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Let us take for instance E&M, in it when we deal with the failure of divergence theorem to give us the right expression when evaluating the volume integral of divergence of electric field over space, we introduce the concept of divergence of unit radial vector by radius squared being actually equal to dirac delta at origin.

Another approach I saw to solving this discrepancy is to say that we are integrating over a volume with the points which electric charge exist as removed. In such a space with holes, the divergence theorem doesn't hold. This second approach seems conceptually much simpler to me because we don't have to deal with the issue of defining a distribution and such.

What are the rigorous differences between these approaches and are they equivalent?

Refer

tryst with freedom
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  • "In such a space with holes, the divergence theorem doesn't hold. This second approach seems conceptually much simpler to me" -- Yes that prevents the need for distributive sense of the equations, but this is somewhat untidy. Real physical space does not have holes, and it is better (even pedagogically) to have theorems that are valid as generally as possible. Extending quantities and theorems to singular distributions is a way to do that.

     – Ján Lalinský May 25 '22 at 13:59
  • "Another approach I saw to solving this discrepancy is to say that we are integrating over a volume with the points which electric charge exist is removed." Where did you see this approach? It sounds like you are referring to a treatment of magnetic monopoles (as e.g. in this answer of mine), but I think you have misunderstood the problem: In that case we don't have the option to use a $\delta$ because $\mathrm{d}F = 0$ is a necessary consequence of the definition $\mathrm{d}A = F$ - we can't have $\mathrm{d}F = \delta$. – ACuriousMind May 25 '22 at 14:29
  • I thought it was being directly being hinted at in Needham's VDG but I suppose it is wrong after discussing with peek aboo @ACuriousMind – tryst with freedom May 25 '22 at 14:43
  • What about the dipole field? Divergence would give you delta' so from div and curl outside the singularity it is indistingushable. – lalala May 25 '22 at 19:08

3 Answers3

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The approach you describe appeared in a recent answer of mine, so I'll explain what I meant a bit more thoroughly.

The typical statement of the divergence theorem in $\mathbb R^3$ goes as follows. Let $B$ be a simple (i.e. non self-intersecting) closed surface which is continuously differentiable (so it has a well-defined outward pointing normal vector $\hat n$ at each point). By the Jordan-Brouwer separation theorem, this surface is the boundary of a bounded, connected interior region $U$. Given a vector field $\mathbf F$ which is continuously differentiable on $U\cup B$, we have that

$$\int_B \big(\mathbf F \cdot \hat n\big) \mathrm dS = \int_U \mathrm{div}(\mathbf F) \mathrm dV\tag{$\star$}$$

Now consider the vector field $\mathbf F = \hat r/r^2$, with $B$ a simple closed surface enclosing the origin and $U$ the region enclosed. Naively, one might compute the divergence of $\mathbf F$ and find that $\mathrm{div}(\mathbf F)= 0$, suggesting that both sides of $(\star)$ vanish. However, this is not so, because $\mathbf F$ is not continuously differentiable (or indeed, even defined) at $r=0$.

As a result, $\mathbf F$ does not satisfy the requirements for the divergence theorem to apply. We could try to fix this problem by removing the origin from $U$ since $\mathbf F$ is continuously differentiable (with vanishing divergence) on $U-\{\mathbf 0\}$, but that still fails to satisfy the conditions of the divergence theorem because $U-\{\mathbf 0\}$ is not the interior of $B$.

This certainly resolves the apparent paradox - that $\mathrm{div}(\mathbf F)$ vanishes on the domain of $\mathbf F$ but the left-hand side of $(\star)$ does not - but it doesn't really tell us how to move forward with our analysis. We're doing physics here, and we'd like to talk about point charges after all. By extending the divergence theorem to include distribution-valued quantities, we can simply say that the divergence of the electric field due to a point charge is proportional to $\delta$, which matches nicely with the intuition from $\nabla \cdot \mathbf E \propto \rho$.

