Euler-Lagrange Equation and Geodesics

[Henrique Mello]

The Euler-Lagrange equation just guarantees that the integral is stationary. In the calculation of geodesics, how do I know that the path I found is actually the shortest distance between two points?

[ajb]

Geodesics are locally distance-minimizing paths, which is what you show using the variational techniques you describe. You minimize the length of the curves locally on some open subset of your manifold. We have local existence and uniqueness theorems, but are you asking about global theorems?

[Henrique Mello]

But my doubt is, for example: The shortest distance between two points in Earth is the arc a great circle. But we actually have two possiblities: To travel from New York to Madrid directly (crossing Atlantic Ocean) or go through Japan, Russia etc until arrive at Spain. The geodesics just says that the path is a great circle, but doesn't say which path is the shortest. You know what I mean?

[ajb]

But my doubt is, for example: The shortest distance between two points in Earth is the arc a great circle. But we actually have two possiblities: To travel from New York to Madrid directly (crossing Atlantic Ocean) or go through Japan, Russia etc until arrive at Spain. The geodesics just says that the path is a great circle, but doesn't say which path is the shortest. You know what I mean?

If I recall correctly, as long as the two points are not antipodal points, the great circle joining two points is unique. If the points ate antipodal, that is directly opposite each other, then there is an infinite number of great circles all of the same length. Edited February 26, 2013 by ajb

[elfmotat]

[math]\delta S=0[/math] doesn't guarantee that the action is minimal, or even maximal, just that it's stationary. Take, for example, the curve y=x³ : the derivative at x=0 is zero, but it's neither a minimum nor a maximum (it's a saddle point).

 

Intuitively we know that geodesics (usually, if not always) correspond to minimal path-length because it's not possible to draw a path with maximum length - you can always draw a crazier, more squiggly path that's longer. I can't, off the top of my head, think of any situations where the length of a geodesic would be a saddle point. The fact that geodesics correspond to curves of stationary length comes from the fact that varying the Lagrangian [math]L=\frac{ds}{d\lambda}[/math] yields the geodesic equation. You can find this worked out here: http://www.mth.uct.ac.za/omei/gr/chap6/node6.html .

 

**Note that in the link I provided, they use the Lagrangian [math]L=\frac{d\tau}{d\lambda}[/math] instead of [math]L=\frac{ds}{d\lambda}[/math]. This corresponds, (usually if not always) to curves of maximum proper time, because [math]ds^2=-d\tau^2[/math]. This should make some intuitive sense because, for example in the Twin Paradox, the twin that goes away and comes back has always aged less than the twin who was sitting around in his inertial reference frame.

 

 

 

But my doubt is, for example: The shortest distance between two points in Earth is the arc a great circle. But we actually have two possiblities: To travel from New York to Madrid directly (crossing Atlantic Ocean) or go through Japan, Russia etc until arrive at Spain. The geodesics just says that the path is a great circle, but doesn't say which path is the shortest. You know what I mean?

 

This has to do with the orientation of the curve. For timelike curves in GR, there is really only one possible orientation because you can't travel backwards in time. For your sphere example, which is a spacelike geodesic, there are two possible orientations. One of them corresponds to the shortest path, and the other doesn't. The geodesic equation doesn't know which two points you're trying to find shortest distance between, so both orientations are possible. I.e. if you're trying to find the shortest path connecting A and B, you use orientation #1. If you then want to find the shortest path between B and A, you need to use orientation #2.

[Henrique Mello]

Thank you, guys