# Is current density (J) in Ampere's law derivable?

In Ampere's law:

$$\nabla\times\mathbf{B}=\mu_0\mathbf{J} +\mu_0\epsilon_0\frac{\partial\mathbf{E}}{\partial t}$$

the current density is listed explicitly as a separate term from the change in electric field. My understanding of the history (perhaps completely wrong), is that the $J$ term was determined first, and then the $E$ term was added later (by Maxwell?) to account for displacement current.

As the $J$ term is physically a set of moving charges, which each produce a time-varying electric field, why isn't the $E$ term sufficient to calculate the magnetic field? That is, to determine the magnetic field from a set of moving charges, couldn't you determine the magnetic field of a single moving charge from:

$$\nabla\times\mathbf{B}=\mu_0\epsilon_0\frac{\partial\mathbf{E}}{\partial t}$$

and then the total magnetic field of a current would be the sum of magnetic fields from many moving charges?

• What do you mean when you say "As the J term is physically a set of moving charges, which each produce an electric field"? Moving charges need an electric field, but they do not produce one. Do you mean magnetic field? Jan 1, 2017 at 20:00
• I meant that each charge produces an electric field, and if the charge is moving, then it produces a time varying electric field. Jan 1, 2017 at 20:20
• I do not understand your question - why would you assume that for a single moving charge, the equation holds without the $J$-term? A single moving charge is still a non-zero current! Jan 1, 2017 at 21:09
• Perhaps this helps - asked a different way: does a single moving charge contribute to the J term (as it is a current) or to the E term (as it is a changing E field)? Jan 2, 2017 at 1:47

Nothing in Maxwell's equations depends on the fact that the charges that we encounter in the real world are all tied to individual particles.

So we can imagine that we have an infinite rod of charge in space and it's moving with uniform velocity along its axis, and Maxwell's equations will still apply to this situation. And here we have moving charge, but with no net change in the configuration of the charge, so no change in the electric field associated with that charge.

In the real world, conduction in a wire, for example, involves the motion of quadrillions of electrons, mostly moving randomly, but with a slight bias in one direction or the other. Even though there is a current through our hypothetical wire, there's no net change in the configuration of the charge over time, except at the microscopic level, and so no significant change in the electric field resulting from the motion of the charged particles.

• I think this answers the majority of my question - that is they are independent terms... And your of an infinite rod provides clarity on this. But does a single moving charge contribute to the J term or the E term? Jan 2, 2017 at 1:50
• @Brian, it contributes to both. But if we're talking about a single point charge moving in space, it's only going to contribute a J term at exactly the point where it is found, and that term will be a singularity. Jan 2, 2017 at 3:51

Just by looking at the two equation that you wrote down, you get immediately that for them to be consistent you need to have $\mathbf{J}=0$. In other words the two systems of equation are not equivalent (unless you are in the trivial case $\mathbf{J}=0$).

Looking at the whole structure of Maxwell's equation + Lorentz forces on the charges, you can see that they consist of a set of coupled equations between EM fields and Matter fields (ie charges and hence currents). In general it is a very difficult problem to solve unless you have additional symmetries (like rotational symmetry or time independence).

It is moreover true that you cannot solve the equations one variable at time independently by each other, because these are coupled differential equations: a change in the charge distribution causes a change in the fields, both E and B, but not in a independent way (they are related!), then this causes a change in the charge distribution and so on. This is why a generic solution is difficult to write (at least explicitly).

However in certain situations it could be meaningful to consider some fields as given, non dynamical, and solve the equations for the other fields, freezing somehow some degrees of freedom.