# Derivation Poisson's gravity equation by divergence theorem [closed]

I'm trying deduce the poisson's equation $\nabla^2\Phi (x)=-4\pi G\sigma(x)$ by divergence theorem

Let $D:x^2+y^2+z^2\leq 1$ and $\sigma:D\to \mathbb{R}$ be the mass density function of $D$ (suppose $\sigma \in C^\infty$). It's define gravity field of $S$ applied in $x\in\mathbb{R}^3-D$ $$g(x)=-G\iiint_D\frac{\sigma(\xi)}{||x-\xi||^3}(x-\xi)\ d\xi$$ This field is conservative: $$g=-\nabla \Phi\ \text{ with }\ \Phi(x)=\iiint_D\frac{\sigma(\xi)}{||x-\xi||}\ d\xi\ \text{ for all }\ x\in\mathbb{R}^3-D$$ Remerber that divergence theorem says:

*Let $V$ be a compact subspace of $\mathbb{R}^3$ and $\vec F:V\to \mathbb{R}^3$ differential function. Then: $$\iiint_V\nabla\cdot \vec F\ dV=\iint_{\partial V} \vec F\cdot dS$$ *

We calculate: $$\nabla^2\Phi =\nabla\cdot (\nabla \Phi)=\nabla_{x}\cdot g =\iiint_D\sigma(\xi)\nabla_x\cdot\Big(\frac{x-\xi}{||x-\xi||}\Big)\ d\xi$$ applying divergence theorem, we get $(S=\partial D)$: $$\nabla_x\cdot g(x)=\iint_S\sigma(\xi)\frac{x-\xi}{||x-\xi||^3}\ d\xi$$ Considering the following parametrization of $S:x^2+y^2+z^2=1$: \begin{align*} \vec r(\theta,\varphi)&=(\cos \theta \sin\varphi,\sin\theta\sin\varphi,\cos\varphi)\qquad \text{for all }\ (\theta,\varphi)\in[0,2\pi]\times [0,\pi]\\ &\frac{\partial \vec r}{\partial \theta}\times\frac{\partial \vec r}{\partial \varphi}=-\sin \varphi \vec r(\theta,\varphi) \end{align*} we get, for $x=(x^1,x^2,x^3)$ $$\nabla_x\cdot g(x)=\int_0^\pi\int_0^{2\pi} \frac{ \sigma(\vec r(\theta,\varphi))}{||x-\vec r (\theta,\varphi)||^3}(x-\vec r(\theta,\varphi))\ d\theta d\varphi$$

1.- When I try to calculate $\frac{ \sigma(\vec r(\theta,\varphi))}{||x-\vec r (\theta,\varphi)||^3}(x-\vec r(\theta,\varphi))$ I get a difficult expresion for integrate. Any help?

2.- If there exists an easily methon for derive this poisson's equation by Newton's mechanics, let me now.

many thanks!!

## closed as off-topic by ACuriousMind♦, garyp, user36790, Gert, CuriousOneJun 5 '16 at 9:43

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I think it is not the fastest solution to try and use the divergence theorem, try to apply $\nabla$ two times.
Take in your notation $$\nabla^2\Phi = (-G) \Delta \iiint_D\frac{\sigma(\xi)}{||x-\xi||}\ d\xi\ = (-G) \iiint_D \sigma(\xi) \Delta \frac{1}{||x-\xi||}\ d\xi\$$ If I remember my field theory lectures correctly $$\Delta \frac{1}{||x-\xi||} = 4\pi \delta( x-\xi)$$ and your Poisson equation follows suit. I do not have a different derivation in mind (doesn't mean there is none).