Integral vs differential forms of Maxwell's equations

Question

As stated in this post, the integral and differential Maxwell equations should be identical. However, in a text I was reading it states that

The integral forms of Maxwell’s equations describe the behaviour of electromagnetic field quantities in all geometric configurations. The differential forms of Maxwell’s equations are only valid in regions where the parameters of the media are constant or vary smoothly i.e. in regions where "$\epsilon(x, y, z, t), μ(x, y, z, t)$ and $\rho(x, y, z, t)$ do not change abruptly. In order for a differential form to exist, the partial derivatives must exist, and this requirement breaks down at the boundaries between different materials.

Doesn't this imply that the integral form of Maxwell's equation is more fundamental than the differential form because it works for all configurations?

Answer 1

The integral and differential versions are equivalent, so it sounds like your text simply doesn't know how to use the differential version in as general a way as your text knows how to use the integral version.

For instance, you do not need to have partial derivatives to define the divergence and/or curl, but if you have the partial derivatives, then there are simple formulas for the divergence and curl in that situation, you might be used to those formulas as the definition, but they are not the proper general definition.

The two forms really are equivalent, and they are equivalent because they aren't really saying anything different whatsoever. The understanding for the integral forms is that they hold for every closed/open surface in any bounded region. The understanding for the differential form is that they hold in any bounded region.

It sounds like your text knows how to setup the integral forms in a fairly general way. The differential forms can be setup in an equally general way since there are very general ways to define the divergence and the curl. One way is to take the divergence theorem and the curl theorem as the definitions of the operators (instead of assuming partial derivatives, then making up random combinations and then making a theorem). For instance

$$\vec{\nabla}\cdot\vec{A}@(x,y,z)=\lim_{DiameterV\rightarrow 0, V\rightarrow\{(x,y,z)\}}\frac{\iint_{\partial V}\hat{n}\cdot\vec{A}dS}{\iiint_VdV}$$

and

$$\hat{S}\cdot\left(\vec{\nabla}\times\vec{A}@(x,y,z)\right)=\lim_{Diameter S\rightarrow 0, S\rightarrow\{(x,y,z)\}}\frac{\int_{\partial S}\vec{A}\cdot d\vec{l}}{\iint_SdS}$$

don't require partial derivatives, only that the fields be nice enough to have limits that allow the divergence and the curl theorems, nothing more, nothing less. But that level of generality is based assuming your integral version is using Riemann integration. That is the simplest version of integration (but not the most general), and I want you to see how we don't have to arbitrarily choose to make one version more general than the other.

What you see from the above is that knowing how to integrate is fully enough to tell us how to differentiate. So we can choose to differentiate as generally as we want to integrate. Let's look at how generally we want to integrate and differentiate.

If you look at an equation like $\vec{\nabla}\cdot\vec{E}=\frac{q}{\epsilon_0}\delta^3(\vec{r}-\vec{r}_0)$ you see that the result of a derivative might be a generalized function. And a generalized functions is defined by what it does inside integrals. So a general derivative is defined by what the result does under integrals.

How is $\delta^3(\vec{r}-\vec{r}_0)$ defined? It is defined by sending any smooth function $f$ that goes to zero (and all its derivatives go to zero) fast enough to avoid boundary terms from integration by parts (called a test function) and sending it to $\iiint f(\vec{r})\delta^3(\vec{r}-\vec{r}_0)dxdydz=f(\vec{r}_0)$. And it is just one of many such maps called distributions that send test functions to numbers in a nice (linear and continuous) way. So if we expect taking derivatives to give us distributions, then we should define them so that they do.

First note that any nice function $g$ is a distribution in the sense that it sends a test function $f$ to the number $\iiint fg dxdydz$, let's denote that by $G:f\mapsto\iiint fg dxdydz=G(f)$ and then note that if the function $g$ had a partial derivative $\partial g$ that was a nice function then $\iiint f\partial g dxdydz$ = $-\iiint g\partial f dxdydz$. This leads to a generalization of partial derivatives. For any distribution $G$ that sends tests functions $f$ to $G(f)$ we can define it's derivative to be the distribution that sends like $f$ to go to $-G(\partial f)$. To any nice function this definition agrees with the original definition of derivatives in the sense that they act the same under integrals. But now any distribution has a derivative.

But even with this generalization, the divergence can be defined more generally. Using $$\int \phi \vec{\nabla}\cdot \vec{A}=\int \vec{\nabla}\cdot (\phi\vec{A})-\int\vec{A}\cdot\nabla\phi,$$ we can throw away the total divergence term (because things go to zero fast enough on the boundary), to get

$$\int \phi \vec{\nabla}\cdot \vec{A}=\int -\vec{A}\cdot\nabla\phi.$$

So for regular functions $\vec{A}$ the divergence turns it into something that acts on test functions $\phi$ by sending them to the number that $\vec{A}$ sends $ -\nabla \phi$. So for every vector distribution $\vec{G}$ that sends vector test functions $\vec{f}$ to $\vec{G}(\vec{f})$ the divergence of $\vec{G}$ is the distribution that sends a scalar test function $\phi$ to $-\vec{G}(\nabla\phi)$. This a very general kind of divergence, both for what it can act on and what it can give.

It's just not fair to extend how general a thing you allow your integrals to do and not to be equally general about what you allow your derivatives to do. But no one truly desires to define their derivatives and integrals to be anything other than equally general, there simply is no point to that.

Answer 2

Dirac delta is the most common distribution in physics. It is defined by an equation: $ \int \limits_{-\infty}^\infty \delta(x) f(x) = f(0) $

One could say that this could be "intuitively: a function, which is $\infty$ at $0$, and $0$ everywhere else, but this makes no sense, since such integral would be $0$ (not $f(0)$), as integral over the set of zero measure (a point).

Using $\delta$, you could write down e.g. Gauss law for a point charge at (0,0,0): $$ \nabla \cdot \vec E(\vec r) = \frac{q}{\epsilon_0}\delta^3 (\vec r).$$ Integrating left-hand side in the usual way and RHS (using aforementioned equation) , we recover an integral form of Gauss law. You could find many examples of this in Griffiths' or Jackson's electrodynamics textbook.