# 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.

• Short answer : No. There is no gravitational energy term in the right hand side of the Einstein equations, which is the covariant stress-energy tensor $T_{ik}$. Pression and density are just components of $T_{ik}$ in some (often cosmological) model (you may have several pressions/densities corresponding to different species - dust, relativistic or non-relativistic matter, dark energy). The only case with a negative pression corresponds to the dark energy (or equivalently, the cosmological constant), with $p=-\rho$. – Trimok Jun 14 '14 at 11:10
• If $V \propto a^3$, I don't understand the equation above for $\frac{\ddot{V}}{V}$. From both FLRW equations, I get this instead (including the cosmological constant and the curvature of space) : \begin{equation}\frac{\ddot{V}}{V} = 12 \pi G (\rho - p) + 3 \Lambda - \frac{6 k}{a^2}.\end{equation} So where does your equation (from Baez ?) above comes from ? – Cham Jun 10 '17 at 22:07

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.
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....