Legendre transform for non-local Lagrangians, or Hamiltonian of non-local Lagrangian and their properties

Question

This is sort of a multi-part question, mostly dealing with how to treat non-local Hamiltonians and how the corresponding properties of Hamiltonians work in a non-local framework. I proposed an example to make explanation easier, but generalized answers are helpful too.

To start:

How does one apply the Legendre transform to a non-local Lagrangian, or get the Hamiltonian of such a Lagrangian?

For example, given the functional: $$ F=\frac{1}{2}\int^t_0 \left(\dot{x}(\tau)\dot{x}(t-\tau)-x(\tau)x(t-\tau)\right)\,\text{d}\tau $$ Which, when varied with respect to the state, is stationary for the system: $$ \ddot{x}(\tau)-x(\tau)=0 $$ Taking the derivative with respect to $\dot{x}$ of $F$ yields: $$ \frac{\partial F}{\partial \dot{x}}=\frac{1}{2}\int^t_0 \dot{x}(\tau)\,\text{d}\tau+\frac{1}{2}\int^t_0 \dot{x}(t-\tau)\,\text{d}\tau=\int^t_0 \dot{x}(\tau)\,\text{d}\tau $$ Would it then be true that?: $$ \frac{\partial F}{\partial \dot{x}}=\int^t_0 p(\tau)\,\text{d}\tau $$ So we can go on to say: $$ \int^t_0 \mathcal{H}\,\text{d}\tau=\int^t_0 p(\tau)p(t-\tau)\,\text{d}\tau-\left.F\right|_{\dot{x}=p} $$ Which gives: $$ \int^t_0 \mathcal{H}\,\text{d}\tau=\frac{1}{2}\int^t_0 p(\tau)p(t-\tau)+x(\tau)x(t-\tau)\,\text{d}\tau $$ So can we say?: $$ \mathcal{H}=\frac{1}{2}\left(p(\tau)p(t-\tau)+x(\tau)x(t-\tau)\right) $$ Additionally: $$ \frac{\text{d}\mathcal{H}}{\text{d}\tau}=\frac{1}{2}\left(\dot{p}(\tau)p(t-\tau)-p(\tau)\dot{p}(t-\tau)+\dot{x}(\tau)x(t-\tau)-\dot{x}(t-\tau)x(\tau)\right) $$ Would it then be correct to substitute?: $$ \dot{p}(\tau)=x(\tau) $$ $$ \dot{p}(t-\tau)=x(t-\tau) $$ $$ \dot{x}(\tau)=p(\tau) $$ $$ \dot{x}(t-\tau)=p(t-\tau) $$ Which gives: $$ \frac{\text{d}\mathcal{H}}{\text{d}\tau}=\frac{1}{2}\left(u(\tau)p(t-\tau)-p(\tau)u(t-\tau)+p(\tau)x(t-\tau)-p(t-\tau)x(\tau)\right) $$ Integrating both sides (and making some integration substitutions) then yields: $$ \int^t_0 \frac{\text{d}\mathcal{H}}{\text{d}\tau} = 0 $$ Would this imply that $\mathcal{H}$ is constant?

Lastly, to get back the equations of motion from $\mathcal{H}$, we have: From: $$ \frac{\partial\mathcal{H}}{\partial p}=\dot{x} $$ and $$ \frac{\partial\mathcal{H}}{\partial x}=-\dot{p} $$ We have $$ \frac{\partial}{\partial p}\int^t_0 \mathcal{H}\,\text{d}\tau = \int^t_0 \dot{x}(\tau)\,\text{d}\tau $$ Or $$ \int^t_0 \dot{x}(\tau)\,\text{d}\tau=\int^t_0 p(\tau)\,\text{d}\tau $$ and $$ \frac{\partial}{\partial x}\int^t_0 \mathcal{H}\,\text{d}\tau = \int^t_0 \dot{p}(\tau)\,\text{d}\tau $$ Or $$ \int^t_0 \dot{p}(\tau)\,\text{d}\tau=-\int^t_0 x(\tau)\,\text{d}\tau $$ This does not match the initial equations of motion, is that due to the non-local nature of the formulation? Would we then say that?:

$$ \frac{\partial\mathcal{H}}{\partial p}=\dot{x} $$ and $$ \frac{\partial\mathcal{H}}{\partial x}=\dot{p} $$

Answer 1

Let here consider point mechanics (as opposed to field theory) for simplicity, i.e. the generalization to field theory is left as an exercise.

I) Bad news. If the Lagrangian action functional $S[q]$ is non-local, the usual definition of Lagrangian momenta as a partial derivative $$p_i~:=~\frac{\partial L}{\partial v^i} \tag{1}$$ does not apply.

II) Good news. If the (possibly non-local) Lagrangian action functional is of the form $$\left. S[q,v]\right|_{v=\dot{q}}, \tag{2}$$ then we can in principle perform a Legendre transformation on the action $S[q,v]$. The Legendre transformation will depend on how we split the action $S[q,v]$ into two independent$^1$ sets of variables $q^i$ and $v^j$. Nevertheless assume that this splitting $S[q,v]$ has been done. Next define the momentum as a functional derivative

$$p_i~:=~\frac{\delta S[q,v]}{\delta v^i} .\tag{3}$$

Now one may in principle perform the Legendre transformation very similar to the standard method outlined in e.g. this Phys.SE post, modulo the usual caveats about a possible singular Legendre transformation. Define the Hamiltonian functional$^2$ as

$$\mathbb{H}[q,p]~:=~\sup_{v}\left( \int_{t_i}^{t_f}\! dt~ p(t) v(t) - S[q,v] \right). \tag{4}$$

The corresponding Hamilton's eqs. become

$$\dot{q}^i~\approx~ \frac{\delta \mathbb{H}}{\delta p_i},\qquad -\dot{p}_i~\approx~ \frac{\delta \mathbb{H}}{\delta q^i}. \tag{5}$$

[Here the $\approx$ symbol means equality modulo eqs. of motion.] We can define a Poisson bracket

$$\{F,G\}_{PB}~:=~\int_{t_i}^{t_f}\! dt~ \frac{\delta F}{\delta q^i(t)} \frac{\delta G}{\delta p_i(t)} -(F\leftrightarrow G). \tag{6}$$

Then Hamilton's eqs. (5) can be written as

$$\dot{z}^I(t)~\approx~ \{z^I(t), \mathbb{H}\}_{PB},\tag{7}$$

where $\{z^I\}=\{q^i;p_j\}$ is a collective notation for the phase space variables.

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$^1$ Note in particular that an action $S[q,v]$ with two independent sets of variables $q^i$ and $v^j$ flies in the face of the usual paradigm that the velocity profile is completely determined by the position profile in the action, cf. e.g. this Phys.SE post.

$^2$ For local theories, the Hamiltonian functional $$\mathbb{H}[q,p]~=~\int_{t_i}^{t_f}\! dt~ H(q(t),p(t),t)\tag{8}$$ is just the integral of the Hamiltonian.