Does mutual gravitational attraction imply a negative pressure? Imagine a small ball of gravitating dust particles initially at rest at the center of a large volume $V$.
Following John Baez, Einstein's field equations say (in units of $c=8 \pi G=1$):
$$\frac{\ddot{V}}{V}  = - \frac{1}{2} (\rho + 3 p)$$
Now traditionally dust particles have a pressure $p=0$.
However the dust particles have a mutual gravitational attraction for each other.
Could this mutual attraction be described within the framework of GR by assigning a negative pressure $p$ to the ball of dust particles?
Another way of looking at it is to say that the ball of dust particles has a certain amount of negative gravitational self-energy which counteracts some of the gravitational effects of its positive rest mass energy.
Apparently the concept of gravitational energy in GR is complex so maybe negative pressure is a better way to think about it.
 A: Based on Baez's simplified presentation of the field equations, all that would matter would be $\rho+3p$, and mutual attraction is described by $\ddot{V}<0$, which only requires $\rho+3p>0$. So if we observe mutual attraction, there is clearly no reason to think it implies $p<0$; making $p$ more positive would increase the attraction.
In general, you're trying to use this simplified version of the field equations to do more than it can do.
Rather than talking about negative pressure, relativists would normally describe this kind of thing in terms of energy conditions like the strong energy condition. Dark energy violates the strong energy condition. It's perfectly possible that we will discover matter fields in the future besides dark energy that also violate the strong energy condition. However, there is presently no evidence from cosmological observations for any such thing.
A: You are right. This is indeed what happens in stars and nebulas...the reason they don't collapse under their own gravity is that they exert thermal pressure(due to nuclear reactions), which counteracts gravity so that the star doesn't "break down" under the gravitational pull. And this continues as long as there is enough fuel for the nuclear reaction. In case the inward pointing gravity is greater than the outward pointing pressure, the star collapses into a neutron star or something like that. 
But there is a condition, only If the particles are made of unstable atoms and have enough energy to produce thermal pressure necessary to counteract gravity can you assign a pressure $p$ which is negative relative to the direction of gravity. And you don't need special constraints in your equations, so it will still be valid in GR,  unless the gravitational field is that of a black hole, as it is a fundamental property of nature. I think gravitational energy will be misleading, as on the quantum scale, you will not have any specific potential due to nuclear reactions and radiation which shall change the field's configuration, not to mention the quantum fluctuations. Also, pressure is a natural way to think about something pushing outward from the center as a sort of "opposite" vector to that of inward pointing gravity. Penrose and Hawking have done a lot of work on this, you can check them out. Hope this helps....
