Massive particles in de Sitter space

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 $$k=\Lambda/3$$.

• Can show the geodesic equation you got for radial component? – Manvendra Somvanshi Feb 22 '20 at 18:24
• Actually, it is not the most general one. I would appreciate to get one, which helps to solve my problem. – user56224 Feb 22 '20 at 19:01
• $\dot{r}$ can be larger than c even in special relativity, this is not the local velocity but the celerity, which is the uncontracted coordinate length by contracted proper time. But in De Sitter space the velocity of a far away object can also be larger than c, particles behind the hubble radius can indeed recede faster than c, relative to an imaginary probe that is stationary to the central observer (behind the hubble radius only tachyons can fulfill the condition to be stationary to the central observer). Locally you can't exceed c though, see bit.ly/2SToXQx – Gendergaga Feb 22 '20 at 19:30
• In the lecture he (Susskind) is telling us (somewhat handwaving) that "without dark energy nothing would really move out of our ability to see...." So, what if there is no dark energy...?!? – user56224 Feb 23 '20 at 18:04
• Maybe we are talking about the same thing and the pass through the horizon is just this simple geodesic completion. – user56224 Feb 23 '20 at 18:27

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 $$\dot{r}=\sinh{(\tau/l)}\mp\frac{a^2}{2}e^{\pm \tau/l}$$
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.
• @kaffeeauf The geodesic equations have been calculated for static coordinates which clearly have a coordinate singularity at $r=l$ which can be escaped. If you want to discuss what happens in global coordinates you must represent the geodesic equation in that coordinate and see what you get. In global coordinates the particle is always on the surface of a 3-sphere and a horizon makes no sense. – Manvendra Somvanshi Feb 23 '20 at 7:11