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Divergence of r / r^3.

  1. Sep 24, 2013 #1
    I have searched the forums and the internet to see various discussions about the divergence of an electric field, or more directly, the divergence of r/r^3. I still don't understand this "spike at r = 0" idea, and really don't believe it. It simply seems to be an idea that fills in the gaps of what really should be a plain definition.

    If I write the radial vector in cartesian coordinates, something of which I am more familiar with, I can by the definition of divergence calculate ∇(r/r^3) for r ≠ 0. It is clear that no matter how small we let r become, the divergence is zero. We cannot say anything about the divergence for r = 0. I understand this.

    We then turn to the divergence theorem. We compute a volume integral of the divergence of r/r^3 where this volume contains r = 0, by computing a closed surface integral where this surface is the volume boundary. We get 4∏ for which certainly is not zero.

    This is where I have seen the "spike at r = 0" idea come in to explain things, to render the previous not a contradiction.

    My view, however, is that the volume integral simply does not make sense when it includes r = 0. It seems undefined. The surface integral is well defined because there is no division by zero.

    Rather than saying there is a spike at r = 0 or the divergence theorem is wrong, couldn't we say that because the volume integral is not well defined and the surface integral is, that we define the the divergence of r/r^3 to be 4∏δ(r)? That is, we leave no room for misleading ideas?

    I haven't looked at a good proof of the divergence theorem, so I ask, need it hold for such cases as this? I am sure there are properties that vector fields must hold in order for the divergence theorem to hold on them. For example, if it holds for continuous vector fields by not necessarily discontinuous vector fields, the above would not necessarily be a contradiction. We could then construct an new object to fill in what is undefined. In this case, we use the Dirac delta function.

    Nevertheless, I just want to make sure that we are defining the volume integral containing r = 0 to be 4∏ because it doesn't seem to give rise to contradictions, not actually saying that the summation of volume elements over ∇(r/r^3) could ever be 4∏. That seems absurd to me.
     
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  3. Sep 25, 2013 #2
    I think the answer to your question is that we actually define the divergence of r/r^3 to be 4∏δ(r). Because Dirac's delta function is not actually a "function" as defined by mathematicians and because it is proven that r/r^3 and 4∏δ(r) are equivalent, we cannot treat the volume integration as a summation of volume elements as we would usually do. Instead, the volume integration is directly linked to the definition of delta function (mainly the integral over all space = 1), which suggests that the divergence of r/r^3 is basically defined to be 4∏δ(r).
     
  4. Sep 25, 2013 #3

    vanhees71

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    The correct expression in the sense of distribution theory is indeed
    [tex]\vec{\nabla} \cdot \frac{\vec{r}}{r^3}=4 \pi \delta^{(3)}(\vec{r}).[/tex]
    The proof is to show that
    [tex]\vec{\nabla} \cdot \frac{\vec{r}}{r^3}=0 \quad \text{for} \quad \vec{r} \neq 0,[/tex]
    which is simply done by brute-force differentiations (most conveniently in spherical coordinates).

    Then let [itex]V[/itex] be any volume containing the origin in its interior. Further let [itex]S_R[/itex] be a ball around the origin fully contained in the interior of [itex]V[/itex]. Then you can show by using Gauß's integral theorem that
    [tex]\int_{\partial V} \mathrm{d}^3 \vec{F} \cdot \frac{\vec{r}}{r^3} = \int_{\partial S_R} \mathrm{d}^3 \vec{F} \cdot \frac{\vec{r}}{r^3}.[/tex]
    Now it's very easy to evaluate the latter surface integral to prove that it indeed equals [itex]4 \pi[/itex].

    This is a good hint for the above equation. To make the argument strict you have to use the test-function definition of distributions, showing, integrating by parts that
    [tex]\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{r} f(r) \vec{\nabla} \cdot \frac{\vec{r}}{r^3} = 4 \pi f(0),[/tex]
    for any test function [itex]f[/itex] (e.g., in the space of quickly vanishing [itex]C^{\infty}(\mathbb{R}^3)[/itex] functions).
     
