Are Maxwell's laws mathematically precise?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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 ,