We have notions of derivative for a continuous and differentiable vector fields. The operations like curl,divergence etc. have well defined precise notions for these fields.

We know electrostatic and magneto static fields aren't actually well behaved. They blow up at the sources, have discontinuities and yet we use the same mathematical formulations for them as we would have done for continuous and differentiable vector field.

Why is this done ? Why are laws of electromagnetism(maxwell's equations) expressed in the so called differential forms when clearly that mathematical theory is not perfectly consistent with the electromagnetic field. Why not use a new mathematical structure ?

Is there a resource which can help me overcome these issues without handwaving at particular instances when the methods seem to give wrong results?

Also one of the major concerns is that, given a charge distributions, the maxwell equations in differential form, will always give a nicely behaved continuous and differentiable vector field solution. But the integral form (alone, not satisfying the differential form) can give a discontinuous solution as well. Leading to two different answers for the same configuration of charges. hence there is an inconsistency. Like there is an discontinuous solution for the boundary condition of 2D surface, the perpendicular component of the electric field is discontinuous. ( May be it is just an approximation) and actually the field is continuous but due to not being able to solve the differential equation we give such an approximation, but this isn't mentioned in the textbooks.

  • $\begingroup$ As in mechanics you make a continuum approximation. This is discussed at some point in every good text. The better ones also discuss the situation in which it breaks down. $\endgroup$ Aug 31, 2014 at 18:45
  • $\begingroup$ "There are other physical conditions which we impose from time to time. " Such as? $\endgroup$
    – akrasia
    Aug 31, 2014 at 19:46
  • $\begingroup$ "it is only after there is an absurdity in a result, do we come to our senses that oh! this is wrong and inconsistent" Could you include an example? It is not clear what do you mean exactly. $\endgroup$ Aug 31, 2014 at 22:23
  • $\begingroup$ @JánLalinský Like getting infinities at places, e.g. in the electrostatic energy derivation given in Griffiths' text $\endgroup$
    – Isomorphic
    Sep 1, 2014 at 14:38
  • 1
    $\begingroup$ you might want to look up Dirac delta function $\endgroup$ Sep 4, 2014 at 17:09

4 Answers 4


We have also the same notions of derivation, curl, etc... for functions that are less regular. When you write Maxwell's equations, you are writing a system of partial differential equations.

To investigate them, you have to specify the type of solution you look for (in the language of PDEs: classic, mild, weak...) and the functional space you set your theory in. A natural space for the electric and magnetic fields is $L^2(\mathbb{R}^3)$, because this is the energy space (where the energy $\int_{\mathbb{R}^3}(E(x)^2+B(x)^2)dx$ is defined). Also more regular subspaces, such as the Sobolev spaces with positive index, or bigger spaces as the Sobolev spaces with negative index are often considered.

These spaces rely on the concept of almost everywhere, i.e. they can behave badly, but only in a set of points with zero measure. Also, the Sobolev spaces generalize, roughly speaking, the concept of derivative. I suggest you take a look at some introductory course in PDEs and functional spaces. A standard reference may be the book by Evans, or also the monumental work by Hörmander.

Comment to the edit: it is not true that

the maxwell equations in differential form, will always give a nicely behaved continuous and differentiable vector field solution

Consider, e.g. the static equation \begin{equation*} \nabla\cdot E=\rho \; . \end{equation*} To investigate this equation, you have to give it a precise meaning. What are $E$ and $\rho$? Let's assume, as you said, that $\rho$ is some discontinuous function. Then it is quite strange to look for solutions of $E$ that are smooth and well behaved! We have mathematical objects that can behave even worse than discontinuous functions, and are called distributions. In particular, we are interested in the distributions dual to functions of rapid decrease, that are called $\mathscr{S}'(\mathbb{R}^3)$. Without entering into details, all functions in $L^p(\mathbb{R}^3)$, $1\leq p \leq \infty$ are distributions in $\mathscr{S}'$, as well as Dirac's delta function and its derivatives. And mathematically, it is perfectly legitimate to look at the divergence equation above in the sense of distributions: i.e. to search a distribution $E\in(\mathscr{S}'(\mathbb{R}^3))^3$ such that its distributional divergence $\nabla\cdot E \in \mathscr{S}'(\mathbb{R}^3)$ is equal to $\rho\in\mathscr{S}'(\mathbb{R}^3)$. Suppose that equation admits a solution, then this solution would not, in general, be a regular function, but a distribution. It may be, for example, a discontinuous function in $L^1$, or a sum of derivatives of the delta function.

Anyways, as I already wrote, it is necessary that you understand better the concept of Cauchy and boundary value problems for PDEs in functional spaces, and also the concept of classical, mild and weak solutions to understand fully the machinery behind Maxwell's equations, and the mathematical meaning of a solution for such a problem.

