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I have a vector field $\mathbf{A}(r) = \frac{1}{r^2}\mathbf{a}_r$. I am interested in finding the flux through a sphere enclosing some volume with radius $R$ and center at $r=0$.

  1. Calculating the divergence of vector field $\bf A$, I get zero divergence.

  2. Integrating the zero divergence over the volume of a sphere with radius $R$, I get zero net flux.

  3. But when I integrate the vector field over the surface, I get $4 \pi$.


  1. $$\Rightarrow \nabla \cdot \mathbf{A} = \frac{1}{r^2} \frac{\partial (r^2 \frac{1}{r^2})}{\partial r} = \frac{1}{r^2}\frac{\partial}{\partial r}(1)= 0$$

$$\int_V \nabla \cdot \mathbf{A}\ dV = \oint_S \mathbf{A}\cdot d\mathbf{S}$$

  1. $$\int_V \nabla \cdot \mathbf{A}\ dV = \int_V 0 \ dV = 0 \,?$$

  2. $$\oint_S \mathbf{A}\cdot d\mathbf{S} = \oint_S A\mathbf{a}_r\cdot \mathbf{n}_s dS = A(r)\oint_S dS$$

Sphere with radius R:

$$A(R)(4\pi R^2) = \frac{1}{R^2}4\pi R^2= 4\pi \neq 0$$

If the volume integral of the divergence of the field is zero, it means that the field is constant inside the sphere, or that the field strength inside the sphere in total grows and shrinks by the same amount.

If the surface flux integral of the vector field is equal to some positive constant, then the magnitude of the field is not constant on the boundary of the sphere. This means that the field at some point has to grow more than it shrinks inside the sphere, resulting in a divergence $\neq 0$.

So obviously $4\pi \neq 0$... What is happening here?

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    $\begingroup$ Your divergence formula holds only at $r \ne 0$. $\endgroup$ – velut luna Jun 5 '17 at 12:52
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The hypotheses of the Divergence Theorem require that the vector field be defined and $C^1$ on all of the solid $V$, but here the vector field ${\bf A} := \frac{1}{r^2} {\bf a}_r$ is not defined at zero (indeed, as $r \to 0$, $\bf A$ is unbounded). Thus, the Divergence Theorem does not apply. The computation $$\iint_S {\bf A} \cdot d{\bf S} = 4 \pi$$ is, by the way, correct (for the usual outward-pointing orientation of the sphere $S$).

There's an analogous example for Green's Theorem: The vector field $${\bf X} := \bigg(\underbrace{-\frac{y}{{x^2 + y^2}}}_P \partial_x + \underbrace{\frac{x}{{x^2 + y^2}}}_Q \partial_y\bigg)$$ satisfies $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0$, but the contour integral along the unit circle (oriented anticlockwise) is $$\oint_{\Bbb S^1} {\bf X} \cdot d{\bf s} = 2 \pi,$$ which does not coincide with $$\iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA = 0,$$ where $D$ represents the unit disk. Again, there is no contradiction, because $\bf X$ is badly behaved near zero and so does not satisfy the hypotheses of Green's Theorem.

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  • $\begingroup$ Thanks! I then wonder how the divergence of the vector field can be zero, when it is decreasing radially outwards. It should be negative then? Is this the reason why we use the surface integral relationship to compensate for these situation with a vector field that is not defined on some point(s) inside the volume enclosed by the surface? $\endgroup$ – Kristian Jun 5 '17 at 13:27
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    $\begingroup$ The divergence can be regarded as a measure of the change of volume under the "flow" of the vector field. If one takes a spherical shell of some positive thickness (and centered at the origin), then as it moves under that flow of $\bf A$, its radius increases, but its thickness decreases at just the rate that its total volume is fixed. I do not understand your second question. $\endgroup$ – Travis Willse Jun 5 '17 at 13:41
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    $\begingroup$ It should also be noted that if you're allowed to use Dirac Deltas, the divergence of the original field is a constant multiple of a Dirac Delta, and so upon volume integration you get the same result as the surface integral. $\endgroup$ – Bob Krueger Jun 5 '17 at 15:43
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    $\begingroup$ @user144342 You're welcome, I'm glad you found it useful! $\endgroup$ – Travis Willse Jun 5 '17 at 16:47
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    $\begingroup$ @Bob1123 (+1) You're quite right that this can be generalized in various ways. Given the presentation of the problem here, though, I thought it was likely that OP is encountering this problem in an introductory multivariable calculus course---where the point of the problem is arguably to highlight exactly the distinction OP is asking about---so it seemed only appropriate to cite the form of the Divergence Theorem usually encountered there. $\endgroup$ – Travis Willse Jun 5 '17 at 16:56

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