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Suppose that $(M,g)$ is a smooth Lorentzian manifold, that $I\subset \mathbb R$ is a closed finite interval and that $\gamma: I \to M$ is a smooth null geodesic on $M$, that is to say, it satisfies $$ \nabla_{\dot{\gamma}(s)}\dot\gamma(s)=0 \quad \text{and} \quad g(\dot{\gamma}(s),\dot{\gamma}(s))=0\quad \text{for all $s \in I$}.$$ Assume also that $\beta: I\to M$ is a smooth curve such that $$g(\dot{\beta}(s),\dot{\beta}(s)) =0 \quad \text{for all $s \in I$},$$ and additionally that $\beta$ is not completely overlapping with $\gamma$ at any point (this just means that there is no non empty open interval $J \subset I$ such that $\beta|_{J}$ is a reparametrization of $\gamma$ on some interval of $I$).

Does it follow that $\gamma$ and $\beta$ can only intersect a finite number of times? If the answer is no in general, can one impose some soft assumptions to make the intersection points finite?

Thanks,

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    $\begingroup$ I might be off the mark, but working in Minkowski space for example, wouldn't $\beta$ be allowed to be any curve in the light cone? $\endgroup$
    – Leo Moos
    Jul 6, 2021 at 19:47
  • $\begingroup$ Not necessarily, just think of a circular spacelike curve on the light cone. In fact, on the light cone the only possibilities for $\beta$ are null geodesics on the light cone itself. $\endgroup$
    – Ali
    Jul 6, 2021 at 19:52
  • $\begingroup$ Oops, thanks for the clarification Ali! $\endgroup$
    – Leo Moos
    Jul 6, 2021 at 19:56
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    $\begingroup$ I gave an answer below; it is a bit clunky and there may be ways of optimizing it. Alternatively you may want to reach out to Ettore Minguzzi who is also on MO and more of an expert on these things than I am. $\endgroup$ Jul 7, 2021 at 15:23

1 Answer 1

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We argue by contradiction. Suppose there are infinitely intersections. By reversing time orientation if necessary, there exists a monotonically increasing sequence of times $s_n \in I$ such that at each $\gamma(s_n)$ the curves $\gamma$ and $\beta$ intersect. Since $I$ is compact $s_n$ has a limit $s'\in I$. For convenience label the points $p_n = \gamma(s_n)$, and $p' = \gamma(s')$.

Since $I$ is compact, we have that the image $\beta$ is compact, and hence closed, in $M$. Since $p_n \to p'$ and $p_n \in \beta$, we have that $p'\in \beta$ also. By taking a subsequence if necessary, we can find a increasing sequence of times $t_n$ such that $\beta(t_n) = p_n$ and $t_n$ converges to $t'$, for which $\beta(t') = p'$.

Case I: suppose there exists some $n_0$ such that for all $n > n_0$, the causal segment $\beta|_{[t_n, t']}$ is geodesic (up to reparametrization).

  • $\dot{\beta}(t')$ cannot be equal to $\dot{\gamma}(s')$: other wise by the uniqueness theorem for geodesics $\beta$ and $\gamma$ must overlap on some interval.
  • Since the tangent vectors are not equal, this means that on every open neighborhood of the origin in $T_{p'}M$, the exponential map will have to be non-injective, which contradicts the theorem on existence of normal neighborhoods.

So case I is ruled out.

Case II: the alternative is that $\beta|_{[t_n,t']}$ is not geodesic for any $n$. This implies that $p_n$ is to the chronological past of $p'$ (See Theorem 8.1.2 and Corollary in Wald's General Relativity.)

Theorem 2.14 from https://arxiv.org/abs/gr-qc/0609119 states that every point in a Lorentzian manifold has a causally convex globally hyperbolic neighborhood. Fix such a neighborhood $V'$ of $p'$. Since $\beta$ is compact, there is some $n_0$ such that $\beta|_{[t_{n_0}, t']}$ and $\gamma|_{[s_{n_0}, s']}$ remain both in $V'$. For each $n > n_0$, that $p_n$ is to the chronological past of $p'$ implies there exists a sequence of timelike tangent vectors $v_n\in T_{p'}M$ with $\exp_{p'}(v_n) = p_n$.

$v_n$ must converge to zero, as the exponential map is a diffeomorphism on a sufficiently small neighborhood of zero, and $p_n\to p'$. But this shows that for any neighborhood of zero there exists simultaneously a timelike and a null vector within that neighborhood both of which exponentiates to $p_n$, contradicting the existence of normal neighborhoods.

This rules out case II.

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