Helmholtz decomposition of electromagnetic fields

Remarks

Proof idea: let $W$ be the Newtonian potential $W(x) = -\frac{1}{4\pi} \int F(x') / \abs{x - x'} \, dx'$, so that $\Delta W = F$ with the Laplacian taken componentwise. The identity $\Delta W = \nabla (\nabla \cdot W) - \nabla \times (\nabla \times W)$ gives the splitting with $\Phi = -\nabla \cdot W$ and $A = -\nabla \times W$, and an integration by parts turns these into the formulas above. For uniqueness, $G_1 - G_2 = S_2 - S_1$ has zero curl and zero divergence, so each component is harmonic; it tends to $0$ at infinity, so it vanishes by Liouville's theorem. The decay condition is needed: the constant field $(1, 0, 0)$ has zero curl and zero divergence, so it can be counted in either part.

In electromagnetism the splitting is applied to the electric field $E$ and the magnetic field $B$. Since $\nabla \cdot B = 0$, the field $B$ has no curl-free part, so $B = \nabla \times A$ for a vector potential $A$. The curl-free part of $E$ is called its longitudinal part and the divergence-free part its transverse part. In the Coulomb gauge $\nabla \cdot A = 0$, one has $E = -\nabla \varphi - \partial_t A$ with $-\nabla \varphi$ longitudinal and $-\partial_t A$ transverse, and $\varphi$ is the instantaneous Coulomb potential of the charge density.

Going deeper

For a student

The two parts are found from the divergence and the curl of the field, through the Newtonian potential. So a field that decays at infinity is fixed by its divergence and curl. Maxwell's equations give exactly these: Gauss's laws give the divergences of $E$ and $B$, and the laws of Faraday and Ampère–Maxwell give their curls. The decomposition is the reason one can always write $B$ as a curl, and it splits $E$ into an instantaneous Coulomb part and a radiating part. On bounded domains the analogous splitting needs boundary conditions and leads to the Hodge decomposition of vector fields.