  • $\begingroup$ I thought in electrodynamics we implicitly assume that our space-time in minkowskian and the vectors are four vectors/euclidean. $\endgroup$
    – Isomorphic
    Sep 4, 2014 at 18:05
  • $\begingroup$ You can work over whatever manifold you like, with whatever objects on that manifold, and the main thrust of what @yuggib says holds true: you can allow these objects to be much less smooth than you might think, and yet still have meaningful definitions of derivatives and so forth. $\endgroup$ Sep 4, 2014 at 18:20
  • 1
    $\begingroup$ @Iota When I mention a functional space like $L^2$, it is intended to be the space to which each component of the vector fields $E$ and $B$ (these components are functions from $\mathbb{R}^3$ to $\mathbb{R}$) belongs; not the space-time of your physical theory. $\endgroup$
    – yuggib
    Sep 5, 2014 at 20:42
  • $\begingroup$ So to say that charge density is a scalar field, by that I mean at each point in space there is a number is wrong ? It is a different mathematical beast altogether which behaves like a real number at some points ? and similarly for the electric field component functions ? $\endgroup$
    – Isomorphic
    Sep 6, 2014 at 20:41
  • $\begingroup$ Also, my background is say upto multivariable calculus, where to pick up all these topics from ? some references will be helpful $\endgroup$
    – Isomorphic
    Sep 6, 2014 at 20:43

If I've understood you correctly, you want rigourous mathematical formalism to treat PDE solutions which are not differentiable or not square-integrable etc. That is, you can have those point charges with fields blown up on them, fields being not differentiable on boundaries and so on.

There exists a rigourous formalism to treat such things. It is called generalized functions, or distributions. They are nifty and mathematicaly correct way to use such a 'functions' as Dirac's delta or Heavyside step function.

To give you a taste of those beasts I'll loosely explain how it's going in 1D. Generaly speaking, generalized function is defined as a linear transform $(f, c)$ on carrier functions $c$, with those carrier functions being zero everywhere except some finite region and having derivatives of any order. An example of generalized function is the integral, $$(a, c) = \int_{-\infty}^{+\infty} \mathrm{d}x\,a(x) c(x),$$ where $a(x)$ is an integrable function. The generalized function which is more useful is $$(\delta, c) = c(0)$$ is the well known Dirac's delta function.

The derivative of a generalized function is defined as $$ (f', c) = (f, -c'). $$ It is easy to see the motivation for such a definition, as $$\int_{-\infty}^{+\infty} dx\,\frac {df(x)}{dx} c(x) = f(x)c(x)\Big|_{-\infty}^{+\infty} - \int_{-\infty}^{+\infty} dx\, f(x) \frac{dc(x)}{dx} = - \int_{-\infty}^{+\infty} dx\, f(x) \frac{dc(x)}{dx}$$ because of $c(x)$ being zero for an infinite $x$.

Using the derivative definition and our first example of generalized function, one finds for the square-integrable Heavyside $\theta(x) = \cases{0,& x < 0 \\ 1/2,& x = 0 \\ 1,& x > 0}$ that $$ (\theta', c) = (\theta, -c') = -\int_0^\infty dx\, \frac{dc(x)}{dx} = c(0) = (\delta, x), $$ i.e., $\theta' = \delta$ in sense of generalized functions.

That is, I've shown how a derivative of a non-differentiable function can be introduced ;) The formalism can be used to find non-differentiable solutions of PDEs, including Green functions.

You can read more of this technique applied to PDEs including electrostatic problems and wave equations in Vladimirov's 'Equations of Mathemetical Physics', and you can read a nice inroduction to generalized functions by Gelfand and Shilov, 'Generalized Functions'. Note that those books are freely available in Russian. Also, I have no doubt you can find other books on the subject more accessible to you.

P.S.: Of course boundary discontinuities are approximations, but they are useful ones as long as you won't dive deep into the microscopic. Also, the technique I was talking about don't really help with fundamental problems of electrodynamics like that blown up potential of a point charge -- it is QED what you need if you are too close to an electron, and even QED is not the ultimate answer to the problem yet, and it may be that there is no ultimate answer at all. But, again, as long as you are calculating macroscopic fields, it is usually OK to use classic electrodynamics.


