# Particle current density expression

I was reading the paper https://doi.org/10.1063/1.1747782 by Irving and kirkwood and came across an expression called particle current density and is given as follows $$j(r,r^{\prime};t)= \sum \sum_{k\neq i} \langle (\frac{p_{k}}{m_{k}} \bigoplus \frac{p_{i}}{m_{i}})\delta(R_{k}-r)\delta(R_{i}-r^{\prime});f \rangle$$. Where $f$ is the probability distribution , $p_{k}$ ,$p_{i}$ are the momentum of particle $k$ and $i$ respectively and $R_{k}$ ,$R_{i}$ are the position vectors of the particles $k$ and $i$ . Can someone please explain me this equation and how it came about especially about the part $\frac{p_{k}}{m_{k}} \bigoplus \frac{p_{i}}{m_{i}}$. It is not clear to me what this means. Why are we considering a direct sum? I am not very familiar with direct sums. Thanks for the answers.

In this formulation, each particle has an associated six-dimensional phase space. The system phase space is the direct sum of all the individual particles' subspace, and this is the space in which the system phase-space density $f$ lives. Before addressing the two-particle density and current, let's first look at the simpler one-particle quantities.
The one-particle density is $$\rho^{(1)}(r;t) = \sum_i \left< \delta(R_i-r); f \right>$$ which lives in $\mathbb{R}^3$ and bears the physical meaning of being the number of particles lying at a point $r$ (recall that the $\delta$ is not the usual Dirac delta but a "smeared-out" distribution, which allows multiple particles to "occupy" a single point $r$). The associated one-particle current is analogously $$j^{(1)}(r;t) = \sum_i \left<\frac{p_i}{m_i} \delta(R_i-r) ; f\right>$$ which is proportional to the momentum density at $r$. These are the microscopic analogues to the more familiar macroscopic mass density and momentum density of hydrodynamics.
Now for two-particle quantities, which will live in $\mathbb R^6$. The pair density is easy enough to write down $$\rho^{(2)}(r,r';t) = \sum_i\sum_{k\ne i} \left< \delta(R_i-r)\delta(R_k-r') ;f\right>$$ which is proportional to the probability that two particles have positions $r$ and $r'$. but the object of your question, the pair current, is a bit trickier to reason about. We want it to be proportional to the total momentum density associated with two particles located at $r$ and $r'$. So we'll start with a skeleton and fill in the blank $$j^{(2)}(r,r';t) = \sum_i\sum_{k\ne i}\left< \underbrace{\qquad}_{?}\delta(R_i-r)\delta(R_k-r') ; f\right>$$ We want some kind of sum of the two momenta, but we can't just add them together, because the momenta of two different particles live in two distinct spaces! The only sensible way (as far as I know) to add two objects belonging to different spaces is their direct sum, so the blank above must involve $p_i\oplus p_k$. We conclude that the natural way to define the pair current is $$j^{(2)}(r,r';t) = \sum_i\sum_{k\ne i}\left< \frac{p_i}{m_i}\oplus \frac{p_k}{m_k}\delta(R_i-r)\delta(R_k-r') ; f\right>$$ So the direct sum is just an artifact of our wanting the particles' coordinates and momenta to be vectors in their own little $\mathbb R^3$. If we instead defined the momenta to be vectors in $\mathbb R^{3N}$ -- e.g., $p_1=(p_{1x}, p_{1y}, p_{1z}, 0, \dots, 0)$, $p_2=(0,0,0,p_{2x}, p_{2y}, p_{2z}, 0, \dots, 0)$, etc. -- then instead of the direct sum, we'd have an ordinary sum. This would come at the cost of some notational inconvenience elsewhere, I imagine.