Momentum is a cotangent vector?

Question

Imagine we have a particle described by $x \in M$, where $M$ is some manifold, then it is very intuitive I think that a velocity is an element of the tangent space at $x$, so $x' \in T_{x}M.$ Thus, by definition of the tangent bundle, we have $(x,x') \in TM$.

Now, in classical mechanics we learn that the conjugated mometum $p(x,x') = \partial_2L(x,x')$ and now I read that this guy is an element of the cotangent space, but I have no idea why.

I mean, to be in the cotangent space, you need to take elements in the tangent space of $x$ which are velocities as arguments and only depend linearly on them. Although it is clear that p takes velocities as arguments which is alright, it is not clear to me at this moment why this should happen linearly. Is this an additional (physical) input at this point or is there a mathematical argument why the momentum is now a linear map?

Answer 1

I) Disclaimer: In this answer we will use the (traditional) physicist's definition of tensors using indices and their transformation properties under coordinate transformations. Moreover, let us suppress time dependence $t$ for simplicity.

II) Let the manifold $Q$ be the configuration space. The Lagrangian $L:TQ\to \mathbb{R}$ transforms as a scalar

$$L~~\longrightarrow~~ L^{\prime}~=~L, \tag{1}$$

the velocity $v^i$ transforms as a vector

$$v^i~~\longrightarrow~~ v^{\prime j}~=~\frac{\partial q^{\prime j}}{\partial q^i}v^i,\tag{2}$$

the Lagrangian canonical momentum

$$p_i~:=~ \frac{\partial L}{\partial v^i}\tag{3}$$

transforms as a covector

$$p_i~~\longrightarrow~~ p^{\prime}_j~=~\frac{\partial q^i}{\partial q^{\prime j}}p_i,\tag{4}$$

under general coordinate transformations

$$ q^i~\longrightarrow ~q^{\prime j}~=~ f^j(q)\tag{5}$$

in the configuration space $Q$. Eq. (4) follows from the chain rule.

III) A point in the tangent bundle is of the form

$$(q,v)~\in~TQ,\qquad v~=~v^i \frac{\partial}{\partial q^i}.\tag{6} $$

Note that the velocity $v$ is an independent variable, which transforms as a vector (2) under general coordinate transformations (5) in the configuration space $Q$.

IV) The Lagrangian canonical momentum (3) can be viewed as a section

$$TQ ~\ni~ (q,v) ~\stackrel{p}{\mapsto} ~(q,v; p_i\mathrm{d}q^i)~\in~T^{\ast}TQ \tag{7}$$

in the bundle $T^{\ast}TQ \to TQ$.

V) Finally, let us for completeness & comparison mention the Hamiltonian canonical momentum (also denoted $p$) in the case where the phase space $M$ is the cotangent bundle $M=T^{\ast}Q$. In the case $M=T^{\ast}Q$, the Hamiltonian canonical momentum $p$ is an independent variable, which transforms as a covector (4) under general coordinate transformations (5) in the configuration space $Q$. A point in the tangent bundle is of the form

$$(q,p)~\in~T^{\ast}Q,\qquad p~=~p_i\mathrm{d}q^i.\tag{8} $$

Answer 2

The momentum is a covector because it is a gradient, and gradients are always covariant. It does what it says on the tin. However, you are right that this is a subtle point and it's not particularly clear at first sight.

For a lagrangian of the form $L=T-V$ with $V$ independent of $\dot q$, the canonical momentum is given by $$ p=\frac{\partial L}{\partial \dot q}=\frac{\partial T}{\partial \dot q}. $$ This derivative measures how much $T$ changes with respect to small changes in $\dot q$, when these changes are small enough that a linear approximation to $T$ suffices. This is exactly the linearity of $p$ as a functional of $\dot q$.

This means that $p$ is a functional over increments in $\dot q$ rather than a functional over $\dot q$ itself. This is of course correct: if you have a configuration space $Q$, then lagrangian mechanics takes place in $M=TQ$, which is the space of all configurations $q$ and the corresponding velocities $\dot q$. Hamiltonian mechanics, on the other hand, takes place in $T^*M$ - the space of linear forms over $TM$. Note here that $TM=TTQ$ is, precisely, the space of increments in velocity (along with the velocities themselves as increments in the position.)