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Electrodynamics makes heavy use of vector calculus, which in turn is about differentiation and integration of scalar and vector fields in $\mathbb{R}^3$. At this point everything seems fine to me, since the physical space is isomorphic to $\mathbb{R}^3$. However, Maxwell's laws rely on mathematical abstractions such as supposing that electric charge is a continuous variable.

For instance: $$\nabla \cdot \mathbf{D}=\rho$$ In the LHS, there is a local quantity, since the divergence of a vector field is defined at every point of space. On the other hand, in the RHS there is a global quantity, i.e., it can only be defined as a mean charge density in volumes much greater than the dimensions of the charge carriers involved. This is because electric charge is quantized. All charged particles have charges that are integer multiples of $\frac{1}{3}e$. Since the charge carriers are generally very small, macroscopically it seems that we can indeed define $\rho$ at every point of space, even though this statement does not make much mathematical sense. For instance, in a point of "empty" space (a point not belonging to the charge carriers), $\rho=0$.

Even though I am aware of the extraordinarily high predictive power of classical electrodynamics, the assumption of considering charges continuous instead of discrete seems pretty far-fetched. I'd like to know if, besides quantum applications, Maxwell's laws show some kind of error due to this assumption.

1__
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    You might want to have a look at Jackson's "Classical Electrodynamics", particularly the chapter "Radiation Damping, Self-Fields of a Particle, Scattering and Absorption of Radiation by a Bound System". It's basically about the many difficulties of modeling electric charge as classical point particles. – John Doty Mar 20 '21 at 14:11
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    By treating point charges as Dirac deltas in $\rho$, you can make Gauss's law distributionally correct with a Green's function. – J.G. Mar 20 '21 at 16:48
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    Some operations from calculus are defined on distributions. This allows for things like the dirac-delta distribution (sometimes called a function, but it shouldn't be). This allows certain generalizations of calculus to accomodate point particles and the like. This is a pretty nuanced topic, but there's plenty of material out there if you're interested (see any grad-level math book that covers Elliptic PDEs). – user37344 Mar 20 '21 at 17:50
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    @JohnDoty I think it's a code of conduct violation to inflict Jackson on the unsuspecting. – Mark H Mar 21 '21 at 08:22
  • Also see: https://en.wikipedia.org/wiki/Jefimenko%27s_equations – Matt Timmermans Mar 21 '21 at 19:49
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    Why do you consider continuity of charge a mathematical abstraction, but not the modeling of physical space by $\Bbb{R}^3$? –  Mar 22 '21 at 01:04
  • @user37344 I'd add to that, that the possible application to such problems of electromagnetism was precisely the reason that de Rham named his generalisation of distributions to differential forms "currents". But in fairness, these are a niche topic even for graduate mathematics. – mlk Mar 22 '21 at 08:41
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    Charge density is not "global" (your usage of this term is non-standard), whether you are considering a mean charge density or a distribution of point charges. "Global" refers to something associated with the entire space; "local" means something associated with arbitrarily small regions of space; "non-local" means everything in between. Both mean charge density and point charge density are local. What you are getting at is that mean charge density is approximate. – Yly Mar 22 '21 at 21:20
  • Precision vs Accuracy. – J... Jun 01 '21 at 10:57

5 Answers5

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The title of the question asks if Maxwell's equations are mathematically precise. The answer is certainly yes. Maxwell's equations are well-posed differential equations. There is no ambiguity about the symbols, and one can unambiguously check if a given set of fields and charge/current distributions satisfy the equations. In fact, Maxwell's equations enjoy many additional nice mathematical features, such as existence and uniqueness theorems, as well as beautiful and non-obvious symmetry properties such as Lorentz invariance and electromagnetic duality.

However the text asks a fundamentally different question: whether Maxwell's equations are a perfect model of the real world. The answer is surely no. Maxwell's equations do not describe many important phenomena, such as gravity. Even more to the point, Maxwell's equations describe a classical field theory, while the world we see around us is quantum.

Having said that, it is essentially a triviality in physics to say that any given model does not perfectly describe reality. Even the Standard Model and General Relativity are not complete theories of reality. The interesting question is not a binary "yes or no" question about whether a given theory is "right or wrong", but rather to understand where a given theory breaks down (see Asimov's brilliant essay The Relativity of Wrong). In the case of classical electromagnetism, it is precisely when dealing with small numbers of fundamental particles (a few photons, or a few electrons, say) that one needs to worry about quantum mechanics, and Maxwell's equations break down.

