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Short summary:

I want to calculate the divergence of $\overrightarrow{A}=\overrightarrow{(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2})}^T=\frac{1}{r} \overrightarrow{e_r} $. So I calculate $\nabla \overrightarrow{A}= \frac{1}{r}\frac{\partial}{\partial r}(r\cdot\frac{1}{r})+\frac{1}{r}\frac{\partial v_\phi}{\partial \phi}=\frac{1}{r}(\frac{r-r}{r^2})$ by using the quotient rule. Then i see that the derivative is always zero except for r=0 where its infinity, does this mean its the dirac delta function? And why do I apply the product rule and dont say $\frac{\partial}{\partial r}(r\cdot\frac{1}{r})=\frac{\partial}{\partial r}1=0$?

Longer background not absolutley necessary for the question:

If I calculate the divergence in cartesian coordinates I get: $\nabla \overrightarrow{A}= \frac{y^2-x^2+x^2-y^2}{(x^2+y^2)^2}$ wich means i cant do this $\frac{\partial}{\partial r}(r\cdot\frac{1}{r})=\frac{\partial}{\partial r}1=0$, but I still dont get the mathematical reason behind it.

The exercise wich made me think about the question was the following exercise about the 2d divergence theorem:

If I calculate the flux of A trough the boundary of the square with the corners $(\pm1,\pm'1)$ I get 2π. Normaly Id need to make a second boundary around r=0 wich would give me -2π cause the normal vector points in the other direction. Therefore the flux of A trough the boundary would be 0, just like the integral of the divergence of A over the area of the square without the pole point. Then the divergence theorem would apply. My exercise wants me to calculate the integral of the divergence of A over the area of the square with the pole point. So i need to know if it is the dirac delta function, wich would make sense since: $\int_0^{2\pi}\int_0^{r(\phi)}\frac{1}{r}\delta(r)r \ dr \ d\phi=2\pi$. But this would mean that the divergence theorem still applies, wich it shouldnt because there is a pole point.

Thanks in advance.

Joe Th
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1 Answers1

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Let $\vec r=\hat x x+\hat yy$ and $r=\sqrt{x^2+y^2}$. In terms of classical real analysis, we have for $\vec A=\frac{\vec r}{r^2}$, $r\ne 0$

$$\nabla \cdot \vec A=0$$

That is to say, that the divergence of $\vec A$ is zero in the domain of $\vec A$.

METHODOLOGY $1$:

As a distribution, we adopt the approach used in THIS ANSWER for the 3-dimensional case, to show that $\nabla \cdot \vec A=2\pi \delta(\vec r)$.

Proceeding, we regularize the function $\left(\frac{\vec r}{r^2}\right)$ in terms of a parameter, say $a$. To that end, let $\vec \psi$ be the regularized function given by

$$\vec \psi(\vec r;a)=\frac{\vec r}{r^2+a^2} \tag 1$$

Taking the divergence of $(1)$ reveals that

$$\nabla \cdot \vec \psi(\vec r; a)=\frac{2a^2}{(r^2+a^2)^2}$$

Now, for any sufficiently smooth test function $\phi$, we have that

$$\begin{align} \lim_{a \to 0}\int_S \nabla \cdot \vec \psi(\vec r; a)\phi(\vec r)\,dS&=\lim_{a \to 0}\int_S \frac{2a^2}{(r^2+a^2)^2}\phi(\vec r)\,dS\\\\ &=0 \end{align}$$

if $S$ does not include the origin.

Now, suppose that $S$ does include the origin. Then, we have

$$\begin{align} \lim_{a \to 0}\int_S \nabla \cdot \vec \psi(\vec r; a)\phi(\vec r)\,dS&=\lim_{a\to 0}\int_{S-S_{\delta}} \frac{2a^2}{(r^2+a^2)^2}\phi(\vec r)\,dS+\lim_{a\to 0}\int_{S_{\delta}} \frac{2a^2}{(r^2+a^2)^2}\phi(\vec r)\,dS\\\\ &=\lim_{a\to 0}\int_{S_{\delta}} \frac{2a^2}{(r^2+a^2)^2}\phi(\vec r)\,dS \end{align}$$

where $S_{\delta}$ is the circular region centered at $\vec r=0$ with radius $\delta$. For any $\epsilon>0$, take $\delta>0$ such that $|\phi(\vec r)-\phi(0)|\le \epsilon/(2\pi)$ whenever $0<|\vec r|< \delta$. Then, we have

$$\begin{align} \lim_{a \to 0}\left|\int_S \nabla \cdot \vec \psi(\vec r; a)(\phi(\vec r)-\phi(0))\,dS\right|&\le \lim_{a\to 0} \int_{S_{\delta}} \left|\phi(\vec r)-\phi(0)\right|\frac{2a^2}{(r^2+a^2)^2}\,dS\\\\ &\le \left(\frac{\epsilon}{2\pi}\,2\pi\right)\,\lim_{a \to 0}\int_{0}^{\infty}\frac{2a^2}{(r^2+a^2)^2}\,r\,dr\\\\ &\le \epsilon \end{align}$$

