Seeking clarification of the distinction between completeness of geodesics/extendibility of curves in GR spacetimes? (Confirm: not the geodesic completeness of a spacetime but the completeness of an individual geodesic.)

I think it's the use of the negative inextendible that may be confusing me.

I am happy with past/future inextendibility for a causal curve as being without a past/future endpoint as defined by Wald (General Relativity, 1984, p193) for a spacetime manifold $M$, paraphrased as,

Let $\gamma$ be a 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 $\gamma(t)\in O\ \forall t>t_{0}$, and similarly for a past endpoint. [Because $M$ is Hausdorff it can have at most one future and one past endpoint.]

Some specific cases:

Consider a causal geodesic $\lambda$ through the (arbitrary) origin of Minkowski space.

Case 1 - Inextendible + Incomplete Remove the point $\vec{0}$ from $\lambda$; $\lambda$ is now in two geodesic pieces; the piece $\lambda(t), t>0$ is past inextendible and the piece $t<0$ is future inextendible (each is an open set and so has no endpoints).

With the Minkowski metric, both pieces seem incomplete, since the affine parameter does not span $(-\infty, \infty)$.

Case 2 - Not Inextendible + Incomplete If, instead of a point, a closed interval on $\lambda$ is removed, both pieces are extendible (they have endpoints) but they also incomplete.

Case 3 - Not Inextendible + Complete $M$ not Minkowski, but with a metric that "pushes $p$ off to infinite distance along the curve".

Inextendible = no endpoint, but as Wald says , "the endpoint need not lie on the curve, i.e., there need not exist a value of $t$ such that $\lambda(t) = p$." so if the affine parameter takes $\lambda$ arbitrarily close to $p$ as it tends to an $\infty$, the geodesic would be complete and not inextendible.

(But definitely $p\in M$ and not not one of those philosophically troublesome "missing points" :)

Case 4 - Inextendible + Complete $M$ Minkowski without excisions, $\lambda$ is infinitely long in both directions: affine parameter spans $(-\infty,\infty)$ and there are no endpoints.

Have I got this right, and what's the picture for spacelike geodesic/curves?

  • $\begingroup$ What does it mean for a single geodesic to be complete? $\endgroup$ – zzz Apr 17 '16 at 18:13
  • $\begingroup$ IF I have understood correctly, per the detail in case 4, a geodesic is complete if and only if its affine parameter spans the interval (−∞,∞). The definition of an affine parameter means that given an affine paramater $\gamma$, $\gamma' = \alpha\gamma + \beta$ (where $\alpha$ and $\beta$ are $\in$ reals) is also an affine parameter, so the specific initial choice of $\gamma$ is immaterial. $\endgroup$ – Julian Moore Apr 18 '16 at 11:05
  • $\begingroup$ and that's different from a geodesic being extensible to $(-\infty, \infty)$ how? $\endgroup$ – zzz Apr 18 '16 at 23:41
  • $\begingroup$ The negative term "inextendible" is preferred, I think; and, yes - a curve is inextendible (in a particular direction) if it has no endpoint that way, i.e. it is extendible if it does have an endpoint. I think the idea is that a curve with an endpoint could be joined to another curve having the same endpoint, but Wald's point that the endpoint need not lie on the curve also confuses me - I'm not sure it's a problem for such a joining (if that is what is meant) but I'm also not sure it isn't. This is why I wrote out all four possible combinations... $\endgroup$ – Julian Moore Apr 19 '16 at 14:47

Your Answer

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

Browse other questions tagged or ask your own question.