We know that when averaging the velocity vector field of a current distribution over a particular volume, we get velocity fields that are slow varying over macroscopically small / infinitesimal distances.

This is helpful as this is primarily the reason why we can write $$\vec j = \vec v \rho$$

that's because we assume all the charges that enter this small area element $dA$ (which is many 1000s molecules in diameter) enter it with the same velocity $\vec v(r)$, and the analysis is carried out in a simple fashion considering all the charges entering this area coming from a volume $ \vec v \cdot d\vec A dt $ and hence charge per unit time through this macroscopically infinitesimal area $$dI = \rho \vec v \cdot d\vec A$$ which is the current, macroscopically infinitesimally.

Suppose I have a current distribution, and I want to know the microscopic current density at a given point through an area $dA$ which is about a handful of molecules in size. Suppose the $i^{th}$ type of particle(s) move(s) with a velocity $\vec v_{i}$, the last particle of the $i^{th}$ type that will enter this area $d\vec A$ is a distance $\vec v dt$ away, and hence if $\rho_{i}$ is the density of the distribution of the $i^{th}$ type of particles, then the amount of charge entering this area solely due to the $ i^{th} $ particle will be $$dQ_{i}=\rho_{i} d\vec A \cdot \vec v_{i} dt $$

Now plugging in charge density of the $i^{th}$ particle, $$\rho_{i}(r')= q_{i} \delta{(r'-r_{i}'(t))}$$

$$I_{i}=\frac{dQ_{i}}{dt}= q_{i} \delta{(r'-r_{i}'(t))} d\vec A \cdot \vec v_{i} $$ and the total current is just the sum of contributions from all types of particles:

$$I=\frac{dQ}{dt}= \sum{q_{i} \delta{(r'-r_{i}'(t))} \vec v_{i} \cdot d\vec A } $$

Giving us $$\vec j = \sum{ q_{i} \delta{(r'-r_{i}'(t))} \vec v_{i} }$$

(This derivation of the $j_{micro} $ expression is my own, and I am highly positive it's correct (mainly because I know the formula is), if not, correct me)

Now my question is, how to go from $$ \vec j_{micro}= \sum{ q_{i} \delta{(r'-r_{i}'(t))} \vec v_{i} }$$ to $$\vec j_{macro}(r)=\rho(r) \vec v(r)$$

We know that Maxwell's laws, especially $$\nabla \times b = \mu_{0} \vec j + \mu_{0} \epsilon_{0} \frac{d\vec e}{dt} $$ can be averaged to get the corresponding macroscopic laws, and the averaging procedure being commutative, commutes through these differential operators to give: $$ \nabla \times <b> = \mu_{0} <\vec j> + \mu_{0} \epsilon_{0} \frac{d<\vec e>}{dt} $$ so I would like to know how one can jump from the microscopic current density to macroscopic current density.

(Note: when I say 1000s of molecules, it's solely for bookkeeping purpose, it could be higher depending on the situation and the kind of averaging we want.

Also, it is not necessary to answer the question based on Lorenz averaging method, any kind is fine as long as it reproduces the correct results, I'm willing to accept it)


1 Answer 1



$$\vec j_{micro}=\sum_{all-charges}{q_{i}}\vec v_{i} \delta(r-r_{i}'(t))$$

consider an appropriate volume to average over, $\Delta V$. I denote the position of the volume element in space with the vector $\vec R$. It is clear that averaging is done over a sphere with centre $\vec R$. when we do the averaging: $$\left< \vec j_{micro} \right>=\vec j_{macro} (\vec R)=\frac{\iiint_{\Delta V}\sum_{all-charges}q_{i} \vec v_{i}\delta(r-r_{i}'(t)) d^3r}{\Delta V}$$

now, if the charge's position is within the volume $\Delta V$, the term pertaining to it survives the integral, while the term pertaining to charges that lie outside the volume perishes. With this in mind, the integral then integrates to:

$$ \left< \vec j_{micro} \right>=\vec j_{macro}(\vec R)=\frac{\sum_{\Delta V} q_{i} \vec v_{i}}{\Delta V} $$

Leaving this here for a moment, we turn our attention to $\rho_{micro}(r)=\sum_{all-charges} q_{i}\delta(r-r_{i}'(t)) $. Averaging this quantity over the same volume $\Delta V$ gives us: $$\left< \rho_{micro} \right>=\rho_{macro}(\vec R)=\frac{\iiint_{\Delta V}\sum_{all-charges} q_{i}\delta(r-r_{i}'(t)) d^3r}{\Delta V}$$ which gives us:

$$\rho_{macro}(\vec R)=\frac{\sum_{\Delta V} q_{i}}{\Delta V}$$

now, we defnie the velocity vector field that we use in our macroscopic description of things as:

$$\vec v_{macro} (\vec R)=\frac{\left< \vec j_{micro} \right>}{\left< \rho_{micro} \right>}=\frac{\sum_{\Delta V} q_{i} \vec v_{i}}{\sum_{\Delta V} q_{i}}$$

This allows us to average the maxwell's equations, in particular, the Ampere-Maxwell equation:

$$\nabla \times \vec b = \mu_{0} \vec j_{micro} + \mu_{0} \epsilon_{0} \frac{d\vec e}{dt}$$ $$\nabla \times \left< \vec b \right> = \mu_{0} \left< \vec j_{micro} \right> + \mu_{0} \epsilon_{0} \frac{d \left< \vec e \right>}{dt}$$ this gives $$\nabla \times \vec B = \mu_{0} \rho_{macro}\vec v_{macro} + \mu_{0} \epsilon_{0} \frac{d\vec E}{dt}$$ and we can safely call $\rho_{macro} \vec v_{macro}$ as the macroscopic current density that we have been using all along. This fits with the theory as the $\rho_{macro}$ appearing here is exactly the same that appears in Guass' Law, keeping things consistent.


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