Thus, we have for any test function $\phi$,

$$\begin{align} \lim_{a \to 0}\int_S \nabla \cdot \vec \psi(\vec r; a)\phi(\vec r)\,dS&=2\pi \phi(0) \end{align}$$ and it is in this sense (i.e., as a distribution) that $$\bbox[5px,border:2px solid #C0A000]{\lim_{a\to 0} \nabla \cdot \vec \psi(\vec r;a)=2\pi \delta(\vec r)}$$

METHODOLOGY $2$:

Alternatively, we write in distribution $$\begin{align} \langle \nabla \cdot \vec A, \phi \rangle&=-\langle \vec A, \nabla \phi \rangle\\\\ &=-\int_0^{2\pi}\int_0^\infty \left(\frac{\vec r}{r^2}\right)\cdot \nabla \phi(\vec r)\,r\,dr\,d\phi\\\\ &=-2\pi \int_0^\infty \frac{\partial \phi(\vec r)}{\partial r}\,dr\\\\ &=2\pi \phi(0) \end{align}$$

as expected!

Mark Viola
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  • Thank you, I understand most of your answer but I dont understand why you can change the integral and the limes. According to the dominated convergence theorem Id need to find an integrable function g wich is bigger or equal than $\nabla \Psi(r,a)$. For a=0 i could use the dirac function, but it wouldnt work for for a>0. – Joe Th Apr 19 '20 at 14:41
  • @JoeTh You're welcome my pleasure. – Mark Viola Apr 19 '20 at 15:07
  • I don't know what you mean by the sentence "I don't understand why you can change the integral and the limes." What integral do think was "changed?" And what are "limes" in the context of this topic? We never invoke the Dominated Convergence Theorem here. The Dirac Delta is NOT a function. So it does not "dominate" any function. The Dirac Delta is a Distribution (Generalized Function). So, I suggest you read about distributions (generalized functions) and regularizations of the Dirac Delta. – Mark Viola Apr 19 '20 at 15:12
  • According to my script it is $\int \delta(x)f(x)dx=\lim_{a->0} \int \omega_a(x)f(x)dx$ (2) with $\omega_a(x)=\frac{1}{\pi}\frac{a}{x^{2}+a^{2}}$. But it also says $\delta(x)=\lim_{a->0}\omega_a(x)$ (2). But if I use (2) in (1) I get: $$\int \lim_{a->0}\omega_a(x)f(x)dx=\lim_{a->0} \int \omega_a(x)f(x)dx$$ But those two arent equal, so my qouestion is why the lim is on the left side of the integral, even tough $\delta(x)=\lim_{a->0}\omega_a(x)$ is inside the integral. Wich rule tells me to pull the limes outside of the integral as soon as i use expression (2)? – Joe Th Apr 19 '20 at 19:16
  • The limit cannot be interchanged with the integral. And the limit of $\nabla \cdot \psi $ is not the Dirac Delta since the Dirac Delta is not even a function. What we mean by $\lim_{x\to a} \nabla\cdot\psi=2\pi \delta(\vec r)$ is interpreted in the sense of distributions. – Mark Viola Apr 19 '20 at 21:21
  • The expleantion Methology 1 was really good, thanks alot. I now understand why $lim_{x→a}∇⋅ψ=2πδ(r⃗ )$ is the same. But im still not sure why $\nabla A=lim_{x→a}∇⋅ψ$. I know that $A=lim_{x→a}ψ$ and therefore $\nabla A=\nabla lim_{x→a}ψ$. But why is $\nabla lim_{x→a}ψ=lim_{x→a}∇⋅ψ$? Is it for the same reason that the lim stands before the integral? – Joe Th Apr 21 '20 at 18:00
  • The explanation is in the sense of distributions$$\nabla \cdot \vec A(\vec r)=\nabla \cdot \left(\lim_{a\to 0}\vec \psi\right)= \lim_{a\to 0}\nabla \cdot \vec \psi = 2\pi \delta(\vec r)\tag 1$$ In the classical sense, $$\lim_{a\to 0}\nabla \cdot \vec \psi\ne \nabla \cdot \left(\lim_{a\to 0}\vec \psi\right)$$But in distribution, we interpret $(1)$ to mean that for test function $\phi$ $$\int_V \phi(\vec r)\nabla\cdot\vec A(\vec r),dV=\lim_{a\to 0}\int_V \nabla \cdot \vec \psi(\vec r;a)\phi(\vec r),dV=2\pi \phi(0)$$if $V$ contains $0$ and $0$ otherwise. – Mark Viola Apr 21 '20 at 19:22
  • Ok, I think I get it. Thanks again for your help. I'll mark my question as answered. – Joe Th Apr 21 '20 at 21:00
  • Pleased to hear Joe. Well done. And you're quite welcome. Thank you for accepting the answer. – Mark Viola Apr 21 '20 at 21:03
  • @moderator Someone down voted this answer along with the linked answer. I view this as completely unfounded and malicious behavior. Would you please investigate this? – Mark Viola Jul 14 '22 at 21:12