In the Feynman Lectures, Chapter 21, I find the statement

We have solved Maxwell's equations. Given the currents and charges in any circumstance, we can find the potentials directly from these integrals and then differentiate and get the fields.

In Purcell's book on electricty and magnetism, I find the statement

Except for the possible addition of a constant field pervading all of space, the conditions $curl({\bf B}) =4\pi{\bf J}/c$ and $div({\bf B})=0$, uniquely determine the magnetic field of a given distribution of currents.

I don't have Griffiths's textbook in front of me at the moment but I'm pretty sure he says something similar.

Clearly all of these statements are false. For example, if the current and charge distributions are identically zero, then I can solve Maxwell's equations by setting ${\bf E}=grad(f)$ and ${\bf B}=grad(g)$ where $f$ and $g$ are arbitrary harmonic functions, so that in particular ${\bf E}$ and ${\bf B}$ are by no means unique (even up to the addition of a constant vector field).

Presumably, then, there is some hypothesis that Feynman, Purcell and others have omitted, possibly because they thought it was too obvious to mention. What is that hypothesis?

  • $\begingroup$ I think they're missing the boundary conditions. $\endgroup$
    – Ryan Unger
    Mar 22 '16 at 20:39
  • $\begingroup$ As a concrete example, on any solution to Maxwell you can add on light waves (EM radiation) or near fields (EM evanescent waves). Now obviously the evanescent waves diverge but light waves can be fully compact. $\endgroup$
    – Nanite
    Apr 8 '16 at 16:13

The assumption missing in all these statements is that there are boundary conditions assumed to be given.

E.g. for Poisson's equation $\Delta f = \rho$, the solution is unique for Dirichlet and/or Neumann boundary conditions, see e.g. section 1.9 in Jackson's "Classical Electrodynamics".

  • $\begingroup$ Thank you. It seems to me to be a bit odd that you'd be able to impose boundary conditions on a field that pervades an isotropic universe. Am I wrong to be troubled by this? $\endgroup$
    – WillO
    Mar 22 '16 at 20:46
  • $\begingroup$ @WillO: What's troubling about being able to make a closed metal surface that separates the field inside from outside (having said that, the closure is not actually necessary to introduce boundary conditions)? That closing this surface can have non-trivial consequences is something we are teaching in high school with electrostatics experiments... $\endgroup$
    – CuriousOne
    Mar 22 '16 at 20:49
  • 2
    $\begingroup$ @WillO: For a field that "pervades the universe", the appropiate conditions are either given by solving the equations inside a box (with zero boundary conditions) and then letting the boundaries of the box tend towards infinity, or by demanding from the start that the field goes to zero towards infinity faster than the surface of a sphere with radius $r$ grows towards infinity.. $\endgroup$
    – ACuriousMind
    Mar 22 '16 at 20:54
  • $\begingroup$ @CuriousOne: Thanks for editing your comment. I didn't understand the first version, but it's much clearer now. $\endgroup$
    – WillO
    Mar 22 '16 at 20:56
  • 1
    $\begingroup$ @WillO: There is a certain surface integral that has to be constrained to vanish by the boundary conditions, cf. the Wikipedia article. $\endgroup$
    – ACuriousMind
    Mar 22 '16 at 21:02

I think that the texts you quote are referring to localized charge and current densities and the fields are defined in the whole space. The natural requirement is that far from the sources the vector fields decay as $1/r^2$ or faster, uniformly in all directions. This is the hypothesis of an induction field which is appropriate for static fields. With this requirement the elactrostatic and magnetostatic equations uniquely determine a solution. Dealing with potential fields the requirement is that they decay as $1/r$ or faster. Your counterexample does not works, as your harmonic functions would be bounded (from below or from above) and thus they must be constant in view of Liouville's theorem for harmonic functions in $R^n$. As they vanish at infinity they must be zero everywhere.

  • $\begingroup$ Thanks. This is very helpful and I'd certainly have accepted it had it arrived before @ACuriousMind's answer. $\endgroup$
    – WillO
    Mar 22 '16 at 21:14
  • $\begingroup$ @Timaeus: I think they both deserve it! $\endgroup$
    – WillO
    Mar 22 '16 at 21:20
  • $\begingroup$ Do not worry, it is ok as it stands. $\endgroup$ Mar 22 '16 at 21:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.