Vipul: 2 revisions

2008-05-18T19:41:22Z

2 revisions

← Older revision	Revision as of 19:41, 18 May 2008
(No difference)

Vipul at 10:49, 5 August 2007

2007-08-05T10:49:10Z

← Older revision		Revision as of 10:49, 5 August 2007
Line 2:		Line 2:

	Let <math>M</math> be a [[complete Riemannian manifold]]. Then the piecewise smooth [[geodesic in a Riemannian manifold\|geodesic]]s in <math>M</math> are precisely the same as the [[critical path of a functional\|critical paths]] of the [[energy functional]].		Let <math>M</math> be a [[complete Riemannian manifold]]. Then the piecewise smooth [[geodesic in a Riemannian manifold\|geodesic]]s in <math>M</math> are precisely the same as the [[critical path of a functional\|critical paths]] of the [[energy functional]].

			==Proof==

			===Geodesic implies critical path of energy functional===

			This is clear -- any geodesic is locally length-minimizing, and hence, locally energy-minimizing, and hence, it must be a critical path of the energy functional.

			===Critical path of energy functional implies geodesic===

			The proof of this follows by cleverly choosing a variation, and then applying the [[first variation formula]] to conclude, from the vanishing of the derivative, that the acceleration vector must vanish at each point.

Vipul at 10:26, 5 August 2007

2007-08-05T10:26:12Z

New page

==Statement==

Let <math>M</math> be a [[complete Riemannian manifold]]. Then the piecewise smooth [[geodesic in a Riemannian manifold|geodesic]]s in <math>M</math> are precisely the same as the [[critical path of a functional|critical paths]] of the [[energy functional]].

Geodesic equals energy critical path - Revision history

Vipul: 2 revisions

Vipul at 10:49, 5 August 2007

Vipul at 10:26, 5 August 2007