There are several explanations on this site [1] [2] [3] about why momentum is a covector while velocity is a vector. This distinction is important for the geometric description of classical mechanics.

However, none of these explanations reconciles this with the seeming contradiction that $p=mv$, which on its face suggests that momentum and velocity are the same type of object. How can we resolve this apparent contradiction?

I would expect that $p=mv$ exploits an isomorphism between the tangent and cotangent spaces, allowing us to represent $p$ as a vector even though it is "naturally" a covector. But if that's the case, how is this isomorphism defined, and where does it enter into the formalism of classical mechanics?


1 Answer 1

  1. Yes, OP points to the fact that classical mechanics typically relies on the existence of a fiducial/distinguished/background metric structure on the configuration manifold so that we can apply the musical isomorphism to raise and lower indices, e.g. between the contravariant velocity vector and the covariant momentum covector.

  2. Example: If $L=\frac{1}{2}m_{ij}v^iv^j-V$, then $p_i=\frac{\partial L}{\partial v^i}= m_{ij}v^j$. Often the mass metric structure is of the form $m_{ij}=m~g_{ij}$.

  3. Concerning the related notion of phase space, see e.g. this Phys.SE post.

  • $\begingroup$ I am curious, though: how "formally" is the musical isomorphism defined in this case? It seems the action (given by the Lagrangian) is what provides the metric, right - esp. noting how Wikipedia suggests that "there are similar isomorphisms on symplectic manifolds", which is what the classical phase space is. However usually we think of position and momentum in CM as just a point and vector in ordinary space, so shouldn't the metric just be the usual Euclidean one? So then why/how does the Lagrangian enter in to "transpose" the vector? Also why is mass seemingly a tensor here and not a scalar? $\endgroup$ Aug 19, 2022 at 7:25
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Aug 19, 2022 at 7:59
  • $\begingroup$ So the metric here is the spatial (i.e. Euclidean 2- or 3-dimensional, for classical mechanics) metric, right? $\endgroup$ Aug 20, 2022 at 4:18
  • 1
    $\begingroup$ $\uparrow$ In many cases. $\endgroup$
    – Qmechanic
    Aug 20, 2022 at 4:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.