1
$\begingroup$

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."

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.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
8
$\begingroup$

Configuration space is, by definition, the position space of your particles. Phase space, on the other hand, is the space of pairs (position, momentum). The latter has a symplectic structure; the former has a Riemannian structure.

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.

Improve this answer
$\endgroup$
4
  • $\begingroup$ I think the confusion may have been because velocity is in the tangent space, not the cotangent space, and one can think of the inner product as being a map from $TM\times TM^*\rightarrow \mathbb{R}$, $g_{ij} v^j \mapsto v_i \mapsto v_i w^i$. But the way we want to think of the inner product in this context is a map from $TM\times TM\rightarrow \mathbb{R}$, which defined a Riemannian structure and literally gives us the kinetic energy. $\endgroup$
    – jeremy
    Commented Jun 8, 2010 at 8:12
  • 1
    $\begingroup$ Another way to say this is that the Riemannian metric on the configuration space $M$ induces an isomorphism between the tangent bundle $TM$ and the cotangent bundle $T^\ast M$. $\endgroup$
    – Kevin H. Lin
    Commented Jun 8, 2010 at 8:58
  • 1
    $\begingroup$ It should be pointed out that there are constrained mechanical systems for which the kinetic term is degenerate and hence the lagrangian does not give rise to an isomorphism $TM \cong T^*M$. In other words, one may not be able to invert the definition of the canonical momenta (i.e., the fibre derivatives of the lagrangian) to express the velocities in terms of the momenta. I'm guessing that the restriction to "rigid" constraints in Deligne and Freed is meant to avoid such pathologies. $\endgroup$
    – José Figueroa-O'Farrill
    Commented Jun 8, 2010 at 11:18
  • $\begingroup$ My confusion resulted from misunderstanding configuration space, exactly as Kevin pointed out. Also for the Riemannian metric to be determined by the kinetic energy, one would need to use the Polarization identity presumably. $\endgroup$
    – John Jiang
    Commented Jun 9, 2010 at 21:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .