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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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2 Answers2

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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 would also like to give another perspective, as I have also thought about this. Take a path $q^i(t)$ on configuration space. At each point of it, it has a velocity $\dot {q^i} (t)$ associated to it and a canonical momentum $p_i (t)$. Now if the velocity is a tangent vector and the canonical momentum is a cotangent vector, how can we have :

$$p_i =m\dot {q^i}$$

The correct way to interpret the above equation is as a relationship between components of $p_i$ and $\dot {q^i}$, i.e. $p_0=\dot {q^0}$, $p_1=\dot {q^1}$....

This relationship between components is of course co ordinate dependent, because $\dot q^i$ and $p _i$ transform differently under co ordinate changes, so the relationship cant hold in arbitrary co ordinates. This relationship holds in co ordinates where the kinetic term looks like $\frac{m}{2}\sum (\dot {q ^i})^2$. If we do the co ordinate transform $q=2q'$, the new kinetic term is $2m \sum (\dot {q'^i})^2$, and the relationship becomes between components of $p_i '$ and $\dot {q'^i}$ is:

$$p'_i = 4m\dot {q'^i}$$

So $p=mv$, when $p$ is interpreted as canonical momentum (as opposed to kinetic momentum), is a co ordinate dependent relationship between the components of $p$ and $mv$

Ryder Rude
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