Can geodesics in a Lorentzian manifold change their character?

Question

From a physics perspective, it's pretty easy to see why a a massive particle will be restricted to timelike paths, etc. but does the math guarantee that on its own or do we have to impose it? More specifically, given an arbitrary smooth Lorentzian manifold, can there be geodesics which change character from spacelike to null to timelike, or timelike to null, etc., and how do we rule these out/why are they naturally ruled out?

Answer 1

There is a conserved quantity for geodesics which comes from the fact that the metric $g_{ab}$ is (trivially) a Killing tensor, i.e. $$\nabla_{(c}g_{ab)} = 0.$$

Any tensor $\xi_{ab}$ that satisfies $\nabla_{(c}\xi_{ab)}=0$ gives rise to the conserved quantity $\epsilon = u^a u^b\xi_{ab}$, which is preserved along geodesics for which $u^a$ is the tangent vector. To see this, we write

$$\frac{d}{d\lambda}\epsilon = u^a\nabla_a\epsilon=u^a\nabla_a(u^bu^c\xi_{bc}) = u^au^bu^c\nabla_a\xi_{bc}+u^c\xi_{bc}u^a\nabla_a u^b+u^b\xi_{bc}u^a\nabla_au^c$$

The three terms on the RHS all vanish. The first term is symmetric in the lower indices, so it is zero because we assumed $\xi_{ab}$ is a Killing tensor. The second two are proportional to the acceleration of $u^a$, which vanishes since we assumed that $u^a$ is tangent to geodesics. So we find $$\frac{d}{d\lambda}\epsilon=0$$

Now taking $\xi_{ab} = g_{ab}$, since we have a Killing tensor, the quantity $\epsilon = g_{ab}u^au^b$ has to be constant. But for our tangent vector normalized to $0$ or $\pm1$, $\epsilon$ is just telling us whether $u^a$ is spacelike, timelike or null. So, mathematically, the tangent vector of a geodesic cannot change it's normalization (and hence cannot switch between spacelike, timelike or null) because it is a conserved quantity along geodesics.