# Can Maxwell's Equations in differential form be viewed as equalities of measures?

What I have in mind is the following -

Suppose we choose to model the universe as a 3 dimensional flat Euclidean space $\mathbb{R}^3$ equipped with the standard topology and the Borel-sigma algebra. We can then define the charge distribution ($\rho$) as a signed measures on ($\mathbb{R}^3,\sigma_{Borel}$) which assigns to each measurable set a charge.

Similarly the divergence of a vector field ($\nabla.\vec{A}$) which is defined as the flux density of a field may also be interpreted as a signed measure on ($\mathbb{R}^3,\sigma_{Borel}$) which assigns to each measurable set a flux (of the field $\vec{A}$ "through it").

Now Maxwell's first equation which reads - $\nabla.\vec{E} = \dfrac{\rho}{\epsilon_{0}}$ can be interpreted as equating the flux measure and the charge measure. The Lebesgue integral of both sides reproduces the integral form of Maxwell's Equation.

Is this a valid point of view? I can't find too much literature online relating to this approach (perhaps because it's wrong?). Moreover if this is valid, this also justifies the use of the dirac delta function for point charges since the charge density for a point charge should reproduce the dirac measure.

Moreover is there a similar way of interpreting the equations for the curl - i.e. the circulation density?

• Yes, sure, I'm not certain what exactly you want from an answer - if you take the standard volume form $\omega$, both $(\nabla\cdot E)\omega$ and $\rho\omega$ define non-standard top-dimensional forms against which you can integrate, so you can also view them as a measure. What's the question? – ACuriousMind Apr 24 '17 at 17:00