2

I know that

$$\nabla \dot{}\frac{\hat{r}}{r^2}=4\pi \delta(x)\delta(y)\delta(z) $$

However I thing this could work too

$$\nabla \dot{}\frac{\hat{r}}{r^2}=\frac{\delta(r)}{r^2}$$

If I integrate these expressions near the origin, both of them work. Is it correct or am I doing a mistake? Can $\nabla \dot{}\frac{\hat{r}}{r^2}=\frac{\delta(r)}{r^2}$ be used?

Gappy Hilmore
  • 4,019
  • What do you define $\delta(r)$ as, considering $r$ is a vector? I'm not familiar with such a thing, though I might have guessed it was shorthand for $\delta(x)\delta(y)\delta(z)$, up to a small multiplier. – qman Oct 15 '16 at 00:29
  • $r$ is a scalar function here, @qman. Seems fine to me. – Ted Shifrin Oct 15 '16 at 00:33
  • Understood. Yet, in that case, we probably have a problem: the Dirac impulse probably does not have a sensible definition when $0$ is an extremum of the domain ($r \ge 0$). Also, a three-dimensional integral of an expression including a one-dimensional Dirac impulse also seems to make no sense. Remember that $\nabla\cdot\frac{\hat{r}}{r^2}$ is a function of $(x,y,z)$. There are just too many implied assumptions to rigorously interpret this. – qman Oct 15 '16 at 00:47
  • I guess I have assumed $\int^\infty_0 \delta(r)dr=1$ when it could very well be 1/2 depending on definition. – Gappy Hilmore Oct 15 '16 at 01:05
  • @grdgfgr: That is a relatively minor issue. How do you express the integral of each of the expressions? Remember, you are trying to suggest that $\frac{δ(r)}{r^2}=4πδ(x)δ(y)δ(z)$. You will find that there is no way that this interpretation can be made to work. – qman Oct 15 '16 at 01:39
  • @qman $\int^pi _{0} \int ^{2 \pi}_0\int^\infty_0 \frac{\delta(r)}{r^2}r^2\sin(\theta)drd\theta d\phi=4\pi$ which gives the same result as the one in cartesian coordinates. – Gappy Hilmore Oct 15 '16 at 01:42
  • @grdgfgr, if that was valid, you'd be able to say $\int_0^{2\pi}\int_{-\pi/2}^{\pi/2}\int_0^∞\frac{\delta(r)}{r^2}\sin \theta d r d\theta d\phi = \int_0^∞\int_0^∞\int_0^∞ \frac{\delta(r)}{r^2} dxdydz$. – qman Oct 15 '16 at 01:54
  • @qman You are right. That's the real reason. – Felix Marin Oct 15 '16 at 02:01
  • My point is that the expression on the right clearly has no interpretation. – qman Oct 15 '16 at 03:42
  • @qman Interestingly, in distribution, we have $$\delta(\vec r-\vec r')=\delta(x-x')\delta(y-y')\delta(z-z')=\frac{\delta(r-r')\delta(\theta-\theta')\delta(\phi-\phi')}{r^2\sin(\theta)}$$Therefore, we can interpret $\delta(\vec r)$ as $$\delta(\vec r)\sim \lim_{\vec r'\to 0}\frac{\delta(r-r')\delta(\theta-\theta')\delta(\phi-\phi')}{r^2\sin(\theta)}$$ – Mark Viola Oct 15 '16 at 05:28
  • @Dr.MV - In your final equation, on the left I see a delta of a vector, not a scalar, and on the right I see three deltas of scalars in orthogonal directions, not one, and so there is no way to tie this to $\frac{\delta(r)}{r^2}$. And then there is a pesky $\sin\theta$ in the denominator. If one wishes to assert the validity of an expression, there must at least be a way to produce a rigorous interpretation. Really, it comes down to whether there exists a way to interpret the expression rigorously, and I doubt we've found it. – qman Oct 15 '16 at 22:14
  • @qman The left-hand side, is the Dirac Delta $\delta(x)\delta(y)\delta(z)$. The expression is correct, I assure you. – Mark Viola Oct 15 '16 at 23:26
  • @Dr.MV - That was how I interpreted it, and I am not disputing your statement. I am saying that neither side of your equivalence is clearly related to the expression $\frac{\delta(r)}{r^2}$. – qman Oct 15 '16 at 23:36

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