While Maxwell's equations are not a perfect representation of reality, I don't accept the OP's explanation for why they are imperfect. Maxwell's equations can accommodate discrete charge distributions -- simply insert a sum of three dimensional delta functions for $\rho$. Now, it is true that there is nothing in Maxwell's equations that says that the quantization of charge is logically required, which you may feel is a deficit in the theory. However, there is nothing in the entire Standard Model that requires charge quantization. So, even in our current best understanding of particle physics, the quantization of charge is an empirical fact, and not a consequence of other theoretical principles.

Furthermore, mathematically it is a valid approximation to replace a dense sum of discrete point charges by a smooth charge distribution. This is not a particularly deep observation. It is quite common to approximate a sum containing a large number of terms by an integral.

Andrew
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    I disagree that substituting a collection of delta functions for a continuous charge distribution is simple. If you stay on the well-trodden paths through this wilderness it works all right, but there are monsters lurking if you stray. Infinite self-energy is only one of them. See my comment on the OP. – John Doty Mar 20 '21 at 14:33
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    @JohnDoty Sure. But my reading of the OP's question is that they are concerned about approximating a static lump of $10^{23}$ charged particles with a continuous charge distribution -- computing the macroscopically observable field by using a smooth charge distribution can be proven to be a good approximation to a sum of point particles using Maxwell's equations. Issues with self-energy, causality, etc are a much more subtle and I don't think were part of the scope of the question. – Andrew Mar 20 '21 at 15:30
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    @Andrew: And note that when we compute (in the modern sense of the word) anything using Maxwell's Equations (or indeed, most other things involving differential equations) we're stuck with various discretizations anyway. See e.g. IEEE floating point standard. – jamesqf Mar 20 '21 at 17:24
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    @JohnDoty Isn't what you mention related to the fact it's a classical theory, and there are things that classical theory cannot describe? – user1079505 Mar 20 '21 at 18:59
  • @user1079505 Every theory has its limits. – John Doty Mar 20 '21 at 20:39
  • "Having said that, it is essentially a triviality in physics to say that any given model does not perfectly describe reality." Could you cite an example of quantum mechanics being an inaccurate model of reality? (i.e. where it predicts something that experiment has proven false) – user541686 Mar 21 '21 at 08:34
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    @user541686 "Quantum mechanics predicting something false" is a strawman for what I was trying to say. The point is that any given theory built by humans always has regions where it is incomplete or untested. We don't have a theory that tells us how Nature incorporates gravity into quantum mechanics. Our current theories don't explain dark matter and dark energy. We have only explicitly tested the Standard Model up to the scale of a TeV so we don't know whether it works well above this scale. All knowledge in science is provisional. – Andrew Mar 21 '21 at 12:12
  • I wasn't attempting to make a strawman. What I was getting at was, I understood the OP's question to be specifically about errors (false predictions), not about incompleteness (lack of predictions). I feel like you're muddying the difference between soundness and completeness in your answer. It's already obvious to everyone that Maxwell's equations are incomplete (they don't even mention gravity, let alone QM...). But while all of our theories are incomplete, but I'm not aware of us knowing they are all erroneous. So the interesting topic here is where theories actually break down – user541686 Mar 21 '21 at 12:40
  • @user541686 I agree this is an interesting question and said this in the answer. But I don't think that the OP identified a place where Maxwell's equations break down, just incomplete in the sense that they don't explain why charge is quantized (but neither does the Standard Model). I think the question itself mixed together a lot of different issues, and I wanted to distinguish at least some of them (not a criticism of the OP, this is subtle). Having said that, it's very possible I have not explained things the best way. If you want to write an answer from your POV, I'd like to read it. – Andrew Mar 21 '21 at 13:02
  • On the topic of our current theories: I would say that while we don't know exactly how the Standard Model (SM) breaks down, we do know it must break down. At the very least, the SM does not contain gravity or (at least as far as I know) a viable dark matter candidate. I would also say the issue is with the Standard Model, not with quantum mechanics, which is a framework and not a specific theory. – Andrew Mar 21 '21 at 13:02
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    I suppose I also don't see as sharp a distinction between "incomplete" and "incorrect" as perhaps you would argue for. In the Standard Model, if you have a collection of matter the size of the earth contained in the radius of the earth, and another collection of matter the size of the moon contained in the radius of the moon, and they were separated by the earth-moon separation, neither would collapse into earth- and moon- like objects and these clouds would not orbit each other. That's an incorrect prediction of the Standard Model. – Andrew Mar 21 '21 at 13:10
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    @user541686 Without a field theory that represents all of the fields in play, quantum mechanics predicts nothing at all. But clearly, we do not have a universal field theory, as we have observed phenomena not predicted by our best candidate. Neutrino oscillations, dark matter, dark energy, ... – John Doty Mar 21 '21 at 13:29
  • -1: This answers a question different than OP's and is misleading. "Maxwell's equations can accommodate discrete charge distributions -- simply insert a sum of three dimensional delta functions for ρ. " While the equations can be solved in the sense of distributions, distributions cannot be used to calculate anything non-linear. Pray tell, what is the electrostatic energy of a Coulomb field? Further, whether the coupled system of Newton's second law and Maxwell's equations with a point source can be self-consistent and physical is a highly non-trivial question. Cf. Rohrlich. – Robin Ekman Mar 23 '21 at 00:46
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Maxwell's equations are sometimes referred to as point equations, meaning that they apply at every point in space. That is true for $$\nabla\cdot\mathbf{D}=\rho , $$ because the divergence is only nonzero at a point in space if there is a nonzero charge density at that point in space.

