Newtonian Dynamics from the Principle of Maximum Caliber

Abstract

The foundations of Statistical Mechanics can be recovered almost in their entirety from the principle of maximum entropy. In this work we show that its non-equilibrium generalization, the principle of maximum caliber (Jaynes, Phys Rev 106:620–630, 1957), when applied to the unknown trajectory followed by a particle, leads to Newton’s second law under two quite intuitive assumptions (both the expected square displacement in one step and the spatial probability distribution of the particle are known at all times). Our derivation explicitly highlights the role of mass as an emergent measure of the fluctuations in velocity (inertia) and the origin of potential energy as a manifestation of spatial correlations. According to our findings, the application of Newton’s equations is not limited to mechanical systems, and therefore could be used in modelling ecological, financial and biological systems, among others.

Appendix 1: Canonical Coordinates and Poisson Brackets

An interesting question is how much of the formalism of classical mechanics we can recover from Eq. 22. The fact that most of the structure of classical mechanics is contained in the definition and properties of the Poisson bracket, motivates us to search for an operation analogous to this bracket under the Maximum Caliber formalism.

For arbitrary differentiable functions $f(x, p)$ and $g(x, p)$ the Poisson bracket is defined as

$$\begin{aligned} \{f, g\} = \frac{\partial f}{\partial x}\frac{\partial g}{\partial p} - \frac{\partial f}{\partial p}\frac{\partial g}{\partial x}, \end{aligned}$$

(24)

and it is such that

$$\begin{aligned} \frac{df}{dt}-\frac{\partial f}{\partial t} = \{f, \mathcal {H}\} \end{aligned}$$

(25)

holds. Let us compute the expectation of the left hand side,

$$\begin{aligned} \Big <\frac{df}{dt}\Big > - \Big <\frac{\partial f}{\partial t}\Big > = \Big <\frac{\partial f}{\partial x_k}\dot{x}_k+\frac{\partial f}{\partial p_k}\dot{p}_k\Big >, \end{aligned}$$

(26)

which using Eq. 21 with $\omega =\partial f/\partial p_k$ can be written as

$$\begin{aligned} \Big <\frac{df}{dt}\Big > - \Big <\frac{\partial f}{\partial t}\Big > = \Big <\frac{\partial f}{\partial x_k}\dot{x}_k - \frac{\partial }{\partial x_k}\Big (\frac{\partial f}{\partial p_k}\Big ) - \frac{\partial f}{\partial p_k}\Phi '(x_k)\Big > \end{aligned}$$

(27)

Now using our classical Hamiltonian (Eq. 17) we recognize its derivatives

$$\begin{aligned} \dot{x}_k = \frac{\partial \mathcal {H}}{\partial p_k}\end{aligned}$$

(28)

$$\begin{aligned} \Phi '(x_k) = \frac{\partial \mathcal {H}}{\partial x_k} \end{aligned}$$

(29)

and, upon replacing, we have

$$\begin{aligned} \Big <\frac{df}{dt}\Big > - \Big <\frac{\partial f}{\partial t}\Big > = \Big <\frac{\partial f}{\partial x_k}\frac{\partial \mathcal {H}}{\partial p_k} - \frac{\partial }{\partial x_k}\Big (\frac{\partial f}{\partial p_k}\Big ) - \frac{\partial f}{\partial p_k}\frac{\partial \mathcal {H}}{\partial x_k}\Big > \end{aligned}$$

(30)

leading finally to

$$\begin{aligned} \Big <\frac{df}{dt}\Big > - \Big <\frac{\partial f}{\partial t}\Big > = \Big <\{f, \mathcal {H}\}\Big > - \Big <\frac{\partial }{\partial x_k}\Big (\frac{\partial f}{\partial p_k}\Big )\Big >. \end{aligned}$$

(31)

So, in expectation we find a Poisson bracket analog with an additional term. For the particular case $f=\mathcal {H}$, we obtain

$$\begin{aligned} \Big <\frac{d\mathcal {H}}{dt}\Big > = -\Big <\frac{\partial }{\partial x_k}\Big (\frac{\partial \mathcal {H}}{\partial p_k}\Big )\Big >, \end{aligned}$$

(32)

which reduces to

$$\begin{aligned} s \Big <\frac{d\mathcal {H}}{dt}\Big > = -\Big <\frac{\partial \dot{x}_k}{\partial x_k}\Big > = 0, \end{aligned}$$

(33)

using the centered difference,

$$\begin{aligned} \dot{a}_i \approx \frac{a_{i+1}-a_{i-1}}{2\Delta t}. \end{aligned}$$

(34)

Therefore we have shown that, for a Hamiltonian with the form given in Eq. 17, the energy is conserved in expectation. Note that using one-sided differences is not appropriate here, as one obtains $\pm 1/\Delta t$ depending on forward or backward.