# Interpretation of horizon in de Sitter

Given the de Sitter metric for spacetime $$ds^2 = \left( 1 - \frac{\Lambda}{3}r^2\right) dt^2 - \frac{1}{\left( 1 - \frac{\Lambda}{3}r^2 \right)}dr^2 - r^2d\Omega^2$$ we understand this is a solution to Einstein's field equations in vacuum with the addition of a cosmological constant which is maximally symmetric. In this static coordinates there is an apparent issue at $$r=\sqrt{3/\Lambda}$$, where the metric becomes singular. My question is (staying within this coordinates) how can I interpret this singularity?

It seems like a horizon since beyond that point the signs in front of the $$t$$ and the $$r$$ coordinate are switched. But how can I make sense of this under the context of an expanding universe. I guess things that are found at a coordinate $$r>\sqrt{3/\Lambda}$$ are receding from the observer faster than light and are therefore casually disconnected. But what about something from beyond this horizon, falling in... I suppose the answer is that to travel into the $$r<\sqrt{3/\Lambda}$$ region they would need to go faster than light and thus nothing happens.

EDIT

I think the horizon issue is clear for geodesics and point particles. But let us consider for example a scalar field, which is supposed to be continuous and differentiable (twice at least) which is supposed to be defined everywhere. Any differential equation for it will, have trouble at the horizon. I am not sure how to handle the two regions. Should I expect a discontinuity in the field or something of that sort?

• Compare that with other coordinates for de Sitter. "Something beyond this horizon falling in... should not be a problem" why do you think it is possible or impossible? – OON Jul 1 '20 at 9:35
• Well somehow my intuition is that an object should not be able to cross the horizon inwards unless the object has a peculiar velocity higher than the expansion, consider expansion that is slower than light. Should light be able to come in? Its geodesic does not seem to show that – ohneVal Jul 1 '20 at 10:30

In de Sitter space the event horizon coincides with the Hubble radius, with the Hubble constant

$$\rm H=c\sqrt{\Lambda/3}$$

so the line element in terms of $$\rm H$$ is

$${\rm ds^2} = g_{\rm tt} \ \rm dt^2 - {...} = \left( 1 - H^2 r^2 / c^2\right) dt^2 - {...}$$

which gives the recessional velocity

$${\rm v = c} \ \sqrt{1-1/g^{\rm tt}} = \rm H \ r$$

If you solve for

$$\rm v=c$$

you get

$$\rm r=\sqrt{3/\Lambda}$$

• Could you comment on how to get this expression for $v$ in your third equation? – ohneVal Jul 2 '20 at 10:05
• In de Sitter space the recessional velocity is derived the same way as you normaly derive the local escape/freefall-velocity, you set $\sqrt{g^{\rm tt}}=1/\sqrt{1-\rm v^2/c^2}$ and solve for $\rm v$ – Gendergaga Jul 2 '20 at 20:37