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So, I was studying green's functions and in general I understood that if I have an operator $\mathscr{O}$ that acts of a function $h_1(\vec{r})$ such that $$\mathscr{O}h_1(\vec{r})=h_2(\vec{r})$$ Then all I need to do is to find the function, $g(\vec{r})$, on which the operator acts to yield the delta function.

Then I can write, $$h_1(\vec{r})=\int h_2(\vec{\tau})g(\vec{\tau}-\vec{r})\mathrm{d}^3\vec{\tau}$$

Reason being $$\mathscr{O}h_1(\vec{r})=\mathscr{O}\int h_2(\vec{\tau})g(\vec{\tau}-\vec{r})\mathrm{d}^3\vec{\tau}=\int h_2(\vec{\tau})\delta(\vec{\tau}-\vec{r})\mathrm{d}^3\vec{\tau}$$ So far, so good, but then in an effort to solve the Poisson's equation, writes $$V(\vec{r})=\frac{1}{4\pi}\int \frac{\rho(\vec{\tau})}{\epsilon}\frac{1}{|\vec{\tau}-\vec{r}|}\mathrm{d}^3\vec{\tau} $$ Because (and I'm back-calculating) $$-\nabla ^2\left(\frac{1}{4\pi|\vec{r}|}\right)=\delta(\vec{r}).$$ I am unable to understand this move. Is there some mathematical basis for this or just to equate the preconceived notion of potential of point charges does this equation hold good?

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  • $\begingroup$ I'm not sure what you're asking. What exactly are you seeking a "mathematical basis" for? Please try to make it more explicit what you want to know, right now you've written down a bunch of equations which all seem more or less correct to me and I don't know what you want to have explained. $\endgroup$ – ACuriousMind Jan 5 '17 at 1:11
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    $\begingroup$ @ACuriousMind I want the last equation mathematically proved $\endgroup$ – ubuntu_noob Jan 5 '17 at 1:14
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    $\begingroup$ That appears to be a pure math question, which in fact has already been answered at math.SE. $\endgroup$ – ACuriousMind Jan 5 '17 at 1:46
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  1. Calculate the Laplacian of $1/r$ using spherical coordinates, you get that it is zero where $r \neq 0$.
  2. Use Green's Theorem to calculate the volume integral of $\nabla^2(1/r)$ in a sphere (of arbitrary radius) about 0, the value of this integral is $-4\pi$ (regardless of the radius of the sphere chosen).

Since the volume integral is nonzero and there is only one point (0) where the function is nonzero, we have a $\delta$-function.

*edited to be less cryptic :)

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    $\begingroup$ Yes, convincing...:) Have mostly dealt with one dimensional delta functions...:) $\endgroup$ – ubuntu_noob Jan 5 '17 at 2:59
  • $\begingroup$ Yeah, it's basically just defined as the thing (not strictly a function) that when integrated over an interval containing 0, gives the value 1. $\endgroup$ – Stephen McAteer Jan 5 '17 at 3:27

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