Clarification of classical field theory lecture notes by P. Deligne and D. Freed - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T17:15:54Z http://mathoverflow.net/feeds/question/27438 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/27438/clarification-of-classical-field-theory-lecture-notes-by-p-deligne-and-d-freed Clarification of classical field theory lecture notes by P. Deligne and D. Freed John Jiang 2010-06-08T07:02:08Z 2010-06-08T07:40:47Z <p>In section 1.1 under the subtitle system of classical particles with potential, the authors claim that "for a system of classical particles with rigid constraints, the configuration space is a Riemannian manifold X with Riemannian structure given by twice the kinetic energy."</p> <p>I don't quite how the configuration space can be given by a Riemannian manifold, as it is more naturally viewed as a symplectic manifold and there appears to be no natural Riemannian structure on a symplectic manifold. Also the relation between the Riemannian structure and the kinetic energy also eludes me. The best interpretation I can think of is to impose a Riemannian structure on the cotangent bundle via Legendre transform, or the specification of a Lagrangian function. But this is not explcitly given.</p> http://mathoverflow.net/questions/27438/clarification-of-classical-field-theory-lecture-notes-by-p-deligne-and-d-freed/27443#27443 Answer by Kevin Lin for Clarification of classical field theory lecture notes by P. Deligne and D. Freed Kevin Lin 2010-06-08T07:40:47Z 2010-06-08T07:40:47Z <p><em>Configuration</em> space is, by definition, the position space of your particles. <em>Phase</em> space, on the other hand, is the space of pairs (position, momentum). The latter has a symplectic structure; the former has a Riemannian structure.</p> <p>Regarding the relationship between kinetic energy and the Riemannian structure: You will recall from your high school physics class that kinetic energy is $\frac{1}{2} mv^2$. Of course the $v^2$ is really the dot product $v \cdot v$, in other words it's $g(v,v)$, where $g$ is the Riemannian metric and $v$ is a tangent vector. The $\frac{1}{2}$ explains the "twice the kinetic energy" part.</p>