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Consider the following classical Lagrangian with an interaction between velocities:

$$\mathcal{L} = \sum_{i} \frac{1}{2}m \mathbf{v}_{i}^{2} + \sum_{i < j} J(r_{ij}) \hat{\mathbf{v}}_{i} \cdot \hat{\mathbf{v}}_{j},$$

where $r_{ij} = |\mathbf{r}_{i} - \mathbf{r}_{j}|$, $\mathbf{v} =\dot{\mathbf{r}}$, and $\hat{\mathbf{v}} = \mathbf{v} / |\mathbf{v}|$. Models like this come up in studies of collective behavior, like bird flocking.

The canonical momentum is

$$\mathbf{p}_{i} = m \mathbf{v}_{i} + \dfrac{1}{v_{i}} \sum_{j \neq i} J(r_{ij}) \left( \hat{\mathbf{v}}_{j} - (\hat{\mathbf{v}}_{j} \cdot \hat{\mathbf{v}}_{i}) \hat{\mathbf{v}}_{i} \right).$$

It may be helpful to define the "averaged velocity" $\mathbf{d}_{i} \equiv \sum_{j \neq i} J(r_{ij}) \hat{\mathbf{v}}_{j}$, in which case we can rewrite the canonical momentum as

$$\mathbf{p}_{i} = m \mathbf{v}_{i} + \dfrac{1}{v_{i}} \left( \mathbf{d}_{i} - (\mathbf{d}_{i} \cdot \hat{\mathbf{v}}_{i}) \hat{\mathbf{v}}_{i} \right).$$

Noticing that the non-standard term in the momentum is orthogonal to the velocity $\mathbf{v}_{i}$, we can easily write the corresponding Hamiltonian in terms of the velocities as

$$\mathcal{H} = \sum_{i} \frac{1}{2}m \mathbf{v}_{i}^{2} - \sum_{i < j} J(r_{ij}) \hat{\mathbf{v}}_{i} \cdot \hat{\mathbf{v}}_{j}.$$

Question

Is it possible to explicitly write the Hamiltonian $\mathcal{H}$ solely in terms of the canonical positions $\mathbf{r}_{i}$ and momenta $\mathbf{p}_{i}$?

Through calculating dot products of $\mathbf{p}_{i}$ with $\mathbf{v}_{i}$, $\mathbf{d}_{i}$ and $\mathbf{p}_{i}$, I have managed to get to the expression

$$\mathcal{H} = \sum_{i} \dfrac{\mathbf{p}_{i}^{2}}{2m} - \sum_{i} \dfrac{1}{2m v_{i}} \mathbf{p}_{i} \cdot \mathbf{d}_{i},$$

but so far I haven't managed to completely eliminate the velocity dependence.

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  • $\begingroup$ I would be concerned that this Lagrangian isn't convex in the velocities; it's pretty obviously non-convex for the simplest case of two particles in 1-D ($\mathcal{L} = \frac12(v_1^2 + v_2^2 + J \sigma(v_1 v_2)$, where $J$ is a constant and $\sigma$ is the sign function.) I have a dim recollection that convexity is required to make the Legendre transform work properly. Some discussion of this issue can be found here. $\endgroup$
    – Michael Seifert
    Commented Feb 1, 2022 at 21:05
  • $\begingroup$ Hi Michael, thanks for your comment. Indeed convexity can be an issue. For 2 particles, you can fix it by taking the mass $m$ to be sufficiently large. Unfortunately, for $N \gg 1$ particles, we need $m$ to be $O(N)$ for this to work for the model in the question. One way out while keeping $m$ to be $O(1)$ is to include a hardcore potential, so each particle has only $O(1)$ neighbors for which $J(r)$ is appreciable (assuming fast decay with $r$). In that case the Lagrangian can be convex for $m$ sufficiently large but still $O(1)$. $\endgroup$
    – anon1802
    Commented Feb 1, 2022 at 22:18

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