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, CuriousOne Jun 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).

I will still take a few minutes to think about the divergence theorem approach.

Edit: I am having two problems with your application of the divergence theorem. First, you forget to project your field in the normal direction of the surface, but that can be done. My second problem is that when I do that, I still need to integrate over the unknown mass distribution function which I do not know. This does not mean this is not a possible approach, but I do not know how to do it.


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