Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Just curious, can anyone show how the integral and differential form of Maxwell's equation is equivalent? (While it is conceptually obvious, I am thinking rigorous mathematical proof may be useful in some occasions..)

share|cite|improve this question
Are you familiar with Green's and Stoke's theorems? – Jerry Schirmer Oct 9 '12 at 16:48
@JerrySchirmer yes. (and I just want to see more rigorous proof; conceptually, this is somehow obvious, I guess.) – Paul Reubens Oct 9 '12 at 17:05
Just integrate the divergences and the curls using green's and stokes theorems and you will get the integral form. – Prathyush Oct 9 '12 at 19:16
Comment to the question(v2): Could you narrow down which part of the standard textbook treatment you don't find rigorous? – Qmechanic Oct 9 '12 at 22:36
Actually, if you treat the differential operators in the classical sense, the integral form and the differential form are not equivalent. The integral form is more general since it is also valid for discontinuous material behavior. – Tobias Apr 12 '14 at 16:29

Well, as the people said in the comments, the Theorems of Green, Stokes and Gauss will do the job, and are about as mathematically rigorous as you could hope for here!

The two different sets of formula follow directly.

I don't want to write all four of them out, you should be able to do them yourself, but for example, let's consider the Gauss Law.

Starting with the integral form, we have (ignoring physical constants)

$$ \int_{\partial \Omega} \vec{E} . d\vec{S} = \int_{\Omega} \rho\space dV$$

Then by Gauss, we have

$$ \int_{V} \mbox{div} \vec{F} \space dV = \int_{S} \vec{F} .d \vec{S} $$

Hence, we can replace

$$ \int_{\partial \Omega} \vec{E} . d\vec{S} \rightarrow \int_{\Omega} \mbox{div} \vec{E} \space dV $$

to give

$$ \int_{\Omega} \mbox{div} \vec{E} \space dV = \int_{\Omega} \rho\space dV $$

or dropping the integrals,

$$ \mbox{div} \vec{E} = \rho\space $$

which is the differential form.

You should try to derive the other three. This may be helpful in showing you where to start, and where you want to get to.

As for proofs of Green's, Stoke's and Guass Theorems, I recall learning them for some maths exams some years ago, but I wouldn't know where to begin now! Look at any differential geometry course or book and they should be somewhere early on. I can assure you though that the mathematicians have rigirous proofs for them, so we do not need to be shy in using the results of the theorems!

share|cite|improve this answer
Let me know if you get the other three out. Another good exercise, when you've finished that, is to derive each of the integral forms of the laws from physical arguement. – Flint72 Apr 12 '14 at 14:45
Doesn't Green's theorem and Gauss' theorem follow as a consequence of Stokes' theorem which in turn follows from Poincaré duality? – JamalS May 14 '14 at 5:31
I think it's the point where you say "dropping the integrals" that is the OP's problem. – Rob Jeffries Nov 28 '14 at 7:41
While this is nice, it is not entirely rigorous yet. Specifically, having $\int_{\Omega} (\text{div} \bf{E} - \rho) d \Omega = 0$ does not necessarily imply $\text{div} \bf{E} = \rho$. Some assumptions must be made about the space $\Omega$ . – Arthur Suvorov Feb 2 '15 at 2:24
Even as a simple example you could have $\Omega = (-1,1)$ with $\bf{E} = 0$ but $\rho(x) = x^3$. So that $\int^{1}_{-1} (0 - x^3) = 0$ – Arthur Suvorov Feb 2 '15 at 2:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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