In electrostatics, we have Maxwell's equations:

$\nabla \cdot E = \rho$

$\nabla \times E = 0$

These four equations (the second line standing for three equations) can also be written in terms of the electrostatic potential:

$ -\nabla^2 V=\rho $

$ E = -\nabla V $

Now if we know the positions of every charge in our system, we can find the electrostatic field (completely and entirely, with no additional information required) using Coulomb's law:

$ E(x) = -\nabla \int \frac{\rho(x')}{4\pi|x-x'|} \mathrm{d}^3 x' $

My question is: to what extent can we do the same with Maxwell's Equations? For instance, whenever Coulomb's Law is derived from Maxwell's equations, an appeal needs to be made to spherical symmetry. Must we do this? Can we not use the vanishing curl somehow to reach the same conclusion? Similarly, if we derive Coulomb's Law from Poisson's equation, we must specify boundary conditions. We must specify that the potential is some constant, say zero, at infinity.

It appears that the information content is less.

I have read a little bit about the Helmholtz decomposition, and it appears that the curl and divergence of a vector field (E, in this case) do completely determine the vector field, providing certain restrictions are placed on the smoothness and decay of the field at infinity. In other words, it appears that in fact Maxwell's Equations (in the context of electrostatics) do have less information content than Coulomb's Law.

  • $\begingroup$ The Maxwell's equations (together with the Lorentz force law) completely specify all of classical electrodynamics. $\endgroup$
    – Danu
    Oct 1, 2013 at 16:33
  • $\begingroup$ If this is so, then why (for instance) does a derivation of Coulomb's Law from these equations seemingly require the assumption of spherical symmetry? $\endgroup$
    – gj255
    Oct 1, 2013 at 18:17
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/44418/2451 and links therein. $\endgroup$
    – Qmechanic
    Oct 1, 2013 at 19:45

1 Answer 1


Mathematically speaking, it is true that Maxwell's equations by themselves aren't the whole story; they're a set of PDEs for which one needs to specify boundary conditions separately if one wants to solve them. Boundary conditions can be well-motivated from a physical perspective in a given scenario, but they do not follow from the equations themselves.

As for Poisson's equation versus Coulomb's Law, no requirement of spherical symmetry is necessary. Start with Poisson's equation, and set the charge density to be that of a point charge, namely \begin{align} \rho(\mathbf x) = q\delta(\mathbf x - \mathbf x_0). \end{align} Next, use the vanishing curl of the electric field (which works in the absence of explicitly time-varying magnetic fields) to write $\mathbf E = -\nabla \Phi$ and plug this into Poisson's equation to obtain \begin{align} \nabla^2\Phi = -\frac{q}{\epsilon_0}\delta(\mathbf x-\mathbf x_0) \end{align} In other words, we want to determine $\Phi$ that is a Green's function for the Laplace equation. The general solution is \begin{align} \Phi = \frac{1}{4\pi\epsilon_0}\frac{q}{|\mathbf x - \mathbf x_0|} + F(\mathbf x) \end{align} where $F$ is a harmonic function, namely one which satisfies Laplace's equation. If we then invoke the boundary condition that the potential vanishes at infinity, then this forces $F$ to be identically zero, and we obtain \begin{align} \Phi = \frac{1}{4\pi\epsilon_0}\frac{q}{|\mathbf x - \mathbf x_0|} \end{align} which, upon taking the gradient, yields he electric field of a point charge which is essentially the content of Coulomb's Law.

  • $\begingroup$ How does it follow exactly, if I may ask? $\endgroup$
    – gj255
    Oct 1, 2013 at 21:36
  • $\begingroup$ See the link that Qmechanic provided. $\endgroup$
    – Danu
    Oct 1, 2013 at 21:42
  • $\begingroup$ That argument rests on the assumption which I, perhaps misleadingly, termed 'spherical symmetry'. One must assert that the electric field is radial to reach Coulomb's Law. What I am trying to ascertain is: is this necessary? Can we not employ the curl perhaps, or appeal to the uniqueness theorem, rather than take this as an additional postulate? $\endgroup$
    – gj255
    Oct 1, 2013 at 21:47
  • $\begingroup$ @gj255 By Poisson's equation, the right hand left hand sides of the equations I wrote down are equal, so the right hand sides are equal too. No assumption about the electric field being radial is being used. $\endgroup$ Oct 1, 2013 at 21:47
  • $\begingroup$ Yes, but the equality of the right hand sides is not Coulomb's Law. It relates the flux through a surface to the charge enclosed. The expression E = q/4pi r^2 does not follow immediately (I believe?). $\endgroup$
    – gj255
    Oct 1, 2013 at 21:50

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