# Validity of microscopic current density formula in Electrodynamics

Consider $$N$$ point particles, $$i^{\text{th}}$$ point particle having charge $$q_{i}$$ each follow a trajectory $$\vec r_{i}(t)$$. Then the charge density is, $$\eta(r)=\sum_{i=1}^{N}q_{i}\delta({\vec r-\vec r_{i}(t)})$$ $$\textbf{Assuming}$$ the continuity equation holds, $$\frac{\partial \rho}{\partial t}=-\nabla \cdot \vec J$$ $$\frac{ \partial(\eta(r))}{\partial t}=\sum_{i=1}^{N}q_{i}\frac{\partial \delta({\vec r-\vec r_{i}(t)})}{\partial t}$$ From https://physics.stackexchange.com/a/313175, we have: $$\partial_{t} \delta(\vec r-\vec r_{i}(t))=\frac{d\vec r_{i}(t)}{dt}\cdot \nabla_{r_{i}} \delta(\vec r-\vec r_{i}(t))=-\frac{d\vec r_{i}(t)}{dt}\cdot \nabla_{r} \delta(\vec r-\vec r_{i}(t))$$

$$\frac{ \partial(\eta(r))}{\partial t}=\sum_{i=1}^{N} q_{i}\partial_{t} \delta(\vec r-\vec r_{i}(t))=-\sum_{i=1}^{N} q_{i} \frac{d\vec r_{i}(t)}{dt}\cdot \nabla_{r} \delta(\vec r-\vec r_{i}(t))$$ $$\implies -\sum_{i=1}^{N} q_{i} \vec v_{i} \cdot \nabla_{r} \delta(\vec r-\vec r_{i}(t))$$ Recalling the product rule $$\nabla \cdot (a\vec C)=a\nabla \cdot \vec C + \vec C \cdot \nabla a$$ and since $$\nabla \cdot \vec v_{i} = 0$$ , $$\frac{ \partial(\eta(r))}{\partial t}=-\nabla_{r} \cdot \sum_{i=1}^{N} q_{i} \vec v_{i} \delta(\vec r-\vec r_{i}(t))$$ Comparing this with the continuity leads us to define the microscopic current density $$j(r)$$ as: $$\vec j(r,t)=\sum_{i=1}^{N} q_{i} \vec v_{i} \delta(\vec r-\vec r_{i}(t))$$

This is how I, and even Andrew Zangwill in his book, "Modern Electrodynamics", come at the formulation of current density.

## My Question:

In the macroscopic case, the current $$I$$ can unambiguously be written using the current density $$J$$ as $$dI=J\cdot d\vec A$$, this works because we assume certain things about the macroscopic charge density $$\rho_{\text{macro}}(r,t)$$ (it being constant over the averaging volume) and that the macroscopic velocity field $$\vec v$$, at a given macroscopic site, remains same. With these assumptions and the formula $$J=\rho \vec v$$, the formula $$dI=\vec J \cdot d\vec A$$ is correct, given that the area, $$d\vec A$$ is macroscopically small, yet microscopically large.

Can we do this, for the microscopic case??

Can we write the current through a microscopically infinitesimal area (nonclosed area), $$d\vec a$$, as $$dI=\vec j \cdot d\vec a$$ where $$\vec j(r,t)=\sum_{i=1}^{N} q_{i} \vec v_{i} \delta(\vec r-\vec r_{i}(t))$$

I basically want to know whether I can say that, the rate of charge flowing through an infinitesimal area is equal to $$\vec j \cdot da$$ I don't think so, because, there must simply be many many considerations coming into picture, regarding when and how a given $$q_{i}$$, following its trajectory $$r_{i}(t)$$, passes through a given $$d\vec a$$ at a point $$r_{0}$$ . We might need more variables and expressions that the formula $$j \cdot da$$ seems to give us by just plugging the above formula for $$j$$. If I can, please tell my why I can do so, clearly. If I cannot, explain that too. I continually see this formula everywhere, all the papers I read etc.

The formula is valid. Your objection does not hold water. Remember that the current given by $$dI=j\cdot dA$$ is instantaneous is time. Thus only the positions of the particles and eventually its higher derivatives can contribute. However, the entire history of their trajectories is irrelevant.
Intuitively, just look at the charges crossing the surface between $$t$$ and $$t+dt$$. At leading order, the trajectories are linear. You just need to count the segments $$[r_i,r_i+v_i dt]$$ crossing the surface and weight them by $$q_i\cos\theta_i$$ with $$\theta_i$$ the angle between $$v_i$$ and $$dA$$. The formula with delta functions is just a convenient bookkeeping formula enabling the link between discrete and continuum theory.
You can build intuition by looking in 1D. In this case, $$j$$ is directly the current: $$j(x,t)=\sum q_iv_i\delta(x-x_i)$$ By fixing $$x$$ you can interpret it as the current through $$x$$: $$I=\sum_i q_i\sum_j \text{sgn}(v_{ij})\delta(t-t_{ij})$$ with $$t_{ij}$$ the $$j$$-th time the $$i$$-th particle crosses $$x$$. (The sum can be empty)