I am studying Poisson's equation for gravity. $$\nabla^2 \varphi = 4\pi G\rho$$ I have read that it is solved analytically using some Green's function, to give the well known formula of potential $$ \varphi(r) = -\frac{Gm}{r} $$ But I have not studied the Green's function. So, can anybody tell me how to solve this Poisson's equation?
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$\begingroup$ There's a good explanation here: farside.ph.utexas.edu/teaching/jk1/lectures/node16.html $\endgroup$– user37222Commented Mar 29, 2022 at 19:50
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1$\begingroup$ If you test this expression for the potential in the Laplacian, the result is zero, not $4\pi G \rho$ $\endgroup$– Claudio SaspinskiCommented Mar 29, 2022 at 21:01
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$\begingroup$ @ClaudioSaspinski that is not possible, have a look for yourself-> [en.wikipedia.org/wiki/… $\endgroup$– Owl0223Commented Mar 30, 2022 at 11:32
2 Answers
I am studying Poisson's equation for gravity. $$\nabla^2 \varphi = 4\pi G\rho$$ I have read that it is solved analytically using some Green's function
The intuition behind Green's functions is that they act as propagators. One wants to write the potential $\varphi$ as
$$\varphi=\int{G(r,r')\rho(r)}d^3x$$
Plugging this into the Poisson equation one can notice that the Green function has to obey the following differential equation
$$\nabla^2 G(r,r') = 4\pi G\delta (r-r')$$
The above equation can be solved by using the Fourier transform technique. You'll get something like
$$G(r,r') \propto -\frac{1}{|r-r'|}$$
Which then yields the desired result
$$\varphi=\int{G(r,r')\rho(r)}d^3x=-\int{\frac{G\rho(r)}{|r-r'|}}d^3x$$
Modifying the general solution for the Poisson equation, presented in the article of Wikipedia, for the gravitational case where: $f(\mathbf{r}') = 4\pi G\rho$ (the density being in general a function of $\mathbf{r}'$)
$$\varphi(\mathbf{r}) = - \iiint \frac{G\rho}{ |\mathbf{r} - \mathbf{r}'|}\, \mathrm{d}^3\! r', $$
This expression can be understood as the sum of the contribution of all mass elements for the total potential on a given coordinate $\mathbf{r}$.
$G\rho \mathrm{d}^3\! r' = G\rho dV = Gdm$
For certain cases, as the field of a spherical symmetric mass over an external point, the integral can be analytically solved, resulting in a function of the distance to the center of the sphere:
$$\varphi(\mathbf{r}) = -\frac{Gm}{r}$$
The only problem is that if we substitute this solution on left side of the Poisson equation the result is zero, not $4\pi G\rho$. It can be interpreted in my opinion as an anomally, due to fact that the field has the same value if all mass is at a singularity in the center. And in this case, $\rho$ is not well defined. It is zero everywhere and infinite at the singularity.