# Why a timelike geodesic maximizes path length?

I'm studying some GR and my book says that in Pseudo-Riemannian manifolds geodesics may even maximize the path locally. That's what happen to the timelike geodesics, for example. My first question: Is this statement easy to prove? I mean, does it follow straightly from the locally minimizing property of geodesics in Riemannian manifolds? If yes, please explain it to me, otherwise suggest me some reference where I can found the proof.

It seems that those minimizing/maximizing properties depend upon the geodesics' causal structure (timelike, spacelike or null-like) and, in general, nothing can be said about those geodesic's properties without knowing its causal structure. Is this correct?

• Crossposted to math.SE. Please do not crosspost. Also, hint: Observe that it is always possible to find a path in a small surrounding of a time-like geodesic (by winding a path around the "tube" around the time-like path) that is shorter in the Lorentzian metric than the time-like path. – ACuriousMind Apr 27 '15 at 16:39
• tackle this problem using calculus of variations, in the pseudo-Riemannian manifolds the second variation is negative for timelike curves (so geodesics are a local maximium) and in usual Riemannian geometry is positive (so a local minimium). In both case, setting the first variation of the length equal zero gives geodesic equation. – Héctor Apr 27 '15 at 16:47
• There is a related question here. – John Rennie Apr 27 '15 at 16:49
• – Qmechanic Apr 27 '15 at 19:36
• @ACuriousMind, I'm not sure if I got it what you're saying, could you elaborate it a little more please? – Mr. K Apr 27 '15 at 20:04

First we sketch a proof that a timelike geodesic is a maximum of proper time. (We exclude saddle points for now.) Let $\gamma$ be a curve satisfying the geodesic equation, i.e. it is an extremum of proper time defined by $\tau[\gamma]:=\int\sqrt{-\langle\dot\gamma,\dot\gamma\rangle}\,\mathrm{d}t$. It is fairly simple to show that there always exists a curve $\mu$ for which $\tau[\mu]<\tau[\gamma]$, implying $\gamma$ is not a minimum. Construct along $\gamma$ a "tube" which is arbitrarily wide. Let $\mu$ be a curve which has the same start and end points as $\gamma$. Let $\mu$ be confined to the tube along $\gamma$. Now wind $\mu$ along the tube so that it is almost null, i.e. the curve's tangent approaches the null cone at every point on the tube. Thus we have constructed a curve with $\tau[\mu]$ arbitrarily close to zero, which can be made less than $\tau[\gamma]$.

This implies that a geodesic is not a minimum, but cannot determine that a timelike geodesic is not a saddle. However, this is not entirely true either. Here we quote Theorem 9.9.3 in [1]$^1$.

Let $\gamma$ be a smooth timelike curve connecting two points $p,q$. Then the necessary and sufficient condition that $\gamma$ locally maximize the proper time between $p$ and $q$ over smooth one parameter variations is that $\gamma$ be a geodesic with no point conjugate to $p$ between $p$ and $q$.

So a timelike geodesic is not necessarily a maximum of proper time. The study of geodesics does tie in to causal structure, Refs. [1] and [2] are highly recommended for this purpose.

Two standard references on causal structure are:

[1] R.M. Wald, General Relativity (1984).

[2] S.W. Hawking & G.F.R. Ellis, The large scale structure of space-time (1973).

$^1$This is in turn quoted from Proposition 4.5.8 in [2], but I prefer [1]'s wording. Note that the full proof is found in [2].

• I didn't understand it. In your first paragraph you're not using the supposition that $\gamma$ is a geodesic. Consequently, what you're proving is that doesn't exist a shortest timelike curve, since we can always find another timelike curve even shorter (arbitrarily close to zero), not that every timelike curve is shorter than a timelike geodesic. What am I missing? – Mr. K Apr 28 '15 at 18:14
• @IberêKuntz The point I'm making is that a geodesic $\gamma$ is not a minimum of proper time, but rather a maximum (or a saddle, this depends on the causal structure). – Ryan Unger Apr 28 '15 at 20:45
• Would you explain why the last equation in page 111 of Ref. [2] is possible? Note that the vector field $K^a$ is orthogonal to a timelike vector field. One can choose a tetrad and let the timelike vector field be one element of the tetrad, then $K^a=(0,K^1,K^2,K^3)$, so the last equation should always be positive (Hawking & Ellis used the mostly-plus convention). – Drake Marquis Nov 5 '18 at 1:00
• @DrakeMarquis As far as I can tell, the only requirement on $K^a$ is that it is a vector field perpendicular to $\gamma$ that vanishes at its endpoints. So if $DW^a/\partial s\ne 0$ at $r$, then of course can construct such a vector field. (Just take it parallel to $DW^a/\partial s$ and scale it appropriately.) So we just have to answer why that derivative is not zero. – Ryan Unger Nov 6 '18 at 2:59
• If the derivative were zero, then $W^a$ and its derivative would be zero at $r$. But $W^a$ satisfies a homogenous second order ODE, so $W^a$ and $DW^a/\partial s$ vanishing at $r$ would mean $W^a\equiv 0$. This is a contradiction. – Ryan Unger Nov 6 '18 at 3:00