# Can someone show me how Green's function would apply for this simple case?

I'm reading up on some stuff on basic electrostatic here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/laplace.html

Can someone use Green's function to show me the form of $V$?

Update:

I have made an attempt to solve for $V$ but I can't understand why it takes the form $$V = \frac{\rho}{4\pi\epsilon_o} \int \frac{1}{|r-r'|}dr^3$$

(can someone check my solution)

I can't recall this solution from anything I've previously encountered. What is the meaning of this $V$. What is the underlying geometry I'm working with? Is anyone familiar with this form of $V$?

Also, how would you derive $E$ from $V$ given the above expression? I think if I see $E$ then it would make more sense.

I'll try to make a simple derivation. Suppose you have a unit point charge located at position $\vec{r}'$. Then the associated charge density is $\delta(\vec{r}-\vec{r'})$, which is a Dirac distribution. The electrostatic potential produced by this charge is given by the Coloumb's law: $$G(|\vec{r}-\vec{r}'|) = \frac{1}{4\pi\varepsilon_0}\cdot\frac{1}{|\vec{r}-\vec{r}'|}.$$ Note that it obviously satisfies the Laplace equation: $$\partial^2 G(|\vec{r}-\vec{r}'|) = \frac{\delta(\vec{r}-\vec{r}')}{\varepsilon_0}.$$ Now suppose we have an arbitrary distribution of charges with density $\rho(\vec{r})$. One can write this density as a sum of point charges: $\rho(\vec{r}) = \int d^3\vec{r}' \rho(\vec{r}')\delta(\vec{r}-\vec{r}')$. Next, we multiply the above equation by $\rho(\vec{r}')$ and integrate over $\vec{r}'$, we obtain $$\partial^2\int d^3\vec{r}' G(|\vec{r}-\vec{r}'|) \rho(\vec{r}') = \frac{\rho(\vec{r})}{\varepsilon_0}.$$ Now by comparison, you find the electrostatic potential due to $\rho(\vec{r})$ is given by $$V(\vec{r}) = \int d^3\vec{r}' G(|\vec{r}-\vec{r}'|)\rho(\vec{r}').$$ This may be what you seek.

• Do you mean 'obviously satisfies Poisson's equation'? I don't personally find this obvious --- first you must show that the Laplacian of $G$ vanishes for $r \neq r'$, and then you must show that the integral of the Laplacian of $G$ over a ball of radius $\epsilon$ centred on $r = r'$ gives you the $1/\epsilon_0$ that you seek. I would also say that the notation $\nabla^2$ is more common than $\partial^2$, and since the questioner used the former in their answer, I would favour it. As a final point, you're missing the $d^3 r'$ on one of your integrals. – gj255 Aug 28 '14 at 9:25
• (1) By 'obviously satisfying ...', I mean it on the physics not mathematics: Coulomb's law is a solution to Possion's equation (thanks for correcting me for using Laplace equation) in the presence of a point charge. (2) I would have used the symbol by the questioner, but I do not know how to make it in Latex. (3) Thanks for pointing out that the missing $d^3\vec{r}'$. – hyd Aug 28 '14 at 10:29
• Ah OK, I see what you're saying. – gj255 Aug 28 '14 at 11:43
• Thanks, physically it makes sense, however, how would you go about deriving $G(|r-r'|)$ from scratch without invoking coloumb's law? – Carlos - the Mongoose - Danger Aug 29 '14 at 1:35
• There are various ways to show that explicitly. You may take the Fourier transform of the Poisson equation. Then you see $\tilde{G}(\vec{k}) = \varepsilon^{-1}_0k^{-2}$, where $\tilde{G}$ is the Fourier transform of $G$ and $k=|\vec{k}|$. Transforming back , you obtain $G(\vec{r}) = \varepsilon^{-1} \int d^3\vec{k} \frac{e^{i\vec{k}\vec{r}}}{k^2}$. To perform this integral, you can write $d^3\vec{k} = k^2\sin(\theta)dk d\theta d\varphi$ and $\vec{k}\vec{r} = kr\cos(\theta)$, where we have chosen the $\vec{r}$ along z-axis. Now you can use standard formula to get the final result. – hyd Aug 29 '14 at 2:37