  5. Sep 25, 2013 #4
    You're correct to feel suspicious but that doesn't mean you are correct that the integral is meaningless. There is a whole topic in mathematics designed to make sense of integrals like this one. Look up theory of distributions.
     
  6. Sep 27, 2013 #5

    arildno

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    Gauss' integral theorem does NOT hold for functions that are not defined at some point on the interior.

    Rather, EVERY SURFACE INTEGRAL around r=0 equals 4pi (readily shown), and every surface integral not including zero yields zero. This MOTIVATES the introduction of the Delta function, so that Gauss' theorem "holds" for some cases it specifically excluded from reasoning about.

    Later developments in mathematical theory, like the theory of distributions, places what we call the Dirac Delta function on a firmer theoretical basis.
     
  7. Sep 27, 2013 #6
    Thanks for the information everyone. I will need to take a look at theory of distributions.
     
  8. Sep 27, 2013 #7
    So Gauss's theorem for continuous distributions, not point charges? Any discrete collection of charges will have the electric field undefined where the charges lie.
     
  9. Sep 27, 2013 #8
    Physicists often skip over the rigorous mathematical justification for some identities that might seem suspicious. There are a lot of instances where

    (a) an expression does not have an obviously well-defined value, but

    (b) if you are careful enough, you can give definitions that make the expression meaningful, and

    (c) the answer you get after this careful work is the same as the answer the physicists got from straightforward but naive arguments.

    For example, if you are trying to integrate the divergence of the E field of a point charge, the integral is not obviously well-defined. One possible way to deal with this is to work with a spherical volume charge of radius R, and then take the limit as R goes to zero. The limit of the integral of the divergence is indeed 4*pi, as the naive application of Gauss's theorem suggested.
     
  10. Sep 27, 2013 #9

    clem

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    The divergence of r/(4 pi r^3) satisfies one definition of the delta function that
    [itex]\int_V\delta({\bf r})d^3 r=1[/itex] if r=0 inside V, and =0 if r does not = 0 inside V.
     
    Last edited: Sep 27, 2013
  11. Sep 27, 2013 #10

    clem

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    Incidentally, it is the 'divergence theorem', and not 'Gauss's theorem'.
    The divergence theorem relates Maxwell's divergence E equation and 'Gauss's law'.
     
  12. Sep 28, 2013 #11

    vanhees71

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    I've never heard about a "divergence" theorem. Gauß's (in the Russian literature the Gauß-Ostrogradsky theorem) states that
    [tex]\int_V \mathrm{d}^3 \vec{x} \vec{\nabla} \cdot \vec{V}(\vec{x})=\int_{\partial V} \mathrm{d}^2 \vec{F} \cdot \vec{V},[/tex]
    and that's what you need to use to define the divergence of the field that's singular at the origin.

    The physical argument is very simple. From Gauß's Law, i.e., the Maxwell equation
    [tex]\vec{\nabla} \cdot \vec{E}=\rho=q \delta^{(3)}(\vec{x}),[/tex]
    and the fact that the electric field of a unit-point charge at the origin is given by
    [tex]\vec{E}=\frac{1}{4 \pi} \frac{\vec{x}}{|\vec{x}|^3},[/tex]
    leads to the relation
    [tex]\vec{\nabla} \cdot \frac{\vec{x}}{|\vec{x}|^3}=4 \pi \delta^{(3)}(\vec{x}).[/tex]
    Of course, that's not a mathematically strict proof of this formula. Particularly the physics derivation of Coulomb's Law from Maxwell's equations for the electrostatic case uses
    [tex]\vec{\nabla} \times \vec{E}=0,[/tex]
    and that's a priori only true for [itex]\vec{x} \neq 0[/itex] for our Coulomb field.

    The mathematical proof must use the definition of the [itex]\delta[/itex] distribution, i.e., one has to show that for any test function
    [tex]\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} f(\vec{x}) \vec{\nabla} \cdot \vec{E}=f(0).[/tex]
    Since [itex]\vec{\nabla} \cdot \vec{E}=0[/itex] for [itex]\vec{x} \neq 0[/itex] you can as well write
    [tex]\int_{B_R} \mathrm{d}^3 \vec{x} f(\vec{x}) \vec{\nabla} \cdot \vec{E}=f(0)[/tex]
    for any ball [itex]B_R[/itex] with radius [itex]R>0[/itex] around the origin.

