# Evolution of mass and velocity distributions under newtonian gravitation

Let $\rho(r,t)$ and $v(r,t)$ be mass and velocity distributions. Given $\rho(r,0)$ and $v(r,0)$ (initial conditions) what is the differential equation that describes the evolution of $\rho(r,t)$ and $v(r,t)$ in space and time, assuming Newtonian gravitation?

You may assume $r$ to be one dimensional, if it makes writing the equations convenient.

EDIT

I think $v(r,t)$ should be replaced with $p(r,t) = \rho(r,t)v(r,t)$ and name it as momentum distribution to avoid some technical faults.

EDIT 2

It seems that to completely characterize an inviscid fluid flow, Euler equations are needed.

The law of conservation of mass is already given in the answer. A component of pressure is missing in the equation related to conservation of momentum.(first equation). Lastly a third equation corresponding to law of conservation of energy should be introduced.(3rd one of the euler equations of inviscid fluid flow).For the third equation we assume that the internal energy per unit volume of the fluid is zero.

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Is this question too basic to be asked here ? –  Rajesh D Nov 29 '10 at 13:50
@Marek: what is your take on this ? –  Rajesh D Nov 29 '10 at 14:08
well... most of your questions are answered in basic physics courses. Personally, I'd suggest you take some books (classical mechanics, continuum mechanics, quantum mechanics, maybe some mathematics for that, e.g. vector calculus) and study yourself and only ask for physical explanations of mathematics you don't understand. If this is what you're already doing then I think your questions are fine. Although maybe it wouldn't hurt if you provided some motivation as to why are you learning this stuff. Context always helps to provide better answers. –  Marek Nov 29 '10 at 15:19
The reason why i am assuming internal energy of the fluid to be zero is not to venture into any other field, also as there is no intention to explain any physical situation as yet....physicists have been doing this for a long time now. Even if the model cannot sustain, we would atleast know what are the drawbacks of such a model completely. If we try to explain a physical situation directly, in most of the cases we run into trouble very soon and introduce theories on ad-hoc basis and what would happen is we will not be able to distinguish between a theory and physical reality ! –  Rajesh D Nov 30 '10 at 13:05
Suddenly this question seems to be very interesting to me. –  Rajesh D Nov 30 '10 at 13:07

Newton's law of gravitation says that to get the acceleration at a given point, you take any distant bit of mass and make a vector pointing towards it with length proportional to the mass and inversely proportional to the square of the distance to it. Integrate that expression over the entire mass distribution.

$\frac{\partial{\vec{v}}(\vec{r},t)}{\partial t} + [\vec{v}(\vec{r},t)\cdot\nabla]\vec{v}(\vec{r},t) = G \int_{\textrm{space}}\frac{\rho(\vec{r'},t)}{|(\vec{r'}-\vec{r})|^3}(\vec{r'}-\vec{r}) \textrm{d}V(\vec{r'})$

$G$ is Newton's gravitational constant, and $\textrm{d}V(\vec{r'})$ is a volume element at the location $\vec{r'}$.

Alternatively, in terms of $\vec{p}$,

$\frac{\textrm{d}\vec{p}}{\textrm{d}t} = G\rho\int_{\textrm{space}}\frac{\rho'}{|\vec{r'}-\vec{r}|^3}(\vec{r'}-\vec{r}) \textrm{d}V(\vec{r}')$

Conservation of mass requires that the net flow of mass into a region of space result in an increase in the density there.

$\frac{\partial\rho(\vec{r},t)}{\partial t} = \nabla \cdot \vec{p}(\vec{r},t)$

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To clarify, this is basically making use of Gauss's law for gravity. –  Noldorin Nov 29 '10 at 14:24
It's just good old F=ma with F given by Newton's gravitational law. –  Mark Eichenlaub Nov 29 '10 at 14:28
@Mark Eichenlaub: request you to kindly edit the answer keeping in line with the new 'EDIT' part in the question. –  Rajesh D Nov 29 '10 at 14:35
@Rajesh Sure. The continuity equation looks better in terms of $p$. However, the equation for the acceleration is easier to parse in terms of $v$ and makes more physical sense to me that way, so I'll leave that one as is. –  Mark Eichenlaub Nov 29 '10 at 14:42
@Mark: true, I suppose, although in the notation I'm most familiar with (and which I think is most common), $\nabla$ only acts on a scalar field, whereas the directional derivative $\vec{v}\cdot\nabla$ can act on anything. –  David Z Nov 29 '10 at 23:05

Gauss law of gravity is what you want wikipeadia link. This will give you $\vec{g}$ at each location.

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This problem is insoluble (since it is not well-posed) without a matter equation of state (and, unless you assume $T\approx 0$, a thermal equation of state). If you specify a non-dust equation of state, it ends up having the Navier-Stokes equations embedded in it. What do you need this for? The above answers discuss some special cases (most specifically, static distributions) well, but this is still an open problem in the general case.

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@Jerry Schirmer: In Navier-Stokes equation if i assume the fluid is nonviscuous, i can get rid of one term.I can also assume that the only bodily forces acting are due to gravity and convert $F$ to be solely due to gravity. Now the equation reduces to the equation given in the above answer. –  Rajesh D Nov 30 '10 at 4:02
You still need a relationship between the pressure (and temperature, if not at absolute zero) and the density, and if the fluid is not at absolute zero, you need a relationship between the internal energy of the fluid and the fluid's temperature. We don't even have an existence and uniqueness theorem for the Navier-Stokes equations, much less the Navier-Stokes equations coupled to gravity –  Jerry Schirmer Nov 30 '10 at 4:07
And sure, you can assume that there is no pressure (the bodily forces only being due to gravity), but that's still a choice of an equation of state (namely, the dust equation of state), and should be stated explicitly in the statement of the problem. –  Jerry Schirmer Nov 30 '10 at 4:09
@Jerry Schirmer: discarding thermodynamic nature of matter i do not understand why we need a relation between $\rho$ and pressure $P$ ? The only force here is due to gravity and we have considered that. –  Rajesh D Nov 30 '10 at 4:12
But the Pressure provides a force. As the density changes, the force due to the pressure will change. Unless you know how the pressure depends on the density, you have no control over how this takes place. Think about solving either of Mark's equations--either equation involves an unknown $\rho$ and another unknown--either $\vec v$ or $\vec p$. Either one is insoluble without another equation, since you need an equal number of equations and unknowns. –  Jerry Schirmer Nov 30 '10 at 4:19