J. Murray
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  • I'm having difficulty reconciling this with peek-a-boo's answer which in my naive view seems to be in direct contradiction to what you are saying. They say that it is not about topological nature of the space at all while you say it is – tryst with freedom May 25 '22 at 17:23
  • @Aplateofmomos I agree that our answers appear to be contradictory. I would be interested to see the conditions which peek-a-boo mentions under which the divergence theorem holds. I suspect one could make the case that if both $\mathbf F$ and $\mathrm{div}(\mathbf F)$ approach well-defined limits as $r\rightarrow 0$, then the hole in $U-{\mathbf 0}$ could be "patched" insofar as the LHS and RHS of $\star$ would be equal; in other words, in order for the divergence theorem to explicitly fail, either $\mathbf F$ or $\mathrm{div}(\mathbf F)$ must have a singularity. – J. Murray May 25 '22 at 17:55
  • @Aplateofmomos If that's the case (which again, I suspect is true but would need to think about it for a bit), then one could make peek-a-boo's case. On the other hand, it's possible that the singularities in question may correspond to a lack of differentiability rather than the fields becoming unbounded, but I can't think of a good example of that off the top of my head. I'll update further after some thought. – J. Murray May 25 '22 at 17:58
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    The version of Stokes/Divergence I had in mind is the following: "Let $M$ be a $C^2$ $n$-dimensional smooth manifold, $\omega$ a $C^1$ differential $(n-1)$-form on $M$ and $A\subset M$ an open set. If $A$ has $C^1$-regular boundary and $\overline{A}\cap \text{support}(\omega)$ is compact, then $\int_Ad\omega=\int_{\partial A}\omega$". Note that there are of course minor variants in how the theorem can be phrased, and that one has to be slightly careful with how to define the boundary etc. Also, we can allow for rougher boundaries with more effort. – peek-a-boo May 25 '22 at 18:04
  • In this formulation, and with your notation the set $U\setminus{0}$ is an open set whose topological boundary is $B\cup{0}$. The surface $B$ is nice, and the singleton ${0}$ is just that: a singleton, whose effects aren't significant (it's a set of atleast $(n-2)$-dimensional measure zero). So the theorem can be applied safely provided we have forms/vector fields which are smooth in a neighborhood of $\overline{U}$ (the closure in the ambient $\Bbb{R}^3$); unfortunately the fields of a point charge do not satisfy this. – peek-a-boo May 25 '22 at 18:04
  • @peek-a-boo Right, okay. So I think our ostensibly contradictory answers are compatible in the following sense. Utilizing your variant of the theorem, the spirit of my answer is that if $\omega$ is not defined at ${0}$, then $\int_{U\backslash{0}} \mathrm d\omega = \int_{B\cup{0}}\omega$ doesn't even make sense to write, so the theorem doesn't work. I believe that your point is that if $\omega$ can be extended in a smooth way to all of $U$, then the integration over the singleton point ${0}$ does not contribute, and so we recover the theorem. Is that accurate? – J. Murray May 25 '22 at 18:17
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    @J.Murray yes precisely, the formulation I have requires it to be smoothly defined in a small neighborhood of $\overline{U}$, so for example a very mild field like $\mathbf{G}(x,y,z)=(x,y^2,z)$ would satisfy this requirement, so the divergence theorem would apply to this field, even though the domain $U$ has a hole (you could even apply it to a spherical annular region, say $1<r<2$; being careful with boundary orientations of course). – peek-a-boo May 25 '22 at 18:21
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    @peek-a-boo Okay. Then I propose we are saying the same thing, using different formulations of the divergence theorem. In my (more restrictive) formulation, the region over which we integrate $\mathrm{div}(\mathbf F)$ must be the interior of a simple closed surface, so the presence of a hole (which is a topological statement) spoils things. In your more general formulation, we permit more general regions as long as $\mathbf F$ and its divergence are smooth in a neighborhood of its closure, so whether the theorem holds is not purely a statement about $U$ [...] – J. Murray May 25 '22 at 18:32
  • [...] but also the behavior of $\mathbf F$ in a neighborhood of $\overline U$. – J. Murray May 25 '22 at 18:32
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    yup we're in agreement here! – peek-a-boo May 25 '22 at 18:37
  • I'm accepting this as it seemed to be the one most closest to what I asked while using a variant of Stoke's theorem which seems closest to the doubt. However points to peek a boo as well for providing the general version. – tryst with freedom May 25 '22 at 20:41
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The second paragraph isn't right. The divergence theorem holds provided all its conditions are satisfied. The problem with the point charges is not that there is a hole in the region. Actually, there is no hole in the region (by region I mean where the point charge lives, and that is at the origin of $\Bbb{R}^3$); maybe you meant there is a hole in the domain ($\Bbb{R}^3\setminus\{0\}$) of the fields involved. However, the divergence theorem doesn't really care about the topology of your ambient space, nor whether or not the domain of the fields has any 'holes'. Rather that the fields blow up near the origin.

