Is there a case when it is better to use the integral form of the Maxwell equations rather than the differential form?

Question

I was wondering if there is a case where the integral form of the Maxwell equations is preferred over the differential form? If you could provide with an example for each one of the equations I would really appreciate that.

Answer 1

The integral forms are useful in (typically static) situations where the charge/current distribution is symmetric enough that you can use a symmetry argument to replace the surface/line integrals with a simple product of a uniform field strength times an area/length of an imaginary closed "Gaussian surface" or "Amperian loop".

They are also useful for figuring out the far-field behavior of a localized charge/current source - e.g. we can use them to show that that, no matter how complicated a static localized charge distribution, very far away it produces an electric field of the form ${\bf E} = (Q/r^2)\, \hat{{\bf r}}$, where $Q$ is the distribution's total electric charge.

A friend of mine once attended a lecture titled "A Defense of the Integral Forms of Maxwell's Equations" in which a distinguished elderly physicist argued that the integral forms are underappreciated.

Answer 2

Take the interface conditions for electromagneticfields, e.g.

$$ \vec n\cdot \Delta\vec D = \vec\rho_s$$

i.e. the normal component of the $\vec D$ field is continuous if no surface charge $\vec\rho_s$ is present.

To derive that, imagine a box (or cylinder, prism etc) that goes through the interface parallel to the normal vector $\vec n$ (sorry, no pic yet). Then decompose the surface integral in Gauss' law such that

$$\begin{align} \newcommand{\oiint}{\oint\!\!\!\!\!\int} % sorry... \iiint \rho\,dV &= \oiint \vec D\,d\vec n \\ &= \iint_\text{top&bottom} \vec D\, d\vec n + \iint_\text{side}\vec D\, d\vec n \\ \iint\int\rho\,dh\,dA &= \iint_A\underbrace{(D_1-D_2)}_{=\Delta\vec D}\, dA + \iint_\text{side}\vec D\, d\vec n \end{align}$$

The last line uses that $\vec n_\text{bottom} = -\vec n_\text{top}$. Now let the height decrease to zero, thus $\iint_\text{side}\vec D\, d\vec n\to 0$ and define $\int\rho\,dh\to\rho_s$. The base area $A$ of top and bottom is arbitrary, thus the integrand is also equal and we just get the interface condition above.

Answer 3

The same principles apply as when you normally want the differential form of something vs. the integral form.

It's like asking when do you want to know the total distance and time of a road trip vs. what the speed was at a particular point.

With a uniformly charged sphere, for example, you can use the integral form to get the total flux at some radius, and then use that to infer what the potential is at any point because of the perfect symmetry of a sphere. Wouldn't work with a cube!

The analogy would be if you know you drove the exact same speed throughout an entire road trip, you could calculate what the speed was by dividing total distance / total time. Not so if there was a lot of starting and stopping!

Answer 4

Dynamical Example

Prove that there are no electromagnetic monopole antennas / sources

A monopole antenna field is a time varying, spherically symmetric EM field, i.e. a field which is invariant to any rotation about the central point.

Any such vector field must be radially directed, and so it must be of the form $\vec{F} = f(r,\,t)\,\mathbf{\hat{r}}$. That holds whether we are talking about electric or magnetic fields (see the footnote at the end for discussion of often glossed-over points in the underlying symmetry argument).

Apply Gauss's law for magnetism using a spherical pillbox: we immediately see that the radial magnetic field must be nought, otherwise there would be a central magnetic charge / spherically symmetric nett magnetic charge distribution.

Do the same for Gauss's law for the electric field: conclude there's a spherically symmetric electric charge distribution.

Now firstly suppose that the Gaussian pillbox is a sphere in freespace outside and enclosing all the spherically symmetric charge and current sources. Then, because there is no way for charge to cross this sphere (by assumption), the total charge within must be constant, by charge conservation so that the electric field cannot be time varying. Thus the only possible monopolar electromagnetic field separated from its sources is the electrostatic field owing to a spherically symmetric charge distribution, and there are no monopolar antennas.

Further investigation along these lines shows up something rather peculiar to this problem. Suppose now that the Gaussian pillbox is a sphere inside the charge / current distribution. Then, the total enclosed charge is $Q=4\,\pi\,\,r^2\,f(r,\,t)$ if the displacement vector is $\vec{D} = f(r,\,t)\,\mathbf{\hat{r}}$, thus there is a radial current density given by:

$$4\,\pi\,r^2\,\vec{J} = -\frac{\mathrm{d}\,Q}{\mathrm{d}\,t}\,\mathbf{\hat{r}}\quad\Rightarrow \quad\vec{J} = -\partial_t f(r,\,t) \mathbf{\hat{r}}$$

Of course we must have $\vec{J}(0,\,t)=\mathbf{0}$. As we have seen, there cannot be any magnetic field present and indeed the current defined above exactly cancels the electric displacement current.

