# Why is the conjugate momentum $p(x,x')=\partial_2 L(x,x')$ an element of the cotangent space?

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?

• Just to clarify: When you say "Is this an additional (physical) input...?", you mean the simple physical input of defining of momentum as the product of mass and velocity?
– Kyle
Apr 28, 2015 at 21:01

The 1-form $p(x,x')$ is linear in the velocities in the following sense: sticking with your notation, the $i$th component of the one form is $$p_i(x,x') = \frac{\partial L}{\partial {x'}^i}(x,x').$$ The 1-form $p(x,x')$ acts linearly on velocities $v\in T_xM$ by $$p(x,x')(v) = p_i(x,x')v^i = \frac{\partial L}{\partial {x'}^i}(x,x')v^i.$$ In other words, the linear dependence comes not through the $x'$ as you seem to believe, but through the $v$.