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In electromagnetism (electrostatics), we often come across the equation $\nabla \cdot \mathbf{E}=\frac{\rho}{\epsilon_0}$.

In order for this equation to be meaningful, $\mathbf{E}$ must be a differentiable vector field. Or at least we need to show that partial derivatives of $\mathbf{E}$ exist if we rotate the Cartesian coordinate system.

In my textbooks none of them has been shown. Yet they introduce the equation $\nabla \cdot \mathbf{E}=\frac{\rho}{\epsilon_0}$. How is it justifiable?

Qmechanic
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N.G.Tyson
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3 Answers3

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In physics we are only concerned about electric fields for which we can sensibly apply Mawell's equations, even if we have to reach for distribution theory to define derivatives when functions are discontinuous.

From a physical perspective, the only real discontinuities we encounter in electric fields are those from ideal point charges, and even those run into complications in quantum field theory. Point being, at the introductory level, there's no harm in imagining that every charge distribution resolves into ones that have finite density, and therefore produce electric fields that are smoothly continuous everywhere.

Sean E. Lake
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The equation is a condition that an electric field $\mathbf{E}$ must meet to be a "valid" field for the situation in question, which is specified (at least in part, though this is tagged "electrostatics") by the charge distribution, $\rho$.

The fact that a differential equation must be satisfied logically implies that the any functions going into it that might be solutions must be differentiable - if it's not, you cannot talk about the derivative to begin with to test if it satisfies the equation.

The same goes with Newton's laws of motion - you don't need to state that the position function be twice differentiable with respect to the time, that comes with the fact that that is necessary for it to even be a candidate to go into the equation in the first place.

Even more elementarily, it goes with algebraic equations:

$$\frac{1}{x} = 5$$

$x = 0$ doesn't solve the equation because the LHS is undefined. Likewise, if $\mathbf{E}$ is something non-differentiable, it does not satisfy

$$\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}$$

because the left-hand side is undefined. Undefined can't equal a defined quantity.

There is no need to state as a postulate what is a logical consequence. You can, but it's redundant and generally speaking, independent axioms/postulates are preferred for the basis of theories.

The_Sympathizer
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  • Thanks for the answer... Your answer seems logical +1. But I need to ensure the equation holds everywhere and nowhere undefined. Showing the existence of partial derivatives seems the only way. Am I right? – N.G.Tyson May 21 '19 at 10:52
  • If it's undefined anywhere, it can't equal the right-hand side. – The_Sympathizer May 21 '19 at 10:53
  • Unless, of course, you restrict the domain under consideration. – The_Sympathizer May 21 '19 at 10:59
  • That said, if you have a candidate solution, then yes, to show that it solves the equation, you technically have to first show that that particular solution is differentiable. But that's different from asking why solutions have to be differentiable, or how it is valid to state the equation as a condition without also explicitly stating differentiability as a requirement alongside it. – The_Sympathizer May 21 '19 at 11:00
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The partial derivatives of the electric field do not always exist. They won't exist, for example, at a point charge, line charge, or charged plane -- situations where there is a singularity in the charge density. In these situations, Gauss's law still holds in integral form, if the Gaussian surface doesn't intersect any singularities.

None of this causes any foundational issues in physics, because we can only verify our physical theories down to some finite scale.