So charge in this peculiar example can be shuttled to and fro along radially directed currents, yet the radiation from the in general accelerated charge is perfectly and precisely absorbed by countervailing, oppositely accelerated charge, and only the static electric field remains outside the time-varying charge / current distribution.

Some General Comments

A few somewhat vague, but I believe important words before I give my example. It's not as though there's a dichotomy between or the two approaches or that one is "better" than the other - one is the limiting form of the other, and I believe that almost always when I think intuitively about electromagnetism - as opposed to doing calculations - I almost always think in terms of the integral forms. When I see a "curl" symbol, I always think of a little loop, with the curl relating the integral around the loop of the operand to the flux of the curl's result through the loop - even if I am working with the differential forms. When I see a "div" symbol, likewise I think of a little bounded, simply connected blob, with the div relating the flux of the operand through the blob's boundary to the volume integral of the div's result over the blob. Electromagnetism is almost always about blobs and loops for me, and very sometimes C++ code (unless I can get someone else to do this!). Later on, if you haven't done so already, you'll learn about the unification of grad, curl and div into the exterior derivative; the former being the one, two and three-dimensional special of the latter (modulo a few minor conventions messed up by the inventors of grad div and curl). We take a little hyperblob - a bounded $n+1$-manifold with nontrivial boundary, and a differential $n$-form which takes as inputs $n$-vectors that define an element of the boundary; we sum up the result for vectors that define a "triangulation" (or a "polytopization"- if that's a word) of the boundary. Then we divide by the volume of the hyperblob, and work out the limit of this process as we shrink the hyperblob's volume to nought. All we have to do is prove that this process is well defined (and it is, for appropriate assumptions of continuity of derivatives of the forms involved) and we have the meaning of the exterior derivative: the limiting form of the sum over the "exterior" of manifold of the form in question divided by the manifold's volume as the volume shrinks to nought (I fancy this is the origin of the name "exterior" derivative, but I've never checked out this fancy's truth). The theorems of Stokes and Gauss are now merely part of the definition, as their generalization in the generalized Fundamental Theorem of Differential Calculus falls out of the well-definedness for free. A curl is literally a limit of an integral around a loop, a divergence a limit of a surface integral, and these are the foremost meanings I believe one should associate with these notions, rather than letting oneself become distracted by the almost unfathomable symbolic mess that is often these things written out as components. I'll leave off my banging on with a quote the title of Section 4.3 in Misner, Thorne and Wheeler, "Gravitation":

"Forms Illuminate Electromagnetism, and Electromagnetism Illuminates Forms"

which I urge you to read as soon as you have a background in forms. The authors clearly have a most vivid, almost poetically-visual way of thinking.

Another document I have come across in recent years on these topics whose clarity has impressed me is chapters I through V of "A Quick and Dirty Introduction to Exterior Calculus", itself part of the beautifully written CalTech lecture course:

Peter Schröder, "Discrete Differential Geometry"

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Footnote: Spherical Symmetry Implies Radial Vector Field in Three Spatial Dimensions

In three dimensions, a spherically symmetric vector field must always be radially directed.

This is because, by invariance wrt rotation, $f$ cannot depend on $\theta$ or $\phi$. If there were a $\boldsymbol{\hat{\phi}}$ or $\boldsymbol{\hat{\theta}}$ component to this $\theta$ and $\phi$-independent field, it would constitute a never vanishing, nonsingular vector field on the surface of the 2-sphere, gainsaying the hairy ball theorem.

We also note an important aside: one can also see that this radially directed condition need not and does not hold in even dimensional spherically symmetric fields. One couldn't draw this conclusion if one had occasion to do four-spatial-dimensional electromagnetism, for example.

But more practically, it doesn't hold in two dimensions either: thus we can have spherically symmetric purely azimuthal magnetic and electric fields as well as radial ones in problems with cylindrical symmetry. We see them all the time around electrical wires!

Answer 5

The integral form of Maxwell's equations is the most general form.

You may ask, why? Aren't both of them equivalent statements. No, they are not, the integral form is more powerful, yet it's underappreciated.

The first condition for a function to be differentiated is differentiable, if there's a single point which is non differentiable within the interval, the whole function cannot be differentiated.

For example, the differential form is useless when you deal with boundaries. If you have a metal ball, you can't just setup a function and differentiate it to get your result, metal boundaries make the function discontinuous. The integral form is not limited to continuous function, as long as the function is piecewise continuous, and integrable, you can integrate it easily with Gauss's surface and Ampèrian loop.

Stoke's theorem and Divergence Theorem can only applied if the loop/surface is simply connected, which means, you can use the Integral form to recover differential form if the function is continuous, or the other way round. However, If the function is discontinuous, you must use integral form, and you can't switch to differential form.

The integral form really is the "true" and most general Maxwell Equation.