As I understand it, a timelike hypersurface is one that has only spacelike normal vectors. But does this not imply that a the geodesic of a particle crossing it must be spacelike at that point? But geodesics of massive particles follow timelike geodesics...I suppose the same question arises for the event horizon of a black hole, which is a null surface, but in that case I just imagine that a particle doesn't have to have a timelike tangent vector at that point since it isn't locally flat.
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No, a geodesic can freely cross a timelike hypersurface. Just consider the surface $x=0$ in minkowski space--obviously, an particle is free to cross this surface--all it needs is an x-component to its velocity.
The crossing geodesic will have a normal component and a tangent component, however--it won't be completely normal.
answered Mar 27, 2013 at 3:41
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1$\begingroup$ Ah, ok I see, that was silly. The normal part will contribute positively to the norm of the vector, but the tangential part will make it overall negative. $\endgroup$– JLACommented Mar 27, 2013 at 3:45