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Calculate the flux of the field $$\mathbf{F} = k\left(\frac{\mathbf{r}}{|\mathbf{r}|^3}\right)$$ where $\mathbf{r}=\langle x,y,z\rangle$ out of an arbitrary closed surface not intersecting $(0,0,0)$.

My attempt

I get $$\operatorname{div} \mathbf{F} = 0$$ Using Gauss’s theorem I get that the flux crossing an arbitrary surface must be $0$ since no flux is produced.

The answer is however $4k\pi$ if the surface envelopes $(0,0,0)$ and otherwise it is $0$. How can this be true? How can any flux pass any surface if no flux is created anywhere? My understanding is obviously flawed, but I can’t pinpoint where.

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  • $\begingroup$ If the surface encloses the origin, then you get a problem that your divergence calculation is not correct at the origin. In fact in the sense of distribution theory there is a delta function divergence at zero, which gives this kind of behavior. This is really not that unusual in physics: a point charge at the origin creates a very similar effect. $\endgroup$ – Ian Aug 7 '17 at 19:56
  • $\begingroup$ You also can see by symmetry considerations that there must be a flux: if your surface is a sphere centered at the origin, note that your field points outward (or inward if $k<0$) at every point, so there must be net flux. $\endgroup$ – Ian Aug 7 '17 at 20:01
  • $\begingroup$ @Ian Thanks, that explains some of it. But how would I arrive at $4k\pi?$ What's the method I would use? $\endgroup$ – Heuristics Aug 7 '17 at 20:07
  • $\begingroup$ Try it for a sphere centered at the origin; in that case you can do a direct calculation. For a more general surface, artificially add another boundary term corresponding to a sphere centered at the origin, then note that the flux outside this sphere vanishes. (The other way to do it is to use distribution theory to identify the divergence of that field as a certain multiple of the Dirac delta function, but that requires some more setup.) $\endgroup$ – Ian Aug 7 '17 at 20:12
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$ \vec{F} = \left( \frac{kx}{(\sqrt{x^2+y^2+z^2})^{3}},\frac{ky}{(\sqrt{x^2+y^2+z^2})^{3}}, \frac{kz}{(\sqrt{x^2+y^2+z^2})^{3}}\right),$ $ (x,y,z)\in R^3\setminus\{0\}.$

We'll use spherical of coordinates:

$ x = r\cos(\phi)\cos(\theta),\ \ y = r\sin(\phi)\sin(\theta), \ \ z = r\sin(\theta).$

$ 0< \phi< 2\pi, \ \ -\frac{1}{2}< \theta < \frac{1}{2}\pi.$

Let

$ I =\left\{(\phi, \theta): 0<\phi< 2\pi, -\frac{1}{2}<\theta<\frac{1}{2}\pi \right\}$

and

$\Phi: I \rightarrow S. $

Differential form of flux

$\omega^2_{F} = \frac{k}{|r|^3} (xdy\wedge dz + ydz\wedge dx + zdx \wedge dy).$

Therefore

$ \Phi^{*}\omega_{F}(\phi,\theta) = \frac{k}{|r|^3} \left|\begin{matrix}x&y&z\\ \frac{\partial x}{\partial \phi}&\frac{\partial y}{\partial \phi} & \frac{\partial z}{\partial \phi} \\ \frac{\partial x}{\partial \theta} & \frac{\partial y}{\partial \theta}&\frac{\partial z}{\partial \theta} \end{matrix}\right|=\frac{k}{|r|^3}\left|\begin{matrix}r\cos(\phi)\cos(\theta)& r\sin(\phi)\cos(\theta)& r\sin(\theta) \\ -r\sin(\phi)\cos(\theta)& r\cos(\phi)\cos(\theta)& 0 \\ -rcos(\phi)\sin(\theta)& -r\sin(\phi)\sin(\theta)& r\cos(\theta) \end{matrix}\right|d\phi \wedge d\theta = k\cos(\phi)d\phi \wedge d\theta.$

$\Phi = \int\int_{(S)}\omega^2_{F}= +\int\int_{I}\Phi^{*}\omega_{F}=\int_{0}^{2\pi}d\phi\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}k \cos(\theta)d\theta = 4\pi k. $

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If a surface $S$ encloses the origin then the flux isn't defined on each point in the enclosed space, so you can't use the Divergence Theorem. To solve for that case draw a sphere $S_1$ around the origin s.t. $S_1$ is inside of $S$ (Since $S$ doesn't pass through the origin this is always possible). Now you can see that in $S-S_1$ the flux is $0$, as it's a closed surface which doesn't enclose the origin. Hence the flux in $S$ and $S_1$ is the same. Now it's fairly easy to calculate the flux in the sphere using the spherical coordinates

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  • $\begingroup$ No, you're wrong. If $S$ encloses the origin (but the origin does not lie on the surface), then certainly the flux is defined. But the hypotheses of the Divergence Theorem do not hold. $\endgroup$ – Ted Shifrin Aug 7 '17 at 23:28
  • $\begingroup$ @TedShifrin The flux function isn't defined at the origin, so you can't calculate the divergence of $F$ at the origin. If you try to exclude the origin for this cause the enclosed space isn't compact, which violates the conditions of the Divergence Theorem. Anyway I was trying to say that one can't use the Divergence Theorem when $S$ encloses the origin. Of course the flux is defined at $S$, which is why this problem is well-defined. $\endgroup$ – Stefan4024 Aug 7 '17 at 23:41
  • $\begingroup$ @TedShifrin We require $F$ to be continuously differentiable at $V$, but it's not even defined at the origin. As I said I was trying to point out why we can't use the Divergence Theorem. $\endgroup$ – Stefan4024 Aug 7 '17 at 23:43
  • $\begingroup$ I'm saying your first sentence is wrong. Read it carefully and read what I wrote carefully. There's no such thing as a "flux function." The flux across the surface $S$ is a numerical quantity, and it makes perfectly fine sense. $\endgroup$ – Ted Shifrin Aug 7 '17 at 23:46
  • $\begingroup$ Yeah I forgot to add on inside the surface S. So thanks for pointing it out. When it comes to terminology since English isn't my first language it might be tricky, but I guess it's clear enough so you can see to what I'm refering to. $\endgroup$ – Stefan4024 Aug 7 '17 at 23:50

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