Given two masses $M$ and $m$ (with $M\gg m$) in a de Sitter background with cosmological constant $\Lambda>0$ and positive spatial curvature ($k=+1$). What is the corresponding (semiclassical "Newtonian") gravitational force between $M$ and $m$?

Form the $g_{00}$ component of the static Schwarzschild-de Sitter solution of the Einstein field equations I would naively guess

$$F\approx -G\frac{Mm}{r^2}+\frac{\Lambda c^2}{3} m \,r,$$

with gravitational constant $G$ and distance $r$. Actually, the second term in this expression is repulsive. Since I have not found any hint in literature I would like to address this question here.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – tpg2114
    Commented Feb 17, 2020 at 14:37

1 Answer 1


Your force is correct, that is also the expression for $\ddot{r}$ in the real Schwarzschild De Sitter metric when you set the first proper time derivatives of the spatial coordinates to zero:

The geodesic equation gives the radial component of the 4-acceleration (in natural units):

$$ \ddot{r}= \color{gray}{ \frac{\left(\Lambda r^3-3\right) \dot{r}^2}{r \left(\Lambda r^3-3 r+6\right)} } -\frac{\left(\Lambda r^3-3 r+6\right) \left(\color{gray}{ 3 r^3 \left(\dot{\theta}^2 +\sin ^2 \theta \ \dot{\phi}^2\right) }+\left(\Lambda r^3-3\right) \dot{t}^2\right)}{9 r^3} $$

where you set $\dot{r}=\dot{\rm \theta}=\dot{\rm \phi}=0$ and plug in

$$ \dot{t}=\sqrt{g^{t t}} \ \color{gray}{ \gamma } = \sqrt{\frac{1 \ / \ (1-2/r-\Lambda r^2/3)}{1-\color{gray}{ v^2 }}}$$

with $v=0$, where $v$ is the velocity measured by local and stationary (relative to the dominant mass) Fidos, then you get

$$\ddot{r} = -\frac{1}{r^2}+\frac{\Lambda r}{3}$$

which is, in natural units, the expression you correctly guessed. The overdot is the differentiation with respect to proper time, but in the newtonian limit the proper and coordinate time are the same.

This equation assumes the dominant mass in the center, for an n-body simulation the $r$ in the $-1/r^2$ term is relative to the masses and the $r$ in the $+\Lambda r/3$ term relative to the center of your coordinate grid from which everything else accelerates away.

  • $\begingroup$ Can you set ${\dot r}(\tau) = 0$ if ${\ddot r}(\tau) \neq 0$? $\endgroup$
    – Prahar
    Commented Feb 17, 2020 at 22:44
  • $\begingroup$ @Prahar - Yes, since for the newtonian limit the $ \ddot{r} $ for a free falling observer equals the force required to stay stationary or at constant velocity and is independend of $ \dot{r} $, but in the strong field you'd have to distinguish between the two. $\endgroup$
    – Yukterez
    Commented Dec 15, 2022 at 17:17
  • $\begingroup$ You can independently set ${\dot r}(\tau)$ only at a specific time $\tau$, not for all $\tau$. If ${\dot r}(\tau) = 0$ for all $\tau$, then it implies that ${\ddot r}(\tau) = 0$. $\endgroup$
    – Prahar
    Commented Dec 15, 2022 at 18:45
  • $\begingroup$ @Prahar - that's when you integrate the equations of motion to get your position by time, not when finding the newtonian limit for $\ddot{r}$ where the term depending on $\dot{r}$ vanishes. I didn't say $\dot{r}$ is actually $0$, but here we don't need it in $\ddot{r}$. $\endgroup$
    – Yukterez
    Commented Dec 15, 2022 at 19:06

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