The fact that charges are physically point charges therefore means that the divergence becomes Dirac delta functions at the locations of the charges and zero everywhere else.

Often, the distribution of point charges is modeled by a smooth charge density. In such cases the divergence will also give a nonzero function in that region.

So, in all cases (whether you consider the exact physical nature of charges or a more tractable model of the charge distribution), Maxwell's equation describe the behavior correctly.

flippiefanus
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  • I understand this. My doubt is whether considering a smooth charge density (which is almost every time done) is mathematically acceptable. – 1__ Mar 20 '21 at 04:52
  • The mathematics for a point charge is awkward. The reason is that one would then often try to do the calculation in spherical coordinates with the point charge at the origin. Well, the funny thing is that the origin is excluded from the domain in spherical coordinate and therefore the calculation does not produce the Dirac delta as expected. The best is to stay in Cartesian coordinates when you do the calculation. That would produce the expected Dirac delta function. – flippiefanus Mar 21 '21 at 04:23
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    @Lemoine A smooth charge distribution is fine from a mathematical point of view. Modeling charge as discrete points, however, is mathematically treacherous. Physicists carefully navigate around the traps that point charges put in their mathematical models while mathematicians insist "we mustn't have traps" and try to come up with rules to disarm them. – John Doty Mar 22 '21 at 12:18
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If you want to object to Maxwell's equations on quantum grounds, you can't cherry-pick just charge quantisation. You should also respect the Heisenberg principle: an electron wavefunction cannot be of zero extent. If you calculate the electron wavefunctions for atoms, molecules, metals, semiconductors, superconductors, etc., they are very much not Dirac deltas.

You say: if we look closely there are discrete particles, and the charge distribution is not a smooth function but a Dirac $\delta(x-x_0)$. I say: if we look even closer, we see that those particles are quantum and the charge density of, say, a hydrogen atom is actually $\sim e^{-|x-x_0|/a}$ outside the nucleus, where $a$ is some constant of the order of the Bohr radius. Inside the nucleus, it's trickier, but the proton is genuinely a finite-size object, so...

What is the charge density of a free electron? It's not $~\delta(x-x_0)$ because that is not allowed by Heisenberg. Position and momentum eigenstates are not physical. The most classical free electron states are coherent states and then you expect the charge density to be a Gaussian of width on the order of the de Broglie length, and not shorter than the Compton length. Newton and Wigner, Rev. Mod. Phys. 21, 400 (1949).

If you really want to be unassailable from quantum objections, you should also treat the electromagnetic field as a quantum field. By Ehrenfest's theorem Maxwell's equations hold for the expectation values of the quantised fields.

Urb
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Robin Ekman
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I think you are confused, the equations are merely mathematical constraints, we are lucky to get to apply. This is just the nature if differential equations.

marshal craft
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No , Maxwell's equation describe the "classical" behavior correctly . because besides quantum applications , classical physic rules with her "simplicity" and "flexibility" ,.. much more than the Maxwell's equations need to be accurate , when you start to think quantum applications you lost immediately this flexibility ,

Anisrael
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