# Gauss divergence theorem (GDT) in physics

Some of the statements for $$GDT$$ which I find in modern textbooks (both electromagnetism and multivariable calculus textbooks) are:

Let $$D$$ be a regular, three dimensional domain whose boundary $$S$$ is an oriented, closed surface with unit normal field $$\mathbf{\hat{N}}$$ pointing out of $$D$$. If $$\mathbf{F}$$ is a smooth vector field defined on $$D$$, then....(the equation)

(2) Calculus: James Stewart

Let $$E$$ be a simple solid region and let $$S$$ be the boundary surface of $$E$$, given with positive (outward) orientation. Let $$\mathbf{F}$$ be a vector field whose component functions have continuous partial derivatives on an open region that contains $$E$$. Then....(the equation)

These multivariable calculus textbooks are imposing strong sufficient conditions. However in the textbooks around Maxwell's time (in the later part of the 19th century), I can find simple statements of $$GDT$$. The following are "statement as well as elementary proof" of GDT from late nineteenth century physics textbooks.

(1) Maxwell's treatise Vol I 1873 Condition: Vector field must be continuous and finite

(2) Heaviside's electrical papers Vol I 1892 Condition: Vector field must be continuous and finite

(3) The mathematical theory of electricity and magnetism 1911 Condition: Vector field must be continuous and finite

So as a Physics student, which statements shall I use? The more difficult modern mathematical one or the one which can be found in the 19th century books I mentioned.

If the statement for $$GDT$$ is really as simple as in old textbooks and those statements are perfectly valid, what is the benefit to complicate it in modern multivariable textbooks?

• So how does maxwell calculate the divergence of a continous non-difgerentiable vector field? – lalala Mar 17 at 20:07
• As I remember, there are a lot formulation of Gauss theorem. I mean that in multivariable calculus books authors usually use that formulation (i.e. conditions) which is the easiest to proof. The criterion of easy proof is the minimal number of required definitions and lemmas. For instance, my lecturer proofs these theorem with continous functions in closure of domain and continous & bounded derivatives in domain for 3D case. The another lecturer proofs generalized theorem (it is called Stoker theorem for differential forms) and use less stronger conditions. – Artem Alexandrov Mar 17 at 20:09
• I mean that as physics student, you usually should not care about the conditions for functions and just use the result of theorem. – Artem Alexandrov Mar 17 at 20:10
• Is Maxwell's statement rigor enough to be applicable to electrostatics whatever be the finite continuous charge density? Please explain why? Please also explain why multi variable statement is not necessarily in electrostatics. – Oliver Mar 17 at 20:57
• It's up to you. You'll never go wrong with the math definition. You will get a lot of math problems wrong if you use the physics definition, but you will get almost no real physics problems wrong. – knzhou Mar 18 at 0:57

The correct conditions to apply Gauß theorem are the ones stated in the mathematics books. Textbooks and articles in physics (especially the old ones) do not generally go through the list of all conditions mainly because

1. Physicists have the (bad) habit of first calculating things and then checking whether they hold true (I say this as a physicist myself)

2. Fields in physics are typically smooth together with their derivatives up to the second order (because they solve second order partial differential equations) and vanish at infinity.

This said, there are classical examples in exercises books where failure of smoothness/boundary conditions lead to contradictions (therefore you learn a posteriori): an example of such a failure should be the standard case of infinitely long plates/charge densities where the total charge is infinite but you may always construct the apparatus so that the divergence of the electric field is finite or zero (due to symmetries), the trick being that for such infinite distributions the fields do not vanish at infinity (and hence Gauß theorem does not hold in the first place).

I think it leads to confusion with Gauss's law to call the divergence theorem 'Gauss's theorem'. Gauss's law does not mention divergence. The divergence theorem was derived by many people, perhaps including Gauss. I don't think it is appropriate to link only his name with it.

Actually all the statements you give for the divergence theorem render it useless for many physical situations, including many implementations of Gauss's law, where E is not finite everywhere. That includes point charges and line currents, or any case where the divergence leads to the Dirac delta function.

A definition of the divergence in terms of the limit of a volume integral allows a divergence theorem that includes the delta function.