I have been wrestling with Maxwell's equations using a source term in which both charge density and current density are time-independent functions of position, and where current is considered to be equal to charge density multiplied by “charge velocity”. Clearly, the divergence of the current density (i.e., the charge density times the charge velocity) must equal zero in order to ensure that charge density is time-independent.
Where I'm running into a problem is in proving that the solution of Maxwell's equations then consists of two components: a time-dependent component that is independent of the source terms, and a time-independent (static) component that depends on the source terms.
I'm sure there are standard, straightforward ways to do this, but my attempts so far are unsuccessful.
What I've come up with are these equations (setting all constants = 1):
$$\nabla^2E – \frac{\partial^{2}}{\partial t^2}E = -\nabla \rho$$ and $$\nabla^2B – \frac{\partial^{2}}{\partial t^2}B = -\nabla \times \vec j)$$
I think separating E and B into time-dependent and time-independent components will show that a time-independent charge and current distribution cannot produce EM radiation.
Is it legitimate in general to write: $$\vec E = \vec E_{rad} + \vec E_{static}$$
where $$\vec E_{rad}$$ satisfies $$\nabla^2 \vec E_{rad} – \frac{\partial^{2}}{\partial t^2} \vec E_{rad} = 0$$
and
$$ \nabla^2 \vec E_{static}(\vec x) = -\nabla \rho (\vec x)?$$
And, if so, then does that prove there is no EM radiation produced by a time-independent distribution of charge and current?
