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How to start with a perfect fluid concept and reach (by approximations through certain mathematically well defined assumptions) to the concept of particle ? Here newtonian gravitation is being assumed. The state equation of matter can be assumed to be that of dust.


Assume any initial mass distribution $\rho(r,t=0)$ and momentum distribution $p(r,t=o)$ and state equation of dust.


I don't intend to throw away the fluid concept,but still want to arrive at a particle as some form of a mass distribution.One example could be $\rho_p(r,t)$ which has a compact support in r describing a particle.


Assume there are a number of such particles as described in the earlier edit part of the question as $\rho_p(\vec{r}-\vec{r_i},t=0)$ for $i = 1,2,3,...$. distributed arbitrarily in space such that no two are overlapping. This could be considered as $\rho(r,t=0) = \displaystyle\sum\limits_{i=0}^n \rho_p(\vec{r}-\vec{r_i},t=0)$. Similarily for momentum distibution at $t=0$ as $\vec{p}(r,t=0) = \displaystyle\sum\limits_{i=0}^n \vec{p_p}(\vec{r}-\vec{r_i},t=0)$. What would eventually happen to such a system under its own newtonian gravity. Just out of curiosity what would happen if GR is assumed.

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Can you improve the question by putting your starting assumptions in equation form? It's hard to tell exactly what you mean here. – Sklivvz Dec 4 '10 at 9:33
@Sklivvz : i have updated the question – Rajesh Dachiraju Dec 4 '10 at 9:38
For what it's worth, Newtonian gravity in the case of fluids with mass less than planetary orders of magnitude is more or less negligible. – Justin L. Dec 5 '10 at 18:24
I think you should take a look at Landau Volume 6, and read the section in chapter 1 which describes the difference between the concept of a point particle and a fluid particle. Keep in mind that a "fluid particle" consists of a very large number of atoms, and the velocity distribution in the NS equation describes the motion of fluid particles, not point particles. – Matt Calhoun Dec 5 '10 at 23:49
Under Newtonian Gravity, I believe this system will begin to rotate and condense; and under GR I believe it will form a black hole in the limit of infinite time. – Matt Calhoun Dec 5 '10 at 23:50
up vote 1 down vote accepted

If I understand your question correctly you want to get from Gauss' gravitational law to Newton's.

$$ \nabla \cdot \textbf{g} = -4 \pi G \rho$$

Integrating both sides over a volume $V$ enclosing a mass $M$ and having surface $S$.

$$\int_{V} \! \nabla \cdot \textbf{g } \mathrm{d}\textbf{V} = -4 \pi G \int_{V} \! \rho\ \mathrm{d}\textbf{V}$$

$$\int_{V} \! \nabla \cdot \textbf{g } \mathrm{d}\textbf{V} = -4 \pi G M$$

Using the divergence theorem: $\int_{V} \! \nabla \cdot \textbf{g } \mathrm{d}\textbf{V} = \oint_{\partial{V}} \textbf{g} \cdot \mathrm{d}\textbf{S}$

$$\oint_{\partial{V}} \! \textbf{g} \cdot \mathrm{d}\textbf{S} = -4\pi GM $$

Assuming that $\textbf{g}$ is constant over $\mathrm{d}\textbf{S}$, and that the force is centripetal we can write $$\textbf{g} = g(r){\textbf{e}_r}$$

where $\textbf{e}_r$ are the unit vectors of the acceleration that point towards the center on the surface $S$ and g(r) gives us the magnitude of the acceleration at radius $r$.

Our equation then becomes:

$$g(r)\oint_{\partial{V}} \! \textbf{e}_r \cdot \mathrm{d}\textbf{S} = -4\pi GM $$

$$g(r) 4\pi r^2 = -4\pi GM $$

$$g(r) = -\frac{GM}{r^2} $$

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Is this what you are looking for? – Sklivvz Dec 4 '10 at 10:31
I'll just add that to get from third to fourth line one needs a Stokes' theorem. – Marek Dec 4 '10 at 10:38
Correct, I forgot to mention it. – Sklivvz Dec 4 '10 at 10:44
@Sklivvz : I don't intend to throw away the fluid concept,but still want to arrive at a particle as some form of a mass distribution.One example could be $\rho_p(r,t)$ which has a compact support in $r$ describing a particle. – Rajesh Dachiraju Dec 4 '10 at 11:01
It's notoriously chaotic and has many different endgames, depending on the starting conditions. See and – Sklivvz Dec 4 '10 at 17:51

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