are geodesic shortest path or quickest path?

Question

I'm a bit confused with geodesics : are they the shortest path (in distance) or the quickest path (in time). For example, Let take a triangle ABC. I'm using a car. I'm in $A$ and I have to go in $B$. The path $AB$ is 2km long, but I can go at 10 km/h only, where as the path that path through C has 4 km length, but it's a free way and I can go at 100 km/h.

Clearly, the path through C is quicker, but the path AB is shorter. What is going to be the Geodesic ? The path through $C$ or the path $AB$ ?

Answer 1

First note that a Geodesic does not have to be the quickest path between two points: For example, there are two geodesics on the sphere from say the north pole to say London. The shortest path is going down the Greenwich meridian from the north pole to London. But going from the north pole to the south pole along that same meridian and then going from the south pole to London along the shortest path (again along the Greenwich meridian) is also a geodesic from the North pole to London.

However, the shortest path between two points (if such a shortest path exists!) always is a geodesic.

Now to your actual question: what is meant by distance? In my example above "distance" meant the usual distance on a sphere. However you can choose what you mean by "distance" by specifying a metric. Roughly speaking a metric lets you measure lengths (and angles, too) in space. Your metric can just measure distance between points but you could modify your metric to take into account that something is slowing down movement.

Answer 2

Is the shortest, but only locally.

Think about $\mathbb{S}^{2}$ with the metric induced by euclidean topology, then great circles are geodesic. Take two points $A=(0,0,1)$, $B=(0,\frac{\sqrt{3}}{2},\frac{1}{2})$ then both $\alpha(t)=(0,\sin(t),\cos(t))$, $t\in[0,\frac{\pi}{3}]$ and $\beta(t)=(0,\sin(t),\cos(t))$, $t\in[0,-\frac{5}{3}\pi]$, are geodesic, but $\alpha$ is shorter than $\beta$.

Moreover, it's easy to show that exist paths shorter than $\beta$ which aren't geodesic.

Answer 3

A geodesic (resp. a curve) $\gamma:[0,T]\to M,t\mapsto\gamma(t)$ can be reparametrized by $\tilde{\gamma}:[0,\frac{T}{c}],t\mapsto\gamma(ct)$, the resulting curve still being a geodesic (resp. a curve) but $c$ times quicker than $\gamma$. So, since talking about "quickest curves" can be confusing since it depends of the parametrizations you choose, you can :

• choose a reparametrization with unit speed (that is $|\gamma'(t)|\equiv 1$, it exists and for geodesics it is even one of the reparametrizations of the form $t\mapsto\gamma(ct)$ since a geodesic always has constant speed: $$D_t|\gamma'|^2=D_t(\gamma',\gamma')=2(\gamma',D_t\gamma')=2(\gamma',0)=0,$$ and so $|\gamma'|$ is constant),

• read the $\tilde T$ obtained in the new interval $\tilde\gamma:[0,\tilde T]\to M$ (which now corresponds to a traveled distance!),

• check that geodesics are (locally) the shortest paths, that is if there is a unit speed curve $\gamma$ joining $a$ and $b$ such that $T_\gamma\leq T_\beta$ for all other unit speed curves $\beta$ joining $a$ and $b$, then $\gamma$ is a geodesic.

Answer 4

(Disclaimer: It should be a comment, but I don't have enough rep yet)

Mathematically the other answers are correct, that the "shortest path" is always a geodesic. But you also asked about "quickest" and if you look at it physically, in GR the "time" of a path as measured by the proper time $d \tau^2 = -ds^2$ (which is the time experienced by an observer traveling in the spacetime), is maximized along a geodesic ($ds^2$ is minimized thus $-ds^2$ is maximized). So a geodesic is not the quickest but the slowest path when it comes to experienced time while traveling along the geodesic.

An easy way to see this is actually in flat Minkowski spacetime (remember that we live on a 4D manifold with a Lorentzian metric.. at least according to GR). In the twin paradox the "stationary" twin (A) is moving along a geodesic (stationary in space, but moving in time) while the twin (B) flying in a spaceship is accelerating and thus not on a geodesic. When both meet, A is older than B. A's proper time was maximized since his path was minimized.

Needless to say, this does not make sense if you consider euklidian space without time, but then again, without time you can not talk about "fastest" or "quickest".