# Metric and the Lagrangian [duplicate]

Does the Lagrangian formalism require a metric on the configuration manifold $Q$ in order to define a Lagrangian $L$ on the tangent bundle $TQ$?

Further, if we specify a metric on the tangent bundle then we can via an isomorphism, move this to the cotangent bundle $T^*Q$. That being said how does this metric structure interplay with the symplectic structure?

## marked as duplicate by ACuriousMind♦, Kyle Kanos, John Rennie, ManishearthApr 2 '15 at 7:55

1. On one hand, let there be given a configuration space $(Q,g)$ endowed with a metric $g$. (As ACuriousMind points out in a comment, there is a 1-1 correspondence between a metric $g$ and the kinetic term in a Lagrangian.) On the other hand, note that the canonical symplectic 2-form $\omega$ on the cotangent bundle $T^{\ast}Q$ does not depend on the metric $g$ at all, cf. e.g. this Phys.SE post. The lesson is to expect no relation between $g$ and $\omega$ generically.
No, but you are most likely to get one from the kinetic term of the Lagrangian itself. In most cases one requires it to be a convex function in the $\dot q$ variables. You then get a metric if such kinetic term is quadratic in $\dot q$ (and of course sensible kinetic energy is positive-definite).