Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider a collection of $n$ point masses evolving under the influence of gravity. For very large $n$, one might be interested in making a continuum approximation of this system (for cosmology, say). However, my naïve attempt to derive a continuum model directly from particle-particle interactions fails. In particular, consider a function $\rho: \Omega \rightarrow \mathbb{R}$ that describes the mass density over some domain $\Omega$. To compute the gravitational force at a point $x \in \Omega$, I might try to simply integrate the force contributed by every other point $y$:

$$ f(x) = \int_\Omega \frac{\rho(x)\rho(y)}{|x-y|^2}\frac{y-x}{|y-x|} $$

Even if we exclude the point $x$ (i.e., we integrate over $\Omega \backslash x$ instead of all of $\Omega$) this integral diverges -- we effectively have particles arbitrarily close to $x$.

So the question is: what's the right way to compute the gravitational force in this situation? It would also be nice to have some pointers to continuum models used in cosmology and their derivation from the particle model (many references simply start with a continuum model and do not explain its origin). By the way, I really only care about classical models -- let's forget about relativity, etc., for now.


share|cite|improve this question

The integral is not divergent, it is linearly convergent. If you do the integral for a constant mass distribution it ends up being

$$ \int {x\over |x|^3} d^3x $$

Which is a convergent integral for the force, as you can see by counting powers of x on the top and bottom--- there is one more power of x on top than on bottom, so the integral is linearly convergent, meaning that the contribution to the force from a ball of radius $\epsilon$ shrinks as the first power of $\epsilon$. This is not really accurate away from a material boundary (like the edge of a solid object), since otherwise the vector contributions are cancelling, and you find that the contribution from a small ball goes to zero as $\epsilon^2$.

The better way is to do the continuum limit is to integrate to find the potential, not the force, an then the potential is the solution to the equation

$$ \nabla^2\phi = 4\pi G\rho $$

Which gives a regular $\phi$ for every regular $\rho$. For the potential, the analogous integral is

$$ \phi(y) = \int {G\rho(x) \over |x-y|} d^3x$$

And this is a convergent integral, and the contribution from a small ball vanishes as $\epsilon^2$ (as you can see explicitly by dividing the ball into radial shells and doing the integral, or just by dimensional analysis as above). So the continuum approximation is valid, and you can do gravity with a continuous fluid.

But what happens is that there is an instability in the continuum equations--- if you don't have pressure, the continuous dust collapses in on itself to make points with infinite density. This is the attractive property of gravity. The way to fix this is to have a temperature and a pressure, and then you can make a self-gravitating fluid stable. The stable solution is localized objects with high pressure surrounded by very low-density gas, and this is stars and planets surrounded by void.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.