One of the major issues that seems to be going on here is the notion of point and surface structures in our 3D world. When we define electrostatic fields by a distribution of point charges, we are being somewhat non-physical. If we keep zooming in on an electron, it's going to start not looking like a point charge anymore. Consider the Darwin Term in the Fine Structure Hamiltonian. The "rapid quantum oscillation smearing out the charge" removes the idea of a stationary point charge (albeit for the proton). What's more important in electrostatics is to say: in what region does our field need to be valid? The answer is only the region in which we're doing physics. To a good approximation, the electron behaves like a point charge as long as you're not on top of. Our point like charge distribution gives a field which is valid and a good approximation pretty much all the way down to the point itself. This doesn't need to be a problem though. Let us compare with an example from GR: In the normal derivation of the Schwarschild Metric in GR, we're only concerned with the region outside the spherical body. If the Schwarschild radius of the body lies outside the physical boundary of the spherical body, then our solution starts producing strange behaviours, and that's great, but we never try to go into the body itself using this metric. There's a region we're concerned with and we stick to it and it's all fine.

There's a similar issue with surface charges. Physically, you cannot confine charge to the plane. You can do a pretty good job approximating the plane, but random quantum behaviour puts a limit in place. We have to realise that the model is not a perfect representation of the world. But, the level we're usually looking at it, the normal E-Field is pretty much discontinuous across a boundary and our theory is the limit that it is discontinuous. That doesn't mean it isn't useful. If we start going right up to that boundary, our model is going to break down. As an aside, a spherical conductor is not a uniform distribution of matter. If it were, it would be a mathematical ball, and the Banch-Tarski paradox would have some very interesting things to say about that conductor. If we're going to say let's throw away this theory because the field isn't defined everywhere, I'd say we should have thrown it away sooner because of Banach-Tarkski. If we stick with Maxwell's Electrodynamics then we need to study it for itself to make sure we're always self consistent.

You mention the electrostatic energy derivation given in Griffiths text in a comment. I think you're talking about the Electric Potential calculation and the choice of reference point. If the charge distribution extends to infinity, we cannot use the point at infinity as the zero reference in calculating potential because the potential blows up at infinity. This is fundamental to the the theory we use. It is equivalent to trying to use the point at a point charge as the zero. We have to use the theory as is. If I remember correctly, Griffiths goes on to say that such problems do not occur in the real world because infinite distributions do not exist, which brings a small amount of peace. But you have to ask yourself if you're really surprised when unhelpful things happen because you playing with mathematical curiosities.

You ask about an alternative that doesn't have these issues? We don't use Maxwell's Electrodynamics to calculate electromagnetic cross-sections when colliding electrons. We use QED. In QED, the electrons don't have an Electric field like they do in Maxwell's. Electrons go in, something happens, electrons come out. That something is the exchange of virtual photons: the first electron excites the background field, and the excitation - the photon - propagates and then interacts with the other electron. There are many different 'paths' via which this can happen and we need to sum over them etc. Let's not get bogged down with Quantum Field Theory though, because you don't need to be an expert to know it's littered with infinities.

So should we use the full standard model lagrangian to do everything? Well no. It's probably worth taking a look at the two big reasons why. Firstly, it's not a theory of everything, it doesn't do gravity. Secondly, the computational demands of the dynamics of the 3 quarks + gluon plasma (+ whatever else is hanging around through pair production) is somewhat vast, never mind what's going on in my glass of water at the quark level. If we want to say something useful about my glass of water, we have a look at what assumptions we can make and find a simpler theory we can actually work with.

Really, what you've stumbled on to is the nasty truth of physics. We're used to hearing it all the time, but usually we don't realise quite what it means and how far reaching it is. Physics is about modelling the universe. Newton's Law of gravity is a model. It works in the weak field limit, but GR is "better". We accept it's not 100% but we know it's pretty darn accurate under certain conditions, and it's a hell of a lot easier to deal with. Here its obvious. But in the same sense, GR is wrong, the standard model of particle physics is wrong etc. There are some fundamental assumptions being made and we have to restrict ourselves to problems where the assumptions hold, or we go and win a noble prize.


Something like $\textbf{E}(\textbf{r}) \propto (\textbf{r}-\textbf{r}^{\prime})/|\textbf{r}-\textbf{r}^{\prime}|^{3}$ due to a static point charge does not belong to the appropriate class of functions (differentiable, square-integrable, etc.) for $\textbf{E}$ fields. In a mathematically precise sense, it is not a solution to Maxwell's equations.

Still, we can consider it only as an intermediate object (Green's function) rather than a final product. To obtain an actual solution, we always make a convolution with a source, viz., \begin{equation} \textbf{E}(\textbf{r}) = \int d^{3}\textbf{r}^{\prime} \rho(\textbf{r}^{\prime})\frac{\textbf{r}-\textbf{r}^{\prime}}{|\textbf{r}-\textbf{r}^{\prime}|^{3}}. \end{equation}

As for discontinuities of fields at boundaries, we can simply allow them to be discontinuous there. They don't lead to a pathological situation, e.g., the energy being infinite. The Maxwell's equations (in differential forms) are valid only in the interior of each region, and the solutions from different regions are matched according to boundary conditions.


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