# How do Maxwell's equations uniquely determine ${\bf E}$ and ${\bf B}$ despite no. of equations exceeding no. of unknowns?

Maxwell's equations in free space are given by $${\bf\nabla}\cdot\textbf{E}=0,~~{\bf\nabla}\cdot\textbf{B}=0$$ and $${\bf\nabla}\times\textbf{E}=-\frac{\partial\textbf{B}}{\partial t},~~{\bf\nabla}\times\textbf{B}=c^{-2}\frac{\partial\textbf{E}}{\partial t}.$$ The first two equations are two scalar equations whereas the second two equations are vector equations each of which gives three independent equations (componentwise)! Therefore, there are $$2+6=8$$ equations while only $$6$$ unknowns: $$(E_x,E_y,E_z)$$ and $$(B_x,B_y,B_z)$$.

Question When we have a larger number of unknowns than the number of equations, we don't, in general, expect to obtain a unique solution. However, given the appropriate boundary conditions, Maxwell's equations work triumphantly and give unique solutions to electric and magnetic fields, I must be overlooking something. What is the resolution to this apparent paradox?

Provided that the first two equations hold true at the initial condition, they are redundant for the time evolution, because $$\nabla \cdot \frac{\partial \mathbf{E}}{\partial t} = \frac{1}{c^2} \nabla \cdot \nabla \times \mathbf{B} = 0$$ and hence $$\nabla \cdot \mathbf{E}$$ is constant, with a similar argument for $$\nabla \cdot \mathbf{B}$$. So we actually only have $$6$$ equations determining the time evolution, which is just the right amount.

• But don't you think that you will face the same problem while dealing with sourced equations i.e. with nonzero $\rho$ and ${\bf J}$? One thing that I notice is that $\rho$ and ${\bf J}$ are related via the continuity equation which is built into Maxwell's equations. Don't know if that's a rescue.
– SRS
Apr 12, 2020 at 7:20
• @SRS Yes, they're redundant by the continuity equation. There's a standard textbook proof that Maxwell's equations imply the continuity equation. You can take precisely the same proof and run it in reverse to show that two of Maxwell's equations plus the continuity equation (which must hold if you have physical sources) imply the other two. Apr 12, 2020 at 7:30

Maxwell equations are partial differential equations, so much of the intuition that one have from dealing with the systems of linear equations or the ordinary differential equations is not applicable here.

More specifically: the solution of the divergence equations are defined up to a curl, i.e. $$\nabla\cdot \mathbf{A} = \nabla\cdot (\mathbf{A} + \nabla\times \mathbf{B}).$$ Likewise the solutions of the curl equations are defined up to a gradient: $$\nabla\times\mathbf{A} = \nabla\times(\mathbf{A} + \nabla f).$$ This lack of definiteness lies at the core of defining the potentials: $$\mathbf{E} = -\nabla\varphi +\frac{1}{c}\frac{\partial \mathbf{A}}{\partial t}, \mathbf{B} = \nabla\times\mathbf{A}.$$ Note that the potentials are not uniquely define, indeed - they need to be supported by an equation fixing the gauge (typically Coulomb or Lorentz gauge).

Finally, the equations in the question do not contain sources (i.e., the electric charge density and the current densities). In fact, Maxwell equations are underdefined, since they do not contain the material equations, specifying how the sources are affected by the electromagnetic field.

For free em fields E and B are mutually dependent so there are really only three equations. For each of the three directions of E or B there are two propagation directions and two phases, which makes 12 independent solutions per frequency, if which there are infinitely many. The total number of possible solutions is infinite. In terms of the vector potential all the free Maxwell's equations do is impose the restriction that $$\omega = kc$$.