3
$\begingroup$

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?

$\endgroup$
1
  • 2
    $\begingroup$ Your divergence formula holds only at $r \ne 0$. $\endgroup$
    – velut luna
    Jun 5, 2017 at 12:52

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ 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, 2017 at 13:27
  • 1
    $\begingroup$ 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. $\endgroup$ Jun 5, 2017 at 13:41
  • 2
    $\begingroup$ 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. $\endgroup$ Jun 5, 2017 at 15:43
  • 1
    $\begingroup$ @user144342 You're welcome, I'm glad you found it useful! $\endgroup$ Jun 5, 2017 at 16:47
  • 2
    $\begingroup$ @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. $\endgroup$ Jun 5, 2017 at 16:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.