L(t, \phi(t), \partial_{\mu}\phi)dt \tag{2}\label{2} \end{eqnarray} where the Lagrangian is given by an integral: \begin{eqnarray} L(t,\phi(t), \partial_{\mu}\phi) := \int \mathscr{L}({\bf{x}}, \phi, \partial_{\mu}\phi)d{\bf{x}} \tag{3}\label{3} \end{eqnarray} with $\mathscr{L}$ being the Lagrangian density.

Question: Is this a closed form for the Hamiltonian $H$ of such a system, given that $H$ is the Legendre transform of $L$? If there is, how to obtain it?

Let us assume that $L$ and $H$ do not depend explicitly on $t$. As far as I understand, $H$ is the Legendre transform of $L$ with respect to the variable $\dot{\phi} = \partial_{4}\phi$, so we have to perform a change $\dot{{\bf{\phi}}} \leftrightarrow {\bf{p}}$. Thus, it is natural to define $H(\phi, \frac{\partial \phi}{\partial {\bf{x}}}, {\bf{p}}) = \max_{\dot{\phi}}(\langle {\bf{p}}, \dot{\phi}\rangle - L(\phi, \frac{\partial \phi}{\partial {\bf{x}}}, \dot{\phi}))$. Using (\ref{1}), we get:

\begin{eqnarray} {\bf{p}} = \frac{\partial L}{\partial {\bf{x}}} = \frac{\partial}{\partial {\bf{x}}}\int \mathscr{L}({\bf{x}}, \phi, \partial_{\mu}\phi)d{\bf{x}} \tag{4}\label{4} \end{eqnarray}

Here is where things become unclear. Physicists usually define: \begin{eqnarray} \pi({\bf{x}}) := \frac{\partial \mathscr{L}({\bf{x}})}{\partial {\bf{x}}} \tag{5}\label{5} \end{eqnarray} so that: \begin{eqnarray} {\bf{p}} = \int \pi({\bf{x}}) d{\bf{x}} \tag{6}\label{6} \end{eqnarray} But the Hamiltonian should become:

\begin{eqnarray} H = \int \pi({\bf{x}})\dot{\phi}({\bf{x}})d{\bf{x}} - L \tag{7}\label{7} \end{eqnarray}

But this is really strange since the $\dot{\phi}$ seems to be being integrated together with $\pi$, and this is not the case if you put (\ref{6}) into $\langle {\bf{p}}, \dot{\phi}\rangle - L(\phi, \frac{\partial \phi}{\partial {\bf{x}}}, \dot{\phi})$.

So, what am I doing wrong?

  • $\begingroup$ Just a general comment, applicable both to this and your previous question. In the study of classical fields, the situation where the numerical value of the action functional $S(\phi)$ and whether it is a true minimum (or even a critical point) at some specific field configuration $\phi$ with respect to a highly technical notion of derivative actually occur very rarely. From personal experience, demands for rigor may really be a manifestation an underlying quest for conceptual clarity, which is inevitably somewhat subjective. ... $\endgroup$ – Igor Khavkine Feb 2 at 22:03
  • $\begingroup$ What can help with conceptual understanding is working under sufficiently liberal hypotheses so that all formal manipulations that you encounter "just work". A continuous variable should be interpreted as an index? Consider an analogous case where it is an index (e.g., spatial discretization). An integral might not converge? Suppose that all fields have compact support. Is it OK to differentiate $n$ times? Suppose everything is smooth and differentiate as much as you like. These are just some examples. $\endgroup$ – Igor Khavkine Feb 2 at 22:08
  • $\begingroup$ @IgorKhavkine thanks! I've had my share of 'hard times' trying to understand some concepts of quantum field theory. In QFT, sometimes is really difficult (if not impossible to these days) to give precise meaning of some objetcs. I was hoping this would be different in the present case, where although a field theory, the 'classical' aspect of it should turn things easier. Actually, I'm convinced after my conversation with Pedro that this is indeed the case. But my previous understanding of the subject, as it is exposed in my post, was not satisfatory. $\endgroup$ – MathMath Feb 2 at 23:00
  • $\begingroup$ My point is just: I agree that being liberal helps a lot understanding the formal manipulations, but it seems that the concepts discussed here are not formal at the end of the day. $\endgroup$ – MathMath Feb 2 at 23:02
  • $\begingroup$ This observation is true for many aspects of classical and quantum fields, more than first meets the eye. However, digging up the "rigor" may be a non-trivial task. To justify the cost, unless you just make it a goal in itself, next time you need a rigorous/highly precise definition or derivation, try to find an important application/situation where it really matters. Exercise: what would it be for your previous question? $\endgroup$ – Igor Khavkine Feb 3 at 10:45

There is a fundamental misunderstanding in your translation of Hamilton's formalism to classical field theory, which pertains to the proper identification of dynamical variables.

