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Suppose $(M,g)$ is a smooth globally hyperbolic Lorentzian manifold of dimension $n$. Let $\beta:I \to M$ be a finite null geodesic in $M$, that is to say: $$ \nabla^g_{\dot{\beta}}\dot{\beta} = 0, \quad g(\dot \beta,\dot \beta)=0 \quad \forall \, s \in I=(s_-,s_+)$$

Is it possible for a null geodesic to satisfy $$ \lim_{s \to s_+} \dot{\beta}(s) =0.$$

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    $\begingroup$ It is not clear to me how to define the limit of a tangent vector, if there is no limiting point $\lim_{s \to s_+}\beta(s)$. But if there is a limiting point, then the geodesic continues past $s=s_+$, and the velocity is parallel transported, so never zero. $\endgroup$
    – Ben McKay
    Commented Oct 18, 2019 at 13:53
  • $\begingroup$ Well, one natural way to define the limit of a vector field along a curve is by parallel-transporting everything to a given point: so if $V$ is a vector field along a curve $\gamma$, then we can talk about $\lim_{s\to s_0} V\circ \gamma$ by parallel transporting $V|_{\gamma(s)}$ to the tangent space of some point along $\gamma$. Of course, in the case of $V$ being the tangent vector of a geodesic $\gamma$, this means (the parallel transport of) $V$ is the constant vector field, so as long as $\gamma$ is non trivial the limit is non-zero. $\endgroup$
    – Willie Wong
    Commented Oct 19, 2019 at 15:56
  • $\begingroup$ Given the title, perhaps the question is asking whether it is possible that given $\beta$ has compact closure in $M$, that $\beta$ is inextensible? The answer is no. $\endgroup$
    – Willie Wong
    Commented Oct 19, 2019 at 16:05

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