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Self-gravity thoughts...

Thoughts about the potential having negative curvature

3D

So in 3D - with radial symmetry - you can have a potential that radially has a negative curvature (ie \phi = -1/r) and still satisfy poisson's equation \nabla^2 \phi = \rho > 0 since \nabla^2 \phi \ne \frac{\partial^2 \phi}{\partial^2 r} but rather \nabla^2 \phi = \frac{1}{r^2}\frac{\partial}{\partial r}r^2\frac{\partial \phi}{\partial r}

So the potential can go like r² inside a uniform sphere and then switch to -1/r outside.

1D

For an infinite slab of uniform density - the potential inside goes like x² and switches to just x¹ outside… - so it has zero curvature outside as expected - and never has negative curvature anywhere.

Thoughts about the 1st derivative of the potential - or the Gravitational Field Strength |g|

Uniform density

Imagine a uniform density sphere in a very low density environment. As you move towards the sphere from infinity - the gravitational field increases because you are getting closer to all of the mass. Now as you enter the mass two competing effects happen. On the one hand you are getting closer to most of the mass, but at the same time you effectively stop feeling the rest of the surrounding mass.

Or in equation form…

\frac{\partial{|g|}}{\partial (-r)} = -\frac{\partial GM_E(r)/r^2}{\partial r} = 2GM_E(r)/r^3 - \frac{G}{r^2}\frac{\partial M_E(r)}{\partial r}

Here the first term represents the fact that you are getting closer to the remaining interior mass - while the second term reflects the fact that the amount of matter inside is decreasing.

For a uniform density mass distribution, M_E(r)=4/3\pi r^3 \rho so we have

\frac{\partial{|g|}}{\partial (-r)} = \frac{8\pi G}{3} - 4\pi G = -\frac{4 \pi G}{3}

So the strengh of the gravitational field peaks at the outer edge of the cloud. This means that the outer material feels a stronger acceleration and falls in faster… This makes sense since if everything accelerated at the same rate - you would get material piling up at the center - and the uniform collapse is uniform. So the uniform collapse is essentially an outward in collapse

1/r mass distribution

If the mass distribution goes like 1/r something very interesting happens…

M_E(r) = C r^2

and we have

\frac{\partial{|g|}}{\partial (-r)} = \frac{2 G C}{r} - \frac{2 G C}{r} = 0

So every point feels the same inward acceleration. Matter will accumulate fastest at the center and you have an inside out collapse..

Point mass

The other extreme is to have a very centrally concentrated mass distribution (ie a point mass). In that case the second term is zero since as you get closer to a point mass - you don't see any less mass - so the strength of the field increases like 2GM/r^3 - and so the acceleration is fastest near the point particle - and you would get an inside out collapse.

BE Sphere

A BE sphere is something in between a uniform density and a 1/r potential… And the inflection point will probably be where the density switches from a constant to 1/r - or it looks like at a non-dimensional radius of ~ 3… Of course we have completely neglected pressure gradients which are extremely important for a BE sphere…

BE sphere in a uniform density background

So now if we imagine a BE sphere surrounded by a uniform density ambient….

First the ambient effectively sees a point mass (provided that the BE sphere is much heavier than the ambient) - so it will want to collapse from the inside out. So a rarefaction should develop in the ambient. Except at the edge of the BE sphere - where the infalling material must pile up (or heat up and re-expand)… In any event this would add additional mass to the outer edge of the BE sphere and send a pressure wave through the cloud. The center of the BE sphere - must be completely ignorant of any additional mass piling up outside - so it will only see pressure waves. If the pressure wave is slow enough - or low enough amplitude - the central region should remain stable while the outside is free to undergo collapse…

astrobear