Massive particles in de Sitter space

Question

Can massive particles in de Sitter space move faster than light?

For the radial coordinate (in static coordinates) I have got the hyperbolic expression

$$r(\tau)\propto \sinh\left(\sqrt{\frac{\Lambda}{3}}\,\, c\, \tau\right)$$

with cosmological constant $\Lambda$, the speed of light $c$ and the proper-time $\tau$ of the particle. Accordingly, $\dot r(\tau)$ should become greater than $c$ after a certain time.

EDIT: The associated geodesic equation is

$$\ddot r-\frac{\Lambda c^2}{3} r=0$$

Metric:

$$ds^2=(1-k r^2)\, c^2 dt^2-\frac{dr^2}{1-k r^2}-r^2 d\Omega^2_2$$

with $k=\Lambda/3$.

Answer 1

Firstly let's find a general geodesic equation (this part is not actually related to the question but you asked for it in the comments). Taking $\phi=const$ and $\theta=\pi/2$ the Lagrangian becomes $$ L=(1-r^2/l^2)\dot{t}^2-\frac{1}{(1-r^2/l^2)}\dot{r}^2$$ Where $l$ is the de sitter horizon. Substituting this in the Euler Lagrange equation we get two equations, one for $t$ and one for $r$. Solving these equations one can write the radial equation as $$(l^2-r^2)\ddot{r}-ra^2+r\dot{r}=0$$ $a$ is a constant you'll get when you solve the time geodesic. Solving this and substituting appropriate initial conditions one gets $$r(\tau)=l\cosh{(\tau/l)}-\frac{la^2}{2}e^{\pm \tau/l}$$ Taking the derivative $$\dot{r}=\sinh{(\tau/l)}\mp\frac{a^2}{2}e^{\pm \tau/l}$$

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Coming to your actual question now. It doesn't matter if the particle is traveling faster than the speed of light until you can observe it doing so. Long before the particle reaches the speed of light it will be beyond the de sitter horizon (you can calculate this using the formulas given above). The exception is when $a=1$. In that case it can be shown that the particle never reaches the speed of light.

Hope this helps.