In classical mechanics, the position variables are dynamical variables, whereas time is the external parameter in terms of which we register the dynamics, i.e. how dynamical variables (position and velocity / momenta, depending on whether you employ the Lagrangian / Hamiltonian formalism) evolve. In classical field theory, position becomes an external parameter in the same footing as time and the dynamical variables are the field values $\phi(t,\mathbf{x})$. In other words, classical field theory is about the space-time evolution of the field values. That is why in classical field theory the Lagrangian involves field derivatives in time and space. Edit: actually, as discussed in the comments below and the ensuing chat discussion, the appearance of spatial derivatives in the Lagrangian is enforced in the case of relativistic fields by the presence of time derivatives and finite speed of propagation for the Euler-Lagrange equation.

Another consequence of this is that the configuration space of classical field theory becomes infinite-dimensional - after all, we are following the time evolution of the field values $\phi(\cdot,\mathbf{x})$ at each point in space, thus comprising a continuum of independent dynamical variables, unlike the $n$ (say) Cartesian position variables in classical mechanics - that is, the "coordinate index" $\mathbf{x}$ no longer takes discrete values $1,\ldots,n$ but the whole of $\mathbb{R}^n$. As such, the field momentum $\pi(t,\mathbf{x})$ at each position $\mathbf{x}$ and time $t$ should be defined (and actually is, in physics textbooks) as $$\pi(t,\mathbf{x})=\frac{\partial\mathscr{L}(t,\mathbf{x},\phi(t,\mathbf{x}),\partial_\mu\phi(t,\mathbf{x}))}{\partial(\partial_t\phi(t,\mathbf{x}))}\ .$$ In other words, we also have a continuum of independent field momentum variables. This is why the time derivative $\dot{\phi}(t,\mathbf{x})=\partial_t\phi(t,\mathbf{x})$ of $\phi$ appears paired with $\pi(t,\mathbf{x})$ inside a spatial integral in the definition of the Hamiltonian $H$ - this is just the appropriate choice of (infinite-dimensional) scalar product.

  • $\begingroup$ Pedro, thanks for the nice answer. This is something I always get confused about, and I think I'm getting to it. Let me see if I understood it correctly: a field is a (infinite) vector indexed by both ${\bf{x}}$ and $t$. Thus, the space of fields should be the space $\mathcal{F}$ of all functions $\phi: \mathbb{R}^{4} \to \mathbb{R}^{n}$. Then, $L$ and $H$ become functions of $t$, $\phi$ and $\partial_{\mu}\phi$ on that space, i.e. $L: \mathbb{R}\times \mathcal{F}\times \mathcal{F}$ and $L=L(t,\phi,\partial_{\mu}\phi)$. Is this correct? $\endgroup$ – MathMath Feb 1 at 22:12
  • $\begingroup$ Or is it just ${\bf{x}\}$ which index $\phi$ and time is a fixed variable? I think I'm confused... $\endgroup$ – MathMath Feb 1 at 22:17
  • $\begingroup$ Strictly speaking, the space of all functions with a certain regularity (since you want to compute their derivatives) and with sufficient decay towards spatial infinity (since otherwise the spatial integrals may not be defined), but yes. Conversely, classical mechanics can be treated as a field theory living in 0-dimensional space (hence, the dynamical variables depend only on time in this case). $\endgroup$ – Pedro Lauridsen Ribeiro Feb 1 at 22:19
  • $\begingroup$ Right! And time goes as an index as well? Or just ${\bf{x}}$? $\endgroup$ – MathMath Feb 1 at 22:21
  • $\begingroup$ For a fixed time $t$ (say, $t=0$), you may consider given values $\phi(t,\mathbf{x})=\phi_0(\mathbf{x})$ at each $\mathbf{x}$ as "initial values" of these dynamical variables at the instant $t(=0)$, just like the position variables $x_1(0),\ldots,x_n(0)$ at time $t=0$ in classical mechanics. $\endgroup$ – Pedro Lauridsen Ribeiro Feb 1 at 22:22

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