# Does an observer inside a black hole experience the de Sitter metric?

My question has emerged, because of the following similarities between the metric an observer experiences in the Schwartzschild space being under the Schwartzschild radius and an observer in the de Sitter space:

1. Both observe an event horizon covering the whole sky. All paths lead to the event horizon.

2. The distance to the event horizon shrinks over time, as well as its area.

3. The event horizon emits radiation with the temperature inversely proportional to its radius.

4. Both observers experience tidal ripping forces.

5. All geodesics end on the future timelike singularity, there is no path to escape.

If in fact both metrics are different, in what ways can the observer determine, in which one of the two he is situated?

• Strictly speaking, the similarity above is at most qualitative: Schwarzschild spacetime, for instance, is not constant curvature at any fixed time-slice while de Sitter is constant curvature surface. Basically, quantitatively the curvatures (local quantities that observers can measure) are different: one can simply check the Riemann tensor. Apr 20, 2019 at 7:59
• @Everiana I think this is an answer
– user107153
Apr 20, 2019 at 11:11
• @Everiana: one can simply check the Riemann tensor. The trouble is that the components of the Riemann tensor will look different in different coordinate systems.
– user4552
Apr 20, 2019 at 19:25
• @BenCrowell You can compute the Riemann tensor to check if it corresponds to a maximally symmetric spacetime. I think that's what Everiana means. Apr 21, 2019 at 23:28
• @BenCrowell yes I mean using Riemann tensor in specific manner. Kretschmann scalar would probably work too, in general, or some other curvature invariants. My point is that maximally symmetric spacetimes are quite different in many ways, even Killing vector counts would differ. Actual tidal force would also differ, since the force itself is coordinate independent. Maximally symmetric spacetimes also have Riemann tensor which depends only on the Ricci scalar and the metric, while for Schwarzschild only Weyl part is nonvanishing. But I like your way haha. Apr 23, 2019 at 4:12

No. The Carminati-McLenaghan invariant $$W_1$$ distinguishes them. The Schwarzschild spacetime has $$W_1=6m^2r^{-6}$$, while the de Sitter spacetime has $$W_1=0$$ everywhere. Because $$W_1$$ is a scalar, this test doesn't depend on the choice of coordinates. The vanishing of the invariant for the de Sitter spacetime is because of this spacetime's higher symmetry: it's isotropic, so there are no tidal forces.