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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?

WillG
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1 Answers1

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  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.

Qmechanic
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  • 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? – The_Sympathizer Aug 19 '22 at 07:25
  • I updated the answer. – Qmechanic Aug 19 '22 at 07:59
  • So the metric here is the spatial (i.e. Euclidean 2- or 3-dimensional, for classical mechanics) metric, right? – The_Sympathizer Aug 20 '22 at 04:18
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    $\uparrow$ In many cases. – Qmechanic Aug 20 '22 at 04:50