# Does a geodesic always extremize its path length? [duplicate]

I've learned that a geodesic maximizes its proper time in Minkowski spacetime. Is this still true in general curved spacetime?

In other words, does the geodesic equation give the globally extremal path from one point in spacetime to another. Assuming the points are time-like separated.

I would think that by the local equivalence theorem, the geodesic equation gives a locally extremal path. Can it proven to be globally extremal?

• I think the must be that not all geodesics do but there will be a geodesic which does. The reason for the qualification is, essentially, gravitational lensing: not all of the images can be global extrema (OK, these are null geodesics, but I think the same would be true for timelike ones). This isn't an answer because I'm just guessing. – tfb Feb 1 '19 at 10:22
• So the globally extremal path must be a geodesic, but different geodesics can exist at the same time? – Timbo Feb 1 '19 at 10:44
• I think that, given suitable smoothness conditions, and assuming there is a global extrema (see my comment to doetoe's answer for a pathological case where there isn't) then it will be a geodesic. But yes, there can be multiple geodesics between two timelike-separated points even without weird topology I think. However my previous example is wrong: those are examples of distinct families of null geodesics which intersect the same pair of worldlines (the object being lensed and our eyes observing it), not the same pair of points. Sorry! – tfb Feb 1 '19 at 11:14

Let $$g$$ be your spacetime metric. We can define the energy functional

$$E(\gamma) = \frac12\int_0^1g(\dot\gamma(t),\dot\gamma(t))dt$$

on curves $$\gamma$$ in your space. The Euler-Lagrange equations for this functional is the geodesic equation. Likewise we can define the length functional

$$\ell(\gamma) = \int_0^1\sqrt{g(\dot\gamma(t),\dot\gamma(t))}dt,$$

which maps a curve to its length. It is not hard to see that curves that extremize the former automatically extremize the latter, for details, see wikipedia. In particular, a geodesic always extremizes the path length.

Depending on the topology of spacetime, it does not have to be a global extremal though. For example, on a cylinder $$\mathbb R\times S^1$$ with the Minkowski metric, consider the points $$(0,P)$$ and $$(1,P)$$ where $$P$$ is any point on the circle $$S^1$$. Then the curve $$\gamma(t) = (t,P)$$ gives a geodesic, but you get another one wrapping around the circle $$n$$ times for every $$n$$. To find these, just unroll the cylinder into $$\mathbb R\times \mathbb R$$ and draw straight lines, which we already know are geodesics in Minkowski space.

• I don't think all members of such a family of geodesics can be timelike, can they? – tfb Feb 1 '19 at 10:29
• @tfb Good remark, thanks. They will be timelike exactly if they are after unrolling. – doetoe Feb 1 '19 at 10:46
• I think that if you construct a spacetime with closed-timelike-curves (say something topologically $\mathbb{R}\times S^1$ but with the metric 'the other way around') then you can probably show that there are infinite families of timelike geodesics between two points. – tfb Feb 1 '19 at 11:01