    I've never seen a strict proof. My suggestion is to regularize the field at the orgin, e.g., by
    [tex]\vec{E}_{\epsilon}=\frac{1}{4 \pi} \frac{\vec{x}}{(\vec{x}^2+\epsilon^2)^{3/2}}.[/tex]
    Then you have
    [tex]\vec{\nabla} \cdot \vec{E}=\frac{3}{4 \pi} \frac{\epsilon^2}{(\vec{x}^2+\epsilon^2)^{5/2}}.[/tex]
    For small [itex]\epsilon[/itex] that's a sharply peaked function around [itex]r=|\vec{x}|=0[/itex]. Thus we can take out the text function out of the integral. Then we find
    [tex]\int_{K_R} \mathrm{d}^3 \vec{x} f(\vec{x}) \vec{\nabla} \cdot \vec{E}_{\epsilon} \simeq f(0) \int_{K_R} \mathrm{d}^3 \vec{x} \vec{\nabla} \cdot \vec{E}_{\epsilon}.[/tex]
    The latter integral is easy to evaluate in spherical coordinates:
    [tex]\int_{K_R} \mathrm{d}^3 \vec{x} \vec{\nabla} \cdot \vec{E}_{\epsilon} = \int_0^{R} \frac{3 \epsilon^2 r^2}{(r^2+\epsilon^2)^{5/2}}=\frac{R^3}{(R^2+\epsilon^2)^{3/2}} \rightarrow 1 \quad \text{for} \quad \epsilon \rightarrow 0.[/tex]
    Of course, for a strict proof you have to make more formal the argument that one can take [itex]f[/itex] of the integral setting it to [itex]f(0)[/itex].
     
  13. Sep 28, 2013 #12

    clem

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    I had only seen it called the 'divergence theorem' in American texts. I see that it is different in Russian literature. Wikipedia gives a history:

    "The theorem was first discovered by Lagrange in 1762,[9] then later independently rediscovered by Gauss in 1813,[10] by Ostrogradsky, who also gave the first proof of the general theorem, in 1826,[11] by Green in 1828,[12] etc.[13] Subsequently, variations on the divergence theorem are correctly called Ostrogradsky's theorem, but also commonly Gauss's theorem, or Green's theorem."

    I guess each country: France, Germany, Russia, England could have a name for it. America came late into the game, so here it is usually called 'Divergence', reminiscent of our politics.
     
  14. Sep 28, 2013 #13

    clem

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    What I meant by my first post about defining the delta function by its operation in an integral, is that you never have to discuss its dependence near the origin, but just how it integrates.
     
  15. Sep 29, 2013 #14

    This is how I was beginning to view it. Of course though, it is easy to see that it works in any particular case. To show that any substitution of divergence of r / (4 pi r^3) for δ(r) never leads to a contradiction, or simply that it represents the same mathematical object in every context where one of the two is applicable, that seems like a difficult task.

    So, is the idea here to think of an electric field (not having a singularity) with the property that it approaches E for ε -> 0, and then using this to show that in the limit we get the same result(s)?

    It certainly seems reasonable. Given a surface integral were we take a spherical surface of radius R > 0 and integrating the flux of E over it, we get something well defined because the singularity is at r = 0. If we take a vector field that is continuously differentiable (as the one you gave for ε > 0) then the divergence theorem applies. Taking a surface integral over this vector field will give a result with the property that for ε -> 0, it becomes closer to that given by the E field with the singularity. However, given this vector field has no singularity, it should certainly be true that the volume integral of the divergence gives the same result. But, given the continuous relationship between an integral and its input (e.g. int[g(x)]dx -> int[f(x)]dx as g(x) -> f(x) ) it seems we could simply take the limit in place of this well defined vector field for one that is not so much, with the idea that at the singularity, it is a limiting case.
     
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