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?


2 Answers 2


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

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

  • $\begingroup$ I am pretty sure that you are aware of the definition of the cotangent space which says that a cotangent space contains linear functionals on the tangent space. My question is: Why is the momentum a linear(!) functional? - See the wikipedia reference for a proper definition en.wikipedia.org/wiki/… I really want to see it according to this definition. $\endgroup$ Commented Apr 16, 2015 at 22:56
  • $\begingroup$ I mean, maybe your answer is perfectly correct, but I am assuming this definition of an element in the cotangent space and would therefore prefer to see it accordingly. $\endgroup$ Commented Apr 16, 2015 at 22:59
  • 3
    $\begingroup$ @Real Analysis: Concerning linear functionals: If you are familiar with how $\mathrm{d}q^i\in T^{\ast}Q$ can be viewed as a linear functional, then you can identify the momentum as the linear functional $p_i\mathrm{d}q^i\in T^{\ast}Q$. $\endgroup$
    – Qmechanic
    Commented Apr 16, 2015 at 23:07
  • $\begingroup$ but I think that is not what the literature always tells us: See also this accepted answer on mathoverflow claiming that $p$ is in some cotangent space mathoverflow.net/q/16460 . $\endgroup$ Commented Apr 16, 2015 at 23:09
  • $\begingroup$ Is the disclaimer because you don't want to bother, or because something actually breaks down when you relax the conditions? $\endgroup$
    – ACuriousMind
    Commented Apr 17, 2015 at 22:37

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.)


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