Maxwell's Equations: are they complete?

[chingkui]

Maxwell's Equations:
\nabla \cdot D= \rho
\nabla \cdot B=0
\nabla \times E=- \partial B/ \partial t
\nabla \times H=J+ \partial D/ \partial t
Together with the continuity eq:
\nabla \cdot J=- \partial \rho / \partial t
There are 9 scalar equations and 16 scalar unknowns (B, E, D, H, J, \rho)
If we are supplied with the relations that relate B to H and E to D (e.g. a linear media relation):
D=f(E)
H=g(B)
we have 6 more scalar equations and therefore 15 equations in total. We are still one equation short of solving the Maxwell Eq, if we are supplied with appropriate B.C. and Initial Conditions, and we do not constrain the current density J and charge density \rho. In that case, how do we solve the Maxwell Equations?

 

[Hans de Vries]

If you describe the EM field by its potentials V and \vec{A}
and the source field by the charge \rho and current \vec{J}

Then you need just eight equations:

\nabla \cdot D= \rho
\nabla \cdot B=0
\nabla \times E=- \partial B/ \partial t \ \ \ \ (=3x)
\nabla \times H=J+ \partial D/ \partial t \ \ \ \ (=3x)Regards, Hans

 

[Galileo]

The continuity equation is not an independent assumption, but follows from Maxwell's equations (apply the curl to the fourth). However, they form an interdependent set of equations, so it's not at all obvious from these equations what the degrees of freedom are, or how to specify the initial state of the system (the field and the particles).

To see the degrees of freedom more clearly, you should use a potential formulation. A statement of a potential function V and a vector potential \vec A will then be necessary and sufficient to give the field everywhere (you also need the positions and velocities of all the particles for the system ofcourse). The choice on V and \vec A is not unique though. There's some freedom of choice which we call gauge freedom.

For the Coulomb gauge, the initial condition you should specify are the positions and velocities of all particles, the field \vec A and its time derivative (since it obeys a wave equation).

 

[Claude Bile]

There are boundary conditions that apply too, for example solving for a field within a waveguide.

Claude.