TrickyDicky
- 3,507
- 28
Well, it is tricky to say, it depends on the coordinates you use, in Schwarzschild coordinates it is a spacelike geodesic, but that seems to be a problem of the coordinates because they are swapped at the EH. In fact a test particle must always be timlike. As you said the norm of its tangent vector can't change sign.DaleSpam said:So, it should be clear that there is no conflict between those restated propositions and the idea that an object free-falling across the EH is represented by a geodesic which is everywhere timelike.
This is taken from wikipedia page:"Geodesics in GR", I would like for anybody disagreeing with any of the quoted (especially the bolded ones) statements to speak up and explain why."With a metric signature of (-+++) being used,
timelike geodesics have a tangent vector whose norm is negative;
null geodesics have a tangent vector whose norm is zero;
spacelike geodesics have a tangent vector whose norm is positive.
Note that a geodesic cannot be spacelike at one point and timelike at another.
An ideal particle (ones whose gravitational field and size are ignored) not subject to electromagnetic forces (or any other non-gravitational force) will always follow timelike geodesics. Note that not all particles follow geodesics, as they may experience external forces, for example, a charged particle may experience an electric field — in such cases, the worldline of the particle will still be timelike, as the tangent vector at any point of a particle's worldline will always be timelike."