As for why textbooks introduce $\delta$, there are several reasons

  • It is convenient mathematically (though depending on the student it may or may not be helpful, depending on how it is presented). Once you learn some basic mechanics of manipulating $\delta$, you can easily arrive at correct answers and usually, ones first intuition is right.
  • More significant I think is it allows us to characterize the nature of the singularity. For example, consider on the real line the functions $f_1,f_2,f_3:\Bbb{R}\setminus\{0\}\to\Bbb{R}$, $f_1(x)=\frac{1}{\sqrt{|x|}}$, $f_2(x)=\frac{1}{x^2}$ and $f_3(x)=\frac{1}{x^4}$. These functions satisfy $\lim\limits_{x\to 0}f_1(x)=\lim\limits_{x\to\infty}f_2(x)= \lim\limits_{x\to 0}f_3(x)=\infty$, but clearly $f_2$ is 'more singular' than $f_1$ (e.g $\int_{-1}^1f_1<\infty$ but $\int_{-1}^1f_2=\infty$), and $f_3$ is more singular than $f_2$. Similarly, if we consider a function like $f(x)=x^3+\frac{1}{x^5}$, then we'd say $f$ has a $4^{th}$ order singularity at the origin (more precisely a pole of order $4$). Knowing the nature of singularities of the objects under consideration is very valuable, and since point charges appear all over the theory, it is convenient to be able to characterize singularities by saying "the electric field diverges like that of a point charge".
peek-a-boo
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Your premise is wrong: Excluding points of charge from the space(time) might seem "simpler" to you (it's certainly mathematically less involved!), but it's really a non-sensical thing to do: Sure, you could say "well, I guess my electric field just isn't defined on the locations of point charges", but that's physically an extremely strange procedure.

What you're saying in that case is that we have $\nabla \cdot E \propto \rho$ as the rule to determine $E$ from $\rho$ as long as the charge density is some nice differentiable function, and then for point charges we pull out an entirely different rule where $E$ is the superposition of $\frac{1}{r^2}$ fields centered around the locations of the point charges but not defined there. That's a really bad physical theory: Why are there two sets of rules for two situations when one seems to be the limit of the other - literally, in the sense that you might think about nascent $\delta$-functions as charge densities $\rho_\epsilon$?

This isn't "simple", it's more complicated - instead of a single equation governing the physics you have a single equation for one kind of situations and a weird ad-hoc rule for the other kind! What matters is the simplicity of the physical theory in terms of special cases and assumptions, not how simple you find the math that's going on.

Physical theories are supposed to be parsimonious in their explanations and special cases (cf. Occam's razor) and it's really much simpler in this case to look at nascent $\delta$-functions $\rho_\epsilon$ and say that the equation $\nabla \cdot E \propto \rho_\epsilon$ still holds as $\nabla \cdot E\propto \delta$ in the distributional sense when $\epsilon\to 0$ and isn't suddenly replaced by a completely different rule.

ACuriousMind
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  • Counterpoint, there are no "ad-hoc rules" in the case excluding distributions. The equation $\nabla\cdot E\sim\rho$ is still satisfied but with $\rho=0$. Instead, charge is defined as $\int_S E\cdot dA$ where $S$ is a closed 2-surface and these are the only rules.

    This is more consistent with a Noether-theoretic approach ("Komar-like" integrals, surface charges etc.) and there are situations where distributions can lead to nonsensical results for example when they need to be multiplied or when other functions are included that are too singular to be distributions. At least I certainly...

     – Bence Racskó May 26 '22 at 15:47
  • ...don't understand the animosity, since the whole "topological defects = sources" paradigm seems to be quite popular in the more technical field theory literature. – Bence Racskó May 26 '22 at 15:48
  • @BenceRacskó 1. Sure, the equation is satisfied but it no longer is useful for determining the field $E$ from a given $\rho$, since the $\rho$ of empty space and of a point charge are the same. 2. In the case of topological defects, the situation I'm describing here is usually not what we have: For instance, I see no issue with describing a magnetic monopole as a defect, because we don't have a large range of observations of fields of continuous magnetic densities of which it should be a limit, meaning the argument I'm giving in my answer does not apply. – ACuriousMind May 26 '22 at 16:03
  • (Also, I'd argue the integral formulation of charge is just the distributional approach: If you say that "the charge" of a point charge is defined by $\int_S E = q$ when its location is inside $S$ and $\int_S E = 0$ when it is outside, that's just more or less the definition of a $\delta$-distribution at that location) – ACuriousMind May 26 '22 at 16:08