Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer
    
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. –  JLA Mar 27 '13 at 3:45
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.