I am reading some lecture notes (the context is the Vlasov–Landau equation but this is irrelevant for this question) and the phrase came up that the equation was "closed by Maxwell's Equations". A quick Google search shows me that this is quite a common phrasing. I can't, however, seem to find an explanation of what it actually means. Therefore my question is this:

What does it mean for an equation to be closed by Maxwell's equations?

  • 2
    $\begingroup$ Hi Q.S., I do believe context is relevant for this question. Do you think you could specify where you read that? Thanks! $\endgroup$ Commented Jul 24, 2017 at 12:46
  • $\begingroup$ @AccidentalFourierTransform The notes themselves arnt' publicly accessible. But the content is simply saying that the Vlasov–Landau equation is closed by Maxwell's Equations. $\endgroup$ Commented Jul 24, 2017 at 12:53
  • $\begingroup$ +1 thanks for asking about this, questions like yours above help with the jargon I am supposed to know. I deleted my guess. $\endgroup$
    – user163104
    Commented Jul 24, 2017 at 13:43
  • $\begingroup$ Are you sure the text says that the equation is closed, as opposed to saying that some differential form appearing in that equation is closed? $\endgroup$
    – WillO
    Commented Jul 24, 2017 at 15:43
  • $\begingroup$ @WillO I am pretty sure it is referring to the equation. The wording in my previous comment is almost verbatim from the text. $\endgroup$ Commented Jul 24, 2017 at 15:58

1 Answer 1


We say that a system of equations is closed if it can be used (at least in principle) to uniquely solve for the unknowns. As an example, Laplace equation is not closed, but if you add appropriate boundary conditions (e.g. Dirichlet or Neumann) it becomes closed system. I think what is meant in the notes you are qutoing is that this Vlasov-Landau equation is not sufficient to solve for the unknonws (too many variables, too few equations), but you should combine it with Maxwell equations to obtain physical predictions.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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