6

The equation of motion (in the center of mass frame) due only to gravitational forces between two point masses is:

$$\frac{d^2r}{dt^2} = -\frac{GM}{r^2}$$

How does the equation get modified when a repulsive force due to vacuum energy/dark energy is included?

DilithiumMatrix
  • 14,136
NeutronStar
  • 5,404
  • 5
  • 36
  • 64

2 Answers2

9

We want the Newtonian limit of the Einstein Field equations for nonzero vacuum energy(=cosmological constant). As $\rho_\mathrm{vac}=\Lambda/4\pi G$ is a mass(=energy) density, Poisson equation is $$ \Delta\Phi=4\pi G\rho(\boldsymbol r)-\Lambda \tag{1} $$

If we assume spherical symmetry, and point-like source $\rho\sim\delta(\boldsymbol r)$, the grativational potential that solves $(1)$ is $$ \Phi(r)=-\frac{GM}{r}-\frac{1}{6}\Lambda r^2\tag{2} $$ so that the gravitational acceleration is given by $$ g=-\partial_r\Phi=-\frac{GM}{r^2}+\frac{1}{3}\Lambda r\tag{3} $$

AccidentalFourierTransform
  • 53,248
  • 20
  • 131
  • 253
  • It is not F for the force but a for the acceleration, but the rest should be correct. – Yukterez May 02 '16 at 20:25
  • I'm not sure I agree with this. When finding the cosmological constant's contribution, what makes the point $r=0$ special? The equation $\nabla^2 \phi = \Lambda$ is inconsistent with the usual boundary condition that the potential goes to zero at infinity. – Javier May 03 '16 at 03:18
  • @Javier I've actually half written an answer to that effect. It's tricky, though, because in fact no experiment, neither local nor global, could locate $r = 0$, since with this particular potential the acceleration pushing any two points apart is exactly proportional to their separation. That is, Newton+metric expansion is self-consistent, it's just the more subtle issue of Newton+nonzero homogeneous gravity source that falls apart on consistency. –  May 03 '16 at 07:34
  • @javier im on my phone now. When i get home ill edit my answer to include more detail. But for now, nothing makes $r=0$ special but the fact that I choose to put a point mass there, that is, i took $\rho\propto\delta(r)$. And yes, poisson eq with a source such that $\rho\not\to 0$ is pathological, but in principle the solution "makes sense" physically speaking, because there is no reason for $g\to\eta$ (that is, $\Phi\to0$) if there is "mass everywhere" (constant vacuum energy). It similar to the divergent electric field of an infinite capacitor. – AccidentalFourierTransform May 03 '16 at 08:28
  • r=0 does not need to be special. It simply is the place where the mass is. This also works in n-body simulations where you have a bunch of r=0 all over the place, they don't need to be at the center of your coordinate system for this to work. – Yukterez May 04 '16 at 22:11
  • Taking $\Lambda = 10^{-29} , \mathrm{g cm}^{-3}$, this means that the cosmological term would match the acceleration from our Sun at a distance of $r \approx 8.4\times 10^{20} , \mathrm{cm} \approx 270 , \mathrm{pc}$. For the entire milky way, the distance would be about $r \approx 3 , \mathrm{Mpc}$. – DilithiumMatrix Dec 16 '16 at 17:16
4

You get an extra term that increases with r:

$$a = -\frac{G\cdot M}{r^2} + j\cdot r$$

with j as the repulsive component.

Yukterez
  • 11,553