4

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 $AB$ ?

Dylan
  • 465
  • They are the shortest path as far as I know. The quickest also depends on velocity, right? – exp ikx Jan 27 '19 at 14:32
  • 1
    It depend on your metric ! If on your chart 1 km represent really 1 km, then AB is the geodesic. If 1 km represent the the distance you make in 1h, then the path through C will be the geodesic. – Surb Jan 27 '19 at 14:34
  • This is a great question, and while the answers address it, you might also be interested in the brachistochrone problem (https://en.wikipedia.org/wiki/Brachistochrone_curve), which uses the same methods we use for studying geodesics, to find "fastest path" curves (in a somewhat different context). – John Hughes Jan 27 '19 at 16:32
  • I believe that general relativity uses geodesics which are the fastest path between two points in spacetime (and hence the path followed by light)—but that's the result of using a metric where distance can be interpreted as time. – timtfj Jan 27 '19 at 16:41

4 Answers4

5

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.

Carlos Esparza
  • 3,587

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

    I think you need some conditions for this. Consider the Euclidean plane with the unit-disk removed. To get from point (5,0) to (-5,0), you have to "go around" the hole. Such a path will not be a geodesic, as it must "turn." I think the condition you need to impose is that the manifold is geodesically complete.

     – jnez71 Jan 23 '21 at 01:41
  • I think if a path minimizing distance between two points on a Riemannian manifold exists, then it must be a geodesic since it will minimize energy. Of course it's crucial that the manifold has no boundary. – Carlos Esparza Jan 23 '21 at 02:41
  • The manifold having no boundary might be synonymous with the geodesic completeness I linked above, though I would say my counterexample doesn't have a boundary in a strict sense because it was defined as ${(x,y) \in \mathbb{R}^2 | x^2+y^2 > 1}$, where the boundary ${(x,y) \in \mathbb{R}^2 | x^2+y^2 = 1}$ is not included. Also see the Hopf–Rinow theorem, which gives the conditions for any two points being connected by a length minimizing geodesic. – jnez71 Jan 23 '21 at 02:56
  • I suppose I am getting at the fact that if the manifold isn't "geodesically complete" then we can have two path-connected points that aren't connected by any geodesics, and thus the shortest path cannot be a geodesic. So perhaps the correct edit to your answer is: "However, the shortest path between two points (if there is at least one geodesic connecting them) is always a geodesic." Or you could just impose the manifold's metric to be complete. – jnez71 Jan 23 '21 at 03:06
  • Also, remember that geodesics don't strictly minimize distance ("energy"), they are just stationary curves of the distance functional. (I believe more generally, they are curves with zero covariant derivative; "a crawling ant wouldn't think it turned"). So maybe an analogy to functions is the question of under what conditions a global extrema isn't a stationary point- the idea of a boundary appears again. – jnez71 Jan 23 '21 at 03:26
  • @jnez 1. There's many manifolds without boundary that aren't geodesically complete (e.g. the example you provided, or just $\mathbb{R}^2 \setminus {0}$. It is also not true that any non-complete Riemannian manifold can be embedded in a complete one. IIRC a counterexample is the universal cover of ${\mathbb R}^2 \setminus {0}$, this is explained somewhere in do Carmo. 2. If two points on a manifold cannot be linked by a geodesic then there won't be a shortest path between them (as is the case in the example you provided). 3. Shortest paths between two points minimize Energy. – Carlos Esparza Jan 23 '21 at 06:48
  • You think there is no shortest path between $(5,0)$ and $(-5,0)$ on the Euclidean submanifold ${(x,y) \in \mathbb{R}^2 | x^2 + y^2 > 1}$? There absolutely is. I agree with #1, which is why I was confused by your first reply where you mentioned boundaries. I never claimed the thing about embedding. I think #2 is the crux of the disagreement. What is your definition of geodesic? As for #3, yes they minimize "energy" (distance), globally. The standard definition for geodesics involves local minimization. They aren't always the same. – jnez71 Jan 23 '21 at 06:55
  • @jnez71 No there isn't. The curve you are probably thinking about that touches the circle doesn't actually lie in the manifold since it has distance $=1$ (and not $>1$) while touching the circle. – Carlos Esparza Jan 23 '21 at 06:59
  • @jnez71 Another example where you can see the same phenomenon more clearly is ${\mathbb R}^2 \setminus 0$, where there is no shortest path between $(-1, 0)$ and $(1, 0)$. Any such path will have to "move around" the origin and you can always find a slightly shorter path. – Carlos Esparza Jan 23 '21 at 07:00
  • Alright alright, then include the boundary of the circle. ${(x,y) \in \mathbb{R}^2 | x^2 + y^2 \geq 1}$. How about now? There is a nongeodesic shortest path from $(5,0)$ to $(-5,0)$, as depicted in my previously linked post. Don't you need a caveat for such spaces in your statement that the shortest path is always a geodesic? What do you think that caveat should be? I think geodesic completeness. – jnez71 Jan 23 '21 at 07:02
  • I take it you think the caveat is "have no boundary", but that isn't true. Consider the Euclidean unit-square $[0,1]^2$. There is a boundary, but every shortest-path is a geodesic. It seems pretty clear from the Hoph-Rinow theorem that geodesic completeness is the key, not the existence of a boundary. – jnez71 Jan 23 '21 at 07:13
  • My statements are referring to manifolds, not manifolds with boundary. Note that Hopf-Rinow fails for manifolds with boundary. Anyway, I don't feel like this discussion is relevant to the original question anymore. – Carlos Esparza Jan 23 '21 at 07:14
  • Hm well I can't think of any boundaryless manifold where an (existing) shortest-path between two points is not a geodesic, so perhaps you're right. I guess another way to put it is that I was supposing the existence of a geodesically incomplete manifold for which every pair of points has a shortest-path. Perhaps that's impossible. Well, for now I'll leave it. Thanks for replying! – jnez71 Jan 23 '21 at 08:04
1

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.

ecrin
  • 1,423
0

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.

Balloon
  • 8,374
0

(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".

Aemmel
  • 1