1
$\begingroup$

Let $S \subset \mathcal{M}$ (where $\mathcal{M}$ is our manifold), the future Cauchy development of $S$ is defined as

$$D^+ (S) := \{p \in \mathcal{M} |\ \text{every past inextendible causal curve through} \ p \ \text{intersects} \ S \}$$

analogously, the same name is used for the following set

$$\tilde{D}^+ (S) := \{p \in \mathcal{M} |\ \text{every past inextendible timelike curve through} \ p \ \text{intersects} \ S \}$$

Both Minguzzi (2019) and Hawking & Ellis (1973) (in Remark 3.2 and Proposition 6.5.1 respectively) state that $D^+ (S) \subset \tilde{D}^+ (S)$. Why is this true? Both sources state it as a direct consequence from the former definitions with no further explanation, but I don't see it. A causal curve is either a null curve or a timelike curve. This means that for $p \in D^+ (S)$, there are more curves that pass through $p$ that intersect $S$ compared to the points in $\tilde{D}^+ (S)$, shouldn't this mean that $\tilde{D}^+ (S) \subset D^+ (S)$? After all, the timelike curves mentioned in the definition of $\tilde{D}^+ (S)$ are included in the definition of $D^+ (S)$.

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ It all causal curves etc passing through $p$ intersect $S$, then, in particular, all timelike curves etc passing through $p$ intersect $S$. In other words, if $p\in D^+(S)$, then $p$ also belongs to the other set. That is just the thesis. $\endgroup$
    – Valter Moretti
    Sep 1, 2021 at 15:17
  • $\begingroup$ But doesn't it work the other way around too? If all timelike curves passing through $p$ intersect $S$ (thus $p \in \tilde{D}^+ (S)$), said curves are also causal by defintion, then $p$ belongs in $D^+ (S)$. I don't see where I'm wrong here, but I must be, otherwise both sets would be the same. $\endgroup$
    – RMC777
    Sep 1, 2021 at 15:22
  • 1
    $\begingroup$ Consider an open spacelike disk $S$ in Minkowski spacetime. Then $\tilde{D}^+(S)$ includes all points in the future conical surface whose basis is $S$. Think of the tip of the cone in particular. These points do not belong to ${D}^+(S)$, so that the inclusion is strict in general. $\endgroup$
    – Valter Moretti
    Sep 1, 2021 at 16:07

1 Answer 1

Reset to default
1
$\begingroup$

Suppose there exists a point $p\in\mathcal{M}$ for which every past inextendible timelike curve through $p$ intersects $S$, but there exist an (inextendible) null curve through $p$ that does not intersect $S$. Then $p \in \tilde{D}^{+}(S)$, but $p \notin D^{+}(S)$.

Improve this answer
$\endgroup$
4
  • $\begingroup$ I see. However, let's put it the other way around. Let $p \in \tilde{D}^+ (S) \Rightarrow$ every timelike curve that passes through $p$ intersects $S$. In particular, since these curves are timelike, they're also causal, doesn't this mean that $p \in D^+ (S)$? $\endgroup$
    – RMC777
    Sep 1, 2021 at 15:26
  • 1
    $\begingroup$ No, because of the potential case above. There could be causal curves through $p$ that do not intersect $S$. $\endgroup$
    – mmeent
    Sep 1, 2021 at 15:28
  • $\begingroup$ Indeed. I was ignoring the requirement that $\underline{\text{every}}$ causal curve through $p$ must intersect $S$ in the definition of $D^+ (S)$. Thank you for your explanation. $\endgroup$
    – RMC777
    Sep 1, 2021 at 15:37
  • $\begingroup$ I explicitly described a counterexample in my comment above. $\endgroup$
    – Valter Moretti
    Sep 1, 2021 at 16:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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