Action of conjugate momentum on $TM$ and explicit form

Question

In hamiltonian mechanics the phase space of a particle is a symplectic manifold. In the case we have a configuration space $M$, that is the manifold describing the possible positions of the particle, we can canonically identify the cotangent bundle $T^*M$ as the phase space. The one-forms in the cotangent space correspond to the canonical momentum.

But the one thing that isn't clear to me is how this one-forms look and how they operate on the elements of the tangent space, corresponding to the velocities. With the basis $\{dq^i\}$ of the cotangent space, every momentum can be written as $p = p_i dq^i$. But how do the coefficients $p_i$ look?

In case of a free particle the classical momentum is give as $p = m \dot q$. Does this translate to $p = \sum m dq^i$, so that $p$ simply assigns every velocity the corresponding momentum?

Or are the $p_i$ given as $p_i = m \dot q^i$, so that $p = \sum m \dot q^i dq^i$? I don't think this is the case, because then the coordinates of the one-form $p$ would depend on elements $\dot q^i$ from the tangent space, which doesn't make sense to me. Also it mixes the elements of the tangent space and the cotangent space which looks really wrong.

On the other side my first idea of $p = \sum m dq^i$ also has some problems. In the case that $p$ doesn't depend linearly on $\dot q$ the resulting "one-form" wouldn't be linear and therefore not a real one-form.

It would be a great help if someone could exemplary show how the momentum one-forms look and accordingly how they act on velocity vectors. In case my first idea was correct, then how can one guarantee the linearity? In case the idea was totally wrong, what have I misunderstood?

I wanna add that I know the strength of differential forms is that one can formulate them without choosing a basis, but I find it helpful to look at explicit examples to get some intuition.

Answer 1

The one-forms in the cotangent space correspond to the canonical momentum.

I'd dispute that statement. Basically the (total space) of the cotangent bundle $T^\ast M$ is the set of all 1-forms of $M$. So a momentum is not an 1-form on $T^\ast M$, it is a point of $M$.

There can be something which may confuse you, which is the tautological 1-form. There is a preferred $\theta\in\Omega^1(T^\ast M)$ 1-form on $T^\ast M$, which, in some sense, corresponds to momenta, however this is not any 1-form on $T^\ast M$, but is a single, specific 1-form on $T^\ast M$.

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A point of $T^\ast M$ may be written as $(q,p)$ which means that this is the point that consists of the momentum p at point $q$. The tautological 1-form acts on an arbitrary $X_{(q,p)}\in T_{(q,p)}T^\ast M$ tangent vector (of $T^\ast M$) as $$ \theta_{(q,p)}(X_{(q,p)})=p(\pi^\ast X_{p,q}). $$ What does this mean? The point $(q,p)$ is the 1-form $p$ at $q$, so $p$ can act on a vector at $q$, but this is a vector of $M$, not $T^\ast M$. So $\theta$ first projects the tangent vector of $T^\ast M$ to a tangent vector of $M$ via the canonical projection $\pi: T^\ast M\rightarrow M$, and then has $p$ act on it.

So what momentum does $\theta$ represent is determined by which point of $T^\ast M$ you evaluate it at.

With the basis ${dq^i}$ of the cotangent space, every momentum can be written as $p=p_idq^i$.

This statement is however correct, but this is not an 1-form on the cotangent bundle, this is an 1-form of the cotangent bundle, so to speak.

But how do the coefficients $p_i$ look?

In Hamiltonian mechanics, any way you want. The point is, the positions and the momenta are independent variables. You are free to specify any position $q$ and any momentum $p$ (at that point $q$ ! ), and, as this provides the proper initial conditions for your equations of motion, the motion of the system is then determined.

Possibly what confuses you here is that if you start off with a Lagrangian formalism, there is a correspondance (which is unique if the Lagrangian satisfies the Hessian condition) between the tuplets $(q,\dot{q})$ and $(q,p)$, which is essentially a vector bundle isomorphism between $TM$ and $T^\ast M$.

For the sake of simplicity, let us assume the Lagrangian is independent of time, which means that the Lagrangian is a function $L:TM\rightarrow\mathbb R$.

Now, fix down a point $q\in M$. If we consider this point fixed, then for an arbitrary $\dot{q}\in T_q M$ velocity (at $q$), the Lagrangian is a function $T_q M\rightarrow\mathbb R$ in the sense that it maps $\dot q\mapsto L(q,\dot{q})\in\mathbb R$. This function is not, in general a linear function, however we can linearize it. Let $v\in T_qM$ be an arbitrary vector, and consider $$ \mathbb{L}_{(q,\dot q)}(v)=\frac{d}{d\epsilon}L(q,\dot{q}+\epsilon v)|_{\epsilon=0}. $$ This is a linear map on $v$ because it is a directional derivative, and it depends on points of $TM$. So, for fixed $(q,\dot q)\in TM$, it is a linear map $T_q M\rightarrow \mathbb R$, hence, it is a covector in $T^\ast_q M$, let us rename it as $p(q,\dot q)$.

In short, in Lagrangian mechanics, the canonical momentum, $p(q,\dot q)\in T^\ast_q M$ associated with the generalized velocity $(q,\dot q)\in TM$ acts on an arbitrary generalized velocity $v\in T_q M$ as $$ p(q,\dot q)(v)=\frac{d}{d\epsilon}L(q,\dot q+\epsilon v)|_{\epsilon=0}. $$ In local coordinates, $$ p(q,\dot q)(v)=\frac{d}{d\epsilon}L(q,\dot q+\epsilon v)|_{\epsilon=0}=\frac{\partial L}{\partial\dot{q}^i}v^i, $$ and so $$ p(q,\dot q)=\frac{\partial L(q,\dot q)}{\partial \dot q ^i}dq^i|_{q}, $$ which is of course the usual definition.

Answer 2

The momentum 1-form acts on the elements of the tangent space $v\in T_qM$ as follows: $$ p(q,\dot q)(v) = p_i(q,\dot q)v^i = \frac{\partial L}{\partial {\dot q}^i}(q,\dot q)v^i $$