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Please help me find this anti-derivative: $$\int{r^2\delta(x)\delta(y)\delta(z)dr}$$, with $\delta$ being the delta function and $r=\sqrt{x^2+y^2+z^2}$

Thank you in advance.

A Slow Learner
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    Note that $\delta (\vec r)=\frac{\delta(r)\delta(\theta)\delta(\phi)}{r^2\sin(\theta)}$. Then, $$\int_0^\infty r^2\delta(x)\delta(y)\delta(z),dr=\int_0^\infty r^2\delta(\vec r),dr=\int_0^\infty r^2\frac{\delta(r)\delta(\theta)\delta(\phi)}{r^2\sin(\theta)},dr=\frac{\delta(\theta)\delta(\phi)}{\sin(\theta)}$$ – Mark Viola Apr 30 '17 at 03:55

1 Answers1

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In transforming to spherical coordinates, we have

$$\delta (\vec r)=\delta(x)\delta(y)\delta(z)=\frac{\delta(r)\delta(\theta)\delta(\phi)}{r^2\sin(\theta)}\tag 1$$

Applying $(1)$ reveals that

$$\begin{align} \int_0^\infty r^2\delta(x)\delta(y)\delta(z)\,dr&=\int_0^\infty r^2\delta(\vec r)\,dr\\\\ &=\int_0^\infty r^2\frac{\delta(r)\delta(\theta)\delta(\phi)}{r^2\sin(\theta)}\,dr\\\\ &=\frac{\delta(\theta)\delta(\phi)}{\sin(\theta)} \end{align}$$

Formally, we can write

$$\int r^2\delta(x)\delta(y)\delta(z)\,dr=\int r^2\frac{\delta(r)\delta(\theta)\delta(\phi)}{r^2\sin(\theta)}\,dr=\frac{\delta(\theta)\delta(\phi)}{\sin(\theta)}\,H(r)+C$$

where $H(r)$ is the Heaviside Function.

Mark Viola
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  • Thank you. Could you help me with another anti-derivative, which actually follows directly from your answer?

    $$\int{\frac{1}{r^2}\frac{\delta(\theta)\delta(\phi)}{sin(\theta)}}dr$$

     – A Slow Learner Apr 30 '17 at 04:08
  • You're welcome! My pleasure. For this last one, $-\frac1r \frac{\delta(\theta)\delta(\phi)}{\sin(\theta)}$. – Mark Viola Apr 30 '17 at 04:11
  • Could you let me know the value of the function$$\frac{1}{r}\frac{\delta(\theta)\delta(\phi)}{sin(\theta)}H(r)$$ at the origin? – A Slow Learner Apr 30 '17 at 04:22
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    This isn't a function, it is a distribution. So, there is no meaning to assigning a value to it at a point. The function $H(r)$ can be defined at $0$ to be $0$. And its derivative (in the sense of distributions) is the distribution $\delta(r)$. HERE is a primer on the Dirac Delta. – Mark Viola Apr 30 '17 at 04:24
  • I ran into this integration when trying to solve the equation $$\nabla^2V=-\frac{q}{\epsilon}\delta^3(\vec{r})$$ for the electric potential due to a point charge at the origin. Would you be able to provide me with some insights regarding the physical meaning of the results you just gave me? – A Slow Learner Apr 30 '17 at 04:31
  • Note for $\vec r\ne 0$, $\nabla ^2 V=0$. And if we apply the Divergence Theorem in the sense of Distributions, we have $$\oint_{|\vec r|=\epsilon}\frac{\partial V}{\partial r},dS=-\frac{q}{\epsilon}$$The formal heuristic way to see this is that $$\int_{|\vec r|\le \epsilon}\delta(\vec r),dV=\int_0^{2\pi}\int_0^\pi \int_0^\epsilon \frac{\delta(r)\delta(\theta)\delta(\phi)}{r^2\sin(\theta)},r^2\sin(\theta),dr,d\theta,d\phi=1$$ – Mark Viola Apr 30 '17 at 04:38
  • Thank you. I understand it when $\vec{r}\neq0$ and in that case I can solve the equation by myself. But what I am interested in is the value of the potential at the location of the charge. I was hoping that solving the Laplacian equation without the assumption of $\vec{r}\neq0$ would answer my question because the solution will be the general solution, which includes the case when $\vec{r}\neq0$. But I have not been able to see any relationships between your results and the result when $\vec{r}\neq0$, i.e., $$V(r)=\frac{q}{4\pi\epsilon r}, r\neq0$$ I hope you could provide me further help – A Slow Learner Apr 30 '17 at 04:54
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    The potential of a point charge at the origin is not defined at the origin. The general solution to Laplace's equation for $r\ne 0$ is $V=A/r$. Then, $V'(r)=-A/r^2$ and $\int_0^{2\pi}\int_0^\pi V'(r),r^2\sin(\theta),d\theta,d\phi=-4\pi A=-q/\epsilon\implies A=q/(4\pi \epsilon)$. And $V=\frac{q}{4\pi \epsilon r}$ – Mark Viola Apr 30 '17 at 04:56