# Why a past-inextendible causal curve can have a starting point?

I have some doubts about the notions of future and past-inextendible causal curves. The definitions of future and past inextendible causal curve that I have are the following:

Let $$\gamma$$ be a future-directed causal curve. Then $$\gamma$$ has a future endpoint $$p\in M$$ if for every neighbourhood $$O$$ of $$p$$ there exists a $$t_{0}$$ such that $$\gamma(t)\in O\ \forall t>t_{0}$$.

Let $$\gamma$$ be a past-directed causal curve. Then $$\gamma$$ has a past endpoint $$p\in M$$ if for every neighbourhood $$O$$ of $$p$$ there exists a $$t_{0}$$ such that $$\gamma(t)\in O\ \forall t.

Intuitively, in the Minkowski spacetime $$\mathcal{M}$$, I understand a past-directed causal curve as a causal curve $$\gamma:[a,b]\rightarrow\mathcal{M}$$ that goes down in a Minkowski diagram when the parameter $$\lambda$$ in $$\gamma(\lambda)$$ increases (with respect to an inertial observer). That is, the coordinate $$t$$ of $$\gamma(\lambda_1)$$ is less than the coordinate $$t$$ of $$\gamma(\lambda_2)$$ when $$\lambda_1<\lambda_2$$.

In the same way, I understand a future-directed causal curve as a causal curve $$\gamma:[a,b]\rightarrow\mathcal{M}$$ that goes up in a Minkowski diagram when the parameter $$\lambda$$ in $$\gamma(\lambda)$$ increases (with respect to an inertial observer).

If I use this understanding, a series of doubts arise:

Why a past-inextendible causal curve can have a starting point?

Is it possible to define the notion of past-starting point for future directed causal curves?

Some people say that, when in a future-directed causal curve you eliminate a point, then the portion of the curve before such a point becomes an future-inextendible causal curve, but I don't understand why. According to me, when you eliminate such a point, that point becomes a future-endpoint of the portion of the curve before it, so the curve is actually extendible

If you can put some figures in your answer, that will help me a lot.

• The definitions you provided are of endpoints, not exactly of future/past-inextendible causal curves, so I'm not fully sure of which definition you're using. If I recall correctly, a future-inextendible causal curve is a future-directed causal curve without a future endpoint, but this definition makes your question "Why a past-inextendible causal curve can have a starting point?" not make sense to me Commented Aug 20, 2022 at 19:23
• It would also be helpful to mention the source of the definitions you're providing. Commented Aug 20, 2022 at 19:26

Why a past-inextendible causal curve can have a starting point?

It can have a starting point as long as that starting point is located to the future of the causal curve. Hence, a "past-inextendible causal curve starting at $$p$$" is everywhere in the causal past of $$p$$.

Is it possible to define the notion of past-starting point for future directed causal curves?

You can consider any future-directed causal curve as a past-directed causal curve by reversing the parameterization (given $$\gamma \colon [a,b] \to \mathcal{M}$$ future directed, define $$\delta \colon [-b,-a] \to \mathcal{M}$$, by $$\delta(\lambda) = \gamma(-\lambda)$$). You can then use the usual definition for past-directed curves. Alternatively,

Let $$\gamma$$ be a future-directed causal curve. Then $$\gamma$$ has a starting point $$p \in \mathcal{M}$$ if for every neighbourhood $$O$$ of $$p$$ there exists a $$t_0$$ such that $$\gamma(t) \in O, \forall t > t_0$$.

Some people say that, when in a future-directed causal curve you eliminate a point, then the portion of the curve before such a point becomes an future-inextendible causal curve, but I don't understand why. According to me, when you eliminate such a point, that point becomes a future-endpoint of the portion of the curve before it, so the curve is actually extendible

Suppose $$\gamma: \mathbb{R} \to \mathcal{M}$$ is a future-inextendible causal curve. Pick $$t_0 \in \mathbb{R}$$, define $$p = \gamma(t_0)$$ and consider the spacetime $$\mathcal{M} \setminus \lbrace p\rbrace$$. Then the curve $$\delta: (-\infty,t_0) \to \mathcal{M} \setminus \lbrace p\rbrace$$ defined by $$\delta(\lambda) = \gamma(\lambda)$$ is future-inextendible. The reason is that you can't make the curve go past the point $$p$$, since it is not in the spacetime anymore. It is a "hole" in spacetime. Rigorously, it is not an endpoint of $$\delta$$ because the only candidate to an endpoint for $$\delta$$ is $$p$$ (since $$\gamma$$ was inextendible). However, $$p$$ can't be an endpoint, because it is not even in the spacetime. In the definition

Let $$\gamma$$ be a future-directed causal curve. Then $$\gamma$$ has a future endpoint point $$p \in \mathcal{M}$$ if for every neighbourhood $$O$$ of $$p$$ there exists a $$t_0$$ such that $$\gamma(t) \in O, \forall t < t_0$$,

you don't have any valid $$t_0$$ (since we excluded $$t_0$$ from the domain of $$\delta$$, and if we didn't we'd not have a function, since $$\delta$$ would run into something that is not in spacetime). Due to this, $$\delta$$ is inextendible (and $$\mathcal{M} \setminus \lbrace p\rbrace$$ is singular, $$p$$ being what we'd call the singularity).

• You say that a past inextendible causal curve doesn't have a starting point, however in the Barrett O'Neil's book "Semi-riemannian geometry" there are many statements where he says: Let $\alpha$ be a past inextendible causal curve starting at p... Commented Aug 20, 2022 at 22:15
• I think the mistake is in the definition of past-endpoint. Instead of $\forall~t<t_0$, the correct is $\forall~t>t_0$ Commented Aug 20, 2022 at 22:17
• @ThomasBelichick Adding O'Neil's statement helped me understand what was going on. Please take a look at the edited version of the answer (I only changed the first bit, about the starting point) Commented Aug 20, 2022 at 22:34